22-Mec-B9 Advanced Engineering Structures · December 2018
Question 2 of 8: Yield prediction under a three-dimensional stress state
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 16-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions are printed and the rubric states that any five constitute a complete exam paper, with all problems of equal total value (20 marks each). All eight are worked here. Marks for the individual parts are those printed inside each problem.
Reference texts. T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. (thin-walled open and closed sections, shear flow, shear centre, multi-cell torsion, crack growth, laminated plates); A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. (yield criteria, unsymmetrical bending, torsion of non-circular sections); N. E. Dowling, Mechanical Behavior of Materials, 4th ed. (Paris law, linear elastic fracture mechanics); R. C. Hibbeler, Mechanics of Materials, 10th ed. (statically indeterminate axial members, thermal stress).
Check: the direction of the 500 N load in Question 3. On the isometric view the 500 N arrow lies antiparallel to the direction in which both flanges project (both make the same 28° angle with the horizontal on the drawing, in opposite senses). The solution therefore takes the 500 N as acting away from the flange tips, so that its bending contribution puts the lower-flange tip A into tension and adds to the contribution of the 900 N load. Had the arrow been read the other way, the stress at A would be 122.5 MPa compressive instead of 155.3 MPa tensile; every section property and both moment components are unchanged, so only the sign of the 500 N load’s contribution (138.9 MPa of the 155.3) moves.
Check: two rubric details not printed on the paper. (i) Question 8 gives no mark split for its parts (a) and (b); an even 10 + 10 is assumed, consistent with every other two-part question on this sitting. (ii) In Question 4(b) “120 mm to the left of the shear center” is taken as 120 mm further forward, i.e. towards the semicircular nose, which is what “left” means on the printed figure. Moving the load the other way would simply reverse the sign of the superposed torsional flow.
Question 2: Yield prediction under a three-dimensional stress state (20 marks)
No shear acts on the $z$ faces — only $\tau_{xy}$ is quoted — so $\tau_{yz} = \tau_{zx} = 0$ and the $z$ direction is already principal.
Find. Whether the state reaches yield under (a) the maximum-shear-stress (Tresca) criterion and (b) the von Mises criterion.
Question 2 — the three Mohr's circles for the given stress state. The governing circle is the one spanning $\sigma_1$ to $\sigma_3$; its radius is the maximum shear stress, which is compared against the shear yield strength $S_Y/2$.
Approach. Reduce the state to its three principal stresses, then evaluate each criterion as an equivalent stress and compare it against the uniaxial yield strength.
Extract the principal stresses. Because $\tau_{yz}=\tau_{zx}=0$, $\sigma_z = 290$ MPa is one principal stress outright, and the other two follow from the in-plane $x$–$y$ state:$$\sigma_{a,b} = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = 45 \pm \sqrt{(-165)^2 + 75^2}$$The radius is $\sqrt{27\,225+5625} = \sqrt{32\,850} = 181.25$ MPa, so $\sigma_a = 226.25$ MPa and $\sigma_b = -136.25$ MPa.
Order them. Ranking the three values gives$$\boxed{\sigma_1 = 290\ \text{MPa},\quad\sigma_2 = 226.25\ \text{MPa},\quad\sigma_3 = -136.25\ \text{MPa}}$$The shear stress does its damage indirectly: it drags the algebraically smallest principal stress from −120 MPa down to −136.25 MPa while pushing the middle one up, widening the spread that both criteria measure.
Apply the maximum-shear-stress criterion (part a). Tresca compares the largest shear stress in the body with the shear stress at yield in a simple tension test, $S_Y/2$:$$\tau_{\max} = \frac{\sigma_1-\sigma_3}{2} = \frac{290-(-136.25)}{2} = 213.12\ \text{MPa}\qquad\text{versus}\qquad \frac{S_Y}{2} = 162.5\ \text{MPa}$$Written as an equivalent stress, $\sigma_{\text{Tresca}} = \sigma_1-\sigma_3 = 426.25$ MPa, so$$\boxed{\sigma_{\text{Tresca}} = 426.25\ \text{MPa} > S_Y = 325\ \text{MPa}\ \Rightarrow\ \text{yielding is predicted}}$$The corresponding factor of safety is $325/426.25 = 0.762$, so the state exceeds the criterion by 31 per cent.
Apply the von Mises criterion (part b). The distortion-energy criterion uses all three differences:$$\sigma_{\text{vM}} = \sqrt{\tfrac{1}{2}\big[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\big]}$$With the differences 63.75, 362.49 and −426.25 MPa this becomes $\sqrt{\tfrac12(4065+131\,401+181\,686)}$, giving$$\boxed{\sigma_{\text{vM}} = 398.21\ \text{MPa} > S_Y = 325\ \text{MPa}\ \Rightarrow\ \text{yielding is predicted}}$$with a factor of safety of $325/398.21 = 0.816$.
Verify with the invariant form. The same number must come out of the components directly, without ever finding the principal stresses:$$\sigma_{\text{vM}} = \sqrt{\sigma_x^2+\sigma_y^2+\sigma_z^2-\sigma_x\sigma_y-\sigma_y\sigma_z-\sigma_z\sigma_x+3\tau_{xy}^2}\ =\ 398.21\ \text{MPa}$$The agreement confirms both the principal stresses and the ordering.
Both criteria reach the same verdict here, which is the physically important outcome: the component yields, and no choice of failure theory rescues it. What the two criteria disagree about is the margin. Tresca always predicts the larger equivalent stress for any state that is not pure tension or pure biaxial-equal tension, and the gap between the two here is 7 per cent (426.25 against 398.21 MPa). That is close to the theoretical maximum discrepancy of 15.5 per cent, which occurs in pure shear, because this state does contain a substantial shear component. For ductile metals von Mises usually matches test data better, so Tresca is the conservative design choice rather than the accurate one.
