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22-Mec-A7 Advanced Strength of Materials · December 2017

Question 8 of 8: Allowable Force on an Overhanging Beam by Castigliano

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

Notes on this paper

Paper format. National Exams, December 2017 — 16-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours; any five of eight problems constitute a complete paper (all problems of equal value). All eight problems are solved 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 8: Allowable Force on an Overhanging 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.

C A B 10 kN/m P M = 39 kN·m 5 m 3 m
Q8. Overhanging beam: pin at C, roller at A, point B 5 m from C. Triangular load 0 at C to 10 kN/m at B, downward P at B, CCW couple M = 39 kN·m at A.

Given. A beam pinned at C and on a roller at A, with point B 5 m from C and 3 m from A.

Given data
Spans$CB = 5\ \text{m}$, $BA = 3\ \text{m}$ (total $CA = 8\ \text{m}$)
Distributed loadtriangular, $0$ at C to $10\ \text{kN/m}$ at B
Applied couple at A$M = 39\ \text{kN}\cdot\text{m}$ (counter-clockwise)
Section$E=195\ \text{GPa}$, $I=975\times10^{6}\ \text{mm}^4$, $EI=1.901\times10^{14}\ \text{N}\cdot\text{mm}^2$

Find. The magnitude and direction of $P$ that limits the deflection at B to 1 mm.

Approach. Carry $P$ (downward, at B) as a symbol. Find the reactions, write $M(x)$ over the two segments, and apply Castigliano $\delta_B=\partial U/\partial P=\frac{1}{EI}\int M\,(\partial M/\partial P)\,dx$. The result is linear in $P$; set $|\delta_B|=1\ \text{mm}$.

  1. Reactions. The triangular load totals $W=\tfrac12(10)(5)=25\ \text{kN}$ at $\bar x=\tfrac23(5)=3.33\ \text{m}$ from C. Moments about C (CCW +) give $R_A=\dfrac{25(3.33)+5P-39}{8}$, and $R_C=25+P-R_A$ (with $P$ in kN).
  2. Bending moment. For $0\le x\le5\,\text{m}$, $M=R_Cx-\dfrac{(10/5)x^3}{6}$; for $5\le x\le8\,\text{m}$, $M=R_Cx-25(x-3.33)-P(x-5)$. The internal moment at A closes to the applied $39\ \text{kN}\cdot\text{m}$, confirming the reactions.
  3. Castigliano integral. Differentiating and integrating $M\,(\partial M/\partial P)/EI$ over both segments yields a linear relation $$\delta_B=1.93\ \text{mm}\ (\downarrow)\ +\ 0.0493\,P\ \text{[kN]}\ \text{mm}.$$
  4. Interpret the limit. With $P=0$ the distributed load and couple alone deflect B by $1.93\ \text{mm}$ downward — already beyond the 1 mm limit — so $P$ must act upward to reduce it.
  5. Solve. Setting $\delta_B=+1.0\ \text{mm}$ (down) with $P$ negative (upward): $$1.0=1.93-0.0493\,P_{\uparrow}\ \Rightarrow\ \boxed{P=18.8\ \text{kN, directed upward}}.$$
Results — allowable force at B
QuantityValue
Deflection at B with $P=0$1.93 mm downward
Required $P$ (for $\delta_B=1$ mm)18.8 kN
Direction of $P$upward
Check: The 1 mm limit is a two-sided bound. An upward $P$ between about 18.8 kN (B deflects 1 mm down) and 59.4 kN (B deflects 1 mm up) keeps $|\delta_B|\le1\ \text{mm}$; the minimum required force is 18.8 kN up. A downward $P$ would only worsen the already-excessive downward deflection.
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