22-Mec-B9 Advanced Engineering Structures · December 2017
Question 4 of 8: Factors of safety of a tubular compression strut
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2017 — 16-Mec-B9
Advanced Engineering Structures. Three hours, open book, any non-communicating
calculator permitted. The paper prints eight questions of equal value (20 marks each) and
states that any five constitute a complete exam paper. All eight are solved below,
because the set is a study resource rather than a sitting.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. —
unsymmetrical bending, shear flow in open and closed thin-walled sections, shear centre,
structural idealisation, single- and multi-cell torsion (Ch. 15, 16, 17, 20, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed.
— three-dimensional states of stress, principal stresses, the Tresca and von Mises yield
criteria (Ch. 1, 4).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. —
shear centre, torsion of multiply connected cells, column stability (Ch. 6, 12).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life
fatigue, the Coffin–Manson relation, the Palmgren–Miner rule and Paris-law crack
growth (Ch. 9, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — Euler buckling of
pin-ended columns, slenderness ratio and the transition slenderness (Ch. 13).
Sign convention used throughout. A single right-handed frame is used
for every thin-walled question: x runs along the span from the root to the free end,
Y is vertically upward and Z horizontally to the right, both measured from the
section centroid. Direct stress is written $\sigma_x = aZ + bY$, and the two coefficients follow
from $aI_{ZZ} + bI_{YZ} = M_Y$ and $aI_{YZ} + bI_{YY} = M_Z$, in which
$I_{ZZ}=\int Z^{2}\,dA$, $I_{YY}=\int Y^{2}\,dA$, $I_{YZ}=\int ZY\,dA$,
$M_Y \equiv \int \sigma_x Z\,dA$ and $M_Z \equiv \int \sigma_x Y\,dA$. This one pair of
equations replaces the memorised unsymmetrical-bending fraction and is self-checking: for a
cantilever carrying a tip load, the fibres on the side the load points toward must go into
compression.
Question 4: Factors of safety of a tubular compression strut (20 marks)
Given. A horizontal beam AB pinned to a wall at A and
propped at its far end B by an inclined strut BC, itself pinned to the wall at C, 2.5 m below A.
The beam carries a uniformly distributed load over its whole 6.0 m span.
Given data — frame and strut
Quantity
Symbol
Value
Beam span, A to B
$L_{AB}$
6.0 m
Vertical offset, A to C
$h$
2.5 m
Uniformly distributed load on AB
$w$
25 kN/m
Strut outer diameter
$D_o$
65 mm
Strut wall thickness
$t$
3 mm
Young's modulus
$E$
200 GPa
Yield strength
$\sigma_Y$
240 MPa
Find. The factor of safety of strut BC against (i) elastic (Euler) buckling
and (ii) yielding in compression.
The frame. BC carries no load along its length and is pinned at both ends, so it
is a two-force member and its force acts along its own axis — which is what makes one
moment equation about A sufficient.
Approach. Take moments about the pin at A to get the axial force in the
two-force strut, evaluate the tube's area, second moment and radius of gyration, confirm from
the slenderness ratio that elastic buckling governs, and form the two ratios.
Recognise BC as a two-force member. BC is pinned at both ends and carries
no load between them, so the force it transmits acts along the line CB. With C 2.5 m below and
6.0 m to the left of B, the strut's length is
$$L_{BC} = \sqrt{6.0^{2}+2.5^{2}} = 6.50\ \text{m},$$
and its direction cosines relative to the horizontal and vertical are $6.0/6.5$ and
$2.5/6.5$.
Resolve the distributed load. The uniformly distributed load totals
$W = wL_{AB} = 25 \times 6.0 = 150$ kN, acting through the mid-span of AB, that is 3.0 m from
the pin at A.
Take moments about A. Only the strut force and the distributed load have a
moment about A. The vertical component of the strut force at B is $F_{BC}(2.5/6.5)$ and acts
6.0 m from A, so
$$F_{BC}\left(\frac{2.5}{6.5}\right)(6.0) = 150 \times 3.0
\quad \Longrightarrow \quad \boxed{F_{BC} = 195.0\ \text{kN (compression)}}$$
The sense is compressive because the strut must push B upward to hold the beam.
Evaluate the tube's section properties. With
$D_i = 65-2(3) = 59$ mm,
$$A = \frac{\pi}{4}\left(D_o^{2}-D_i^{2}\right) = 584.34\ \text{mm}^{2}, \qquad
I = \frac{\pi}{64}\left(D_o^{4}-D_i^{4}\right) = 2.8143\times10^{5}\ \text{mm}^{4},$$
so the radius of gyration is $r = \sqrt{I/A} = 21.95$ mm.
Confirm that elastic buckling governs. The slenderness ratio is
$L/r = 6500/21.95 = 296.2$, against a transition slenderness
$\sqrt{2\pi^{2}E/\sigma_Y} = \sqrt{2\pi^{2}(200\,000)/240} = 128.3$. Because
$296 \gg 128$, the strut is firmly in the long-column range and the Euler formula applies
without an inelastic correction.
Compute the elastic buckling load. With pins at both ends the effective
length equals the true length, so
$$P_{cr} = \frac{\pi^{2}EI}{L_{BC}^{2}}
= \frac{\pi^{2}(200\,000)(2.8143\times10^{5})}{6500^{2}} = 13.15\ \text{kN}
\qquad (\sigma_{cr} = 22.5\ \text{MPa}).$$
Form the factor of safety against buckling. Dividing by the applied axial
force,
$$\boxed{n_{\text{buckling}} = \frac{P_{cr}}{F_{BC}} = \frac{13.15}{195.0} = 0.067}$$
Form the factor of safety against yielding. The squash load is
$P_Y = \sigma_Y A = 240 \times 584.34 = 140.24$ kN, and the actual compressive stress is
$195\,000/584.34 = 333.7$ MPa, so
$$\boxed{n_{\text{yielding}} = \frac{P_Y}{F_{BC}} = \frac{140.24}{195.0} = 0.719}$$
State the engineering conclusion. Both factors are below unity, so the
strut as specified is not merely under-designed but incapable of carrying the load at all: it
would buckle elastically at about 7 % of the applied force, long before the material reached
yield. The ratio of the two factors, $0.719/0.067 \approx 11$, is the quantitative statement
that stability — not strength — is what governs a member this slender.
Check: the answer is a failed member, and it should be reported
as such. A factor of safety below one is a legitimate outcome of the arithmetic and
must not be adjusted to look acceptable. If a redesign is wanted, keeping the same 3 mm wall
and the same steel, the outer diameter would have to rise to roughly 195 mm to reach
$n_{\text{buckling}} = 2$; a more economical route is to shorten the effective length —
bracing BC at mid-length quadruples $P_{cr}$ to 52.6 kN — or to change to a section with a
much larger radius of gyration for the same area.
Final results — Question 4
Quantity
Value
Strut length
$L_{BC} = 6.50$ m
Axial force in BC
195.0 kN compression
Tube area / second moment
$584.34$ mm$^{2}$ / $2.8143\times10^{5}$
mm$^{4}$
Radius of gyration / slenderness
$21.95$ mm / $L/r = 296$
Euler buckling load
$P_{cr} = 13.15$ kN
Squash (yield) load
$P_Y = 140.24$ kN
Factor of safety, elastic buckling
0.067
Factor of safety, yielding
0.719
Verdict
strut inadequate; buckling governs by a factor of about 11
