22-Mec-A7 Advanced Strength of Materials · May 2015
Question 2 of 8: Overhang-Beam Tip-Deflection Limit
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2015 — 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours. Eight problems of equal value; any five constitute a complete paper. All eight are solved here.
Reference texts. Ugural & Fenster, Advanced Strength and Applied Elasticity (4th ed.); Boresi & Schmidt, Advanced Mechanics of Materials (6th ed.); Hibbeler, Mechanics of Materials (10th ed.); Timoshenko & Goodier, Theory of Elasticity.
Given. A determinate overhang beam (pin at C, roller at B) with a downward load P on the overhang and a clockwise tip couple.
Given data
Backspan C–B, L
2.0 m
Overhang B–A, e
2.0 m
Position of P from B, c
1.0 m
Tip couple, MA
15 000 N·m (clockwise)
Flexural rigidity, EI
180×109 × 805×10-6 = 1.449×105 kN·m2
Deflection limit at A
7 mm (down)
Find. The magnitude and direction of P so that the downward tip deflection at A equals its 7 mm limit.
Overhang beam: simple supports at C and B, overhang tip A carrying the clockwise couple; P acts 1 m past B.
Approach. Both P and the couple deflect the tip; superpose their tip deflections (each found by the unit-load method on the released beam) and set the sum equal to 7 mm down. A clockwise tip couple and a downward overhang load both push A down.
Tip deflection from the overhang load P. Carrying P to B produces a hogging moment that rotates the backspan (slope $\theta_B=\tfrac{PcL}{3EI}$) and the overhang bends as a cantilever; combining, $$\delta_A^{P}=\frac{P\,[\,2cLe+c^2(3e-c)\,]}{6EI}=\frac{P\,[\,2(1)(2)(2)+1^2(6-1)\,]}{6EI}=\frac{13P}{6EI}\;(\text{down}).$$
Tip deflection from the couple. A tip couple $M_A$ carries a constant moment along the overhang, giving $$\delta_A^{M}=\frac{M_A\,e\,(2L+3e)}{6EI}=\frac{M_A(2)(4+6)}{6EI}=\frac{20\,M_A}{6EI}\;(\text{down}).$$ With $M_A=15\,000$ N·m this is $\delta_A^{M}=0.345$ mm down.
Impose the deflection limit. Setting $\delta_A^{P}+\delta_A^{M}=7$ mm down, $$\frac{13P}{6EI}+0.345\times10^{-3}=7\times10^{-3}\ \Rightarrow\ P=\frac{6EI\,(6.655\times10^{-3})}{13}.$$ With $EI=1.449\times10^{5}$ kN·m$^2$, $$\boxed{P=445\ \text{kN, directed downward.}}$$ P must act downward: the couple alone deflects A only 0.345 mm down, so a downward force is required to reach the full 7 mm.
