Two-hinged parabolic arch: BM at D

A two-hinged arch has one redundant, the horizontal thrust $H$. With $I_x = I_c\sec\theta$ we have $ds/I = dx/I_c$, so the thrust is

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

where $M_0$ is the bending moment in the equivalent simply supported beam of the same span.

Data and arch profile

Span $L = 60$ m, rise $h = 10$ m, A at the left support (origin). The UDL acts on $0 \le x \le 30$ m and the 50 kN load acts at $x = 50$ m (10 m from B).

$$y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{90}$$

Reactions of the simple beam

UDL resultant $= 30 \times 30 = 900$ kN at $x = 15$ m.

$$V_B = \frac{900(15) + 50(50)}{60} = 266.67\ \text{kN}, \qquad V_A = 950 - 266.67 = 683.33\ \text{kN}$$

Free bending moment $M_0$

Integrals

$$\begin{aligned} \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 3200 \\ \int_0^{30} M_0 y\,dx &= 1347500, \quad \int_{30}^{50} M_0 y\,dx = 855556, \quad \int_{50}^{60} M_0 y\,dx = 51852 \\ \int_0^{60} M_0 y\,dx &= 2254907 \end{aligned}$$ $$H = \frac{2254907}{3200} = 704.66\ \text{kN}$$

Bending moment at D ($x = 50$ m)

$$\begin{aligned} y_D &= \frac{50(10)}{90} = 5.556\ \text{m} \\ M_0 &= V_B \times 10 = 2666.67\ \text{kNm} \\ M_D &= M_0 - H\,y_D = 2666.67 - 704.66(5.556) = -1248.10\ \text{kNm} \end{aligned}$$

Answer: $H = 704.66$ kN and the bending moment at D is $M_D = -1248.10$ kNm (hogging, tension on the top/extrados).