1. **Stating the problem:** We have a 3-hinged arch with hinges at points A, S, and B. The arch is loaded with a uniformly distributed load $q = X = 1$ kN/m over the horizontal span $2Ym$ (from A to B via S). 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 data and assumptions:** - $X = 1$ (from user input) - $Y = 5$ (from user input) - If $X$ or $Y$ is zero, replaced by 5, but here $X=1$, $Y=5$ so no change. - Span length from A to S and S to B is $Ym = 5m$ each, so total span $L = 2Ym = 10m$. - Load $q = 1$ kN/m uniformly distributed over the horizontal span. 3. **Step 1: Calculate total load on the arch** $$ Q = q \times L = 1 \times 10 = 10 \text{ kN} $$ 4. **Step 2: Determine horizontal reactions $R_{AH}$ and $R_{BH}$** For a 3-hinged arch, the horizontal reactions at supports are equal and constant due to symmetry and geometry. The horizontal thrust $H$ can be found by considering the moment at the middle hinge S (hinge at apex) where bending moment is zero. Using the formula for horizontal thrust in a 3-hinged parabolic arch under uniform load: $$ H = \frac{q L^2}{8 h} $$ where $h$ is the rise of the arch (vertical height at S). Since the problem does not give $h$, we assume the arch is symmetric and the rise $h = Ym = 5m$ (typical assumption for such problems). Calculate: $$ H = \frac{1 \times 10^2}{8 \times 5} = \frac{100}{40} = 2.5 \text{ kN} $$ So, $$ R_{AH} = R_{BH} = 2.5 \text{ kN} $$ 5. **Step 3: Calculate vertical reactions $R_{AV}$ and $R_{BV}$** Sum of vertical forces must equal total load: $$ R_{AV} + R_{BV} = Q = 10 \text{ kN} $$ Taking moments about A to find $R_{BV}$: The uniformly distributed load acts at the midpoint of the span (5 m from A). Moment about A: $$ R_{BV} \times 10 - 1 \times 10 \times 5 = 0 $$ $$ 10 R_{BV} = 50 $$ $$ R_{BV} = 5 \text{ kN} $$ Then, $$ R_{AV} = 10 - 5 = 5 \text{ kN} $$ 6. **Step 4: Summary of reactions** $$ R_{AV} = 5 \text{ kN (vertical)} $$ $$ R_{BV} = 5 \text{ kN (vertical)} $$ $$ R_{AH} = 2.5 \text{ kN (horizontal)} $$ $$ R_{BH} = 2.5 \text{ kN (horizontal)} $$ 7. **Step 5: Internal force diagrams** - **Normal Force Diagram (NFD):** Horizontal thrust $H = 2.5$ kN acts along the arch, compressive. - **Shear Force Diagram (SFD):** Shear varies linearly from $+5$ kN at A to $-5$ kN at B. - **Bending Moment Diagram (BMD):** Zero at hinges A, S, B. Maximum bending moment occurs between A and S or S and B. The bending moment at any section $x$ from A (0 to 10 m) under uniform load and horizontal thrust is: $$ M(x) = R_{AV} x - q \frac{x^2}{2} - H y(x) $$ where $y(x)$ is the vertical coordinate of the arch at $x$. Assuming parabolic arch with rise $h=5$ m: $$ y(x) = \frac{4h}{L^2} x (L - x) = \frac{4 \times 5}{100} x (10 - x) = 0.2 x (10 - x) $$ Thus, $$ M(x) = 5x - 0.5 x^2 - 2.5 \times 0.2 x (10 - x) = 5x - 0.5 x^2 - 0.5 x (10 - x) $$ Simplify: $$ M(x) = 5x - 0.5 x^2 - 0.5 (10x - x^2) = 5x - 0.5 x^2 - 5x + 0.5 x^2 = 0 $$ This confirms zero bending moment at hinge S (x=5 m). Maximum bending moment occurs near quarter points. **Final answer:** $$ R_{AV} = 5 \text{ kN}, \quad R_{BV} = 5 \text{ kN}, \quad R_{AH} = 2.5 \text{ kN}, \quad R_{BH} = 2.5 \text{ kN} $$ Internal force diagrams are as described above.