Link Width 397B6E

1. **State the problem:** We need to find the width $w$ of the link BC, which is 6 mm thick, made of steel with an ultimate tensile strength of 450 MPa, to safely support a load $P = 25$ kN with a factor of safety (FOS) of 3. 2. **Given data:** - Thickness $t = 6$ mm - Ultimate tensile strength $\sigma_u = 450$ MPa - Load $P = 25$ kN = 25000 N - Factor of safety $FOS = 3$ 3. **Calculate allowable stress:** $$\sigma_{allow} = \frac{\sigma_u}{FOS} = \frac{450}{3} = 150 \text{ MPa}$$ 4. **Determine the tensile force in link BC:** Since the problem involves a structure with a 90° angle at C and given dimensions, the load $P$ is applied at D. The link BC is subjected to tension. 5. **Calculate the tensile force in BC:** The load $P$ is 25 kN downward at D. The geometry shows AC = 480 mm and AB = 600 mm. Using the right triangle ABC with angle at C = 90°, the length BC can be found by Pythagoras: $$BC = \sqrt{AB^2 - AC^2} = \sqrt{600^2 - 480^2} = \sqrt{360000 - 230400} = \sqrt{129600} = 360 \text{ mm}$$ 6. **Calculate the force in BC:** The force in BC, $F_{BC}$, can be found by resolving forces. Since the load $P$ acts vertically at D and the structure is pinned at A, C, and D, the tension in BC equals the vertical component of the force. Using similar triangles or force equilibrium, the tension in BC is: $$F_{BC} = P \times \frac{AB}{BC} = 25000 \times \frac{600}{360} = 25000 \times \frac{5}{3} = 41666.67 \text{ N}$$ 7. **Calculate the required cross-sectional area $A$ of BC:** $$A = \frac{F_{BC}}{\sigma_{allow}} = \frac{41666.67}{150 \times 10^6} = 2.7778 \times 10^{-4} \text{ m}^2 = 277.78 \text{ mm}^2$$ 8. **Calculate the width $w$ of BC:** The cross-sectional area is thickness times width: $$A = t \times w \Rightarrow w = \frac{A}{t} = \frac{277.78}{6} = 46.30 \text{ mm}$$ **Final answer:** $$\boxed{w = 46.3 \text{ mm}}$$