Question 1 of 8: Composite Rod-and-Pipe Assembly (Statically Indeterminate)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Basic Studies, 04-BS-06 Mechanics of Materials, 2018-May. Closed book; one hand-written aid sheet permitted. The exam instructs "any FIVE of the eight questions constitute a complete paper," but every question is solved below as a full study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial deformation ch.4; shear/moment diagrams ch.6; beam deflection by integration ch.9; stress transformation & Mohr's circle ch.9; transformed-section composite beams ch.6; torsion of circular shafts ch.5; column buckling ch.13).
Given. Rod $AB$ (steel, above the plate): $A=500\text{ mm}^2$, $E=200\text{ GPa}$, $L=250\text{ mm}$, $\sigma_y=350\text{ MPa}$. Pipe $BC$ (aluminum, below the plate): $A=2{,}000\text{ mm}^2$, $E=70\text{ GPa}$, $L=500\text{ mm}$, $\sigma_y=300\text{ MPa}$. Applied load $P=70\text{ kN}$ at the rigid plate B; A and C are fixed supports.
Find. Part (a): forces $F_{AB}$, $F_{BC}$. Part (b): the downward displacement $\delta_B$ of plate B.
Rigid plate B loaded between rod AB (tension) and pipe BC (compression), both fixed at their outer ends.
Approach. The rod (tension) and pipe (compression) act in parallel between two fixed points, so this is statically indeterminate: combine force equilibrium at the plate with the compatibility condition that both members share the same displacement at B, then solve the two equations together.
Part (a) — equilibrium at the plate. Vertical equilibrium of the rigid plate gives $F_{AB}+F_{BC}=P$, one equation in two unknowns — the system is indeterminate to the first degree.
Axial stiffness of each member. $k=\dfrac{AE}{L}$: $k_{AB}=\dfrac{500\times200{,}000}{250}=400{,}000\text{ N/mm}$, $k_{BC}=\dfrac{2{,}000\times70{,}000}{500}=280{,}000\text{ N/mm}$.
Compatibility. Plate B moves down by $\delta_B$; the rod stretches by $\delta_B$ (tension) and the pipe shortens by $\delta_B$ (compression), so $\dfrac{F_{AB}}{k_{AB}}=\dfrac{F_{BC}}{k_{BC}}=\delta_B$. Substituting into equilibrium, $F_{BC}\left(\dfrac{k_{AB}}{k_{BC}}+1\right)=P$, so $F_{BC}=\dfrac{70}{400/280+1}=28.82\text{ kN}$ and $F_{AB}=70-28.82=\boxed{41.18\text{ kN (tension)}}$, $F_{BC}=\boxed{28.82\text{ kN (compression)}}$.
Check against yield. $\sigma_{AB}=41{,}180/500=82.4\text{ MPa}\lt350\text{ MPa}$; $\sigma_{BC}=28{,}820/2{,}000=14.4\text{ MPa}\lt300\text{ MPa}$ — both members stay elastic, so the linear stiffness analysis is valid.
Part (b) — displacement. $\delta_B=\dfrac{F_{AB}}{k_{AB}}=\dfrac{41{,}180}{400{,}000}=\boxed{0.103\text{ mm downward}}$ (matches $F_{BC}/k_{BC}=28{,}820/280{,}000=0.103\text{ mm}$, confirming compatibility).
