22-Mec-A4 Design and Manufacture of Machine Elements · May 2018
Question 4 of 6: Maximum stress and factor of safety of a cantilevered box section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination 16-Mec-A4, Design and Manufacture of Machine Elements, May 2018. Three hours, open book, non-communicating calculator permitted. Six questions in two parts: Part A (Q1-Q3, manufacturing processes) and Part B (Q4-Q6, machine-element design). Candidates answer two questions from each part; four questions constitute a complete paper and all questions carry equal value (25 % each). All six questions are worked here.
Reference texts for this subject.
S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. (polymer processing: thermoforming, compression and transfer moulding, extrusion and die swell).
M. P. Groover, Fundamentals of Modern Manufacturing: Materials, Processes and Systems, 7th ed. (shaping processes for plastics).
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 10th ed. (Ch. 5 failure prevention, Ch. 9 welded joints, Ch. 16 brakes and clutches).
R. C. Hibbeler, Mechanics of Materials, 10th ed. (combined loading, transverse shear).
CSA S16 / CSA G40.21 for Canadian structural steel grades; ASTM A36 is quoted in Q4 as printed on the paper, and the Canadian equivalent 300W is noted where it matters.
Check: Part B is figure-driven; every dimension used below was read from the printed figures. Where a figure is not dimensionally self-consistent (Q5, the drum is drawn oversized relative to the lever dimensions) the reading adopted and its effect on the answer are stated explicitly in that question.
Question 4: Maximum stress and factor of safety of a cantilevered box section
Find. The maximum stress in the member and the factor of safety against yielding, identifying which point of which section governs.
Figure 4.1 - Cantilever, loading and hollow rectangular section. C is the top fibre at the built-in end (maximum bending stress); D lies on the neutral axis (maximum transverse shear).
Approach. Find the internal shear and moment at the built-in end, where both are largest; evaluate the bending stress at the extreme fibre (point C) and the transverse shear stress at the neutral axis (point D); reduce each to an equivalent (von Mises) stress and take the smaller factor of safety.
Section properties of the hollow rectangle. The section is the difference of two solid rectangles, outer $b \times h$ and inner $(b-2t) \times (h-2t) = 38 \times 88$ mm:$$I = \frac{bh^{3} - b_{i}h_{i}^{3}}{12} = \frac{50(100)^{3} - 38(88)^{3}}{12} = \frac{50.000\times10^{6} - 25.896\times10^{6}}{12} = 2.009\times10^{6}\ \text{mm}^{4}$$The area is $A = 50(100) - 38(88) = 1656\ \text{mm}^{2}$, and the section modulus is $S = I/c = 2.009\times10^{6}/50 = 40.17\times10^{3}\ \text{mm}^{3}$.
Internal actions at the built-in end. Both the tip load and the self weight act downwards over the full span, so the wall section carries the largest shear and the largest moment:$$V = P + wL = 13\,000 + 230(1.5) = 13\,345\ \text{N}$$$$M = PL + \frac{wL^{2}}{2} = 13\,000(1.5) + \frac{230(1.5)^{2}}{2} = 19\,500 + 258.75 = 19\,758.75\ \text{N}\cdot\text{m}$$The self weight contributes only 1.3 % of the moment, but it is retained because the question supplies it.
Bending stress at point C (top fibre, built-in end). The flexure formula with $c = h/2 = 50$ mm gives$$\sigma_{C} = \frac{Mc}{I} = \frac{19.759\times10^{6}\ \text{N}\cdot\text{mm} \times 50\ \text{mm}}{2.009\times10^{6}\ \text{mm}^{4}} = \boxed{491.8\ \text{MPa}}$$At C the transverse shear is zero (the first moment of the area outside the top fibre is zero), so this is a state of pure uniaxial tension and the von Mises stress equals 491.8 MPa.
Transverse shear at point D (neutral axis). The first moment of the area above the neutral axis is the outer half-rectangle less the inner half-rectangle:$$Q = b\frac{h}{2}\frac{h}{4} - b_{i}\frac{h_{i}}{2}\frac{h_{i}}{4} = 50(50)(25) - 38(44)(22) = 62\,500 - 36\,784 = 25\,716\ \text{mm}^{3}$$Two webs of 6 mm each carry the shear, so the effective width is $2t = 12$ mm:$$\tau_{D} = \frac{VQ}{I\,(2t)} = \frac{13\,345(25\,716)}{2.009\times10^{6}(12)} = 14.24\ \text{MPa}$$Point D lies on the neutral axis, so the bending stress there is zero and the state is pure shear; the equivalent stress is $\sigma'_{D} = \sqrt{3}\,\tau_{D} = 24.66$ MPa.
Factors of safety. Comparing each equivalent stress with the yield strength,$$n_{C} = \frac{S_{y}}{\sigma_{C}} = \frac{250}{491.8} = \boxed{0.51}, \qquad n_{D} = \frac{250}{24.66} = 10.1$$Bending at the built-in end governs by a factor of twenty, which is the expected result for a slender cantilever: transverse shear is never critical unless the span-to-depth ratio falls below about five.
The governing factor of safety is less than unity. The member as dimensioned does not merely fail to meet a design margin - it yields under the stated load, since the required stress of 491.8 MPa is very nearly twice the 250 MPa yield strength of A36. The correct answer to the question as asked is therefore the stress and the number 0.51, together with the statement that the section is inadequate; the arithmetic should not be adjusted to force a passing result.
Check: The result $n = 0.51 < 1$ is the honest consequence of the printed data (13 kN on a 1.5 m cantilever of a 50 x 100 x 6 box). It is a legitimate examination answer - the examiner asks for the factor of safety, not for a design that passes. If a design revision were wanted, the required section modulus is $S \geq M/S_{y} = 19.759\times10^{6}/250 = 79.0\times10^{3}\ \text{mm}^{3}$, i.e. about double the 40.17 x 103 mm3 available: a 50 x 150 x 6 box or an increase in wall thickness to about 12 mm would be needed, before any margin is added. Tip deflection is also excessive at 36.8 mm (about $L/41$ against a normal $L/240$ limit), confirming that the section is undersized rather than that the stress calculation is in error.
