1. **Problem Statement:** Calculate the net change in length of a prismatic bar of diameter 10 mm subjected to axial forces along three segments of lengths 3 m, 2 m, and 1 m respectively. The forces are 50 kN leftwards at A, 10 kN rightwards at B, 20 kN rightwards at C, and 80 kN rightwards at D. The modulus of elasticity is $E = 200000$ N/mm². 2. **Formula Used:** The axial deformation $\Delta L$ of a segment is given by: $$\Delta L = \frac{PL}{AE}$$ where $P$ is the axial force, $L$ is the length of the segment, $A$ is the cross-sectional area, and $E$ is the modulus of elasticity. 3. **Calculate Cross-sectional Area:** Diameter $d = 10$ mm $$A = \frac{\pi d^2}{4} = \frac{\pi \times 10^2}{4} = 78.54\ \text{mm}^2$$ 4. **Determine axial force in each segment:** - Segment 1 (A to B, length 3 m = 3000 mm): Force $P_1 = 50$ kN leftwards = $-50000$ N (tension is positive, compression negative; assume leftwards force is compression) - Segment 2 (B to C, length 2 m = 2000 mm): Forces at B and C are 10 kN and 20 kN rightwards, net force in segment 2 is $P_2 = 10 + 20 = 30$ kN rightwards = $+30000$ N - Segment 3 (C to D, length 1 m = 1000 mm): Forces at C and D are 20 kN and 80 kN rightwards, net force in segment 3 is $P_3 = 20 + 80 = 100$ kN rightwards = $+100000$ N 5. **Calculate deformation for each segment:** $$\Delta L_1 = \frac{P_1 L_1}{AE} = \frac{-50000 \times 3000}{78.54 \times 200000} = \frac{-150000000}{15708000} = -9.55\ \text{mm}$$ $$\Delta L_2 = \frac{P_2 L_2}{AE} = \frac{30000 \times 2000}{78.54 \times 200000} = \frac{60000000}{15708000} = 3.82\ \text{mm}$$ $$\Delta L_3 = \frac{P_3 L_3}{AE} = \frac{100000 \times 1000}{78.54 \times 200000} = \frac{100000000}{15708000} = 6.37\ \text{mm}$$ 6. **Calculate net change in length:** $$\Delta L_{net} = \Delta L_1 + \Delta L_2 + \Delta L_3 = -9.55 + 3.82 + 6.37 = 0.64\ \text{mm}$$ 7. **Interpretation:** The positive net change means the bar elongates by 0.64 mm under the given loading conditions.