22-Mec-A4 Design and Manufacture of Machine Elements · December 2016

Question 4 of 6

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

Notes on this paper

Paper format. National Examinations, December 2016 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3) is qualitative manufacturing-process theory, Part B (Q4–Q6) is quantitative machine-element design. The candidate answers two from Part A and two from Part B; four questions of equal value (25 % each) constitute a complete paper. All six are solved here, because the set is a study resource rather than an exam script.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. — Part A (sheet-metal forming, casting, machine-tool selection).
M. P. Groover, Fundamentals of Modern Manufacturing: Materials, Processes and Systems, 7th ed. — fine blanking, die casting, turning-machine classes.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. — Part B (welded joints Ch. 9, shafts Ch. 7, bolted and riveted joints Ch. 8).
R. L. Norton, Machine Design: An Integrated Approach, 6th ed. — corroborating weld and riveted-joint treatments.

Check — weld allowable stress basis (Q4). The paper gives an electrode ultimate strength (E60, S_u = 60 ksi) and a safety factor of 3.0, but not the strength theory. This solution uses the distortion-energy shear strength of the deposited metal, S_su = 0.577 S_u, divided by the given factor — the treatment used throughout Shigley Ch. 9. The alternative code route (AISC allowable 0.30 S_u, which already embeds its own reserve) is worked out at the end of Q4 and gives a smaller leg. State whichever basis you adopt; the marker is looking for the weld-as-a-line method, not the code table.

Question 4 (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.

Total load on the bracket	60 kN, vertical, downward
Number of plates sharing the load	2 (identical weld groups)
Weld pattern per plate	two horizontal welds b = 75 mm, one vertical weld d = 100 mm
Load line to the vertical weld	55 mm + 75 mm = 130 mm
Electrode	E60 series, S_u = 60 ksi = 413.7 MPa
Required factor of safety	n = 3.0

Find. The fillet-weld leg size h to be specified on the drawing, such that the most highly stressed point in the weld group satisfies the factor of safety of 3.0 against shear failure of the deposited metal.

Figure 4.1 — Weld group on one plate, treated as a line. The centroid G lies 22.5 mm out from the vertical weld, so the moment arm is 130 − 22.5 = 107.5 mm. The corner A at the far end of a horizontal weld has the largest radius from G and the most unfavourable direction, and governs.

Approach. Treat the fillet weld as a line of unit throat, share the 60 kN equally between the two identical plates, resolve the eccentric in-plane load at each weld group into a direct shear through the centroid plus a couple, superpose the two shear intensities vectorially at each candidate corner, and size the throat so the largest resultant meets the allowable shear stress.

Load per weld group. The two plates are identical and symmetrically placed, so each carries half the bracket load: $$V=\frac{60\ \text{kN}}{2}=30\ \text{kN}$$
Weld-line length and centroid. With the vertical weld on the wall line taken as $x=0$ and the horizontal welds running outward to $x=-b$, $$A_{u}=2b+d=2(75)+100=250\ \text{mm}$$ $$\bar{x}=\frac{2b\left(-\tfrac{b}{2}\right)+d(0)}{A_{u}}=\frac{2(75)(-37.5)}{250}=-22.5\ \text{mm}$$ The centroid therefore lies 22.5 mm out from the vertical weld, and by symmetry on the mid-height line.
Unit second polar moment of the weld line. Taking the two horizontal welds and the vertical weld about the centroid, $$I_{xu}=\frac{d^{3}}{12}+2b\left(\frac{d}{2}\right)^{2}=\frac{100^{3}}{12}+2(75)(50)^{2}=458\,333\ \text{mm}^{3}$$ $$I_{yu}=d\,\bar{x}^{2}+2\left[\frac{b^{3}}{12}+b\left(\frac{b}{2}-|\bar{x}|\right)^{2}\right]=154\,688\ \text{mm}^{3}$$ so that $$J_{u}=I_{xu}+I_{yu}=613\,021\ \text{mm}^{3}$$ Shigley's closed form for this three-sided pattern, $J_{u}=\left(8b^{3}+6bd^{2}+d^{3}\right)/12-b^{4}/(2b+d)$, returns the same 613 021 mm³, which confirms both the geometry and the centroid.
Moment about the weld-group centroid. The load line stands 130 mm out from the vertical weld, and the centroid is 22.5 mm out, so $$e=130-22.5=107.5\ \text{mm}\qquad M=Ve=30\,000(107.5)=3.225\times10^{6}\ \text{N}\!\cdot\!\text{mm}$$
Primary (direct) shear intensity. Spread uniformly over the weld line, $$f'=\frac{V}{A_{u}}=\frac{30\,000}{250}=120\ \text{N/mm}$$ acting vertically downward at every point.
Secondary (torsional) shear intensity at the critical corner. The furthest point from the centroid is the outer end of a horizontal weld, at $(x,y)=(-75,\,\pm50)$ relative to the vertical weld, hence at $$r=\sqrt{(75-22.5)^{2}+50^{2}}=\sqrt{52.5^{2}+50^{2}}=72.5\ \text{mm}$$ from G, giving $$f''=\frac{Mr}{J_{u}}=\frac{3.225\times10^{6}(72.5)}{613\,021}=381.4\ \text{N/mm}$$ directed perpendicular to r, that is with components 263.0 N/mm horizontally and 262.5 N/mm vertically downward at the upper outer corner.
Resultant at the critical point. Adding the two contributions as vectors at that corner, $$f=\sqrt{\left(f''_{x}\right)^{2}+\left(f'+f''_{y}\right)^{2}}=\sqrt{263.0^{2}+(120+262.5)^{2}}=475.6\ \text{N/mm}$$ The same construction at the end of the vertical weld gives only 263.0 N/mm, so $$\boxed{f_{\max}=475.6\ \text{N/mm per unit throat, at the outer corner of a horizontal weld}}$$
Allowable shear stress in the deposited metal. The distortion-energy shear strength of an E60 electrode, divided by the specified factor of safety, is $$S_{su}=0.577S_{u}=0.577(413.7)=238.7\ \text{MPa}\qquad \tau_{\text{all}}=\frac{S_{su}}{n}=\frac{238.7}{3.0}=79.6\ \text{MPa}$$
Required leg size. For a fillet weld the throat is 0.707h, so the stress is $\tau=f_{\max}/(0.707h)$ and $$h=\frac{f_{\max}}{0.707\,\tau_{\text{all}}}=\frac{475.6}{0.707(79.6)}=8.45\ \text{mm}$$ Rounding up to the next standard fillet leg, $$\boxed{h=10\ \text{mm fillet, all round the three-sided pattern on both plates}}$$ At 10 mm the actual weld stress is 475.6/(0.707 × 10) = 67.3 MPa, giving an achieved factor of safety of 238.7/67.3 = 3.55 against the electrode shear strength — comfortably above the required 3.0.

Check — alternative code basis. If instead the AISC allowable for fillet welds is used, τ_all = 0.30 S_u = 124.1 MPa (that allowable already contains its own reserve against the electrode ultimate), the required leg becomes 475.6/(0.707 × 124.1) = 5.42 mm, so a 6 mm fillet would be specified. The two answers bracket normal practice; the higher one is quoted as the design because the question explicitly imposes a safety factor of 3.0 on the given ultimate strength. Note also that this solution assumes the 55 mm dimension runs from the load line to the near end of the 75 mm welds, so that the load stands 130 mm from the wall; that reading is consistent with the scaled figure.

Q4 — Weld group results (per plate)
Quantity	Symbol	Value
Load per plate	V	30.0 kN
Unit weld length	A_u	250 mm
Centroid offset from the vertical weld	x̄	22.5 mm
Unit second polar moment	J_u	613 021 mm³
Moment arm to the centroid	e	107.5 mm
Moment on the weld group	M	3.225 × 10⁶ N·mm
Primary shear intensity	f′	120.0 N/mm
Secondary shear intensity at the critical corner	f″	381.4 N/mm
Resultant shear intensity	f_max	475.6 N/mm
Allowable shear stress (0.577 S_u/n)	τ_all	79.6 MPa
Required leg	h	8.45 mm
Specified fillet leg	h	10 mm

Topic: Eccentrically loaded fillet-weld groups — the weld-as-a-line method.

Key relations: Direct shear intensity $f'=V/A_u$; torsional shear intensity $f''=Mr/J_u$; resultant by vector addition; stress $\tau=f/(0.707h)$ with the throat measured on the 45° plane of the fillet. All weld-line properties are computed per unit throat and scaled by 0.707h at the end, which is what makes the leg size fall out of one division.

Why this works: A fillet weld is thin compared with its length, so its cross-section can be idealized as a line of unit width. Every area property then reduces by one power of length — area becomes length, polar moment becomes a third-power term — and because the throat is the same everywhere it factors straight out of the stress expression. The pattern here is loaded in its own plane, so the couple produces a torsional (rotational) shear distribution about the centroid, not a bending distribution; that is why J_u rather than I_u appears.

Common pitfall: Taking the moment arm to the wall (130 mm) instead of to the weld-group centroid (107.5 mm), which overstates the couple by 21 %; and assuming that the point furthest from the centroid automatically governs. Here it does, but only after checking — the governing corner is the one where the secondary shear has the largest component in line with the primary shear, which is not always the most distant one.

Source: Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., Sec. 9-3 and 9-4 (torsion in welded joints; Table 9-1 torsional properties of fillet welds) and Table 9-6 (minimum fillet sizes and allowable stresses); Norton, Machine Design, 6th ed., Sec. 15.9.

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