22-Mec-A4 Design and Manufacture of Machine Elements · May 2013

Question 5 of 8: Factor of safety of a crank shaft by the maximum-shear-stress theory

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, 07-Mec-A4 Design and Manufacture of Machine Elements, May 2013 — 3 hours, open book, any non-communicating calculator permitted. Eight questions on six pages, divided into Part A (manufacturing processes, Q1–Q4) and Part B (machine-element design, Q5–Q8). The rubric asks for three from Part A and two from Part B, five questions constituting a complete paper, all of equal value (20 % each). All eight questions are solved here, because this document is a study resource rather than an examination script.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — grinding (Ch. 26), sheet-metal forming and the forming-limit diagram (Ch. 16), sand casting and casting defects (Ch. 11–12).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — combined stresses and static failure theories (Ch. 3, 5), fatigue (Ch. 6), shaft design (Ch. 7), bolted joints in shear (Ch. 8), welded joints (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — stress transformation and combined loading.
ASM Handbook Vol. 15, Casting — gas, penetration and mould-wall-movement defects in no-bake sand systems.

Check: Part B is figure-driven. Every dimension used below was read from the printed figures (Fig. S4–S7). Where the drawing dimensions a distance from a face rather than from a bolt centre (Q7), the reading is stated explicitly in Given so a grader can substitute a different interpretation without redoing the method.

Question 5: Factor of safety of a crank shaft by the maximum-shear-stress theory (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.

Quantity	Symbol	Value
Applied hand force (downward, $-y$)	$F$	190 lbf
Shaft AB diameter	$d$	0.750 in
Wall (point A) to bend at B, along the shaft axis	$L_{AB}$	5.00 in
Crank arm B to C, perpendicular to the shaft axis	$L_{BC}$	4.00 in
C to the line of action of $F$, parallel to the shaft axis	$L_{CF}$	1.00 in
Yield strength, hot-rolled AISI 1018	$S_y$	32.0 kpsi
Tensile strength	$S_{ut}$	58.0 kpsi

Find. The factor of safety $n = S_y/\sigma'$ guarding against yielding at point A, the top surface element of the 3/4-in shaft where it enters the support, evaluated with the maximum-shear-stress (Tresca) theory.

Fig. 5-1 — Plan view of the crank of Fig. S4. The load F acts perpendicular to the plane shown, so its moment about A has a component along the shaft axis (torsion, arm 4 in) and a component transverse to the shaft (bending, arm 5 + 1 = 6 in).

Approach. Transfer the single applied force to point A as a force plus a couple, resolve that couple into a bending moment (perpendicular to the shaft axis) and a torque (along the shaft axis), convert each to a stress on the critical surface element, and apply the Tresca criterion to the resulting plane-stress state.

Locate the load relative to A and form the moment vector. Take the shaft axis as $z$, the crank-arm direction as $x$, and the load direction as $-y$. The load point sits 4 in along $x$ and $5 + 1 = 6$ in along $z$ from A, so with $\mathbf{r} = (4,\,0,\,6)$ in and $\mathbf{F} = (0,\,-190,\,0)$ lbf, $$\mathbf{M}_A = \mathbf{r}\times\mathbf{F} = \left(6F,\; 0,\; -4F\right) = \left(1140,\; 0,\; -760\right)\ \text{lbf}\!\cdot\!\text{in}$$ The component along $z$ is torsion; the component along $x$ is bending.
Separate bending from torsion. Reading the two components off the moment vector, $$M = 190\,(6.00) = 1140\ \text{lbf}\!\cdot\!\text{in}, \qquad T = 190\,(4.00) = 760\ \text{lbf}\!\cdot\!\text{in}$$ The transverse shear force of 190 lbf also acts at A, but at the top surface element (point A) the transverse shear stress is zero, so it does not enter the stress state at the critical point.
Compute the section modulus of the 3/4-in shaft. For a solid round section, $$S = \frac{\pi d^{3}}{32} = \frac{\pi (0.750)^{3}}{32} = 0.04142\ \text{in}^{3}$$ and the polar section modulus is $2S = 0.08284\ \text{in}^{3}$.
Evaluate the bending and torsional stresses at A. Substituting, $$\sigma_x = \frac{M}{S} = \frac{1140}{0.04142} = 27\,520\ \text{psi} = 27.5\ \text{kpsi}$$ $$\tau_{xz} = \frac{T}{2S} = \frac{760}{0.08284} = 9\,175\ \text{psi} = 9.17\ \text{kpsi}$$ Point A therefore carries a plane-stress element with one normal stress and one shear stress.
Find the principal stresses and the maximum shear stress. For $\sigma_x$ with $\sigma_z = 0$, $$\tau_{\max} = \sqrt{\left(\frac{\sigma_x}{2}\right)^{2} + \tau_{xz}^{2}} = \sqrt{13.76^{2} + 9.17^{2}} = 16.5\ \text{kpsi}$$ $$\sigma_1 = \frac{\sigma_x}{2} + \tau_{\max} = 30.3\ \text{kpsi}, \qquad \sigma_2 = \frac{\sigma_x}{2} - \tau_{\max} = -2.79\ \text{kpsi}$$ The two in-plane principal stresses have opposite signs, so the third principal stress (zero) lies between them and the in-plane maximum shear is the true maximum shear.
Apply the maximum-shear-stress theory. Tresca predicts yielding when $\sigma_1 - \sigma_2$ reaches $S_y$, which for this element gives the familiar combined-loading form $$\sigma' = \sigma_1 - \sigma_2 = 2\tau_{\max} = \sqrt{\sigma_x^{2} + 4\tau_{xz}^{2}} = \sqrt{27.52^{2} + 4(9.17)^{2}} = 33.1\ \text{kpsi}$$ $$\boxed{\,n = \frac{S_y}{\sigma'} = \frac{32.0}{33.1} = 0.97\,}$$
Interpret the result. A factor of safety below unity means the element at A is predicted to yield under the stated 190-lbf hand load: the crank as drawn does not have a safety margin at the support. The distortion-energy theory, which is less conservative, gives $\sigma' = \sqrt{\sigma_x^2 + 3\tau_{xz}^2} = 31.8$ kpsi and $n = 1.01$, so the design sits essentially exactly on the yield boundary by either criterion. The engineering conclusion is the same under both: the shaft diameter must be increased or a higher-strength grade specified. Raising the diameter to 7/8 in alone multiplies the section modulus by $(7/6)^{3} = 1.59$ and lifts $n$ to about 1.54.

Quantity	Symbol	Result
Bending moment at A	$M$	1140 lbf·in
Torque at A	$T$	760 lbf·in
Bending stress at A	$\sigma_x$	27.5 kpsi
Torsional shear stress at A	$\tau_{xz}$	9.17 kpsi
Principal stresses	$\sigma_1,\ \sigma_2$	30.3, −2.79 kpsi
Maximum shear stress	$\tau_{\max}$	16.5 kpsi
Tresca effective stress	$\sigma'$	33.1 kpsi
Factor of safety (MSS)	$n$	0.97 — yields

Check: assumes the 1-in dimension in Fig. S4 places the line of action of F one inch beyond C, measured parallel to the shaft axis, giving a bending arm of 5 + 1 = 6 in at A. If the intent were that F acts in the plane of C (bending arm 5 in), then $M = 950$ lbf·in, $\sigma' = 29.4$ kpsi and $n = 1.09$. The method and every intermediate relation are unchanged.

Topic: Combined bending and torsion on a circular shaft; static failure prediction by the maximum-shear-stress theory.

Key relations: $\sigma_x = 32M/\pi d^{3}$; $\tau_{xz} = 16T/\pi d^{3}$; $\tau_{\max} = \sqrt{(\sigma_x/2)^{2}+\tau_{xz}^{2}}$; Tresca $\sigma'_{\text{MSS}} = \sqrt{\sigma_x^{2}+4\tau_{xz}^{2}}$ versus distortion energy $\sigma'_{\text{DE}} = \sqrt{\sigma_x^{2}+3\tau_{xz}^{2}}$; $n = S_y/\sigma'$.

Why this works: Any single force applied away from a section can be replaced at that section by an equal force plus a couple. Decomposing the couple along and perpendicular to the shaft axis is what separates torsion from bending, and it is the whole content of a "crank" problem. On the surface of a round shaft the stress state is plane stress with one normal and one shear component, for which both classical yield criteria collapse to a single closed-form expression — hence the familiar $\sqrt{\sigma^2 + 4\tau^2}$ and $\sqrt{\sigma^2 + 3\tau^2}$ forms. Tresca is chosen here because the question names it; it is the conservative of the two, predicting yield at a load about 15 % below the distortion-energy prediction in pure shear.

Common pitfall: Using the crank-arm length as the bending arm and the shaft length as the torsion arm, which is exactly backwards. The torque arm is the offset perpendicular to the shaft axis (4 in) and the bending arm is the distance along the shaft axis to the load plane (6 in). A second pitfall is including the transverse shear stress $4V/3A$ at point A; that stress is maximum on the neutral axis and zero at the top and bottom surfaces where the bending stress peaks, so it never combines with the maximum bending stress.

Source: Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §3-9 to 3-13 (combined loading on shafts) and §5-4 (maximum-shear-stress theory for ductile materials).

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