Question 7 of 8: Truss With Euler-Buckling Compression Members
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 7: Truss With Euler-Buckling Compression Members (20 marks)
Given. W-shaped truss A(pin)-B-C-D-E(roller), span 4×3 m, height 4 m, downward loads P at B and D, all members 50×50 mm steel, SF=2 on Euler buckling, no SF on yield.
Quantity
Value
Panel width
3 m (×4 = 12 m span)
Height
4 m
Section
50×50 mm square, all members
E, σy
200 GPa, 240 MPa
Buckling SF
2 (no SF on yield)
Find. The largest load P.
W-truss: pin at A, roller at E, loads P at B and D.
Approach. Solve every member force in terms of P by the method of joints, then check each compression member against both its Euler buckling capacity (with SF=2) and yield, and each tension member against yield, taking the smallest resulting P.
Method of joints (unit load P). Solving the full 10-equation joint system (5 joints × 2 equations = 7 member forces + 3 reactions) gives
$$F_{AB}=F_{DE}=-1.25P,\quad F_{BD}=-0.75P,\quad F_{AC}=F_{CE}=+0.75P,\quad \boxed{F_{BC}=F_{CD}=0}$$
BC and CD are the zero-force members the hint refers to (confirmed by the determinacy check m+r=2j=10=10).
Section properties. 50×50 mm square, same I about any centroidal axis (so "in-plane buckling only" needs no special treatment for this shape):
$$A=2500\text{ mm}^2,\qquad I=\frac{50^4}{12}=5.2083e+05\text{ mm}^4$$
Compression members: buckling vs. yield. AB and DE (5 m, force 1.25P) and BD (6 m, force 0.75P) are all in compression:
$$P_{cr}=\frac{\pi^2EI}{L^2},\qquad P_{cr,allow}=\frac{P_{cr}}{2}$$
For AB/DE: $P_{cr}=41123$ N, $P_{cr,allow}=20562$ N, giving $P_{max}=P_{cr,allow}/1.25=\boxed{16449\text{ N}}$ from buckling (vs. 480,000 N from yield — buckling governs by a wide margin). BD checks similarly to a higher $P_{max}$, so AB/DE govern.
Tension members. AC and CE (0.75P) only need checking against yield: $P_{max}=\sigma_yA/0.75=800{,}000$ N, far above the compression limit.
Governing load.
$$\boxed{P_{max}=16449.34\text{ N}=16.449\text{ kN}}\ \text{(Euler buckling of AB and DE)}$$
