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22-Mec-A7 Advanced Strength of Materials · May 2015

Question 5 of 8: Welded Rod Assembly Between Rigid Walls

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.

Question 5: Welded Rod Assembly Between Rigid Walls (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 collinear rods A–B–C–D fixed at both walls (uA = uD = 0), loaded axially at the interior joints.

Given data
k1 = E1A1/L1145 GPa × 12×103 mm2 / 1.5 m = 1160 MN/m
k2 = E2A2/L277 GPa × 20×103 mm2 / 2.0 m = 770 MN/m
k3 = E3A3/L31160 MN/m
Load at B300 kN (←)
Load at C450 kN (→)

Find. The axial displacements uB and uC.

(1)(2)(3)ABCD300 kN450 kN
Rods 1–3 in series between fixed walls; the welded block carries the interior loads at B and C.

Approach. Treat each rod as an axial spring; with both walls fixed only uB and uC are unknown, so assemble a 2×2 stiffness system and solve.

  1. Equilibrium of the two free joints. Taking rightward as positive, $$(k_1+k_2)u_B-k_2u_C=-300\ \text{kN},\qquad -k_2u_B+(k_2+k_3)u_C=+450\ \text{kN}.$$
  2. Insert stiffnesses. $$\begin{bmatrix}1.93&-0.77\\-0.77&1.93\end{bmatrix}\!\times\!10^{9}\begin{Bmatrix}u_B\\u_C\end{Bmatrix}=\begin{Bmatrix}-300\\450\end{Bmatrix}\!\times\!10^{3}\ \text{N}.$$
  3. Solve. $$\boxed{u_B=-0.0742\ \text{mm (left)},\qquad u_C=+0.204\ \text{mm (right)}.}$$ Joint B moves slightly toward the left wall against the stiff rod 1, while C moves right; the net stretch of the soft middle rod 2 absorbs most of the relative motion.
Final results — Question 5
JointDisplacement
B−0.0742 mm (left)
C+0.204 mm (right)