Two-hinged arch: BM at 10 m intervals

A two-hinged arch is statically indeterminate to the first degree. The horizontal thrust $H$ is the redundant, found from the condition that the horizontal movement of the hinge B is zero:

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

For $I=I_c\sec\theta$ we have $ds/I = dx/I_c$ ($ds=dx\sec\theta$), so $H=\int M_0 y\,dx\,/\int y^2dx$.

Data

Span $L=60$ m, rise $h=8$ m, load $P=80$ kN at $x=20$ m from A. Parabola:

$$y=\frac{4h\,x(L-x)}{L^2}=\frac{4\times8\,x(60-x)}{3600}$$

Reactions and simple-beam moment $M_0$

$$V_A=\frac{80\times40}{60}=53.33\ \text{kN},\qquad V_B=80-53.33=26.67\ \text{kN}$$

$M_0 = 53.33x$ for $x\le20$ and $M_0=26.67(60-x)$ for $x\ge20$.

Horizontal thrust

$$\int_0^{60} M_0\,y\,dx=208592.6\ \text{kN m}^3,\qquad \int_0^{60} y^2dx=\frac{8h^2L}{15}=2048.0\ \text{m}^3$$ $$H=\frac{208592.6}{2048.0}=101.85\ \text{kN}$$

(Check with the standard result $H=\dfrac{5PL\,\alpha(1-2\alpha^2+\alpha^3)}{8h}$ with $\alpha=1/3$: same value.)

Bending moment $M=M_0-Hy$ at 10 m intervals

Sign: positive = sagging (tension on the lower side).

The BM is zero at both hinges, reaches its maximum +342.4 kNm under the load, changes sign at about x = 29.5 m (where $M=(60-x)(26.667-0.9053x)=0$) and is hogging beyond that, with a numerical maximum of about -211 kNm near x = 44.7 m. The values at 10 m intervals plot the BM diagram: sagging on the left of x = 29.5 m (apart from a negligible hogging moment, less than 0.3 kNm, within about 1.1 m of A) and hogging on the right.

Answer: $H=101.85$ kN; BM at 0, 10, 20, 30, 40, 50, 60 m = 0.00, 80.66, 342.39, -14.81, -190.95, -186.01, 0.00 kNm.