Advanced Strength and Applied Elasticity, Fourth Edition

Answers to Selected Problems

Chapter 1 Analysis of Stress

1.1. P = 4.27 kN

1.4. P = 32.5 kN, θ = 26.56°

1.5. σ_x′ = −13.15 MPa, τ_{x′ y′} = −15.67 MPa

1.6. F = 67.32 kN/m³

1.9. F_x = F_y = F_z = 0

1.10. (a) σ₁ = 121 MPa, σ = −71 MPa, τ_max = 96 MPa

σ₁ = 200 MPa, σ₂ = −50 MPa, τ_max = 125 MPa

(b)

1.15. (a)σ_x = 38.68 MPa, σ_y = 12.12 MPa, τ_xy = 7MPa

(b)σ₁ = 40.41 MPa, σ₂ = 10.39 MPa,

1.18. P = 3πpr²

1.24. (a)σ = −0.237τ_o, τ = 0.347τ_o

(b)σ₁ = 3.732τ_o,σ₂ = −0.268τ_o,

1.26. σ₁ = −σ₂ = 51.96 MPa,

1.28. σ₁ = 46.17 MPa, σ₂ = 13.83 MPa,

1.30. (a) ...