Civil war Day 4(sep 18)-PWD

A composite bar is made of Steel & Aluminium strips each having 3 cm² area of cross-section. The composite bar is subjected to an axial load of 12000 N. If E_steel = 3E_Al, the stress in steel is

• 10 N/mm²
• 20 N/mm²
• 30 N/mm²
• None of the above

Answer explanation

Composite bar of steel + aluminium (equal length, equal area).

Given:

A_s = A_al = 3 cm² = 300 mm²

P = 12000 N

E_s = 3 E_al

Strain compatibility:

σ_s / E_s = σ_al / E_al ⇒ σ_s = 3 σ_al

Equilibrium:

P = σ_s A_s + σ_al A_al

12000 = 300 (σ_s + σ_al)

Substitute σ_s = 3 σ_al:

12000 = 300 (3 σ_al + σ_al)

12000 = 300 (4 σ_al)

12000 = 1200 σ_al

σ_al = 10 N/mm²

Therefore,

σ_s = 3 × 10 = 30 N/mm²

Final Answer: Stress in steel = 30 N/mm²

Two bars of different materials and same size are subjected to the same tensile force. If the bars have unit elongation in the ratio of 2:5 then, the ratio of modulus of elasticity of the two materials will

• 1:5

• 5:2

• 1:1

• 2:5

Answer explanation

Elongation of a bar under tensile force:

ΔL = PL / (AE)

Since P, L and A are same for both bars,

ΔL ∝ 1/E

Given ΔL₁ : ΔL₂ = 2 : 5

So,

E₁ : E₂ = 1/ΔL₁ : 1/ΔL₂ = 1/2 : 1/5 = 5 : 2

Two circular bars A and B of same material are subjected to uniaxial loading. If the diameter and load acting on bar B is double than that of A, the ratio of safety of bar B to A is

• 2
• 1/2
• 1/4
• 4

Answer explanation

Factor of safety ∝ 1 / stress

Stress = P / A = 4P / (π d²)

For bar A: diameter = d, load = P

σA = 4P / (π d²)

For bar B: diameter = 2d, load = 2P

σB = 2P / (π d²)

Ratio of stresses = σB / σA = (2P / π d²) ÷ (4P / π d²) = 1/2

Therefore,

FOSB / FOSA = σA / σB = 2

Answer: 2 : 1

The Poisson's ratio of a material whose Young's modulus is 1.2 × 10⁵ N/mm² and modulus of rigidity is 4.8 × 10⁴ N/mm², is

• 2.25
• 1.25
• 0.25
• 0.5

Answer explanation

E = 2G(1+ν) → ν = 120000/(2×48000) − 1 = 0.25.

For a given material assume, Young’s modulus E = 300 GN/m² and Modulus of rigidity G = 150 GN/m². Its Bulk modulus K will be

• 120 GN/m²
• 100 GN/m²
• 200 GN/m²
• 250 GN/m²

Answer explanation

E = 9KG/(3K+G) → K = 100 GN/m².

If all dimensions of a prismatic bar of square cross section suspended freely from the ceiling of a roof is doubled, the total elongation due to self weight will become

• 8 times
• 4 times
• 2 times
• 16 times

Answer explanation

δ_self ∝ L². Doubling dimensions → L doubles → δ ×4.

The ratio of elongation of a conical bar due to its own weight and that of a prismatic bar of same length is

• 3
• 1/2
• 1/3
• 2

Answer explanation

Prismatic: δ=ρgL²/(2E); Conical: δ=ρgL²/(6E) → ratio = 1/3.

If the external load tries to push the cross section of a bar, it will be under

• tensile stress
• compressive stress
• shear stress
• volumetric stress

Answer explanation

Push (shortening) → compression.

The change in length due to a tensile or compressive force acting on a body is given by

• PA/LE
• PL/AE
• AE/PL
• E/PLA

Answer explanation

Hooke: δ = PL/AE.

Material with higher value of coefficient of thermal expansion will be having ...... change in length due to temperature change

• less
• more
• change in length does not depend on value of coefficient of thermal expansion
• cannot be predicted

Answer explanation

ΔL = αLΔT → higher α → more ΔL.