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# MCQs on Stresses in Members-2

**MCQs on 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: cExplanation: 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: aExplanation: 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 the stresses in the two sections respectively in N/mm2?

strength-materials-questions-answers-stress-varying-bars-q3

a) 111.408 and 49.5146

b) 111.408 and 17.85

c) 97.465 and 49.5146

d) 97.465 and 34.263

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Answer: aExplanation: The stress = P/A Where P = 35000N and A is the respective cross section area of the sections. advertisement

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?

strength-materials-questions-answers-stress-varying-bars-q4

a) 0.127 and 0.0031

b) 0.034 and 0.0045

c) 0.153 and 0.003

d) 0.124 and 0.124

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Answer: aExplanation: 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?

strength-materials-questions-answers-stress-varying-bars-q5

a) 0.153mm

b) 0.183mm

c) 0.197mm

d) 0.188mm

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Answer: bExplanation: 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?

a) 0.200mm

b) 0.345mm

c) 0.509mm

d) 0.486mm

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Answer: cExplanation: 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: dExplanation: 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: aExplanation: 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: bExplanation: 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: aExplanation: The stress = P/A Where P = 15000N and A is the respective cross section area of the sections.

**MCQs on 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: bExplanation: 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: aExplanation: 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: cExplanation: 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. advertisement

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: aExplanation: 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: cExplanation: 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: dExplanation: 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: cExplanation: 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: bExplanation: 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: dExplanation: 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: bExplanation: 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.

**MCQs on Bars of Composite Sections – 1**

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

a) P/E x (L1 + L2)

b) P/E x (A1/L1 + A2/ L2)

c) P/E x (L1/A1 + L2/A2)

d) E/P x (L1/A1 + L2/A2)

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Answer: cExplanation: Change in length of section 1 = PL1/EA1 Change in length of section 2 = PL2/EA2 Total change in length of bar = PL1/EA1 + PL2/EA2.

How does the elastic constant varys with the elongation of body?

a) The elastic constant is directly proportional to the elongation

b) The elastic constant is directly proportional to the elongation

c) The elongation does not depends on the elastic constant

d) None of these

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Answer: bExplanation: Elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2) E is inversely proportional to bar elongation.

A composite rod is 1000mm long, its two ends are 40 mm2 and 30mm2 in area and length are 400mm and 600mm respectively. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation(E = 200GPa)?

a) 0.130m

b) 0.197mm

c) 0.160mm

d) 0.150mm

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Answer: aExplanation: As elongation of a composite bar of two sections = P/E x (L1/A1 + L2/A2) Putting L1, L2, A1 and A2 400mm2, 600mm2, 40mm2 and 30mm2 and P = 1000 and E = 200 x 103. advertisement

A mild steel wire 5mm in diameter and 1m ling. If the wire is subjected to an axial tensile load 10kN what will be its extension?

a) 2.55mm

b) 3.15mm

c) 2.45mm

d) 2.65mm

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Answer: aExplanation: As change in length = PL/AE P = 10x 1000N, L = 1m, A = πd2/4 = 1.963 x 10-5 m2, E = 200 x 109 N/m2.

A composite rod is 1000mm long, its two ends are 40mm2 and 30mm2 in area and length are 300mm and 200mm respectively. The middle portion of the rod is 20mm2 in area. If the rod is subjected to an axial tensile load of 1000N, what will be its total elongation (E = 200GPa)?

a) 0.145mm

b) 0.127mm

c) 0.187mm

d) 0.196mm

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Answer: dExplanation: P = 1000N, Area A1 = 40mm2, A2 = 20mm2, A30 = 30mm2 Length, L1 = 300mm, L2 = 500mm, L3 = 200mm E = 200GPa = 200x 1000 N/mm2 Total extension = P/E x (L1/A1 + L2/A2 + L3/A3).

A rod of two sections of area 625mm2 and 2500mm2 of length 120cm and 60cm respectively. If the load applied is 45kN then what will be the elongation (E = 2.1x 105 N/mm2)?

a) 0.462mm

b) 0.521mm

c) 0.365mm

d) 0.514mm

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Answer: aExplanation: P = 45,000N, E =2.1x 105 N/mm2, Area, A1 = 625mm2, A2 = 2500mm2, Length, L1 = 1200mm, L2 = 600mm Elongation = P/E x (L1/A1 + L2/A2).

What will be the elongation of a bar of 1250mm2 area and 90cm length when applied a force of 130kN if E = 1.05x 105 N/mm2?

a) 0.947mm

b) 0.891mm

c) 0.845mm

d) 0.745mm

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Answer: bExplanation: As change in length = PL/AE P = 130x 1000N, L = 900mm, A = 1250 mm2, E = 1.05 x 105 N/m2.

A bar shown in diagram is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the diameter of the middle portion?

strength-materials-questions-answers-bars-composite-section-1-q8

a) 3.456 cm

b) 3.685 cm

c) 4.524 cm

d) 4.124 cm

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Answer: bExplanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions. For middle portion area = load / stress This gives area by which diameter can be calculated.

A steel bar of 20mm x 20mm square cross-section is subjected to an axial compressive load of 100kN. If the length of the bar is 1m and E=200GPa, then what will be the elongation of the bar?

a) 1.25mm

b) 2.70mm

c) 5.40mm

d) 4.05mm

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Answer: aExplanation: Elongation in bar = PL/ AE = (100x1000x1) / (0.2×0.2x200x106) = 1.25mm.

A solid uniform metal bar is hanging vertically from its upper end. Its elongation will be _________

a) Proportional to L and inversely proportional to D2

b) Proportional to L2 and inversely proportional to D

c) Proportional of U but independent of D

d) Proportional of L but independent of D

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Answer: aExplanation: Elongation = WL / 2AE = 4WL / 2πD2E α L/D2.

**MCQs on Bars of Composite Sections – 2**

A member ABCD is subjected to points load P1=45kN, P2, P3=450kN and P4=130kN. what will be the value of P necessary for equilibrium?

strength-materials-questions-answers-bars-composite-section-2-q1

a) 350kN

b) 365kN

c) 375kN

d) 400kN

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Answer: bExplanation: On resolving forces P1 + P3 = P2 + P4 So P2 = 45 + 450 – 130 I.e. P2 = 365kN.

A member ABCD is subjected to points load P1=45kN, P2, P3=450kN and P4=130kN. What will be the total elongation of the member, assuming the modulus of elasticity to be 2.1x105N/mm2. The cross sectional area is 625mm, 2500mm, 1250mm respectively.

a) 0.4914mm

b) 0.4235mm

c) 0.4621mm

d) 0.4354mm

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Answer: aExplanation: First of all the fores will be calculated on resolving forces P1 + P3 = P2 + P4 So P2 = 45 + 450 – 130 I.e. P2 = 365kN So forces on three sections will be 45kN, 320kN and 130kN respectively. After that increase in length = PL/AE for all three sections will be calculated.

A tensile rod of 40kN is acting on a rod of diameter 40mm and of length 4m. a bore of diameter 20mm is made centrally on the rod. To what length the rod should be bored so that the total extension will increase 30% under the same tensile load if E = 2×105 N/mm2?

strength-materials-questions-answers-bars-composite-section-2-q2

a) 2m

b) 2.7m

c) 3.2m

d) 3.6m

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Answer: dExplanation: The extension = PL / AE = 2/π mm Extension after the bore is made = 1.3x 2/π mm = 2.6/π mm The extension after the bore is made, is also obtained by finding the extension of the un bored length and bored length. Stress = load / area So total extension after bore is made can have two equations which can be put equal and the length the rod should be bored up is calculated. advertisement

A bar is subjected to a tensile load of 150kN. If the stress in the middle portion is limited to 160 N/mm2, what will be the diameter of the middle portion of the total elongation of the bar is 0.25cm (E= 2 x 105)?

strength-materials-questions-answers-bars-composite-section-2-q4

a) 3cm

b) 3.45cm

c) 3.85cm

d) 4cm

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Answer: bExplanation: Total extension = P/E x (L1/A1 + L2/A2 + L3/A3 ) Only variable in the equation is A2. after getting this the diameter of the section can be calculated.

A rod, which tapers uniformly from 5cm diameter to 3cm diameter in a length of 50cm, is subjected to an axial load of 6000N. if E = 2,00,000 N/mm2, what will be the extension of the rod?

a) 0.00114cm

b) 0.00124cm

c) 0.00127cm

d) 0.00154cm

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Answer: cExplanation: The extension in the rod = PL / Et(a-b) x loge (a/b) Where a = 50mm, b = 30mm.

A bar is in two sections having equal lengths. The area of cross section of 1st is double that of 2nd. if the bar carries an axial load of P, then what will be the ratio of elongation in section 2nd to section 1st ?

a) 1/2

b) 2

c) 4

d) 1/4

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Answer: bExplanation: Ratio of elongation in 2nd / ratio of elongation in 1st = L2/L1 x A2/A1 Since L1 = L2 and A1 = 2A2 Therefore, ratio = 1 x 2/1 = 2.

A round bar made of same material consists of 4 parts each of 100mm length having diameters of 40mm, 50mm, 60mm and 70mm, respectively. If the bar is subjected to an axial load of 10kN, what will be the total elongation of the bar in mm?

a) 0.4/πE ( 1/16 + 1/25 + 1/36 + 1/49)

b) 4/πE ( 1/16 + 1/25 + 1/36 + 1/49)

c) 2/πE ( 1/16 + 1/25 + 1/36 + 1/49)

d) 40/πE ( 1/16 + 1/25 + 1/36 + 1/49)

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Answer: dExplanation: Total elongation = 4PL/πE ( 1/d12 + 1/d22 + 1/d32 + 1/d42) = 4x10x100/πEx100 ( 1/16 + 1/25 + 1/36 + 1/49) mm = 40/πE ( 1/16 + 1/25 + 1/36 + 1/49).

A bar shown in the diagram below is subjected to load 160kN. If the stress in the middle portion Is limited to 150N/mm2, what will be the length of the middle portion, if the total elongation of the bar is to be 0.2mm? Take E = 2.1 x 105 N/mm2.

strength-materials-questions-answers-bars-composite-section-2-q8

a) 18.45cm

b) 17.24cm

c) 16.45cm

d) 20.71cm

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Answer: dExplanation: Let L2 and D2 be the dimensions of the middle portion and L1 and D2 be the end portion dimensions. For middle portion area = load / stress This gives area by which diameter can be calculated. As Total extension = P/E x (L1/A1 + L2/A2) This gives the value of L2.

A composite bar consists of a bar enclosed inside a tune of another material when compressed under a load as whole through rigid collars at the end of the bar. What will be the equation of compatibility?

a) W1 + W2 = W

b) W1 + W2 = constant

c) W1/A1E1 = W2/A2E2

d) W1/A1E2 = W2/A2E1

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Answer: aExplanation: Compatibility equation insists that the change in length of the bar must be compatible with the boundary conditions. Here W1 + W2 = W it is also correct but it is equilibrium equation.