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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  GATE Civil Engineering (CE) 2025 Mock Test Series  >  Test: Solid Mechanics- 5 - Civil Engineering (CE) MCQ

Test: Solid Mechanics- 5 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Solid Mechanics- 5

Test: Solid Mechanics- 5 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Solid Mechanics- 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Solid Mechanics- 5 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Solid Mechanics- 5 below.
Solutions of Test: Solid Mechanics- 5 questions in English are available as part of our GATE Civil Engineering (CE) 2025 Mock Test Series for Civil Engineering (CE) & Test: Solid Mechanics- 5 solutions in Hindi for GATE Civil Engineering (CE) 2025 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt Test: Solid Mechanics- 5 | 10 questions in 30 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study GATE Civil Engineering (CE) 2025 Mock Test Series for Civil Engineering (CE) Exam | Download free PDF with solutions
Test: Solid Mechanics- 5 - Question 1
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Shear center of the T-section can be shown by point S and center of gravity by point G. The correct representation of S and G for the section is :

Detailed Solution for Test: Solid Mechanics- 5 - Question 1

Shear centre is the point from which if a concentrated load passes then, there will be only bending and no twisting. It is also called centre of flexure. It is that point through which resultant of shear passes.

The shear centre (S) and centre of gravity (G) of some sections is shown below.

Test: Solid Mechanics- 5 - Question 2
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Two equal length beams are fixed at their ends. One carries a distributed load and other carries same load but concentrated in the middle. The ratio of maximum deflections will be

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Test: Solid Mechanics- 5 - Question 3
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A composite section made of two materials has moduli of elasticity in the ratio 1 : 2 and lengths in the ratio 2 : 1. The ratio of the corresponding deflection under a direct load is

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Test: Solid Mechanics- 5 - Question 4
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A cantilever beam of span L and uniform flexular rigidity ‘EI’ is loaded with a downward U.D.L of w kN/m and upward force ‘P’ at the free end. The deflection at the free end will be zero, if
Detailed Solution for Test: Solid Mechanics- 5 - Question 4



Using Principal of superposition 

Test: Solid Mechanics- 5 - Question 5
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If the depth of a simply supported beam carrying an isolated load at its center, is doubled the deflection of the beam at the center will be changed by a factor of
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The central deflection in a simply supported beam due to point load,

As the depth of the beam is double

Test: Solid Mechanics- 5 - Question 6
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A force P is applied at a distance x from the end of the beam as shown in the figure. What would be the value of x so that the displacement at ‘A’ is equal to zero?
Detailed Solution for Test: Solid Mechanics- 5 - Question 6

Load at point A is shown in figure below


⇒ x = L/3 = 0.33 L

Test: Solid Mechanics- 5 - Question 7
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If the deflection at free end of a uniformly loaded cantilever beam is 18 mm and the slope of the deflection curve at the free end is 0.02 then the length of the beam is
Detailed Solution for Test: Solid Mechanics- 5 - Question 7


_______1
_______2
From (2) ⇒
 0.02 × 6 = 0.12
substitute in equation (1)

⇒ l = 1.2 m

Test: Solid Mechanics- 5 - Question 8
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A beam 40mm wide, 80mm deep, is freely supported over a span of 2m. A weight of 50 N is dropped on to the middle of the beam from a height of 40mm. Calculate the maximum instantaneous stress and deflection. E=200 GPa.
Detailed Solution for Test: Solid Mechanics- 5 - Question 8

Let W be equivalent static load that will produce same deflection as dynamic load applied from a height of 40 mm. Strain energy in beam due to dropping of 50 N load = 50 (40 + δ)

Strain energy in the beam due to load W applied gradually = ½ Wδ

50(40 + δ) = ½ Wδ                                            (1)

Substituting in (1)

*Answer can only contain numeric values
Test: Solid Mechanics- 5 - Question 9
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Locate the shear center for the channel section as shown in the figure in mm?

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Test: Solid Mechanics- 5 - Question 10
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The equation for the deflected shape of a beam carrying a uniformly distributed load and simply supported at the ends is given by

What is the intensity of uniformly distributed load (in kN/m)? (The unit of EI is kN-m2 & it is a flexural rigidity)
Detailed Solution for Test: Solid Mechanics- 5 - Question 10


But we know that  (Bending moment atany section)

⇒ w = 4 kN/m (Numerically)

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