Two-hinged parabolic arch: horizontal reaction, symmetric loads

Method: a two-hinged arch is indeterminate to the first degree; the horizontal thrust $H$ is the redundant. Using the condition that the horizontal displacement of the support is zero (strain energy, bending only):

$$H=\frac{\int M_0\,y\,\dfrac{ds}{EI}}{\int y^2\,\dfrac{ds}{EI}}$$

With the secant variation $I=I_c\sec\theta$, $ds/I=dx/I_c$, so

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

Data

$L=80$ m, rise $h=5$ m, parabolic axis $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{20\,x(80-x)}{6400}$. Loads (kN) at 10 m spacing: 4, 10, 8, 10, 8, 10, 4 at $x=10,20,\ldots,70$ m (total 54 kN).

Reactions

By symmetry $V_A=V_B=27$ kN.

Simple-beam moment $M_0$ and ordinate $y$

$M_0$ is linear between the loads, for example $M_0(10)=27\times10=270$, $M_0(20)=27\times20-4\times10=500$, $M_0(30)=27\times30-4\times20-10\times10=630$, $M_0(40)=680$.

Integrals

$$\int_0^{80}M_0\,y\,dx=140562.5\ \text{kN m}^3,\qquad \int_0^{80}y^2dx=\frac{8h^2L}{15}=\frac{8\times25\times80}{15}=1066.67\ \text{m}^3$$ $$H=\frac{140562.5}{1066.67}=131.78\ \text{kN}$$

The bending moment at the crown is $M_C=680-131.78\times5=21.11$ kNm (small compared with $M_0$, which shows how effectively the thrust reduces the moments).

Answer: horizontal reaction $H=131.78$ kN at each support (inward); $V_A=V_B=27$ kN.