22-Mec-A7 Advanced Strength of Materials · December 2016
Question 1 of 8: Displacement of a three-element pin-jointed truss
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 · 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 as a study resource.
Reference texts. Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed.; Ugural & Fenster, Advanced Strength and Applied Elasticity, 4th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Timoshenko & Gere, Theory of Elastic Stability.
Check: the figure for Question 1 labels the applied load P = 7000 N, whereas the printed text reads “7800 N”. The drawn value on the exam figure governs, so 7000 N is used throughout.
Question 1: Displacement of a three-element pin-jointed truss (20 marks)
Q1 — three members meet at the single free joint B: AB (0.80 m, horizontal), CB (1.00 m diagonal) and DB (0.60 m, vertical). A, C, D are pinned supports.
Given. Three members frame the one free joint B; supports at A, C, D are pinned.
Given data
Applied load at B
P = 7000 N (horizontal, +x)
Member area (all)
A = 8.8 cm² = 8.8×10−⁴ m²
Modulus (all)
E = 88 GPa → EA = 7.744×10⁷ N
Geometry
A(−0.80, 0), C(−0.80, −0.60), D(0, −0.60), B(0, 0) m
Find. The horizontal (u) and vertical (v) displacement of joint B.
Approach. Only joint B can move, so the pin-jointed structure reduces to a 2×2 stiffness problem: each bar contributes an axial stiffness $EA/L$ projected onto the global axes through its direction cosines, and $\mathbf{K}\,\mathbf{d}=\mathbf{F}$ is solved for $d=(u,v)$.
Direction cosines and bar stiffnesses. For a bar from B toward its support, $(c,s)$ is its unit vector and $k=EA/L$. AB: $(c,s)=(-1,0)$, $k_{AB}=EA/0.8=9.68\times10^{7}$; CB: $(c,s)=(-0.8,-0.6)$, $k_{CB}=EA/1.0=7.744\times10^{7}$; DB: $(c,s)=(0,-1)$, $k_{DB}=EA/0.6=1.291\times10^{8}$ N/m.
Assemble the joint stiffness. Each bar adds $k\begin{bmatrix}c^{2}&cs\\ cs&s^{2}\end{bmatrix}$. Summing the three bars, $$\mathbf{K}=\begin{bmatrix}1.4636\times10^{8} & 3.717\times10^{7}\\ 3.717\times10^{7} & 1.5695\times10^{8}\end{bmatrix}\ \text{N/m}.$$
Solve for the displacements. With $\mathbf{F}=(7000,\,0)^{\mathsf T}$ N, $\mathbf{d}=\mathbf{K}^{-1}\mathbf{F}$ gives $$\boxed{u=+0.0509\ \text{mm},\qquad v=-0.0121\ \text{mm}.}$$ The positive u is in the direction of P (rightward); the small negative v means B settles very slightly, drawn down by the inclined member CB.
