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# Strength of Materials (MCQs) Stress due to Materials Used and Their Applications

Which test is conducted to measure the ability of a material to resist scratching, abrasion, deformation and indentation?

a) Creep test

b) Fatigue test

c) Hardness test

d) Compression test

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Answer: c

Explanation: The ability of a material to resist scratching, abrasion, deformation and indentation is called hardness. So to measure this hardness test is used. it is generally expressed by Brinell, Rockwell or Vickers hardness numbers.

Which test is conducted to measure the endurance limit of the material?

a) Creep test

b) Fatigue test

c) Compression test

d) Hardness test

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Answer: b

Explanation: The fatigue test is used to design components subjected to varying load. It experimentally determines the endurance limit of the material.

What is the process in which the metal is cooled rapidly in water after heating the metal above the lower critical temperature to increase the hardness of the material?

a) Quenching

b) Tampering

c) Hardening

d) Annealing

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Answer: a

Explanation: Quenching is the process in which the metal is cooled rapidly in water after heating the metal above the lower critical temperature to increase the hardness of the material. Hardness is achieved during the quenching process depends on the amount of carbon content and cooling rate.

What is the process of heating the metal in the furnance to a temperature slightly above the upper critical temperature and cooling slowly In the furnance.

a) Quenching

b) Tampering

c) Annealing

d) Normalizing

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Answer: c

Explanation: Annealing is the process of heating the metal in the furnance to a temperature slightly above the upper critical temperature and cooling slowly In the furnance. It produces an even grain structure, reduces hardness and increases ductility usually at a reduction of strength.

Photo stress method is ___________

a) Stress analysis method

b) Creep test

c) Ultra violet test

d) None of the mentioned

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Answer: a

Explanation: Photo stress is a widely used full field technique for accurately measuring surface strains to determine the stresses in a part or structure during static of dynamic testing.

What is the factor of safety?

a) The ratio of total stress to the permissible stress

b) The ratio of ultimate stress to the permissible stress

c) The ratio of ultimate stress to the applied stress

d) The ratio of ultimate stress to the modulus of elasticity

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Answer: b

Explanation: The ratio of ultimate stress to the permissible or working stress is called the factor of safety. This factor of safety is kept in mind in designing any structure.

Which one of the following has the largest value of thermal coefficient?

a) Brass

b) Copper

c) Steel

d) Aluminium

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Answer: d

Explanation: Aluminium has the large value of thermal coefficient among them of value 24 x 10-6. whereas brass and copper has 19×10-6 and 17×10.

Identify which factor may cause a lowered body temperature:

a) Infection

b) Stress

c) Shock

d) Exercise

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Answer: c

Explanation: Shock can cause the body temperature to drop, and so the cause of shock must be found. Other factors that can cause a lowered body temperature include: very young/old, serious haemorrhage, recovery from anesthesis and poisons.

### Strength of Materials MCQs Bars of varying sections

If a bar of two different length are in a line and P load is acting axially on them then what will be the change in length of the bar if the radius of both different lengths is same?

a) P/E x (L1 + L2)

b) PA/E x (L1 + L2)

c) P/EA x (L1 + L2)

d) E/PA x (L1 + L2)

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Answer: c

Explanation: Change in length of section 1 = PL1/EA1

Change in length of section 2 = PL2/EA2

Since diameter is same for both the sections, the respective area will be the same

Total change in length of bar = PL1/EA1 + PL2/EA2 = P/EA x (L1 + L2).

If a bar of two sections of different diameters of same length are in a line and P load is acting axially on them then what will be the change in length of the bar?

a) PL/E x (1/A1 + 1/A2)

b) P/E x (1/A1 + 1/A2)

c) P/EL x (1/A1 + 1/A2)

d) PE/L x (1/A1 + 1/A2)

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Answer: a

Explanation: Change in length of section 1 = PL1/EA1

Change in length of section 2 = PL2/EA2

Since length is same for both the sections,

Total change length of bar = PL/E x (1/A1 + 1/A2).

An axial pull of 35000 N is acting on a bar consisting of two lengths as shown with their respective dimensions. What will be thestresses in the two sections respectively in N/mm2?<br/>

a) 111.408 and 49.5146<br/>

b) 111.408 and 17.85<br/>

c) 97.465 and 49.5146<br/>

d) 97.465 and 34.263

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Answer: a

Explanation: The stress = P/A

Where P = 35000N and A is the respective cross section area of the sections.

An axial pull of 1kN is acting on a bar of consisting two equal lengths as shown but of dia 10cm and 20cm respectively. What will be the stresses in the two sections respectively in N/mm2?<br/>

a) 0.127 and 0.0031<br/>

b) 0.034 and 0.0045<br/>

c) 0.153 and 0.003<br/>

d) 0.124 and 0.124

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Answer: a

Explanation: The stress = P/A

Where P = 1000N and A is the respective cross section area of the sections.

An axial pull of 35000 N on a bar consisting of two lengths as shown with their respective dimensions. What will be the total extension of the bar if the young’s modulus = 2.1 x 105?<br/>

a) 0.153mm<br/>

b) 0.183mm<br/>

c) 0.197mm<br/>

d) 0.188mm

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Answer: b

Explanation: The total extension in the bar = P/E x ( L1/A1 + L1/A1 )

Where P = 35000 N, E = 2.1 x 105 N/mm2, L1 and L2 are the 20cm and 25cm respectively and A1 and A2 are the area of both the sections respectively.

An axial pull of 20 kN on a bar of two equal lengths of 20cm as shown with their respective dimensions. What will be the total extension of the bar if the young’s modulus = 2×105?<br/>

a) 0.200mm<br/>

b) 0.345mm<br/>

c) 0.509mm<br/>

d) 0.486mm<br/>

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Answer: c

Explanation: The total extension in the bar = P/E x (L1/A1 + L1/A1)

Where P = 2 kN, E = 2 x 105 N/mm2, L1 and L2 are same of 20cm and A1and A2 are the area of both the sections respectively.

Does the value of stress in each section of a composite bar is constant or not?

a) It changes in a relationship with the other sections as well

b) It changes with the total average length

c) It is constant for every bar

d) It is different in every bar in relation with the load applied and the cross sectional area

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Answer: d

Explanation: The value of stress in every section of a composite bar is given by P/A which is it is dependent on the load applied and the cross sectional area of the section. The value of stress in a section does not depend on the dimensions of other sections in the bar.

A composite bar of two sections of equal length and equal diameter is under an axial pull of 10kN. What will be the stresses in the two sections?

a) 3.18 N/mm2

b) 2.21 N/mm2

c) 3.45 N/mm2

d) 2.14 N/mm2

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Answer: a

Explanation: The stress = P/A

Where P = 1000N and A is the respective cross section area of the sections. Here the stress will be equal in both the sections as the dimensions are the same.

A composite bar of two sections of unequal length and equal diameter is under an axial pull of 10kN. What will be the stresses in the two sections?

a) 2.145 N/mm2

b) 3.18 N/mm2

c) 1.245 N/mm2

d) 2.145 N/mm2

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Answer: b

Explanation: The stress = P/A

Where P = 1000N and A is the respective cross section area of the sections. Here the stress will be equal in both the sections as the diameter is the same for both the sections. Even if the length is the variable it will not alter the stress value as the length does not depend on the stress.

A composite bar of two sections of equal length and given diameter is under an axial pull of 15kN. What will be the stresses in the two sections in N/mm2?

a) 190.9 and 84.88

b) 190.9 and 44.35

c) 153.45 and 84088

d) 153045 and 44.35

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Answer: a

Explanation: The stress = P/A

Where P = 15000N and A is the respective cross section area of the sections.

### Strength of Materials MCQs Principle of Superposition

Which law states the when a number of loads are acting on a body, the resulting strain, according to principle of superposition, will be the algebraic sum of strains caused by individual loads?

a) Hooke’s law

b) Principle of superposition

c) Lami’s theorem

d) Strain law

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Answer: b

Explanation: The principle of superposition says that when a number of loads are acting on a body, the resulting strain, according to the principle of superposition, will be the algebraic sum of strains caused by individual loads.

How the total strain in any body subjected to different loads at different sections can be calculated?

a) The resultant strain is the algebraic sum of the individual strain

b) The resultant strain calculated by the trigonometry

c) The resultant will be through Lame’s theorem

d) None of the mentioned

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Answer: a

Explanation: In a bar of different sections, the resultant strain is the algebraic sum of the individual stresses.

Three sections in a beam are of equal length of 100mm. All three sections are pulled axially with 50kN and due to it elongated by 0.2mm. What will be the resultant strain in the beam?

a) 0.002

b) 0.004

c) 0.006

d) 0.020

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Answer: c

Explanation: The strain = dL / L = 0.2/100 = 0.002

This strain will be for one section. By the principle of superposition the resultant strain will be the algebraic sum of individual strains I.e. = 0.002 + 0.002 + 0.002 = 0.006.

Two sections in a bar of length 10cm and 20cm respectively are pulled axially. It causes an elongation of 0.2mm and 0.4mm respectively in each section. What will be the resultant strain in the bar?

a) sd0.004

b) 0.002

c) 0.003

d) 0.006

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Answer: a

Explanation: The strain = dL / L

In column 1, strain = 0.2/100 = 0.002

In column 2, strain = 0.4/200 = 0.002

Resultant strain = 0.002 + 0.002 = 0.004.

A composite bar have four sections each of length 100mm, 150mm, 200mm, 250mm. When force is applied, all the sections causes an elongation of 0.1mm. What will the resultant strain in the bar?

a) 0.0012

b) 0.00154

c) 0.00256

d) 0.0020

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Answer: c

Explanation: Strain in section 1 = 0.1/100

Strain in section 2 = 0.1/150

Strain In section 3 = 0.1/200

Strain in section 4 = 0.1/250

Resultant strain = 0.001+0.0006+0.0005+0.0004 = 0.00256.

A brass bar, having cross sectional area of 100mm2, is subjected to axial force of 50kN. The length of two sections is 100mm and 200mm respectively. What will be the total elongation of bar if E = 1.05 x 105 N/mm2 ?

a) 1.21mm

b) 2.034mm

c) 2.31mm

d) 1.428mm

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Answer: d

Explanation: Elongation in section 1 = P/AE x L = 50,000/(100×1.05×100,000) x 100 = 0.476mm

Elongation In section 2 = P/AE x L = 50,000/(100×1.05×100,000) x 200 = 0.952mm

Total elongation = 0.476 + 0.952 = 1.428mm.

A composite bar having two sections of cross-sectional area 100mm2 and 200mm2 respectively. The length of both the sections is 100mm. What will be the total elongation of bar if it is subjected to axial force of 100kN and E = 105 N/mm2?

a) 1.0

b) 1.25

c) 1.5

d) 2.0

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Answer: c

Explanation: Elongation in section 1 = 100,000 x 100 / 100000 x100 = 1

Elongation in section 2 = 100,000 x 100 / 100000x 200 = 0.5

Total elongation = 1 + 0.5 = 1.5mm.

A bar having two sections of cross sectional area of 100mm2 and 200mm2 respectively. The length of both the sections is 200mm. What will be the total strain in the bar if it is subjected to axial force of 100kN and E = 105 N/mm2?

a) 0.010

b) 0.015

c) 0.020

d) 0.030

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Answer: b

Explanation: Strain in section 1 = P/AE = 100,000 / 100×100000 = 0.010

Strain is section 2 = P / AE = 100,000 / 200×100000 = 0.005

Resultant strain in the bar = 0.010 + 0.005 = 0.015mm.

A brass bar, having cross sectional area of 150mm2, is subjected to axial force of 50kN. What will be the total strain of bar if E= 1.05 x 104 N/mm2?

a) 0.062mm

b) 0.025mm

c) 0.068mm

d) 0.054mm

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Answer: d

Explanation: Strain in section 1 = P/AE = 50,000/(100×1.05×100,000) = 0.031mm

Strain In section 2 = P/AE = 50,000/(100×1.05×100,000) = 0.031mm

Resultant strain = 0.031 + 0.031 = 0.062mm.

Here the calculation of strain does not requires the value of lengths of the sections.

A composite bar of two sections of each of length 100mm, 150mm. When force is applied, all the sections causes an elongation of 0.1mm. What will the resultant strain in the bar?

a) 0.0016

b) 0.00154

c) 0.00256

d) 0.0020

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Answer: a

Explanation: Strain in section 1 = 0.1/100

Strain in section 2= 0.1/150

Resultant strain = 0.001+0.0006 = 0.0016.

If the given forces P1, P2, P3, P4,and P5 which are co planar and concurrent are such that the force polygon does not close, then the system will

a) Be in equilibrium

b) Always reduce to a resultant force

c) Always reduce to a couple

d) Always be in equilibrium and will always reduce to a couple

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Answer: b

Explanation: For a system to be in equilibrium force polygon and funicular polygon must close. If the force polygon does not close then the forces will reduce to a resultant force. If funicular polygon does not close, then there is resultant moment on the system.