Two-hinged arch: horizontal reaction, UDL on half

For a two-hinged arch the thrust $H$ makes the horizontal displacement of the support zero. With $I=I_c\sec\theta$ (so $ds/I=dx/I_c$):

$$H=\frac{\int_0^L M_0\,y\,dx}{\int_0^L y^2\,dx}$$

Data

$L=50$ m, $h=8$ m, $w=12$ kN/m on the left half (0 to 25 m). Parabola: $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{32x(50-x)}{2500}$.

Reactions and $M_0$

Total load $=12\times25=300$ kN acting at 12.5 m from A.

$$V_A=\frac{300\times(50-12.5)}{50}=225\ \text{kN},\qquad V_B=75\ \text{kN}$$

• For $0\le x\le25$: $M_0=225x-6x^2$
• For $25\le x\le50$: $M_0=75(50-x)$

Integrals

$$\int_0^{50}M_0\,y\,dx=400000\ \text{kN m}^3,\qquad \int_0^{50}y^2dx=\frac{8h^2L}{15}=\frac{8\times64\times50}{15}=1706.67\ \text{m}^3$$ $$H=\frac{400000}{1706.67}=234.375\ \text{kN}$$

Check: by symmetry a UDL on the half span gives half of the full-span thrust, $\tfrac12\cdot\dfrac{wL^2}{8h}=\tfrac12\times\dfrac{12\times2500}{64}=234.375$ kN. ✓

Answer: horizontal reaction $H=234.38$ kN at each support (inward). Vertical reactions: $V_A=225$ kN and $V_B=75$ kN.