1. **Problem Statement:** Determine the normal stress in cables AB, CD, EF and cable A, and the extension of point E for a rigid bar ACE supported by three steel cables AB, CD, EF. 2. **Given Data:** - Cross-sectional area of each cable, $A = 0.025\,m^2$ - Modulus of elasticity, $E = 210\,MPa = 210 \times 10^6\,Pa$ - Length of each cable, $L = 0.5\,m$ - Horizontal distances: $AC = 0.4\,m$, with $AB$, $CD$, $EF$ vertical - Downward force at midpoint between A and C, $F = 15000\,N$ 3. **Step 1: Calculate reactions at supports** Since the bar is rigid and horizontal, the vertical force is balanced by the tension in the three cables. 4. **Step 2: Determine cable forces using equilibrium** Let tensions in cables AB, CD, EF be $T_{AB}$, $T_{CD}$, $T_{EF}$. Sum of vertical forces: $$T_{AB} + T_{CD} + T_{EF} = 15000$$ Taking moments about point A (left end): $$T_{CD} \times 0.2 + T_{EF} \times 0.4 = 15000 \times 0.2$$ Taking moments about point C (middle): $$T_{AB} \times 0.2 = T_{EF} \times 0.2$$ From moment about C: $$T_{AB} = T_{EF}$$ Substitute $T_{AB} = T_{EF} = T$: From vertical forces: $$T + T_{CD} + T = 15000 \Rightarrow 2T + T_{CD} = 15000$$ From moment about A: $$T_{CD} \times 0.2 + T \times 0.4 = 3000$$ 5. **Step 3: Solve system of equations** From moment about A: $$0.2 T_{CD} + 0.4 T = 3000$$ From vertical forces: $$T_{CD} = 15000 - 2T$$ Substitute into moment equation: $$0.2 (15000 - 2T) + 0.4 T = 3000$$ $$3000 - 0.4 T + 0.4 T = 3000$$ This simplifies to: $$3000 = 3000$$ This means the system is statically indeterminate; however, since the bar is rigid and cables have equal length, tensions in AB and EF are equal, and CD carries the remaining load. Assuming equal elongation in cables (compatibility), tensions can be found by: 6. **Step 4: Calculate normal stresses** Normal stress in a cable is: $$\sigma = \frac{T}{A}$$ Since $T_{AB} = T_{EF} = T$, and $T_{CD} = 15000 - 2T$, we need $T$. 7. **Step 5: Calculate extension of cables** Extension $\delta$ is: $$\delta = \frac{T L}{A E}$$ Since the bar is rigid, extensions of cables AB and EF are equal, and extension of CD is equal as well. Set extensions equal: $$\delta_{AB} = \delta_{CD} = \delta_{EF}$$ So: $$\frac{T L}{A E} = \frac{(15000 - 2T) L}{A E}$$ Simplifies to: $$T = 15000 - 2T \Rightarrow 3T = 15000 \Rightarrow T = 5000\,N$$ Then: $$T_{AB} = T_{EF} = 5000\,N$$ $$T_{CD} = 15000 - 2 \times 5000 = 5000\,N$$ 8. **Step 6: Calculate normal stresses:** $$\sigma_{AB} = \sigma_{CD} = \sigma_{EF} = \frac{5000}{0.025} = 200000\,Pa = 0.2\,MPa$$ 9. **Step 7: Calculate extension at point E (cable EF):** $$\delta_E = \frac{T_{EF} L}{A E} = \frac{5000 \times 0.5}{0.025 \times 210 \times 10^6} = \frac{2500}{5.25 \times 10^6} = 4.76 \times 10^{-4} m = 0.476\,mm$$ **Final answers:** - Normal stress in cables AB, CD, EF: $0.2\,MPa$ - Normal stress at cable A (same as AB): $0.2\,MPa$ - Extension of point E: $0.476\,mm$