Strength of Materials - 2 - Mechanical Engineering MCQ

\(\sigma = \frac{P}{A} = \frac{{40 \times 1000}}{{1000}} = 4\;MPa\)

Normal stress on inclined plane:

σ_n = BM' = R + R cos 60° = R(1 + cos 60)

σ_n = 2(1 + 0.5) = 3 MPa

Some more important calculations which may be asked in exams:

Shear stress on inclined plane:

\({\tau _s} = R\sin 60^\circ = \frac{{2 \times \sqrt 3 }}{2} = 1.732\;MPa\)

Maximum principal stress, σ₁ = 4 MPa (θ = 0°)

Minimum principal stress, σ₂ = 0 MPa (θ = 90°)

Radius of the Mohr circle R = σ₁/2 = 2 MPa

Maximum shear stress, τ_max = R = 2 MPa (θ = 45°)

For thin cylinder:

For thin spherical shell:

τ_max = 1.5 τ_avg

Young's Modulus of Rigidity is the ratio of longitudinal stress to longitudinal strain.

Shear Modulus or Modulus of rigidity is the shear stress to shear strain.

Possion's ratio is the ratio of lateral strain to longitudinal stress.

Bulk Modulus of elasticity is the ratio of volumetric stress to volumetric strain.

Using Flexural Formula

M ∝ Z

d is constant

M ∝ b (width)

According to maximum shear stress criteria.

Deflection of cantilever beam having concentrated load at the free end is given by

Power=

Strain rosette can be defined as the arrangement strain gauge oriented at different angles. It only measures strain in one direction, to get principal strains, it is necessary to use a strain rosette.

Depending on the arrangement of strain gauges, strain rosettes are classified in to:-

1. Rectangular strain gauge rosette

2. Delta strain gauge rosette

3. Star strain gauge rosette

K₁N₁ = K₂N₂

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.

Stress= =25N/mm2=25MPa$

For rectangular section:

The beam is strongest if section modulus is maximum

Dimensions of strongest beam

Given condition is “Effective length of a column effectively held in position and restrained in direction at one but neither held in position nor restrained in direction at the other end”

This implies one end fixed, other end free

l_eFF = 2L

In case of plane stress problem, Shear stress is

Allowable stress in spherical pressure vessel:

In beam there is only stress in x and y direction, there is no stress in z direction.

σ₁ + σ₂ = σ_x + σ_y

100 + 50 = 65 + σ_y

σ_y = 150 – 65 = 85 N/mm²