APSC AE Mechanical - Strength of Materials MCQs
1. What is the unit of stress?
Correct Answer: b) N/m²
Stress is defined as force per unit area. The SI unit of force is Newton (N), and area is in square meters (m²), so stress is measured in N/m², also known as Pascal (Pa).
2. A bar of 2 m length elongates by 2 mm under a tensile load. What is the strain?
Correct Answer: a) 0.001
Strain is the ratio of elongation to original length. Here, elongation = 2 mm = 0.002 m, original length = 2 m. Strain = 0.002 / 2 = 0.001 (dimensionless).
3. Which of the following is true for Young’s modulus?
Correct Answer: b) It is the ratio of longitudinal stress to longitudinal strain.
Young’s modulus (E) is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit, applicable for tensile or compressive loading.
4. A rod of 10 mm diameter is subjected to a tensile load of 10 kN. Calculate the stress.
Correct Answer: a) 127.3 MPa
Stress = Force / Area. Force = 10 kN = 10,000 N. Diameter = 10 mm, so radius = 5 mm = 0.005 m. Area = π × (0.005)² = 7.854 × 10⁻⁵ m². Stress = 10,000 / 7.854 × 10⁻⁵ = 127.3 × 10⁶ N/m² = 127.3 MPa.
5. Poisson’s ratio is defined as:
Correct Answer: b) Ratio of lateral strain to longitudinal strain.
Poisson’s ratio (ν) is the ratio of lateral strain to longitudinal strain, typically negative as materials contract laterally when stretched longitudinally.
6. A material has Young’s modulus of 200 GPa and Poisson’s ratio of 0.3. What is the shear modulus?
Correct Answer: a) 76.92 GPa
Shear modulus (G) is related to Young’s modulus (E) and Poisson’s ratio (ν) by the formula: G = E / [2(1 + ν)]. Given E = 200 GPa, ν = 0.3, G = 200 / [2(1 + 0.3)] = 200 / 2.6 = 76.92 GPa.
7. The maximum shear stress in a circular shaft under torsion is given by:
Correct Answer: a) τ = Tr/J
For a circular shaft under torsion, the maximum shear stress (τ) occurs at the outer radius (r) and is given by τ = Tr/J, where T is the torque and J is the polar moment of inertia.
8. A beam is subjected to a bending moment of 1000 N·m. If the section modulus is 2 × 10⁵ mm³, calculate the maximum bending stress.
Correct Answer: a) 5 MPa
Bending stress (σ) = M/Z, where M is the bending moment and Z is the section modulus. M = 1000 N·m = 1000 × 10³ N·mm, Z = 2 × 10⁵ mm³. σ = (1000 × 10³) / (2 × 10⁵) = 5 N/mm² = 5 MPa.
9. Which of the following statements is true for a cantilever beam?
Correct Answer: b) Maximum bending moment occurs at the fixed end.
In a cantilever beam, the maximum bending moment occurs at the fixed end due to the reaction forces and moments resisting the applied loads.
10. The slenderness ratio of a column is defined as:
Correct Answer: b) Effective length / Radius of gyration
Slenderness ratio (λ) = Effective length (Lₑ) / Radius of gyration (k). It determines the buckling behavior of columns.
11. A rectangular beam has a width of 100 mm and depth of 200 mm. Calculate the section modulus.
Correct Answer: a) 6.67 × 10⁵ mm³
Section modulus (Z) for a rectangular section = (b × d²) / 6, where b = 100 mm, d = 200 mm. Z = (100 × 200²) / 6 = (100 × 40,000) / 6 = 6.67 × 10⁵ mm³.
12. The relationship between Young’s modulus (E), shear modulus (G), and Poisson’s ratio (ν) is:
Correct Answer: a) E = 2G(1 + ν)
This is a fundamental relationship in elasticity, derived from the definitions of E, G, and ν for isotropic materials.
13. A column is subjected to an axial load of 500 kN. If the cross-sectional area is 0.01 m², what is the compressive stress?
Correct Answer: a) 50 MPa
Compressive stress = Force / Area = 500 × 10³ N / 0.01 m² = 50 × 10⁶ N/m² = 50 MPa.
50. The maximum deflection of a simply supported beam with a central point load is given by:
Correct Answer: a) δ = WL³ / (48EI)
For a simply supported beam with a central point load W, length L, modulus of elasticity E, and moment of inertia I, the maximum deflection at the center is δ = WL³ / (48EI).