22-Mec-B9 Advanced Engineering Structures · May 2013
Question 2 of 8: Factor of safety against elastic buckling of the strut BC
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013 — 07-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — open and closed thin-walled sections, shear flow, structural idealisation, shear centre (Ch. 16, 17, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — unsymmetrical bending, yield criteria, torsion of non-circular bars (Ch. 4, 5, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of rectangular sections, columns and elastic stability (Ch. 6, 12).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — linear elastic fracture mechanics, the Paris law, strain-life and Miner's rule (Ch. 8, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — Euler buckling and combined loading (Ch. 8, 13).
Axes and sign convention used throughout. A right-handed set $(X, Y, Z)$ is used for every thin-walled question: $X$ runs along the span from the free end toward the root, $Y$ is vertical (upward) in the cross-section, and $Z$ is horizontal, positive to the right when the section is viewed looking along $+X$. Direct stress is written $\sigma = aZ + bY$ with $Z$ and $Y$ measured from the section centroid, so that the two constants follow from the stress resultants the section must carry, $M_{Y} = \int \sigma Z\,\mathrm{d}A$ and $M_{Z} = \int \sigma Y\,\mathrm{d}A$:
$$a I_{ZZ} + b I_{YZ} = M_{Y}, \qquad a I_{YZ} + b I_{YY} = M_{Z}$$
This form is used in preference to the memorised unsymmetrical-bending fraction because it is self-checking: the physical sign of every answer can be confirmed from the simple rule that, for a cantilever carrying a tip load, the fibres on the side the load points toward go into compression.
Question 2: Factor of safety against elastic buckling of the strut BC (20 marks)
Given. A wall-mounted bracket: the horizontal beam AB is pinned to the wall at A and carries a uniformly distributed load; the tubular strut BC runs from a second wall pin at C, one metre below A, up to the free end B.
Quantity
Symbol
Value
Span of beam AB
$L_{AB}$
3.0 m
Vertical offset A above C
$h$
1.0 m
Distributed load on AB
$w$
10 kN/m
Tube outside diameter / wall
$D_{o}$ / $t$
50 mm / 4 mm
Young's modulus
$E$
200 GPa
Yield strength
$\sigma_{Y}$
320 MPa
Find. The ratio of the Euler buckling load of BC to the compressive force the frame actually delivers to it.
Frame ABC. Beam AB is pinned at A and propped at B by the two-force strut BC; the strut is inclined at $\arctan(1/3) = 18.43^\circ$ to the horizontal.
Approach. Treat BC as a two-force member, take moments about the pin A to size the strut force, then compare that force with the Euler load of a pin-ended tube of the same length, checking that the critical stress is still below yield so the elastic formula applies.
Set out the strut geometry. With C as the origin, B lies 3.0 m across and 1.0 m up, so$$L_{BC} = \sqrt{3.0^{2} + 1.0^{2}} = 3.1623\ \text{m},\qquad \sin\theta = \frac{1.0}{3.1623} = 0.31623$$Because BC is pinned at both ends and carries no load between them, its force must act along CB.
Take moments about A for the beam AB. The distributed load resolves to $wL_{AB} = 30$ kN acting at mid-span, 1.5 m from A. Only the vertical component of the strut force has a moment arm about A along the beam:$$F_{BC}\sin\theta \times L_{AB} = wL_{AB}\times \frac{L_{AB}}{2}\;\Longrightarrow\; F_{BC} = \frac{30 \times 1.5}{0.31623\times 3.0} = \boxed{47.43\ \text{kN (compression)}}$$
Compute the tube section properties. The bore is $D_{i} = 50 - 2(4) = 42$ mm, so$$I = \frac{\pi}{64}\left(50^{4} - 42^{4}\right) = 1.5405 \times 10^{5}\ \text{mm}^{4},\qquad A = \frac{\pi}{4}\left(50^{2}-42^{2}\right) = 578.1\ \text{mm}^{2}$$A tube is used here precisely because it puts material at large radius: the same 578 mm$^{2}$ of steel as a solid 27 mm bar carries about four times the second moment.
Evaluate the Euler load. Both ends are pinned, so the effective length equals the true length, $L_{e} = 3\,162$ mm:$$P_{cr} = \frac{\pi^{2}EI}{L_{e}^{2}} = \frac{\pi^{2}\left(200\,000\right)\left(1.5405 \times 10^{5}\right)}{\left(3\,162\right)^{2}} = \boxed{30.41\ \text{kN}}$$
Confirm that the elastic formula is legitimate. The critical stress is$$\sigma_{cr} = \frac{P_{cr}}{A} = 52.6\ \text{MPa}$$which is far below the 320 MPa yield strength; the slenderness ratio $L_{e}/r = 194$ (with $r = \sqrt{I/A} = 16.32$ mm) is well above the transition value of roughly 110 for this steel. Euler buckling, not squashing, governs.
Form the factor of safety.$$\text{FoS} = \frac{P_{cr}}{F_{BC}} = \frac{30.41}{47.43} = \boxed{0.641}$$The factor is less than unity, so the strut as dimensioned buckles before the stated load is reached. That is the answer the data give, and it should be reported as such rather than forced into a comfortable number.
Quantify the redesign. Restoring a factor of safety of 2.0 needs $P_{cr} = 2F_{BC} = 94.9$ kN, hence $I \ge P_{cr}L_{e}^{2}/\pi^{2}E = 4.806 \times 10^{5}$ mm$^{4}$. Holding the 4 mm wall, the required outside diameter is $D_{o} = 71.3$ mm — a 72 mm tube is the smallest standard size that answers.
Quantity
Result
Strut length $L_{BC}$
3.162 m
Compressive force in BC
47.43 kN
Second moment of area $I$
$1.541 \times 10^{5}\ \text{mm}^{4}$
Euler load $P_{cr}$ (pin-ended)
30.41 kN
Critical stress $\sigma_{cr}$
52.6 MPa (elastic, below yield)
Factor of safety against elastic buckling
0.641 — the strut buckles
Outside diameter for FoS = 2.0 at $t$ = 4 mm
71.3 mm
Check
Both ends of BC are drawn as pins, so the effective-length factor is taken as $K = 1.0$; if the wall connection at C were treated as fixed the buckling load would rise and the factor of safety with it. Buckling is assessed in the plane of the frame about the tube's (equal) principal axes, self-weight is neglected, and the beam AB is assumed stiff enough that the load path is the statically determinate one shown.
