1. **Stating the problem:** We have a three-hinged arch with hinges at points A, B, and S. The arch is loaded with a uniformly distributed load $q = X$ kN/m over the horizontal span $AB$ of length $2Y$ meters. The height from $AB$ to the apex $S$ is $Y$ meters. We need to find the reaction forces at supports $A$ and $B$: vertical reactions $R_{AV}$ and $R_{BV}$, horizontal reactions $R_{AH}$ and $R_{BH}$, and then draw the internal force diagrams (Normal Force Diagram (NFD), Shear Force Diagram (SFD), and Bending Moment Diagram (BMD)). 2. **Given values:** - $X = 5$ (since original $X=0$ replaced by 5) - $Y = 5$ (since original $Y=0$ replaced by 5) - Uniform load $q = 5$ kN/m - Span $AB = 2Y = 10$ m 3. **Step 1: Calculate horizontal reactions $R_{AH}$ and $R_{BH}$** For a three-hinged arch, the horizontal reactions at supports are equal and opposite: $$R_{AH} = R_{BH} = H$$ The horizontal thrust $H$ can be found by taking moment about the apex $S$ and using equilibrium. 4. **Step 2: Calculate vertical reactions $R_{AV}$ and $R_{BV}$** Sum of vertical forces: $$R_{AV} + R_{BV} = q \times AB = 5 \times 10 = 50 \text{ kN}$$ 5. **Step 3: Moment equilibrium about point A** Taking moments about $A$ (counterclockwise positive): $$\sum M_A = 0 = -q \times AB \times \frac{AB}{2} + R_{BV} \times AB$$ $$-5 \times 10 \times 5 + R_{BV} \times 10 = 0$$ $$-250 + 10 R_{BV} = 0$$ $$10 R_{BV} = 250$$ $$R_{BV} = 25 \text{ kN}$$ 6. **Step 4: Calculate $R_{AV}$** $$R_{AV} = 50 - 25 = 25 \text{ kN}$$ 7. **Step 5: Calculate horizontal reaction $H$ using moment about $S$** Taking moments about $S$ (apex): The horizontal reactions create moments due to vertical distances. Sum of moments about $S$: $$\sum M_S = 0 = R_{AV} \times Y - R_{BV} \times Y - q \times AB \times \frac{AB}{2}$$ But this is not correct because the horizontal reactions create moments about $S$ due to vertical components. Actually, for three-hinged arches, the horizontal thrust $H$ is found by the moment equilibrium about the hinge $S$ considering the vertical load and geometry: $$H = \frac{q L^2}{8 h}$$ Where: - $L = AB = 10$ m - $h = Y = 5$ m Calculate: $$H = \frac{5 \times 10^2}{8 \times 5} = \frac{5 \times 100}{40} = \frac{500}{40} = 12.5 \text{ kN}$$ So, $$R_{AH} = R_{BH} = 12.5 \text{ kN}$$ 8. **Step 6: Summary of reactions** - $R_{AV} = 25$ kN (upward) - $R_{BV} = 25$ kN (upward) - $R_{AH} = 12.5$ kN (to the right at A) - $R_{BH} = 12.5$ kN (to the left at B) 9. **Step 7: Internal force diagrams** - **Normal Force Diagram (NFD):** Horizontal thrust $H = 12.5$ kN compresses the arch. - **Shear Force Diagram (SFD):** Shear varies linearly from $+25$ kN at A to $-25$ kN at B due to uniform load. - **Bending Moment Diagram (BMD):** Maximum bending moment occurs at the hinge $S$ and is zero at supports. The bending moment at $S$ is zero because it is a hinge. --- **Final answers:** $$R_{AV} = 25 \text{ kN}, \quad R_{BV} = 25 \text{ kN}, \quad R_{AH} = 12.5 \text{ kN}, \quad R_{BH} = 12.5 \text{ kN}$$