1. **Problem Statement:** Determine the maximum normal stress in bar ABCD with cross-sectional area $600\ \text{mm}^2$ under applied loads 25 kN at B, 20 kN at C, and 30 kN at D. 2. **Given Data:** - Cross-sectional area, $A = 600\ \text{mm}^2 = 600 \times 10^{-6}\ \text{m}^2 = 6 \times 10^{-4}\ \text{m}^2$ - Loads: $P_B = 25\ \text{kN} = 25000\ \text{N}$, $P_C = 20\ \text{kN} = 20000\ \text{N}$, $P_D = 30\ \text{kN} = 30000\ \text{N}$ - Distances: $AB = 1.5\ \text{m}$, $BC = 3\ \text{m}$, $CD = 3\ \text{m}$ 3. **Approach:** Calculate the internal axial force $N(x)$ at sections along the bar and find the maximum magnitude. 4. **Step 1: Calculate reactions at supports if any.** Since the bar is loaded only by axial forces and no supports are specified, we consider equilibrium from left to right. 5. **Step 2: Calculate axial force in each segment:** - Segment AB (from A to B): No load applied between A and B, so axial force $N_{AB} = 0$ (assuming free end at A). - At B, load $25\ \text{kN}$ acts downward (tensile or compressive depends on direction). Assuming tensile positive, the force just right of B is $N_{B^+} = -25000\ \text{N}$ (compression). - Segment BC (between B and C): Axial force $N_{BC} = -25000\ \text{N}$ (constant between B and C). - At C, load $20\ \text{kN}$ acts downward, so just right of C: $$N_{C^+} = N_{B^+} - 20000 = -25000 - 20000 = -45000\ \text{N}$$ - Segment CD (between C and D): Axial force $N_{CD} = -45000\ \text{N}$. - At D, load $30\ \text{kN}$ acts downward, so just right of D: $$N_{D^+} = N_{C^+} - 30000 = -45000 - 30000 = -75000\ \text{N}$$ 6. **Step 3: Identify maximum axial force magnitude:** Maximum magnitude is $|N_{D^+}| = 75000\ \text{N}$. 7. **Step 4: Calculate maximum normal stress using formula:** $$\sigma = \frac{N}{A}$$ 8. **Step 5: Substitute values:** $$\sigma_{max} = \frac{75000}{6 \times 10^{-4}} = 125000000\ \text{Pa} = 125\ \text{MPa}$$ 9. **Answer:** The maximum normal stress in the bar is $125\ \text{MPa}$ (compressive).