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# Strength of Materials MCQs Sudden Loading

What is the relation between maximum stress induced due to sudden loading to maximum stress the gradual loading?

a) Maximum stress in sudden load is equal to the maximum stress in gradual load

b) Maximum stress in sudden load is half to the maximum stress in gradual load

c) Maximum stress in sudden load is twice to the maximum stress in gradual load

d) Maximum stress in sudden load is four times to the maximum stress in gradual load

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Answer: c

Explanation: Maximum stress in sudden loading = 2P/A

Maximum stress in gradual loading = P/A.

What is the strain energy stored in a body when the load is applied suddenly?

a) σE/V

b) σE2/V

c) σV2/E

d) σV2/2E

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Answer: d

Explanation: Strain energy in gradual loading = σ2V/2E.

A tensile load of 60kN is suddenly applied to a circular bar of 4cm diameter. What will be the maximum instantaneous stress induced?

a) 95.493 N/mm2

b) 45.25 N/mm2

c) 85.64 N/mm2

d) 102.45 N/mm2

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Answer: a

Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2.

A tensile load of 60kN is suddenly applied to a circular bar of 4cm and 5m length. What will be the strain energy absorbed by the rod if E=2×105 N/mm2?

a) 140.5 N-m

b) 100 N-m

c) 197.45 N-m

d) 143.2 N-m

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Answer: d

Explanation: Maximum instantaneous stress induced = 2P/A = 2×60000/400π = 95.49 N/mm2

Strain energy = σ2V/2E = 95.492 x 2×106π / (2x2x105) = 143238 N-mm = 143.23 N-m.

A tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm. What will be the instantaneous stress in bar?

a) 100 N/mm2

b) 120 N/mm2

c) 150 N/mm2

d) 250 N/mm2

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Answer: d

Explanation: Stress = 2x load / area = 2×100,000/ (20×40) = 250 N/mm2.

2 tensile load of 100kN is suddenly applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy absorbed in the bar if E=1×105 N/mm2?

a) 312.5 N-m

b) 314500 N-mm

c) 1250 N-m

d) 634 N-m

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Answer: c

Explanation: Stress = 2xload / area = 2×100,000/ (20×40) = 250 N/mm2

Strain energy = σ2V/2E = 250x250x20x40x5000/ (2×100,000) = 1250000 N-mm = 1250 N-m.

A steel rod is 2m long and 50mm in diameter. A axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous stress induced in the rod?

a) 101.89 N/mm2

b) 94.25 N/mm2

c) 130.45 N/mm2

d) 178.63 N/mm2

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Answer: a

Explanation: Area = π/4 d2 = 625π

Load = 100kN = 100×1000 N

Stress = 2 x load / area = 2x100x1000 / (625π) = 101.86 N/mm2.

A steel rod is 2m long and 50mm in diameter. An axial pull of 100kN is suddenly applied to the rod. What will be the instantaneous elongation produced in the rod if E=22GN/m2?

a) 0.0097 mm

b) 1.0754 mm

c) 1.6354 mm

d) 1.0186 mm

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Answer: d

Explanation: Area = π/4 d2 = 625π

Load = 100kN = 100×1000 N

E=22GN/m2 = 200 x 109 / 106 = 200,000 N/mm2

Stress = 2 x load / area = 2x100x1000 / (625 π )

Elongation = stress x length / E = 101.86×2000 / 200000 = 1.0186 mm.

What will be the amount of axial pull be applied on a a 4cm diameter bar to get an instantaneous stress value of 143 N/mm2?

a) 50kN

b) 60kN

c) 70kN

d) 80kN

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Answer: b

Explanation: Instantaneous stress = 2 x load / area

Load = instantaneous stress x area / 2

= 143 x 400×3.14 / 2 = 60kN.

What will be the instantaneous stress produced in a bar 10cm2 in area ans 4m long by the sudden application of tensile load of unknown magnitude, if the extension of the bar due to suddenly applied load is 1.35mm if E = 2×105 N/mm2?

a) 67.5 N/mm2

b) 47 N/mm2

c) 55.4 N/mm2

d) 78.5 N/mm2

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Answer: a

Explanation: The value of stress = load / area where area is 10cm2 and load can be calculated by stress strain equation.

### Strength of Materials MCQs Gradual Loading

What is the strain energy stored in a body when the load is applied gradually?

a) σE/V

b) σE2/V

c) σV2/E

d) σV2/2E

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Answer: d

Explanation: Strain energy in gradual loading = σ2V/2E.

What is strain energy?

a) The work done by the applied load In stretching the body

b) The strain per unit volume

c) The force applied in stretching the body

d) The stress per unit are

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Answer: a

Explanation: The strain energy stored in a body is equal to the work done by the applied load in stretching the body.

What is the relation between maximum stress induced due to gradual load to maximum stress the sudden load?

a) Maximum stress in gradual load is equal to the maximum stress in sudden load

b) Maximum stress in gradual load is half to the maximum stress in sudden load

c) Maximum stress in gradual load is twice to the maximum stress in sudden load

d) Maximum stress in gradual load is four times to the maximum stress in sudden load

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Answer: b

Explanation: Maximum stress in gradual loading = P/A

Maximum stress in sudden loading = 2P/A.

A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5cm long. What will be the stress in the rod if E=1×105 N/mm2?

a) 47.746 N/mm2

b) 34.15 N/mm2

c) 48.456 N/mm2

d) 71.02 N/mm2

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Answer: a

Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 47.746 N/mm2.

A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the stress in the rod if E=1×105 N/mm2?

a) 1.19mm

b) 2.14mm

c) 3.45mm

d) 4.77mm

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Answer: d

Explanation: Stress = Load/ area = 60,000 / (π/4 D2) = 47.746 N/mm2

So stretch = stress x length / E = 4.77mm.

A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm. What will be the stress in bar?

a) 100 N/mm2

b) 120 N/mm2

c) 125 N/mm2

d) 150 N/mm2

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Answer: c

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm2.

A tensile load of 100kN is gradually applied to a rectangular bar of dimension 2cmx4cm and length of 5m. What will be the strain energy in the bar if E=1×105 N/mm2?

a) 312.5 N-m

b) 314500 N-mm

c) 245.5 N-m

d) 634 N-m

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Answer: a

Explanation: Stress = load / area = 100,000/ (20×40) = 125 N/mm2

Strain energy = σ2V/2E = 125x125x20x40x5000/ (2×100,000) = 312500 N-mm = 312.5N-m.

A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 10m long. What will be the strain energy absorbed by the rod if E=1×105 N/mm2?

a) 100 N-m

b) 132 N-m

c) 148 N-m

d) 143.2 N-m

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Answer: d

Explanation: Stress = 60,000 / 400π = 47.746

Strain energy = σ2V/2E = 47.746×47.746×12,566,370 / (2×100000) = 143,236.54 N-mm = 143.2N-m.

A uniform bar has a cross sectional area of 700mm and a length of 1.5m. if the stress at the elastic limit is 160 N/mm, what will be the value of gradually applied load which will produce the same extension as that produced by the suddenly applied load above?

a) 100kN

b) 110kN

c) 112kN

d) 120kN

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Answer: c

Explanation: For gradually applied load, stress = load / area

Load = stress x area = 160 x 700 = 112000 N = 112kN.

A tension bar 6m long is made up of two parts, 4m of its length has cross sectional area of 12.5cm while the remaining 2m has 25cm. An axial load 5tonnes is gradually applied. What will be the total strain energy produced if E = 2 x 106 kgf/cm2?

a) 240kgf/cm

b) 242kgf/cm

c) 264kgf/cm

d) 270kgf/cm

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Answer: b

Explanation: First stress = load /area, then the strain energy will be calculated as

Strain energy = σ2V/2E.

Strength of Materials MCQs Impact Loading

What is the strain energy stored in a body when the load is applied with impact?

a) σE/V

b) σE2/V

c) σV2/E

d) σV2/2E

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Answer: d

Explanation: Strain energy in impact loading = σ2V/2E.

What is the value of stress induced in the rod due to impact load?

a) P/A (1 + (1 + 2AEh/PL)1/2)

b) P/A (2 + 2AEh/PL)

c) P/A (1 + (1 + AEh/PL)1/2)

d) P/A ((1 + 2AEh/PL)1/2)

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Answer: a

Explanation: The value of stress is calculated by equating the strain energy equation and the work done equation.

What will be the stress induced in the rod if the height through which load is dropped is zero?

a) P/A

b) 2P/A

c) P/E

d) 2P/E

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Answer: b

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)

Putting h=0, we get stress = 2P/A.

A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the instantaneous stress (E=210GPa)?

a) 149.4 N/mm2

b) 179.24 N/mm2

c) 187.7 N/mm2

d) 156.1 N/mm2

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Answer: c

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)

Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2.

A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous stress induced in the vertical bar if E = 200GPa?

a) 50.87 N/mm2

b) 60.23 N/mm2

c) 45.24 N/mm2

d) 63.14 N/mm2

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Answer: b

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )

Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.

A weight of 10kN falls by 30mm on a collar rigidly attached to a vertical bar 4m long and 1000mm2 in section. What will be the strain (E=210GPa)?

a) 0.00089

b) 0.0005

c) 0.00064

d) 0.00098

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Answer: a

Explanation: As stress = P/A (1 + (1 + 2AEh/PL)1/2)

Putting P = 10,000, h = 30, L = 4000, A = 1000, E = 210,000 we will get stress = 187.7 N/mm2

As strain = stress / E, thus, strain = 187.7 / 210,000 = 0.00089.

A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the maximum instantaneous elongation in the vertical bar if E = 200GPa?

a) 0.245mm

b) 0.324mm

c) 0.452mm

d) 0.623mm

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Answer: c

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )

Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2

Elongation = stress x length / E = 60.23 x 1500 / 200,000 = 0.452mm.

A load of 100N falls through a height of 2cm onto a collar rigidly attached to the lower end of a vertical bar 1.5m long and of 105cm2 cross- sectional area. The upper end of the vertical bar is fixed. What is the strain energy stored in the vertical bar if E = 200GPa?

a) 2.045 N-m

b) 3.14 N-m

c) 9.4 N-mm

d) 2.14 N-m

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Answer: a

Explanation: As stress = P/A ( 1 + ( 1 + 2AEh/PL)1/2 )

Putting P = 100, h = 20, L = 1500, A = 150, E = 200,000 we will get stress = 60.23 N/mm2.

Strain energy stored = stress2 x volume / 2E = 60.232 x 2525000 / (2×200,000) = 2.045 N-m.

The maximum instantaneous extension, produced by an unknown falling weight in a vertical bar of length 3m. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?

a) 100 N/mm2

b) 110 N/mm2

c) 120 N/mm2

d) 140 N/mm2

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Answer: d

Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.

The maximum instantaneous extension, produced by an unknown falling weight through a height of 4cm in a vertical bar of length 3m and of cross section area 5cm2. what will be the instantaneous stress induced in the vertical bar and the value of unknown weight if E = 200GPa?

a) 1700 N

b) 1459.4 N

c) 1745.8 N

d) 1947.5 N

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Answer: c

Explanation: Instantaneous stress = E x instantaneous strain = E x δL/L = 200,000x 2.1 / 3000 = 140N/mm2.

As, P( h + δL) = σ2/2E x V

So P = 1745.8 N.

An unknown weight falls through a height of 10mm on a collar rigidly attached to a lower end of a vertical bar 500cm long. If E =200GPa what will be the value of stress?

a) 50 N/mm2

b) 60 N/mm2

c) 70 N/mm2

d) 80 N/mm2

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Answer: d

Explanation: Stress = E x strain = E x δL/L = 200,000 x 2 /5000 = 80 N/mm2.