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Test: Solid Mechanics- 2 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Solid Mechanics- 2

Test: Solid Mechanics- 2 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Solid Mechanics- 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Solid Mechanics- 2 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- 2 below.
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Test: Solid Mechanics- 2 - Question 1

The shear stress at the centre of a circular shaft under torsion is

Detailed Solution for Test: Solid Mechanics- 2 - Question 1

The simple torsion equation is written as

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft and the following is the stress distribution

Hence the maximum shear stress occurs on the outer surface of the shaft where r = R and at the centre the shear stress is zero.

Test: Solid Mechanics- 2 - Question 2

Match List-I with List-II and select the correct answer using the code given below the lists.

List-I (Loaded Beam)

A.


B.


C.
 

List-II (Maximum Bending moment)
1. 
2. 
3. 
4. 

Detailed Solution for Test: Solid Mechanics- 2 - Question 2

A-3, B-1, C-4

Beam A is cantilever so maximum bending moment occurs at the fixed support

Beam B is simply supported so maximum bending moment occurs at midspan

Beam C is cantilever so maximum bending moment occurs at fixed support
MC

Test: Solid Mechanics- 2 - Question 3

A solid circular shaft of diameter d is subjected to a combined effect of bending moment M and torque T. The relation is used  for designing the shaft using which one of this material property

Detailed Solution for Test: Solid Mechanics- 2 - Question 3



Ssy = Torsional yield strength

Test: Solid Mechanics- 2 - Question 4

A solid steel shaft is surrounded by a copper shaft, such that, Isteel = 1/2 Icopper. If Gcopper = 1/2 Gsteel, what is the ratio of Ts/Tc if the composite shaft is subjected to a twisting moment?

Detailed Solution for Test: Solid Mechanics- 2 - Question 4


Both shaft will rotate through same degree.

Hence 

Test: Solid Mechanics- 2 - Question 5

Couple M is applied at C on a simply supported beam AB. The maximum shear force on AC will be

Detailed Solution for Test: Solid Mechanics- 2 - Question 5


R1 + R2 = 0

∑MA = 0

⇒ R2 × 3 = -M


So maximum shear force = M/3

Test: Solid Mechanics- 2 - Question 6

A hollow steel shaft of external diameter 100 mm and internal diameter 50 mm is to be replaced by a solid allow shaft. Assuming the same value of polar modulus for both, the diameter of the solid allows shaft will be-

Detailed Solution for Test: Solid Mechanics- 2 - Question 6

Polar modulus, 

Let the diameter of solid shaft is ‘d’

*Answer can only contain numeric values
Test: Solid Mechanics- 2 - Question 7

For a loaded cantilever beam of uniform cross-section, the bending moment (in N-mm) along the length is M(x) = 5x2 + 10x, where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x = 10 mm is ________.


Detailed Solution for Test: Solid Mechanics- 2 - Question 7

M = 5x2 + 10x

Shear Force = dM/dx = 10x + 10

Shear force at X = 10 mm

= 10 x 10 + 10 = 110 N

Test: Solid Mechanics- 2 - Question 8

S.F.D for the beam is:

Detailed Solution for Test: Solid Mechanics- 2 - Question 8





 

Test: Solid Mechanics- 2 - Question 9

The equation for the deflected shape of a beam carrying a U.D.L and simply supported at ends is given below  The load carrying capacity of the beam is

Detailed Solution for Test: Solid Mechanics- 2 - Question 9


Using the equation of deflection:

W = 40 kN/m

*Answer can only contain numeric values
Test: Solid Mechanics- 2 - Question 10

For the given figure, calculate the bending moment (in kN-m) at 2 m from the left end?


Detailed Solution for Test: Solid Mechanics- 2 - Question 10


∑FV = 0

RA + RB = 3 + 2 × 2

RA + RB = 7 kN

∑FH = 0

HA = 0

∑Mz = 0

RA × 4 – 3 × 3 - 2 × 2 × 1= 0

RA = 3.25 kN

RB = 3.75 kN

BMD = R× 2 - 3 × 1

BMD = 3.25 × 2 - 3 = 3.5 kN

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