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22-Mec-A7 Advanced Strength of Materials · May 2014

Question 7 of 8: Joint Displacements of a Three-Bar Truss

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

Notes on this paper

Paper format: National Exams, May 2014 — 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours; any five of the eight problems constitute a complete paper and all problems are of equal value. All eight problems are solved.

Reference texts: A. C. Ugural & S. K. Fenster, Advanced Strength and Applied Elasticity, 4th ed.; A. P. Boresi & R. J. Schmidt, Advanced Mechanics of Materials, 6th ed.; R. C. Hibbeler, Mechanics of Materials, 10th ed.; S. P. Timoshenko & J. M. Gere, Theory of Elastic Stability, 2nd ed.

Question 7: Joint Displacements of a Three-Bar Truss (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. Three pin-jointed bars meeting at the free joint B; $A=5$ cm$^2=5\times10^{-4}$ m$^2$; $a=0.75$ m, $b=0.90$ m; $E=200$ GPa; horizontal load $P=22\,000$ N at B.

Given data
MembersAB (horizontal, $L=a$), BD (vertical, $L=b$), BC (diagonal, $L=\sqrt{a^2+b^2}$)
SupportsA (horizontal restraint), C and D (pinned to ground)
Axial rigidity$EA=200\times10^{9}(5\times10^{-4})=1.0\times10^{8}$ N
ABCD P = 22 000 N a = 75 cm b = 90 cm
Fig. 7—Three-bar pin-jointed truss; only joint B is free. Coordinates A(0,b), B(a,b), C(0,0), D(a,0).

Find. The horizontal ($u$) and vertical ($v$) displacements of joint B.

Approach. Only joint B moves, but three bars meet there against two equilibrium equations, so the joint is statically indeterminate to the first degree. Use the stiffness (displacement) method: each bar contributes $ (EA/L)$ times the outer product of its direction cosines to a $2\times2$ joint stiffness, and $\mathbf{K}\{u,v\}=\{P,0\}$.

  1. Member geometry. Lengths $L_{AB}=0.75$ m, $L_{BD}=0.90$ m, $L_{BC}=\sqrt{0.75^2+0.90^2}=1.172$ m; direction cosines from B toward each far end are $(-1,0)$, $(0,-1)$, and $(-0.640,-0.768)$.
  2. Assemble the joint stiffness. Summing $\dfrac{EA}{L}\!\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$ over the three bars (units MN/m), $$\mathbf{K}=\begin{bmatrix}168.3&41.98\\41.98&161.5\end{bmatrix}\ \text{MN/m}.$$
  3. Solve for the displacements. With $\mathbf{K}\{u,v\}^{\mathsf T}=\{22\,000,\,0\}^{\mathsf T}$ N, $$\boxed{u=0.140\ \text{mm (right)},\qquad v=-0.036\ \text{mm (down)}}.$$
  4. Interpretation. The horizontal load produces mainly a horizontal movement; the diagonal BC couples in a small downward vertical component. Bar forces recovered from the displacements are $N_{AB}=-18.6$ kN (C), $N_{BD}=+4.0$ kN (T), $N_{BC}=-5.3$ kN (C), which satisfy joint equilibrium at B.
QuantityResult
Horizontal displacement $u$$0.140$ mm (right)
Vertical displacement $v$$-0.036$ mm (down)