Two-hinged arch: thrust with rib shortening, support yield

A two-hinged arch has one redundant, the horizontal thrust $H$. Taking $I_x = I_0\sec\alpha$, the compatibility equation with rib shortening and yielding of a support is

$$H = \frac{\displaystyle\int \frac{M_0\,y}{EI_0}\,dx - \delta}{\displaystyle\int \frac{y^2}{EI_0}\,dx + \int \frac{\cos^2\alpha}{EA_c}\,dx}$$

where $\delta$ is the outward yield of support B. (The small axial-force term of the free structure is neglected, as is usual.)

Data (converted to kN and m)

• $L = 60$ m, $h = 6$ m, parabola $y = \dfrac{4h\,x(L-x)}{L^2} = \dfrac{x(60-x)}{150}$
• $EI_0 = 10\,\text{kN/mm}^2 \times 6\times10^3\,\text{cm}^4 = 10^7 \times 6\times10^{-5} = 600$ kNm$^2$
• $EA_c = 10^7 \times 0.1 = 10^6$ kN, $\delta = 6$ mm $= 0.006$ m
• $V_A = V_B = 20$ kN; $M_0 = 20x$ for $0 \le x \le 30$ (peak 600 kNm at the crown)

Integrals

$$\begin{aligned} \int_0^{60} M_0 y\,dx &= 2\int_0^{30} 20x\cdot\frac{x(60-x)}{150}\,dx = 90000 \\ \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 1152 \\ \int_0^{60} \cos^2\alpha\,dx &= 57.08 \end{aligned}$$

($\tan\alpha = \dfrac{dy}{dx} = \dfrac{4h(L-2x)}{L^2}$, so $\cos^2\alpha = \dfrac{1}{1+\tan^2\alpha}$.)

$$\begin{aligned} \frac{\int M_0 y\,dx}{EI_0} &= \frac{90000}{600} = 150.000 \\ \frac{\int y^2\,dx}{EI_0} &= \frac{1152}{600} = 1.9200 \\ \frac{\int \cos^2\alpha\,dx}{EA_c} &= \frac{57.08}{10^6} = 5.708\times10^{-5} \end{aligned}$$

Horizontal thrust

$$H = \frac{150.000 - 0.006}{1.9200 + 5.708\times10^{-5}} = 78.12\ \text{kN}$$

For comparison: without rib shortening and yielding, $H = \dfrac{25PL}{128h} = 78.125$ kN. With the printed $I_0$ the arch is very flexible in bending, so rib shortening and a 6 mm yield change the thrust only slightly. (A stiffer rib, say $I_0 = 6\times10^6$ cm$^4$, would give $H = 72.83$ kN.)

Bending moment $M = M_0 - H y$

The BMD is symmetric about the crown: zero at the hinges A and B, hogging (negative) about the quarter points, and sagging with a peak of 131.28 kNm under the load at the crown.

Answer: $H \approx 78.12$ kN; crown BM $= 131.28$ kNm (sagging), BM at $x = 15$ m $= -51.54$ kNm.