Strength of Materials Past Paper PDF

NAME: OPOKU MAXWELL INDEX NUMBER: FMM.41.004.170.23 COURSE: STRENGTH OF MATERIALS CLASS: MR 1A DATE: 28TH JUNE, 2024 1. A copper wire of original length 1 meter is stretched to 1.01 meters by a force of 500 N. Calculate the strain in the wire. Solution: Strain, ε = (Change i...

NAME: OPOKU MAXWELL INDEX NUMBER: FMM.41.004.170.23 COURSE: STRENGTH OF MATERIALS CLASS: MR 1A DATE: 28TH JUNE, 2024 1. A copper wire of original length 1 meter is stretched to 1.01 meters by a force of 500 N. Calculate the strain in the wire. Solution: Strain, ε = (Change in Length) / (Original Length) ε = (1.01 m - 1 m) / 1 m ε = 0.01m 2. A steel rod of length 2 meters and cross-sectional area 0.01 square meters is subjected to an axial tensile force of 100 kN. Calculate the axial stress in the rod. Solution: Axial stress, σ = Force / Area σ = 100,000 N / 0.01 m² σ = 10,000,000 N/m² or 10 MPa 3. A cylindrical wire of cross-sectional area 5 mm² is stretched by a force of 600 N. Calculate the stress in the wire. Solution: Area (A) = 5 mm² = 5 × 10⁻⁶ m². Stress (σ) = Force / Area = 600 N / 5 × 10⁻⁶ m² = 120,000,000 Pa or 120 MPa. 4. If a metal rod with a modulus of elasticity of 210 GPa and a length of 2 m is subjected to a tensile force causing an elongation of 0.001 m, calculate the stress. Solution: Strain (ε) = Change in Length / Original Length = 0.001 m / 2 m = 0.0005. Stress (σ) = Young's Modulus × Strain = 210 GPa × 0.0005 = 105 MPa. 5. A metal wire of cross-sectional area 3 mm² is stretched by a force of 500 N. Calculate the stress in the wire. Solution: Area (A) = 3 mm² = 3 × 10⁻⁶ m². Stress (σ) = Force / Area = 500 N / 3 × 10⁻⁶ m² = 166,666,667 Pa or 166.67 MPa. 6. A brass rod of diameter 9 mm elongates by 0.5 mm when subjected to a tensile load. If the original length is 300 mm, calculate the strain. Solution: Strain (ε) = Change in Length / Original Length = 0.5 mm / 300 mm = 0.00167. 7. A cylindrical rod of bronze with a diameter of 18 mm and a Young's modulus of 95 GPa is subjected to a compressive load of 7500 N. The rod shortens by 0.18 mm. Calculate the original length of the rod. Solution: Area (A) = πr² = π × (0.009 m)² = 2.544 × 10⁻⁴ m². Stress (σ) = Force / Area = 7500 N / 2.544 × 10⁻⁴ m² = 29.48 MPa. Strain (ε) = Stress / Young's Modulus = 29.48 MPa / 95 GPa = 0.00031. Original Length (L₀) = Change in Length / Strain = 0.18 mm / 0.00031 = 0.18 × 10⁻³ m / 0.00031 = 0.581 m. 8. A cylindrical bar of radius 0.1 m and length 2 m is subjected to a force of 4000 N. Calculate the stress. Solution: Area (A) = πr² = π × (0.1 m)² = 0.0314 m². Stress (σ) = Force / Area = 4000 N / 0.0314 m² = 127,389 Pa. 9. A wire of diameter 0.005 meters is tested to its breaking point by applying a tensile force of 2 kN. Calculate the tensile strength of the wire. Solution: Cross-Sectional Area, A = π * (Diameter/2)², A = π * (0.005 m / 2)², A = 0.0000196 m² Tensile Strength, σ_t = Force / Area, σ_t = 2,000 N / 0.0000196 m², σ_t ≈ 102,040,816 N/m² or 102.04 MPa 10. A metal plate of thickness 0.01 meters is subjected to a tensile force of 20 kN along its length of 2 meters. If the width of the plate is 0.1 meters, calculate the longitudinal strain if the Young's Modulus is 70 GPa. Solution: Cross-Sectional Area, A = Width * Thickness, A = 0.1 m * 0.01 m, A = 0.001 m² Stress, σ = Force / Area, σ = 20,000 N / 0.001 m², σ = 20,000,000 N/m² or 20 MPa Strain, ε = σ / Young's Modulus, ε = 20 MPa / 70 GPa, ε = 0.0002857m