Minimum square section under combined axial load and torque: 22-Mec-B9 Advanced Engineering Structures, May 2013, Q6

Question 6: Minimum square section under combined axial load and torque (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A solid square bar carrying a constant axial force and a constant torque, to be sized against yield with a safety factor of three.

Quantity	Symbol	Value
Axial force (compressive)	$P$	250 kN
Torque	$T$	13 kN$\cdot$m
Yield strength	$\sigma_{Y}$	280 MPa
Safety factor	$N$	3
Torsion constant for a square	$\tau_{\max} = T/(0.208w^{3})$	coefficient 0.208

Find. The smallest side length $w$ that satisfies (a) the maximum-shear-stress (Tresca) criterion and (b) the distortion-energy (von Mises) criterion.

The bar and the critical surface element, which sits at the mid-point of a face — the location of $\tau_{\max}$ for a square shaft, not the corner. The element carries a uniaxial direct stress and a shear stress; the free surface makes the third principal stress zero.

Approach. Write both stresses in terms of $w$, form each yield criterion as an equation in $w$ alone, and solve numerically because the axial and torsional terms scale with different powers of $w$.

Locate the critical point and write the stresses. The direct stress is uniform over the section; the torsional shear stress in a solid square shaft is greatest at the mid-point of each side, where the standard result is $\tau_{\max} = T/(0.208w^{3})$. Both maxima therefore coincide at that point, which is also a free surface, so$$\sigma = \frac{P}{w^{2}} = \frac{250\times10^{3}}{w^{2}}, \qquad \tau = \frac{T}{0.208w^{3}} = \frac{13\times10^{6}}{0.208\,w^{3}} = \frac{6.250 \times 10^{7}}{w^{3}}\ \text{(N and mm)}$$
Form the principal stresses. For an element carrying one direct stress and a shear stress on a free surface,$$\sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^{2} + \tau^{2}}, \qquad \sigma_{3} = 0$$Because the radical always exceeds $|\sigma/2|$, one in-plane principal stress is tensile and the other compressive; $\sigma_{3} = 0$ therefore lies between them and plays no part in the Tresca criterion.
Apply the maximum-shear-stress criterion (part a). Tresca requires $\sigma_{1} - \sigma_{3,\min} = \sigma_{Y}/N$, and here that difference is $2\sqrt{(\sigma/2)^{2}+\tau^{2}}$, so$$\sqrt{\left(\frac{\sigma}{2}\right)^{2} + \tau^{2}} = \frac{\sigma_{Y}}{2N} = \frac{280}{6} = 46.667\ \text{MPa}$$Substituting the two expressions in $w$ gives$$\frac{1.5625 \times 10^{10}}{w^{4}} + \frac{3.9062 \times 10^{15}}{w^{6}} = 2177.8$$
Solve for $w$. The two terms scale as $w^{-4}$ and $w^{-6}$, so no closed form exists; bisection between 20 mm and 400 mm converges to$$\boxed{w = 111.1\ \text{mm}}$$At that size $\sigma = 20.2$ MPa and $\tau = 45.6$ MPa, so the torque supplies almost all of the demand — the axial term contributes barely five per cent of the radical.
Apply the distortion-energy criterion (part b). For this stress state von Mises reduces to the familiar combined-loading form$$\sqrt{\sigma^{2} + 3\tau^{2}} = \frac{\sigma_{Y}}{N} = 93.33\ \text{MPa}$$and the same bisection gives$$\boxed{w = 106.1\ \text{mm}}$$with $\sigma = 22.2$ MPa and $\tau = 52.3$ MPa.
Compare the two answers. Von Mises permits a section 4.7 per cent smaller in side length, which is about 8.8 per cent less material. That gap is close to the theoretical maximum: for pure shear the two criteria differ by the factor $2/\sqrt{3} = 1.155$ in allowable stress, and this problem is nearly pure shear. Tresca is the conservative choice and the one the question specifies for part (a).
Note the insensitivity to the sign of $P$. Neither criterion changes if the 250 kN is tensile rather than compressive: the Tresca radical contains $(\sigma/2)^{2}$ and the von Mises expression contains $\sigma^{2}$. Only a stability check would distinguish the two cases, and no length is given for the cantilever, so buckling cannot be assessed.

Quantity	Result
Allowable Tresca difference $\sigma_{Y}/N$	93.33 MPa
(a) Minimum $w$, maximum-shear-stress criterion	111.1 mm
stresses at that size	$\sigma$ = 20.2 MPa, $\tau$ = 45.6 MPa
(b) Minimum $w$, von Mises criterion	106.1 mm
stresses at that size	$\sigma$ = 22.2 MPa, $\tau$ = 52.3 MPa
Difference in side length	4.7 per cent

Check

The torsional coefficient 0.208 is the standard tabulated value for a solid square shaft with $\tau_{\max}$ at the mid-side; using the circular-shaft formula instead would understate the shear stress badly. Transverse shear from any lateral load is absent because none is applied, and stress concentrations at the built-in end are neglected. No length is given, so column buckling under the 250 kN compression is not checked; for a slender member it could well govern over yield.

Concept & Theory

Topic. Yield criteria applied to combined direct and shear stress, and torsion of a non-circular section.

Key relations. $\sigma = P/A$; $\tau_{\max} = T/(\alpha_{1}wb^{2})$ with $\alpha_{1} = 0.208$ for a square; Tresca $\sigma_{1}-\sigma_{3} = \sigma_{Y}/N$, giving $2\sqrt{(\sigma/2)^{2}+\tau^{2}} = \sigma_{Y}/N$; von Mises $\sqrt{\sigma^{2}+3\tau^{2}} = \sigma_{Y}/N$.

Why this works. Both criteria say that yield depends only on the deviatoric part of the stress, but they measure it differently: Tresca takes the largest shear stress, von Mises takes the root-mean-square of the differences between the three principal stresses. The Tresca hexagon is inscribed in the von Mises ellipse, touching at the six points of uniaxial and equal-biaxial stress and departing furthest in pure shear, where Tresca predicts yield at $0.5\sigma_{Y}$ and von Mises at $0.577\sigma_{Y}$. A square shaft in torsion is not a circular shaft: plane sections warp, the corners carry no shear at all because two free surfaces meet there, and the peak occurs at the mid-side.

Common pitfall. The classic mistake is to place the maximum torsional shear at the corner of the square, by analogy with the circular shaft where the maximum is at the largest radius. The second is to solve the Tresca equation as though it were a quadratic in $w^{2}$; the two terms carry different powers of $w$ and the equation must be solved numerically or by iteration.

Source. Ugural and Fenster, Advanced Strength and Applied Elasticity, 5th ed., §4.6–4.7 (Tresca and von Mises) and §6.5 (torsion of rectangular bars); Boresi and Schmidt, Advanced Mechanics of Materials, 6th ed., §6.4.

Question 6 of 8: Minimum square section under combined axial load and torque

Question 6: Minimum square section under combined axial load and torque (20 marks)