GATE Past Year Questions: Stress & Strain | Strength of Materials (SOM) - Mechanical Engineering PDF Download

E_{mild steel} = 206 GPa
E_{cast iron} = 100 GPa
Now, Elongation in cast iron,

Elongation in mild steel,

At section 55 : -
The left side of rod

The right side of rod

∑ fx = 0
⇒ R₂ – P = 0
⇒ R₂ = P

∴ Hence normal = P/A

For a plane stress criteria.
Normal stress in Z direction = 0.

Brittle material breaks without plastic deformation.

Volume remain same
L₁ = 20 mm, L₂ = 40 mm, L₃ = 10mm

True strain = 

Using similar Δ,s in fig (2)

Putting above eqⁿ in eqⁿ (ii)

As we know G = E / 2(1 +μ) so this gives the ratio of E to G = 2(1 + μ).

Here, P_x = P_y = P_z = – 15 × 10⁶ Pa (compression)
Now, 
where, m = Poisson’s ratio

∴ Change in volume = Total strain × Original = 90 m³

Pressure at mid – depth = ρgh
= 10³ × 10 × 1 = 10⁴ N /m² 
∴ 
= 5 × 10⁶ N/mm² = 5 MPa

Axial strain in the cylinder would be produced due to the axial (longitudinal) stress and due to the lateral stress i .e. hoop stress, e = 

Total change in length

=  –1 × 10^–5 m = –10μm

Let R_A & R_B be the reation at the supports A & B. For the equilibrium of the bar these reaction must act towards left. so that; 

For thermal stress to be devel oped there must be constraint in the system to oppose. So strain develops but there is no thermal stress.

Rod is not restrained but completely free

Since rod is free to expand, therefore
ΔL = elongation = LαΔT
∴ 
Thermal stress = E α Δt
Since rod is fixed at both ends, so thermal strain will be zero but there will be thermal stresses.

Since one end of the rod is fixed and other is free to expand. Hence the temperature stresses is zero

Since all the faces are free to expand the stresses due to temperature rise is equal to 0.