Test: Solid Mechanics- 1 - Civil Engineering (CE) MCQ

# Test: Solid Mechanics- 1 - Civil Engineering (CE) MCQ

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## 20 Questions MCQ Test Mock test series of SSC JE Civil Engineering 2025 - Test: Solid Mechanics- 1

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

### The maximum positive bending moment in a fixed beam of span 8m and subjected to a central point load of 120 kN is (in kN-m)

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

Bending moment at centre of the span

= (W×L)/4      ie (120 kn × 8 m )/4

= 240 kn-m.

Since the load acting in the middle of span. Hence the load will be equally distributed at both endsTherefore max bending moment in fixed beam

= 240 kn-m / 2= 120 kn-m .

Test: Solid Mechanics- 1 - Question 2

### The major and minor principal stresses at a point are 3Mpa and -3Mpa respectively. The maximum shear stress at the point is

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

Maximum shear stress (τ) is given by:

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Test: Solid Mechanics- 1 - Question 3

### A point within the cross sectional plane of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross-sectional of the beam is called:

Detailed Solution for Test: Solid Mechanics- 1 - Question 3

Shear centre is a point from which 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 the resultant of shear passes. To avoid twisting and only cause bending, it is necessary for the forces to act through the particular point, which may not coincide with the centroid.

Test: Solid Mechanics- 1 - Question 4

A rectangular beam is 24 cm wide and 50 cm deep, its section modulus is given by:

Detailed Solution for Test: Solid Mechanics- 1 - Question 4

Section modulus is defined as the ratio of moment of inertia of a beam about its C.G to the maximum distance of extreme x-section of the beam (Ymax).

Test: Solid Mechanics- 1 - Question 5

The fixed support in a real beam becomes in the conjugate beam as:-

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

• Real Beam: In a real beam, a fixed support is a type of support that prevents rotation and translation of the beam at that point. It provides both vertical and horizontal support to the beam.

• Conjugate Beam: In the conjugate beam theory, we replace the real beam with an imaginary beam called the conjugate beam. The conjugate beam has the same dimensions as the real beam but the supports and loading conditions are different.

• Transformation of Supports: When we convert a real beam to its conjugate beam, the fixed support in the real beam becomes a free support in the conjugate beam. This means that the fixed support in the real beam, which provides both vertical and horizontal support, is transformed into a support that only provides vertical support in the conjugate beam.

• Free Support in Conjugate Beam: The free support in the conjugate beam allows the beam to rotate but prevents translation. It only provides vertical support to the beam, similar to a roller support in a real beam.

Test: Solid Mechanics- 1 - Question 6

The shear stress on a beam section is maximum:-

Detailed Solution for Test: Solid Mechanics- 1 - Question 6
Shear Stress on a Beam Section

• Definition: Shear stress is the force per unit area acting tangentially to the beam cross-section.

• Maximum Shear Stress Location: The shear stress on a beam section is maximum at the neutral axis of the section.

• Explanation: When a beam is subjected to bending, shear stresses are generated along its cross-section. These shear stresses are maximum at the neutral axis of the section.

• Neutral Axis: The neutral axis is a line through the cross-section of the beam where the internal bending stress is zero.

• Shear Stress Distribution: The distribution of shear stress varies linearly from zero at the neutral axis to a maximum value at the top and bottom surfaces of the beam.

• Free Edges: While shear stresses are also present at the free edges of the beam, they are not maximum at these locations.

Therefore, the correct answer is B: At the neutral axis of the section, where the shear stress on a beam section is maximum.

Test: Solid Mechanics- 1 - Question 7

A simple supported beam with rectangular cross section is subjected to a concentrated load at mid span. If the width and depth of beam are doubled, the deflection at mid span will be reduced to:-

Detailed Solution for Test: Solid Mechanics- 1 - Question 7

Test: Solid Mechanics- 1 - Question 8

A tensile test is performed on a round bar. After fracture it has been found that the diameter remains approximately same at fracture. The material under test was

Detailed Solution for Test: Solid Mechanics- 1 - Question 8

In Brittle materials under tension test undergoes brittle fracture i.e there failure plane is 90° to the axis of load and there is no elongation in the rod that’s why the diameter remains same before and after the load. Example: Cast Iron, concrete etc

But in case of ductile materials, material first elongate and then fail, their failure plane is 45° to the axis of the load. After failure cup-cone failure is seen. Example Mild steel, high tensile steel etc.

Test: Solid Mechanics- 1 - Question 9

Consider the following statements:

1. On a Principal plane, only normal stress acts

2. On a Principal plane, both normal and shear stresses acts

3. On a Principal plane, only shear stress acts

4. Isotropic state of stress is independent of frame of reference

Which of the above statements is/are correct?

Detailed Solution for Test: Solid Mechanics- 1 - Question 9

Principal planes are the planes at which only normal stress is acting and shear stress is zero at that plane,

Option 2 and 3 are wrong. So, correct option is 1 and 4

Test: Solid Mechanics- 1 - Question 10

The ratio of crippling loads of a column having both the ends fixed to the column having both the ends hinged, is

Detailed Solution for Test: Solid Mechanics- 1 - Question 10

For both ends fixed,

For both ends hinged, leff = l

Test: Solid Mechanics- 1 - Question 11

What is the limit to Poisson's ratio?

Detailed Solution for Test: Solid Mechanics- 1 - Question 11

Limit of poisons ratio varies between -1 to 0.5.

μ = 0.5 for rubber.

Test: Solid Mechanics- 1 - Question 12

Match the following beams with their bending moment diagrams:

Detailed Solution for Test: Solid Mechanics- 1 - Question 12

Bending moment for two point load in simply supported beam will be trapezoidal.

Bending moment diagram for UDL in simply supported beam will be parabolic.

BMD for UDL in cantilever beam as shown in figure (ii)

Test: Solid Mechanics- 1 - Question 13

The crippling load taken by a column with both ends hinged is 100 kN. The crippling load taken by the same column with one end fixed and other end free will be:

Detailed Solution for Test: Solid Mechanics- 1 - Question 13

if all other parameters remain constant

Test: Solid Mechanics- 1 - Question 14

Modulus of rigidity is defined as the ratio of ________.

Detailed Solution for Test: Solid Mechanics- 1 - Question 14

The ratio of tensile stress or compressive stress to the corresponding strain is a constant. This ratio is known as Young’s Modulus or Modulus of Elasticity and is denoted by E.

The ratio of shear stress to the corresponding shear strain within the elastic limit, is known as Modulus of Rigidity or Shear Modulus. This is denoted by C or G or N.

Test: Solid Mechanics- 1 - Question 15

The Mohr’s circle given below corresponds to which one of the following stress condition.

Detailed Solution for Test: Solid Mechanics- 1 - Question 15

Radius of Mohr’s circle = 100 MPa

And centre of Mohr’s circle is at a distance,

from the origin. Here it is in origin.

both the normal stress may be zero which is a pure shear case or opposite in nature which don’t exist in any of the options. So it is the pure shear case, where the radius is

Test: Solid Mechanics- 1 - Question 16

If maximum principal stress σ1 = 60 N/mm2, σ2 and σ3 of value zero act on a cube of unit dimensions, then the maximum shear strain energy stored in it would be

Detailed Solution for Test: Solid Mechanics- 1 - Question 16

Shear strain energy

Test: Solid Mechanics- 1 - Question 17

If the deflection and slope at the free end of a cantilever beam subjected to the uniformly distributed load are 15 mm and 0.02 radians respectively. The length of the beam is _______.

Detailed Solution for Test: Solid Mechanics- 1 - Question 17

Deflection at the free end of a cantilever beam subjected to udl,

Slope at free end subjected to udl,

Test: Solid Mechanics- 1 - Question 18

The stress in a body due to suddenly applied load compared to when it is applied gradually is _______.

Detailed Solution for Test: Solid Mechanics- 1 - Question 18

here, work done is given as (F δL) / 2 and strain energy stored = (σ2 / 2E) AL

Work done is equal to the strain energy stored.

(F δL) / 2 = (σ2 / 2E) AL

Therefore, σ = (F/A) ----------- (1)

Suddenly applied load is given as σ = (2F/A), here work done = (F δL)

(F δL) = (σ2 / 2E) AL

Therefore, σ = (2F/A) ------------- (2)

From (1) and (2), it can be concluded that

Test: Solid Mechanics- 1 - Question 19

If the deflection at the free end of a cantilever beam of span 1 m, subjected UDL over entire span is 15 mm, then the slope at the free end is ________

Detailed Solution for Test: Solid Mechanics- 1 - Question 19

Deflection at the free end of a cantilever beam subjected to udl,

Slope at free end subjected to udl,

Test: Solid Mechanics- 1 - Question 20

For a given material, the modulus of rigidity is 120 GPa and Poisson’s ratio is 0.4. The value of modulus of elasticity in GPa is

Detailed Solution for Test: Solid Mechanics- 1 - Question 20

E = 2G (1 + μ) = 2 × 120 (1 + 0.4) = 336 GPa

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