Overhang Beam — Force for a Limiting Tip Deflection: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q8

Question 8: Overhang Beam — Force for a Limiting Tip Deflection (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. $E=200$ GPa, $I=805\times10^{6}$ mm$^4$ ($EI=1.61\times10^{8}$ N·m$^2$); pin C at $x=0$, roller B at $x=4$ m, downward load $P$ at $x=6$ m, free tip A at $x=8$ m; clockwise couple $M_A=17\,000$ N·m at A; limit $\delta_A=3$ mm downward.

Find. The magnitude and direction of $P$.

Fig. 8—Beam with pin at C ($x=0$) and roller at B ($x=4$ m); overhang B–A carries $P$ at $x=6$ m and a clockwise couple $M_A$ at the tip A ($x=8$ m).

Approach. The tip deflection is a linear superposition of the effect of $M_A$ and of the unknown $P$. Find the support reactions, then obtain $\delta_A$ by the unit-load (virtual-work) method; set $\delta_A=3$ mm downward and solve for $P$.

Reactions (clockwise $M_A$). Taking counter-clockwise positive and moments about C, $R_B(4)-P(6)-M_A=0$, so $$R_B=\tfrac{6P+17\,000}{4}=1.5P+4250\ \text{N},\qquad R_C=P-R_B=-0.5P-4250\ \text{N}.$$
Tip deflection by superposition. Applying a unit downward virtual load at A and integrating $\int M\,m/EI\,dx$ over the two supported bays and the overhang, the downward tip deflection is linear in $P$: $$\delta_A=\underbrace{1.408\ \text{mm}}_{\text{from }M_A}+\;\big(1.077\times10^{-4}\ \text{mm/N}\big)\,P\quad(\text{downward}).$$ The clockwise couple alone already deflects A downward by $1.408$ mm.
Impose the deflection limit. Setting $\delta_A=3$ mm downward, $$3.000=1.408+1.077\times10^{-4}\,P\;\Rightarrow\;P=\frac{1.592}{1.077\times10^{-4}}=1.479\times10^{4}\ \text{N}.$$
Result. The required load is $$\boxed{P\approx 14.8\ \text{kN, directed downward}.}$$ A downward $P$ on the overhang adds to the downward deflection already caused by $M_A$; a much smaller force than a naive estimate because $M_A$ has done most of the work.

Check / figure reading. The couple $M_A$ is clockwise as drawn at A. A clockwise tip couple deflects A downward, which is why only $\approx14.8$ kN of additional downward $P$ is needed to reach the $3$ mm limit. Had $M_A$ been counter-clockwise it would lift A, and $P$ would need to be $\approx40.9$ kN downward instead.

Quantity	Result
Deflection from $M_A$ alone	$1.41$ mm (down)
Required force $P$	$14.8$ kN
Direction of $P$	Downward

Topic: Deflection of an overhanging beam and inverse design for a limiting displacement.

Key relations: Unit-load theorem $\delta=\int M\,m/(EI)\,dx$; superposition of $M_A$- and $P$-effects; $EI=1.61\times10^{8}$ N·m$^2$.

Why this works: Linear elastic beams superpose, so the tip deflection is an affine function of the single unknown $P$; one evaluation of the $M_A$ effect and one of the $P$ effect (both by the unit-load integral over the real and virtual moment diagrams) give the whole relationship. Setting it equal to the allowable deflection then solves for $P$ directly, without iterating.

Common pitfall: Getting the sense of $M_A$ wrong (it flips the required direction of $P$), or omitting the overhang segment from the moment integral. Also keep consistent units—$EI$ in N·m$^2$ with lengths in metres, so that a metre-based integral yields a deflection in metres.

Source: Hibbeler, Mechanics of Materials §8.4 and Ch. 14 (energy methods, unit-load); Ugural & Fenster Ch. 5 (beam deflections).

Question 8 of 8: Overhang Beam — Force for a Limiting Tip Deflection

Question 8: Overhang Beam — Force for a Limiting Tip Deflection (20 marks)