22-Mec-B9 Advanced Engineering Structures · December 2016
Question 5 of 8: Sizing a square bar under combined axial force and torque
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2016 — 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. — thin-walled open and closed sections, shear flow, shear centre, structural idealisation, single- and multi-cell torsion (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield criteria, torsion of non-circular and thin-walled members, unsymmetrical bending (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of multiply connected cells and of solid non-circular shafts (Ch. 6).
R. M. Jones, Mechanics of Composite Materials, 2nd ed. — orthotropic lamina stiffness, the reduced stiffness matrix and its transformation (Ch. 2).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — strain-life fatigue, the Coffin–Manson relation and the Palmgren–Miner rule (Ch. 9, 14).
Question 5: Sizing a square bar under combined axial force and torque (20 marks)
Given. A solid square prismatic bar carrying an axial force and a torque simultaneously, to be sized against a yield strength with a factor of safety of three.
Given data — loads and material
Quantity
Symbol
Value
Axial force (compressive)
$P$
$225\times10^{3}$ N
Applied torque
$T$
$22\times10^{3}$ N·m = $22\times10^{6}$ N·mm
Yield strength
$\sigma_Y$
330 MPa
Factor of safety
$n$
3
Torsion shape factor, square section
—
$\tau_{\max} = T/(0.208\,w^{3})$
Find. The smallest side dimension $w$ that satisfies first the maximum-shear-stress (Tresca) criterion and then the distortion-energy (von Mises) criterion.
Figure 5.1 — the cantilevered square bar carrying an axial force and a torque at its free end. The critical point is the midpoint of a side, where the torsional shear peaks.
Approach. Identify the critical point — the mid-side of the square, where the torsional shear peaks and the uniform axial stress also acts — write the two-dimensional stress state there, and impose each yield criterion at the allowable stress $\sigma_Y/n$.
Set the allowable stress. With a factor of safety of three on a yield strength of 330 MPa, the design equivalent stress is$$\sigma_{\text{allow}} = \frac{\sigma_Y}{n} = \frac{330}{3} = 110\ \text{MPa}.$$Both criteria will be evaluated against this single number, so the whole comparison in part (b) comes down to how each criterion combines $\sigma$ and $\tau$.
Locate the critical point and write its stress state. For a solid square shaft the torsional shear stress is greatest at the midpoint of each side (not at the corners, where it is zero), and there$$\tau = \frac{T}{0.208\,w^{3}} = \frac{22\times10^{6}}{0.208\,w^{3}} .$$The axial force is uniform over the whole section, so at that same point$$\sigma = \frac{P}{w^{2}} = \frac{225\times10^{3}}{w^{2}}\ \ (\text{compressive}).$$The mid-side point therefore carries the worst combination of the two, and it is a plane-stress state: $\sigma_x = -\sigma$, $\sigma_y = 0$, $\tau_{xy} = \tau$.
Form the principal stresses. For this state,$$\sigma_{1,2} = -\frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^{2} + \tau^{2}},\qquad \sigma_3 = 0,$$with $\sigma_1 > 0 > \sigma_2$ whenever any shear is present. The intermediate principal stress is the out-of-plane zero, so Tresca will use $\sigma_1 - \sigma_2$ directly.
Apply the maximum-shear-stress criterion (part a). Yielding by Tresca requires $\sigma_1 - \sigma_2 = \sigma_{\text{allow}}$, and the radical above gives $\sigma_1 - \sigma_2 = \sqrt{\sigma^{2} + 4\tau^{2}}$. Hence$$\sqrt{\left(\frac{225\times10^{3}}{w^{2}}\right)^{2} + 4\left(\frac{22\times10^{6}}{0.208\,w^{3}}\right)^{2}} = 110 .$$Solving this single equation in $w$ numerically,$$\boxed{w_{\text{Tresca}} = 124.7\ \text{mm}} .$$At that size $\sigma = 14.5$ MPa and $\tau = 54.5$ MPa, so the torque supplies almost all of the demand — the axial force contributes under 1 % of the equivalent stress (about 1.7 % of its square).
Apply the distortion-energy criterion (part b). For the same plane-stress state the von Mises equivalent stress reduces to$$\sigma_{\text{vM}} = \sqrt{\sigma^{2} + 3\tau^{2}} = 110 ,$$the only change being the factor 3 in place of 4 on the shear term. Solving again,$$\boxed{w_{\text{von Mises}} = 119.0\ \text{mm}} ,$$which is 4.6 % smaller than the Tresca answer, or 9.0 % lighter in cross-sectional area. That is the familiar result that Tresca is the conservative of the two, and it is at its most conservative — a factor of $2/\sqrt3 = 1.155$ on the shear term — exactly in the pure-shear-dominated regime this bar occupies.
Note what does not matter. Both criteria depend on $\sigma$ only through $\sigma^{2}$, so the answer is unchanged if the axial force is tensile rather than compressive. Reversing the sign of $P$ swaps which principal stress is the larger but leaves $\sigma_1-\sigma_2$ and $\sigma_{\text{vM}}$ untouched. That is a useful check on the algebra, and it also means a real design would need a separate buckling check, which these two criteria cannot supply.
