Let's break down the problem step by step: ### **Given Data** - \( E = 200 \) GPa \( = 200 \times 10^3 \) MPa - \( \alpha = 11.7 \times 10^{-6} \) /°C - Dimensions in mm (see image) - External forces at A, B, and D - No elongation (\(\Delta L_{total} = 0\)) - Consider temperature and external load effects ### **Step 1: Forces and Sections** #### **Forces Applied** - At A: 120 kN (to the left) - At B: 220 kN (to the right) - At D: 360 kN (to the right) - At D: Unknown force \( F_D \) (to the left, to ensure equilibrium and zero elongation) #### **Sections and Lengths** - AB: 300 mm - BC: 350 mm - CD: 400 mm ### **Step 2: No Elongation Condition** The total elongation due to load and temperature must be zero: \[ \Delta L_{total} = \Delta L_{load} + \Delta L_{temp} = 0 \] #### **Elongation Due to Load:** \[ \Delta L_{load} = \sum \frac{P_i L_i}{A_i E} \] #### **Elongation Due to Temperature:** \[ \Delta L_{temp} = \alpha \Delta T \sum L_i \] ### **Step 3: Force Equilibrium** Sum of forces (assuming right is positive): \[ -120 + 220 + 360 + F_D = 0 \implies F_D = -460 \text{ kN} \] So, a 460 kN force to the left is required at D for equilibrium. ### **Step 4: Stress Calculation** #### **Find Internal Forces in Sections:** **Section AC (A to C):** - At A: -120 kN (left) - At B: +220 kN (right) - At C: Internal force balances previous ones **Section CD (C to D):** - At D: -460 kN (left) - At C: Internal force Let’s use the method of sections: - **Section AC (left of B):** Force = -120 kN (compression) - **Section BC (between B and C):** Force = -120 + 220 = 100 kN (tension) - **Section CD (right of C):** Force = -120 + 220 + 360 = 460 kN (tension) #### **Stresses:** \[ \sigma = \frac{P}{A} \] where \(A = 80 \times 120 = 9600\, \mathrm{mm}^2\) - **Section AC:** \(P = -120,000\, \mathrm{N}\) - **Section CD:** \(P = 460,000\, \mathrm{N}\) \[ \sigma_{AC} = \frac{-120,000}{9600} = -12.5\, \mathrm{MPa} \] \[ \sigma_{CD} = \frac{460,000}{9600} = 47.92\, \mathrm{MPa} \] ### **Step 5: Effect of Temperature** If the temperature changes by \(\Delta T\), thermal expansion/contraction occurs. For no elongation: \[ \frac{P_{eq} L_{total}}{A E} + \alpha \Delta T L_{total} = 0 \] \[ P_{eq} = -A E \alpha \Delta T \] This negative sign indicates the force needed to counteract the thermal expansion (if temperature increases, force should be compressive). ### **Summary of Answers** #### **a) Force at D:** \[ F_D = -460\, \mathrm{kN} \quad (\text{to the left for equilibrium}) \] #### **b) Stresses:** - **Section AC:** \(-12.5\, \mathrm{MPa}\) (compression) - **Section CD:** \(47.92\, \mathrm{MPa}\) (tension) #### **c) Temperature Effects:** - To ensure zero elongation, the force at D must also compensate for thermal expansion/contraction: \[ F_D = -460\, \mathrm{kN} - (A E \alpha \Delta T) \] where \(\Delta T\) is the change in temperature. --- **If you provide the value of \(\Delta T\), the exact force to be applied at D can be calculated considering temperature.**