Largest Load P Ec3135

1. **Problem Statement:** Determine the largest load $P$ applied at point $A$ in the steel structure with given pin diameters and stress limits, considering a factor of safety of $3.0$. 2. **Given Data:** - Diameter of pin at $C$: $d_C = 6$ mm - Diameter of pins at $B$ and $D$: $d_B = d_D = 10$ mm - Ultimate shearing stress: $\tau_u = 152.0$ MPa - Ultimate normal stress in link $BD$: $\sigma_u = 400$ MPa - Factor of safety: $FS = 3.0$ - Distances: $AB = 160$ mm, $BC = 120$ mm, $BD = 18$ mm (thickness assumed from context) 3. **Step 1: Calculate allowable stresses** $$\tau_{allow} = \frac{\tau_u}{FS} = \frac{152.0}{3.0} = 50.667 \text{ MPa}$$ $$\sigma_{allow} = \frac{\sigma_u}{FS} = \frac{400}{3.0} = 133.333 \text{ MPa}$$ 4. **Step 2: Calculate shear areas for pins** Shear area for pin $C$ (double shear assumed): $$A_{shear,C} = 2 \times \frac{\pi}{4} d_C^2 = 2 \times \frac{\pi}{4} (6)^2 = 2 \times 28.274 = 56.548 \text{ mm}^2$$ Shear area for pins $B$ and $D$ (single shear each): $$A_{shear,B} = A_{shear,D} = \frac{\pi}{4} d_B^2 = \frac{\pi}{4} (10)^2 = 78.540 \text{ mm}^2$$ 5. **Step 3: Calculate maximum shear forces at pins using allowable shear stress** $$V_C = \tau_{allow} \times A_{shear,C} = 50.667 \times 56.548 = 2866.7 \text{ N}$$ $$V_B = V_D = 50.667 \times 78.540 = 3980.2 \text{ N}$$ 6. **Step 4: Calculate normal stress in link $BD$ due to axial load** Cross-sectional area of link $BD$ (assuming thickness $t = 18$ mm and diameter hole $d = 10$ mm): $$A_{BD} = t \times (\text{width} - d)$$ Since width is not given, assume width equals pin diameter $d_B = 10$ mm (common assumption for link thickness), so: $$A_{BD} = 18 \times (10 - 10) = 0 \text{ mm}^2$$ This is zero, so we must consider the net area around the hole. Usually, the net area is: $$A_{net} = t \times (\text{length} - d)$$ Length of link $BD$ is $18$ mm, so: $$A_{net} = 18 \times (18 - 10) = 18 \times 8 = 144 \text{ mm}^2$$ 7. **Step 5: Calculate maximum axial force in link $BD$ using allowable normal stress** $$N_{BD} = \sigma_{allow} \times A_{net} = 133.333 \times 144 = 19200 \text{ N}$$ 8. **Step 6: Analyze equilibrium to relate $P$ to forces at pins and link** Assuming the structure is statically determinate and the load $P$ causes shear at pins and axial load in link $BD$. - Shear at pin $C$ is related to $P$ by geometry. - Shear at pins $B$ and $D$ also relate to $P$. - Axial force in link $BD$ relates to $P$. Using moments about point $C$ or force equilibrium, the largest $P$ is limited by the smallest of these calculated maximum forces. 9. **Step 7: Calculate $P$ from shear at pin $C$** Shear force at $C$ is equal to $P$ (assuming direct load): $$P_{max,C} = V_C = 2866.7 \text{ N} = 2.867 \text{ kN}$$ 10. **Step 8: Calculate $P$ from shear at pins $B$ and $D$** Assuming load distribution, the shear at $B$ or $D$ is proportional to $P$ by geometry. For simplicity, assume: $$P_{max,B} = V_B = 3980.2 \text{ N} = 3.980 \text{ kN}$$ 11. **Step 9: Calculate $P$ from axial load in link $BD$** Assuming axial force in $BD$ equals $P$: $$P_{max,BD} = N_{BD} = 19200 \text{ N} = 19.200 \text{ kN}$$ 12. **Step 10: Determine the limiting load $P$** The smallest of the three is the limiting load: $$P_{max} = \min(2.867, 3.980, 19.200) = 2.867 \text{ kN}$$ **Final answer:** The largest load $P$ that can be applied at $A$ is approximately **2.867 kN**.