1. **Problem statement:** A three-hinged parabolic arch with a span of 50 m and a central rise of 4 m is subjected to a uniformly distributed load (UDL) of 20 KN/m over half the span (25 m). We need to find the horizontal moment (H-moment) produced at a distance of 10 m from the left support. 2. **Given data:** - Span, $L = 50$ m - Central rise, $f = 4$ m - UDL, $w = 20$ KN/m over half span (25 m) - Distance from left support, $x = 10$ m 3. **Step 1: Calculate the horizontal thrust $H$ of the arch.** For a three-hinged parabolic arch with a uniformly distributed load over half the span, the horizontal thrust is given by: $$H = \frac{w a^2}{8 f}$$ where $a$ is the length over which the load acts (half span = 25 m). Calculate $H$: $$H = \frac{20 \times 25^2}{8 \times 4} = \frac{20 \times 625}{32} = \frac{12500}{32} = 390.625 \text{ KN}$$ 4. **Step 2: Calculate the bending moment at distance $x=10$ m.** The bending moment $M(x)$ at a distance $x$ from the left support for a three-hinged parabolic arch under partial UDL is: $$M(x) = H y - R_A x + M_{load}$$ where: - $y$ is the rise of the arch at $x$ - $R_A$ is the reaction at the left support - $M_{load}$ is the moment due to the applied load about the point at $x$ 5. **Step 3: Calculate the rise $y$ at $x=10$ m.** The parabolic arch equation (with vertex at mid-span) is: $$y = \frac{4f}{L^2} x (L - x)$$ Calculate $y$: $$y = \frac{4 \times 4}{50^2} \times 10 \times (50 - 10) = \frac{16}{2500} \times 10 \times 40 = \frac{16}{2500} \times 400 = 2.56 \text{ m}$$ 6. **Step 4: Calculate the reaction at left support $R_A$.** Total load $W = w \times a = 20 \times 25 = 500$ KN acting at the midpoint of the loaded length (12.5 m from left support). Taking moments about right support: $$R_A \times 50 = 500 \times (50 - 12.5) = 500 \times 37.5 = 18750$$ $$R_A = \frac{18750}{50} = 375 \text{ KN}$$ 7. **Step 5: Calculate the moment due to the load $M_{load}$ at $x=10$ m.** Since the load acts from 0 to 25 m, and $x=10$ m lies within the loaded region, the load to the left of $x$ is: $$w \times x = 20 \times 10 = 200 \text{ KN}$$ This load acts at $x/2 = 5$ m from the left support. Moment about point at $x=10$ m due to this load is: $$M_{load} = 200 \times (10 - 5) = 200 \times 5 = 1000 \text{ KNm}$$ 8. **Step 6: Calculate the bending moment $M(x)$ at $x=10$ m.** $$M(10) = H y - R_A x + M_{load} = 390.625 \times 2.56 - 375 \times 10 + 1000$$ Calculate each term: $$390.625 \times 2.56 = 1000$$ $$375 \times 10 = 3750$$ So, $$M(10) = 1000 - 3750 + 1000 = -1750 \text{ KNm}$$ The negative sign indicates the direction of the moment. 9. **Step 7: Interpret the result.** The magnitude of the horizontal moment at 10 m is 1750 KNm, which is closest to option (a) 2000 KNm. **Final answer:** 2000 KNm (option a)