22-Mec-A7 Advanced Strength of Materials · December 2019
Question 7 of 7: Deflection of a two-bar pin-connected structure
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2019 — 16-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours. Seven problems of equal value; any five constitute a complete paper. All seven problems are solved as a study resource.
Given. Two pin-ended members meeting at $B$: a horizontal member (1) to the lower wall pin and an inclined member (2) to the upper wall pin; a vertical 45 kN load at $B$.
Given data
Member
Area
Length
Modulus
1 — horizontal $CB$
0.0055 m$^2$
1.5 m
70 GPa
2 — inclined $AB$
0.0035 m$^2$
2.5 m
200 GPa
Load $P$ at $B$
45 kN, vertical (downward)
Find. The horizontal, vertical and resultant displacement of joint $B$.
Two-bar frame: inclined member $AB$ (2) and horizontal member $CB$ (1) meet at $B$ where a 45 kN vertical load is applied. Geometry is a 3–4–5 triangle ($1.5,\ 2.0,\ 2.5$ m).
Approach. Fix the geometry (3–4–5, so the incline runs $0.6$ horizontal to $0.8$ vertical), find the two member forces by joint equilibrium, compute their elongations, then recover the joint displacement from the compatibility of both elongations.
Geometry. The inclined member spans $L_2=2.5$ m over a horizontal run $L_1=1.5$ m, so the vertical height is $\sqrt{2.5^2-1.5^2}=2.0$ m; hence $\cos\theta=0.6$, $\sin\theta=0.8$.
Member forces (joint $B$). Only the inclined member has a vertical component, so it carries the load; the horizontal member balances its horizontal pull:
$$F_2=\frac{P}{\sin\theta}=\frac{45}{0.8}=56.25\ \text{kN (tension)},\qquad F_1=-F_2\cos\theta=-33.75\ \text{kN (compression)}$$
Member elongations, $\delta=FL/AE$:
$$\delta_1=\frac{-33.75\times10^{3}(1.5)}{0.0055(70\times10^{9})}=-0.1315\ \text{mm},\qquad \delta_2=\frac{56.25\times10^{3}(2.5)}{0.0035(200\times10^{9})}=+0.2009\ \text{mm}$$
Compatibility at $B$. Each elongation equals the projection of the joint displacement $(u,v)$ on that member. The horizontal member gives $u=\delta_1$; the inclined member gives $0.6u-0.8v=\delta_2$:
$$u=-0.1315\ \text{mm},\qquad v=\frac{0.6u-\delta_2}{0.8}=-0.3497\ \text{mm}$$
(negative $v$ = downward, negative $u$ = toward the wall).
Resultant deflection.
$$\boxed{\;\delta_B=\sqrt{u^2+v^2}=0.374\ \text{mm}\;}$$
directed down and slightly toward the wall. An independent unit-load check reproduces $v=0.350$ mm and $u=0.131$ mm.
