1. **Stating the problem:** We have a timber member fixed at point A with three forces acting along its length at various points (B, C, D). Given the cross-sectional area $A = 1750$ mm² and modulus of elasticity $E = 12$ GPa, we need to find the change in the total length of the member after these loads are applied. 2. **Convert Units:** - Cross-sectional area: $A = 1750$ mm² = $1750 \times 10^{-6}$ m² = $0.00175$ m² - Modulus of elasticity: $E = 12$ GPa = $12 \times 10^{9}$ Pa 3. **Calculate the axial force in each segment:** The member is fixed at A, so let's analyze axial forces from left to right. - At point B (1.5 m): Force $+40$ kN (to the right) - At point C (4.5 m): Force $-35$ kN (to the left) - At point D (7.5 m): Force $+20$ kN (to the right) Starting from point A to B (segment AB): Only the 40 kN force acts, so internal force $P_{AB} = 40$ kN = $40000$ N. Segment BC (1.5 m to 4.5 m): Forces to the right minus forces to the left up to that segment: $P_{BC} = 40kN - 35kN = 5$ kN = $5000$ N (tension). Segment CD (4.5 m to 7.5 m): Forces to the right minus left: $P_{CD} = 40kN - 35kN + 20kN = 25$ kN = $25000$ N. 4. **Lengths of the segments:** - $L_{AB} = 1.5$ m - $L_{BC} = 3$ m - $L_{CD} = 3$ m 5. **Calculate elongation in each segment using:** $$\Delta L = \frac{P L}{A E}$$ where $P$ in newtons, $L$ in meters, $A$ in square meters, and $E$ in pascals. - $\Delta L_{AB} = \frac{40000 \times 1.5}{0.00175 \times 12 \times 10^{9}} = \frac{60000}{21000000} = 0.002857$ m = 2.857 mm - $\Delta L_{BC} = \frac{5000 \times 3}{0.00175 \times 12 \times 10^{9}} = \frac{15000}{21000000} = 0.0007143$ m = 0.7143 mm - $\Delta L_{CD} = \frac{25000 \times 3}{0.00175 \times 12 \times 10^{9}} = \frac{75000}{21000000} = 0.003571$ m = 3.571 mm 6. **Total elongation:** $$\Delta L_{total} = 2.857 + 0.7143 + 3.571 = 7.1423 \text{ mm}$$ **Answer:** The total change in length of the timber member is approximately **7.14 mm**.