Systems & Control – Structures — Paper 4 · IGCSE Design and Technology 0445

1 Rigid PVC is a plastic often used for the manufacture of door and window frames. (a) Give two advantages of using rigid PVC as a structural material when compared to wood. 1 … … 2 … … [2] (b) Give two advantages of using wood as a structural material when compared to plastics. 1 … … 2 … … [2]

4 marks

Mark scheme: 1(a) Advantages of plastics could be: • Rigid PVC will not rot and is resistant to insect attack • Lightweight using hollow sections • Can be easily joined by welding • Long lasting • Does not warp • No natural defects 2 Allow other suitable advantages, comparisons must be against wood. Allow water resistance. Allow higher strength / weight ratio 1(b) Advantages of wood could be: • Renewable resource • Easily obtainable • Can be joined with temporary or permanent joints • Higher tensile strength than rigid PVC • Different types available in different dimensions • Higher compressive strength 2 Allow other suitable advantages, comparisons can be against any plastics Allow stronger joints possible and aesthetic reasons Allow bio-degradable Allow ‘resists heat’

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2 Fig. 1 shows two different types of structural beam used to span an opening in a building. hollow section steel concrete Fig. 1 (a) Give two reasons for hollow section steel beams often being used in preference to solid concrete beams. 1 … … 2 … … [2] (b) Use sketches and notes to show what is meant by a ‘laminated wooden beam’. [2] (c) Give one reason for choosing a laminated wooden beam rather than concrete or hollow section steel. … … [1]

5 marks

Mark scheme: 2(a) Reasons for using hollow section steel will include: • Lighter than concrete • More precise dimensions • More of the beam can be hidden from view • Longer lengths easier to handle • Stronger in tension • Higher strength / weight ratio 2 Allow other valid reasons for use of hollow sections in beams Allow resistant to shearing 2(b) Sketches to show beam made from three or more horizontal layers [1] Notes indicating gluing together [1] 2 Beam may be curved or a shape other than straight If there is an indication that the beam is short enough to be made from single lengths of lamination allow second mark Allow other valid reasons 2(c) Laminated beams can be produced in much greater lengths Easier to produce curves 1 Allow aesthetic and other valid reasons Allow ‘renewable resource’

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7 Fig. 3 shows a 12-way rotary switch. common terminal flat Fig. 3 (a) Describe the operation of a 12-way rotary switch. … … … [2] (b) State the purpose of the flat on the spindle of the rotary switch. … [1]

3 marks

Mark scheme: 7(a) In a rotary switch a number of terminals [1] are joined in turn to a common terminal [1] as the switch shaft is rotated 2 7(b) The flat is to allow a control knob to be attached without rotating on the spindle 1 Total: 25

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8 (a) Fig. 4 shows a steel strip in a test rig. C steel strip load weights Fig. 4 (i) Give the name of electronic component C. … [1] (ii) State the physical property that will change in component C when the load is increased. … [1] (iii) Component C could be attached to either the top or bottom face of the steel strip. State the force being measured according to which face of the steel strip is used. top face … … bottom face … … [2] (iv) Use sketches and notes to show how a non-electronic method could be used to test the amount of movement in the steel strip when a load is applied. [3] (b) Fig. 5 shows packaging for a delicate electronic device. The moulded inner packaging fits inside the assembled outer. When the box is assembled no glue is used. outer packaging - flat corrugated card moulded inner packaging Fig. 5 (i) Give two reasons why the assembled packaging provides protection to the contents when it is being transported. 1 … 2 … [2] (ii) The card used in the construction of the outer packaging is corrugated, a shape that can also be applied to plastic or metal sheet. Use sketches and notes to show what is meant by the term ‘corrugated’. [2] (iii) Fig. 6 shows a frame for a carport. The roof is to be covered using six uncut corrugated plastic sheets 2600 mm × 800 mm. m 2.5 wall 80° 4 m driveway Fig. 6 Draw on Fig. 6 to show two of the corrugated plastic sheets correctly positioned on the roof frame. [2] (c) (i) Fig. 7a and Fig. 7b show two views of a beam used to support a pulley for raising loads on a building site. Fig. 7a Fig. 7b Draw the following methods of reinforcement: • a gusset plate on Fig. 7a • a brace on Fig. 7b [3] (ii) Fig. 7c shows an alternative method of supporting the beam. 800 30° 200N Fig. 7c Draw a scaled force triangle in the space below to calculate the magnitude of the force acting on the chain in Fig. 7c. The diagram has been started for you. 30° Force acting on the chain … N [3] (d) Tall garden plants will often need supporting. Use sketches and notes to show a structure supporting a plant that has reached a height of 1.5 m. Give details of suitable materials for the structure and for any joints used. [4] (e) Explain, using an example, what is meant by torsion in a structure. … … … … [2]

25 marks

Mark scheme: 8(a)(i) Strain gauge 1 8(a)(ii) Very small changes in resistance will be registered by the gauge when it is stretched 1 Allow mark for Resistance without reference to the level of change 8(a)(iii) The steel strip will register tension on the top face and compression on the bottom face 2 Both points must be mentioned in the explanation for 2 marks 8(a)(iv) The non-electronic method must use a dial gauge to register movement when the steel strip is loaded Dial gauge [1] in a suitable position [1] fastening method to the test rig indicated [1] 3 8(b)(i) Reasons for strength of packaging will include: • Folds on long edges, double triple thickness • Locking tabs used with cut outs for edges to fit • Flat sides are supported by strengthened edges • One piece of card used, no extra joints • Folded edges on lids • Inner package is moulded, curves giving strength to the shape • Ribs used on the moulded inner packaging • Absorb shock / impact 2 Allow other valid reasons 1 mark for each valid reason 2 × 1 marks 8(b)(ii) Regular shape / uniform pitch [1] and height [1] of corrugation Accept either with outer sheets or without, corrugations can be square 2 Accept triangular corrugations 8(b)(iii) Corrugations running from wall to edge [1] overlap visible [1] 2 Question Answer Marks Guidance 8(c)(i) Gusset plate principle [1] Brace principle [1] Functional proportions on both methods [1] 3 8(c)(ii) Scaled line to represent 200 N [1] Triangle completed [1] Force in chain taken from drawing – 400 N [1] 3 Allow 1 mark for force calculated rather than using diagram 8(d) Structure stable in three dimensions [1] Height of structure 1.5 m or greater [1] Materials noted [1] Joints described or clearly drawn [1] 4 8(e) Explanation to include the term ’twisting’ or similar term [1] Valid example given [1] 2 Total: 25 R S

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1 Rigid PVC is a plastic often used for the manufacture of door and window frames. (a) Give two advantages of using rigid PVC as a structural material when compared to wood. 1 … … 2 … … [2] (b) Give two advantages of using wood as a structural material when compared to plastics. 1 … … 2 … … [2]

4 marks

Mark scheme: 1(a) Advantages of plastics could be: • Rigid PVC will not rot and is resistant to insect attack • Lightweight using hollow sections • Can be easily joined by welding • Long lasting • Does not warp • No natural defects 2 Allow other suitable advantages, comparisons must be against wood. Allow water resistance. Allow higher strength / weight ratio 1(b) Advantages of wood could be: • Renewable resource • Easily obtainable • Can be joined with temporary or permanent joints • Higher tensile strength than rigid PVC • Different types available in different dimensions • Higher compressive strength 2 Allow other suitable advantages, comparisons can be against any plastics Allow stronger joints possible and aesthetic reasons Allow bio-degradable Allow ‘resists heat’

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2 Fig. 1 shows two different types of structural beam used to span an opening in a building. hollow section steel concrete Fig. 1 (a) Give two reasons for hollow section steel beams often being used in preference to solid concrete beams. 1 … … 2 … … [2] (b) Use sketches and notes to show what is meant by a ‘laminated wooden beam’. [2] (c) Give one reason for choosing a laminated wooden beam rather than concrete or hollow section steel. … … [1]

5 marks

Mark scheme: 2(a) Reasons for using hollow section steel will include: • Lighter than concrete • More precise dimensions • More of the beam can be hidden from view • Longer lengths easier to handle • Stronger in tension • Higher strength / weight ratio 2 Allow other valid reasons for use of hollow sections in beams Allow resistant to shearing 2(b) Sketches to show beam made from three or more horizontal layers [1] Notes indicating gluing together [1] 2 Beam may be curved or a shape other than straight If there is an indication that the beam is short enough to be made from single lengths of lamination allow second mark Allow other valid reasons 2(c) Laminated beams can be produced in much greater lengths Easier to produce curves 1 Allow aesthetic and other valid reasons Allow ‘renewable resource’

This question in 0445/43 May/June 2017

7 Fig. 3 shows a 12-way rotary switch. common terminal flat Fig. 3 (a) Describe the operation of a 12-way rotary switch. … … … [2] (b) State the purpose of the flat on the spindle of the rotary switch. … [1]

3 marks

Mark scheme: 7(a) In a rotary switch a number of terminals [1] are joined in turn to a common terminal [1] as the switch shaft is rotated 2 7(b) The flat is to allow a control knob to be attached without rotating on the spindle 1 Total: 25

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8 (a) Fig. 4 shows a steel strip in a test rig. C steel strip load weights Fig. 4 (i) Give the name of electronic component C. … [1] (ii) State the physical property that will change in component C when the load is increased. … [1] (iii) Component C could be attached to either the top or bottom face of the steel strip. State the force being measured according to which face of the steel strip is used. top face … … bottom face … … [2] (iv) Use sketches and notes to show how a non-electronic method could be used to test the amount of movement in the steel strip when a load is applied. [3] (b) Fig. 5 shows packaging for a delicate electronic device. The moulded inner packaging fits inside the assembled outer. When the box is assembled no glue is used. outer packaging - flat corrugated card moulded inner packaging Fig. 5 (i) Give two reasons why the assembled packaging provides protection to the contents when it is being transported. 1 … 2 … [2] (ii) The card used in the construction of the outer packaging is corrugated, a shape that can also be applied to plastic or metal sheet. Use sketches and notes to show what is meant by the term ‘corrugated’. [2] (iii) Fig. 6 shows a frame for a carport. The roof is to be covered using six uncut corrugated plastic sheets 2600 mm × 800 mm. m 2.5 wall 80° 4 m driveway Fig. 6 Draw on Fig. 6 to show two of the corrugated plastic sheets correctly positioned on the roof frame. [2] (c) (i) Fig. 7a and Fig. 7b show two views of a beam used to support a pulley for raising loads on a building site. Fig. 7a Fig. 7b Draw the following methods of reinforcement: • a gusset plate on Fig. 7a • a brace on Fig. 7b [3] (ii) Fig. 7c shows an alternative method of supporting the beam. 800 30° 200N Fig. 7c Draw a scaled force triangle in the space below to calculate the magnitude of the force acting on the chain in Fig. 7c. The diagram has been started for you. 30° Force acting on the chain … N [3] (d) Tall garden plants will often need supporting. Use sketches and notes to show a structure supporting a plant that has reached a height of 1.5 m. Give details of suitable materials for the structure and for any joints used. [4] (e) Explain, using an example, what is meant by torsion in a structure. … … … … [2]

25 marks

Mark scheme: 8(a)(i) Strain gauge 1 8(a)(ii) Very small changes in resistance will be registered by the gauge when it is stretched 1 Allow mark for Resistance without reference to the level of change 8(a)(iii) The steel strip will register tension on the top face and compression on the bottom face 2 Both points must be mentioned in the explanation for 2 marks 8(a)(iv) The non-electronic method must use a dial gauge to register movement when the steel strip is loaded Dial gauge [1] in a suitable position [1] fastening method to the test rig indicated [1] 3 8(b)(i) Reasons for strength of packaging will include: • Folds on long edges, double triple thickness • Locking tabs used with cut outs for edges to fit • Flat sides are supported by strengthened edges • One piece of card used, no extra joints • Folded edges on lids • Inner package is moulded, curves giving strength to the shape • Ribs used on the moulded inner packaging • Absorb shock / impact 2 Allow other valid reasons 1 mark for each valid reason 2 × 1 marks 8(b)(ii) Regular shape / uniform pitch [1] and height [1] of corrugation Accept either with outer sheets or without, corrugations can be square 2 Accept triangular corrugations 8(b)(iii) Corrugations running from wall to edge [1] overlap visible [1] 2 Question Answer Marks Guidance 8(c)(i) Gusset plate principle [1] Brace principle [1] Functional proportions on both methods [1] 3 8(c)(ii) Scaled line to represent 200 N [1] Triangle completed [1] Force in chain taken from drawing – 400 N [1] 3 Allow 1 mark for force calculated rather than using diagram 8(d) Structure stable in three dimensions [1] Height of structure 1.5 m or greater [1] Materials noted [1] Joints described or clearly drawn [1] 4 8(e) Explanation to include the term ’twisting’ or similar term [1] Valid example given [1] 2 Total: 25 R S

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6 Draw and name a natural frame structure. [1]

1 marks

Mark scheme: 6 Any natural frame structure, 1 mark 1 No marks for man- made structures

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7 Draw and name a natural shell structure. [1]

1 marks

Mark scheme: 7 Any natural shell structure, 1 mark 1 No marks for man- made structures

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8 Fig. 4 shows a bracket made from square steel tube. welded joint Fig. 4 Use sketches and notes on Fig. 4 to show a method of reinforcing the welded joint in the bracket. [3]

3 marks

Mark scheme: 8 Gusset, brace or tie used 1 mark. Correct position, e.g. tie used above joint, brace below joint, gusset either above or below joint, 1 mark. Clear sketches / notes to show fixing method / how the reinforcement would work, 1 mark. 3

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9 Describe what is meant by equilibrium in a structure. … … … [2]

2 marks

Mark scheme: 9 Description could relate to: • clockwise moment = anticlockwise moment, opposing forces being equal or a state of balance, 1 mark • Stability or no movement, 1 mark 2

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10 Fig. 5 shows a roof truss. C B A Fig. 5 (a) Use the terms given below and information from Fig. 5 to complete the description. torsion compression tension bending A B Part … is a strut, which is placed there to resist … . Part … is a tie which will resist … . When the roof covering is added, part C will have to resist a … force. [5] (b) Explain the meaning of the following terms that appear on a stress / strain graph for mild steel. (i) Elastic deformation … … … [2] (ii) Elastic limit … … … [2] (iii) Plastic deformation … … … [2] (c) Fig. 6 shows a flagpole made from aluminium tube. The flagpole can be rotated about a pivot for maintenance of the pulley at the top. The raised flagpole is held in position by a locking pin. Cables are used to stabilise the raised flagpole in high winds. One cable is shown in position. pulley cable flagpole in lowered position pivot hole for locking pin locking pin Fig. 6 (i) State the minimum number of cables that should be used to stabilise the flagpole. … [1] (ii) Use sketches and notes to show a method of tensioning the cables that allows for adjustment. [3] (iii) State the force that will be applied to the pivot when the flagpole is raised and the cables are tight. … [1] (iv) Fig. 7 shows the flagpole in the raised and lowered positions. To lower the flagpole the locking pin is removed and the flagpole rotates about the pivot. 5.1 m flagpole pivot 0.9 m locking pin counterweight 25 N 125 N X 0.45 m 2.55 m 2.55 m 0.45 m Fig. 7 The 25 N and 125 N forces represent the distributed load of the aluminium tube. Calculate the value of counterweight X that will be required to keep the pole in equilibrium. … … … … … [4] (d) Fig. 8 shows a truss bridge. Fig. 8 Describe, using examples, the difference between static (stationary) loads and dynamic (moving) loads on the structure of the bridge. … … … … … … [4] (e) Fig. 9 shows a piece of aluminium honeycomb sheet of the type commonly used in aircraft manufacture. 0.75 mm aluminium top and bottom skins 10 aluminium honeycomb core Fig. 9 Give one reason why this material is suitable for aircraft manufacture. … … [1]

25 marks

Mark scheme: 10(a) Part … A … is a strut, which is placed there to resist compression Part … B … is a tie which will resist tension When the roof covering is added part C will have to resist a bending force. 5 1 mark for each term correctly placed 10(b)(i) Elastic deformation allows the material to go back to its original shape / length [1] after the loading is removed [1] 2 Allow 1 mark for some understanding shown. 10(b)(ii) Elastic limit is the maximum that a material can be stretched [1] without any permanent change to its shape / length [1]. 2 Allow 1 mark for some understanding shown. Question Answer Marks Guidance 10(b)(iii) Plastic deformation is permanent deformation of the material [1] without any fracture occurring [1]. 2 Allow 1 mark for some understanding shown. 10(c)(i) 3 / three cables is the minimum, 1 mark. 1 10(c)(ii) Functional method [1] Adjustment possible [1] Clear understandable sketch / notes [1]. 3 10(c)(iii) Shear force, 1 mark. 1 10(c)(iv) (0.9 × X) + (0.45 × 25) = 2.55 × 125, 1 mark 0.9X + 11.25 = 318.75, 1 mark X = (318.75 – 11.25) / 0.9, 1 mark X = 341.66 N, 1 mark 4 Award 4 marks for correct answer with no working. 10(d) Static loads are those that do not change [1] made up of construction materials used in the building of the bridge [1] Dynamic loads are changing values [1] made up of vehicles, pedestrians, animals or the loading caused by changing weather conditions. [1] 4 For changing weather conditions allow: High winds, snow, heavy rain, earthquake. For static loads allow any item described as stationary. 10(e) Reasons for using aluminium honeycomb could include: • Low weight / high strength • Resistance to twisting / torsion • Moisture and corrosion resistance • High thermal conductivity 1 Do not allow marks for ‘strong’ with no justification

This question in 0445/41 Oct/Nov 2017

6 Draw and name a natural frame structure. [1]

1 marks

Mark scheme: 6 Any natural frame structure, 1 mark 1 No marks for man- made structures

This question in 0445/42 Oct/Nov 2017

7 Draw and name a natural shell structure. [1]

1 marks

Mark scheme: 7 Any natural shell structure, 1 mark 1 No marks for man- made structures

This question in 0445/42 Oct/Nov 2017

8 Fig. 4 shows a bracket made from square steel tube. welded joint Fig. 4 Use sketches and notes on Fig. 4 to show a method of reinforcing the welded joint in the bracket. [3]

3 marks

Mark scheme: 8 Gusset, brace or tie used 1 mark. Correct position, e.g. tie used above joint, brace below joint, gusset either above or below joint, 1 mark. Clear sketches / notes to show fixing method / how the reinforcement would work, 1 mark. 3

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9 Describe what is meant by equilibrium in a structure. … … … [2]

2 marks

Mark scheme: 9 Description could relate to: • clockwise moment = anticlockwise moment, opposing forces being equal or a state of balance, 1 mark • Stability or no movement, 1 mark 2

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10 Fig. 5 shows a roof truss. C B A Fig. 5 (a) Use the terms given below and information from Fig. 5 to complete the description. torsion compression tension bending A B Part … is a strut, which is placed there to resist … . Part … is a tie which will resist … . When the roof covering is added, part C will have to resist a … force. [5] (b) Explain the meaning of the following terms that appear on a stress / strain graph for mild steel. (i) Elastic deformation … … … [2] (ii) Elastic limit … … … [2] (iii) Plastic deformation … … … [2] (c) Fig. 6 shows a flagpole made from aluminium tube. The flagpole can be rotated about a pivot for maintenance of the pulley at the top. The raised flagpole is held in position by a locking pin. Cables are used to stabilise the raised flagpole in high winds. One cable is shown in position. pulley cable flagpole in lowered position pivot hole for locking pin locking pin Fig. 6 (i) State the minimum number of cables that should be used to stabilise the flagpole. … [1] (ii) Use sketches and notes to show a method of tensioning the cables that allows for adjustment. [3] (iii) State the force that will be applied to the pivot when the flagpole is raised and the cables are tight. … [1] (iv) Fig. 7 shows the flagpole in the raised and lowered positions. To lower the flagpole the locking pin is removed and the flagpole rotates about the pivot. 5.1 m flagpole pivot 0.9 m locking pin counterweight 25 N 125 N X 0.45 m 2.55 m 2.55 m 0.45 m Fig. 7 The 25 N and 125 N forces represent the distributed load of the aluminium tube. Calculate the value of counterweight X that will be required to keep the pole in equilibrium. … … … … … [4] (d) Fig. 8 shows a truss bridge. Fig. 8 Describe, using examples, the difference between static (stationary) loads and dynamic (moving) loads on the structure of the bridge. … … … … … … [4] (e) Fig. 9 shows a piece of aluminium honeycomb sheet of the type commonly used in aircraft manufacture. 0.75 mm aluminium top and bottom skins 10 aluminium honeycomb core Fig. 9 Give one reason why this material is suitable for aircraft manufacture. … … [1]

25 marks

Mark scheme: 10(a) Part … A … is a strut, which is placed there to resist compression Part … B … is a tie which will resist tension When the roof covering is added part C will have to resist a bending force. 5 1 mark for each term correctly placed 10(b)(i) Elastic deformation allows the material to go back to its original shape / length [1] after the loading is removed [1] 2 Allow 1 mark for some understanding shown. 10(b)(ii) Elastic limit is the maximum that a material can be stretched [1] without any permanent change to its shape / length [1]. 2 Allow 1 mark for some understanding shown. Question Answer Marks Guidance 10(b)(iii) Plastic deformation is permanent deformation of the material [1] without any fracture occurring [1]. 2 Allow 1 mark for some understanding shown. 10(c)(i) 3 / three cables is the minimum, 1 mark. 1 10(c)(ii) Functional method [1] Adjustment possible [1] Clear understandable sketch / notes [1]. 3 10(c)(iii) Shear force, 1 mark. 1 10(c)(iv) (0.9 × X) + (0.45 × 25) = 2.55 × 125, 1 mark 0.9X + 11.25 = 318.75, 1 mark X = (318.75 – 11.25) / 0.9, 1 mark X = 341.66 N, 1 mark 4 Award 4 marks for correct answer with no working. 10(d) Static loads are those that do not change [1] made up of construction materials used in the building of the bridge [1] Dynamic loads are changing values [1] made up of vehicles, pedestrians, animals or the loading caused by changing weather conditions. [1] 4 For changing weather conditions allow: High winds, snow, heavy rain, earthquake. For static loads allow any item described as stationary. 10(e) Reasons for using aluminium honeycomb could include: • Low weight / high strength • Resistance to twisting / torsion • Moisture and corrosion resistance • High thermal conductivity 1 Do not allow marks for ‘strong’ with no justification

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1 Fig. 1 shows the rear derailleur gear on a cycle with a close up view of the gear cable. outer gear cable inner gear cable outer gear cable inner gear cable nylon lining Fig. 1 (a) Name the force that has been applied, when manufacturing the inner gear cable to keep the strands of cable in position. … [1] (b) The outer gear cable has a nylon lining. Give two benefits of using the nylon lining. 1 … 2 … [2] (c) Explain why small radius bends should be avoided in the gear cable when it is attached to the cycle. … … … [2]

5 marks

Mark scheme: 1(a) Torque or torsion 1 Allow twisting 1(b) Benefits of the nylon lining could include: • Reduce friction • Reduce the amount of lubrication required • Prevent water getting to inner cable • Prevent inner cable rubbing on outer 2 × 1 marks for suitable benefits 2 Allow any other valid benefit. E.g. lightweight, waterproof, flexible. 1(c) Small radius bends will restrict the cable and prevent efficient movement of the inner cable. Permanent distortion of the outer cable could result. Possibility of cable snapping. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark.

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2 (a) Explain why steel is used for control cables on a cycle. … … … [2] (b) State the force being used when a control cable is operated. … [1]

3 marks

Mark scheme: 2(a) Explanation should mention that steel is durable, ductile, strong in tension, can bend and is able to return to same position. Low cost material that can also be recycled after use. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark. 2(b) Tension 1

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10 Fig. 5 shows the initial design for a scaffold to give safe access while building work is carried out. X Y Fig. 5 (a) (i) Name the type of structure used in scaffolding. … [1] (ii) Draw on Fig. 5 three additional scaffolding poles to make the structure rigid and safe from collapse. [3] (iii) Explain the purpose of features X and Y at the base of the scaffold. … … … [2] (iv) When scaffolding is designed, a factor of safety for the structure is considered. Explain what is meant by a factor of safety. … … … [2] (b) Workers on a construction site have to wear ‘hard hats’ as shown in Fig. 6. Fig. 6 (i) State the type of structure used in the hard hat. … [1] (ii) Explain how the strength of the hard hat has been improved without adding any extra material. … … … [2] (c) Fig. 7 shows the axes for a stress/strain graph for mild steel. The position of three features that will appear on the graph are marked. C B A – elastic limit A B – upper yield point C – ultimate stress point stress strain Fig. 7 (i) Draw the shape of the graph on Fig. 7. [3] (ii) Explain what is meant by ‘elastic limit’ on the graph. … … … … [2] (d) Fig. 8a shows a folding clothes airer in the assembled position. Fig 8a Fig. 8b (i) Name the strengthening feature on the clothes airer shown in Fig. 8b. … [1] (ii) Give two reasons for using the type of strengthening feature shown in Fig. 8b. 1 … … 2 … … [2] (iii) The load carrying ability of each rail in the clothes airer is to be tested. Use sketches and notes on Fig. 9 to show how a dial gauge could be fixed to give a positive reading on the gauge when a load is added to the centre of the Ø6 rail. 750 Ø6 Ø12 [3] Fig. 9 Question 10 continues on the next page (e) Fig. 10 shows two concrete pillars. 7 kN 8 kN Ø200 175 × 175 pillar A pillar B Fig. 10 Calculate which pillar is subject to the greatest stress when the loads are applied. forceUse the formula: stress = cross sectional area … … … … [3]

25 marks

Mark scheme: 10(a)(i) Scaffolding is a frame structure 1 10(a)(ii) Each face of the structure to be made rigid, minimum of three poles. 1 mark for each pole in the correct position. 3 Maximum of 2 marks for multiple added poles in any one plane. 10(a)(iii) Features X and Y are to spread the load of the scaffold [1] and prevent it from sinking into soft ground. [1] They can also be used for levelling on uneven ground. [1] 2 1 mark each for any two correct points included. 10(a)(iv) A factor of safety is the load carrying capacity of a structure [1] beyond the expected loads. [1] The factor of safety is a number that is multiplied by the total expected load giving a load that design of structure must carry. [1] 2 One mark for each valid point made, allow examples in the explanation. Allow 2 marks for one point fully explained. 10(b)(i) The hard hat is a shell structure 1 10(b)(ii) Ribs have been moulded into the shell [1] to make the structure more rigid / prevent any flexing [1]. 2 Both points needed for 2 marks. 10(c)(i) 3 Smooth rise to elastic limit A[1] Sharp dip after upper yield point B[1] Gradual fall after ultimate stress point C[1] 10(c)(ii) Material will stretch / elongate until this point. [1] Elastic limit is the point beyond which the material will not return to its original length and is permanently deformed. [1] Stress is directly proportional to strain up to the elastic limit. [1] 2 Two points needed for 2 marks. 10(d)(i) Gusset plate 1 A B C stress strain A - elastic limit B - upper yield point C - ultimate stress point Question Answer Marks Guidance 10(d)(ii) Reasons will include: • To keep parts of the structure at 90° to each other • Reinforce the corner • Does not take up as much room as a strut or tie • Does not interfere with clothing on the rack • Lightweight • Low cost 2 × 1 marks 2 Allow other valid reasons. 10(d)(iii) • Dial gauge taking reading from bottom of rail [1] • Tip of dial gauge in contact with centre of rail [1] • Method of supporting dial gauge shown [1] 3 × 1 marks 3 10(e) Stress on pillar A = 7000 / 31415.9 mm2 = 7000 / 0.0314159 m2 = 222817 N/m2 1 mark = 0.223 N/mm2 Stress on pillar B = 8000 / 30625 mm2 = 8000 / 0.030625 m2 = 261224 N/m2 1 mark = 0.261 N/mm2 Pillar B is subject to greatest stress, 1 mark 3 Stress must be calculated for each pillar to gain marks. No ecf.1

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1 Fig. 1 shows the rear derailleur gear on a cycle with a close up view of the gear cable. outer gear cable inner gear cable outer gear cable inner gear cable nylon lining Fig. 1 (a) Name the force that has been applied, when manufacturing the inner gear cable to keep the strands of cable in position. … [1] (b) The outer gear cable has a nylon lining. Give two benefits of using the nylon lining. 1 … 2 … [2] (c) Explain why small radius bends should be avoided in the gear cable when it is attached to the cycle. … … … [2]

5 marks

Mark scheme: 1(a) Torque or torsion 1 Allow twisting 1(b) Benefits of the nylon lining could include: • Reduce friction • Reduce the amount of lubrication required • Prevent water getting to inner cable • Prevent inner cable rubbing on outer 2 × 1 marks for suitable benefits 2 Allow any other valid benefit. E.g. lightweight, waterproof, flexible. 1(c) Small radius bends will restrict the cable and prevent efficient movement of the inner cable. Permanent distortion of the outer cable could result. Possibility of cable snapping. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark.

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2 (a) Explain why steel is used for control cables on a cycle. … … … [2] (b) State the force being used when a control cable is operated. … [1]

3 marks

Mark scheme: 2(a) Explanation should mention that steel is durable, ductile, strong in tension, can bend and is able to return to same position. Low cost material that can also be recycled after use. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark. 2(b) Tension 1

This question in 0445/42 May/June 2018

10 Fig. 5 shows the initial design for a scaffold to give safe access while building work is carried out. X Y Fig. 5 (a) (i) Name the type of structure used in scaffolding. … [1] (ii) Draw on Fig. 5 three additional scaffolding poles to make the structure rigid and safe from collapse. [3] (iii) Explain the purpose of features X and Y at the base of the scaffold. … … … [2] (iv) When scaffolding is designed, a factor of safety for the structure is considered. Explain what is meant by a factor of safety. … … … [2] (b) Workers on a construction site have to wear ‘hard hats’ as shown in Fig. 6. Fig. 6 (i) State the type of structure used in the hard hat. … [1] (ii) Explain how the strength of the hard hat has been improved without adding any extra material. … … … [2] (c) Fig. 7 shows the axes for a stress/strain graph for mild steel. The position of three features that will appear on the graph are marked. C B A – elastic limit A B – upper yield point C – ultimate stress point stress strain Fig. 7 (i) Draw the shape of the graph on Fig. 7. [3] (ii) Explain what is meant by ‘elastic limit’ on the graph. … … … … [2] (d) Fig. 8a shows a folding clothes airer in the assembled position. Fig 8a Fig. 8b (i) Name the strengthening feature on the clothes airer shown in Fig. 8b. … [1] (ii) Give two reasons for using the type of strengthening feature shown in Fig. 8b. 1 … … 2 … … [2] (iii) The load carrying ability of each rail in the clothes airer is to be tested. Use sketches and notes on Fig. 9 to show how a dial gauge could be fixed to give a positive reading on the gauge when a load is added to the centre of the Ø6 rail. 750 Ø6 Ø12 [3] Fig. 9 Question 10 continues on the next page (e) Fig. 10 shows two concrete pillars. 7 kN 8 kN Ø200 175 × 175 pillar A pillar B Fig. 10 Calculate which pillar is subject to the greatest stress when the loads are applied. forceUse the formula: stress = cross sectional area … … … … [3]

25 marks

Mark scheme: 10(a)(i) Scaffolding is a frame structure 1 10(a)(ii) Each face of the structure to be made rigid, minimum of three poles. 1 mark for each pole in the correct position. 3 Maximum of 2 marks for multiple added poles in any one plane. 10(a)(iii) Features X and Y are to spread the load of the scaffold [1] and prevent it from sinking into soft ground. [1] They can also be used for levelling on uneven ground. [1] 2 1 mark each for any two correct points included. 10(a)(iv) A factor of safety is the load carrying capacity of a structure [1] beyond the expected loads. [1] The factor of safety is a number that is multiplied by the total expected load giving a load that design of structure must carry. [1] 2 One mark for each valid point made, allow examples in the explanation. Allow 2 marks for one point fully explained. 10(b)(i) The hard hat is a shell structure 1 10(b)(ii) Ribs have been moulded into the shell [1] to make the structure more rigid / prevent any flexing [1]. 2 Both points needed for 2 marks. 10(c)(i) 3 Smooth rise to elastic limit A[1] Sharp dip after upper yield point B[1] Gradual fall after ultimate stress point C[1] 10(c)(ii) Material will stretch / elongate until this point. [1] Elastic limit is the point beyond which the material will not return to its original length and is permanently deformed. [1] Stress is directly proportional to strain up to the elastic limit. [1] 2 Two points needed for 2 marks. 10(d)(i) Gusset plate 1 A B C stress strain A - elastic limit B - upper yield point C - ultimate stress point Question Answer Marks Guidance 10(d)(ii) Reasons will include: • To keep parts of the structure at 90° to each other • Reinforce the corner • Does not take up as much room as a strut or tie • Does not interfere with clothing on the rack • Lightweight • Low cost 2 × 1 marks 2 Allow other valid reasons. 10(d)(iii) • Dial gauge taking reading from bottom of rail [1] • Tip of dial gauge in contact with centre of rail [1] • Method of supporting dial gauge shown [1] 3 × 1 marks 3 10(e) Stress on pillar A = 7000 / 31415.9 mm2 = 7000 / 0.0314159 m2 = 222817 N/m2 1 mark = 0.223 N/mm2 Stress on pillar B = 8000 / 30625 mm2 = 8000 / 0.030625 m2 = 261224 N/m2 1 mark = 0.261 N/mm2 Pillar B is subject to greatest stress, 1 mark 3 Stress must be calculated for each pillar to gain marks. No ecf.1

This question in 0445/42 May/June 2018

1 Fig. 1 shows the rear derailleur gear on a cycle with a close up view of the gear cable. outer gear cable inner gear cable outer gear cable inner gear cable nylon lining Fig. 1 (a) Name the force that has been applied, when manufacturing the inner gear cable to keep the strands of cable in position. … [1] (b) The outer gear cable has a nylon lining. Give two benefits of using the nylon lining. 1 … 2 … [2] (c) Explain why small radius bends should be avoided in the gear cable when it is attached to the cycle. … … … [2]

5 marks

Mark scheme: 1(a) Torque or torsion 1 Allow twisting 1(b) Benefits of the nylon lining could include: • Reduce friction • Reduce the amount of lubrication required • Prevent water getting to inner cable • Prevent inner cable rubbing on outer 2 × 1 marks for suitable benefits 2 Allow any other valid benefit. E.g. lightweight, waterproof, flexible. 1(c) Small radius bends will restrict the cable and prevent efficient movement of the inner cable. Permanent distortion of the outer cable could result. Possibility of cable snapping. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark.

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2 (a) Explain why steel is used for control cables on a cycle. … … … [2] (b) State the force being used when a control cable is operated. … [1]

3 marks

Mark scheme: 2(a) Explanation should mention that steel is durable, ductile, strong in tension, can bend and is able to return to same position. Low cost material that can also be recycled after use. 2 Full explanation of a single point, 2 marks. Explanation that includes two points, 2 marks. One point mentioned, 1 mark. 2(b) Tension 1

This question in 0445/43 May/June 2018

10 Fig. 5 shows the initial design for a scaffold to give safe access while building work is carried out. X Y Fig. 5 (a) (i) Name the type of structure used in scaffolding. … [1] (ii) Draw on Fig. 5 three additional scaffolding poles to make the structure rigid and safe from collapse. [3] (iii) Explain the purpose of features X and Y at the base of the scaffold. … … … [2] (iv) When scaffolding is designed, a factor of safety for the structure is considered. Explain what is meant by a factor of safety. … … … [2] (b) Workers on a construction site have to wear ‘hard hats’ as shown in Fig. 6. Fig. 6 (i) State the type of structure used in the hard hat. … [1] (ii) Explain how the strength of the hard hat has been improved without adding any extra material. … … … [2] (c) Fig. 7 shows the axes for a stress/strain graph for mild steel. The position of three features that will appear on the graph are marked. C B A – elastic limit A B – upper yield point C – ultimate stress point stress strain Fig. 7 (i) Draw the shape of the graph on Fig. 7. [3] (ii) Explain what is meant by ‘elastic limit’ on the graph. … … … … [2] (d) Fig. 8a shows a folding clothes airer in the assembled position. Fig 8a Fig. 8b (i) Name the strengthening feature on the clothes airer shown in Fig. 8b. … [1] (ii) Give two reasons for using the type of strengthening feature shown in Fig. 8b. 1 … … 2 … … [2] (iii) The load carrying ability of each rail in the clothes airer is to be tested. Use sketches and notes on Fig. 9 to show how a dial gauge could be fixed to give a positive reading on the gauge when a load is added to the centre of the Ø6 rail. 750 Ø6 Ø12 [3] Fig. 9 Question 10 continues on the next page (e) Fig. 10 shows two concrete pillars. 7 kN 8 kN Ø200 175 × 175 pillar A pillar B Fig. 10 Calculate which pillar is subject to the greatest stress when the loads are applied. forceUse the formula: stress = cross sectional area … … … … [3]

25 marks

Mark scheme: 10(a)(i) Scaffolding is a frame structure 1 10(a)(ii) Each face of the structure to be made rigid, minimum of three poles. 1 mark for each pole in the correct position. 3 Maximum of 2 marks for multiple added poles in any one plane. 10(a)(iii) Features X and Y are to spread the load of the scaffold [1] and prevent it from sinking into soft ground. [1] They can also be used for levelling on uneven ground. [1] 2 1 mark each for any two correct points included. 10(a)(iv) A factor of safety is the load carrying capacity of a structure [1] beyond the expected loads. [1] The factor of safety is a number that is multiplied by the total expected load giving a load that design of structure must carry. [1] 2 One mark for each valid point made, allow examples in the explanation. Allow 2 marks for one point fully explained. 10(b)(i) The hard hat is a shell structure 1 10(b)(ii) Ribs have been moulded into the shell [1] to make the structure more rigid / prevent any flexing [1]. 2 Both points needed for 2 marks. 10(c)(i) 3 Smooth rise to elastic limit A[1] Sharp dip after upper yield point B[1] Gradual fall after ultimate stress point C[1] 10(c)(ii) Material will stretch / elongate until this point. [1] Elastic limit is the point beyond which the material will not return to its original length and is permanently deformed. [1] Stress is directly proportional to strain up to the elastic limit. [1] 2 Two points needed for 2 marks. 10(d)(i) Gusset plate 1 A B C stress strain A - elastic limit B - upper yield point C - ultimate stress point Question Answer Marks Guidance 10(d)(ii) Reasons will include: • To keep parts of the structure at 90° to each other • Reinforce the corner • Does not take up as much room as a strut or tie • Does not interfere with clothing on the rack • Lightweight • Low cost 2 × 1 marks 2 Allow other valid reasons. 10(d)(iii) • Dial gauge taking reading from bottom of rail [1] • Tip of dial gauge in contact with centre of rail [1] • Method of supporting dial gauge shown [1] 3 × 1 marks 3 10(e) Stress on pillar A = 7000 / 31415.9 mm2 = 7000 / 0.0314159 m2 = 222817 N/m2 1 mark = 0.223 N/mm2 Stress on pillar B = 8000 / 30625 mm2 = 8000 / 0.030625 m2 = 261224 N/m2 1 mark = 0.261 N/mm2 Pillar B is subject to greatest stress, 1 mark 3 Stress must be calculated for each pillar to gain marks. No ecf.1

This question in 0445/43 May/June 2018

7 Fig. 4 shows a commonly used method of reinforcing a joint on a structure made from steel tube. welded joints Fig. 4 (a) Name the method of reinforcement. … [1] (b) Use sketches and notes to show a different method of reinforcing the joint. [2]

3 marks

Mark scheme: 7(a) The joint is reinforced with a gusset plate. 1 7(b) Accept any functional method such as an angled tie or strut or any form of triangulation. Functional method, 1 mark. Clear drawing / notes, 1 mark. 2 Allow gusset plate if 7(a) is incorrect.

This question in 0445/41 Oct/Nov 2018

8 (a) State the force that is applied to a screw as it is inserted into a piece of wood. … [1] (b) Explain why more care has to be taken when inserting brass screws into a piece of wood compared to steel screws. … … … [2]

3 marks

Mark scheme: 8(a) Torsion / torque is applied to a screw that is being inserted. 1 8(b) 1 mark each for any two of the factors below: • Brass is a softer / more malleable material than steel • The screw slot or pozidrive slot can easily be damaged • The body of the screw can fail / shear due to excessive torsion. 2 Allow ‘not as strong as steel’.

This question in 0445/41 Oct/Nov 2018

9 Glass reinforced plastic (GRP) is a material that can be used for vehicle and boat construction. It consists of layers of chopped or woven glass strands in a polyester resin. (a) Name the class of materials to which GRP belongs. … [1] (b) State one advantage of using plastics in a structure rather than natural timber. … … [1]

2 marks

Mark scheme: 9(a) Composite material 1 9(b) Advantages could include: • Not damaged by damp / wet conditions • More durable • Not damaged by insect attack • No grain structure to cause a weak point • Plastics are generally lighter than timber • Can be manufactured into different shapes. 1 Accept any other valid advantage. Do not accept any cost related advantages.

This question in 0445/41 Oct/Nov 2018

10 (a) Draw and name one example of the structures named in each of the boxes below. natural frame man-made frame natural shell man-made shell [4] (b) Concrete is used extensively in the construction industry. (i) Describe how the tensile strength of concrete can be improved when it is used for making a beam. … … … [2] (ii) Give one reason why the compressive strength of concrete does not need improving. … … [1] (iii) Name one other force that concrete in a structure may have to withstand. … [1] (c) Fig. 5 shows a concrete beam with a 9 kN load acting on it. 9 kN R1 R2 2.25 m 3.75 m Fig. 5 Calculate the reactions at each end of the beam. … … … Reaction at R1 … kN Reaction at R2 … kN [4] (d) Fig. 6 shows an axle stand and ramps, both used to support the weight of a car. pin X ramps axle stand Fig. 6 (i) Circle an area on Fig. 6 where triangulation has been used to increase strength. [1] (ii) Pin X is used to adjust the height of the axle stand. State the force that will be acting on pin X when a car is being supported. … [1] (iii) Give two changes to pin X that will allow for an increased load. 1 … 2 … [2] (iv) Explain why steel angle as shown in Fig. 7 has been used to construct the ramps rather than flat section material. steel angle 25 × 25 × 3 Fig. 7 … … … [2] (v) The car ramps have to withstand both stationary and moving loads. Add notes to the drawing below to show how the moving load of a car ascending the ramps will affect the structure. [3] (vi) Give two reasons why welding has been chosen as the joining method on both the axle stand and the ramps. 1 … 2 … [2] (vii) Explain why the designers and manufacturers of the axle stand and ramps will have considered Factor of Safety in the design. … … … [2]

25 marks

Mark scheme: 10(a) 1 mark for each type of structure. E.g. spider’s web, ladder, snail shell, moulded toy. 4 labels to indicate what they are. 10(b)(i) Description to include the use of steel rod / re-bar / plastic fibres in the mix. 2 First mark for naming material, second mark for describing method. 10(b)(ii) Concrete is naturally strong in compression 1 Allow mark for understanding shown. 10(b)(iii) Torsion, [1] bending [1] or shear [1]. 1 10(c) Moments at R1 = 0 = (6 × R2) – (2.25 × 9) [1] R2 = 20.25 / 6, [1] R2 = 3.375 kN, [1] R1 = 9 – 3.375 = 5.625 kN, [1] 4 Award 4 marks for correct results with no working shown. Allow ecf for second value. Award 1 mark for correct numerical values incorrectly matched to R1 and R2. 10(d)(i) Triangular shape of the axle stand, 1 mark. 1 Accept wedge shape on ramp. 10(d)(ii) Shear, 1 mark 1 10(d)(iii) Increase diameter of pin, 1 mark Use higher tensile steel for pin, 1 mark. 2 Accept use of a stronger metal 10(d)(iv) Angle iron is used to: • increase resistance to torsion / bending / compression • increase rigidity or stiffness • Keep wheels / tyres in line with ramp 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. Allow ‘increased strength’. Question Answer Marks Guidance 10(d)(v) The moving car will try to push the ramp forward. The material of the ramp will affect each section of the length of the ramp as the car moves up. 3 2 marks for two points indicated on sketches or in notes, 1 mark for clarity of sketches / notes. Allow compression on top face, tension on bottom face as car ascends ramp. 10(d)(vi) Reasons for using welding include: • No need for disassembly • Joints are permanent • Strength of joint • Cannot work loose like nuts / bolts • Low cost as no extra components / drilling needed. 2 2 marks for any two valid reasons. Allow any other valid reason. 10(d)(vii) Reasons for consideration of Factor of Safety include: • Varying weights of vehicle • Conditions of use cannot be enforced after manufacture • Protection of manufacturer in the event of an accident • To enable a safe working load to be specified for users. 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. bending

This question in 0445/41 Oct/Nov 2018

7 Fig. 4 shows a commonly used method of reinforcing a joint on a structure made from steel tube. welded joints Fig. 4 (a) Name the method of reinforcement. … [1] (b) Use sketches and notes to show a different method of reinforcing the joint. [2]

3 marks

Mark scheme: 7(a) The joint is reinforced with a gusset plate. 1 7(b) Accept any functional method such as an angled tie or strut or any form of triangulation. Functional method, 1 mark. Clear drawing / notes, 1 mark. 2 Allow gusset plate if 7(a) is incorrect.

This question in 0445/42 Oct/Nov 2018

8 (a) State the force that is applied to a screw as it is inserted into a piece of wood. … [1] (b) Explain why more care has to be taken when inserting brass screws into a piece of wood compared to steel screws. … … … [2]

3 marks

Mark scheme: 8(a) Torsion / torque is applied to a screw that is being inserted. 1 8(b) 1 mark each for any two of the factors below: • Brass is a softer / more malleable material than steel • The screw slot or pozidrive slot can easily be damaged • The body of the screw can fail / shear due to excessive torsion. 2 Allow ‘not as strong as steel’.

This question in 0445/42 Oct/Nov 2018

9 Glass reinforced plastic (GRP) is a material that can be used for vehicle and boat construction. It consists of layers of chopped or woven glass strands in a polyester resin. (a) Name the class of materials to which GRP belongs. … [1] (b) State one advantage of using plastics in a structure rather than natural timber. … … [1]

2 marks

Mark scheme: 9(a) Composite material 1 9(b) Advantages could include: • Not damaged by damp / wet conditions • More durable • Not damaged by insect attack • No grain structure to cause a weak point • Plastics are generally lighter than timber • Can be manufactured into different shapes. 1 Accept any other valid advantage. Do not accept any cost related advantages.

This question in 0445/42 Oct/Nov 2018

10 (a) Draw and name one example of the structures named in each of the boxes below. natural frame man-made frame natural shell man-made shell [4] (b) Concrete is used extensively in the construction industry. (i) Describe how the tensile strength of concrete can be improved when it is used for making a beam. … … … [2] (ii) Give one reason why the compressive strength of concrete does not need improving. … … [1] (iii) Name one other force that concrete in a structure may have to withstand. … [1] (c) Fig. 5 shows a concrete beam with a 9 kN load acting on it. 9 kN R1 R2 2.25 m 3.75 m Fig. 5 Calculate the reactions at each end of the beam. … … … Reaction at R1 … kN Reaction at R2 … kN [4] (d) Fig. 6 shows an axle stand and ramps, both used to support the weight of a car. pin X ramps axle stand Fig. 6 (i) Circle an area on Fig. 6 where triangulation has been used to increase strength. [1] (ii) Pin X is used to adjust the height of the axle stand. State the force that will be acting on pin X when a car is being supported. … [1] (iii) Give two changes to pin X that will allow for an increased load. 1 … 2 … [2] (iv) Explain why steel angle as shown in Fig. 7 has been used to construct the ramps rather than flat section material. steel angle 25 × 25 × 3 Fig. 7 … … … [2] (v) The car ramps have to withstand both stationary and moving loads. Add notes to the drawing below to show how the moving load of a car ascending the ramps will affect the structure. [3] (vi) Give two reasons why welding has been chosen as the joining method on both the axle stand and the ramps. 1 … 2 … [2] (vii) Explain why the designers and manufacturers of the axle stand and ramps will have considered Factor of Safety in the design. … … … [2]

25 marks

Mark scheme: 10(a) 1 mark for each type of structure. E.g. spider’s web, ladder, snail shell, moulded toy. 4 labels to indicate what they are. 10(b)(i) Description to include the use of steel rod / re-bar / plastic fibres in the mix. 2 First mark for naming material, second mark for describing method. 10(b)(ii) Concrete is naturally strong in compression 1 Allow mark for understanding shown. 10(b)(iii) Torsion, [1] bending [1] or shear [1]. 1 10(c) Moments at R1 = 0 = (6 × R2) – (2.25 × 9) [1] R2 = 20.25 / 6, [1] R2 = 3.375 kN, [1] R1 = 9 – 3.375 = 5.625 kN, [1] 4 Award 4 marks for correct results with no working shown. Allow ecf for second value. Award 1 mark for correct numerical values incorrectly matched to R1 and R2. 10(d)(i) Triangular shape of the axle stand, 1 mark. 1 Accept wedge shape on ramp. 10(d)(ii) Shear, 1 mark 1 10(d)(iii) Increase diameter of pin, 1 mark Use higher tensile steel for pin, 1 mark. 2 Accept use of a stronger metal 10(d)(iv) Angle iron is used to: • increase resistance to torsion / bending / compression • increase rigidity or stiffness • Keep wheels / tyres in line with ramp 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. Allow ‘increased strength’. Question Answer Marks Guidance 10(d)(v) The moving car will try to push the ramp forward. The material of the ramp will affect each section of the length of the ramp as the car moves up. 3 2 marks for two points indicated on sketches or in notes, 1 mark for clarity of sketches / notes. Allow compression on top face, tension on bottom face as car ascends ramp. 10(d)(vi) Reasons for using welding include: • No need for disassembly • Joints are permanent • Strength of joint • Cannot work loose like nuts / bolts • Low cost as no extra components / drilling needed. 2 2 marks for any two valid reasons. Allow any other valid reason. 10(d)(vii) Reasons for consideration of Factor of Safety include: • Varying weights of vehicle • Conditions of use cannot be enforced after manufacture • Protection of manufacturer in the event of an accident • To enable a safe working load to be specified for users. 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. bending

This question in 0445/42 Oct/Nov 2018

7 Fig. 4 shows a commonly used method of reinforcing a joint on a structure made from steel tube. welded joints Fig. 4 (a) Name the method of reinforcement. … [1] (b) Use sketches and notes to show a different method of reinforcing the joint. [2]

3 marks

Mark scheme: 7(a) The joint is reinforced with a gusset plate. 1 7(b) Accept any functional method such as an angled tie or strut or any form of triangulation. Functional method, 1 mark. Clear drawing / notes, 1 mark. 2 Allow gusset plate if 7(a) is incorrect.

This question in 0445/43 Oct/Nov 2018

8 (a) State the force that is applied to a screw as it is inserted into a piece of wood. … [1] (b) Explain why more care has to be taken when inserting brass screws into a piece of wood compared to steel screws. … … … [2]

3 marks

Mark scheme: 8(a) Torsion / torque is applied to a screw that is being inserted. 1 8(b) 1 mark each for any two of the factors below: • Brass is a softer / more malleable material than steel • The screw slot or pozidrive slot can easily be damaged • The body of the screw can fail / shear due to excessive torsion. 2 Allow ‘not as strong as steel’.

This question in 0445/43 Oct/Nov 2018

9 Glass reinforced plastic (GRP) is a material that can be used for vehicle and boat construction. It consists of layers of chopped or woven glass strands in a polyester resin. (a) Name the class of materials to which GRP belongs. … [1] (b) State one advantage of using plastics in a structure rather than natural timber. … … [1]

2 marks

Mark scheme: 9(a) Composite material 1 9(b) Advantages could include: • Not damaged by damp / wet conditions • More durable • Not damaged by insect attack • No grain structure to cause a weak point • Plastics are generally lighter than timber • Can be manufactured into different shapes. 1 Accept any other valid advantage. Do not accept any cost related advantages.

This question in 0445/43 Oct/Nov 2018

10 (a) Draw and name one example of the structures named in each of the boxes below. natural frame man-made frame natural shell man-made shell [4] (b) Concrete is used extensively in the construction industry. (i) Describe how the tensile strength of concrete can be improved when it is used for making a beam. … … … [2] (ii) Give one reason why the compressive strength of concrete does not need improving. … … [1] (iii) Name one other force that concrete in a structure may have to withstand. … [1] (c) Fig. 5 shows a concrete beam with a 9 kN load acting on it. 9 kN R1 R2 2.25 m 3.75 m Fig. 5 Calculate the reactions at each end of the beam. … … … Reaction at R1 … kN Reaction at R2 … kN [4] (d) Fig. 6 shows an axle stand and ramps, both used to support the weight of a car. pin X ramps axle stand Fig. 6 (i) Circle an area on Fig. 6 where triangulation has been used to increase strength. [1] (ii) Pin X is used to adjust the height of the axle stand. State the force that will be acting on pin X when a car is being supported. … [1] (iii) Give two changes to pin X that will allow for an increased load. 1 … 2 … [2] (iv) Explain why steel angle as shown in Fig. 7 has been used to construct the ramps rather than flat section material. steel angle 25 × 25 × 3 Fig. 7 … … … [2] (v) The car ramps have to withstand both stationary and moving loads. Add notes to the drawing below to show how the moving load of a car ascending the ramps will affect the structure. [3] (vi) Give two reasons why welding has been chosen as the joining method on both the axle stand and the ramps. 1 … 2 … [2] (vii) Explain why the designers and manufacturers of the axle stand and ramps will have considered Factor of Safety in the design. … … … [2]

25 marks

Mark scheme: 10(a) 1 mark for each type of structure. E.g. spider’s web, ladder, snail shell, moulded toy. 4 labels to indicate what they are. 10(b)(i) Description to include the use of steel rod / re-bar / plastic fibres in the mix. 2 First mark for naming material, second mark for describing method. 10(b)(ii) Concrete is naturally strong in compression 1 Allow mark for understanding shown. 10(b)(iii) Torsion, [1] bending [1] or shear [1]. 1 10(c) Moments at R1 = 0 = (6 × R2) – (2.25 × 9) [1] R2 = 20.25 / 6, [1] R2 = 3.375 kN, [1] R1 = 9 – 3.375 = 5.625 kN, [1] 4 Award 4 marks for correct results with no working shown. Allow ecf for second value. Award 1 mark for correct numerical values incorrectly matched to R1 and R2. 10(d)(i) Triangular shape of the axle stand, 1 mark. 1 Accept wedge shape on ramp. 10(d)(ii) Shear, 1 mark 1 10(d)(iii) Increase diameter of pin, 1 mark Use higher tensile steel for pin, 1 mark. 2 Accept use of a stronger metal 10(d)(iv) Angle iron is used to: • increase resistance to torsion / bending / compression • increase rigidity or stiffness • Keep wheels / tyres in line with ramp 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. Allow ‘increased strength’. Question Answer Marks Guidance 10(d)(v) The moving car will try to push the ramp forward. The material of the ramp will affect each section of the length of the ramp as the car moves up. 3 2 marks for two points indicated on sketches or in notes, 1 mark for clarity of sketches / notes. Allow compression on top face, tension on bottom face as car ascends ramp. 10(d)(vi) Reasons for using welding include: • No need for disassembly • Joints are permanent • Strength of joint • Cannot work loose like nuts / bolts • Low cost as no extra components / drilling needed. 2 2 marks for any two valid reasons. Allow any other valid reason. 10(d)(vii) Reasons for consideration of Factor of Safety include: • Varying weights of vehicle • Conditions of use cannot be enforced after manufacture • Protection of manufacturer in the event of an accident • To enable a safe working load to be specified for users. 2 Allow 2 marks for a single point well explained. 2 marks for two points in less depth. bending

This question in 0445/43 Oct/Nov 2018

3 Fig. 2 shows two types of joint that can be used for joining a chair rail to a leg. mortise and tenon dowel joint with wedges A B Fig. 2 (a) Explain why joint A will resist tensile force better than joint B. … … … … [2] (b) Explain why joint A will resist shear force better than joint B. … … … … [2]

4 marks

Mark scheme: 3(a) Explanation to focus on: • Use of wedges • The widening of the mortise at the back • Dovetail effect when the glue has hardened • Larger surface area for adhesive. Explanation including any two valid points 2 marks. 2 3(b) Explanation to focus on: • Larger cross sectional area of tenon • Tenon is in one piece with the rail • Dowels may be a weaker timber than the rail • Larger cross sectional area in joint A Explanation including any two valid points 2 marks. 2

This question in 0445/41 May/June 2019

4 Steel plates on a building structure can be joined together using temporary or permanent methods. (a) Give one temporary method of joining two steel plates together. … [1] (b) Give one permanent method of joining two steel plates together. … [1]

2 marks

Mark scheme: 4(a) Nut and bolt, self tapping screw, 1 mark. 1 Allow any method that can be taken apart. No mark if temporary and permanent method used. 4(b) Welding, riveting, brazing, epoxy resin adhesive, 1 mark. 1 Allow any method that does not allow taking apart. No mark if temporary and permanent method used.

This question in 0445/41 May/June 2019

5 State what is meant by ‘static load’ on a bridge. … [1]

1 marks

Mark scheme: 5 The static load is any load on the bridge that will not move, e.g. weight of bridge components, 1 mark. 1

This question in 0445/41 May/June 2019

9 (a) Fig. 6 shows a model truss bridge that has been made in a school to span a 270 mm gap. 300 80 Fig. 6 Use sketches and notes to show how the model truss bridge can be tested. [3] (b) Fig. 7 shows an arch bridge built using stone. Fig. 7 (i) State the name of this type of structure. … [1] (ii) Describe, using an example, how this type of structure gains its strength. … … … [2] (iii) Fig. 8 shows an arch similar to those in Fig. 7. keystone Fig. 8 State the purpose of the keystone. … … [1] (iv) Stone is a natural material that is used in construction work. Give two advantages of stone compared to concrete. 1 … 2 … [2] (v) Give two advantages of concrete compared to stone. 1 … 2 … [2] (c) Fig. 9 shows part of the underside of a wooden aircraft wing with the canvas covering removed from the lower surface of the wing. wires wooden ribs canvas covering Fig. 9 (i) State two methods of strengthening that can be seen in Fig. 9. 1 … 2 … [2] (ii) Explain why wood is a suitable choice of material for the wing structure. … … … [2] (iii) Describe two factors that should be considered when selecting wood for any structure. … … … … [2] (d) Fig. 10 shows a car park barrier that lifts to allow a car to enter. 0.5 m 1.25 m 0.25 m X 50 N 10 N Fig. 10 Calculate the value of force X that will keep the barrier in equilibrium. … … … … [3] (e) A steel cable used to pull a car onto the back of a breakdown truck has an original length of 10 m. The cable stretches by 1.3 mm when the cable starts to pull. (i) Calculate the strain in the cable. … … … [2] (ii) State the type of strain that occurs in the cable. … [1] (iii) The ‘elastic limit’ of the cable is important to the manufacturer when deciding the maximum load that can be placed on the cable. Explain what is meant by elastic limit. … … … [2]

25 marks

Mark scheme: 9(a) Testing should include: Placing on supports 270 mm apart, 1 mark Adding load to bridge, 1 mark Load distributed across span of bridge or point load placed in centre, 1 mark 3 9(b)(i) The bridge is a mass structure 1 9(b)(ii) Strength is gained from the large amount of material used, the weight of the structure will provide strength, e.g. brick structures, concrete structures, dams 2 1 mark for example 1 mark for understanding of where the strength comes from. 9(b)(iii) The keystone will lock the two halves of the structure together preventing the arch from collapsing. The keystone is wedge shaped. 1 1 mark for understanding shown. 9(b)(iv) Advantages of stone will include: • Aesthetic qualities • Strength in tension compared to concrete with no reinforcement • Durable • Weather resistant • 2 × 1 marks for suitable advantages. 2 Allow any other valid advantages. 9(b)(v) Advantages of concrete will include: • Can be formed into different shapes • Reinforcement can be included • Texture can be cast into the concrete • Easier to transport in the raw material form than stone. 2 Allow any other valid advantages. Accept ‘cheaper’. 9(c)(i) Visible strengthening methods are: • Gusset plates • Triangulated struts • Wire ties. 2 × 1 marks 2 Question Answer Marks Guidance 9(c)(ii) Reasons for suitability include: • Light weight • Ease of bending • Will flex to accommodate dynamic loading • Readily available • Durable 2 Explanation with two points, 2 marks Allow 2 marks for clear explanation of a single point. 9(c)(iii) Factors to be considered include; • Natural defects in the timber, E.g. knots, splits • Insect damage • Moisture content • Strength class of the wood • Rate of growth • Stability 2 Description that includes two valid points, 2 marks. Accept other valid alternatives. 9(d) 50 × 1.25 = (10 × 0.25) + 0.5 × X [1] 62.5 – 2.5 = 0.5X [1] X = 120N [1] 3 Award 3 marks for correct answer with no working shown. 9(e)(i) Strain = 1.3 / 10000 [1] Strain = 0.00013 [1] 2 Award 2 marks for correct answer with no working shown. 9(e)(ii) Tensile strain. 1 9(e)(iii) The elastic limit is the amount that the cable can be stretched [1] before permanent distortion takes place [1]. i.e. if the load is removed before the elastic limit is reached the cable will return to the original length. 2 Both points to be included for 2 marks.

This question in 0445/41 May/June 2019

3 (a) The body of a tumble drier is normally made from sheet steel. State the name given to this type of structure. … [1] (b) Use sketches and notes to show how a flat sheet of steel can be made more resistant to bending and torsion without adding any extra parts to it.

3 marks

Mark scheme: 3(a) Shell structure, 1 mark. 1 3(b) Sketches to show folds, creases or indentations to stiffen the sheet steel, 1 mark. Notes on suitable feature included, 1 mark. 2 No marks for constructing a box or similar

This question in 0445/42 May/June 2019

8 (a) Fig. 4 shows a table of materials that can be used for a load bearing beam and a list of properties that are required in the beam. (i) For each property place one tick (✓) in the material column in the table that best matches the property required. One has been done for you. laminated Property required wood steel concrete wood ✓ renewable natural resource resists torsion and bending long lasting and no maintenance needed can be easily formed into curves Fig. 4 [3] (ii) A beam made from wood could contain natural defects. State two natural defects that can occur in wood. 1 … 2 … [2] (b) Fig. 5 shows examples of three structures. A C B Fig. 5 (i) Circle the type of structure used in each example. A frame structure shell structure mass structure B frame structure shell structure mass structure C frame structure shell structure mass structure [3] (ii) State two methods that have been used to strengthen structure A. 1 … 2 … [2] (c) Fig. 6 shows a folding aluminium work platform used for building work or decorating. Fig. 6 (i) The platform is designed to take stationary and moving loads. Explain, using examples, what is meant by each type of load. … … … … [4] (ii) Fig. 7 shows part of the safety label attached to the leg of the work platform. Fig. 7 Explain how the manufacturer has considered Factor of Safety when designing the work platform. … … … … [2] (iii) Fig. 8 shows the work platform with the loads from a user shown in position. 1000 200 300 400 N 400 N R1 R2 Fig. 8 Calculate the reactions at R1 and R2. … … … … [3] (iv) Describe how the stress in a structure can be calculated. … … … [2] (d) Use sketches and notes to show what is meant by the following forces in a structure: Tension Compression Shear Torsion [4]

25 marks

Mark scheme: 8(a)(i) 1 mark for each row correct. 3 x 1 marks 3 8(a)(ii) Defects wood will include: • Knots • Shakes • Splits • Warping • Rot / decay 2 × 1 marks for suitable defects. 2 Allow insect / termite damage Allow mould 8(b)(i) (a) Frame structure [1] (b) Shell structure [1] (c) Mass structure [1] 3 8(b)(ii) Gusset plate, 1 mark Triangulation, 1 mark Strut, 1 mark 2 × 1 marks. 2 renewable natural resource resists torsion and bending long lasting and no maintenance needed can be easily formed into curves Property required steel concrete wood laminated wood 9 9 9 9 Question Answer Marks Guidance 8(c)(i) • Stationary load will not move and consists of the materials making up the structure plus and non-moving materials placed on the structure. • Moving load will be any load such as a person or vehicle that moves across or on the structure. Explanation must include both types of load for two marks and a valid example for each. 4 8(c)(ii) By indicating that only one person should use the platform and that the maximum loading is 150kg the manufacturer has taken the likely load from one person and added the equipment / tools that are likely to be carried. To this figure an additional loading has been estimated and the design will have been tested to that loading. 1 mark for each valid point in explanation. 2 Allow 2 marks for a single point well explained.. 8(c)(iii) (0.2 × 400) + (0.5 × 400) = R2 × 1 [1] 80 + 200 = R2 = 280 N [1] R1 = 800 – 280 R1 = 520 N [1] 3 3 marks for correct answers with no working. Maximum 2 marks if reactions reversed 8(c)(iv) Stress formula = Force [1] / Cross sectional area [1] Stress at each end is divided between the two legs. [1] 2 × 1 marks 2 Allow mark for reference to the given structure. 8(d) Notes and sketches to show clearly what is meant by each force. 4 × 1 marks. 4 Accept arrows to indicate each force.

This question in 0445/42 May/June 2019

3 Fig. 2 shows two types of joint that can be used for joining a chair rail to a leg. mortise and tenon dowel joint with wedges A B Fig. 2 (a) Explain why joint A will resist tensile force better than joint B. … … … … [2] (b) Explain why joint A will resist shear force better than joint B. … … … … [2]

4 marks

Mark scheme: 3(a) Explanation to focus on: • Use of wedges • The widening of the mortise at the back • Dovetail effect when the glue has hardened • Larger surface area for adhesive. Explanation including any two valid points 2 marks. 2 3(b) Explanation to focus on: • Larger cross sectional area of tenon • Tenon is in one piece with the rail • Dowels may be a weaker timber than the rail • Larger cross sectional area in joint A Explanation including any two valid points 2 marks. 2

This question in 0445/43 May/June 2019

4 Steel plates on a building structure can be joined together using temporary or permanent methods. (a) Give one temporary method of joining two steel plates together. … [1] (b) Give one permanent method of joining two steel plates together. … [1]

2 marks

Mark scheme: 4(a) Nut and bolt, self tapping screw, 1 mark. 1 Allow any method that can be taken apart. No mark if temporary and permanent method used. 4(b) Welding, riveting, brazing, epoxy resin adhesive, 1 mark. 1 Allow any method that does not allow taking apart. No mark if temporary and permanent method used.

This question in 0445/43 May/June 2019

5 State what is meant by ‘static load’ on a bridge. … [1]

1 marks

Mark scheme: 5 The static load is any load on the bridge that will not move, e.g. weight of bridge components, 1 mark. 1

This question in 0445/43 May/June 2019

9 (a) Fig. 6 shows a model truss bridge that has been made in a school to span a 270 mm gap. 300 80 Fig. 6 Use sketches and notes to show how the model truss bridge can be tested. [3] (b) Fig. 7 shows an arch bridge built using stone. Fig. 7 (i) State the name of this type of structure. … [1] (ii) Describe, using an example, how this type of structure gains its strength. … … … [2] (iii) Fig. 8 shows an arch similar to those in Fig. 7. keystone Fig. 8 State the purpose of the keystone. … … [1] (iv) Stone is a natural material that is used in construction work. Give two advantages of stone compared to concrete. 1 … 2 … [2] (v) Give two advantages of concrete compared to stone. 1 … 2 … [2] (c) Fig. 9 shows part of the underside of a wooden aircraft wing with the canvas covering removed from the lower surface of the wing. wires wooden ribs canvas covering Fig. 9 (i) State two methods of strengthening that can be seen in Fig. 9. 1 … 2 … [2] (ii) Explain why wood is a suitable choice of material for the wing structure. … … … [2] (iii) Describe two factors that should be considered when selecting wood for any structure. … … … … [2] (d) Fig. 10 shows a car park barrier that lifts to allow a car to enter. 0.5 m 1.25 m 0.25 m X 50 N 10 N Fig. 10 Calculate the value of force X that will keep the barrier in equilibrium. … … … … [3] (e) A steel cable used to pull a car onto the back of a breakdown truck has an original length of 10 m. The cable stretches by 1.3 mm when the cable starts to pull. (i) Calculate the strain in the cable. … … … [2] (ii) State the type of strain that occurs in the cable. … [1] (iii) The ‘elastic limit’ of the cable is important to the manufacturer when deciding the maximum load that can be placed on the cable. Explain what is meant by elastic limit. … … … [2]

25 marks

Mark scheme: 9(a) Testing should include: Placing on supports 270 mm apart, 1 mark Adding load to bridge, 1 mark Load distributed across span of bridge or point load placed in centre, 1 mark 3 9(b)(i) The bridge is a mass structure 1 9(b)(ii) Strength is gained from the large amount of material used, the weight of the structure will provide strength, e.g. brick structures, concrete structures, dams 2 1 mark for example 1 mark for understanding of where the strength comes from. 9(b)(iii) The keystone will lock the two halves of the structure together preventing the arch from collapsing. The keystone is wedge shaped. 1 1 mark for understanding shown. 9(b)(iv) Advantages of stone will include: • Aesthetic qualities • Strength in tension compared to concrete with no reinforcement • Durable • Weather resistant • 2 × 1 marks for suitable advantages. 2 Allow any other valid advantages. 9(b)(v) Advantages of concrete will include: • Can be formed into different shapes • Reinforcement can be included • Texture can be cast into the concrete • Easier to transport in the raw material form than stone. 2 Allow any other valid advantages. Accept ‘cheaper’. 9(c)(i) Visible strengthening methods are: • Gusset plates • Triangulated struts • Wire ties. 2 × 1 marks 2 Question Answer Marks Guidance 9(c)(ii) Reasons for suitability include: • Light weight • Ease of bending • Will flex to accommodate dynamic loading • Readily available • Durable 2 Explanation with two points, 2 marks Allow 2 marks for clear explanation of a single point. 9(c)(iii) Factors to be considered include; • Natural defects in the timber, E.g. knots, splits • Insect damage • Moisture content • Strength class of the wood • Rate of growth • Stability 2 Description that includes two valid points, 2 marks. Accept other valid alternatives. 9(d) 50 × 1.25 = (10 × 0.25) + 0.5 × X [1] 62.5 – 2.5 = 0.5X [1] X = 120N [1] 3 Award 3 marks for correct answer with no working shown. 9(e)(i) Strain = 1.3 / 10000 [1] Strain = 0.00013 [1] 2 Award 2 marks for correct answer with no working shown. 9(e)(ii) Tensile strain. 1 9(e)(iii) The elastic limit is the amount that the cable can be stretched [1] before permanent distortion takes place [1]. i.e. if the load is removed before the elastic limit is reached the cable will return to the original length. 2 Both points to be included for 2 marks.

This question in 0445/43 May/June 2019

1 Models are often used when designing a structure. Models can be manufactured or they can be computer generated. (a) Give two benefits of using a manufactured model. 1 … 2 … [2] (b) Give two benefits of using a computer generated model. 1 … 2 … [2] (c) State one other use for a computer in manufacturing. … [1]

5 marks

Mark scheme: 1(a) Benefits of using a model could include: • Can be tested • View all-round the model • Dimensions / proportions can be checked • Different materials can be tried • Shows more than a drawing • Easier for a client to understand 2 × 1 marks 2 Accept any other valid benefit 1(b) Benefits of a computer model could include: • Can be viewed three dimensionally • Accurate testing can be carried out quickly • Variables can be changed quickly E.g. dimensions of components • Model can be sent to other people easily • Can be stored easily / does not take up physical space • No cost for materials involved 2 × 1 marks 2 Accept any other valid benefit Reference to speed must be justified 1(c) Other uses include: • Stock control • CAD / CAM • Quality control • Internet research 1 mark for suitable use 1 Accept any other valid use

This question in 0445/41 Oct/Nov 2019

2 Explain what is meant by equilibrium in a structure. … … … [2]

2 marks

Mark scheme: 2 Explanation could include the following points: • Opposing forces are balanced • A state of rest / structure is stable • Structure is not moving • Clockwise moment = anticlockwise moment 2 Explanation that includes two points, two marks 1 mark for one point in explanation. Allow two marks for single point fully explained

This question in 0445/41 Oct/Nov 2019

3 State the meaning of the term ‘moment’ in a structure. … [1]

1 marks

Mark scheme: 3 1

This question in 0445/41 Oct/Nov 2019

9 (a) Fig. 3 shows scaffolding on the outside of a building. Fig. 3 (i) State the type of structure used in scaffolding. … [1] (ii) Identify two safety features of the scaffolding that have been added to protect the public and workers using the scaffolding. 1 … 2 … [2] (iii) State the reason for using triangulation in a scaffolding structure. … [1] (iv) Describe the difference between a strut and a tie in a structure. … … … [2] (v) The lower part of a vertical scaffold pole is shown in Fig. 4. ground level Fig. 4 Use sketches and notes to show how the vertical load from the scaffold pole to the ground can be spread over a greater area. [3] (b) Fig. 5 shows a concrete barrier built across a disused dock to hold back sea water. walkway A B ground level wire face A face B wire fittings sea water concrete barrier Fig. 5 (i) State the force that will be acting on face A, the sea water side, and face B, the disused dock side of the barrier. Force at A … Force at B … [2] (ii) Give one reason why stainless steel would be more suitable than mild steel for the wire and fittings. … [1] (iii) Movement in the wall can be monitored by using a strain gauge. Describe the method of fixing a strain gauge to the wall. … … … [2] (c) Fig. 6 shows a design idea for a barrow for moving heavy building materials. Two methods of cutting the square steel tube to form a corner joint are shown. corner joint method 1 method 2 barrow Fig. 6 (i) Give one benefit of each method of cutting the tube ready for joining. Method 1 … Method 2 … [2] (ii) Describe how the tubes in method 1 could be joined together. … … … [2] (iii) Use sketches and notes to show how a gusset plate could be fitted to the corner joint. [2] (iv) Fig. 7 shows two views of the barrow with a load in place. 850 200 N 175 Fig. 7 Calculate the effort needed to pull back on the handles and lift the load. … … … [3] (v) Explain why the effort needed will reduce as the handles are pulled back. … … … [2]

25 marks

Mark scheme: 9(a)(i) Frame structure, 1 mark 1 9(a)(ii) Safety features include: • Wrapping around lower part of poles at street level • Mesh fencing at each of upper levels • Vertical board on edge at each upper level • Upper levels boarded across full width 2 × 1 marks for features identified 2 Accept any other valid points related to arrangement of poles. 9(a)(iii) Triangulation is used to hold the structure rigid and prevent any flexing or movement away from the wall 1 Award mark for understanding shown. 9(a)(iv) A strut will resist compression, 1 mark A tie will resist tension, 1 mark 2 9(a)(v) Method of spreading the load could include larger steel feet that fit into the bottom of vertical pole or boards placed below Functional method 1 mark Clear illustration / notes, 1 mark Method of fixing to vertical pole clearly shown, 1 mark 3 9(b)(i) Force at A is compression, 1 mark Force at B is tension, 1 mark 2 9(b)(ii) Non rusting / higher tensile strength / will last longer, 1 mark 1 Question Answer Marks Guidance 9(b)(iii) Attached to the wall using glue, 1 mark. Suitable adhesive, e.g. epoxy resin or cyanoacrylate, 1 mark Alignment arrows should be used to attach square to the wall with longer face horizontal, 1 mark 2 × 1 marks 2 Accept superglue 9(c)(i) Benefit of method 1: • Straight cut at 90º, easy to align for joining Benefit of method 2: • Tube remains sealed when joint is complete 2 Accept other valid benefits Allow ease of cutting at 90º Allow greater surface area 9(c)(ii) Joining technique could be: • Aligning joint accurately, 1 mark Followed by: • Welding / brazing • Bolt and captive nut inside one tube 1 mark for either 2 9(c)(iii) Gusset plate shown / described, 1 mark Clear drawing notes of positioning, 1 mark 2 9(c)(iv) Substitution into formula, 200 × 175 = effort × 850, 1 mark 35 000 / 850 = effort, 1 mark Effort = 41.176 , 1 mark 3 Award 3 marks for correct answer with no working shown. Accept 41.2N or 41.18 Question Answer Marks Guidance 9(c)(v) With the barrow at rest the load is acting 175 mm away from the fulcrum, 1 mark. The effort will reduce when the barrow is rotated because the perpendicular distance from the load to the fulcrum decreases. 1 mark 2 Allow marks for understanding shown.

This question in 0445/41 Oct/Nov 2019

1 Models are often used when designing a structure. Models can be manufactured or they can be computer generated. (a) Give two benefits of using a manufactured model. 1 … 2 … [2] (b) Give two benefits of using a computer generated model. 1 … 2 … [2] (c) State one other use for a computer in manufacturing. … [1]

5 marks

Mark scheme: 1(a) Benefits of using a model could include: • Can be tested • View all-round the model • Dimensions / proportions can be checked • Different materials can be tried • Shows more than a drawing • Easier for a client to understand 2 × 1 marks 2 Accept any other valid benefit 1(b) Benefits of a computer model could include: • Can be viewed three dimensionally • Accurate testing can be carried out quickly • Variables can be changed quickly E.g. dimensions of components • Model can be sent to other people easily • Can be stored easily / does not take up physical space • No cost for materials involved 2 × 1 marks 2 Accept any other valid benefit Reference to speed must be justified 1(c) Other uses include: • Stock control • CAD / CAM • Quality control • Internet research 1 mark for suitable use 1 Accept any other valid use

This question in 0445/42 Oct/Nov 2019

2 Explain what is meant by equilibrium in a structure. … … … [2]

2 marks

Mark scheme: 2 Explanation could include the following points: • Opposing forces are balanced • A state of rest / structure is stable • Structure is not moving • Clockwise moment = anticlockwise moment 2 Explanation that includes two points, two marks 1 mark for one point in explanation. Allow two marks for single point fully explained

This question in 0445/42 Oct/Nov 2019

3 State the meaning of the term ‘moment’ in a structure. … [1]

1 marks

Mark scheme: 3 1

This question in 0445/42 Oct/Nov 2019

9 (a) Fig. 3 shows scaffolding on the outside of a building. Fig. 3 (i) State the type of structure used in scaffolding. … [1] (ii) Identify two safety features of the scaffolding that have been added to protect the public and workers using the scaffolding. 1 … 2 … [2] (iii) State the reason for using triangulation in a scaffolding structure. … [1] (iv) Describe the difference between a strut and a tie in a structure. … … … [2] (v) The lower part of a vertical scaffold pole is shown in Fig. 4. ground level Fig. 4 Use sketches and notes to show how the vertical load from the scaffold pole to the ground can be spread over a greater area. [3] (b) Fig. 5 shows a concrete barrier built across a disused dock to hold back sea water. walkway A B ground level wire face A face B wire fittings sea water concrete barrier Fig. 5 (i) State the force that will be acting on face A, the sea water side, and face B, the disused dock side of the barrier. Force at A … Force at B … [2] (ii) Give one reason why stainless steel would be more suitable than mild steel for the wire and fittings. … [1] (iii) Movement in the wall can be monitored by using a strain gauge. Describe the method of fixing a strain gauge to the wall. … … … [2] (c) Fig. 6 shows a design idea for a barrow for moving heavy building materials. Two methods of cutting the square steel tube to form a corner joint are shown. corner joint method 1 method 2 barrow Fig. 6 (i) Give one benefit of each method of cutting the tube ready for joining. Method 1 … Method 2 … [2] (ii) Describe how the tubes in method 1 could be joined together. … … … [2] (iii) Use sketches and notes to show how a gusset plate could be fitted to the corner joint. [2] (iv) Fig. 7 shows two views of the barrow with a load in place. 850 200 N 175 Fig. 7 Calculate the effort needed to pull back on the handles and lift the load. … … … [3] (v) Explain why the effort needed will reduce as the handles are pulled back. … … … [2]

25 marks

Mark scheme: 9(a)(i) Frame structure, 1 mark 1 9(a)(ii) Safety features include: • Wrapping around lower part of poles at street level • Mesh fencing at each of upper levels • Vertical board on edge at each upper level • Upper levels boarded across full width 2 × 1 marks for features identified 2 Accept any other valid points related to arrangement of poles. 9(a)(iii) Triangulation is used to hold the structure rigid and prevent any flexing or movement away from the wall 1 Award mark for understanding shown. 9(a)(iv) A strut will resist compression, 1 mark A tie will resist tension, 1 mark 2 9(a)(v) Method of spreading the load could include larger steel feet that fit into the bottom of vertical pole or boards placed below Functional method 1 mark Clear illustration / notes, 1 mark Method of fixing to vertical pole clearly shown, 1 mark 3 9(b)(i) Force at A is compression, 1 mark Force at B is tension, 1 mark 2 9(b)(ii) Non rusting / higher tensile strength / will last longer, 1 mark 1 Question Answer Marks Guidance 9(b)(iii) Attached to the wall using glue, 1 mark. Suitable adhesive, e.g. epoxy resin or cyanoacrylate, 1 mark Alignment arrows should be used to attach square to the wall with longer face horizontal, 1 mark 2 × 1 marks 2 Accept superglue 9(c)(i) Benefit of method 1: • Straight cut at 90º, easy to align for joining Benefit of method 2: • Tube remains sealed when joint is complete 2 Accept other valid benefits Allow ease of cutting at 90º Allow greater surface area 9(c)(ii) Joining technique could be: • Aligning joint accurately, 1 mark Followed by: • Welding / brazing • Bolt and captive nut inside one tube 1 mark for either 2 9(c)(iii) Gusset plate shown / described, 1 mark Clear drawing notes of positioning, 1 mark 2 9(c)(iv) Substitution into formula, 200 × 175 = effort × 850, 1 mark 35 000 / 850 = effort, 1 mark Effort = 41.176 , 1 mark 3 Award 3 marks for correct answer with no working shown. Accept 41.2N or 41.18 Question Answer Marks Guidance 9(c)(v) With the barrow at rest the load is acting 175 mm away from the fulcrum, 1 mark. The effort will reduce when the barrow is rotated because the perpendicular distance from the load to the fulcrum decreases. 1 mark 2 Allow marks for understanding shown.

This question in 0445/42 Oct/Nov 2019

1 Models are often used when designing a structure. Models can be manufactured or they can be computer generated. (a) Give two benefits of using a manufactured model. 1 … 2 … [2] (b) Give two benefits of using a computer generated model. 1 … 2 … [2] (c) State one other use for a computer in manufacturing. … [1]

5 marks

Mark scheme: 1(a) Benefits of using a model could include: • Can be tested • View all-round the model • Dimensions / proportions can be checked • Different materials can be tried • Shows more than a drawing • Easier for a client to understand 2 × 1 marks 2 Accept any other valid benefit 1(b) Benefits of a computer model could include: • Can be viewed three dimensionally • Accurate testing can be carried out quickly • Variables can be changed quickly E.g. dimensions of components • Model can be sent to other people easily • Can be stored easily / does not take up physical space • No cost for materials involved 2 × 1 marks 2 Accept any other valid benefit Reference to speed must be justified 1(c) Other uses include: • Stock control • CAD / CAM • Quality control • Internet research 1 mark for suitable use 1 Accept any other valid use

This question in 0445/43 Oct/Nov 2019

2 Explain what is meant by equilibrium in a structure. … … … [2]

2 marks

Mark scheme: 2 Explanation could include the following points: • Opposing forces are balanced • A state of rest / structure is stable • Structure is not moving • Clockwise moment = anticlockwise moment 2 Explanation that includes two points, two marks 1 mark for one point in explanation. Allow two marks for single point fully explained

This question in 0445/43 Oct/Nov 2019

3 State the meaning of the term ‘moment’ in a structure. … [1]

1 marks

Mark scheme: 3 1

This question in 0445/43 Oct/Nov 2019

9 (a) Fig. 3 shows scaffolding on the outside of a building. Fig. 3 (i) State the type of structure used in scaffolding. … [1] (ii) Identify two safety features of the scaffolding that have been added to protect the public and workers using the scaffolding. 1 … 2 … [2] (iii) State the reason for using triangulation in a scaffolding structure. … [1] (iv) Describe the difference between a strut and a tie in a structure. … … … [2] (v) The lower part of a vertical scaffold pole is shown in Fig. 4. ground level Fig. 4 Use sketches and notes to show how the vertical load from the scaffold pole to the ground can be spread over a greater area. [3] (b) Fig. 5 shows a concrete barrier built across a disused dock to hold back sea water. walkway A B ground level wire face A face B wire fittings sea water concrete barrier Fig. 5 (i) State the force that will be acting on face A, the sea water side, and face B, the disused dock side of the barrier. Force at A … Force at B … [2] (ii) Give one reason why stainless steel would be more suitable than mild steel for the wire and fittings. … [1] (iii) Movement in the wall can be monitored by using a strain gauge. Describe the method of fixing a strain gauge to the wall. … … … [2] (c) Fig. 6 shows a design idea for a barrow for moving heavy building materials. Two methods of cutting the square steel tube to form a corner joint are shown. corner joint method 1 method 2 barrow Fig. 6 (i) Give one benefit of each method of cutting the tube ready for joining. Method 1 … Method 2 … [2] (ii) Describe how the tubes in method 1 could be joined together. … … … [2] (iii) Use sketches and notes to show how a gusset plate could be fitted to the corner joint. [2] (iv) Fig. 7 shows two views of the barrow with a load in place. 850 200 N 175 Fig. 7 Calculate the effort needed to pull back on the handles and lift the load. … … … [3] (v) Explain why the effort needed will reduce as the handles are pulled back. … … … [2]

25 marks

Mark scheme: 9(a)(i) Frame structure, 1 mark 1 9(a)(ii) Safety features include: • Wrapping around lower part of poles at street level • Mesh fencing at each of upper levels • Vertical board on edge at each upper level • Upper levels boarded across full width 2 × 1 marks for features identified 2 Accept any other valid points related to arrangement of poles. 9(a)(iii) Triangulation is used to hold the structure rigid and prevent any flexing or movement away from the wall 1 Award mark for understanding shown. 9(a)(iv) A strut will resist compression, 1 mark A tie will resist tension, 1 mark 2 9(a)(v) Method of spreading the load could include larger steel feet that fit into the bottom of vertical pole or boards placed below Functional method 1 mark Clear illustration / notes, 1 mark Method of fixing to vertical pole clearly shown, 1 mark 3 9(b)(i) Force at A is compression, 1 mark Force at B is tension, 1 mark 2 9(b)(ii) Non rusting / higher tensile strength / will last longer, 1 mark 1 Question Answer Marks Guidance 9(b)(iii) Attached to the wall using glue, 1 mark. Suitable adhesive, e.g. epoxy resin or cyanoacrylate, 1 mark Alignment arrows should be used to attach square to the wall with longer face horizontal, 1 mark 2 × 1 marks 2 Accept superglue 9(c)(i) Benefit of method 1: • Straight cut at 90º, easy to align for joining Benefit of method 2: • Tube remains sealed when joint is complete 2 Accept other valid benefits Allow ease of cutting at 90º Allow greater surface area 9(c)(ii) Joining technique could be: • Aligning joint accurately, 1 mark Followed by: • Welding / brazing • Bolt and captive nut inside one tube 1 mark for either 2 9(c)(iii) Gusset plate shown / described, 1 mark Clear drawing notes of positioning, 1 mark 2 9(c)(iv) Substitution into formula, 200 × 175 = effort × 850, 1 mark 35 000 / 850 = effort, 1 mark Effort = 41.176 , 1 mark 3 Award 3 marks for correct answer with no working shown. Accept 41.2N or 41.18 Question Answer Marks Guidance 9(c)(v) With the barrow at rest the load is acting 175 mm away from the fulcrum, 1 mark. The effort will reduce when the barrow is rotated because the perpendicular distance from the load to the fulcrum decreases. 1 mark 2 Allow marks for understanding shown.

This question in 0445/43 Oct/Nov 2019

2 Fig. 2.1 shows a set of tools for removing the screws used to secure plastic casings. Fig. 2.1 Fig. 2.2 (a) Give one reason why a slotted head screw would not be used in a plastic casing for an electronic product. … [1] (b) Give two benefits of the tools shown in Fig. 2.2. 1 … … 2 … … [2]

3 marks

Mark scheme: 2(a) Reasons could include: 1 • No special tools needed, manufacturer could want to restrict access • Slot can easily be deformed • Screwdriver can slip causing injury 2(b) Benefits could be: 2 • Will not slip easily • Positive fixing in screw • Special tool is needed to undo screw • Difficult access for untrained people. 2 × 1 mark

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3 Three examples of structures are shown in Fig. 3.1. B C A Fig. 3.1 Add ticks (✓) to complete Table 3.1 to identify the type of structure and whether it is natural or man-made. Two have been completed. Table 3.1 shell mass frame natural man-made A ✓ B C ✓ [4]

4 marks

Mark scheme: 3 4 shell mass frame natural man-made A   B   C   1 mark for each correct 4 × 1.

This question in 0445/41 May/June 2020

4 State the meaning of the term ‘static load’ in structural design. … [1]

1 marks

Mark scheme: 4 A static load is any load that is applied to a structure but does not move or 1 change in force or direction.

This question in 0445/41 May/June 2020

10 (a) (i) Describe what is meant by a ‘moment’ in a structure. … … … [2] (ii) State the unit that is used in calculations to define a moment. … [1] (b) Fig. 10.1 shows a see-saw with the position of two loads marked. 320 N 250 N 750 1000 Fig. 10.1 Calculate the moments and state the effect on the see-saw when the loads are applied. … … … … [3] (c) Use sketches and notes to describe the difference between a strut and a tie in a structure. [3] (d) Fig. 10.2 shows sections through three different beams used in structures. steel rods A B C steel box section steel ‘I’ section concrete Fig. 10.2 (i) Give one different benefit of using each beam. A … … B … … C … … [3] (ii) Explain why the beams in Fig. 10.2 are placed on their narrow edge when in use. … … … [2] (iii) Explain why the concrete beam needs steel rods cast into it. … … … … [3] (e) Fig. 10.3 shows a bridge constructed from steel tubes. Ø170 X Ø115 10 m Fig. 10.3 (i) State the most suitable method of joining the tubes. … [1] (ii) Use sketches and notes to show two methods of reinforcing the joint at point X. [4] (iii) Explain why a Factor of Safety will have been considered by designers of the bridge. … … … … [3]

25 marks

Mark scheme: 10(a)(i) A moment is force [1] × distance [1] 2 10(a)(ii) Newton metre (Nm) 1 10(b) LH moment is 0.75 × 320 = 240 Nm, 1 mark 3 RH moment is 1.0 × 250 = 250 Nm, 1 mark The effect will be to rotate slowly clockwise, 1 mark 10(c) A strut will resist compression 3 A tie will resist tension. Sketches to show both 2 × 1 mark indication of forces resisted, 1 mark. 10(d)(i) A Steel box section is 3 • Light • Resist twisting, tension and compression B Available in long lengths • Resists compression well • Will not bend easily C Resistant to tension and compression • Can be cast in situ • Does not corrode easily 1 mark for a valid benefit for each. 10(d)(ii) Explanation to include: 2 Beams are placed on their narrow edge to increase resistance to bending, 1 mark and increase the load carrying capacity of the beam 1 mark. 10(d)(iii) Concrete is naturally weak in tension, 1 mark 3 The lower part of the beam will be in tension, 1 mark Steel is strong in tension and will help to resist cracking/failure of the beam, 1 mark 10(e)(i) Welding, 1 mark 1 10(e)(ii) 4 Use of gusset plate, 1 mark Suitable position, 1 mark Use of strut,1 mark Suitable position, 1 mark 10(e)(iii) Factor of Safety will be considered for the benefit of those who will use the 3 bridge as it is difficult to predict exactly how many people / what traffic will be on the bridge at one time. Dynamic (moving) loads can be un-predictable e.g. high winds, snow. The bridge will be designed to carry a greater load than it is likely to encounter, this is the factor of safety.

This question in 0445/41 May/June 2020

1 Give two advantages of using plastics rather than wood for making window and door frames. 1 … … 2 … … [2]

2 marks

Mark scheme: Question Answer Marks 1 Advantages of plastics could be: 2 • Not affected by wet weather • Will not rot • Self-finishing • Can be moulded / extruded

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2 Fig. 2.1 shows a wooden door frame before it is positioned in a building. Fig. 2.1 Draw on Fig. 2.1 to show how triangulation can be used to prevent the frame from being distorted while it is being installed in the building. Give details of any joining methods used. [3]

3 marks

Mark scheme: 2 3 Triangulated brace used [1] Suitable size and position [1] Fixing method [1]

This question in 0445/42 May/June 2020

5 State the meaning of ‘torsion’ in a shaft. … … [1]

1 marks

Mark scheme: 5 Turning or twisting force, 1

This question in 0445/42 May/June 2020

10 (a) Beams made from a single piece of wood are limited in length to the size of the tree that they are cut from. (i) Use sketches and notes to show how the length of a wooden beam can be increased by laminating. [3] (ii) Give one reason why laminated beams are often used in large public buildings. … … … [1] (b) Fig. 10.1 shows a child’s rocking toy. Fig. 10.1 (i) Name the type of structure that has been used for the toy. … [1] (ii) Explain how the shape of the toy adds strength to the structure. … … … … [2] (c) Fig. 10.2 shows a single span suspension bridge crossing a river estuary. hanger cables suspension cable tower Fig. 10.2 (i) Give one benefit of designing the bridge with a single span. … … [1] (ii) Give two stationary loads and two moving loads that the bridge will have to withstand. Stationary loads 1 … 2 … Moving loads 1 … 2 … [4] (iii) The bridge deck is a welded steel hollow box section. Use sketches and notes to show one other method of joining steel sheet together. [2] (iv) Fig. 10.3 shows a section through the bridge deck with the positions of two loads shown. hanger cable hanger cable 22 m 17.5 m 4.5 m 13000 N 16000 N section through bridge deck R1 R2 Fig. 10.3 Calculate the reactions at R1 and R2. … … … … … [4] (v) The towers in the bridge are made from reinforced concrete. Explain why reinforced concrete has been used. … … … … [2] (vi) The suspension cables use a large number of Ø 5 mm steel wires spun or twisted together to give a Ø 680 mm cable as shown in Fig. 10.4. Fig. 10.4 Explain why each suspension cable is made from spun steel wires. … … … … [2] (vii) The stress in each Ø 5 mm wire is 1500 N / mm2. Calculate the force acting on each wire. force Use the formula: stress = cross sectional area … … … [3]

25 marks

Mark scheme: 10(a)(i) Notes and sketches to show: 3 • At least 4 pieces laminated, 1 mark • Staggered joints, 1 mark • Length of beam increased, 1 mark. 10(a)(ii) Reasons could include: 1 • Large gaps can be spanned • No intrusion into space below • Economical compared to other methods • Laminated beams are extremely stable • Renewable resource used 10(b)(i) Shell structure, 1 mark. 1 10(b)(ii) The cross-section shape varies giving rigidity to the moulding, 1 mark 2 There are several curves in the moulding which will increase the resistance to bending, torsion and distorting of the moulding, 1 mark. 10(c)(i) Benefits of the single span will include: 1 No obstruction to shipping passing under the bridge • Only two structures to be built in the water • Reduced cost • Aesthetically more pleasing. 10(c)(ii) Stationary loads will include: 4 • Materials used, concrete, steel cables, deck, road surface. 2 × 1 mark Moving loads will include: • Vehicles on the bridge • Wind, snow, rain 2 × 1 mark 10(c)(iii) Sketches / notes to show rivets or bolts, 1 mark. 2 Steel sheets overlapped, 1 mark. 10(c)(iv) Moments at R1 (4.5 × 13000) + (17.5 × 16000) = R2 × 22 [1] 4 58500 + 280000 = R2 × 22, [1] R2 = 338500 / 22 = 15386 N, [1] R1 = (13000 + 16000) – 15386 = 13614 N, [1] 10(c)(v) Concrete is used as it is very resistant to compression, Reinforcement will 2 increase resistance to torsion, tension and bending. It is also resistant to the effects of bad weather and needs little maintenance. 10(c)(vi) Steel is resistant to tension and can readily be drawn out into wire and spun 2 into a cable that is flexible and resistant to stretching. 10(c)(vii) Rearrangement of formula force = stress × cross-sectional area, [1] 3 Cross-sectional area = πr2 = 3.14159 × 2.52 = 19.635, [1] Force = 1500 × 19.64 = 29452.4 N, [1]

This question in 0445/42 May/June 2020

11 (a) Fig. 11.1 shows two uses for bevel gears. driven bevel gear driver bevel gear handle 15t idler gear operating handle fixing point 56t gear 15t gear chuck Dockyard crane Hand drill Fig. 11.1 (i) Use the terms below to complete the description of what is happening in the two examples in Fig. 11.1. 90º 45º slower decrease faster increase 180º remain the same The driven gear in the hand drill will rotate … than the driving gear and the torque will … . The driven gear in the crane will rotate … than the driving gear and the torque will … . In both cases the drive will have moved through … . [5] (ii) Calculate the rotational speed of the handle on the hand drill if the chuck rotates at 475 rpm. … … … … [3] (iii) In both examples in Fig. 11.1 the operational conditions could affect the working life of the gears. Describe what could be done to overcome this problem. … … … … [2] (b) Fig. 11.2 shows a cam profile and follower. 15 25 Fig. 11.2 (i) State the conversion of motion that results from using a cam. … to … [2] (ii) Mark the direction of rotation of the cam on Fig. 11.2. [1] (iii) Give the vertical distance that the follower is lifted in each revolution of the cam in Fig. 11.2. … [1] (iv) Use sketches and notes to show the profile of a cam that will lift a follower twice for each revolution and will have no dwell. [2] (c) Fig. 11.3 shows a heavy packing crate. X edge X floor level 100 Fig. 11.3 Edge X of the packing crate needs to be lifted by 100 mm to the position shown. Draw a design on Fig. 11.3 for a lever system that will allow the edge to be lifted by 100 mm. [3] (d) Fig. 11.4 shows details of a screw thread that is used in a woodworking vice. 6 25 Fig. 11.4 (i) Calculate the mechanical advantage of the screw thread. circumference Use the formula: MA = pitch … … … [2] (ii) The end of the woodworking vice handle applies a load 16 times greater than the effort. Calculate the maximum force applied by the screw thread for an effort of 20 N at the end of the handle. … … [2] (iii) Explain why the actual load applied by the screw thread will be reduced. … … … [2]

25 marks

Mark scheme: 11(a)(i) The driven gear in the hand drill will rotate faster than the driving gear and 5 the torque will decrease. The driven gear in the crane will rotate slower than the driving gear and the torque will increase. In both cases the drive will have moved through 90º. 1 mark for each correct. 11(a)(ii) Gear ratio = 3.73 : 1 [1] 3 475 / 3.73, [1] = 127.3 rpm [1] 11(a)(iii) Lubrication, 1 mark 2 Covering / protecting the gears to prevent ingress of abrasive substances, 1 mark. 11(b)(i) Rotary motion to reciprocating motion, 1 mark for each. 2 11(b)(ii) 1 1 mark for anticlockwise rotation 11(b)(iii) 10 mm 1 11(b)(iv) 2 Two lifts on cam profile, 1 mark No dwell, 1 mark. 11(c) 3 Lever will fit under packing case, 1 mark Handle long enough to decrease effort needed, 1 mark Will provide 100mm lift when rotated, 1 mark 11(d)(i) Circumference = 3.14159 × 25 = 78.54 2 MA = 78.54 / 6 = 13.09 11(d)(ii) Total MA = 13.09 × 16 = 209.44, 1 mark 2 209.44 × 20 = 4189 N, 1 mark 11(d)(iii) Loss of efficiency, 1 mark due to friction between male and female thread, 2 1 mark.

This question in 0445/42 May/June 2020

2 Fig. 2.1 shows a set of tools for removing the screws used to secure plastic casings. Fig. 2.1 Fig. 2.2 (a) Give one reason why a slotted head screw would not be used in a plastic casing for an electronic product. … [1] (b) Give two benefits of the tools shown in Fig. 2.2. 1 … … 2 … … [2]

3 marks

Mark scheme: 2(a) Reasons could include: 1 • No special tools needed, manufacturer could want to restrict access • Slot can easily be deformed • Screwdriver can slip causing injury 2(b) Benefits could be: 2 • Will not slip easily • Positive fixing in screw • Special tool is needed to undo screw • Difficult access for untrained people. 2 × 1 mark

This question in 0445/43 May/June 2020

3 Three examples of structures are shown in Fig. 3.1. B C A Fig. 3.1 Add ticks (✓) to complete Table 3.1 to identify the type of structure and whether it is natural or man-made. Two have been completed. Table 3.1 shell mass frame natural man-made A ✓ B C ✓ [4]

4 marks

Mark scheme: 3 4 shell mass frame natural man-made A   B   C   1 mark for each correct 4 × 1.

This question in 0445/43 May/June 2020

4 State the meaning of the term ‘static load’ in structural design. … [1]

1 marks

Mark scheme: 4 A static load is any load that is applied to a structure but does not move or 1 change in force or direction.

This question in 0445/43 May/June 2020

10 (a) (i) Describe what is meant by a ‘moment’ in a structure. … … … [2] (ii) State the unit that is used in calculations to define a moment. … [1] (b) Fig. 10.1 shows a see-saw with the position of two loads marked. 320 N 250 N 750 1000 Fig. 10.1 Calculate the moments and state the effect on the see-saw when the loads are applied. … … … … [3] (c) Use sketches and notes to describe the difference between a strut and a tie in a structure. [3] (d) Fig. 10.2 shows sections through three different beams used in structures. steel rods A B C steel box section steel ‘I’ section concrete Fig. 10.2 (i) Give one different benefit of using each beam. A … … B … … C … … [3] (ii) Explain why the beams in Fig. 10.2 are placed on their narrow edge when in use. … … … [2] (iii) Explain why the concrete beam needs steel rods cast into it. … … … … [3] (e) Fig. 10.3 shows a bridge constructed from steel tubes. Ø170 X Ø115 10 m Fig. 10.3 (i) State the most suitable method of joining the tubes. … [1] (ii) Use sketches and notes to show two methods of reinforcing the joint at point X. [4] (iii) Explain why a Factor of Safety will have been considered by designers of the bridge. … … … … [3]

25 marks

Mark scheme: 10(a)(i) A moment is force [1] × distance [1] 2 10(a)(ii) Newton metre (Nm) 1 10(b) LH moment is 0.75 × 320 = 240 Nm, 1 mark 3 RH moment is 1.0 × 250 = 250 Nm, 1 mark The effect will be to rotate slowly clockwise, 1 mark 10(c) A strut will resist compression 3 A tie will resist tension. Sketches to show both 2 × 1 mark indication of forces resisted, 1 mark. 10(d)(i) A Steel box section is 3 • Light • Resist twisting, tension and compression B Available in long lengths • Resists compression well • Will not bend easily C Resistant to tension and compression • Can be cast in situ • Does not corrode easily 1 mark for a valid benefit for each. 10(d)(ii) Explanation to include: 2 Beams are placed on their narrow edge to increase resistance to bending, 1 mark and increase the load carrying capacity of the beam 1 mark. 10(d)(iii) Concrete is naturally weak in tension, 1 mark 3 The lower part of the beam will be in tension, 1 mark Steel is strong in tension and will help to resist cracking/failure of the beam, 1 mark 10(e)(i) Welding, 1 mark 1 10(e)(ii) 4 Use of gusset plate, 1 mark Suitable position, 1 mark Use of strut,1 mark Suitable position, 1 mark 10(e)(iii) Factor of Safety will be considered for the benefit of those who will use the 3 bridge as it is difficult to predict exactly how many people / what traffic will be on the bridge at one time. Dynamic (moving) loads can be un-predictable e.g. high winds, snow. The bridge will be designed to carry a greater load than it is likely to encounter, this is the factor of safety.

This question in 0445/43 May/June 2020

3 (a) Fig 3.1 shows a concrete beam used in a road bridge. concrete beam Fig. 3.1 Give two properties of concrete that make it suitable for this purpose. 1 … 2 … [2] (b) (i) Use sketches and notes to show how steel can be used to reinforce a concrete beam. [2] (ii) Name the main force resisted by steel in a concrete beam. … [1]

5 marks

Mark scheme: 3(a) Suitable properties of concrete will include: • Hard • Weather resistant • Easily moulded • Low cost material • Can be moulded on site • High compressive strength • Low maintenance material. Allow any other valid alternative. 3(b)(i) Steel reinforcement rods shown, 1 mark. Below horizontal centre line of beam, 1 mark. 2 Accept any other recognised use of steel in reinforcement. 3(b)(ii) Tension 1

This question in 0445/42 Oct/Nov 2020

9 Fig. 9.1 shows two lengths of softwood that are to be joined at 90° to each other. 80 35 Fig. 9.1 (a) Use sketches and notes to show two different methods of joining the lengths of softwood. [4] (b) Corrugated plastic sheet as shown in Fig. 9.2 is often used as roofing on small buildings. 1800 762 Fig. 9.2 (i) Describe one structural benefit of having corrugations on the sheet. … … … [2] (ii) Draw on Fig. 9.2 to show where support beams should be placed to prevent the corrugated sheet from bending when it is used on a roof. [2] (c) Fig. 9.3 shows a model girder bridge that will span a gap of 400 mm. The model is to be made using resistant materials. 400 supports for testing the bridge Fig. 9.3 (i) Name one resistant material that would be suitable for the model and give a reason for your choice. … … [2] (ii) Describe how the material named in part (c)(i) could be joined. … … … … [2] (iii) Use sketches and notes to show how the model could be tested and evaluated. [3] (d) Fig. 9.4 shows two different types of structure used to support wind turbines. generator blades tower A B Fig. 9.4 (i) State the type of structure used for the tower on each of the turbines. Turbine A … Turbine B … [2] (ii) Name one stationary load that will act on the base of each tower when the blades are at rest. … [1] (iii) Name two moving loads that will act on the wind turbines when they are generating. 1 … 2 … [2] (iv) Explain how the turbine towers in Fig. 9.4 maintain equilibrium. … … … [2] (e) A tie bar in a roof truss is made from a piece of steel rod Ø12 × 1.5 m long. When a load of 15 kN is applied the tie bar extends by 0.35 mm. Calculate the strain in the tie bar. change in length Use the formula: Strain = original length … … … … Strain = … [3]

25 marks

Mark scheme: 9(a) Recognised functional method used, could be halving, mortise and tenon, bridle, gusset plate, 2 × 1 mark. Clear description in sketch / note form, 2 × 1 mark. joint. 9(b)(i) Benefits of corrugation could include: • Increase stiffness / rigidity in one direction • Flexibility in the other direction • Allow watertight joints when overlap is used • Allow increased load to be supported. 2 Description that includes two points, 1 mark for each point included. Allow 2 marks for one fully justified benefit. Question Answer Marks Guidance 9(b)(ii) Support beams across corrugations, 1 mark. At least two beams shown, 1 mark. 2 9(c)(i) Modelling materials could be: • Balsa / softwood / hardwood • Mild steel strip • Aluminium • Plastic strip • Laser cut plastic sheet 1 mark for suitable material. Reason for choice, 1 mark. 2 Allow other suitable alternatives. 9(c)(ii) Named adhesives, epoxy resin, PVA cyanoacrylate (superglue), Tensol Solder for steel strip Rivets, screws, pins. 2 × 1 marks for suitable joining methods. 2 Joining methods must be suitable for the materials named. Allow aluminium solder. Either two methods given, or one method described in detail. 9(c)(iii) Testing must show: • Load being applied, either point load or spread across deck of bridge, • Method of measuring load indicated, • Method of assessing deflection of bridge, • Determination of point(s) of failure • Measurement of distance spanned. 3 × 1 points included, 1 mark each. 3 Allow other suitable tests. Award up to 3 marks for full details of a single test.3 Question Answer Marks Guidance 9(d)(i) Turbine A is a shell structure, turbine B is a frame structure, 2 × 1 marks. 2 Accept ‘mass’ for A. 9(d)(ii) Weight of materials in the tower, weight of the turbine blades. 1 mark for either. 1 9(d)(iii) Moving loads could be: • Wind acting on the tower • Movement of the turbine blades • Mass of generator moving to face the wind • Ice build-up in winter • Tidal / wave forces on offshore turbines. 2 × 1 marks 2 Allow other valid alternatives 9(d)(iv) Points in explanation may include: • Reference to foundation • Secure fixing of tower to the ground. • Equal force acting on each side of the tower • Greater width at the base. 2 Two points included, 2 marks Allow 2 marks for one point fully explained. 9(e) Conversion of rod length / extension to same units, 1 mark Strain = 0.35 1500 , 1 mark Strain = 0.00023, 1 mark 3

This question in 0445/42 Oct/Nov 2020

3 (a) Fig 3.1 shows a concrete beam used in a road bridge. concrete beam Fig. 3.1 Give two properties of concrete that make it suitable for this purpose. 1 … 2 … [2] (b) (i) Use sketches and notes to show how steel can be used to reinforce a concrete beam. [2] (ii) Name the main force resisted by steel in a concrete beam. … [1]

5 marks

Mark scheme: 3(a) Suitable properties of concrete will include: • Hard • Weather resistant • Easily moulded • Low cost material • Can be moulded on site • High compressive strength • Low maintenance material. Allow any other valid alternative. 3(b)(i) Steel reinforcement rods shown, 1 mark. Below horizontal centre line of beam, 1 mark. 2 Accept any other recognised use of steel in reinforcement. 3(b)(ii) Tension 1

This question in 0445/43 Oct/Nov 2020

9 Fig. 9.1 shows two lengths of softwood that are to be joined at 90° to each other. 80 35 Fig. 9.1 (a) Use sketches and notes to show two different methods of joining the lengths of softwood. [4] (b) Corrugated plastic sheet as shown in Fig. 9.2 is often used as roofing on small buildings. 1800 762 Fig. 9.2 (i) Describe one structural benefit of having corrugations on the sheet. … … … [2] (ii) Draw on Fig. 9.2 to show where support beams should be placed to prevent the corrugated sheet from bending when it is used on a roof. [2] (c) Fig. 9.3 shows a model girder bridge that will span a gap of 400 mm. The model is to be made using resistant materials. 400 supports for testing the bridge Fig. 9.3 (i) Name one resistant material that would be suitable for the model and give a reason for your choice. … … [2] (ii) Describe how the material named in part (c)(i) could be joined. … … … … [2] (iii) Use sketches and notes to show how the model could be tested and evaluated. [3] (d) Fig. 9.4 shows two different types of structure used to support wind turbines. generator blades tower A B Fig. 9.4 (i) State the type of structure used for the tower on each of the turbines. Turbine A … Turbine B … [2] (ii) Name one stationary load that will act on the base of each tower when the blades are at rest. … [1] (iii) Name two moving loads that will act on the wind turbines when they are generating. 1 … 2 … [2] (iv) Explain how the turbine towers in Fig. 9.4 maintain equilibrium. … … … [2] (e) A tie bar in a roof truss is made from a piece of steel rod Ø12 × 1.5 m long. When a load of 15 kN is applied the tie bar extends by 0.35 mm. Calculate the strain in the tie bar. change in length Use the formula: Strain = original length … … … … Strain = … [3]

25 marks

Mark scheme: 9(a) Recognised functional method used, could be halving, mortise and tenon, bridle, gusset plate, 2 × 1 mark. Clear description in sketch / note form, 2 × 1 mark. joint. 9(b)(i) Benefits of corrugation could include: • Increase stiffness / rigidity in one direction • Flexibility in the other direction • Allow watertight joints when overlap is used • Allow increased load to be supported. 2 Description that includes two points, 1 mark for each point included. Allow 2 marks for one fully justified benefit. Question Answer Marks Guidance 9(b)(ii) Support beams across corrugations, 1 mark. At least two beams shown, 1 mark. 2 9(c)(i) Modelling materials could be: • Balsa / softwood / hardwood • Mild steel strip • Aluminium • Plastic strip • Laser cut plastic sheet 1 mark for suitable material. Reason for choice, 1 mark. 2 Allow other suitable alternatives. 9(c)(ii) Named adhesives, epoxy resin, PVA cyanoacrylate (superglue), Tensol Solder for steel strip Rivets, screws, pins. 2 × 1 marks for suitable joining methods. 2 Joining methods must be suitable for the materials named. Allow aluminium solder. Either two methods given, or one method described in detail. 9(c)(iii) Testing must show: • Load being applied, either point load or spread across deck of bridge, • Method of measuring load indicated, • Method of assessing deflection of bridge, • Determination of point(s) of failure • Measurement of distance spanned. 3 × 1 points included, 1 mark each. 3 Allow other suitable tests. Award up to 3 marks for full details of a single test.3 Question Answer Marks Guidance 9(d)(i) Turbine A is a shell structure, turbine B is a frame structure, 2 × 1 marks. 2 Accept ‘mass’ for A. 9(d)(ii) Weight of materials in the tower, weight of the turbine blades. 1 mark for either. 1 9(d)(iii) Moving loads could be: • Wind acting on the tower • Movement of the turbine blades • Mass of generator moving to face the wind • Ice build-up in winter • Tidal / wave forces on offshore turbines. 2 × 1 marks 2 Allow other valid alternatives 9(d)(iv) Points in explanation may include: • Reference to foundation • Secure fixing of tower to the ground. • Equal force acting on each side of the tower • Greater width at the base. 2 Two points included, 2 marks Allow 2 marks for one point fully explained. 9(e) Conversion of rod length / extension to same units, 1 mark Strain = 0.35 1500 , 1 mark Strain = 0.00023, 1 mark 3

This question in 0445/43 Oct/Nov 2020

10 (a) Fig. 10.1 shows a building under construction. cladding Fig. 10.1 (i) State the type of structure that is being used for the building. … [1] (ii) Describe the purpose of the cladding material. … … … [2] (iii) State two stationary loads and two moving loads that the finished building will have to withstand. Stationary Loads 1 … 2 … Moving Loads 1 … 2 … [4] (b) Fig. 10.2 shows part of a structure made from steel box section material 50 × 50. 1000 mm 50 mm X mm 650 Fig. 10.2 Draw on Fig. 10.2 to show how corner X could be strengthened using a gusset. Add notes to indicate how the gusset would be joined to the structure. [2] (c) (i) Fig. 10.3 shows a cabinet made from four pieces of wood that are joined together. Use notes and sketches to show one suitable joint on the enlarged view of corner A. corner A Fig. 10.3 [3] (ii) Name two natural defects that can occur in wood. 1 … 2 … [2] (d) Fig. 10.4 shows the jib of a crane with an enlarged view of the base. The crane is in equilibrium. heavy steel counterweights extended legs Fig. 10.4 (i) State what is meant by equilibrium. … … [1] (ii) Describe how the crane is kept in equilibrium when a load is being lifted. … … … … … [3] (iii) Components of the crane will be affected by compression, tension and torsion. Give one example of where each of these forces occurs when the crane is operating. Compression … Tension … Torsion … [3] (e) Fig. 10.5 shows a beam with two loads applied to it. 4 m 1 m 1.5 m 600 N 850 N R1 R2 Fig. 10.5 Calculate the reaction at R1 and R2. … … … … … … [4]

25 marks

Mark scheme: 10(a)(i) Frame structure, 1 mark. 1 10(a)(ii) The cladding is used to: • Fix to the frame to make the building watertight/weathertight • Provide insulation • Provide a weatherproof finish • Allows different colours to be used 2 Allow other valid points 2 marks for two points mentioned 2 marks for one point described in depth. 10(a)(iii) Stationary loads: 2 × 1 marks All materials used in construction Additional partitions inside the building Equipment / furniture of a permanent nature stored in the building Moving loads: 2 × 1 marks Wind/rain Snow Earthquake People inside the building Vehicles entering the storage building. 4 Allow any other valid responses Question Answer Marks Guidance 10(b) Gusset plate drawn on any corner, 1 mark. Indication of fixing method, bolts, welding, rivets, 1 mark. 2 10(c)(i) Suitable joint shown, e.g. housing, comb, dovetail, 1 mark. Accurate drawing of details, 1 mark. Indication of how the joint will be fixed into position, 1 mark. 3 Allow KD fittings/nails/screws 10(c)(ii) Natural defects will include: • Knots / shakes / wind cracks • Woodworm / termites / beetles • Warping / twisting • Splitting 2 × 1 marks. 2 Allow any other recognised defect that occurs naturally. Do not allow rotting. 10(d)(i) A state of rest or balance because the opposing forces are equal 1 10(d)(ii) A steel counterweight that is on the back of the crane base, preventing the jib from tipping forward, 1 mark. Extending legs on either side of crane, preventing the crane from rolling to the side, 1 mark. The safe working load of the crane, 1 mark. is checked against the load to be lifted, 1 mark. 3 First two points must be justified for marks. Any three points mentioned for 3 marks. 2 marks for a single point fully explained. 10(d)(iii) • Compression will be applied to the vertical parts of the crane structure, from the load being lifted and the weight of the crane itself. • Compression of the extended legs will occur • Tension will be applied to the cables carrying out the lift. • Torsion will occur in the tower when the crane is turning whilst carrying a load. • Torsion will occur as a result of wind blowing the jib. 3 Any three points mentioned for 3 marks. 2 marks for a single point fully explained. Allow BOD for compression of wheels/tyres. 10(e) Taking moments about R1 (600 × 1) + (2.5 × 850) = R2 × 4, [1] R2 = (600 + 2125) / 4 = 2725 / 4 [1] R2 = 681.25 N [1] R1 = (600 + 850) – 681.25 = 768.75 N [1] 4 Allow ecf for last part. Allow calculation of R1 first.

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1 A list of materials that can be used in a structure is given below. Softwood Concrete Steel For each material give one different property that makes it suitable for use in a structure. Softwood … Concrete … Steel … [3]

3 marks

Mark scheme: 1 Properties that make the material suitable for a structure will include: Softwood • Plentiful supply • Low cost • Can be joined easily • Renewable resource • Resists tension, compression, bending and torsion Concrete • Strong in compression • Can be strengthened in tension • Low cost • Long lasting • Can be cast in situ Steel • Low cost • High tensile strength • Durable / long lasting • Range of sizes / sections available • Easily joined by temporary or permanent fastenings. 3 × 1 mark 3 Allow any valid alternatives Allow – Can be used to reinforce concrete. Plentiful supply

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2 Use sketches and notes to show how triangulation can be used to prevent a rectangular window frame from distorting while it is being fitted. [2]

2 marks

Mark scheme: 2 Drawing / notes showing either a brace, tie or gusset plate that will keep the frame rigid Suitable position / size, 1 mark Suitable fixing method, 1 mark. 2 Allow 1 mark for a line drawing showing no width of brace/tie

This question in 0445/43 May/June 2021

4 Spur gears can be made from brass, nylon or steel. (a) Give one different reason for choosing each of these materials. Brass … Nylon … Steel … [3] (b) Name the applied load that could cause teeth on a spur gear to break off. … [1]

4 marks

Mark scheme: 4(a) Reasons for using materials will include: Brass • Will not corrode • Good bearing qualities • Teeth can be cut accurately Nylon • Low cost • Can be injection moulded rather than cut • No lubrication needed/low coefficient of friction • Can be used as ‘sacrificial’ gear Steel • Hard wearing / durable • Lower cost than brass • Readily available 3 × 1 mark Allow brass as self-lubricating. Allow durable only if it is not used for steel. Allow reference to the strength of steel. Question Answer Marks Guidance 4(b) Shear, 1 mark. 1

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10 (a) Fig. 10.1 shows a 27 m high structure used to support an illuminated star. Fig. 10.1 (i) Name two stationary and two moving loads that will affect the stability of the structure. Stationary Loads 1 … 2 … Moving Loads 1 … 2 … [4] (ii) Ties and struts are used to keep the illuminated star structure rigid. State one force resisted by a tie and one force resisted by a strut. Tie … Strut … [2] (iii) Give one advantage and one disadvantage of using struts, rather than ties in the structure. Advantage … Disadvantage … [2] (b) Fig. 10.2 shows part of a structure that has been joined using three bolts and nuts. steel strip 70 x 10 3 x M10 bolts x 30 long Fig. 10.2 (i) Explain why more than one bolt is required in the joint. … … … … [2] (ii) When the nut is fully tight, the 30 mm length of bolt passing through the plate has stretched to 30.1 mm. Calculate the strain in the bolt. change in length Use the formula: Strain = original length … … … … [2] (iii) Explain how torsion and shear could affect the bolt and its threads if the nut is over-tightened. … … … … … [4] (iv) Give two possible reasons for joining the materials with bolts and nuts rather than welding the materials together. 1 … 2 … [2] (c) Fig. 10.3 shows two methods of extending the length of a wooden beam. Method A − cut joint with reinforcing bolts Method B − beam ends cut square with bolted plates either side Fig. 10.3 (i) Give one reason why extending a wooden beam may be necessary. … [1] (ii) Describe one advantage of method B when compared to method A. … … … [2] (iii) Laminating is another way of making a long beam. Use sketches and notes to show a laminated beam. [2] (d) Fig. 10.4 shows a playground see-saw mounted on a hard rubber spring. hard rubber spring Fig. 10.4 Explain why the see-saw may not be in a state of equilibrium when it is in use. … … … [2]

25 marks

Mark scheme: 10(a)(i) Stationary loads will include: • Material used in the supporting structure • Star • Lighting on the star Moving loads will include: • Wind • Snow / ice / rain • Construction / maintenance workers on the structure 2 × 2 marks 4 Allow other valid responses for either type of load. 10(a)(ii) A tie will resist tension, 1 mark A strut will resist compression, bending and torsion, 1 mark 2 Allow a single resisted force for strut. 10(a)(iii) Advantage – a strut will resist different types of force, whereas a tie will only resist tension, 1 mark. Disadvantage – increased weight in the structure as a result of using struts The material for a strut is likely to be more expensive than a tie, 1 mark. 2 Question Answer Marks Guidance 10(b)(i) • A single bolt will allow the joint to rotate • At least two bolts are needed to prevent this • To overcome failure of a single bolt • Three bolts are more likely to provide accurate alignment. 2 Two points mentioned for 2 marks. Allow 2 marks for a single point fully explained. 10(b)(ii) Change in length is 0.1 mm,1 mark Strain = 0.1 / 30 = 0.003, 1 mark 2 Award both marks if correct answer is given with no evidence of working. 10(b)(iii) Torsion will cause the bolt to turn with the possibility of the bolt failing. This will only occur if the bolt is over-tightened, in which case the head of the bolt will break away from the body. Shear force will be applied to the threads on the bolt causing the threads to break away. 2 × 2 marks. 4 1 mark for a single point recognised for each force. 2 marks if the response is justified. 10(b)(iv) Reasons for using bolts and nuts could include: • Possible need to disassemble the structure • Ease of transport to site • No chance of distortion of the materials • Different materials / non-metals being joined • Three bolts will spread the load across full width of joint • Fine adjustments to position can be made before final tightening. 2 2 marks for two points given in description. Allow 2 marks for a single point fully justified. 10(c)(i) The span needed is greater than the length of timber available. Access to position of beam may be limited, 1 mark for either point 1 10(c)(ii) Advantages of method B include • No complex cutting and fitting of joint • Plates at side give lateral stability and increased strength • More bolts used that will spread the load. 2 Allow 2 marks for fully justified response. Allow 1 mark for a single mentioned Allow reference to comparative strength of the joints 10(c)(iii) Sketch to show at least three layers of ‘beam width’ laminates, 1 mark Joints at ends of laminate staggered, 1 mark 2 Question Answer Marks Guidance 10(d) • The purpose of the see-saw is to allow two users to rise and fall • The movement of users means that it is not in equilibrium • Different weights of user may mean that it is in equilibrium at the extent of its travel. • Hard rubber spring will generate movement 2 Recognition that it is the users who cause movement and loss of equilibrium, 1 mark. Recognition of effect of imbalance in weight of users, 1 mark.

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1 Fig. 1.1 shows a satellite dish for a television. dish Fig. 1.1 (a) State the type of structure that has been used for the dish. … [1] (b) Use sketches and notes to show a different example of the type of structure in your answer to Part (a). [2]

3 marks

Mark scheme: 1(a) Shell structure. 1 1(b) Valid example of any shell structure, or valid example of structure given in 1(a), 1 mark Clear drawing / notes, 1 mark. 2 Sketch must relate to answer from 1(a)

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2 Fig. 2.1 shows three types of spring. A B C Fig. 2.1 State the force that each spring will resist. A … B … C … [3]

3 marks

Mark scheme: 2 Spring A will resist compression Spring B will resist tension Spring C will resist torsion, 1 mark for each correct answer. 3 Accept terminology with the same meaning as answer. No mark if more than one force given.

This question in 0445/42 Oct/Nov 2021

9 (a) Fig. 9.1 shows part of a plastic case to hold a circuit board. Y X 90 55 Fig. 9.1 (i) Draw a rib in the case between points X and Y. [2] (ii) Add four gussets to the round pillar. [2] (iii) Describe the purpose of the rib and gussets. … … … [2] (iv) Give two reasons why the designer of the casing should consider disassembly of the product. 1 … … 2 … … [2] (b) Fig. 9.2 shows two steel tubes each 2 m long which need to be joined end-to-end. (i) Use sketches and notes to show one permanent method of joining the tubes. Give details of any additional materials used. steel tube 20 × 20 × 2 Fig. 9.2 [3] (ii) Use sketches and notes to show one temporary method of joining the tubes. Give details of any additional materials or components used. [3] (c) Fig. 9.3 shows details of a tower crane in equilibrium, which is carrying a load of 5500 N. Two counterweights A and B each producing a force of 6200 N are used to balance the load. 6 m 3.6 m X A B 5500 N Fig. 9.3 (i) The crane is in a state of equilibrium. Give the meaning of equilibrium. … [1] (ii) Calculate distance X to the centre of counterweight A. … … … … [4] (iii) Give one static load and one moving load that the crane must resist. Static load … Moving load … [2] (iv) The tower crane is designed as a number of prefabricated parts that are bolted together. Give one reason for this method of construction. … … [1] (v) Explain the importance of ‘Factor of Safety’ in crane design. … … … … [3]

25 marks

Mark scheme: 9(a)(i) Rib in correct position, 1 mark. Functional shape, 1 mark. 2 9(a)(ii) Triangular gussets supporting pillar, 1 mark. Four gussets spaced around the pillar, 1 mark. 2 Accept 3 or more gussets. 9(a)(iii) • The rib and gussets are to provide support / reinforcement, 1 mark. • To add strength at a given point without significant increase in weight, 1 mark. • To prevent distortion 2 Allow marks for understanding shown. 9(a)(iv) Reasons for consideration of disassembly could be: • Ease of access for repair • Recycling of usable part in the circuit • Removal of parts to allow for recycling / disposal of case. 2 × 1 marks 2 Allow other valid reasons. Question Answer Marks Guidance 9(b)(i) Support either inside or outside of tubes, 1 mark. Indication of methods – welding / brazing, pop rivets, 1 mark. Permanent, functional method, 1 mark. 3 9(b)(ii) Support either inside or outside of tubes, 1mark. Indication of methods – bolts / screws / dowel pins, 1 mark. Temporary functional method, 1 mark. 3 9(c)(i) Definition of equilibrium – balanced, stable, all opposing forces are balanced, 1 mark. 1 Award mark for understanding shown. 9(c)(ii) 5500 × 6 = (X × 6200) + (3.6 × 6200), 1 mark 33 000 = (6200 X ) + 22 320, 1 mark 6200 X = 33 000 – 22 320 = 10 680, 1 mark X = 10 680 / 6200 = 1.7226 m, 1 mark 4 Award 4 marks for correct answer with no working. Question Answer Marks Guidance 9(c)(iii) Static load is: • Load from any of the crane components • Counterweights Moving load is: • Wind, snow, any other weather related condition • Load from items being lifted • Torque caused by rotation of the tower • Movement / swinging of the load being lifted 2 × 1 marks 1 mark for each 2 9(c)(iv) Reason for prefabrication will include: • Can be constructed and then moved to site • Ease of storage • Ease of transport • Crane structure is temporary. 1 mark. 1 Do not allow cost related reason. Allow other valid reasons. 9(c)(v) Points in explanation could include: • Unknown loads and conditions of use • Designer must consider all who will be working on the crane or in the area • Unknown stability of foundations where the base will rest. • Compliance with safety legislation. 3 1 mark for each point mentioned. Allow 2 marks for a fully justified point.

This question in 0445/42 Oct/Nov 2021

1 Name three types of man-made structure. 1 … 2 … 3 … [3]

3 marks

Mark scheme: 1 Frame structure, 1 mark. Shell structure, 1 mark. Mass structure, 1 mark. 3 Allow examples of the types of structure Any examples must only be applicable to a single type of structure.

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2 Fig. 2.1 shows a wooden support structure for a section of railway track. Fig. 2.1 (a) Describe the methods used to ensure that the structure can safely support a heavy load. … … … [2] (b) Give two reasons for choosing wood as the material for the structure. 1 … 2 … [2]

4 marks

Mark scheme: 2(a) Methods visible are:  Heavy section timber  Cross braces  Triangulation  Solid foundations  Increased width at base  Bolts are used to join sections of timber. 2  1 marks. 2 Allow any other valid response. Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 2(b) Reasons for choosing wood will include:  Local availability of timber  Available in large section  Slightly flexible compared to other materials  Easily repaired / maintained  Durable  Renewable resource  Resists forces well e.g. compression / tension / torsion  Easily joined, nails, screws, bolts, mechanical joints. 2  1 marks. 2 AOVR Do not allow ‘cheap’ or ‘low cost’ unless it is justified.

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3 Window frames can be made from wood or plastic. (a) Give two advantages of using plastic rather than wood for window frames. 1 … … 2 … … [2] (b) Give two disadvantages of using plastic rather than wood for window frames. 1 … … 2 … … [2]

4 marks

Mark scheme: 3(a) Advantages of plastics for window frames include:  Better weather resisting properties  Better insulation  Reduces maintenance  Plastic can be extruded / moulded to a shape  Long lasting / durable  Lighter weight to handle when fitting  Dimensionally more stable. 2  1 marks. 2 Allow any other valid response. Only allow ‘Cost’ if justified. 3(b) Disadvantages of plastics include:  Reduced aesthetic qualities  Can discolour when exposed to UV from sun  Can become brittle.  Environmentally unsound  Not as resistant to compression. 2  1 marks. 2 Allow any other valid response. Do not allow cost of materials.

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10 Fig. 10.1 shows views of a wooden aircraft propeller made from a number of layers of hardwood glued together. Fig. 10.1 (a) (i) Name the method of construction that has been used to produce the propeller. … [1] (ii) Describe the properties that are required in the adhesive that is used to glue the hardwood layers together. … … … [2] (iii) Give two benefits to the manufacturer of using this method of construction to produce the propeller. 1 … … 2 … … [2] (b) Fig. 10.2 shows three methods of extending the length of a piece of steel tube. W X Y Fig. 10.2 (i) Complete Table 10.1 by adding ticks (3), to show two forces that, when applied to the extended length of tube, will be best resisted by each method. Table 10.1 Method Bending Torsion Compression Tension W X Y [6] (ii) For one of the methods used in part (b)(i), use sketches and notes to show an improvement that will allow it to resist greater force. Identify the greater forces being resisted. [3] (c) Fig. 10.3 shows two views of a wall bracket to hold a hanging basket for flowers. A B Fig. 10.3 (i) On view A draw a tie on the bracket to prevent the arm from bending. [2] (ii) On view B draw a strut that will prevent the arm from bending. [2] (iii) Add a gusset plate to view B to increase resistance to bending. [2] (d) Fig. 10.4 shows a car park barrier that can be raised or lowered by an electric motor to allow cars through. 3 m 0.8 m limit stop X 245 N Fig. 10.4 (i) Explain why a limit stop is used on the barrier. … … … … [2] (ii) Calculate the force required at point X needed to keep the barrier in equilibrium. … … … … [3]

25 marks

Mark scheme: 10(a)(i) Lamination is the method of construction 1 10(a)(ii) Properties of adhesive could include:  Must be waterproof / water resistant  Gap filling  Resistant to movement of the wood  High tensile strength  Resistant to shear. 2 Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 10(a)(iii) Benefits of laminating will include:  Large stable section of wood is produced  Increased resistance to bending  Improved precision  Natural defects can be avoided  Consistent results in a batch. 2  1 marks 2 Allow other valid benefits. Do not allow cheap, easy. 10(b)(i) 6  1 marks 6 For Y allow compression. No mark for a row with more than two ticks. 10(b)(ii) Improvement identified in notes / sketches, 1 mark Force being resisted clearly identified, 1 mark Functional method, 1 mark. 3 Method Bending Compression Torsion Tension W X Y       Question Answer Marks Guidance 10(c)(i) Tie in suitable position, 1 mark. Suitable section of material shown, 1 mark. 2 10(c)(ii) Strut in suitable position, 1 mark. Suitable section of material shown, 1 mark. 2 10(c)(iii) Suitable shape of plate, 1 mark. Suitable position, 1 mark. 2 Accept either position Question Answer Marks Guidance 10(d)(i) The limit stop is used to support the barrier in the down position, 1 mark. To avoid any strain on the motor, 1 mark. 2 Accept any other valid response. 10(d)(ii) 3  X = 0.8  245, 1 mark X = 196 / 3, 1 mark X = 65.3 N, 1 mark. 3 Award 3 marks for answer with no working.

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11 Fig. 11.1 shows a hand-operated whisk. turning handle 12t gears beaters 60t gear Fig. 11.1 (a) Materials used in the whisk include stainless steel and nylon. (i) Give one property of each material that makes it suitable for the whisk. stainless steel … … nylon … … [2] (ii) The whisk turning handle operates a 60t bevel gear connected to drive two 12t bevel gears, each attached to a beater. Name the mechanism used in the turning handle. … [1] (iii) State the velocity ratio of the gears. … [2] (iv) The whisk turning handle can be operated comfortably at 30 rpm. Calculate the speed of each beater when the turning handle is operated at 30 rpm. … … … [2] (v) Describe the movement of one beater relative to the other. … … [2] (b) (i) Name three types of bearing commonly used in mechanisms. 1 … 2 … 3 … [3] (ii) Describe the purpose of a bearing in a mechanism. … … … [2] (iii) Use sketches and notes to show how a bearing can control end-to-end movement (thrust) on a shaft. [3] (c) (i) Describe the difference between a first, second and third order lever. … … … … [3] (ii) Explain why levers are important in hand operated machinery. … … … [2] (d) Fig. 11.2 shows a hoist that uses a compound pulley system to lift heavy loads. Fig. 11.2 (i) State the mechanical advantage of the pulley system. … [1] (ii) Calculate the force, in Newtons, needed to raise a mass of 200 kg using the pulley system. 1 kg = 9.81 N … … … [2]

25 marks

Mark scheme: 11(a)(i) Stainless steel properties - hygienic, will not rust or corrode, long lasting. Nylon – self-lubricating bearing / gear material, can be easily moulded, durable. 2 x 1 marks. 2 Allow ‘durable’ for stainless steel. 11(a)(ii) The mechanism in the turning handle is a crank. 1 11(a)(iii) Velocity ratio = 60 / 12 [1] = 5:1[1] Numerical answer, 1 mark. Given as a ratio, 1 mark. 2 5:1 2 marks 1:5 no mark unless the working is correct. 11(a)(iv) Speed of beaters = 30  5, 1 mark = 150 rpm 1 mark. 2 Award 2 marks for correct answer with no working. Allow ecf from (ii) 11(a)(v) Each beater will turn at the same speed as the other one, 1 mark They will turn in opposite directions, 1 mark. 2 11(b)(i) Types of bearings include, plain, ball, roller, 1 mark for each. 3 Accept ceramic, needle or self-aligning bearings. Question Answer Marks Guidance 11(b)(ii) The purpose if a bearing is to:  Provide support to a moving part of a mechanism.  Reduce friction between moving parts  Reduce wear in the mechanism parts  Support both radial and axial loads.  Reduce contact area. 2 Allow other relevant points. Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 11(b)(iii) Shaft shown or described in suitable position, 1 mark. Recognised type of thrust bearing used, ball, roller, plain, 1 mark Functional method used, 1 mark. 3 11(c)(i) Description to use relative positions of fulcrum, load and effort.  First Order is E F L with fulcrum in the centre.  Second order is E L F with load in the centre  Third order is effort in the centre, F E L 1 mark for each correct statement. 3  1 marks 3 11(c)(ii) Reasons for importance of levers in hand operated machinery will include:  To make operation of the machine easier, requiring less effort.  To provide a mechanical advantage  To reduce fatigue in workers  To make operation of the machine safer.  To allow more precise control  Gives feedback to user. 2  1 marks 2 Explanation with two points, 2 marks. Allow 2 marks for single fully justified point. 11(d)(i) The mechanical advantage is 8. 1 11(d)(ii) Effort needed to raise a mass of 200 kg is: 200 / 8, = 25 kg, 1 mark 25 kg = 25  9.81 = 245.25 N, 1 mark. 2 Allow ecf from (d)(i). Award 2 marks for answer with no working. ball bearing plain bearing

This question in 0445/41 May/June 2022

1 Fig. 1.1 shows an adjustable height axle stand used to support the weight of a vehicle. central pin column base Fig. 1.1 (a) State three properties of mild steel that make it suitable for use in the base of the axle stand. 1 … 2 … 3 … [3] (b) Describe the type of structure used in the base of the axle stand. … … … [2] (c) The pin used to support the central column must resist shear force. Use sketches and notes to describe what is meant by ‘shear force’. [2]

7 marks

Mark scheme: 1(a) Suitable properties include:  Strong in tension  Resists torsion  Resists compression  Malleable / ductile  Resists bending  Can be joined easily  Low cost / readily available  Can be easily recycled. 3  1 marks. 3 Allow other valid properties. E.g. durable. Do not allow ‘strong’ unless justified. 1(b) Description may include the following:  Frame structure  Welded joints  Ties on each face  Triangulated. 2 2 marks for a description that includes two valid points. Allow 2 marks for a single point fully explained. 1(c) Sketches / notes to show two opposing compressive / parallel forces, [1] pushing in opposite directions at the point of contact, [1]. 2

This question in 0445/42 May/June 2022

9 Fig. 9.1 shows a wooden chair. Fig. 9.1 (a) (i) State three forces that may affect the stability of the structure when a person sits on the chair. 1 … 2 … 3 … [3] (ii) Fig. 9.2 shows the laminated top rails of the chair. 3 laminations in each top rail Fig. 9.2 Describe the benefits of using lamination to manufacture the top rails. … … … … … [3] (b) Fig. 9.3 shows a joint that could be used in a frame construction. Fig. 9.3 (i) Give one benefit of cutting the joint using CAM technology. … … [1] (ii) Give one benefit of cutting the joint using hand tools. … … [1] (iii) The joint in Fig. 9.3 will be assembled using adhesive. State one property that the adhesive will need. … [1] (c) (i) Use sketches and notes to describe the differences between the following structural members. beam strut tie [6] (ii) Suitable materials for structural members must be selected carefully. Give two examples of defects that should be avoided when selecting wood to be used in structures. 1 … 2 … [2] (iii) Explain, using an example, how composite materials can be used to improve the performance of a structure. … … … … … [3] (d) Fig. 9.4 shows a workshop hoist with a counterbalance weight. 1200 710 extendable arm 220 kg support pillar Fig. 9.4 (i) Label the counterbalance weight on Fig. 9.4. [1] (ii) Calculate the weight of the counterbalance which will keep the workshop hoist structure in equilibrium while lifting the load, with the minimum stress on the structure. … … … … [2] (iii) Describe the considerations that should be taken into account when calculating a Factor of Safety for the workshop hoist. … … … … [2]

25 marks

Mark scheme: 9(a)(i) Any three forces from: tension, compression, torsion, shear, bending, 3  1 marks. 3 9(a)(ii) Benefits of lamination will include:  Retaining shape during assembly  Defects in wood can be avoided  Short grain avoided when shaping  Less waste than cutting from a solid piece  Curves can be formed. 1 mark for each benefit in description, maximum 3. 3 Allow other valid benefits Allow 2 marks for a full description of a single point. 9(b)(i) Benefits of CAM technology include:  Accuracy  Repeatability / all joints will be the same size,  Speed of production.  Reduction in manual labour required 1 mark for valid benefit. 1 9(b)(ii) Benefits of hand methods include:  Low cost of tools required  No time spent in setting up a machine  No electricity needed  No specialist machine knowledge / expertise required  No problem with fitting work onto a machine.  Errors can be identified quickly and corrected 1 mark for valid benefit. 1 Accept other valid benefits. Do not allow ‘faster’ unless it is justified. Question Answer Marks Guidance 9(b)(iii) Properties of adhesive will include:  Setting time must allow for assembly of frame,  May need to be water resistant,  Must be strong in tension,  Should dry clear / not leave any marks on the wood. 1 mark for valid property. 1 9(c)(i) 2 marks for differences identified in each item, 3  2 marks Beam, freestanding with supports at both or either ends6, [1] will resist bending [1] There will be compression on the top face and tension on bottom face. [1] Strut, fixed at two points in a frame. [1] Will resist compression and help to maintain the shape of the frame / made from a rigid material. [1] Tie, fixed at two points in a frame. [1] Structure resisting tension, can be made from a flexible material. [1] 6 Points made must relate to differences between the items not just a description of the individual item. 9(c)(ii) Natural defects in wood will include: large knots, shakes, splits, warping, bending, insect / termite damage, excessive moisture, 2  1 marks. 2 9(c)(iii) Suitable example e.g. GRP, concrete, carbon fibre, [1]. Reference to materials in the composite, [1]. Improvements made in performance of structure as a result, [1]. 3 Accept other composites 9(d)(i) 1 mark for label in suitable position. 1 1200 710 support pillar 220 kg counterbalance weight Question Answer Marks Guidance 9(d)(ii) 220  1200 = 710  counterbalance, [1]. 220  1200 / 710 = 371.8 kg, [1]. 2 Award 2 marks for correct answer with no working. Allow rounding errors, e.g. 374 kg. 9(d)(iii) Factors to be considered will include:  How far can the lifting arm be extended?  The breaking strain of the lift cable  Strength of the materials used  Labelling on hoist to inform user of safe working loads  Percentage reduction from the absolute maximum that can be lifted. 2 1 mark each for 2 points mentioned. Allow 2 marks for a single point fully explained / justified.

This question in 0445/42 May/June 2022

1 Name three types of man-made structure. 1 … 2 … 3 … [3]

3 marks

Mark scheme: 1 Frame structure, 1 mark. Shell structure, 1 mark. Mass structure, 1 mark. 3 Allow examples of the types of structure Any examples must only be applicable to a single type of structure.

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2 Fig. 2.1 shows a wooden support structure for a section of railway track. Fig. 2.1 (a) Describe the methods used to ensure that the structure can safely support a heavy load. … … … [2] (b) Give two reasons for choosing wood as the material for the structure. 1 … 2 … [2]

4 marks

Mark scheme: 2(a) Methods visible are:  Heavy section timber  Cross braces  Triangulation  Solid foundations  Increased width at base  Bolts are used to join sections of timber. 2  1 marks. 2 Allow any other valid response. Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 2(b) Reasons for choosing wood will include:  Local availability of timber  Available in large section  Slightly flexible compared to other materials  Easily repaired / maintained  Durable  Renewable resource  Resists forces well e.g. compression / tension / torsion  Easily joined, nails, screws, bolts, mechanical joints. 2  1 marks. 2 AOVR Do not allow ‘cheap’ or ‘low cost’ unless it is justified.

This question in 0445/43 May/June 2022

3 Window frames can be made from wood or plastic. (a) Give two advantages of using plastic rather than wood for window frames. 1 … … 2 … … [2] (b) Give two disadvantages of using plastic rather than wood for window frames. 1 … … 2 … … [2]

4 marks

Mark scheme: 3(a) Advantages of plastics for window frames include:  Better weather resisting properties  Better insulation  Reduces maintenance  Plastic can be extruded / moulded to a shape  Long lasting / durable  Lighter weight to handle when fitting  Dimensionally more stable. 2  1 marks. 2 Allow any other valid response. Only allow ‘Cost’ if justified. 3(b) Disadvantages of plastics include:  Reduced aesthetic qualities  Can discolour when exposed to UV from sun  Can become brittle.  Environmentally unsound  Not as resistant to compression. 2  1 marks. 2 Allow any other valid response. Do not allow cost of materials.

This question in 0445/43 May/June 2022

10 Fig. 10.1 shows views of a wooden aircraft propeller made from a number of layers of hardwood glued together. Fig. 10.1 (a) (i) Name the method of construction that has been used to produce the propeller. … [1] (ii) Describe the properties that are required in the adhesive that is used to glue the hardwood layers together. … … … [2] (iii) Give two benefits to the manufacturer of using this method of construction to produce the propeller. 1 … … 2 … … [2] (b) Fig. 10.2 shows three methods of extending the length of a piece of steel tube. W X Y Fig. 10.2 (i) Complete Table 10.1 by adding ticks (3), to show two forces that, when applied to the extended length of tube, will be best resisted by each method. Table 10.1 Method Bending Torsion Compression Tension W X Y [6] (ii) For one of the methods used in part (b)(i), use sketches and notes to show an improvement that will allow it to resist greater force. Identify the greater forces being resisted. [3] (c) Fig. 10.3 shows two views of a wall bracket to hold a hanging basket for flowers. A B Fig. 10.3 (i) On view A draw a tie on the bracket to prevent the arm from bending. [2] (ii) On view B draw a strut that will prevent the arm from bending. [2] (iii) Add a gusset plate to view B to increase resistance to bending. [2] (d) Fig. 10.4 shows a car park barrier that can be raised or lowered by an electric motor to allow cars through. 3 m 0.8 m limit stop X 245 N Fig. 10.4 (i) Explain why a limit stop is used on the barrier. … … … … [2] (ii) Calculate the force required at point X needed to keep the barrier in equilibrium. … … … … [3]

25 marks

Mark scheme: 10(a)(i) Lamination is the method of construction 1 10(a)(ii) Properties of adhesive could include:  Must be waterproof / water resistant  Gap filling  Resistant to movement of the wood  High tensile strength  Resistant to shear. 2 Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 10(a)(iii) Benefits of laminating will include:  Large stable section of wood is produced  Increased resistance to bending  Improved precision  Natural defects can be avoided  Consistent results in a batch. 2  1 marks 2 Allow other valid benefits. Do not allow cheap, easy. 10(b)(i) 6  1 marks 6 For Y allow compression. No mark for a row with more than two ticks. 10(b)(ii) Improvement identified in notes / sketches, 1 mark Force being resisted clearly identified, 1 mark Functional method, 1 mark. 3 Method Bending Compression Torsion Tension W X Y       Question Answer Marks Guidance 10(c)(i) Tie in suitable position, 1 mark. Suitable section of material shown, 1 mark. 2 10(c)(ii) Strut in suitable position, 1 mark. Suitable section of material shown, 1 mark. 2 10(c)(iii) Suitable shape of plate, 1 mark. Suitable position, 1 mark. 2 Accept either position Question Answer Marks Guidance 10(d)(i) The limit stop is used to support the barrier in the down position, 1 mark. To avoid any strain on the motor, 1 mark. 2 Accept any other valid response. 10(d)(ii) 3  X = 0.8  245, 1 mark X = 196 / 3, 1 mark X = 65.3 N, 1 mark. 3 Award 3 marks for answer with no working.

This question in 0445/43 May/June 2022

11 Fig. 11.1 shows a hand-operated whisk. turning handle 12t gears beaters 60t gear Fig. 11.1 (a) Materials used in the whisk include stainless steel and nylon. (i) Give one property of each material that makes it suitable for the whisk. stainless steel … … nylon … … [2] (ii) The whisk turning handle operates a 60t bevel gear connected to drive two 12t bevel gears, each attached to a beater. Name the mechanism used in the turning handle. … [1] (iii) State the velocity ratio of the gears. … [2] (iv) The whisk turning handle can be operated comfortably at 30 rpm. Calculate the speed of each beater when the turning handle is operated at 30 rpm. … … … [2] (v) Describe the movement of one beater relative to the other. … … [2] (b) (i) Name three types of bearing commonly used in mechanisms. 1 … 2 … 3 … [3] (ii) Describe the purpose of a bearing in a mechanism. … … … [2] (iii) Use sketches and notes to show how a bearing can control end-to-end movement (thrust) on a shaft. [3] (c) (i) Describe the difference between a first, second and third order lever. … … … … [3] (ii) Explain why levers are important in hand operated machinery. … … … [2] (d) Fig. 11.2 shows a hoist that uses a compound pulley system to lift heavy loads. Fig. 11.2 (i) State the mechanical advantage of the pulley system. … [1] (ii) Calculate the force, in Newtons, needed to raise a mass of 200 kg using the pulley system. 1 kg = 9.81 N … … … [2]

25 marks

Mark scheme: 11(a)(i) Stainless steel properties - hygienic, will not rust or corrode, long lasting. Nylon – self-lubricating bearing / gear material, can be easily moulded, durable. 2 x 1 marks. 2 Allow ‘durable’ for stainless steel. 11(a)(ii) The mechanism in the turning handle is a crank. 1 11(a)(iii) Velocity ratio = 60 / 12 [1] = 5:1[1] Numerical answer, 1 mark. Given as a ratio, 1 mark. 2 5:1 2 marks 1:5 no mark unless the working is correct. 11(a)(iv) Speed of beaters = 30  5, 1 mark = 150 rpm 1 mark. 2 Award 2 marks for correct answer with no working. Allow ecf from (ii) 11(a)(v) Each beater will turn at the same speed as the other one, 1 mark They will turn in opposite directions, 1 mark. 2 11(b)(i) Types of bearings include, plain, ball, roller, 1 mark for each. 3 Accept ceramic, needle or self-aligning bearings. Question Answer Marks Guidance 11(b)(ii) The purpose if a bearing is to:  Provide support to a moving part of a mechanism.  Reduce friction between moving parts  Reduce wear in the mechanism parts  Support both radial and axial loads.  Reduce contact area. 2 Allow other relevant points. Description that includes two points, 2 marks. Allow 2 marks for single fully justified point. 11(b)(iii) Shaft shown or described in suitable position, 1 mark. Recognised type of thrust bearing used, ball, roller, plain, 1 mark Functional method used, 1 mark. 3 11(c)(i) Description to use relative positions of fulcrum, load and effort.  First Order is E F L with fulcrum in the centre.  Second order is E L F with load in the centre  Third order is effort in the centre, F E L 1 mark for each correct statement. 3  1 marks 3 11(c)(ii) Reasons for importance of levers in hand operated machinery will include:  To make operation of the machine easier, requiring less effort.  To provide a mechanical advantage  To reduce fatigue in workers  To make operation of the machine safer.  To allow more precise control  Gives feedback to user. 2  1 marks 2 Explanation with two points, 2 marks. Allow 2 marks for single fully justified point. 11(d)(i) The mechanical advantage is 8. 1 11(d)(ii) Effort needed to raise a mass of 200 kg is: 200 / 8, = 25 kg, 1 mark 25 kg = 25  9.81 = 245.25 N, 1 mark. 2 Allow ecf from (d)(i). Award 2 marks for answer with no working. ball bearing plain bearing

This question in 0445/43 May/June 2022

1 Fig. 1.1 shows two methods of packaging drinks. plastic glass bottle bottle Fig. 1.1 (a) Give one environmental reason for using glass bottles. … [1] (b) Give one functional reason for using plastic bottles. … [1] (c) Name the type of structure used in both bottles in Fig. 1.1. … [1]

3 marks

Mark scheme: Question Answer Marks Guidance 1(a) Reasons for using glass will include: 1 Allow any other valid reason. • Easy recycling • Hygienic, can be used multiple times • The raw materials used are not based on fossil fuels • Reusable • Less pollution caused. 1(b) Functional reasons for plastics will include: 1 Do not allow unjustified reasons related to • Lighter to transport / lightweight cost. • Easily resealable Allow any other valid reason. • Plastic bottles do not break easily / more durable • Handles can easily be moulded into the shape • Suitable for bulk manufacture. 1(c) Both bottles are a shell structure, 1 mark. 1

This question in 0445/42 Oct/Nov 2022

2 Fig. 2.1 shows a skeleton leaf. Fig. 2.1 State the type of structure that forms the leaf. … [1]

1 marks

Mark scheme: 2 The skeleton leaf is a frame structure, 1 mark 1

This question in 0445/42 Oct/Nov 2022

3 Most electricity pylons are made from steel. Give three reasons why steel is a suitable material for the manufacture of electricity pylons. 1 … … 2 … … 3 … … [3]

3 marks

Mark scheme: 3 Reasons for steel being suitable for pylons include: 3 Allow other valid reasons. • Relatively low cost Do not allow reference to resisting • Plentiful supply corrosion. • Can be recycled. • Strong in tension • Strong in compression • Resists torsion / bending • Steel is a durable material • Can be produced in several suitable sections, e.g. angle section. • Easily joined by mechanical means or heat processes. 3  1 marks

This question in 0445/42 Oct/Nov 2022

10 Fig. 10.1 shows an archery bow. … … … Fig. 10.1 (a) (i) Label the area of the bow that is: • in compression • in tension • the neutral axis. [3] (ii) Give two benefits of using composite laminated materials, rather than a single piece of wood, to make a bow. 1 … … 2 … … [2] (iii) Give one property of an adhesive needed when joining the materials in a composite laminated bow. … … [1] (iv) Name two composites and the materials that are used in their manufacture. Name of composite 1 … Materials used in manufacture … Name of composite 2 … Materials used in manufacture … [6] (b) Fig. 10.2a shows a structure used in a multi-storey building to support window openings. The structure includes ties and struts. steel tube steel tube steel rods Fig. 10.2a Fig. 10.2b (i) Add labels for a tie and a strut on Fig. 10.2b. [2] (ii) Ties are often made adjustable in length. Use sketches and notes to show one way of making a tie adjustable in length. [3] (c) (i) Fig. 10.3 shows two steel beams of equal length. A B Fig. 10.3 Explain why beam A can withstand more load than beam B without bending in the centre. … … … [2] (ii) Calculate the reactions R1 and R2 when beam A is loaded as shown in Fig. 10.4. 5.2 m 1.8 m 2 m 1200 N 1200 N R1 R2 Fig. 10.4 … … … … … … [4] (iii) Describe how Factor of Safety contributes to the design of a structure that uses beams. … … … [2]

25 marks

Mark scheme: 10(a)(i) 3 1 mark for each correct label. 10(a)(ii) Benefits of composites will include: 2 Allow any other valid benefits, e.g. • Material is more stable reference to ‘aesthetic qualities’, ‘increased • Natural defects in the materials can be avoided strength’. • Greater range of designs are possible • Increased durability of the composite 10(a)(iii) Adhesive properties will include: 1 Allow other valid properties. • Flexibility will be needed • Moisture resistance / waterproof • Unaffected by change in temperature • Resistant to applied forces e.g. shear / tension • ‘Gap filling’ qualities. 1 mark 10(a)(iv) Examples of composites include: 6 Allow other composites such as Kevlar and Resin based, ceramic composites. GRP – polyester resin and glass fibres Allow Fibreglass for GRP Carbon fibre – graphite fibres bonded with resin Concrete – Portland cement and aggregate Reinforced concrete – Concrete with the addition of reinforcing bars that are strong in tension 1 mark for names 2  1 2  2 marks for constituent materials. 10(b)(i) 2 1 mark for each correct label, 2  1 marks 10(b)(ii) Change in length provided, 1 mark e.g. left and right hand threads on a 3 connector. Adjustable method, capable of being tightened and loosened 1 mark Clear sketches / notes, 1 mark 10(c)(i) The following points could be included: 2 One mark for each point mentioned. • Increased depth of the web will resist bending on beam A. Allow two marks for a single point fully • The top and bottom flanges are supporting the central web explained. • In beam B there is less depth to resist the bending caused by a load on the beam. 10(c)(ii) Taking moments about R1 4 Award full marks for correct answer with no (1.8  1200) + (3.8  1200) = R2  5.2, 1 mark working. 2160 + 4560 = R2  5.2, 1 mark 6720 / 5.2 = R2 = 1292.3 N, 1 mark The reaction at R1 is 2400 – 1292.3 = 1107.7 N, 1 mark 10(c)(iii) • The factor of safety of a structural design will be calculated from the 2 One mark for each point mentioned. failure loading divided by a number, the factor of safety. Allow two marks for a single point fully • This means that even under exceptional conditions the structure should explained. not fail. • A higher factor of safety number will result in the design load that the structure can take being lower. • Accurate calculation and use of the factor of safety means that there is reduced chance of the structure collapsing.

This question in 0445/42 Oct/Nov 2022

1 Fig. 1.1 shows two methods of packaging drinks. plastic glass bottle bottle Fig. 1.1 (a) Give one environmental reason for using glass bottles. … [1] (b) Give one functional reason for using plastic bottles. … [1] (c) Name the type of structure used in both bottles in Fig. 1.1. … [1]

3 marks

Mark scheme: Question Answer Marks Guidance 1(a) Reasons for using glass will include: 1 Allow any other valid reason. • Easy recycling • Hygienic, can be used multiple times • The raw materials used are not based on fossil fuels • Reusable • Less pollution caused. 1(b) Functional reasons for plastics will include: 1 Do not allow unjustified reasons related to • Lighter to transport / lightweight cost. • Easily resealable Allow any other valid reason. • Plastic bottles do not break easily / more durable • Handles can easily be moulded into the shape • Suitable for bulk manufacture. 1(c) Both bottles are a shell structure, 1 mark. 1

This question in 0445/43 Oct/Nov 2022

2 Fig. 2.1 shows a skeleton leaf. Fig. 2.1 State the type of structure that forms the leaf. … [1]

1 marks

Mark scheme: 2 The skeleton leaf is a frame structure, 1 mark 1

This question in 0445/43 Oct/Nov 2022

3 Most electricity pylons are made from steel. Give three reasons why steel is a suitable material for the manufacture of electricity pylons. 1 … … 2 … … 3 … … [3]

3 marks

Mark scheme: 3 Reasons for steel being suitable for pylons include: 3 Allow other valid reasons. • Relatively low cost Do not allow reference to resisting • Plentiful supply corrosion. • Can be recycled. • Strong in tension • Strong in compression • Resists torsion / bending • Steel is a durable material • Can be produced in several suitable sections, e.g. angle section. • Easily joined by mechanical means or heat processes. 3  1 marks

This question in 0445/43 Oct/Nov 2022

10 Fig. 10.1 shows an archery bow. … … … Fig. 10.1 (a) (i) Label the area of the bow that is: • in compression • in tension • the neutral axis. [3] (ii) Give two benefits of using composite laminated materials, rather than a single piece of wood, to make a bow. 1 … … 2 … … [2] (iii) Give one property of an adhesive needed when joining the materials in a composite laminated bow. … … [1] (iv) Name two composites and the materials that are used in their manufacture. Name of composite 1 … Materials used in manufacture … Name of composite 2 … Materials used in manufacture … [6] (b) Fig. 10.2a shows a structure used in a multi-storey building to support window openings. The structure includes ties and struts. steel tube steel tube steel rods Fig. 10.2a Fig. 10.2b (i) Add labels for a tie and a strut on Fig. 10.2b. [2] (ii) Ties are often made adjustable in length. Use sketches and notes to show one way of making a tie adjustable in length. [3] (c) (i) Fig. 10.3 shows two steel beams of equal length. A B Fig. 10.3 Explain why beam A can withstand more load than beam B without bending in the centre. … … … [2] (ii) Calculate the reactions R1 and R2 when beam A is loaded as shown in Fig. 10.4. 5.2 m 1.8 m 2 m 1200 N 1200 N R1 R2 Fig. 10.4 … … … … … … [4] (iii) Describe how Factor of Safety contributes to the design of a structure that uses beams. … … … [2]

25 marks

Mark scheme: 10(a)(i) 3 1 mark for each correct label. 10(a)(ii) Benefits of composites will include: 2 Allow any other valid benefits, e.g. • Material is more stable reference to ‘aesthetic qualities’, ‘increased • Natural defects in the materials can be avoided strength’. • Greater range of designs are possible • Increased durability of the composite 10(a)(iii) Adhesive properties will include: 1 Allow other valid properties. • Flexibility will be needed • Moisture resistance / waterproof • Unaffected by change in temperature • Resistant to applied forces e.g. shear / tension • ‘Gap filling’ qualities. 1 mark 10(a)(iv) Examples of composites include: 6 Allow other composites such as Kevlar and Resin based, ceramic composites. GRP – polyester resin and glass fibres Allow Fibreglass for GRP Carbon fibre – graphite fibres bonded with resin Concrete – Portland cement and aggregate Reinforced concrete – Concrete with the addition of reinforcing bars that are strong in tension 1 mark for names 2  1 2  2 marks for constituent materials. 10(b)(i) 2 1 mark for each correct label, 2  1 marks 10(b)(ii) Change in length provided, 1 mark e.g. left and right hand threads on a 3 connector. Adjustable method, capable of being tightened and loosened 1 mark Clear sketches / notes, 1 mark 10(c)(i) The following points could be included: 2 One mark for each point mentioned. • Increased depth of the web will resist bending on beam A. Allow two marks for a single point fully • The top and bottom flanges are supporting the central web explained. • In beam B there is less depth to resist the bending caused by a load on the beam. 10(c)(ii) Taking moments about R1 4 Award full marks for correct answer with no (1.8  1200) + (3.8  1200) = R2  5.2, 1 mark working. 2160 + 4560 = R2  5.2, 1 mark 6720 / 5.2 = R2 = 1292.3 N, 1 mark The reaction at R1 is 2400 – 1292.3 = 1107.7 N, 1 mark 10(c)(iii) • The factor of safety of a structural design will be calculated from the 2 One mark for each point mentioned. failure loading divided by a number, the factor of safety. Allow two marks for a single point fully • This means that even under exceptional conditions the structure should explained. not fail. • A higher factor of safety number will result in the design load that the structure can take being lower. • Accurate calculation and use of the factor of safety means that there is reduced chance of the structure collapsing.

This question in 0445/43 Oct/Nov 2022

1 (a) Name one example of each of the following natural structures. Natural frame structure … Natural shell structure … Natural mass structure … [3] (b) Sketch and name one example of a man-made mass structure. [1]

4 marks

Mark scheme: 1(a) Any recognisable natural frame structure [1] Any recognisable natural shell structure [1] Any recognisable natural mass structure [1] 3 Named structures must be natural. Accept ‘Bird’s nest’ as a frame structure Accept ‘Cave’ and ‘Tree trunk’ as mass 1(b) Any recognisable man-made mass structure [1] 1 If sketch is clearly recognisable without the name award mark

This question in 0445/41 May/June 2023

2 Fig. 2.1 shows a diving springboard. X springboard Fig. 2.1 (a) Name one force that will act on the springboard, causing it to deform, when a diver jumps up and lands at point X. … [1] (b) The motion of the diver jumping and landing at point X will be converted into movement of the springboard. State the conversion of motion that will be caused by the diver jumping up and landing at point X. … motion to … motion. [2]

3 marks

Mark scheme: 2(a) Bending force [1] 1 2(b) Reciprocating motion [1] to oscillating motion [1]. 2

This question in 0445/41 May/June 2023

10 Fig. 10.1 shows a concrete beam with two reinforcing rods running through it. Fig. 10.1 (a) (i) Give one reason for using reinforcing rods in this concrete beam. … … [1] (ii) The reinforcing rods are held in tension until the concrete has set. When the concrete is fully set the tension on the rods is released. Explain why this is a better method than laying the reinforcing rods in the concrete without tension on them. … … … … [3] (iii) The strain on one tensioning rod is calculated to be 0.00025 when the rod has extended by 0.3 mm. Calculate the original length of the rod. change in length Use the formula Strain = original length … … … … [2] (iv) Fig. 10.1 shows the marking on the top face of the beam. Explain the reason for the marking. … … … … [2] (b) Fig. 10.2 shows three methods of joining lengths of timber. steel plates glued joint nails screws bolts A B C Fig. 10.2 (i) Give one benefit for each method. A … … B … … C … … [3] (ii) Give one drawback for each method. A … … B … … C … … (iii) Describe how the fittings used for connecting the timber could be made resistant to moisture. … … … [2] (c) Joints in structures often need to be reinforced. (i) Use sketches and notes to show the use of the following reinforcement methods: • gussets • ribs • braces. [6] (ii) Fig. 10.3 shows a method of providing a right-angled joint in the framework of a timber building. Hardwood dowels are used to secure the joint. X force hardwood dowels Fig. 10.3 Name the force that will act on the dowels when force X is applied to the horizontal piece. … [1] (iii) Use sketches and notes to show one way of holding the parts of the joint securely in place without using screws, nails or dowels. [2]

25 marks

Mark scheme: 10(a)(i) Reinforcing rods are used to increase the strength in tension 1 10(a)(ii) When the tension is released the rods contract [1] As the rods are held inside the concrete they will be pulling the concrete with them [1] When tension is applied to the concrete the compression caused by the rods contracting will cause the whole beam to resist rather than just the steel reinforcement [1] 3 Award marks for understanding shown Explanation with three points made [3] Allow 2 marks for a fully justified point 10(a)(iii) Rearrangement of formula: original length =  change in length 0.3 0.00025 0.00025 [1] Original length = 1200 mm (1.2 m) [1] 2 Award 2 marks for correct answer with no working 10(a)(iv) The reasons for the marking should include: The marking is to show which way up the beam should be installed [1] The reinforcing rods are positioned below the centre [1] line, which is where the tension will occur when the beam is loaded [1] 2 Award marks for understanding shown 10(b)(i) Any 1 benefit for each method: Method A: The steel plates will hold the two timbers in line [1] Steel is stronger [1] Bolts can be tightened if shrinkage occurs [1] Joint can be taken apart [1] Method B: Quick to join [1] Precise measurement not needed [1] Method C: More stable/Less space taken up [1] Timbers automatically held in line [1] Permanent joint [1] [3 x 1], 1 mark for each method 3 Allow any valid alternatives Question Answer Marks Guidance 10(b)(ii) Any 1 drawback for each method: Method A: More expensive(cost of plates and bolts) [1] Increases the width of joint [1] Method B: Timber can split during nailing [1] Movement can occur during nailing [1] Method C: Mechanically weaker [1] Smaller surface area of joint [1] Permanent joint, i.e. cannot take apart Joints must be accurately cut [3 x 1], 1 mark for each method 3 Allow any valid alternatives 10(b)(iii) Steel plates could be galvanised, dip coated or painted [1] Screws and nails, and bolts could be zinc plated or stainless steel [1] 2 Allow use of waterproof adhesive for method C Description to include two points or a single point described in depth 10(c)(i) Sketch of gusset [1], rib [1] and brace [1] Notes to indicate position / fixing method / material [3 x 1] 6 10(c)(ii) Shear will act on the dowels 1 10(c)(iii) Functional method used [1] Clear sketch / notes [1] 2 Allow PVA glue Allow the use of bolts

This question in 0445/41 May/June 2023

3 Name the types of structure shown in Fig. 3.1. 5 section telescopic jib extension jib reservoir overflow steps mobile crane Fig. 3.1 reservoir overflow steps … 5 section telescopic jib … extension jib … [3]

3 marks

Mark scheme: 3 Reservoir overflow steps – mass structure [1] 5 section telescopic jib – shell structure [1] Extension jib – frame structure [1] 3

This question in 0445/42 May/June 2023

4 Fig. 4.1 shows a beam that will be used in a structure. B A Fig. 4.1 Name the forces acting on the beam indicated by the arrows in Fig. 4.1. A … B … [2]

2 marks

Mark scheme: 4 Force A - torsion [1] Force B - tension [1] 2 Allow ‘torque’ for torsion

This question in 0445/42 May/June 2023

10 Fig. 10.1 shows an adjustable steel trestle used by builders to reach heights safely. When in use wooden planks are rested on a pair of trestles. brackets to hold wooden planks in position A sliding pin to lock height adjustment details of sliding pin B trestle legs sliding joint trestle feet Fig. 10.1 (a) (i) State two design features that ensure the trestle is safe and secure. 1 … 2 … [2] (ii) The trestle is made from mild steel. State one method of protecting mild steel from corrosion. … [1] (iii) Each of the trestle legs is removable. Give two benefits of having removable legs on the trestle. 1 … 2 … [2] (iv) A part view of joints A and B of the trestle frame is shown in Fig. 10.2. Use sketches to show a gusset plate at point A and a strut at point B, positioned to reduce any movement in the frame of the trestle. Add notes to show how the gusset plate and strut will be fixed in position. A B Fig. 10.2 [4] (v) The trestle is designed for use on level ground. Use sketches and notes to show a change to the design of the trestle feet that will provide adjustment so that the trestle can remain stable on uneven ground. [3] (vi) Fig. 10.3 shows two trestles with loads acting on the planks spanning the trestles. 0.875 m 1.75 m 0.375 m 400 N 750 N R1 R2 Fig. 10.3 Calculate the reaction at R1 and R2 on the trestles. … … … … … [4] (b) Fig. 10.4 shows a hollow steel lintel used to provide support above an opening in a building. front edge supports brickwork Fig. 10.4 (i) Give two benefits of using a hollow steel lintel. 1 … … 2 … … [2] (ii) Fig. 10.5 shows a section of brickwork supported by the lintel. lintel Fig. 10.5 Explain why the stationary load on the lintel is restricted to the shaded area of brickwork. … … … [2] (c) Fig. 10.6 shows part of a fence structure made from steel wires, which will be held under tension. The top wire is shown in position before it is tensioned. Ø5 steel wire concrete posts Fig. 10.6 (i) Use sketches and notes to show one method of applying tension to the wire when installing it. [2] (ii) Calculate the stress in the Ø5 steel wire when a tensile force of 3 kN is exerted on it. force Use the formula: Stress = cross sectional area … … … … … [3]

25 marks

Mark scheme: 10(a)(i) Any 2 methods such as: Locking pins [1] Triangulation on legs [1] Brackets to stop planks from sliding [1] Width of feet to spread the load [1] the trestle legs. 10(a)(ii) Any 1 suitable finishing method that will prevent water / moisture from coming into contact with the frame such as: Paint, dip coating, electroplating, powder coating, galvanizing 1 Do not allow coating in oil or grease. 10(a)(iii) Any 2 benefits of removable legs such as: Storage and transport [1] Taking up far less room [1] Each trestle folds / packs flat [1] Easily repaired by changing damaged part [1] 2 Allow other valid alternatives. 10(a)(iv) Gusset plate position [1] suitable fixing method [1] Strut position [1] suitable fixing method [1] 4 Fixing methods could include bolts, rivets, welding. 10(a)(v) Method of adjusting height of at least one foot [1] Suitable size for adjustment piece [1] Suitable position on the foot [1] 3 10(a)(vi) Taking moments about R1 (0.875  400) + (2.625  750) = R2  3, [1] 350 + 1968.75 = R2  3, [1] 2318.75 / 3 = R2 = 772.92 N, [1] The reaction at R1 is 1150 – 772.92 = 377.08 N. [1] 4 Award full marks for correct answers with no working shown. Question Answer Marks Guidance 10(b)(i) Any 2 benefits of a hollow steel lintel such as: Lighter than concrete [1] More precise dimensions [1] Easier to transport / move into position [1] Quicker to manufacture [1] Better strength : weight ratio [1] 2 Allow other valid benefits. Do not allow reference to cost. 10(b)(ii) Line of force goes up at 45° from each end of the lintel through the courses of bricks [1] Only those bricks within the triangle formed are directly loading the lintel [1] 2 1 mark for each point mentioned. Allow two marks for a full explanation of a single point. 10(c)(i) Any 2 methods of tensioning such as: Use of threaded eyes through which the wire is fixed before using the thread to apply tension [1] Wire pulled through holes in concrete and then bent back on itself before twisting to lock in place [1] Threaded tensioning device with using left and right hand threads to pull in the steel wire [1] 2 Allow any other valid method. 1 mark for functional principle of method 1 mark for relevant sketches and notes to show application of method. 10(c)(ii) Calculation of cross-sectional area of wire using Πr2 [1] 3.14159  2.52 = 19.639 [1] 3000 19.639 = 152.759 [1] 3 Allow any errors caused by rounding. Award 3 marks for correct answer with no working shown.

This question in 0445/42 May/June 2023

1 (a) Name one example of each of the following natural structures. Natural frame structure … Natural shell structure … Natural mass structure … [3] (b) Sketch and name one example of a man-made mass structure. [1]

4 marks

Mark scheme: 1(a) Any recognisable natural frame structure [1] Any recognisable natural shell structure [1] Any recognisable natural mass structure [1] 3 Named structures must be natural. Accept ‘Bird’s nest’ as a frame structure Accept ‘Cave’ and ‘Tree trunk’ as mass 1(b) Any recognisable man-made mass structure [1] 1 If sketch is clearly recognisable without the name award mark

This question in 0445/43 May/June 2023

2 Fig. 2.1 shows a diving springboard. X springboard Fig. 2.1 (a) Name one force that will act on the springboard, causing it to deform, when a diver jumps up and lands at point X. … [1] (b) The motion of the diver jumping and landing at point X will be converted into movement of the springboard. State the conversion of motion that will be caused by the diver jumping up and landing at point X. … motion to … motion. [2]

3 marks

Mark scheme: 2(a) Bending force [1] 1 2(b) Reciprocating motion [1] to oscillating motion [1]. 2

This question in 0445/43 May/June 2023

10 Fig. 10.1 shows a concrete beam with two reinforcing rods running through it. Fig. 10.1 (a) (i) Give one reason for using reinforcing rods in this concrete beam. … … [1] (ii) The reinforcing rods are held in tension until the concrete has set. When the concrete is fully set the tension on the rods is released. Explain why this is a better method than laying the reinforcing rods in the concrete without tension on them. … … … … [3] (iii) The strain on one tensioning rod is calculated to be 0.00025 when the rod has extended by 0.3 mm. Calculate the original length of the rod. change in length Use the formula Strain = original length … … … … [2] (iv) Fig. 10.1 shows the marking on the top face of the beam. Explain the reason for the marking. … … … … [2] (b) Fig. 10.2 shows three methods of joining lengths of timber. steel plates glued joint nails screws bolts A B C Fig. 10.2 (i) Give one benefit for each method. A … … B … … C … … [3] (ii) Give one drawback for each method. A … … B … … C … … (iii) Describe how the fittings used for connecting the timber could be made resistant to moisture. … … … [2] (c) Joints in structures often need to be reinforced. (i) Use sketches and notes to show the use of the following reinforcement methods: • gussets • ribs • braces. [6] (ii) Fig. 10.3 shows a method of providing a right-angled joint in the framework of a timber building. Hardwood dowels are used to secure the joint. X force hardwood dowels Fig. 10.3 Name the force that will act on the dowels when force X is applied to the horizontal piece. … [1] (iii) Use sketches and notes to show one way of holding the parts of the joint securely in place without using screws, nails or dowels. [2]

25 marks

Mark scheme: 10(a)(i) Reinforcing rods are used to increase the strength in tension 1 10(a)(ii) When the tension is released the rods contract [1] As the rods are held inside the concrete they will be pulling the concrete with them [1] When tension is applied to the concrete the compression caused by the rods contracting will cause the whole beam to resist rather than just the steel reinforcement [1] 3 Award marks for understanding shown Explanation with three points made [3] Allow 2 marks for a fully justified point 10(a)(iii) Rearrangement of formula: original length =  change in length 0.3 0.00025 0.00025 [1] Original length = 1200 mm (1.2 m) [1] 2 Award 2 marks for correct answer with no working 10(a)(iv) The reasons for the marking should include: The marking is to show which way up the beam should be installed [1] The reinforcing rods are positioned below the centre [1] line, which is where the tension will occur when the beam is loaded [1] 2 Award marks for understanding shown 10(b)(i) Any 1 benefit for each method: Method A: The steel plates will hold the two timbers in line [1] Steel is stronger [1] Bolts can be tightened if shrinkage occurs [1] Joint can be taken apart [1] Method B: Quick to join [1] Precise measurement not needed [1] Method C: More stable/Less space taken up [1] Timbers automatically held in line [1] Permanent joint [1] [3 x 1], 1 mark for each method 3 Allow any valid alternatives Question Answer Marks Guidance 10(b)(ii) Any 1 drawback for each method: Method A: More expensive(cost of plates and bolts) [1] Increases the width of joint [1] Method B: Timber can split during nailing [1] Movement can occur during nailing [1] Method C: Mechanically weaker [1] Smaller surface area of joint [1] Permanent joint, i.e. cannot take apart Joints must be accurately cut [3 x 1], 1 mark for each method 3 Allow any valid alternatives 10(b)(iii) Steel plates could be galvanised, dip coated or painted [1] Screws and nails, and bolts could be zinc plated or stainless steel [1] 2 Allow use of waterproof adhesive for method C Description to include two points or a single point described in depth 10(c)(i) Sketch of gusset [1], rib [1] and brace [1] Notes to indicate position / fixing method / material [3 x 1] 6 10(c)(ii) Shear will act on the dowels 1 10(c)(iii) Functional method used [1] Clear sketch / notes [1] 2 Allow PVA glue Allow the use of bolts

This question in 0445/43 May/June 2023

1 Fig. 1.1 shows part of the steel frame structure of an industrial building. Fig. 1.1 (a) State one reason for using triangulation in the steel frame structure. … [1] (b) The individual parts of the steel frame structure are all fabricated off-site before being assembled at the site. Give two benefits of using this method for the frame structure of a building. 1 … … 2 … … [2] (c) Give one drawback of using concrete for the framework of this type of structure. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Triangulation is used to: 1 Award mark for understanding shown. Allow reference to increased strength • Maintain 90° angles in the frame • Increase rigidity / stability 1(b) Benefits of offsite manufacture include: 2 Allow other valid benefits • Controlled conditions – weather – lighting – assembly Do not allow • Speed or ease of assembly ‘Ease of transport’ or ‘Ease of storage’ • Accuracy of dimensions • Ready access to power – machinery • Individual sections can be tested before moving to site • Less risk of damage to parts. • Easy to replace parts [2  1] 1(c) Drawbacks of concrete: 1 Do not allow cost related drawbacks Reasons for weakness must be qualified • Concrete is weak in tension / bending • Time taken for the concrete to cure • The framework cannot be taken apart • Adds weight • Heavy – difficult to manoeuvre • Not so easy to join • Slow to assemble parts.

This question in 0445/41 Oct/Nov 2023

3 Describe, using an example, the meaning of static load in a structure. … … … [2]

2 marks

Mark scheme: 3 Description to include: 2 Static load is that placed on a structure which does not move [1] Suitable example named [1]

This question in 0445/41 Oct/Nov 2023

9 Fig. 9.1 shows part of a model of a roof truss. Joints at X and Y have been glued together. apex R S T X Y 350 Fig. 9.1 (a) (i) Name the force that will act on beam T when a load is placed at the apex of the truss. … [1] (ii) Describe the likely effect on the glued joints X and Y with the load in place. … … … [2] (iii) Use sketches and notes to show one method of strengthening joints X and Y without adding any extra materials or components to the joints. [3] (iv) Fig. 9.2 shows weights that will be used to test the completed model. Fig. 9.2 To test the completed model the weights will be suspended from the apex where the rafters R and S meet allowing gradual, measured increase in the load. Sketch a design for a simple test rig that will allow the weights to be applied gradually. [3] (v) Name one static and one moving load that a full-size roof truss must withstand. static load … moving load … [2] (b) Fig. 9.3 shows two structures used to access a building under construction. Both structures use linkages to achieve vertical movement. scissor lift aerial work platform Fig. 9.3 (i) Give two advantages of using the scissor lift rather than an aerial work platform. 1 … … 2 … … [2] (ii) State one disadvantage of using the scissor lift rather than an aerial work platform. … … [1] (iii) Both types of lifting equipment will require Factor of Safety to be considered. State who will be responsible for identifying the safety considerations in any lifting equipment used. … [1] (iv) Describe three features of the equipment in use that will have been considered when deciding on the Factor of Safety. … … … … [3] (c) Fig. 9.4 shows a beam that is loaded on both sides of a central support. all dimensions are in metres 3.5 3.5 2 3 1 100 N 250 N X beam central support R1 R2 Fig. 9.4 (i) Calculate the clockwise moment on the right-hand side of the beam. … … … … [3] (ii) Calculate the load that must be placed at X to keep the beam in equilibrium. … … … … [2] (iii) State the reaction at R1 and R2 when the beam is in equilibrium. … [2]

25 marks

Mark scheme: 9(a)(i) Tension 1 9(a)(ii) Joints X and Y are likely to slide [1], horizontally away from each other [1] 2 Award marks for understanding shown. Caused by shear force [1] 9(a)(iii) 3 Allow any mechanical method of combining the two pieces which prevents movement, reducing tension on the glued joint. Joints prevented from sliding [1], no extra materials used [1] notes to indicate how the method will work [1] 9(a)(iv) Test rig must: 3 • Allow weights to hang from the apex [1] • Provide support under joints X and Y for the model [1] • Allow extra weights to be easily added [1] 9(a)(v) Static load could be roof covering, roofing timbers bearing on the truss [1] 2 Allow birds as moving load Moving load could be person walking on the roof, weather related – wind / snow / rain [1] 9(b)(i) Advantages of the scissor lift: 2 Accept any valid alternative. • More stable Accept safety related advantages • Larger base surface area • More people can have access to work area • Can carry greater load • Less chance of joints breaking • Larger working area when lift is raised. [2  1] 9(b)(ii) Disadvantage could relate to: 1 Accept any valid alternative • Difficulty of access to building with the equipment • More limited height and reach • Takes longer to manoeuvre to a different work position 9(b)(iii) The designer and manufacturer of the equipment will be responsible for initial 1 safety of the equipment. Construction company will be responsible for implementing the safety precautions. 1 mark for identified person / company. 9(b)(iv) Features assessed will include: 3 1 mark for each point included up to three. Allow 2 marks for a single point described • Height to be reached in depth. • Extension length • Angle of lifting arm • Load to be carried • Prevention of wheeled items moving • Prevention of tipping sideways • Barriers to prevent falling. 9(c)(i) Clockwise moment = (1  100) + (2  250 N), [1] 3 3 marks for correct answer with no = 100 + 500 [1] working. = 600 Nm [1] 9(c)(ii) 3  X = 600 [1] 2 X = 200 N [1] 9(c)(iii) With beam in equilibrium reaction at R1 = 0 [1] reaction at R2 = 0 [1] 2 Allow equal reactions 1 mark for anything other than 0.

This question in 0445/41 Oct/Nov 2023

1 Fig. 1.1 shows part of the steel frame structure of an industrial building. Fig. 1.1 (a) State one reason for using triangulation in the steel frame structure. … [1] (b) The individual parts of the steel frame structure are all fabricated off-site before being assembled at the site. Give two benefits of using this method for the frame structure of a building. 1 … … 2 … … [2] (c) Give one drawback of using concrete for the framework of this type of structure. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Triangulation is used to: 1 Award mark for understanding shown. Allow reference to increased strength • Maintain 90° angles in the frame • Increase rigidity / stability 1(b) Benefits of offsite manufacture include: 2 Allow other valid benefits • Controlled conditions – weather – lighting – assembly Do not allow • Speed or ease of assembly ‘Ease of transport’ or ‘Ease of storage’ • Accuracy of dimensions • Ready access to power – machinery • Individual sections can be tested before moving to site • Less risk of damage to parts. • Easy to replace parts [2  1] 1(c) Drawbacks of concrete: 1 Do not allow cost related drawbacks Reasons for weakness must be qualified • Concrete is weak in tension / bending • Time taken for the concrete to cure • The framework cannot be taken apart • Adds weight • Heavy – difficult to manoeuvre • Not so easy to join • Slow to assemble parts.

This question in 0445/42 Oct/Nov 2023

3 Describe, using an example, the meaning of static load in a structure. … … … [2]

2 marks

Mark scheme: 3 Description to include: 2 Static load is that placed on a structure which does not move [1] Suitable example named [1]

This question in 0445/42 Oct/Nov 2023

9 Fig. 9.1 shows part of a model of a roof truss. Joints at X and Y have been glued together. apex R S T X Y 350 Fig. 9.1 (a) (i) Name the force that will act on beam T when a load is placed at the apex of the truss. … [1] (ii) Describe the likely effect on the glued joints X and Y with the load in place. … … … [2] (iii) Use sketches and notes to show one method of strengthening joints X and Y without adding any extra materials or components to the joints. [3] (iv) Fig. 9.2 shows weights that will be used to test the completed model. Fig. 9.2 To test the completed model the weights will be suspended from the apex where the rafters R and S meet allowing gradual, measured increase in the load. Sketch a design for a simple test rig that will allow the weights to be applied gradually. [3] (v) Name one static and one moving load that a full-size roof truss must withstand. static load … moving load … [2] (b) Fig. 9.3 shows two structures used to access a building under construction. Both structures use linkages to achieve vertical movement. scissor lift aerial work platform Fig. 9.3 (i) Give two advantages of using the scissor lift rather than an aerial work platform. 1 … … 2 … … [2] (ii) State one disadvantage of using the scissor lift rather than an aerial work platform. … … [1] (iii) Both types of lifting equipment will require Factor of Safety to be considered. State who will be responsible for identifying the safety considerations in any lifting equipment used. … [1] (iv) Describe three features of the equipment in use that will have been considered when deciding on the Factor of Safety. … … … … [3] (c) Fig. 9.4 shows a beam that is loaded on both sides of a central support. all dimensions are in metres 3.5 3.5 2 3 1 100 N 250 N X beam central support R1 R2 Fig. 9.4 (i) Calculate the clockwise moment on the right-hand side of the beam. … … … … [3] (ii) Calculate the load that must be placed at X to keep the beam in equilibrium. … … … … [2] (iii) State the reaction at R1 and R2 when the beam is in equilibrium. … [2]

25 marks

Mark scheme: 9(a)(i) Tension 1 9(a)(ii) Joints X and Y are likely to slide [1], horizontally away from each other [1] 2 Award marks for understanding shown. Caused by shear force [1] 9(a)(iii) 3 Allow any mechanical method of combining the two pieces which prevents movement, reducing tension on the glued joint. Joints prevented from sliding [1], no extra materials used [1] notes to indicate how the method will work [1] 9(a)(iv) Test rig must: 3 • Allow weights to hang from the apex [1] • Provide support under joints X and Y for the model [1] • Allow extra weights to be easily added [1] 9(a)(v) Static load could be roof covering, roofing timbers bearing on the truss [1] 2 Allow birds as moving load Moving load could be person walking on the roof, weather related – wind / snow / rain [1] 9(b)(i) Advantages of the scissor lift: 2 Accept any valid alternative. • More stable Accept safety related advantages • Larger base surface area • More people can have access to work area • Can carry greater load • Less chance of joints breaking • Larger working area when lift is raised. [2  1] 9(b)(ii) Disadvantage could relate to: 1 Accept any valid alternative • Difficulty of access to building with the equipment • More limited height and reach • Takes longer to manoeuvre to a different work position 9(b)(iii) The designer and manufacturer of the equipment will be responsible for initial 1 safety of the equipment. Construction company will be responsible for implementing the safety precautions. 1 mark for identified person / company. 9(b)(iv) Features assessed will include: 3 1 mark for each point included up to three. Allow 2 marks for a single point described • Height to be reached in depth. • Extension length • Angle of lifting arm • Load to be carried • Prevention of wheeled items moving • Prevention of tipping sideways • Barriers to prevent falling. 9(c)(i) Clockwise moment = (1  100) + (2  250 N), [1] 3 3 marks for correct answer with no = 100 + 500 [1] working. = 600 Nm [1] 9(c)(ii) 3  X = 600 [1] 2 X = 200 N [1] 9(c)(iii) With beam in equilibrium reaction at R1 = 0 [1] reaction at R2 = 0 [1] 2 Allow equal reactions 1 mark for anything other than 0.

This question in 0445/42 Oct/Nov 2023

1 Fig. 1.1 shows part of the steel frame structure of an industrial building. Fig. 1.1 (a) State one reason for using triangulation in the steel frame structure. … [1] (b) The individual parts of the steel frame structure are all fabricated off-site before being assembled at the site. Give two benefits of using this method for the frame structure of a building. 1 … … 2 … … [2] (c) Give one drawback of using concrete for the framework of this type of structure. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Triangulation is used to: 1 Award mark for understanding shown. Allow reference to increased strength • Maintain 90° angles in the frame • Increase rigidity / stability 1(b) Benefits of offsite manufacture include: 2 Allow other valid benefits • Controlled conditions – weather – lighting – assembly Do not allow • Speed or ease of assembly ‘Ease of transport’ or ‘Ease of storage’ • Accuracy of dimensions • Ready access to power – machinery • Individual sections can be tested before moving to site • Less risk of damage to parts. • Easy to replace parts [2  1] 1(c) Drawbacks of concrete: 1 Do not allow cost related drawbacks Reasons for weakness must be qualified • Concrete is weak in tension / bending • Time taken for the concrete to cure • The framework cannot be taken apart • Adds weight • Heavy – difficult to manoeuvre • Not so easy to join • Slow to assemble parts.

This question in 0445/43 Oct/Nov 2023

3 Describe, using an example, the meaning of static load in a structure. … … … [2]

2 marks

Mark scheme: 3 Description to include: 2 Static load is that placed on a structure which does not move [1] Suitable example named [1]

This question in 0445/43 Oct/Nov 2023

9 Fig. 9.1 shows part of a model of a roof truss. Joints at X and Y have been glued together. apex R S T X Y 350 Fig. 9.1 (a) (i) Name the force that will act on beam T when a load is placed at the apex of the truss. … [1] (ii) Describe the likely effect on the glued joints X and Y with the load in place. … … … [2] (iii) Use sketches and notes to show one method of strengthening joints X and Y without adding any extra materials or components to the joints. [3] (iv) Fig. 9.2 shows weights that will be used to test the completed model. Fig. 9.2 To test the completed model the weights will be suspended from the apex where the rafters R and S meet allowing gradual, measured increase in the load. Sketch a design for a simple test rig that will allow the weights to be applied gradually. [3] (v) Name one static and one moving load that a full-size roof truss must withstand. static load … moving load … [2] (b) Fig. 9.3 shows two structures used to access a building under construction. Both structures use linkages to achieve vertical movement. scissor lift aerial work platform Fig. 9.3 (i) Give two advantages of using the scissor lift rather than an aerial work platform. 1 … … 2 … … [2] (ii) State one disadvantage of using the scissor lift rather than an aerial work platform. … … [1] (iii) Both types of lifting equipment will require Factor of Safety to be considered. State who will be responsible for identifying the safety considerations in any lifting equipment used. … [1] (iv) Describe three features of the equipment in use that will have been considered when deciding on the Factor of Safety. … … … … [3] (c) Fig. 9.4 shows a beam that is loaded on both sides of a central support. all dimensions are in metres 3.5 3.5 2 3 1 100 N 250 N X beam central support R1 R2 Fig. 9.4 (i) Calculate the clockwise moment on the right-hand side of the beam. … … … … [3] (ii) Calculate the load that must be placed at X to keep the beam in equilibrium. … … … … [2] (iii) State the reaction at R1 and R2 when the beam is in equilibrium. … [2]

25 marks

Mark scheme: 9(a)(i) Tension 1 9(a)(ii) Joints X and Y are likely to slide [1], horizontally away from each other [1] 2 Award marks for understanding shown. Caused by shear force [1] 9(a)(iii) 3 Allow any mechanical method of combining the two pieces which prevents movement, reducing tension on the glued joint. Joints prevented from sliding [1], no extra materials used [1] notes to indicate how the method will work [1] 9(a)(iv) Test rig must: 3 • Allow weights to hang from the apex [1] • Provide support under joints X and Y for the model [1] • Allow extra weights to be easily added [1] 9(a)(v) Static load could be roof covering, roofing timbers bearing on the truss [1] 2 Allow birds as moving load Moving load could be person walking on the roof, weather related – wind / snow / rain [1] 9(b)(i) Advantages of the scissor lift: 2 Accept any valid alternative. • More stable Accept safety related advantages • Larger base surface area • More people can have access to work area • Can carry greater load • Less chance of joints breaking • Larger working area when lift is raised. [2  1] 9(b)(ii) Disadvantage could relate to: 1 Accept any valid alternative • Difficulty of access to building with the equipment • More limited height and reach • Takes longer to manoeuvre to a different work position 9(b)(iii) The designer and manufacturer of the equipment will be responsible for initial 1 safety of the equipment. Construction company will be responsible for implementing the safety precautions. 1 mark for identified person / company. 9(b)(iv) Features assessed will include: 3 1 mark for each point included up to three. Allow 2 marks for a single point described • Height to be reached in depth. • Extension length • Angle of lifting arm • Load to be carried • Prevention of wheeled items moving • Prevention of tipping sideways • Barriers to prevent falling. 9(c)(i) Clockwise moment = (1  100) + (2  250 N), [1] 3 3 marks for correct answer with no = 100 + 500 [1] working. = 600 Nm [1] 9(c)(ii) 3  X = 600 [1] 2 X = 200 N [1] 9(c)(iii) With beam in equilibrium reaction at R1 = 0 [1] reaction at R2 = 0 [1] 2 Allow equal reactions 1 mark for anything other than 0.

This question in 0445/43 Oct/Nov 2023

1 Fig. 1.1 shows an aluminium stepladder and a softwood bench. Fig. 1.1 (a) Name the type of structure used in both items. … [1] (b) State the method that is used to make the structures rigid. … [1]

2 marks

Mark scheme: 1(a) Frame Structure [1] 1 1(b) Triangulation [1] 1

This question in 0445/41 May/June 2024

2 (a) Beams used in a building can be made from wood, steel, concrete or composite materials. Give one different benefit for each material. Wood … … Steel … … Concrete … … Composite … … [4] (b) Fig. 2.1 shows a wooden beam supported at one end in the wall of a building. X Fig. 2.1 Describe the effect on the beam when a load is applied at point X. … … … [2]

6 marks

Mark scheme: 2(a) Wood – renewable resource, high availability, can be worked / shaped easily, aesthetic properties Steel – resists most forces, can be formed into different shapes / sections. Concrete – can be cast into shapes on site or preformed, strong in compression can be reinforced with steel, durable Composite – does not require long lengths of timber. Properties of different materials can be included. Complex shapes can be formed. [41] Do not allow cost related benefits. 2(b) Description to include:  Bending of the beam  Tension on the top face of the beam  Compression on the bottom face of the beam 2 1 mark for each valid point included Allow 2 marks for a point that is fully explained.

This question in 0445/41 May/June 2024

10 (a) Concrete is commonly used in structures. (i) Name three materials that are used to make concrete. 1 … 2 … 3 … [3] (ii) State the force that concrete will naturally have most resistance to. … [1] (iii) Use sketches and notes to show how resistance to other forces can be increased when using concrete. [2] (iv) Fig. 10.1 shows a design for a concrete pillar that could be used to resist load in a structure. 8.5 kN Ø 0.3 m Fig. 10.1 Calculate the stress that the pillar is subject to when a compressive force of 8.5 kN is applied. force (N) Use the formula: stress = cross-sectional area (m2) … … … … … [3] (b) Fig. 10.2 shows a beam in a roof with loads applied to it. The beam is supported at each end by a wall. 5.75 m 4.0 m 1.5 m 1200 N 2000 N R1 R2 Fig. 10.2 (i) The beam is in equilibrium. State what is meant by the term ‘equilibrium’. … … [1] (ii) Calculate the reactions at R1 and R2. Reaction at R1 … … … Reaction at R2 … … … [4] (iii) Give two examples of natural defects that could be present in a wooden beam. 1 … 2 … [2] (iv) Fig. 10.3 shows part of a roof truss. X timber section 100 × 60 Fig. 10.3 Use sketches and notes to show a suitable method of joining the timber at X. A method of preventing movement in the joint must be shown. [3] (c) Using an example of equipment used in building construction, explain two ‘Factor of Safety’ measures. … … … … [3] (d) Fig. 10.4 shows three different methods of joining metals in a structure. Give one different reason for using each method. nut and bolt countersunk rivet welded joint Fig. 10.4 Nut and bolt … Countersunk rivet … Welded joint … [3]

25 marks

Mark scheme: 10(a)(i) The materials in concrete are:  Sand and gravel  Ballast  Cement powder  Water [31] 3 Award 1 mark for sand and gravel. No marks for stones or rocks Allow addition of PVA or other plasticiser. 10(a)(ii) Compression [1] 1 10(a)(iii) Sketches to show reinforcing bar used in the concrete to resist tension. Use of reinforcing bar [1] In a suitable position [1] 2 Accept any other valid alternatives 10(a)(iv) Pillar cross sectional area = 3.142 (Π) x 0.152 = 0.0706858 [1] Stress = 8500 / 0.0706858 [1] Stress = 120250 N/m2 or 120.25 kNm2 [1] 3 Award 3 marks for correct answer with no working. Allow ecf for incorrect area. Allow rounding errors in area calculation. 10(b)(i) Equilibrium is a state where opposing forces are balanced [1] 1 Accept words to that effect. 10(b)(ii) (1.5 x 1200) + (4 x 2000) = R2 x 5.75 [1] R2 = 1800 + 8000 / 5.75 [1] R2 = 1704.348 N [1] R1 = 3200 – 1704.348 = 1495.642 [1] 4 Allow rounding errors in answers. 10(b)(iii) Any two defects such as include: knots, shakes, twisting / warping, bending, termite damage, rot. [21] 2 Accept any other valid defects. Question Answer Marks Guidance 10(b)(iv) Any one suitable joining method [1] such as:  Mechanical method, mortise and tenon, tee halving, bridle.  Nail plate / gusset plate  Nails / bolts / screws  Adhesive Clear sketches notes [1] Method of holding joint in place e.g. dowels / bolts / nails / wedges / adhesive [1] 3 10(c) Suitable example [1] Description of two factor of safety measures that apply to example:  Could be Safe Working Load, extension of crane / hoist  Safety of electrical devices  Warning of overhead cables  Training of employees [21] 3 Accept any other valid safety measures. Allow 2 marks for full description of one measure. Do not allow PPE . 10(d) Reasons for each method will include:  Nut and bolt – temporary fastening can be removed or adjusted if necessary  Countersunk rivet – permanent method, keeps the surfaces clean and flush  Welded joint – permanent method, joint strong and resists compression / tension / torsion, allows flush surface, no additional materials needed. [31] 3 Accept any other valid response

This question in 0445/41 May/June 2024

1 Fig. 1.1 shows four examples of timber structures used in a building. A B C D Fig. 1.1 (a) Name the type of structures shown in Fig. 1.1. … [1] (b) Give two reasons why structure B will be the least rigid of the four structures shown. 1 … 2 … [2] (c) Fig. 1.2 shows structure C with a force acting on it. force Fig. 1.2 Describe the effect of the force acting on the structure. … … … [2]

5 marks

Mark scheme: 1(a) Frame structures 1 1(b) Any two reasons. Could include:  No support at ends of horizontals  Will only resist compressive force  Will not resist any force that is not vertical  Only two points of connection  No triangulation, struts or ties [2  1] 2 Accept any other valid points 1(c) The force will be spread equally between the two triangles [1] The triangles will spread the force evenly across the three contact areas in lower horizontal [1] 2 Accept any other valid points e.g. bending in the top piece, compression on the sides of triangles.

This question in 0445/42 May/June 2024

2 State the meaning of static load on a structure. … [1]

1 marks

Mark scheme: 2 Static load is the sum of all loads acting on the structure that do not move, change in magnitude or change over time [1] 1 Allow mark for understanding shown.

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3 A list of materials and properties is given below. Draw lines to link the material to the property. One has been done for you. Material Property hardwood can be reformed with heat aluminium strong in compression polyvinyl chloride (PVC) renewable / sustainable glass reinforced plastic (GRP) resistant to corrosion concrete lightweight stainless steel can be moulded into irregular shape [4]

4 marks

Mark scheme: 3 2 lines correct [2] 3 lines correct [3] 4 or 5 lines correct [4]

This question in 0445/42 May/June 2024

10 (a) Students have been set the task of designing a model suspension bridge to span a gap of 400 mm. The following materials are available for the task: • softwood strips 250 × 10 × 6 • steel wire 1 mm diameter • adhesive (i) Fig. 10.1 shows the gap with steel eyes screwed in to anchor the suspension wires. Using sketches and notes, draw on Fig. 10.1 to show a possible design for the model suspension bridge. Use only the available materials. steel eye wood base 400 gap softwood strip 250 × 10 × 6 drawn to scale Fig. 10.1 [5] (ii) Name the force that will act on the suspension wires. … [1] (iii) Give two examples of moving loads that act on a full size suspension bridge. 1 … 2 … [2] (b) Fig. 10.2 shows lengths of timber for a framed building. Fig. 10.2 (i) Explain why the timber should be carefully selected before use. … … … [2] (ii) Wood is often identified as a sustainable resource. State the meaning of the term ‘sustainable resource’. … … [1] (iii) Fig. 10.3 shows two joints that could be used in the timber framed building. Explain why the mortise and tenon joint will be a better choice than a nailed joint. wedges mortise and tenon nailed Fig. 10.3 … … … … [3] (iv) Fig. 10.4 shows braces used on the mortise and tenon joint. brace brace Fig. 10.4 Give two reasons for using this type of brace. 1 … … 2 … … [2] (c) Use sketches and notes to show what is meant by the following structural members. (i) Strut [2] (ii) Tie [2] (iii) Name the forces resisted by a strut and a tie. Force resisted by a strut … Force resisted by a tie … [2] (d) Calculate the stress on a steel rod made of 9 mm diameter wire when a force of 1750 N is applied to it. … … … … [3]

25 marks

Mark scheme: 10(a)(i) Design is a suspension bridge [1]  Two or more softwood strips used to form bridge deck [1]  Strips used vertically to form towers at each end [1]  Suspension wires correctly placed [1]  Joining methods noted [1]  Only available materials used [1] [5  1] 5 10(a)(ii) The force is tension. 1 10(a)(iii) Moving loads on a bridge could be:  Traffic  People  Wind [2  1] 2 Accept any other valid responses 10(b)(i) Explanation should include:  Moisture content of timber  Visible natural defects, splits, shakes, knots, warping  Insect / termite damage  Rot 2 1 mark for each valid point included Allow 2 marks for a point that is fully explained. 10(b)(ii) A sustainable resource is one that can renew itself, be harvested and regrown 1 Award mark for understanding shown Question Answer Marks Guidance 10(b)(iii) Joint A (Mortise and tenon) is fully supported and provides a good mechanical joint which does not rely on adhesive. Pegs or wedges can hold it in place, no additional clamps required, is more durable Joint B (nailed joint) no lateral strength, reliant on strength of nails and adhesive, or clamps to hold in place while being assembled. Nails are going into end grain so reduced holding power. Can cause splitting n the timber. Nails can corrode 3 Accept any other valid response e.g. M&T has a larger cross sectional area, M&T is more durable 1 mark for each valid point included, allow two marks for a point that is fully explained. 10(b)(iv) Reasons for using braces to include:  Provide additional support  Prevent distortion of a frame  Allow reduced sections of timber to be used on framing  Reduce the forces acting directly on a joint  Uses triangulation. [2  1] 2 Allow marks for understanding shown. 10(c)(i) Example of strut shown [1] supporting existing structure [1] 2 10(c)(ii) Example of tie shown [1] supporting existing structure [1] 2 10(c)(iii) A strut resists compression [1] A tie resists tension [1] 2 10(d) Use of correct formula: force / cross sectional area [1] Calculation of cross sectional area Π  4.52 = 63.62 mm2 [1] 1750 / 63.62 = 27.51 N / mm2 [1] 3

This question in 0445/42 May/June 2024

11 (a) Fig. 11.1 shows a gardening tool used for cutting branches from trees. It uses a lever combined with a gear mechanism to provide the cutting force. larger gear smaller gear lever A lever A Fig. 11.1 (i) Add the letters ‘L’ ‘F’ and ‘E’ to the correct circles to show the position of the load (L), fulcrum (F) and effort (E) for lever A. [3] (ii) Explain how the gear mechanism reduces the effort needed to operate the gardening tool when cutting a branch. … … … … [3] (iii) Give one reason for loss of efficiency when using the gardening tool. … … [1] (b) Using examples, describe why the following types of gear would be used. (i) Bevel gears … … … [2] (ii) Worm gear … … … [2] (iii) Rack and pinion gears … … … [2] (c) Fig. 11.2 shows the drive system on a lawnmower. The drive uses toothed and vee belt pulley systems to reduce the speed of rotation provided by the electric motor. toothed belt motor drive casing lawnmower vee belt Fig. 11.2 (i) Give one reason for using a toothed belt rather than a flat belt for the drive from the motor. … … [1] (ii) The final drive to the cutting blades is provided by a vee belt. Explain how the use of a vee belt could prevent damage to the motor. … … … [2] (iii) Fig. 11.3 shows the method of tensioning the vee belt. tensioning pulley adjustment slot Fig. 11.3 Give two reasons why adjustment to the tension of the vee belt may be needed. 1 … … 2 … … [2] (iv) Fig. 11.4 shows the belt and pulley system between the motor and the final drive. Calculate the speed of rotation of the final drive when the motor turns at 1000 rpm. Ø75 Ø50 motor tensioning pulley Ø100 Ø12.5 final drive Fig. 11.4 … … … … [4] (d) (i) State what is meant by the ‘pitch’ of a screw thread on a bolt. … … … [1] (ii) Give two ways, apart from pitch, that can be used to specify a screw thread on a bolt. 1 … 2 … [2]

25 marks

Mark scheme: 11(a)(i) 1 mark for load identified, 1 mark for fulcrum identified, 1 mark for effort identified 3 11(a)(ii) The use of gear mechanism will reduce effort / small to larger gear [1] The lever effect of the long handle is multiplied [1] by the gear ratio [1] to give overall reduced effort 2 1 mark for each valid point. Award 2 marks for a single point fully explained. 11(a)(iii) Lost efficiency will be due to friction in the mechanism. 1 11(b)(i) Bevel gears are used where the input and output are in different orientations normally 90 to each other [1] VR needs altering [1] e.g. Drill chuck key [1] [2  1] 2 Accept any other valid response 11(b)(ii) A worm gear will give a large reduction in speed [1] cannot rotate in reverse [1] so provides a self-locking facility [1] e.g. winch, elevator motor drive [1] small space used [1] [2  1, any two points made] 2 Accept any other valid response, e.g. changing axes of rotation by 90 Question Answer Marks Guidance 11(b)(iii) Rack and pinion gears are used to convert rotary motion into linear or reciprocating motion [1], e.g. car steering, drill table rise and fall [1] 2 Accept any other valid response, e.g. lifting mechanisms, positioning mechanisms. 11(c)(i) A toothed belt will not slip, it is possible to keep pulleys in the same relative positions. 1 11(c)(ii) The vee belt will allow slipping [1] if the blades get jammed [1] 2 11(c)(iii) Tension of the vee belt may need adjusting:  To allow a new belt to be fitted.  To prevent slipping on the pulleys  To compensate for wear in the belt as it stretches.  To prevent wear in the belt. [2  1] 2 Accept any other valid reasons. 11(c)(iv) Reduction from motor to intermediate pulley is 6:1 [1] Reduction from intermediate pulley to final drive is 2:1 [1] Total reduction is 12:1[1] Final speed = 83.3 rpm [1] 4 Award 4 marks for correct answer with no working. 11(d)(i) The pitch of a thread is:  The distance that it moves in for each rotation  or  It is measured from the top of one thread to the top of the next 1 Allow mark for understanding shown. 11(d)(ii) The following methods can be used to specify a screw thread: length, diameter, material, finish, head shape [2  1] 2 Allow different thread profiles, e.g. acme, square, LH thread.

This question in 0445/42 May/June 2024

1 Fig. 1.1 shows an aluminium stepladder and a softwood bench. Fig. 1.1 (a) Name the type of structure used in both items. … [1] (b) State the method that is used to make the structures rigid. … [1]

2 marks

Mark scheme: 1(a) Frame Structure [1] 1 1(b) Triangulation [1] 1

This question in 0445/43 May/June 2024

2 (a) Beams used in a building can be made from wood, steel, concrete or composite materials. Give one different benefit for each material. Wood … … Steel … … Concrete … … Composite … … [4] (b) Fig. 2.1 shows a wooden beam supported at one end in the wall of a building. X Fig. 2.1 Describe the effect on the beam when a load is applied at point X. … … … [2]

6 marks

Mark scheme: 2(a) Wood – renewable resource, high availability, can be worked / shaped easily, aesthetic properties Steel – resists most forces, can be formed into different shapes / sections. Concrete – can be cast into shapes on site or preformed, strong in compression can be reinforced with steel, durable Composite – does not require long lengths of timber. Properties of different materials can be included. Complex shapes can be formed. [4  1] Do not allow cost related benefits. 2(b) Description to include:  Bending of the beam  Tension on the top face of the beam  Compression on the bottom face of the beam 2 1 mark for each valid point included Allow 2 marks for a point that is fully explained.

This question in 0445/43 May/June 2024

10 (a) Concrete is commonly used in structures. (i) Name three materials that are used to make concrete. 1 … 2 … 3 … [3] (ii) State the force that concrete will naturally have most resistance to. … [1] (iii) Use sketches and notes to show how resistance to other forces can be increased when using concrete. [2] (iv) Fig. 10.1 shows a design for a concrete pillar that could be used to resist load in a structure. 8.5 kN Ø 0.3 m Fig. 10.1 Calculate the stress that the pillar is subject to when a compressive force of 8.5 kN is applied. force (N) Use the formula: stress = cross-sectional area (m2) … … … … … [3] (b) Fig. 10.2 shows a beam in a roof with loads applied to it. The beam is supported at each end by a wall. 5.75 m 4.0 m 1.5 m 1200 N 2000 N R1 R2 Fig. 10.2 (i) The beam is in equilibrium. State what is meant by the term ‘equilibrium’. … … [1] (ii) Calculate the reactions at R1 and R2. Reaction at R1 … … … Reaction at R2 … … … [4] (iii) Give two examples of natural defects that could be present in a wooden beam. 1 … 2 … [2] (iv) Fig. 10.3 shows part of a roof truss. X timber section 100 × 60 Fig. 10.3 Use sketches and notes to show a suitable method of joining the timber at X. A method of preventing movement in the joint must be shown. [3] (c) Using an example of equipment used in building construction, explain two ‘Factor of Safety’ measures. … … … … [3] (d) Fig. 10.4 shows three different methods of joining metals in a structure. Give one different reason for using each method. nut and bolt countersunk rivet welded joint Fig. 10.4 Nut and bolt … Countersunk rivet … Welded joint … [3]

25 marks

Mark scheme: 10(a)(i) The materials in concrete are:  Sand and gravel  Ballast  Cement powder  Water [3  1] 3 Award 1 mark for sand and gravel. No marks for stones or rocks Allow addition of PVA or other plasticiser. 10(a)(ii) Compression [1] 1 10(a)(iii) Sketches to show reinforcing bar used in the concrete to resist tension. Use of reinforcing bar [1] In a suitable position [1] 2 Accept any other valid alternatives 10(a)(iv) Pillar cross sectional area = 3.142 (Π) x 0.152 = 0.0706858 [1] Stress = 8500 / 0.0706858 [1] Stress = 120250 N/m2 or 120.25 kNm2 [1] 3 Award 3 marks for correct answer with no working. Allow ecf for incorrect area. Allow rounding errors in area calculation. 10(b)(i) Equilibrium is a state where opposing forces are balanced [1] 1 Accept words to that effect. 10(b)(ii) (1.5 x 1200) + (4 x 2000) = R2 x 5.75 [1] R2 = 1800 + 8000 / 5.75 [1] R2 = 1704.348 N [1] R1 = 3200 – 1704.348 = 1495.642 [1] 4 Allow rounding errors in answers. 10(b)(iii) Defects will include: knots, shakes, twisting / warping, bending, termite damage, rot. [2  1] 2 Accept any other valid defects. Question Answer Marks Guidance 10(b)(iv) Any one suitable joining method [1] such as:  Mechanical method, mortise and tenon, tee halving, bridle.  Nail plate / gusset plate  Nails / bolts / screws  Adhesive Clear sketches notes [1] Method of holding joint in place e.g. dowels / bolts / nails / wedges / adhesive [1] 3 10(c) Suitable example [1] Description of two factor of safety measures that apply to example [2  1]  Could be Safe Working Load, extension of crane / hoist  Safety of electrical devices  Warning of overhead cables  Training of employees 3 Accept any other valid safety measures. Allow 2 marks for full description of one measure. Do not allow PPE . 10(d) Reasons for each method will include:  Nut and bolt – temporary fastening can be removed or adjusted if necessary  Countersunk rivet – permanent method, keeps the surfaces clean and flush  Welded joint – permanent method, joint strong and resists compression / tension / torsion, allows flush surface, no additional materials needed. [3  1] 3 Accept any other valid response

This question in 0445/43 May/June 2024

1 Fig. 1.1 shows a wooden door frame ready to be placed in a building. A Fig. 1.1 (a) Use words from the list to complete the description of the wooden door frame. nails glue triangulation flexible rigid Joints at each end of part A are temporary. They are made using … Part A uses … to ensure that the wooden door frame is … . [3] (b) Name the force indicated by the arrow that will act on the wooden door frame when the building is complete. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Joints at each end of part A are temporary, they are made using nails. Part A 3 uses triangulation to ensure that the frame is rigid 1 mark for each word correctly inserted. 1(b) Compression 1 Do not allow bending

This question in 0445/41 Oct/Nov 2024

2 Composite materials can be used in the construction of beams. (a) Give one example of a composite material. … [1] (b) State two benefits of using composite materials. 1 … 2 … [2]

3 marks

Mark scheme: 2(a) 1 mark for a suitable example; concrete, GRP, carbon fibre, CFRP, fibreglass. 1 Allow kevlar Allow any other valid composite. 2(b) Benefits will include: properties of constituent materials can be combined, 2 Allow other valid benefits weakness of individual materials can be avoided, increased strength, reduced weight, increased corrosion resistance, increased durability, complex shapes 1 mark for each benefit stated up to 2 can be achieved [2  1] marks.

This question in 0445/41 Oct/Nov 2024

9 Fig. 9.1 shows a building with scaffolding in position. A Fig. 9.1 (a) (i) Give one example of each type of manufactured structure that is visible in Fig. 9.1. Frame structure … Mass Structure … Shell structure … [3] (ii) Describe how movement in the scaffolding has been prevented. … … … [2] (iii) Give one reason for using the wooden boards marked A. … [1] (iv) Describe two safety features that can be seen in the construction of the scaffolding. 1 … 2 … [2] (b) Fig. 9.2 shows a road bridge with a concrete beam supported by columns. crack beam column column Fig. 9.2 (i) State the force that is the likely cause of the cracking on the concrete beam. … [1] (ii) Add arrows to Fig. 9.2 to show the force acting on the beam. [1] (iii) Explain how small cracks in structural concrete can cause more serious damage over several years. … … … … [3] (iv) Fig. 9.3 shows two beams used to span a gap with a central support. Each beam weighs 200 kg. The load from the beams is equally divided between the supports. m 2 m 2 m 0.2 0.3 m Fig. 9.3 Calculate the stress in the central support. 1 kg = 9.81 N force Use the formula stress = cross sectional area Include the correct units in your answer. … … … … [4] (c) Use sketches and notes to show the following structures: (i) A shelf supported at each end by struts. [2] (ii) A shelf supported at each end by ties. [2] (iii) Fig. 9.4 shows a softwood shelf with a bend along its length caused by excessive load in the centre. 900 20 × 50 8 × 25 20 × 20 A B C extra materials available Fig. 9.4 Use sketches and notes to show a method of supporting the shelf to remove the bend. Include details of how any extra materials used are fixed in place. [4]

25 marks

Mark scheme: 9(a)(i) Frame structure – ladder or scaffolding [1] 3 Allow other valid alternatives. E.g. Mass structure – building / house [1] windows, door frames. Shell structure – car body or scaffold poles [1] Do not allow ‘car’ without ‘body’. 9(a)(ii) Description to include: triangulation formed by diagonal poles at front and 2 1 mark for each valid point up to maximum sides, securely held by extending over ridge of roof. short length of poles, poles 2. held firmly by strong rigid joints, some loading transferred to the roof, square Allow 2 marks for a single point fully plates at base of verticals justified. 9(a)(iii) Boards marked A will spread the vertical load across a wider area, protecting 1 Award mark for understanding shown. the roof tiles. A flat / level surface is available for scaffold to rest on 9(a)(iv) Safety features include: boards on edge to stop items falling, boards provide a 2 Allow 2 marks for a single point fully solid base, gates at the two horizontal platform entrances, ladders secured in described. position, horizontal poles on walkways 9(b)(i) Shear 1 9(b)(ii) 1 Arrows almost in vertical alignment [1] 9(b)(iii) Small cracks will allow water to get in, leading to expansion of cracks [1] 3 Award marks for understanding shown When the water freezes in winter it will cause the exposed reinforcement to 1 mark for each valid point up to maximum expand, causing further cracking [1] 3. Pieces of the outer skin of concrete will drop off. [1] Allow 2 marks for a single point fully Structure can eventually collapse [1] justified. 9(b)(iv) Load from each beam = 100 kg acting on central support 4 Award 3 marks for a correct numerical Total load on centre support = 200 kg  9.81 = 1962 N [1] answer with no working. Area of centre support = 0.3  0.2 = 0.06 m2 [1] Units must be there for 4th mark. Stress = 1962 / 0.06 = 32 700 [1] Nm2 [1] 9(c)(i) 2 Strut correctly drawn at each end [1] Suitable proportions [1] 9(c)(ii) 2 Tie correctly drawn at each end [1] Suitable proportions [1] 9(c)(iii) Section B chosen for supporting material [1] 4 Section B piece fixed on edge [1] Suitable position for support [1] Method of fixing indicated [1]

This question in 0445/41 Oct/Nov 2024

1 Fig. 1.1 shows a wooden door frame ready to be placed in a building. A Fig. 1.1 (a) Use words from the list to complete the description of the wooden door frame. nails glue triangulation flexible rigid Joints at each end of part A are temporary. They are made using … Part A uses … to ensure that the wooden door frame is … . [3] (b) Name the force indicated by the arrow that will act on the wooden door frame when the building is complete. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Joints at each end of part A are temporary, they are made using nails. Part A 3 uses triangulation to ensure that the frame is rigid 1 mark for each word correctly inserted. 1(b) Compression 1 Do not allow bending

This question in 0445/42 Oct/Nov 2024

2 Composite materials can be used in the construction of beams. (a) Give one example of a composite material. … [1] (b) State two benefits of using composite materials. 1 … 2 … [2]

3 marks

Mark scheme: 2(a) 1 mark for a suitable example; concrete, GRP, carbon fibre, CFRP, fibreglass. 1 Allow kevlar Allow any other valid composite. 2(b) Benefits will include: properties of constituent materials can be combined, 2 Allow other valid benefits weakness of individual materials can be avoided, increased strength, reduced weight, increased corrosion resistance, increased durability, complex shapes 1 mark for each benefit stated up to 2 can be achieved [2  1] marks.

This question in 0445/42 Oct/Nov 2024

9 Fig. 9.1 shows a building with scaffolding in position. A Fig. 9.1 (a) (i) Give one example of each type of manufactured structure that is visible in Fig. 9.1. Frame structure … Mass Structure … Shell structure … [3] (ii) Describe how movement in the scaffolding has been prevented. … … … [2] (iii) Give one reason for using the wooden boards marked A. … [1] (iv) Describe two safety features that can be seen in the construction of the scaffolding. 1 … 2 … [2] (b) Fig. 9.2 shows a road bridge with a concrete beam supported by columns. crack beam column column Fig. 9.2 (i) State the force that is the likely cause of the cracking on the concrete beam. … [1] (ii) Add arrows to Fig. 9.2 to show the force acting on the beam. [1] (iii) Explain how small cracks in structural concrete can cause more serious damage over several years. … … … … [3] (iv) Fig. 9.3 shows two beams used to span a gap with a central support. Each beam weighs 200 kg. The load from the beams is equally divided between the supports. m 2 m 2 m 0.2 0.3 m Fig. 9.3 Calculate the stress in the central support. 1 kg = 9.81 N force Use the formula stress = cross sectional area Include the correct units in your answer. … … … … [4] (c) Use sketches and notes to show the following structures: (i) A shelf supported at each end by struts. [2] (ii) A shelf supported at each end by ties. [2] (iii) Fig. 9.4 shows a softwood shelf with a bend along its length caused by excessive load in the centre. 900 20 × 50 8 × 25 20 × 20 A B C extra materials available Fig. 9.4 Use sketches and notes to show a method of supporting the shelf to remove the bend. Include details of how any extra materials used are fixed in place. [4]

25 marks

Mark scheme: 9(a)(i) Frame structure – ladder or scaffolding [1] 3 Allow other valid alternatives. E.g. Mass structure – building / house [1] windows, door frames. Shell structure – car body or scaffold poles [1] Do not allow ‘car’ without ‘body’. 9(a)(ii) Description to include: triangulation formed by diagonal poles at front and 2 1 mark for each valid point up to maximum sides, securely held by extending over ridge of roof. short length of poles, poles 2. held firmly by strong rigid joints, some loading transferred to the roof, square Allow 2 marks for a single point fully plates at base of verticals justified. 9(a)(iii) Boards marked A will spread the vertical load across a wider area, protecting 1 Award mark for understanding shown. the roof tiles. A flat / level surface is available for scaffold to rest on 9(a)(iv) Safety features include: boards on edge to stop items falling, boards provide a 2 Allow 2 marks for a single point fully solid base, gates at the two horizontal platform entrances, ladders secured in described. position, horizontal poles on walkways 9(b)(i) Shear 1 9(b)(ii) 1 Arrows almost in vertical alignment [1] 9(b)(iii) Small cracks will allow water to get in, leading to expansion of cracks [1] 3 Award marks for understanding shown When the water freezes in winter it will cause the exposed reinforcement to 1 mark for each valid point up to maximum expand, causing further cracking [1] 3. Pieces of the outer skin of concrete will drop off. [1] Allow 2 marks for a single point fully Structure can eventually collapse [1] justified. 9(b)(iv) Load from each beam = 100 kg acting on central support 4 Award 3 marks for a correct numerical Total load on centre support = 200 kg  9.81 = 1962 N [1] answer with no working. Area of centre support = 0.3  0.2 = 0.06 m2 [1] Units must be there for 4th mark. Stress = 1962 / 0.06 = 32 700 [1] Nm2 [1] 9(c)(i) 2 Strut correctly drawn at each end [1] Suitable proportions [1] 9(c)(ii) 2 Tie correctly drawn at each end [1] Suitable proportions [1] 9(c)(iii) Section B chosen for supporting material [1] 4 Section B piece fixed on edge [1] Suitable position for support [1] Method of fixing indicated [1]

This question in 0445/42 Oct/Nov 2024

1 Fig. 1.1 shows a wooden door frame ready to be placed in a building. A Fig. 1.1 (a) Use words from the list to complete the description of the wooden door frame. nails glue triangulation flexible rigid Joints at each end of part A are temporary. They are made using … Part A uses … to ensure that the wooden door frame is … . [3] (b) Name the force indicated by the arrow that will act on the wooden door frame when the building is complete. … [1]

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) Joints at each end of part A are temporary, they are made using nails. Part A 3 uses triangulation to ensure that the frame is rigid 1 mark for each word correctly inserted. 1(b) Compression 1 Do not allow bending

This question in 0445/43 Oct/Nov 2024

2 Composite materials can be used in the construction of beams. (a) Give one example of a composite material. … [1] (b) State two benefits of using composite materials. 1 … 2 … [2]

3 marks

Mark scheme: 2(a) 1 mark for a suitable example; concrete, GRP, carbon fibre, CFRP, fibreglass. 1 Allow kevlar Allow any other valid composite. 2(b) Benefits will include: properties of constituent materials can be combined, 2 Allow other valid benefits weakness of individual materials can be avoided, increased strength, reduced weight, increased corrosion resistance, increased durability, complex shapes 1 mark for each benefit stated up to 2 can be achieved [2  1] marks.

This question in 0445/43 Oct/Nov 2024

9 Fig. 9.1 shows a building with scaffolding in position. A Fig. 9.1 (a) (i) Give one example of each type of manufactured structure that is visible in Fig. 9.1. Frame structure … Mass Structure … Shell structure … [3] (ii) Describe how movement in the scaffolding has been prevented. … … … [2] (iii) Give one reason for using the wooden boards marked A. … [1] (iv) Describe two safety features that can be seen in the construction of the scaffolding. 1 … 2 … [2] (b) Fig. 9.2 shows a road bridge with a concrete beam supported by columns. crack beam column column Fig. 9.2 (i) State the force that is the likely cause of the cracking on the concrete beam. … [1] (ii) Add arrows to Fig. 9.2 to show the force acting on the beam. [1] (iii) Explain how small cracks in structural concrete can cause more serious damage over several years. … … … … [3] (iv) Fig. 9.3 shows two beams used to span a gap with a central support. Each beam weighs 200 kg. The load from the beams is equally divided between the supports. m 2 m 2 m 0.2 0.3 m Fig. 9.3 Calculate the stress in the central support. 1 kg = 9.81 N force Use the formula stress = cross sectional area Include the correct units in your answer. … … … … [4] (c) Use sketches and notes to show the following structures: (i) A shelf supported at each end by struts. [2] (ii) A shelf supported at each end by ties. [2] (iii) Fig. 9.4 shows a softwood shelf with a bend along its length caused by excessive load in the centre. 900 20 × 50 8 × 25 20 × 20 A B C extra materials available Fig. 9.4 Use sketches and notes to show a method of supporting the shelf to remove the bend. Include details of how any extra materials used are fixed in place. [4]

25 marks

Mark scheme: 9(a)(i) Frame structure – ladder or scaffolding [1] 3 Allow other valid alternatives. E.g. Mass structure – building / house [1] windows, door frames. Shell structure – car body or scaffold poles [1] Do not allow ‘car’ without ‘body’. 9(a)(ii) Description to include: triangulation formed by diagonal poles at front and 2 1 mark for each valid point up to maximum sides, securely held by extending over ridge of roof. short length of poles, poles 2. held firmly by strong rigid joints, some loading transferred to the roof, square Allow 2 marks for a single point fully plates at base of verticals justified. 9(a)(iii) Boards marked A will spread the vertical load across a wider area, protecting 1 Award mark for understanding shown. the roof tiles. A flat / level surface is available for scaffold to rest on 9(a)(iv) Safety features include: boards on edge to stop items falling, boards provide a 2 Allow 2 marks for a single point fully solid base, gates at the two horizontal platform entrances, ladders secured in described. position, horizontal poles on walkways 9(b)(i) Shear 1 9(b)(ii) 1 Arrows almost in vertical alignment [1] 9(b)(iii) Small cracks will allow water to get in, leading to expansion of cracks [1] 3 Award marks for understanding shown When the water freezes in winter it will cause the exposed reinforcement to 1 mark for each valid point up to maximum expand, causing further cracking [1] 3. Pieces of the outer skin of concrete will drop off. [1] Allow 2 marks for a single point fully Structure can eventually collapse [1] justified. 9(b)(iv) Load from each beam = 100 kg acting on central support 4 Award 3 marks for a correct numerical Total load on centre support = 200 kg  9.81 = 1962 N [1] answer with no working. Area of centre support = 0.3  0.2 = 0.06 m2 [1] Units must be there for 4th mark. Stress = 1962 / 0.06 = 32 700 [1] Nm2 [1] 9(c)(i) 2 Strut correctly drawn at each end [1] Suitable proportions [1] 9(c)(ii) 2 Tie correctly drawn at each end [1] Suitable proportions [1] 9(c)(iii) Section B chosen for supporting material [1] 4 Section B piece fixed on edge [1] Suitable position for support [1] Method of fixing indicated [1]

This question in 0445/43 Oct/Nov 2024

2 Name the type of structure shown in Fig. 2.1. Fig. 2.1 … [1]

1 marks

Mark scheme: 2 Frame structure, 1 mark. 1

This question in 0445/41 May/June 2025

3 Give one reason for using triangulation in a structure. … [1]

1 marks

Mark scheme: 3 Triangulation will hold parts of a frame in the strongest 1 Allow mark for understanding shown. position, giving stability / rigidity. Do not allow ‘stronger’ without qualification.

This question in 0445/41 May/June 2025

4 Describe, using examples, the following forces that can act on a structure. tension … … … torsion … … … static load … … … [6]

6 marks

Mark scheme: 4 Tension is a pulling force, [1] examples could be wires / 6 Accept any other valid response for examples. cables in a suspension bridge. [1] Torsion is a twisting force, [1] example could be torsion spring on a car. [1] Static load is a load applied to a structure, [1] that does not move or change. [1]

This question in 0445/41 May/June 2025

10 Fig. 10.1 shows a concrete canopy over the front of a warehouse building. The canopy is supported either side by steel structures that resist forces caused by the weight of the concrete canopy. A detailed view of one of the steel structures is shown. link threaded sleeve steel rods concrete canopy steel X poles concrete pile 1.5 m deep Fig. 10.1 (a) (i) Name the force acting on each of the steel rods. … [1] (ii) Explain how the threaded sleeve can be used to provide adjustment on the steel rods. … … … … [3] (iii) Part X in Fig. 10.1 is one of four features at the point where the steel poles are joined. Name part X and explain the part that it plays in supporting the concrete canopy. … … … … [3] (iv) State the purpose of the links at each end of the steel rods. … … [1] (v) A concrete pile 1.5 m deep is used as the foundation for the steel structure. Give two reasons why concrete is used as the foundation. 1 … 2 … [2] (b) Fig. 10.2 shows the same wheelbarrow loaded with bricks in two different ways. X A B Fig. 10.2 (i) State the order of lever that is used in a wheelbarrow. … [1] (ii) Name the type of structural member at X. … [1] (iii) Explain the change in efficiency caused by the different methods of loading the wheelbarrow. … … … … [3] (iv) Describe how equilibrium is maintained when a wheelbarrow is in use. … … … [2] (v) Fig. 10.3 shows wheelbarrow A loaded with 38 bricks, each brick weighing 2.1 kg. Calculate the effort required to lift the loaded wheelbarrow. 1 kg = 9.81 N 38 bricks 980 240 Fig. 10.3 … … … … [3] (c) Fig. 10.4 shows a roof truss made of softwood. enlarged view of nail plate Fig. 10.4 (i) The softwood is joined using nail plates. Give two advantages of using nail plates rather than traditional joints. 1 … … 2 … … [2] (ii) Describe the defects that a manufacturer would look for when selecting lengths of softwood for use in the roof truss. … … … … [3]

25 marks

Mark scheme: 10(a)(i) Each of the steel rods is in tension. 1 10(a)(ii) The sleeve can be used to reduce or increase the effective 3 Award marks for understanding shown. length of the rods, [1] using a right-hand thread in one and a Allow 2 marks for a single point fully explained. left-hand thread in the other [1]. Turning the sleeve will act on both rods at the same time. [1] 10(a)(iii) Part X is a gusset plate, [1]. The plates reduce / prevent 3 Allow reference to the gusset plate supporting the joint or movement [1], which could distort the two poles. [1] distributing the load. 10(a)(iv) The links will allow some movement in the rods. 1 10(a)(v) • Concrete is strong in compression 2 No mark for ‘strong material’ with no indication that it is • It can be poured on site ‘strong in compression’. • Concrete provides a solid foundation with no movement • The concrete pile will be below the frost line so has little chance of moving or being damaged by frost • Will resist water damage. • Durable / no maintenance required [2  1] 10(b)(i) The wheelbarrow is a second order / class 2 lever. 1 10(b)(ii) The structural member at X is a strut. 1 Allow ‘brace’. 10(b)(iii) The explanation should include: 3 1 mark for each valid point made. Award 2 marks for a • In wheelbarrow A the load is closer to the fulcrum single point fully explained. • The lever will give increased mechanical advantage • This will increase the efficiency • Wheelbarrow B is less efficient. 10(b)(iv) • When the wheelbarrow is stationary, resting on the 2 Accept any other valid response ground it will be in equilibrium, it is effectively a tripod Any two valid points mentioned in the description. • When the handles of the wheelbarrow are lifted the Allow 2 marks for a single point fully described. device will be unstable • The user’s hands pulling / pushing on either handle maintain equilibrium in a moving wheelbarrow. 10(b)(v) The load is 38  2.1  9.81 = 782.8 N [1] 3 Award full marks for correct answer with no working. 240  782.8 = 1220  Effort [1] Effort = 240  782.8 / 1220 = 153.9 N [1] 10(c)(i) Advantages of nail plates will include: 2 Allow any other valid alternatives Allows offsite construction • Low cost • Secure method • Can be applied using a press • Less skill required • Increased speed of production. [2  1] 10(c)(ii) Defects that can be avoided during selection include: 3 Award marks for understanding shown. • Bent / warped pieces 1 mark for each valid point in description. • Large knots Allow 2 marks for a single point fully described • Splits / shakes Allow mark for checking moisture level. • Signs of insect damage • Signs of rot.

This question in 0445/41 May/June 2025

1 Add two ticks (✓) to each item, A, B and C, in Fig. 1.1 to identify the type of structure and if it is man‑made or natural. A B C frame shell mass man‑made natural Fig. 1.1 [6]

6 marks

Mark scheme: Question Answer Marks Guidance Section A 1 6 No marks for each item with more than two ticks added unless extra ticks are crossed out.6 1 mark for each correct tick

This question in 0445/42 May/June 2025

2 Suspension bridges are often supported by steel cables. Give two properties of steel that make it suitable for use in cables. 1 … 2 … [2]

2 marks

Mark scheme: 2 Properties of steel will include: 2 Allow other valid alternative properties. • ductile • durable Do not allow: • strength in tension • ‘strong’ unless qualified • crush resistant • Rigid • toughness • Flexible • relative low cost. • Corrosion resistant [2  1] • Malleable.

This question in 0445/42 May/June 2025

3 Fig. 3.1 shows a wooden structure. State the method that has been used in the wooden structure to ensure rigidity. Fig. 3.1 … [1]

1 marks

Mark scheme: 3 Triangulation. 1

This question in 0445/42 May/June 2025

9 (a) (i) State what is meant by a ‘moment’ in a structure. … … [1] (ii) State the unit of measurement applied in moments calculations. … [1] (iii) Fig. 9.1 shows a fire door in a building that uses a door closer to ensure that the fire door remains closed when not in use. door closer door frame hinge fire door opening force X 680 580 150 closing force 20 N Fig. 9.1 Calculate the minimum value of force X that will be necessary to open the door. … … … [3] (b) Fig. 9.2 shows three methods of joining timber to extend the length. Give one benefit and one drawback for each method. woodscrews hardwood wedges bolts A B C Fig. 9.2 Method A benefit … … Method A drawback … … Method B benefit … … Method B drawback … … Method C benefit … … Method C drawback … … [6] (c) (i) Explain what is meant by the term ‘laminating’ in a structure. … … … [2] (ii) Name two materials that can be used in a laminated structure. 1 … 2 … [2] (d) (i) Fig. 9.3 shows roof supports in a building. Fig. 9.3 Name the method that has been used to join the parts of the roof supports. … [1] (ii) Explain why welding is used in modern building construction rather than the method shown in Fig. 9.3. … … … [2] (e) (i) State what is meant by the term ‘equilibrium’ in a structure. … … [1] (ii) Fig. 9.4 shows a flagpole that needs to remain vertical. Use sketches and notes to show one method of supporting the flagpole in a vertical position. flagpole height 5 m ground level Fig. 9.4 [3] (f) Fig. 9.5 shows a steel cable on a crane arm with a load attached to it. A 15 m length of cable is found to stretch by 2 mm when the load is raised. crane arm steel cable 20 mm diameter 500 N Fig. 9.5 Calculate the strain in the steel cable. change in length Use the formula: strain = original length … … … … [3]

25 marks

Mark scheme: 9(a)(i) A moment is defined as: 1 Allow marks for understanding shown. • The turning effect of a force. • Distance  Magnitude of force 9(a)(ii) Nm or Newton-metre 1 9(a)(iii) 0.150  20 = 0.680  X, [1] 3 Award full marks if correct answer is given with no working. X = 3 / 0.680 = 4.41 N [1] Any force greater than 4.41 N will cause door to open. [1] 9(b) A 6 Allow other valid benefits / drawbacks Benefit no increased width to joint, no fixings to corrode, strong in tension, mechanical integrity. Drawback time taken to produce, over time wedges could work loose. B Benefit will align horizontally with 5 bolts preventing movement, simple joint can be produced on site, easily taken apart if necessary. Drawback increased width to joint / parts not in line, bolts become loose if wood shrinks. C Benefit pieces are in line, vertical movement prevented Drawback woodscrews could corrode or break, will resist compression but not tension. [6  1] 9(c)(i) Laminating is: 2 1 mark for each valid point in the explanation. Allow 2 marks • The use of multiple layers of materials which could for a single point fully explained either be the same or different • Normally fixed with adhesive Do not accept reference to decorative laminates. • Strength and stability are increased • Used to produce a rigid structure • Used to increase sound insulation. 9(c)(ii) Timber layers, plastic laminate sheeting, carbon fibre, 2 Allow any materials that are commonly combined in a named metals, GRP, paper or card. structure. [2  1] Do not allow the adhesives used. 9(d)(i) Riveting has been used to join the roof supports. 1 9(d)(ii) Welding is: 2 Accept any other valid response • Faster 1 mark for each valid point in the explanation. Allow 2 marks • Lower cost for a single point fully explained. • Suitable for pre-fabrication Do not allow ‘stronger’ • Precise method of joining 9(e)(i) Equilibrium is a state when all forces or moments are 1 Allow mark for understanding shown. balanced or equal. 9(e)(ii) 3 or 4 cables / fixings spaced around the flagpole [1] 3 Accept any other valid method Fixings at ground level [1] Flagpole must be supported from all directions for full marks Functional method. [1] 9(f) Change in length = 2 mm 3 Award full marks if correct answer is given with no working. Original length = 15 000 mm 1 mark for substitutions Strain = 2 / 15 000 [1] = 0.000133 [1] or 1.33  10-4

This question in 0445/42 May/June 2025

2 Name the type of structure shown in Fig. 2.1. Fig. 2.1 … [1]

1 marks

Mark scheme: 2 Frame structure, 1 mark. 1

This question in 0445/43 May/June 2025

3 Give one reason for using triangulation in a structure. … [1]

1 marks

Mark scheme: 3 Triangulation will hold parts of a frame in the strongest 1 Allow mark for understanding shown. position, giving stability / rigidity. Do not allow ‘stronger’ without qualification.

This question in 0445/43 May/June 2025

4 Describe, using examples, the following forces that can act on a structure. tension … … … torsion … … … static load … … … [6]

6 marks

Mark scheme: 4 Tension is a pulling force, [1] examples could be wires / 6 Accept any other valid response for examples. cables in a suspension bridge. [1] Torsion is a twisting force, [1] example could be torsion spring on a car. [1] Static load is a load applied to a structure, [1] that does not move or change. [1]

This question in 0445/43 May/June 2025

10 Fig. 10.1 shows a concrete canopy over the front of a warehouse building. The canopy is supported either side by steel structures that resist forces caused by the weight of the concrete canopy. A detailed view of one of the steel structures is shown. link threaded sleeve steel rods concrete canopy steel X poles concrete pile 1.5 m deep Fig. 10.1 (a) (i) Name the force acting on each of the steel rods. … [1] (ii) Explain how the threaded sleeve can be used to provide adjustment on the steel rods. … … … … [3] (iii) Part X in Fig. 10.1 is one of four features at the point where the steel poles are joined. Name part X and explain the part that it plays in supporting the concrete canopy. … … … … [3] (iv) State the purpose of the links at each end of the steel rods. … … [1] (v) A concrete pile 1.5 m deep is used as the foundation for the steel structure. Give two reasons why concrete is used as the foundation. 1 … 2 … [2] (b) Fig. 10.2 shows the same wheelbarrow loaded with bricks in two different ways. X A B Fig. 10.2 (i) State the order of lever that is used in a wheelbarrow. … [1] (ii) Name the type of structural member at X. … [1] (iii) Explain the change in efficiency caused by the different methods of loading the wheelbarrow. … … … … [3] (iv) Describe how equilibrium is maintained when a wheelbarrow is in use. … … … [2] (v) Fig. 10.3 shows wheelbarrow A loaded with 38 bricks, each brick weighing 2.1 kg. Calculate the effort required to lift the loaded wheelbarrow. 1 kg = 9.81 N 38 bricks 980 240 Fig. 10.3 … … … … [3] (c) Fig. 10.4 shows a roof truss made of softwood. enlarged view of nail plate Fig. 10.4 (i) The softwood is joined using nail plates. Give two advantages of using nail plates rather than traditional joints. 1 … … 2 … … [2] (ii) Describe the defects that a manufacturer would look for when selecting lengths of softwood for use in the roof truss. … … … … [3]

25 marks

Mark scheme: 10(a)(i) Each of the steel rods is in tension. 1 10(a)(ii) The sleeve can be used to reduce or increase the effective 3 Award marks for understanding shown. length of the rods, [1] using a right-hand thread in one and a Allow 2 marks for a single point fully explained. left-hand thread in the other [1]. Turning the sleeve will act on both rods at the same time. [1] 10(a)(iii) Part X is a gusset plate, [1]. The plates reduce / prevent 3 Allow reference to the gusset plate supporting the joint or movement [1], which could distort the two poles. [1] distributing the load. 10(a)(iv) The links will allow some movement in the rods. 1 10(a)(v) • Concrete is strong in compression 2 No mark for ‘strong material’ with no indication that it is • It can be poured on site ‘strong in compression’. • Concrete provides a solid foundation with no movement • The concrete pile will be below the frost line so has little chance of moving or being damaged by frost • Will resist water damage. • Durable / no maintenance required [2  1] 10(b)(i) The wheelbarrow is a second order / class 2 lever. 1 10(b)(ii) The structural member at X is a strut. 1 Allow ‘brace’. 10(b)(iii) The explanation should include: 3 1 mark for each valid point made. Award 2 marks for a • In wheelbarrow A the load is closer to the fulcrum single point fully explained. • The lever will give increased mechanical advantage • This will increase the efficiency • Wheelbarrow B is less efficient. 10(b)(iv) • When the wheelbarrow is stationary, resting on the 2 Accept any other valid response ground it will be in equilibrium, it is effectively a tripod Any two valid points mentioned in the description. • When the handles of the wheelbarrow are lifted the Allow 2 marks for a single point fully described. device will be unstable • The user’s hands pulling / pushing on either handle maintain equilibrium in a moving wheelbarrow. 10(b)(v) The load is 38  2.1  9.81 = 782.8 N [1] 3 Award full marks for correct answer with no working. 240  782.8 = 1220  Effort [1] Effort = 240  782.8 / 1220 = 153.9 N [1] 10(c)(i) Advantages of nail plates will include: 2 Allow any other valid alternatives Allows offsite construction • Low cost • Secure method • Can be applied using a press • Less skill required • Increased speed of production. [2  1] 10(c)(ii) Defects that can be avoided during selection include: 3 Award marks for understanding shown. • Bent / warped pieces 1 mark for each valid point in description. • Large knots Allow 2 marks for a single point fully described • Splits / shakes Allow mark for checking moisture level. • Signs of insect damage • Signs of rot.

This question in 0445/43 May/June 2025

1 Fig. 1.1 shows two structures made from composite materials. edging block boat hull Fig. 1.1 (a) (i) Name a suitable composite material for the edging block. … [1] (ii) Name one other composite material that can be used in a structure. … [1] (b) Name the type of structure used for the boat hull. … [1]

3 marks

Mark scheme: Question Answer Marks Guidance 1(a)(i) The composite material is concrete. 1 Accept reinforced concrete 1(a)(ii) Laminated wood / plywood / chipboard 1 Allow GRP, carbon fibre, fibreglass, Kevlar and reinforced bamboo. 1(b) The boat hull is a shell structure 1

This question in 0445/41 Oct/Nov 2025

2 Fig. 2.1 shows a rigid gate structure made from welded steel tubes. Z Y X steel tube Fig. 2.1 (a) Explain why the structure in Fig. 2.1 will remain rigid when force X is applied. … … … [2] (b) Name force Y in Fig. 2.1. … [1] (c) Name force Z in Fig. 2.1. … [1]

4 marks

Mark scheme: 2(a) The structure will remain rigid because of triangulation [1] 2 Triangulation must be included for 2 marks. The diagonal will reinforce the structure preventing any distortion [1] 2(b) Force Y is bending. 1 2(c) Force Z is shear. 1

This question in 0445/41 Oct/Nov 2025

10 Fig. 10.1 shows a peg puller for removing large tent pegs from the ground. gripping jaws side supports A baseplate 200 × 120 steel lever arm tent peg Fig. 10.1 (a) (i) State the order of lever used in the peg puller. … [1] (ii) Give the name of feature A on the baseplate in Fig. 10.1. … [1] (iii) State the purpose of feature A. … [1] (iv) Explain why the base plate is needed at the point of contact with the ground. … … … [2] (v) Name the force that will act on the ground beneath the base plate. … [1] (vi) Fig. 10.2 shows a dimensioned view of the peg puller. steel lever arm X Y steel side plates 1500 steel lever arm 130 Fig. 10.2 Name the forces that will act on the upper surface X and the lower surface Y of the steel lever arm when it is operated. X … Y … [2] (vii) State one method that could be used to permanently join the steel side plates to the steel lever arm. … [1] (viii) Calculate the approximate force that is available to pull the tent peg when an effort of 200 N is applied to the end of the lever arm. Include units in your answer. … … … … [3] (b) (i) Using sketches and notes, complete Fig. 10.3 to show a beam supported by walls A and B. A B Fig. 10.3 [2] (ii) Using sketches and notes, complete Fig. 10.4 to show a tie at joint C. C Fig. 10.4 [2] (iii) Using sketches and notes, complete Fig. 10.5 to show a strut at joint D. D Fig. 10.5 [2] (c) Fig. 10.6 shows the layout of timber in the base of a small wooden building. spacing pieces Fig. 10.6 (i) Give two features of the timber used in the layout that will affect performance of the structure. 1 … 2 … [2] (ii) Give two reasons for using the spacing pieces in Fig. 10.6. 1 … 2 … [2] (iii) Use sketches and notes to show one way of securing the spacing pieces in position. List any fixings that will be used. [3]

25 marks

Mark scheme: 10(a)(i) First order lever 1 10(a)(ii) Feature A is a gusset plate 1 10(a)(iii) Feature A is used to support the joints between the 1 Award marks for understanding shown. baseplate and side supports. 10(a)(iv) If the ground is soft and if there were no baseplate the 2 Allow any other valid points. device would sink into the ground [1] Explanation to include any two valid points, 2 marks. The baseplate provides a larger surface area which is less Award 2 marks for a full explanation of a single point. likely to sink when pressure is applied to the lever arm [1] Efficiency of the peg puller is improved [1] The baseplate acts as a fulcrum [1] 10(a)(v) The force acting on the ground is compression. 1 10(a)(vi) The force acting on surface X is tension [1] 2 The force acting on surface Y is compression [1] 10(a)(vii) Welding is used to join the steel side plates to the steel 1 Allow brazing. Do not allow riveting. lever arm. 10(a)(viii) force  130 = 200  1500 [1] 3 Award 3 marks for correct answer with no working. force = 300000 / 130 [1] force = 2307.7 N [1] 10(b)(i) Accurately sketch / notes [1] 2 Proportions suitable [1] 10(b)(ii) Accurately sketch / notes / drawn in correct position [1] 2 If both strut and tie are shown 1 mark only Proportions suitable [1] 10(b)(iii) Accurately sketch / notes / drawn in correct position [1] 2 No mark for strut drawn with a single line. Proportions suitable [1] If both strut and tie are shown 1 mark only 10(c)(i) Features affecting performance will include: 2 Accept valid alternatives. • Knots • Shakes • Splits • Twisting • Finish or treatment needed on the timber [2  1] 10(c)(ii) The spacing pieces will: 2 • Help to prevent twisting / warping of the joists [1] • Provide extra support to floor [1] • Maintain spacing between joists. [1] [2  1] 10(c)(iii) The spacing pieces are staggered across the width to allow 3 Example method screws / nails to be fitted. • Functional method [1] • Clear sketch / notes showing method [1] • List of fixings screws / nails / joist hangers [1] • Mechanical joint used / mortise and tenon / bridle joint.[1]

This question in 0445/41 Oct/Nov 2025

11 Fig. 11.1 shows an air compressor driven by an electric motor. electric motor air compressor vee belt Fig. 11.1 (a) The electric motor uses a vee belt to connect to the air compressor. (i) Name two other types of belt that could be used. 1 … 2 … [2] (ii) Explain why the vee belt is the most suitable method of connecting the electric motor to the air compressor. … … … [2] (b) Fig. 11.2 shows the main moving parts of the air compressor. expanding rings used to form a tight seal B 50 diameter 12.5 12.5 B Fig. 11.2 (i) Use words from the list below to complete the description of the air compressor operating. Not all of the words will be used. bearing connecting crankshaft cylinder gear piston pulley The vee … transfers motion to the … causing it to rotate. This in turn causes the … rod to move the … . [4] (ii) Name the type of mechanism used to convert motion in the air compressor. … [2] (iii) Calculate the volume of air that is displaced for each rotation of part B. The formula for the area of a circle is A = π r2. … … … … mm3 [3] (c) (i) Fig. 11.3 shows the small oil holes used to lubricate the plain bearings. oil holes plain bearings Fig. 11.3 Explain why grease would not be suitable for lubricating the plain bearings. … … … [2] (ii) Name two alternative types of bearing that could have been used. 1 … 2 … [2] (iii) Give one practical benefit of using plain bearings in the compressor. … … [1] (d) Fig. 11.4 shows two hexagonal head bolts of the same length, diameter and material. 1.5 0.75 A B Fig. 11.4 (i) Give one difference that will be noticed when fitting the bolts into a threaded hole. … [1] (ii) Explain why bolt B, which has a smaller thread pitch, will resist higher torque when being tightened. … … … [2] (e) Fig. 11.5 shows a gear system. motor speed = 850 rpm motor 30t 40t 15t Fig. 11.5 (i) State the velocity ratio of the worm gear and 30t gear. … [1] (ii) Calculate the speed of the 40t gear when the motor is turning. … … … … [3]

25 marks

Mark scheme: 11(a)(i) Other types of belt could be flat, toothed or round. 2 Accept ‘plain belt’. Sketches must be identified with a name for mark to be [2  1] awarded. 11(a)(ii) Reasons for the vee belt being most suitable are: 2 Allow any other valid points. • Transmit power very well Explanation to include any two valid points, 2 marks. • Limited slip between belt and pulleys / high friction force Award 2 marks for a full explanation of a single point. • Low noise and vibration levels • Easy to install and maintain. 11(b)(i) The vee pulley transfers motion to the crankshaft causing it 4 1 mark for each word correctly used to rotate. This in turn causes the connecting rod to move the piston . [4  1] 11(b)(ii) Crank [1] / slider [1] mechanism. 2 11(b)(iii) Area of piston = Π r2 = 3.142  252 = 1963.75 (1963.495 on 3 Award 3 marks for correct answer with no working. calculator) [1] Throw of piston = 12.5  2 = 25 [1] Volume displaced = 1963.75  25 = 49093.75 mm3 or 49 087.39 [1] 11(c)(i)) Grease would not be suitable because: 2 Allow any other valid points. • It is too thick / viscous to flow through the oil holes. Explanation to include any two valid points, 2 marks. • The bearing also uses oil to carry away any particles Award 2 marks for a full explanation of a single point. from the bearing. • Grease would keep the particles in the bearing, allowing metal to metal contact. 11(c)(ii) Roller or ball bearings could have been used. 2 Accept needle bearings [2  1] 11(c)(iii) Benefits of plain bearings include: 1 Allow any other valid points. • Smaller diameter • Low maintenance • Plain bearings support higher loads. 11(d)(i) Differences in use include: 1 AOVR • More difficult to start bolt B in threaded hole / nut • Easier to start bolt A in threaded hole • Easy to cross thread bolt B when inserting • More turns needed to move bolt B through a set linear distance • Faster to insert bolt A. 11(d)(ii) • The core diameter of bolt B is larger than for bolt A due 2 Allow any other valid points. to smaller pitch and lower depth of thread Explanation to include any two valid points, 2 marks. • Larger diameter will have greater resistance to shear Award 2 marks for a full explanation of a single point. forces • Higher torque can be applied before the bolt will fail / shear. • More precise in transmitting motion 11(e)(i) The velocity ratio of the worm gear is 30:1 1 11(e)(ii) Worm gear ratio 30:1 15 t:40 t ratio 2.666:1 [1] 3 Allow ecf from (d)(ii). Total reduction ratio = 30  2.666 = 80:1 [1] Speed of 40 t gear = 850 / 80 = 10.625 rpm [1] Award 3 marks for correct answer with no working.

This question in 0445/41 Oct/Nov 2025

1 Fig. 1.1 shows two structures made from composite materials. edging block boat hull Fig. 1.1 (a) (i) Name a suitable composite material for the edging block. … [1] (ii) Name one other composite material that can be used in a structure. … [1] (b) Name the type of structure used for the boat hull. … [1]

3 marks

Mark scheme: Question Answer Marks Guidance 1(a)(i) The composite material is concrete. 1 Accept reinforced concrete 1(a)(ii) Laminated wood / plywood / chipboard 1 Allow GRP, carbon fibre, fibreglass, Kevlar and reinforced bamboo. 1(b) The boat hull is a shell structure 1

This question in 0445/42 Oct/Nov 2025

2 Fig. 2.1 shows a rigid gate structure made from welded steel tubes. Z Y X steel tube Fig. 2.1 (a) Explain why the structure in Fig. 2.1 will remain rigid when force X is applied. … … … [2] (b) Name force Y in Fig. 2.1. … [1] (c) Name force Z in Fig. 2.1. … [1]

4 marks

Mark scheme: 2(a) The structure will remain rigid because of triangulation [1] 2 Triangulation must be included for 2 marks. The diagonal will reinforce the structure preventing any distortion [1] 2(b) Force Y is bending. 1 2(c) Force Z is shear. 1

This question in 0445/42 Oct/Nov 2025

10 Fig. 10.1 shows a peg puller for removing large tent pegs from the ground. gripping jaws side supports A baseplate 200 × 120 steel lever arm tent peg Fig. 10.1 (a) (i) State the order of lever used in the peg puller. … [1] (ii) Give the name of feature A on the baseplate in Fig. 10.1. … [1] (iii) State the purpose of feature A. … [1] (iv) Explain why the base plate is needed at the point of contact with the ground. … … … [2] (v) Name the force that will act on the ground beneath the base plate. … [1] (vi) Fig. 10.2 shows a dimensioned view of the peg puller. steel lever arm X Y steel side plates 1500 steel lever arm 130 Fig. 10.2 Name the forces that will act on the upper surface X and the lower surface Y of the steel lever arm when it is operated. X … Y … [2] (vii) State one method that could be used to permanently join the steel side plates to the steel lever arm. … [1] (viii) Calculate the approximate force that is available to pull the tent peg when an effort of 200 N is applied to the end of the lever arm. Include units in your answer. … … … … [3] (b) (i) Using sketches and notes, complete Fig. 10.3 to show a beam supported by walls A and B. A B Fig. 10.3 [2] (ii) Using sketches and notes, complete Fig. 10.4 to show a tie at joint C. C Fig. 10.4 [2] (iii) Using sketches and notes, complete Fig. 10.5 to show a strut at joint D. D Fig. 10.5 [2] (c) Fig. 10.6 shows the layout of timber in the base of a small wooden building. spacing pieces Fig. 10.6 (i) Give two features of the timber used in the layout that will affect performance of the structure. 1 … 2 … [2] (ii) Give two reasons for using the spacing pieces in Fig. 10.6. 1 … 2 … [2] (iii) Use sketches and notes to show one way of securing the spacing pieces in position. List any fixings that will be used. [3]

25 marks

Mark scheme: 10(a)(i) First order lever 1 10(a)(ii) Feature A is a gusset plate 1 10(a)(iii) Feature A is used to support the joints between the 1 Award marks for understanding shown. baseplate and side supports. 10(a)(iv) If the ground is soft and if there were no baseplate the 2 Allow any other valid points. device would sink into the ground [1] Explanation to include any two valid points, 2 marks. The baseplate provides a larger surface area which is less Award 2 marks for a full explanation of a single point. likely to sink when pressure is applied to the lever arm [1] Efficiency of the peg puller is improved [1] The baseplate acts as a fulcrum [1] 10(a)(v) The force acting on the ground is compression. 1 10(a)(vi) The force acting on surface X is tension [1] 2 The force acting on surface Y is compression [1] 10(a)(vii) Welding is used to join the steel side plates to the steel 1 Allow brazing. Do not allow riveting. lever arm. 10(a)(viii) force  130 = 200  1500 [1] 3 Award 3 marks for correct answer with no working. force = 300000 / 130 [1] force = 2307.7 N [1] 10(b)(i) Accurately sketch / notes [1] 2 Proportions suitable [1] 10(b)(ii) Accurately sketch / notes / drawn in correct position [1] 2 If both strut and tie are shown 1 mark only Proportions suitable [1] 10(b)(iii) Accurately sketch / notes / drawn in correct position [1] 2 No mark for strut drawn with a single line. Proportions suitable [1] If both strut and tie are shown 1 mark only 10(c)(i) Features affecting performance will include: 2 Accept valid alternatives. • Knots • Shakes • Splits • Twisting • Finish or treatment needed on the timber [2  1] 10(c)(ii) The spacing pieces will: 2 • Help to prevent twisting / warping of the joists [1] • Provide extra support to floor [1] • Maintain spacing between joists. [1] [2  1] 10(c)(iii) The spacing pieces are staggered across the width to allow 3 Example method screws / nails to be fitted. • Functional method [1] • Clear sketch / notes showing method [1] • List of fixings screws / nails / joist hangers [1] • Mechanical joint used / mortise and tenon / bridle joint.[1]

This question in 0445/42 Oct/Nov 2025

11 Fig. 11.1 shows an air compressor driven by an electric motor. electric motor air compressor vee belt Fig. 11.1 (a) The electric motor uses a vee belt to connect to the air compressor. (i) Name two other types of belt that could be used. 1 … 2 … [2] (ii) Explain why the vee belt is the most suitable method of connecting the electric motor to the air compressor. … … … [2] (b) Fig. 11.2 shows the main moving parts of the air compressor. expanding rings used to form a tight seal B 50 diameter 12.5 12.5 B Fig. 11.2 (i) Use words from the list below to complete the description of the air compressor operating. Not all of the words will be used. bearing connecting crankshaft cylinder gear piston pulley The vee … transfers motion to the … causing it to rotate. This in turn causes the … rod to move the … . [4] (ii) Name the type of mechanism used to convert motion in the air compressor. … [2] (iii) Calculate the volume of air that is displaced for each rotation of part B. The formula for the area of a circle is A = π r2. … … … … mm3 [3] (c) (i) Fig. 11.3 shows the small oil holes used to lubricate the plain bearings. oil holes plain bearings Fig. 11.3 Explain why grease would not be suitable for lubricating the plain bearings. … … … [2] (ii) Name two alternative types of bearing that could have been used. 1 … 2 … [2] (iii) Give one practical benefit of using plain bearings in the compressor. … … [1] (d) Fig. 11.4 shows two hexagonal head bolts of the same length, diameter and material. 1.5 0.75 A B Fig. 11.4 (i) Give one difference that will be noticed when fitting the bolts into a threaded hole. … [1] (ii) Explain why bolt B, which has a smaller thread pitch, will resist higher torque when being tightened. … … … [2] (e) Fig. 11.5 shows a gear system. motor speed = 850 rpm motor 30t 40t 15t Fig. 11.5 (i) State the velocity ratio of the worm gear and 30t gear. … [1] (ii) Calculate the speed of the 40t gear when the motor is turning. … … … … [3]

25 marks

Mark scheme: 11(a)(i) Other types of belt could be flat, toothed or round. 2 Accept ‘plain belt’. Sketches must be identified with a name for mark to be [2  1] awarded. 11(a)(ii) Reasons for the vee belt being most suitable are: 2 Allow any other valid points. • Transmit power very well Explanation to include any two valid points, 2 marks. • Limited slip between belt and pulleys / high friction force Award 2 marks for a full explanation of a single point. • Low noise and vibration levels • Easy to install and maintain. 11(b)(i) The vee pulley transfers motion to the crankshaft causing it 4 1 mark for each word correctly used to rotate. This in turn causes the connecting rod to move the piston . [4  1] 11(b)(ii) Crank [1] / slider [1] mechanism. 2 11(b)(iii) Area of piston = Π r2 = 3.142  252 = 1963.75 (1963.495 on 3 Award 3 marks for correct answer with no working. calculator) [1] Throw of piston = 12.5  2 = 25 [1] Volume displaced = 1963.75  25 = 49093.75 mm3 or 49 087.39 [1] 11(c)(i)) Grease would not be suitable because: 2 Allow any other valid points. • It is too thick / viscous to flow through the oil holes. Explanation to include any two valid points, 2 marks. • The bearing also uses oil to carry away any particles Award 2 marks for a full explanation of a single point. from the bearing. • Grease would keep the particles in the bearing, allowing metal to metal contact. 11(c)(ii) Roller or ball bearings could have been used. 2 Accept needle bearings [2  1] 11(c)(iii) Benefits of plain bearings include: 1 Allow any other valid points. • Smaller diameter • Low maintenance • Plain bearings support higher loads. 11(d)(i) Differences in use include: 1 AOVR • More difficult to start bolt B in threaded hole / nut • Easier to start bolt A in threaded hole • Easy to cross thread bolt B when inserting • More turns needed to move bolt B through a set linear distance • Faster to insert bolt A. 11(d)(ii) • The core diameter of bolt B is larger than for bolt A due 2 Allow any other valid points. to smaller pitch and lower depth of thread Explanation to include any two valid points, 2 marks. • Larger diameter will have greater resistance to shear Award 2 marks for a full explanation of a single point. forces • Higher torque can be applied before the bolt will fail / shear. • More precise in transmitting motion 11(e)(i) The velocity ratio of the worm gear is 30:1 1 11(e)(ii) Worm gear ratio 30:1 15 t:40 t ratio 2.666:1 [1] 3 Allow ecf from (d)(ii). Total reduction ratio = 30  2.666 = 80:1 [1] Speed of 40 t gear = 850 / 80 = 10.625 rpm [1] Award 3 marks for correct answer with no working.

This question in 0445/42 Oct/Nov 2025

1 Fig. 1.1 shows two structures made from composite materials. edging block boat hull Fig. 1.1 (a) (i) Name a suitable composite material for the edging block. … [1] (ii) Name one other composite material that can be used in a structure. … [1] (b) Name the type of structure used for the boat hull. … [1]

3 marks

Mark scheme: Question Answer Marks Guidance 1(a)(i) The composite material is concrete. 1 Accept reinforced concrete 1(a)(ii) Laminated wood / plywood / chipboard 1 Allow GRP, carbon fibre, fibreglass, Kevlar and reinforced bamboo. 1(b) The boat hull is a shell structure 1

This question in 0445/43 Oct/Nov 2025

2 Fig. 2.1 shows a rigid gate structure made from welded steel tubes. Z Y X steel tube Fig. 2.1 (a) Explain why the structure in Fig. 2.1 will remain rigid when force X is applied. … … … [2] (b) Name force Y in Fig. 2.1. … [1] (c) Name force Z in Fig. 2.1. … [1]

4 marks

Mark scheme: 2(a) The structure will remain rigid because of triangulation [1] 2 Triangulation must be included for 2 marks. The diagonal will reinforce the structure preventing any distortion [1] 2(b) Force Y is bending. 1 2(c) Force Z is shear. 1

This question in 0445/43 Oct/Nov 2025

10 Fig. 10.1 shows a peg puller for removing large tent pegs from the ground. gripping jaws side supports A baseplate 200 × 120 steel lever arm tent peg Fig. 10.1 (a) (i) State the order of lever used in the peg puller. … [1] (ii) Give the name of feature A on the baseplate in Fig. 10.1. … [1] (iii) State the purpose of feature A. … [1] (iv) Explain why the base plate is needed at the point of contact with the ground. … … … [2] (v) Name the force that will act on the ground beneath the base plate. … [1] (vi) Fig. 10.2 shows a dimensioned view of the peg puller. steel lever arm X Y steel side plates 1500 steel lever arm 130 Fig. 10.2 Name the forces that will act on the upper surface X and the lower surface Y of the steel lever arm when it is operated. X … Y … [2] (vii) State one method that could be used to permanently join the steel side plates to the steel lever arm. … [1] (viii) Calculate the approximate force that is available to pull the tent peg when an effort of 200 N is applied to the end of the lever arm. Include units in your answer. … … … … [3] (b) (i) Using sketches and notes, complete Fig. 10.3 to show a beam supported by walls A and B. A B Fig. 10.3 [2] (ii) Using sketches and notes, complete Fig. 10.4 to show a tie at joint C. C Fig. 10.4 [2] (iii) Using sketches and notes, complete Fig. 10.5 to show a strut at joint D. D Fig. 10.5 [2] (c) Fig. 10.6 shows the layout of timber in the base of a small wooden building. spacing pieces Fig. 10.6 (i) Give two features of the timber used in the layout that will affect performance of the structure. 1 … 2 … [2] (ii) Give two reasons for using the spacing pieces in Fig. 10.6. 1 … 2 … [2] (iii) Use sketches and notes to show one way of securing the spacing pieces in position. List any fixings that will be used. [3]

25 marks

Mark scheme: 10(a)(i) First order lever 1 10(a)(ii) Feature A is a gusset plate 1 10(a)(iii) Feature A is used to support the joints between the 1 Award marks for understanding shown. baseplate and side supports. 10(a)(iv) If the ground is soft and if there were no baseplate the 2 Allow any other valid points. device would sink into the ground [1] Explanation to include any two valid points, 2 marks. The baseplate provides a larger surface area which is less Award 2 marks for a full explanation of a single point. likely to sink when pressure is applied to the lever arm [1] Efficiency of the peg puller is improved [1] The baseplate acts as a fulcrum [1] 10(a)(v) The force acting on the ground is compression. 1 10(a)(vi) The force acting on surface X is tension [1] 2 The force acting on surface Y is compression [1] 10(a)(vii) Welding is used to join the steel side plates to the steel 1 Allow brazing. Do not allow riveting. lever arm. 10(a)(viii) force  130 = 200  1500 [1] 3 Award 3 marks for correct answer with no working. force = 300000 / 130 [1] force = 2307.7 N [1] 10(b)(i) Accurately sketch / notes [1] 2 Proportions suitable [1] 10(b)(ii) Accurately sketch / notes / drawn in correct position [1] 2 If both strut and tie are shown 1 mark only Proportions suitable [1] 10(b)(iii) Accurately sketch / notes / drawn in correct position [1] 2 No mark for strut drawn with a single line. Proportions suitable [1] If both strut and tie are shown 1 mark only 10(c)(i) Features affecting performance will include: 2 Accept valid alternatives. • Knots • Shakes • Splits • Twisting • Finish or treatment needed on the timber [2  1] 10(c)(ii) The spacing pieces will: 2 • Help to prevent twisting / warping of the joists [1] • Provide extra support to floor [1] • Maintain spacing between joists. [1] [2  1] 10(c)(iii) The spacing pieces are staggered across the width to allow 3 Example method screws / nails to be fitted. • Functional method [1] • Clear sketch / notes showing method [1] • List of fixings screws / nails / joist hangers [1] • Mechanical joint used / mortise and tenon / bridle joint.[1]

This question in 0445/43 Oct/Nov 2025

11 Fig. 11.1 shows an air compressor driven by an electric motor. electric motor air compressor vee belt Fig. 11.1 (a) The electric motor uses a vee belt to connect to the air compressor. (i) Name two other types of belt that could be used. 1 … 2 … [2] (ii) Explain why the vee belt is the most suitable method of connecting the electric motor to the air compressor. … … … [2] (b) Fig. 11.2 shows the main moving parts of the air compressor. expanding rings used to form a tight seal B 50 diameter 12.5 12.5 B Fig. 11.2 (i) Use words from the list below to complete the description of the air compressor operating. Not all of the words will be used. bearing connecting crankshaft cylinder gear piston pulley The vee … transfers motion to the … causing it to rotate. This in turn causes the … rod to move the … . [4] (ii) Name the type of mechanism used to convert motion in the air compressor. … [2] (iii) Calculate the volume of air that is displaced for each rotation of part B. The formula for the area of a circle is A = π r2. … … … … mm3 [3] (c) (i) Fig. 11.3 shows the small oil holes used to lubricate the plain bearings. oil holes plain bearings Fig. 11.3 Explain why grease would not be suitable for lubricating the plain bearings. … … … [2] (ii) Name two alternative types of bearing that could have been used. 1 … 2 … [2] (iii) Give one practical benefit of using plain bearings in the compressor. … … [1] (d) Fig. 11.4 shows two hexagonal head bolts of the same length, diameter and material. 1.5 0.75 A B Fig. 11.4 (i) Give one difference that will be noticed when fitting the bolts into a threaded hole. … [1] (ii) Explain why bolt B, which has a smaller thread pitch, will resist higher torque when being tightened. … … … [2] (e) Fig. 11.5 shows a gear system. motor speed = 850 rpm motor 30t 40t 15t Fig. 11.5 (i) State the velocity ratio of the worm gear and 30t gear. … [1] (ii) Calculate the speed of the 40t gear when the motor is turning. … … … … [3]

25 marks

Mark scheme: 11(a)(i) Other types of belt could be flat, toothed or round. 2 Accept ‘plain belt’. Sketches must be identified with a name for mark to be [2  1] awarded. 11(a)(ii) Reasons for the vee belt being most suitable are: 2 Allow any other valid points. • Transmit power very well Explanation to include any two valid points, 2 marks. • Limited slip between belt and pulleys / high friction force Award 2 marks for a full explanation of a single point. • Low noise and vibration levels • Easy to install and maintain. 11(b)(i) The vee pulley transfers motion to the crankshaft causing it 4 1 mark for each word correctly used to rotate. This in turn causes the connecting rod to move the piston . [4  1] 11(b)(ii) Crank [1] / slider [1] mechanism. 2 11(b)(iii) Area of piston = Π r2 = 3.142  252 = 1963.75 (1963.495 on 3 Award 3 marks for correct answer with no working. calculator) [1] Throw of piston = 12.5  2 = 25 [1] Volume displaced = 1963.75  25 = 49093.75 mm3 or 49 087.39 [1] 11(c)(i)) Grease would not be suitable because: 2 Allow any other valid points. • It is too thick / viscous to flow through the oil holes. Explanation to include any two valid points, 2 marks. • The bearing also uses oil to carry away any particles Award 2 marks for a full explanation of a single point. from the bearing. • Grease would keep the particles in the bearing, allowing metal to metal contact. 11(c)(ii) Roller or ball bearings could have been used. 2 Accept needle bearings [2  1] 11(c)(iii) Benefits of plain bearings include: 1 Allow any other valid points. • Smaller diameter • Low maintenance • Plain bearings support higher loads. 11(d)(i) Differences in use include: 1 AOVR • More difficult to start bolt B in threaded hole / nut • Easier to start bolt A in threaded hole • Easy to cross thread bolt B when inserting • More turns needed to move bolt B through a set linear distance • Faster to insert bolt A. 11(d)(ii) • The core diameter of bolt B is larger than for bolt A due 2 Allow any other valid points. to smaller pitch and lower depth of thread Explanation to include any two valid points, 2 marks. • Larger diameter will have greater resistance to shear Award 2 marks for a full explanation of a single point. forces • Higher torque can be applied before the bolt will fail / shear. • More precise in transmitting motion 11(e)(i) The velocity ratio of the worm gear is 30:1 1 11(e)(ii) Worm gear ratio 30:1 15 t:40 t ratio 2.666:1 [1] 3 Allow ecf from (d)(ii). Total reduction ratio = 30  2.666 = 80:1 [1] Speed of 40 t gear = 850 / 80 = 10.625 rpm [1] Award 3 marks for correct answer with no working.

This question in 0445/43 Oct/Nov 2025