22-Mec-A7 Advanced Strength of Materials · December 2019
Question 5 of 7: Stepped composite bar with an end gap
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. A two-segment axial bar fixed at the left, free (with a 0.1 mm gap) at the right, loaded by 200 kN at the step.
Given data
Segment
Diameter
Area
Length
1 (left)
40 mm
1257 mm$^2$
600 mm
2 (right)
30 mm
707 mm$^2$
600 mm
Applied load $P$
200 kN at the step (toward the right wall)
Right-hand gap
0.1 mm
Find. The reactions at the left and right walls.
Fixed–left, gapped–right stepped bar; 200 kN applied at the 40/30 mm step. Both segments are 600 mm long.
Approach. First check whether the 200 kN load closes the 0.1 mm gap; if it does, the problem is statically indeterminate — solve by compatibility (the step displacement and the prescribed 0.1 mm right-end movement) with the stiffnesses of the two segments.
Segment stiffnesses (assume both segments are steel, $E=200$ GPa):
$$k_1=\frac{A_1E}{L_1}=\frac{1257(200\,000)}{600}=4.189\times10^{5}\ \text{N/mm},\quad k_2=\frac{A_2E}{L_2}=2.356\times10^{5}\ \text{N/mm}$$
Does the gap close? Without the right wall, segment 2 carries no load, so the free right end simply follows the step:
$$\delta_{\text{free}}=\frac{P}{k_1}=\frac{200\,000}{4.189\times10^{5}}=0.477\ \text{mm}\gt 0.1\ \text{mm}$$
The gap closes and the right wall reacts — the bar is now indeterminate.
Compatibility with the closed gap. Let $u_B$ be the step displacement and impose the right end at the wall, $u_C=0.1$ mm. Equilibrium of the step node gives
$$k_1u_B+k_2\,(u_B-u_C)=P\ \Rightarrow\ u_B=\frac{P+k_2u_C}{k_1+k_2}=\frac{200\,000+2.356\!\times\!10^{5}(0.1)}{6.545\times10^{5}}=0.3416\ \text{mm}$$
Left reaction = tension in segment 1 (force in segment 2, $k_2(u_C-u_B)\lt 0$, is compression):
$$R_L=k_1u_B=4.189\times10^{5}(0.3416)=143.1\ \text{kN (tension)}$$
Right reaction. Segment 2 carries $k_2(u_C-u_B)$, a compressive force that pushes on the right wall:
$$\boxed{R_L=143.1\ \text{kN},\qquad R_R=56.9\ \text{kN}}$$
Check: $R_L+R_R=200$ kN $=P$. ✓
Check: The word "composite" is taken as a stepped bar of a single material; no modulus is printed, so both segments are assumed steel ($E=200$ GPa). $E$ does not cancel. The share of $P$ taken by each wall is fixed by $A_1/A_2$ ($R_L=128.0$ kN from that split alone), but the extra force needed to push the right end through the 0.1 mm gap scales with $E$ (15.1 kN for steel). For aluminium ($E=70$ GPa), for example, the reactions would be $R_L=133.3$ kN and $R_R=66.7$ kN. State the assumed material in the answer.
