Compatibility equation for two-hinged arch

A two-hinged arch is indeterminate to the first degree. Take the horizontal thrust $H$ as the redundant, release the hinge at B (roller), and write that the net horizontal movement of B is equal to the yield of the support.

          C
       .-----.
     /         \
   A o           o B  <-- H (redundant)
   |<----- L ----->|

Displacement of B in the released structure (outward positive)

1. External loads: $\Delta_{10} = \displaystyle\int \frac{M_0\,y}{EI}\,ds$ ($M_0$ = BM of the equivalent simple beam, $y$ = ordinate of the arch axis). The axial-force term is small and is usually neglected.
2. Temperature rise $t$: $\Delta_{1t} = \alpha\,t\,L$ (a fall in temperature gives a negative value).
3. Redundant $H$ (unit horizontal force at B gives $m = -y$ and $N = -\cos\theta$):

$$\Delta_{1H} = -H\,f_{11}, \qquad f_{11} = \int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\theta}{AE}\,ds$$

The first term of $f_{11}$ is bending, and the second is the rib shortening due to the axial thrust $H\cos\theta$.

4. Yielding of support: B moves outward by $\delta$ (positive outward).

Compatibility equation

The final horizontal movement of B must equal the support yield $\delta$:

$$\Delta_{10} + \alpha\,t\,L - H\,f_{11} = \delta$$ $$\boxed{H = \frac{\displaystyle\int \frac{M_0\,y}{EI}\,ds + \alpha\,t\,L - \delta}{\displaystyle\int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\theta}{AE}\,ds}}$$

Remarks

• For $I = I_c\sec\theta$ the bending integrals become $\dfrac{1}{EI_c}\int M_0\,y\,dx$ and $\dfrac{1}{EI_c}\int y^2\,dx$, and with $A = A_c\sec\theta$ the rib shortening term is $\dfrac{1}{EA_c}\int\cos^2\theta\,dx$.
• A temperature rise or an inward movement of B increases $H$. An outward yield and rib shortening reduce $H$ (rib shortening acts like an elastic spring in the denominator).
• After $H$ is found, the moment at any section is $M = M_0 - Hy$, the normal thrust is $N = H\cos\theta + V_0\sin\theta$ and the radial shear is $V_r = V_0\cos\theta - H\sin\theta$.