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22-Mec-A7 Advanced Strength of Materials · Undated paper

Question 1 of 8: Semicircular curved beam by Castigliano

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exam — 16-Mec-A7 Advanced Strength of Materials (May 2019). Open-book; eight problems of equal value; any five constitute a complete paper. Every problem is solved as a study resource.

Reference texts: Boresi & Schmidt, Advanced Mechanics of Materials (6th ed.); Ugural & Fenster, Advanced Strength and Applied Elasticity; Timoshenko & Goodier, Theory of Elasticity; Hibbeler, Mechanics of Materials.

Question 1: Semicircular curved beam by Castigliano (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 curved cantilever, fixed at end B and free at end A, carries a horizontal force P at A. Its horizontal extension at A is limited to 0.1 mm.

Given data
Mean radius, R510 mm
Second moment of area, I$815\times10^{6}\ \text{mm}^4$
Young's modulus, E205 GPa
Load linehorizontal at free end A
Limit on horizontal deflection at A0.1 mm

Find. (a) the largest admissible P; (b) the magnitude and direction of the accompanying vertical deflection at A.

PABR
Semicircular curved beam, fixed at B, horizontal load P at free end A.

Approach. Use Castigliano's second theorem — only bending energy is retained for a slender arc — taking the bending moment as a function of the polar angle and integrating around the semicircle; a dummy vertical load at A extracts the vertical deflection.

  1. Bending moment around the arc. With the origin at the centre and the angle $\theta$ measured from B, the horizontal load P at A gives an internal moment $$M(\theta)=P R\sin\theta,\qquad 0\le\theta\le\pi,\qquad ds=R\,d\theta.$$
  2. Horizontal deflection (Castigliano). The deflection at A along P is $$\delta_H=\frac{1}{EI}\int_0^{\pi} M\,\frac{\partial M}{\partial P}\,R\,d\theta=\frac{PR^3}{EI}\int_0^{\pi}\sin^2\theta\,d\theta=\frac{\pi P R^{3}}{2EI}.$$
  3. Solve for the allowable load. Setting $\delta_H=0.1\ \text{mm}$ with $EI=(205\times10^{3})(815\times10^{6})=1.671\times10^{14}\ \text{N}\cdot\text{mm}^2$ and $R^{3}=1.327\times10^{8}\ \text{mm}^3$, $$P=\frac{2EI\,\delta_H}{\pi R^{3}}=\boxed{80.2\ \text{kN}}.$$
  4. Vertical deflection. Adding a dummy vertical load Q at A gives $\partial M/\partial Q=-R(1+\cos\theta)$, so $$\delta_V=\frac{PR^3}{EI}\int_0^{\pi}\sin\theta\,(1+\cos\theta)\,d\theta=\frac{2PR^{3}}{EI}=\boxed{0.127\ \text{mm (upward)}}.$$

The two deflections are locked in the ratio $\delta_V/\delta_H=4/\pi=1.273$, so limiting the horizontal movement automatically fixes the vertical one. The vertical component comes out positive in the direction of the upward dummy load, i.e. point A rises as the arc straightens under the pull.

Final results
Allowable load, P80.2 kN
Vertical deflection at A0.127 mm, upward
Ratio $\delta_V/\delta_H$$4/\pi=1.273$
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