Stress, Strain, and Strain Energy Old exam Questions...

Drayton Manor High School

Stress, Strain, and Strain Energy Old exam Questions

Q1. The graph shows the variation of stress with strain for a ductile alloy when a specimen is slowly stretched to a maximum strain of εm and the stress is then slowly reduced to zero.

The shaded area

A represents the work done per unit volume when stretching the specimen

B represents the energy per unit volume recovered when the stress is removed

C represents the energy per unit volume which cannot be recovered

D has units of J m–1

(Total 1 mark)

Q2. A stone is projected horizontally by a catapult consisting of two rubber cords. The cords, which obey Hooke’s law, are stretched and released. When each cord is extended by x, the stone is projected with a speed v. Assuming that all the strain energy in the rubber is transferred to the stone, what is the speed of the stone when each cord is extended by 2x?

A v

B

C 2v

D 4v (Total 1 mark)

Q3. The force on a sample of a material is gradually increased and then decreased. The graph of force against extension is shown in the diagram.


The increase in thermal energy in the sample is represented by area

A R

B P + Q

C P + Q + R

D P + Q − R (Total 1 mark)

Q4. A rubber cord is used to provide mechanical resistance when performing fitness exercises. A scientist decided to test the properties of the cord to find out how effective it was for this purpose. The graph of load against extension is shown in the figure below for a 0.50 m length of the cord.

Curve A shows loading and curve B shows unloading of the cord.

(a) State which feature of this graph confirms that the rubber cord is elastic.

...................................................................................................................... (1)


(b) Explaining your method, use the graph (curve A) to estimate the work done in producing an extension of 0.30 m.

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answer = ................................... J (3)

(c) Assuming that line A is linear up to an extension of 0.040 m, calculate the Young modulus of the rubber for small strains.

The cross-sectional area of the cord = 5.0 × 10–6 m2

The unstretched length of the cord = 0.50 m

answer = ................................ Pa (3)

(d) The scientist compared this cord with a steel spring that reached the same extension for the same maximum load without exceeding its limit of proportionality.

(i) On the figure above, draw the load-extension line for this spring up to a load of 50 N and label it C.

(1)

(ii) With reference to the spring, explain what is meant by limit of proportionality.

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............................................................................................................. (1)

(Total 9 marks)


Q5. An aerospace engineer has built two differently designed wings. One wing is made from an aluminium alloy and the other is made from a carbon fibre composite.

The engineer tests a sample of each material by applying a varying stress.

(a) Tick (✓) two of the boxes in the table below to indicate which are properties of the material from which the wing is made.

breaking stress

stiffness constant, k

tensile strain

tensile stress

Young modulus

(1)

(b) Below is the stress−strain graph that the engineer obtains for the aluminium alloy.

(i) The engineer has labelled a point Q on the graph. This is a point beyond which the behaviour of the material changes irreversibly. State the name for this point.

............................................................................................................... (1)


(ii) Use the graph to determine the Young modulus of the aluminium alloy. Show your working.

Young modulus = ................................................ Pa (2)

(c) The engineer who carried out the experiment to obtain the stress−strain graph decided to stretch another sample to a strain of 0.10. She then gradually reduced the stress to zero.

Show by drawing on the graph how you would expect the stress to vary with strain as the stress is reduced.

(2)

(d) Calculate the volume of 25.0 kg of the aluminium alloy.

density of aluminium alloy = 2.78 × 103 kg m–3.

volume = ................................................ m3

(1)

(e) 1.28% of the aluminium alloy’s volume is copper. Calculate the mass of pure aluminium needed to make 25.0 kg of the aluminium alloy.

density of pure aluminium = 2.70 × 103 kg m–3.

mass of pure aluminium = ................................................ kg (2)

(Total 9 marks)


Q6. A student investigated how the extension of a rubber cord varied with the force used to extend it. She measured the extension for successive increases of the force and then for successive decreases. The diagram below shows a graph of her results.

(a) (i) Give a reason why the graph shows the rubber cord does not obey Hooke’s law.

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............................................................................................................. (1)

(ii) Give a reason why the graph shows the rubber cord does not exhibit plastic behaviour.

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............................................................................................................. (1)

(iii) What physical quantity is represented by the area shaded on the graph between the loading curve and the extension axis?

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............................................................................................................. (1)

(b) Describe, with the aid of a diagram, the procedure and the measurements you would make to carry out this investigation.

The quality of your written answer will be assessed in this question.

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...................................................................................................................... (6)

(Total 9 marks)

Q7. Figure 1 and Figure 2 both show the side view of a steel bolt.

Figure 1 Figure 2

(a) Show on Figure 1 forces acting on the bolt which would produce a tensile strain. (1)

(b) The ultimate tensile stress of steel is 5.0 × 108 Pa, the elastic limit is 2.5 × 108 Pa and the Young modulus of steel is 2.0 × 1011 Pa.

Defining any terms used, state what is meant by:

(i) tensile stress;

...............................................................................................................

............................................................................................................... (1)

(ii) tensile strain;

...............................................................................................................

............................................................................................................... (1)

(iii) the elastic limit.

...............................................................................................................

...............................................................................................................(1)


(c) When the main engines of a space shuttle are fired, they develop a total thrust of 4.5 × 106 N. In a test firing the shuttle is held to the launch pad by 8 steel bolts each of diameter 9.0 × 10–2 m. Using data given in (b):

(i) calculate the strain for each bolt during the test; (4)

(ii) determine the minimum number of bolts that could have been used when carrying out the test.

(3) (Total 11 marks)

Q8. (a) State Hooke’s law for a material in the form of a wire and state the conditions under which this law applies.

......................................................................................................................

...................................................................................................................... (2)

(b) A length of steel wire and a length of brass wire are joined together. This combination is suspended from a fixed support and a force of 80 N is applied at the bottom end, as shown in the figure below.

Each wire has a cross-sectional area of 2.4 × 10–6 m2.

length of the steel wire = 0.80 m length of the brass wire = 1.40 m the Young modulus for steel = 2.0 × 1011 Pa the Young modulus for brass = 1.0 × 1011 Pa

(i) Calculate the total extension produced when the force of 80 N is applied.

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(ii) Show that the mass of the combination wire = 4.4 × 10–2 kg.

density of steel = 7.9 × 103 kg m–3

density of brass = 8.5 × 103 kg m–3

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............................................................................................................. (7)

(c) A single brass wire has the same mass and the same cross-sectional area as the combination wire described in part (b). Calculate its length.

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...................................................................................................................... (2)

(Total 11 marks)

Q9. If lengths of rail track are laid down in cold weather, they may deform as they expand when the weather becomes warmer. Therefore, when rails are laid in cold weather they are stretched and fixed into place while still stretched. This is called pre-straining.

The following data is typical for a length of steel rail:

Young modulus of steel = 2.0 × 1011 Pa cross sectional area of a length of rail = 7.5 × 10–3 m2

amount of pre-strain = 2.5 × 10–5 for each kelvin rise in temperature the rail is expected to experience.

A steel rail is laid when the temperature is 8 °C and the engineer decides to use a pre-strain of 3.0 × 10–4.

(a) Calculate the tensile force required to produce the pre-strain in the rail required by the engineer.

tensile force = ............................... N (3)


(b) Calculate the elastic strain energy stored in a rail of unstressed length 45 m when pre-strained as in part (a)

elastic strain energy = ............................... J (2)

(c) Calculate the temperature at which the steel rail becomes unstressed.

temperature = ............................... °C (2)

(d) Explain why the engineer does not use the highest observed temperature at the location of the railway track to determine the amount of pre-strain to use.

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........................................................................................................................ (2)

(Total 9 marks)

Q10. (a) State what is meant by the yield stress of a material.

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........................................................................................................................ (3)

(b) A steel piano wire has a diameter of 1.8 × 10−3 m and a length of 1.55 m. When tightened to emit a note of the required frequency it extends by 1.3 × 10−3 m. The Young modulus of the steel is 2.1 × 1011 Pa.

(i) Calculate the force exerted on the frame of the piano by this wire. (3)

(ii) Calculate the strain energy stored in this stretched wire. (2)

(Total 8 marks)


ANSWERS

M1.C [1]

M2.C [1]

M3.B [1]

M4. (a) returns to original length/shape/position/state/zero extension/no permanent extension (1)

1

(b) (12 to 14 big squares/318 small squares ± 8 area of 1 big square = 10N × 0.05m = 0.50/small = 2 × 0.01 = 0.02)

statement of method that refers to area (1)

accept triangle if area is mentioned

5.0 to 8.0 (J) or clear attempt to calculate correct area (1) triangle OK here 5.1 (J) for single triangle is max 2

6.0 to 7.0 (J) (1) 3

(c) (E =) (1)

(=) (1) also gets first mark

incorrect values of F and ∆L get first mark only

2.5 × 107 (Pa) (1)

or (stress = F/A =) 10/5 × 10–6 (= 2.0 × 106 Pa) (1)

(strain = ∆L/L =) 0.04/0.5 (= 0.08) (1)

2 × 106/0.08 gets both marks

E correctly evaluated from incorrect value of stress and incorrect value of strain gets 1 mark only

use of 50 (N) and 0.04 (m) giving 1.25 × 108 (Pa) gains 2 marks

use of 5(N) and 0.4 (m) is max 2

2.5 × 107 (Pa) (1) 3

(d) (i) straight line through origin finishing at the same point as the rubber ± 1 small division (1)

1

(ii) point beyond which


graph is no longer linear

or force no longer proportional to extension

or Hooke’s law limit (1) 1

[9]

M5.(a)

breaking stress ✓

stiffness constant, k

tensile strain

tensile stress

Young modulus ✓

1

(b) (i) elastic limit ✓

only one attempt at the answer is allowed 1

(ii) ( E = 300 × 106 / 4 × 10-2 = 7.5 × 109 ) 7.5 (Pa) ✓ allow 7.4 to 7.6 (Pa) × 109 ✓

first mark is for most significant digits ignoring the power of 10. E.g. 7500 gains mark

2

(c) straight line beginning on existing line at a strain of 0.10 and hitting the strain axis at a lower non-zero value ✓ line that ends on the x -axis with strain between 0.045 and 0.055 ✓(only allow if first mark is given)

ie accuracy required ± one division 2

(d) 8.99 × 10-3 (m3) ✓ condone 1 sig fig

allow 9.00 × 10-3

1

(e) 0.9872 × 8.99 × 10-3 or = 8.8749 × 10-3 (m3) ✓ allow CE from 4d

(m = ρV )= 2700 × 8.8749 × 10-3 = 24 (kg) ✓ (23.962 kg) allow CE from first part, e.g. if 1.28% was used gives 0.311 kg

V =0.9872 × (d)

m = 2.665 × (d)

1.28% of vol = 1.15 × 10-4 m3

2 [9]


M6. (a) (i) the lines are not straight (owtte) (1)

(ii) there is no permanent extension (1) (or the overall/final extension is zero or the unloading curve returns to zero extension)

(iii) (area represents) work done (on or energy transfer to the rubber cord) or energy (stored) (1) not heat/thermal energy

3

(b) the mark scheme for this part of the question includes an overall assessment for the Quality of Written Communication

QWC descriptor mark range

good- excellent

The candidate provides a comprehensive and coherent description which includes nearly all the necessary procedures and measurements in a logical order. The descriptions should show awareness of how to apply a variable force. They should know that measurements are to be made as the force is increased then as it is decreased. In addition, they should know how to calculate/measure the extension of the cord. At least five different masses/’large number’ of masses are used. Minimum 7 masses to reach 6 marks. The diagram should be detailed.

5-6

modest- adequate

The description should include most of the necessary procedures including how to apply a variable force and should include the necessary measurements. They may not have described the procedures in a logical order. They may not appreciate that measurements are also to be made as the cord is unloaded. They should know that the extension of the cord must be found and name a suitable measuring instrument (or seen in diagram – label need not be seen)/how to calculate. The diagram may lack some detail.

3-4

poor- limited

The candidate knows that the extension or cord length is to be measured for different forces – may be apparent from the diagram. They may not appreciate that measurements are also to be made as the cord is unloaded. They may not state how to calculate the extension of the cord. The diagram may not have been drawn.

1-2

incorrect, inappropriate

or no response

No answer at all or answer refers to unrelated, incorrect or inappropriate physics.

0

The explanation expected in a competent answer should include a coherent selection of the following physics ideas.

diagram showing rubber cord fixed at one end supporting a weight at the other end or pulled by a force (1)

means of applying variable force drawn or described (eg use of standard


masses or a newtonmeter) (1)

means of measuring cord drawn or described (1)

procedure

measured force applied ( or known weights used) (1)

cord extension measured or calculated (1)

repeat for increasing then decreasing length (or force/weight) (1)

extension calculated from cord length – initial length (1) [9]

M7.(a) two forces shown producing tension (see examples below) B1

(2)

(b) (i) tension / area

or force / area (perpendicular to the surface on which the force acts)

or F / A with terms defined (condone loose definition of area) B1

(2)

(ii) extension / original length

or extension per unit length B1

(2)

(iii) the point / stress / strain / force or extension up to which

either all energy supplied in stretching is returned when (deforming) forces are removed

or the object returns to its original shape when (deforming forces) are removed

or beyond which further stress causes plastic deformation or permanent


deformation B1

(2)

(c) (i) force on each bolt = 0.56 MN (ie total force divided by 8)

or

cross-sectional area each bolt = πr2 plus correct substitution 6.4 × 10–3 m2

C1

stress = 8.8 × 107 or

stress / strain = Young modulus C1

strain = 4.4 × 10–4

A1 (4)

(ii) area of steel needed = maximum force / UTS

or maximum force / elastic limit C1

4.5 × 106 / 5.0 × 108 = 9.0 × 10–3 m2

or 4.5 × 106 / 2.5 × 108 = 18.0 × 10–3 m2

C1

number of bolts = 9.0 × 10–3 / area of a bolt (allow e.c.f.)

= 1.4 bolts

or = 2.8 bolts A1

answer = 2 bolts (c.a.o.)

or answer = 3 bolts (c.a.o.) B1

(3) [11]

M8. (a) Hooke’s law: the extension is proportional to the force applied (1) up to the limit of proportionality or elastic limit [or for small extensions] (1)

2

(b) (i) (use of E = gives) ∆Ls = (1)

= 1.3 × 10–4 (m) (1) (1.33 × 10–4 (m))

∆Lb = = 4.7 × 10–4 (m) (1) (4.66 × 10–4 (m))


total extension = 6.0 × 10–4 m (1)

(ii) m = ρ × V (1) ms = 7.9 × 103 × 2.4 × 10–6 × 0.8 = 15.2 × 10–3 (kg) (1) mb = 8.5 × 103 × 2.4 × 10–6 × 1.4 = 28.6 × 10–3 (kg) (1) (to give total mass of 44 or 43.8 × 10–3 kg)

7

(c) (use of m = ρAl gives) l = (1)

= 2.2 m (1) (2.16 m)

(use of mass = 43.8 × 10–3 kg gives 2.14 m) 2

[11]

M9.(a) Use of Young Modulus = ✓

The first mark is for calculating the tensile stress 1

To give tensile stress = 2 × 1011 × 3.0 × 10-4 = 6.0 × 107✓

The second mark is substituting into the tensile force equation

1

Use of tensile stress =

To give tensile force = 6.0 × 107 × 7.5 × 10-3 = 4.5 × 105 N ✓

The third mark is for the correct answer 1

(b) Use of strain = extension / original length

To give extension = 3.0 × 10-4 × 45 = 1.4 × 10-2 m

(1.35 × 10-2) ✓

The first mark is for calculating the extension 1

Use of energy stored = ½ F e

To give

Energy stored = ½ × 4.5 × 105 × 1.4 × 10-2

= 3.2 × 103 J ✓

(3.04 × 103)

The second mark is for the final answer 1


(c) Temperature change = pre-strain / pre-strain per K

= 3.0 × 10-4 / 2.5 × 10-5 = 12 K✓

The first mark is for the temperature change 1

Temperature = 8°C + 12 = 20 °C ✓

The second mark is for the final answer 1

(d) So that the rail is not always under stress✓ 1

as the rail spends little time at the highest temperature✓

Or

To reduce the average stress the rail is under ✓

as zero stress will occur closer to average temperature / the rail will be under compressive / tensile stress at different times✓

1 [9]

M10. (a) The force per unit area B1

at which the material extends considerably / a lot / plastically / or strain increases considerably etc NOT doesn’t return to its original shape / permanently deformed

B1 for no (or a small) increase in) force / stress

B1 (3)

(b) (i) strain = 8.4 × 10–4 (1.3 × 10–3 / 1.55 seen) (allow if in E = FL / A∆L) B1

or area of cross section = 2.54 × 10–6

or π (0.9 × 10–3)2

stress = E × strain (explicit or numerically) and stress = F / A or E = FL / AL

C1 force = 440 – 450 N(cao)

A1 (3)

(ii) Energy = ½ F ∆ l or ½ stress × strain × volume C1

0.29 J ecf for F from (b)(i) A1

(2) [8]

Stress, Strain, and Strain Energy Old exam Questions...

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