22-Mec-B9 Advanced Engineering Structures · May 2014
Question 4 of 8: Tresca and von Mises predictions for a triaxial stress state
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2014 — 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, shear centre, structural idealisation (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield and fracture criteria, torsion of thin-walled members, statically indeterminate axial members (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of open and closed sections, multiply connected cells (Ch. 6).
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. — thermal stress in restrained members and combined loading (Ch. 4, 8).
Axes and sign convention used throughout. For every thin-walled question a right-handed set is used: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section, and $Z$ is horizontal. A vertical shear $S_y$ acting on a section whose product of inertia vanishes produces the open-section shear flow $q_s = -\dfrac{S_y}{I_{zz}}\displaystyle\int_0^s y\,t\,\mathrm{d}s$, measured positive in the direction of increasing $s$. Closed cells add the constant redundant flow $q_{s,0}$ fixed either by zero twist (shear-centre problems) or by moment equilibrium (a load of known line of action). All shear flows below are quoted with respect to a stated walking direction, and each question closes with the two statical checks — the resolved vertical force must return $S_y$ and the moment of the flows about a chosen origin must return the moment of the applied load.
Question 4: Tresca and von Mises predictions for a triaxial stress state (20 marks)
Given. A general three-dimensional stress state in which the $z$ direction carries no shear, so it is already a principal direction.
Quantity
Symbol
Value
Normal stress, x direction
$\sigma_x$
−95 MPa (compressive)
Normal stress, y direction
$\sigma_y$
+110 MPa (tensile)
Normal stress, z direction
$\sigma_z$
+100 MPa (tensile)
Shear stress on the x-y planes
$\tau_{xy}$
60 MPa
Yield strength of the material
$\sigma_Y$
170 MPa
Find. (a) whether the maximum shear stress (Tresca) criterion predicts yielding; (b) whether the von Mises criterion predicts yielding.
Left: the stress element, with the out-of-plane direct stress noted. Right: the three Mohr circles. The outer circle spans the largest and smallest principal stresses and its radius is the maximum shear stress in the solid.
Approach. The $z$ axis is free of shear, so it is a principal direction and the remaining two principal stresses follow from the ordinary two-dimensional transformation on the $x$-$y$ plane; both criteria are then evaluated directly from the three ordered principal stresses.
Part (a) — identify the principal directions. The stress matrix has no $\tau_{yz}$ or $\tau_{zx}$ component, $$\boldsymbol{\sigma} = \begin{bmatrix} -95 & -60 & 0 \\ -60 & 110 & 0 \\ 0 & 0 & 100 \end{bmatrix}\ \text{MPa}$$ so the $z$ direction is principal with $\sigma_z = 100$ MPa, and the other two principal stresses come from the in-plane block alone. Note that the sign of $\tau_{xy}$ never affects the answer, because it enters only as $\tau_{xy}^{2}$.
In-plane principal stresses by the standard transformation. $$\sigma_{1,2}^{\,\text{(in-plane)}} = \frac{\sigma_x+\sigma_y}{2} \pm\sqrt{\left(\frac{\sigma_y-\sigma_x}{2}\right)^{2}+\tau_{xy}^{2}} = \frac{-95+110}{2}\pm\sqrt{\left(\frac{110+95}{2}\right)^{2}+60^{2}}$$$$= 7.5 \pm \sqrt{(102.5)^{2}+3600} = 7.5 \pm 118.77\ \text{MPa}$$ giving 126.27 MPa and -111.27 MPa.
Order the three principal stresses. Collecting the in-plane pair with $\sigma_z$ and sorting them, $$\begin{aligned}\sigma_1 &= 126.27\ \text{MPa}, \\ \sigma_2 &= 100.00\ \text{MPa}, \\ \sigma_3 &= -111.27\ \text{MPa}\end{aligned}$$ Ordering matters: Tresca depends only on the extreme pair, and getting the middle value into that pair is the commonest arithmetic slip in this question.
Apply the maximum shear stress criterion. Tresca predicts yielding when the largest shear stress in the solid reaches half the yield strength, equivalently when $$\begin{aligned}\sigma_1-\sigma_3 \ge \sigma_Y: \\ 126.27 - \bigl(-111.27\bigr) &= \boxed{237.54\ \text{MPa}} \;>\; 170\ \text{MPa}\end{aligned}$$ The maximum shear stress is $\tau_{\max} = (\sigma_1-\sigma_3)/2 = 118.77$ MPa against the allowable $\sigma_Y/2 = 85$ MPa, so the solid yields. The equivalent safety factor is $170/237.54 = 0.716$.
Part (b) — apply the von Mises criterion. The distortion energy criterion compares the effective stress with the yield strength: $$\sigma_e = \sqrt{\tfrac12\left[(\sigma_1-\sigma_2)^{2}+(\sigma_2-\sigma_3)^{2}+(\sigma_3-\sigma_1)^{2}\right]}$$$$\sigma_e = \sqrt{\tfrac12\left[(26.27)^{2}+(211.27)^{2}+(-237.54)^{2}\right]} = \sqrt{50\,875} = \boxed{225.55\ \text{MPa}}\;>\;170\ \text{MPa}$$ so von Mises also predicts yielding, with a safety factor of 0.754.
Cross-check the effective stress in the original components. The same result must follow without ever forming the principal stresses: $$\sigma_e = \sqrt{\tfrac12\left[(\sigma_x-\sigma_y)^{2}+(\sigma_y-\sigma_z)^{2}+(\sigma_z-\sigma_x)^{2}+6\tau_{xy}^{2}\right]}$$$$= \sqrt{\tfrac12\left[(-205)^{2}+(10)^{2}+(195)^{2}+6(60)^{2}\right]} = 225.55\ \text{MPa}\ \checkmark$$ The agreement confirms both the principal stresses and the criterion arithmetic.
Compare the two verdicts. Both criteria predict yielding, and Tresca is the more severe of the two, as it always is: 237.5 MPa against 225.6 MPa, a difference of 5.3 per cent. Since Tresca is never unconservative relative to von Mises, agreement between them means the verdict is not sensitive to the choice of criterion — the material yields under this stress state either way.
