1. **Problem Statement:** Determine the elongation of wire DE when an 80-kg man sits on seat C. The wire has a diameter of 5 mm and is made from A-36 steel. 2. **Given Data:** - Diameter of wire, $d = 5$ mm - Weight of man, $W = 80 \times 9.81 = 784.8$ N - Lengths: $AB = 800$ mm, $BC = 600$ mm, vertical height $AE = 600$ mm 3. **Step 1: Calculate the length of wire DE.** Point D is at the end of beam at $x = AB + BC = 1400$ mm horizontally, and point E is 600 mm vertically above A. Using Pythagoras theorem: $$L = \sqrt{(1400)^2 + (600)^2} = \sqrt{1960000 + 360000} = \sqrt{2320000} = 1523.16 \text{ mm}$$ 4. **Step 2: Calculate the force in wire DE due to the man's weight.** The man applies a downward force at C. The wire DE supports part of this load. Using static equilibrium, sum moments about A: Moment due to man at C: $$M_C = W \times (AB + BC) = 784.8 \times 1400 = 1,098,720 \text{ Nmm}$$ The wire DE applies a tension force $T$ at D, with a vertical component balancing the moment. Vertical component of tension: $$T_y = T \times \frac{600}{1523.16}$$ Moment arm for wire DE about A is horizontal distance $1400$ mm. Moment due to wire tension: $$M_T = T_y \times 1400 = T \times \frac{600}{1523.16} \times 1400$$ Set moments equal for equilibrium: $$M_T = M_C$$ $$T \times \frac{600}{1523.16} \times 1400 = 1,098,720$$ Solve for $T$: $$T = \frac{1,098,720}{\frac{600}{1523.16} \times 1400} = \frac{1,098,720}{\frac{600 \times 1400}{1523.16}} = \frac{1,098,720 \times 1523.16}{840,000} = 1991.5 \text{ N}$$ 5. **Step 3: Calculate elongation $\delta$ of wire DE.** Use formula: $$\delta = \frac{PL}{AE}$$ Where: - $P = T = 1991.5$ N - $L = 1.52316$ m - $A = \pi \times (d/2)^2 = \pi \times (0.005/2)^2 = 1.9635 \times 10^{-5} \text{ m}^2$ - $E$ for A-36 steel is approximately $200 \times 10^9$ Pa Calculate elongation: $$\delta = \frac{1991.5 \times 1.52316}{1.9635 \times 10^{-5} \times 200 \times 10^9} = \frac{3032.5}{3.927 \times 10^6} = 0.000772 \text{ m} = 0.772 \text{ mm}$$ **Final answer:** The elongation of wire DE is approximately **0.772 mm**.