Question 3 of 8: Semicircular Curved Beam by Castigliano’s Theorem

Question 3: Semicircular Curved Beam by Castigliano’s Theorem (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A semicircular beam fixed at B, loaded by a horizontal force $P$ at the free end A.

Find. (a) allowable $P$; (b) magnitude and direction of the vertical deflection at A.

Approach. For a slender curved beam only bending energy is significant. Parametrize the arc by the polar angle $\theta$ (from B at $\theta=0$ to A at $\theta=\pi$); write $M(\theta)$ from the end loads and apply Castigliano, using a dummy vertical load $Q$ at A for the vertical component.

1. Bending moment. A section at angle $\theta$ carries only the end load $P$ (horizontal, at A): $M(\theta)=-P R\sin\theta$. Adding a dummy vertical $Q$ at A gives $M=-PR\sin\theta-QR(1+\cos\theta)$.
2. Horizontal deflection (Castigliano). With $ds=R\,d\theta$ and $\partial M/\partial P=-R\sin\theta$, $$\delta_H=\frac{1}{EI}\int_0^{\pi}\!M\frac{\partial M}{\partial P}R\,d\theta=\frac{PR^3}{EI}\int_0^{\pi}\sin^2\theta\,d\theta=\frac{\pi P R^3}{2EI}.$$
3. Allowable force (part a). Setting $\delta_H=0.1\ \text{mm}$: $$P=\frac{2EI\,\delta_H}{\pi R^3}=\frac{2(1.62\times10^{14})(0.1)}{\pi(500)^3}=\boxed{82.5\ \text{kN}}$$
4. Vertical deflection (part b). With $\partial M/\partial Q=-R(1+\cos\theta)$ evaluated at $Q=0$, $$\delta_V=\frac{1}{EI}\int_0^{\pi}\!(-PR\sin\theta)\bigl[-R(1+\cos\theta)\bigr]R\,d\theta=\frac{2PR^3}{EI}.$$
5. Evaluate. $\delta_V=\dfrac{2(82\,500)(500)^3}{1.62\times10^{14}}=\boxed{0.127\ \text{mm, directed upward}}$ (the $+Q$ sense), i.e. $\delta_V=\tfrac{4}{\pi}\delta_H$.