Two-hinged arch: BM, thrust, radial shear, support yield

A two-hinged arch has one redundant, the horizontal thrust $H$. With $I = I_c\sec\theta$ the compatibility equation (support B yields outward by $\delta$) is

$$H = \frac{\displaystyle\frac{1}{EI_c}\int M_0\,y\,dx - \delta}{\displaystyle\frac{1}{EI_c}\int y^2\,dx} = H_{load} - \frac{\delta\,EI_c}{\int y^2\,dx}$$

Data

Span $L = 120$ m, rise $h = 20$ m, $EI_c = 100\times10^6$ kNm$^2$, $\delta = 10$ mm $= 0.01$ m; UDL $10$ kN/m on CB (60 m to 120 m).

$$y = \frac{4h\,x(L-x)}{L^2} = \frac{x(120-x)}{180},\qquad \tan\theta = \frac{dy}{dx} = \frac{120 - 2x}{180}$$

Simple-beam reactions and $M_0$

UDL resultant $= 10\times60 = 600$ kN at 90 m: $V_B = \dfrac{600(90)}{120} = 450$ kN, $V_A = 150$ kN.

• $0 \le x \le 60$: $M_0 = 150\,x$
• $60 \le x \le 120$: $M_0 = 150\,x - 5(x-60)^2$

Thrust

$$\int_0^{120} y^2\,dx = \frac{8h^2L}{15} = 25600,\qquad \int_0^{120} M_0\,y\,dx = 11520000$$ $$H_{load} = \frac{11520000}{25600} = 450.00\ \text{kN}$$ $$H_{yield} = -\frac{\delta\,EI_c}{\int y^2dx} = -\frac{0.01\times100\times10^6}{25600} = -39.06\ \text{kN}$$ $$H = 450.00 + (-39.06) = 410.94\ \text{kN}$$

Section D ($x = 30$ m)

$$\begin{aligned} y_D &= \frac{30(90)}{180} = 15\ \text{m},\qquad \tan\theta_D = \frac{120-60}{180} = \frac13,\ \ \theta_D = 18.43^\circ \\ M_0 &= V_A(30) = 4500\ \text{kNm},\qquad V_0 = V_A = 150\ \text{kN} \end{aligned}$$

Bending moment

$$M_D = M_0 - Hy_D = 4500 - 410.94(15) = -1664.06\ \text{kNm}$$

(Due to the load alone: $-2250.00$ kNm; the yield adds $+585.94$ kNm.)

Normal thrust

$$N_D = H\cos\theta + V_0\sin\theta = 410.94(0.9487) + 150(0.3162) = 437.28\ \text{kN}$$

Radial shear

$$Q_D = V_0\cos\theta - H\sin\theta = 150(0.9487) - 410.94(0.3162) = 12.35\ \text{kN}$$

Bending moment diagram $M = M_0 - Hy$ (kNm)

The BMD is hogging for $0 < x < 54.3$ m (largest hogging about $-1683$ kNm near $x = 27$ m) and sagging for the rest, with the largest sagging moment about $2852$ kNm near $x = 88$ m, under the loaded half. It is zero at both hinges.

Answer: at D, $M = -1664.06$ kNm (hogging), $N = 437.28$ kN (compression), radial shear $Q = 12.35$ kN; $H = 410.94$ kN.