Two-hinged arch: horizontal reaction, BM at C

A two-hinged arch has one redundant, the horizontal reaction (thrust) $H$. With $I = I_c\sec\theta$, $ds/I = dx/I_c$, so

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

Data and profile

Span $L = 60$ m, rise $h = 5$ m, UDL 50 kN/m on the right half CB ($30 \le x \le 60$, with A at the origin).

$$y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{180}$$

Reactions of the simple beam

UDL resultant $= 50\times30 = 1500$ kN at $x = 45$ m.

$$V_B = \frac{1500(45)}{60} = 1125\ \text{kN},\qquad V_A = 1500 - 1125 = 375\ \text{kN}$$

Free bending moment

• $0 \le x \le 30$: $M_0 = 375\,x$
• $30 \le x \le 60$: $M_0 = 375\,x - 25(x-30)^2$

Integrals

$$\begin{aligned} \int_0^{60}y^2dx &= \frac{8h^2L}{15} = \frac{8(25)(60)}{15} = 800 \\ \int_0^{30}M_0y\,dx &= 703125, \qquad \int_{30}^{60}M_0y\,dx = 1096875 \\ \int_0^{60}M_0y\,dx &= 1800000 \end{aligned}$$

Horizontal reaction

$$H = \frac{1800000}{800} = 2250.0\ \text{kN}$$

Bending moment at the crown C ($x = 30$ m)

$$\begin{aligned} M_0 &= V_A(30) = 375(30) = 11250\ \text{kNm} \\ y_C &= h = 5\ \text{m} \\ M_C &= M_0 - H\,y_C = 11250 - 2250.0(5) = 0.0\ \text{kNm} \end{aligned}$$

The crown moment is zero for this loading, which is a known property of a parabolic two-hinged arch carrying a UDL over half the span.

Answer: $H = 2250.0$ kN; $M_C = 0.0$ kNm.