Thermal

For a thermal gradient $\varphi$ applied to a beam the curvature is

$$\kappa = \alpha\varphi$$

So the radius of curvature is

$$r = \frac{1}{\kappa}$$

Assuming a circular arc the rotation at the end is perpendicular to the radial lien of the arc. This means that the angle of the radial line to the beam original configuration is

$$\begin{aligned} \cos\beta &= \frac{\frac{l}{2}}{r} \\ \beta &= \cos^{- 1}\left( \frac{l\kappa}{2} \right) \end{aligned}$$

And

$$\phi = \frac{\pi}{2} - \cos^{- 1}\left( \frac{l\kappa}{2} \right)$$

$$\phi = \frac{\pi}{2} - \cos^{- 1}\left( \frac{l\alpha\varphi}{2} \right)$$