Test: Solid Mechanics- 1 - SSC JE MCQ

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 .

Maximum shear stress (τ) is given by:

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.

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 (Y_max).

• 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.

• 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.

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.

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

Crippling load is given by:

For both ends fixed, 

For both ends hinged, l_eff = l

Limit of poisons ratio varies between -1 to 0.5.

μ = 0.5 for rubber.

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)

The crippling load carrying capacity, 

 if all other parameters remain constant

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.

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

Shear strain energy

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

Slope at free end subjected to udl,

  Gradually applied load is given as σ = (F/A) ------ (F is the gradually applied load)

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
Hence, suddenly applied load is twice the gradually applied load.

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

Slope at free end subjected to udl, 

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