22-Mec-A4 Design and Manufacture of Machine Elements · December 2019

Question 4 of 6: Crank Under Combined Bending and Torsion — Factor of Safety at A

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

Question 4: Crank Under Combined Bending and Torsion — Factor of Safety at A (25 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.

Find. The factor of safety $n$ against yielding at point A — the point on the top surface of the ¾-in shaft where it enters the support at the origin — using the maximum-shear-stress theory.

Approach. Transfer the crank-pin load back to point A as a shear force, a bending moment and a torque; convert those to the bending and torsional stresses on an element on the top surface at A; then apply the maximum-shear-stress criterion, $n = (S_y/2)/\tau_{max}$.

1. Resolve the load at A into a force–couple system. The position of the crank pin C relative to A is $\vec{r} = (L,\,h,\,a)$ where $h$ is the vertical rise of the arm, and the load is $\vec{F} = (0,\,-F,\,0)$. The moment at A is

   $$\vec{M}_A = \vec{r}\times\vec{F} = (L,\,h,\,a)\times(0,\,-F,\,0) = (F a,\;0,\;-F L)$$

   The $h$ term drops out entirely, because a vertical offset is parallel to a vertical force and can generate no moment — which is why the 1-in rise and the ¼-in pin projection shown in the figure do not enter the calculation.

2. Identify the torque and the bending moment. The $x$-component of $\vec{M}_A$ acts about the shaft axis and is therefore the torque, while the $z$-component acts about a transverse axis and is the bending moment:

   $$T = F a = 200 \times 4 = 800\ \text{lbf}\cdot\text{in}, \qquad M = F L = 200 \times 5 = 1000\ \text{lbf}\cdot\text{in}$$

   There is in addition a transverse shear force $V = F = 200$ lbf at A. Point A lies on the top surface of the shaft, i.e. on the extreme fibre of the bending distribution, and the transverse-shear stress $\tau = VQ/Ib$ is zero at the extreme fibre. It is therefore correctly omitted, and A is simultaneously the point of maximum bending stress.

3. Compute the bending stress at A. For a solid round shaft the section modulus is $I/c = \pi d^3/32$, so

   $$\sigma_x = \frac{32M}{\pi d^3} = \frac{32(1000)}{\pi (0.75)^3} = \frac{32\,000}{1.3254} = 24\,144\ \text{psi} = 24.14\ \text{kpsi}$$

   Point A is on the top surface and the moment puts the top fibre in tension, so this is a tensile normal stress along the shaft axis.

4. Compute the torsional shear stress at A. The polar section modulus is $J/c = \pi d^3/16$, giving

   $$\tau_{xz} = \frac{16T}{\pi d^3} = \frac{16(800)}{\pi (0.75)^3} = \frac{12\,800}{1.3254} = 9658\ \text{psi} = 9.66\ \text{kpsi}$$

   This shear acts on the same element as the bending stress, so the element at A carries a plane stress state $\sigma_x = 24.14$ kpsi, $\sigma_z = 0$, $\tau_{xz} = 9.66$ kpsi.

5. Find the maximum in-plane shear stress. From Mohr's circle, the radius of the circle for this state is

   $$\tau_{max} = \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xz}^2} = \sqrt{(12.072)^2 + (9.658)^2} = \sqrt{145.74 + 93.27} = 15.46\ \text{kpsi}$$

   The corresponding principal stresses, the circle's centre plus and minus its radius, are

   $$\sigma_1 = 12.07 + 15.46 = 27.53\ \text{kpsi}, \qquad \sigma_2 = 12.07 - 15.46 = -3.39\ \text{kpsi}$$

   Because $\sigma_1$ and $\sigma_2$ have opposite signs, the third principal stress (zero, the free surface) lies between them, and the maximum shear stress in three dimensions is the in-plane value just computed — so $\tau_{max} = 15.46$ kpsi is the correct quantity for the MSS criterion.

6. Apply the maximum-shear-stress theory. MSS predicts yielding when the maximum shear stress reaches the shear yield strength, which the theory takes as $S_{sy} = S_y/2$. Hence

   $$n = \frac{S_{sy}}{\tau_{max}} = \frac{S_y/2}{\tau_{max}} = \frac{30/2}{15.46} = \frac{15.0}{15.46}$$

   $$\boxed{\,n = 0.97\,}$$

The factor of safety is slightly less than unity, so on the maximum-shear-stress criterion the crank is predicted to yield at point A under the full 200 lbf hand load. This is not an arithmetic accident — it is the answer the question is looking for. MSS is the conservative of the two ductile static theories; the distortion-energy theory applied to the same element gives $\sigma' = \sqrt{\sigma_x^2+3\tau_{xz}^2} = \sqrt{24.14^2+3(9.66)^2} = 29.4$ kpsi and hence $n_{DE} = 30/29.4 = 1.02$, marginally above one. The engineering conclusion is the same either way: the design has no useful margin, the shaft diameter should be increased (going from ¾ in to 7/8 in raises $n$ by the cube of the diameter ratio, to about 1.54), or a higher-strength steel specified.