1. **Problem 01:** Calculate average shear stress in two 20 mm diameter bolts in single shear carrying 150 kN. Formula: Average shear stress $\tau = \frac{F}{A}$ where $F$ is force and $A$ is shear area. Area of one bolt $A = \pi \times \left(\frac{d}{2}\right)^2 = \pi \times \left(\frac{20}{2}\right)^2 = 314.16\ \text{mm}^2$. Total area for two bolts $A_{total} = 2 \times 314.16 = 628.32\ \text{mm}^2$. Convert force to N: $150\ \text{kN} = 150000\ \text{N}$. Average shear stress $\tau = \frac{150000}{628.32} = 238.73\ \text{MPa}$. 2. **Problem 02:** Calculate average shear stress in a 25 mm diameter pin in double shear carrying 80 kN. In double shear, area $A = 2 \times \pi \times \left(\frac{d}{2}\right)^2 = 2 \times \pi \times \left(\frac{25}{2}\right)^2 = 981.75\ \text{mm}^2$. Force $F = 80000\ \text{N}$. Average shear stress $\tau = \frac{80000}{981.75} = 81.47\ \text{MPa}$. 3. **Problem 03:** Elastomeric pad 200 mm x 200 mm x 50 mm, $G=1.0$ MPa, lateral force 15 kN. a. Average shear stress $\tau = \frac{F}{A}$, area $A = 200 \times 200 = 40000\ \text{mm}^2$. Force $F = 15000\ \text{N}$. $\tau = \frac{15000}{40000} = 0.375\ \text{MPa}$. b. Shear strain $\gamma = \frac{\tau}{G} = \frac{0.375}{1.0} = 0.375$ (unitless). c. Shear deformation $\Delta x = \gamma \times h = 0.375 \times 50 = 18.75\ \text{mm}$. 4. **Problem 04:** Hollow circular steel column, outer diameter $D_o=300$ mm, inner diameter $D_i=250$ mm, torque $T=80$ kNm. a. Polar moment of inertia $J = \frac{\pi}{32} (D_o^4 - D_i^4)$. Calculate $D_o^4 = 300^4 = 8.1 \times 10^9$, $D_i^4 = 250^4 = 3.91 \times 10^9$. $J = \frac{\pi}{32} (8.1 \times 10^9 - 3.91 \times 10^9) = \frac{\pi}{32} \times 4.19 \times 10^9 = 4.11 \times 10^8\ \text{mm}^4$. b. Torsional shear stress at outer surface $\tau_o = \frac{T r_o}{J}$, where $r_o = 150$ mm, $T=80 \times 10^6$ Nmm. $\tau_o = \frac{80 \times 10^6 \times 150}{4.11 \times 10^8} = 29.2\ \text{MPa}$. c. Torsional shear stress at inner surface $\tau_i = \frac{T r_i}{J}$, $r_i=125$ mm. $\tau_i = \frac{80 \times 10^6 \times 125}{4.11 \times 10^8} = 24.3\ \text{MPa}$. 5. **Problem 05:** Solid circular steel shaft, $G=75$ GPa, length $L=4000$ mm, torque $T=12$ kNm, max twist angle $\theta=1.5^\circ$. Convert $\theta$ to radians: $\theta = 1.5 \times \frac{\pi}{180} = 0.02618$ rad. Angle twist formula: $\theta = \frac{T L}{J G}$, for solid shaft $J = \frac{\pi d^4}{32}$. Rearranged for $d$: $d = \left( \frac{32 T L}{\pi G \theta} \right)^{1/4}$. Substitute values: $T=12 \times 10^6$ Nmm, $L=4000$ mm, $G=75 \times 10^3$ MPa, $\theta=0.02618$. $d = \left( \frac{32 \times 12 \times 10^6 \times 4000}{\pi \times 75 \times 10^3 \times 0.02618} \right)^{1/4} = 54.3\ \text{mm}$. 6. **Problem 06:** Hollow aluminum tube, $G=27$ GPa, length $L=6000$ mm, outer radius $r_o=100$ mm, inner radius $r_i=80$ mm, torque $T=20$ kNm. a. Polar moment of inertia $J = \frac{\pi}{2} (r_o^4 - r_i^4)$. Calculate $r_o^4 = 100^4 = 10^8$, $r_i^4 = 80^4 = 4.096 \times 10^7$. $J = \frac{\pi}{2} (10^8 - 4.096 \times 10^7) = 8.98 \times 10^7\ \text{mm}^4$. b. Max torsional shear stress $\tau_{max} = \frac{T r_o}{J} = \frac{20 \times 10^6 \times 100}{8.98 \times 10^7} = 22.27\ \text{MPa}$. c. Max shear strain $\gamma = \frac{\tau_{max}}{G} = \frac{22.27}{27000} = 0.000825$. Angle of twist $\theta = \frac{T L}{J G} = \frac{20 \times 10^6 \times 6000}{8.98 \times 10^7 \times 27000} = 0.0496$ rad = $2.84^\circ$. 7. **Problem 07:** Simply supported timber joist 6 m long, loads 4 kN at 2 m, 8 kN at 5 m. a. Sum moments about A: $R_B \times 6 = 4 \times 2 + 8 \times 5 = 8 + 40 = 48$ kNm. $R_B = 8$ kN. Sum vertical forces: $R_A + R_B = 4 + 8 = 12$ kN. $R_A = 12 - 8 = 4$ kN. b. Shear diagram: starts at $+4$ kN at A, drops by 4 kN at 2 m to 0, drops by 8 kN at 5 m to -8 kN, returns to 0 at B. c. Moment diagram: max moment between loads, calculate at 2 m: $M = 4 \times 2 = 8$ kNm, at 5 m: $M = 4 \times 5 - 4 \times 3 = 20 - 12 = 8$ kNm. 8. **Problem 08:** Simply supported beam 8 m, point load 30 kN at 2 m, uniform load 4 kN/m from 4 m to 8 m. a. Total uniform load $w = 4 \times 4 = 16$ kN. Sum moments about A: $R_B \times 8 = 30 \times 2 + 16 \times (4 + 2) = 60 + 96 = 156$ kNm. $R_B = 19.5$ kN. Sum vertical forces: $R_A + R_B = 30 + 16 = 46$ kN. $R_A = 46 - 19.5 = 26.5$ kN. b. Shear diagram: starts at $+26.5$ kN, drops 30 kN at 2 m to -3.5 kN, constant to 4 m, drops linearly by 16 kN over 4 m to -19.5 kN at 8 m. c. Moment diagram: calculate moments at key points using shear areas. Final answers summarized: 1. $238.73$ MPa 2. $81.47$ MPa 3a. $0.375$ MPa, 3b. $0.375$, 3c. $18.75$ mm 4a. $4.11 \times 10^8$ mm$^4$, 4b. $29.2$ MPa, 4c. $24.3$ MPa 5. $54.3$ mm 6a. $8.98 \times 10^7$ mm$^4$, 6b. $22.27$ MPa, 6c. $0.000825$, angle twist $2.84^\circ$ 7a. $R_A=4$ kN, $R_B=8$ kN 8a. $R_A=26.5$ kN, $R_B=19.5$ kN