# Strength of Materials Questions and Answers – Principle of Superposition

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Principle of Superposition”.

1. 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

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.

2. 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

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

3. 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

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.

4. 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

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.

5. 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

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.
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6. 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

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.

7. 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

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.

8. 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

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.

9. 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

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.

10. 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

Explanation: Strain in section 1 = 0.1/100
Strain in section 2= 0.1/150
Resultant strain = 0.001+0.0006 = 0.0016.

11. 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

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.

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