Normal Shear Bearing 7Dc4B7

1. **Problem Statement:** Determine the normal stress on gross and net areas, shear stress on the pin, and bearing stress on the steel plate given: - Pin diameter $d = 20$ mm - Plate width $b = 60$ mm - Plate thickness $t = 8$ mm - Load $P = 70$ kN 2. **Normal Stress on Gross Area:** The gross cross-sectional area $A_{gross}$ is the full width times thickness: $$A_{gross} = b \times t = 60 \text{ mm} \times 8 \text{ mm} = 480 \text{ mm}^2$$ Normal stress on gross area is: $$\sigma_{gross} = \frac{P}{A_{gross}} = \frac{70,000 \text{ N}}{480 \text{ mm}^2} = 145.83 \text{ MPa} \approx 145.8 \text{ MPa}$$ 3. **Normal Stress on Net Area:** The net area excludes the pin hole diameter: $$A_{net} = (b - d) \times t = (60 - 20) \text{ mm} \times 8 \text{ mm} = 320 \text{ mm}^2$$ Normal stress on net area is: $$\sigma_{net} = \frac{P}{A_{net}} = \frac{70,000 \text{ N}}{320 \text{ mm}^2} = 218.75 \text{ MPa} \approx 219 \text{ MPa}$$ 4. **Shear Stress on Pin:** Shear stress $\tau$ on the pin is the load divided by the pin's cross-sectional area: $$A_{pin} = \frac{\pi}{4} d^2 = \frac{\pi}{4} (20)^2 = 314.16 \text{ mm}^2$$ $$\tau = \frac{P}{A_{pin}} = \frac{70,000}{314.16} = 222.82 \text{ MPa}$$ 5. **Bearing Stress on Steel Plate:** Bearing stress $\sigma_b$ is the load divided by the projected bearing area (pin diameter times thickness): $$A_b = d \times t = 20 \text{ mm} \times 8 \text{ mm} = 160 \text{ mm}^2$$ $$\sigma_b = \frac{P}{A_b} = \frac{70,000}{160} = 437.5 \text{ MPa}$$ **Final answers:** - Normal stress on gross area: $145.8$ MPa - Normal stress on net area: $219$ MPa - Shear stress on pin: $222.8$ MPa - Bearing stress on steel plate: $437.5$ MPa These stresses help assess the safety and design adequacy of the connection.