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22-Mec-A7 Advanced Strength of Materials · May 2018

Question 7 of 8: Overhang Beam — Sizing a Load by Castigliano’s Theorem

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

Notes on this paper

Paper format. National Exams, May 2018 — 16-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours, eight problems of equal value; a complete paper is any five. All eight problems are solved here as a study resource.

Reference texts. Boresi & Schmidt, Advanced Mechanics of Materials (6th); Ugural & Fenster, Advanced Mechanics of Materials and Applied Elasticity (5th); Timoshenko & Goodier, Theory of Elasticity (3rd); Hibbeler, Mechanics of Materials (10th).

Question 7: Overhang Beam — Sizing a Load 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.

CBA P M₀ = 45000 N·m (CW) 3 m 3 m 5.0 m (to P)
Q7. Simply supported span C–B (pin–roller) with a B–A overhang. P acts 5.0 m from C (2 m past B), and a clockwise couple MA acts at the free tip A.

Given. Pin at C, roller at B (3 m right of C), free tip A (3 m past B, i.e. 6 m from C); downward P at 5.0 m from C; clockwise couple $M_A=45000\ \text{N}\cdot\text{m}$ at A.

Given data
Spans$C\!\to\!B=3\ \text{m}$, $B\!\to\!A=3\ \text{m}$, $P$ at $x=5\ \text{m}$
Tip couple$M_A=45000\ \text{N}\cdot\text{m}$ (clockwise)
Stiffness$E=200\ \text{GPa}$, $I=250\times10^{7}\ \text{mm}^4=2.5\times10^{-3}\ \text{m}^4$; $EI=5.0\times10^{8}\ \text{N}\cdot\text{m}^2$
Limit$\delta_A\le 15\ \text{mm}$ downward

Find. The magnitude and direction of $P$ that brings the tip deflection to exactly 15 mm down.

Approach. Deflection at A is linear in the two loads, $\delta_A=a\,P+b\,M_A$. Use the unit-load form of Castigliano ($\delta_A=\int M\,m\,dx/EI$, with $m$ from a unit downward load at A) to get the two coefficients, then solve $\delta_A=+15\ \text{mm}$ for $P$.

  1. Reactions (in terms of P). Taking the clockwise tip couple as negative and moments about C, $R_B=\dfrac{5P+45000}{3}$ and $R_C=\dfrac{-2P-45000}{3}$; the internal moment closes correctly at the tip, $M(6^-)=-45000\ \text{N}\cdot\text{m}$.
  2. Unit-load moment. A unit downward load at A gives reactions $r_C=-1$, $r_B=+2$, so $m(x)=-x$ on $0\le x\lt 3$ and $m(x)=x-6$ on $3\le x\le6$.
  3. Deflection from the couple alone. Integrating $M\,m/EI$ with $P=0$: $\delta_A^{(M)}=+0.675\ \text{mm}$ (downward). A clockwise tip couple therefore already deflects A down, but by far less than the 15 mm limit.
  4. Deflection coefficient of P. The same integral gives $\partial\delta_A/\partial P=2.133\times10^{-8}\ \text{m/N}$ (downward per newton of downward P).
  5. Solve for P. Setting $\delta_A=0.015\ \text{m}$ down: $P=\dfrac{0.015-0.000675}{2.133\times10^{-8}}$, so $$\boxed{P=671\ \text{kN}\ \text{(downward)}}$$ Because the couple alone gives only 0.7 mm, P must act downward to reach the 15 mm limit.
Results
QuantityValue
Tip deflection from $M_A$ alone0.675 mm (down)
Required force $P$671 kN
Direction of $P$downward
Check: The couple sense controls the sign of the answer. Tracing the drawn arc (tail at top, arrowhead pointing left at the lower right) confirms MA is clockwise, which deflects A downward; a counter-clockwise reading would instead need a larger downward P (about 735 kN). The large P simply reflects the very stiff section ($I=2.5\times10^{-3}\ \text{m}^4$).