22-Mec-A4 Design and Manufacture of Machine Elements · December 2017

Question 6 of 6: Fillet-Weld Group Under an Eccentric Load

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

Notes on this paper

Paper format. National Examinations, December 2017 — 16-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, non-communicating calculator permitted. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element statics); candidates answer two from each part, and all questions carry equal value (25 %). All six questions are solved here.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology — sheet-metal forming, welding metallurgy, polymer processing.
M. P. Groover, Fundamentals of Modern Manufacturing: Materials, Processes and Systems — drawing, moulding and tooling design rules.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design — columns, welded joints, static failure theories.
R. C. Hibbeler, Mechanics of Materials — free-body diagrams, buckling.

Note on units and material data. The paper is metric throughout. Where a property is needed but not printed on the exam (the modulus of elasticity of steel, the reduction of area of 1015 steel, the minimum yield strength of E60 filler metal), the standard handbook value is used and flagged at the point of use, as the paper's own Note 1 invites.

Question 6: Fillet-Weld Group Under an Eccentric Load (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.

Quantity	Symbol	Value
Total load on the bracket	$V_{\text{tot}}$	60 kN, vertically down
Number of plates sharing the load	—	2
Horizontal welds per plate, length	$b$	2 welds at 75 mm
Vertical weld per plate, length	$d$	1 weld at 100 mm
Pin centre out from the beam face (weld line)	$e_{p}$	55 mm
Factor of safety	$n$	3.0
E60 filler metal, minimum yield strength	$S_{y}$	345 MPa
Throat-to-leg relation	$t$	$0.707h$

Find. The fillet-weld leg size $h$ that must be specified so that the most highly stressed point of the weld group satisfies a factor of safety of 3.0 against shear failure of the filler metal.

The three welds of one plate form a C, opening away from the beam. Its centroid G sits 22.5 mm out from the vertical weld, on the load side, so the moment arm is only 32.5 mm — not the 55 mm measured to the beam face.

Approach. Treat the welds by the weld-as-a-line method: compute the unit area and unit polar second moment of the C-shaped pattern, locate its centroid, resolve the eccentric load into a direct shear through the centroid plus a torsional moment about it, superpose the two shear-flow components at each candidate point to find the governing one, and finally size the throat so that the resulting stress meets the allowable at a factor of safety of 3.0.

Share the load between the two plates. The note states that the pattern of two 75 mm welds and one 100 mm weld belongs to each plate, and the isometric view shows two plates straddling the beam, so the 60 kN is carried equally by two identical weld groups: $$V = \frac{V_{\text{tot}}}{2} = \frac{60\ \text{kN}}{2} = 30\ \text{kN per plate}$$ Missing this halves nothing and doubles everything downstream, so it is worth stating explicitly before any geometry is computed.
Compute the unit properties of the C-shaped weld pattern. With $b = 75\ \text{mm}$ for each of the two horizontal welds and $d = 100\ \text{mm}$ for the vertical weld, the standard weld-as-a-line results give the total unit length, the centroid offset from the vertical weld, and the unit polar second moment: $$A_{u} = 2b + d = 2(75) + 100 = 250\ \text{mm}$$ $$\bar{x} = \frac{b^{2}}{2b+d} = \frac{75^{2}}{250} = 22.5\ \text{mm}$$ $$J_{u} = \frac{8b^{3} + 6bd^{2} + d^{3}}{12} - \frac{b^{4}}{2b+d} = 739\,583 - 126\,563$$ $$J_{u} = 613\,021\ \text{mm}^{3}$$ The centroid lies 22.5 mm out from the beam face because the two horizontal welds pull the centre of the pattern away from the vertical one.
Find the true moment arm. The 55 mm dimension on the drawing runs from the pin centre to the face of the beam, which is where the vertical weld lies — but torsion of a weld group is taken about its centroid, not about the wall. Since the centroid is 22.5 mm out on the same side as the load, $$e = e_{p} - \bar{x} = 55 - 22.5 = 32.5\ \text{mm}$$ $$M = Ve = (30\,000)(32.5) = 975\,000\ \text{N}\cdot\text{mm} = 975\ \text{N}\cdot\text{m}$$ Using 55 mm here instead of 32.5 mm would overstate the torsional term by 69 %, so this is the step to get right.
Resolve the load into its two shear-flow components. The primary component is the direct shear, uniformly distributed round the whole pattern and acting in the direction of the load: $$f' = \frac{V}{A_{u}} = \frac{30\,000}{250} = 120\ \text{N/mm downward}$$ The secondary component is torsional, growing linearly with distance from the centroid and acting perpendicular to the radius: $$f'' = \frac{M r}{J_{u}} = \frac{975\,000}{613\,021}\, r = 1.590\, r \ \ \text{N/mm}$$
Test every candidate point rather than assuming the farthest one governs. Measured from the centroid, the two corners at the beam face are at $(-22.5, \pm 50)$ and the two free ends of the horizontal welds are at $(+52.5, \pm 50)$. Superposing the components vectorially at each,

Point Radius $r$ (mm) $f''$ components (N/mm) Resultant $f$ (N/mm)

Corners at the beam face 54.8 (79.5, +35.8) 115.8

Free ends of the horizontal welds 72.6 (79.5, -83.5) 218.5

The free ends govern, and by a wide margin. The reason is not simply that their radius is larger: at the free ends the vertical component of the torsional flow points the same way as the primary shear and the two add arithmetically, whereas at the beam-face corners it points the opposite way and partly cancels it. $$\boxed{f_{\max} = 218.5\ \text{N/mm}}$$
Set the allowable shear stress for E60 filler at the required safety factor. E60xx electrodes have a minimum yield strength of 345 MPa. Fillet welds fail in shear on the throat, so the distortion-energy shear yield strength is $S_{sy} = 0.577 S_{y}$, and applying the stated factor of safety, $$\tau_{\text{all}} = \frac{0.577 S_{y}}{n} = \frac{0.577(345)}{3.0} = 66.36\ \text{MPa}$$
Size the weld. The shear flow acts on the throat, whose width is $t = 0.707h$, so $$\tau = \frac{f_{\max}}{0.707 h} \le \tau_{\text{all}} \quad \Longrightarrow \quad h \ge \frac{f_{\max}}{0.707\,\tau_{\text{all}}} = \frac{218.5}{0.707(66.36)}$$ $$h \ge 4.66\ \text{mm}$$ Rounding up to the next standard fillet size, $$\boxed{\text{Specify a 5 mm fillet weld all round}}$$ Checking back, a 5 mm leg gives a throat stress of $218.5/(0.707 \times 5) = 61.8\ \text{MPa}$ and a realised factor of safety of $0.577(345)/61.8 = 3.22$, comfortably above the required 3.0.

Point	Radius $r$ (mm)	$f''$ components (N/mm)	Resultant $f$ (N/mm)
Corners at the beam face	54.8	(79.5, +35.8)	115.8
Free ends of the horizontal welds	72.6	(79.5, -83.5)	218.5

Check: basis of the allowable stress. The paper specifies the electrode class and a factor of safety but not which allowable-stress convention to apply, and the two common bases differ substantially. The solution above uses the Shigley distortion-energy basis, $\tau_{\text{all}} = 0.577 S_{y}/n$, which is the appropriate reading when a design factor is given explicitly, and it gives $h = 4.66\ \text{mm}$, specified as 5 mm. The AISC allowable-stress convention instead uses $\tau_{\text{all}} = 0.30 S_{ut} = 0.30(427) = 128.1\ \text{MPa}$, in which the factor of safety is already embedded; carrying that value through gives $h = 2.41\ \text{mm}$, or a 3 mm fillet. The 5 mm answer is the conservative and, given the explicit factor of safety of 3.0, the intended one. A minimum leg size governed by the plate thickness may in practice require more than either.

Quantity	Symbol	Value
Load per plate	$V$	30 kN
Unit weld length	$A_{u}$	250 mm
Centroid offset from the vertical weld	$\bar{x}$	22.5 mm
Unit polar second moment	$J_{u}$	613 021 mm³
Moment arm to the weld-group centroid	$e$	32.5 mm
Torsional moment on the group	$M$	975 N·m
Primary (direct) shear flow	$f'$	120 N/mm
Governing resultant shear flow	$f_{\max}$	218.5 N/mm, at the free ends
Allowable shear stress, E60 at $n = 3.0$	$\tau_{\text{all}}$	66.36 MPa
Required fillet leg	$h$	4.66 mm
Weld size specified	—	5 mm fillet, realised $n = 3.22$

Topic. Eccentrically loaded fillet-weld groups in plane torsion, analysed by the weld-as-a-line method.

Key relations. Unit weld length $A_{u} = 2b + d$; centroid $\bar{x} = b^{2}/(2b+d)$; unit polar second moment $J_{u} = (8b^{3}+6bd^{2}+d^{3})/12 - b^{4}/(2b+d)$; primary shear flow $f' = V/A_{u}$; secondary shear flow $f'' = Mr/J_{u}$; resultant by vector addition; throat $t = 0.707h$; allowable $\tau_{\text{all}} = 0.577S_{y}/n$.

Why this works. A fillet weld's throat is thin compared with every other dimension of the joint, so the shear stress can be taken as uniform across it. That allows the throat width to be factored out of the whole analysis: compute the geometry once for a weld of unit width, obtaining $A_{u}$ and $J_{u}$ with units of length and length cubed, and only at the last step divide by the actual throat to convert a shear flow in newtons per millimetre into a stress. The mechanics is then identical to torsion of a thin section — a force through the centroid plus a couple about it — and the two shear flows superpose as vectors. The reward is that changing the required weld size does not require redoing any geometry.

Common pitfall. Three errors dominate this problem type, and this paper sets all three. The first is forgetting that two plates share the load. The second is measuring the moment arm from the load to the wall instead of to the weld-group centroid; here that would inflate the arm from 32.5 to 55 mm. The third is assuming that the point of largest radius must be the governing one; the true criterion is the vector sum, and a nearer point where the two components align can beat a farther point where they oppose — so evaluate every corner and free end.

Source. Budynas and Nisbett, Shigley's Mechanical Engineering Design, Ch. 9 (welding, bonding and the design of permanent joints): Sec. 9-3 on stresses in loaded welded joints, Table 9-1 for the torsional properties of fillet-weld groups, and Table 9-3 for the minimum strengths of E-series electrodes.

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