Question 1 of 8: Composite Column With an Initial Gap
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2017 — 04-BS-6: Mechanics of Materials (3 hours, closed book, one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete paper on the official exam; every question is answered below. A wide-flange (W-shape) section-property table is attached at the end of the official exam; every question below supplies its own built-up or standard section directly, so the table is not needed for any of the eight solutions.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial members with initial gaps, shear/moment diagrams, deflection by direct integration, transformation of stress via Mohr's circle, combined axial+bending on eccentrically-loaded columns, torsion of stepped shafts, Euler column buckling, unsymmetric-section flexure of brittle materials).
Question 1: Composite Column With an Initial Gap (20 marks)
Given. Two steel posts (each A=100 mm², E=200 GPa, σy=240 MPa) and one central aluminum post (A=300 mm², E=70 GPa, σy=210 MPa), all of length L=500 mm, support a rigid plate. The aluminum post is 0.2 mm SHORT of the plate under no load (steel posts touch the plate directly). An applied load P=30 kN is centred on the plate for parts (a) and (b).
Quantity
Value
Steel (×2 posts)
A=100 mm², E=200 GPa, σy=240 MPa
Aluminum (1 post)
A=300 mm², E=70 GPa, σy=210 MPa
Length, L
500 mm (all posts)
Initial gap
0.2 mm (aluminum post short)
Applied load, P
30 kN (parts a, b)
Find. (a) the force in each post at P=30 kN, (b) the plate's downward displacement, (c) the maximum P before either material yields.
Rigid plate on two steel posts and a central aluminum post, 0.2 mm short of the plate.
Approach. Compute the load Pgap that closes the 0.2 mm gap using the steel posts alone; below that load only steel resists P, above it the plate moves in a stiffer composite phase with both materials sharing any additional load in proportion to their axial stiffnesses AE/L.
Stiffnesses and the gap-closing load. $k_{steel}=2A_sE_s/L=2(100)(200{,}000)/500=80{,}000$ N/mm and $k_{al}=A_{al}E_{al}/L=(300)(70{,}000)/500=42{,}000$ N/mm. The load that closes the gap using steel alone is
$$P_{gap}=k_{steel}\times(\text{gap})=80{,}000(0.2)=\boxed{16{,}000\text{ N}}=16\text{ kN}$$
Since the applied P=30 kN exceeds Pgap=16 kN, the aluminum post is already engaged at full load.
Composite phase beyond the gap. The remaining $\Delta P=30{,}000-16{,}000=14{,}000$ N is shared by compatible displacement across both stiffnesses in parallel:
$$\Delta\delta=\frac{\Delta P}{k_{steel}+k_{al}}=\frac{14{,}000}{122{,}000}=0.11475\text{ mm}$$
giving additional forces $\Delta F_{steel}=k_{steel}\Delta\delta=9180.3$ N and $F_{al}=k_{al}\Delta\delta=4819.7$ N.
(a) Total forces at P=30 kN.
$$F_{steel,total}=16{,}000+9180.3=\boxed{25{,}180.3\text{ N}}\;(12{,}590\text{ N each post}),\quad F_{al}=\boxed{4819.7\text{ N}}$$
Check: $25{,}180.3+4819.7=30{,}000$ N = P. Stresses $\sigma_{steel}=125.9$ MPa and $\sigma_{al}=16.1$ MPa are both well under yield, confirming the elastic assumption.
(b) Plate displacement. The plate moves the 0.2 mm gap plus the composite-phase stretch:
$$\delta=0.2+0.11475=\boxed{0.3148\text{ mm}}$$
(c) Maximum P before yield. Beyond the gap, additional steel force is limited to $\Delta F_{s,max}=\sigma_{y,s}(2A_s)-P_{gap}=240(200)-16{,}000=32{,}000$ N. Since $\Delta F_{steel}=\dfrac{k_{steel}}{k_{steel}+k_{al}}\Delta P$,
$$\Delta P_{max}=32{,}000\times\frac{122{,}000}{80{,}000}=48{,}800\text{ N}\ \Rightarrow\ P_{max}=16{,}000+48{,}800=\boxed{64{,}800\text{ N}}=64.8\text{ kN}$$
Checking aluminum at this load: $\Delta F_{al}=\tfrac{42{,}000}{122{,}000}(48{,}800)=16{,}800$ N, so $\sigma_{al}=56.0$ MPa, far below its 210 MPa yield — steel governs, not aluminum.
