22-Mec-A4 Design and Manufacture of Machine Elements · May 2014

Question 8 of 8: Fillet-weld size for the lever–boss joint

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 2014 — 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. — sand casting and casting defects (Ch. 10–12), sheet-metal shearing and fine blanking (Ch. 16), bulk deformation and hydrostatic extrusion (Ch. 15), fusion welding and weld defects (Ch. 30–31).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8), statically indeterminate beams and deflection superposition (Ch. 4), welded joints in torsion (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — combined loading, transverse shear and stress transformation / Mohr's circle (Ch. 7–9).
AWS D1.1 Structural Welding Code — Steel and CSA W59 — preheat, hydrogen control and minimum fillet-weld sizes (Canadian practice).
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot tearing, solidification cracking and riser/chill practice.

Check: Part B is entirely figure-driven. Every number below was read from the printed figures (Figures A, B, C and S7). Two readings are worth stating explicitly so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the rivet group is five rivets in the top row plus one rivet 200 mm below, the lower rivet lying on the same vertical line as the third top rivet; (ii) in Figure B the 67 500 N horizontal force acts on the centroidal axis of the section, so it produces pure tension and no additional bending.

PART A — Manufacturing Processes

Question 8: Fillet-weld size for the lever–boss joint (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. A hand lever welded to a tubular boss at A:

Quantity	Value
Hand force \(F\)	100 lbf
Distance from \(F\) to the boss centre A	16 in
Boss (tube) dimensions	½-in ID × 1-in OD × 2 in long
Lever bar	\(t = 0.5\) in thick, \(w = 2\) in wide
Allowable shear stress in the weld throat	3 000 psi
Fillet welds	two circular welds, one on each face of the bar, around the 1-in OD

Find. A safe fillet-weld leg size \(h\), rounded up to a standard increment.

Figure 8.1 — The hand load applied 16 in from the boss produces a torque about the tube axis, so the circular fillet welds carry torsional shear plus a small direct shear.

Approach. Transfer the hand load to the weld group: it becomes a direct shear \(V\) through the weld centroid plus a moment. Because the lever swings in a plane perpendicular to the tube axis, that moment vector lies along the tube axis, so the circular welds are loaded in torsion, not bending. Treat the welds by the unit-weld method — compute the throat area and the polar second moment per unit leg — add the direct and torsional shears where they are collinear, and solve for the leg size that brings the throat shear to the 3 000 psi allowable.

Loads transferred to the weld group. The hand force acts 16 in from the boss centre, so at the weld \[V = F = 100\ \text{lbf}, \qquad T = F\,\ell = 100 \times 16 = \boxed{1600\ \text{lbf}\cdot\text{in}}\] The moment is about the axis of the tube, which is perpendicular to the plane of the two circular welds — the definition of torsional loading of a weld group.
Weld geometry and the unit-weld properties. Each weld is a complete circle of diameter \(d = 1.0\ \text{in}\) laid around the boss. For a fillet weld the effective throat is \(0.707h\), so with the unit-weld convention (properties computed for a line of unit width, then multiplied by the throat): \[A_u = \pi d = \pi(1.0) = 3.1416\ \text{in}, \qquad J_u = \frac{\pi d^{3}}{4} = \frac{\pi (1.0)^{3}}{4} = 0.7854\ \text{in}^{3}.\] With \(n = 2\) welds — one on each face of the lever bar where it meets the boss — the throat area and polar second moment of the group are \[A = n\,(0.707h)A_u = 2(0.707h)(3.1416) = 4.443h\ \text{in}^2,\] \[J = n\,(0.707h)J_u = 2(0.707h)(0.7854) = 1.1107h\ \text{in}^4.\]
Primary (direct) shear in the throat. The transferred force is shared by the whole throat area: \[\tau' = \frac{V}{A} = \frac{100}{4.443h} = \frac{22.51}{h}\ \text{psi (with } h \text{ in inches)}.\]
Secondary (torsional) shear in the throat. The torsional shear is greatest at the largest radius, which for a circular weld is everywhere on the circle, \(r = d/2 = 0.5\ \text{in}\): \[\tau'' = \frac{T r}{J} = \frac{1600 \times 0.5}{1.1107h} = \frac{720.3}{h}\ \text{psi}.\] The torsional component is some thirty times the direct shear, which confirms that this joint is a torsion problem with a small shear correction rather than the reverse.
Combine at the critical point. The direct shear is uniform and parallel to \(F\); the torsional shear is tangential to the circle. On a circular weld there are always two diametrically opposite points at which the tangent is parallel to \(F\), and at one of them the two components point the same way and add directly: \[\tau_{\max} = \tau' + \tau'' = \frac{22.51 + 720.3}{h} = \frac{742.8}{h}\ \text{psi}.\]
Solve for the required leg size. Setting \(\tau_{\max}\) equal to the 3 000 psi allowable, \[h \ge \frac{742.8}{3000} = 0.2476\ \text{in}.\]
Specify a standard weld size. Rounding up to the nearest standard fillet leg, \[\boxed{h = \tfrac{1}{4}\ \text{in fillet weld, all round, both faces of the lever at A}}\] Checking back, a ¼-in leg gives \(\tau_{\max} = 742.8/0.25 = 2971\ \text{psi} < 3000\ \text{psi}\), so the specification is satisfied with a small margin. A ¼-in fillet is also comfortably above the AWS D1.1 / CSA W59 minimum fillet size for a ½-in-thick member, so the weld is acceptable on code grounds as well as on stress grounds.

Check: The calculation assumes two circular fillet welds — one where each face of the 0.5-in lever bar meets the 1-in OD boss, which is the natural reading of Figure S7 (the note "2 required" on the drawing refers to the two tubular bosses at A and B, not to two welds per boss). If the joint is instead detailed with a weld at each end of each of the two bosses, \(n = 4\), every stress halves, and the required leg drops to \(h = 0.124\ \text{in}\), i.e. a ⅛-in fillet. The two-weld result is the conservative and the conventional one, and is the answer specified above.

Quantity	Value
Direct shear \(V\)	100 lbf
Torque on the weld group \(T\)	1 600 lbf·in
Throat area (2 welds)	\(A = 4.443h\) in²
Polar second moment (2 welds)	\(J = 1.1107h\) in⁴
Primary shear \(\tau'\)	22.5/\(h\) psi
Secondary shear \(\tau''\)	720.3/\(h\) psi
Combined \(\tau_{\max}\)	742.8/\(h\) psi
Required leg size	0.248 in
Specified weld	¼-in fillet all round, both faces (\(\tau = 2\,971\) psi)

Topic. Fillet welds loaded in torsion; the unit-weld (treating the weld as a line) method.

Key relations. Throat \(= 0.707h\); primary shear \(\tau' = V/A\); torsional shear \(\tau'' = Tr/J\); unit-weld properties for a circular weld of diameter \(d\): \(A_u = \pi d\), \(J_u = \pi d^{3}/4\); \(A = 0.707hA_u\), \(J = 0.707hJ_u\); resultant by vector addition at the critical point.

Why this works. A fillet weld fails in shear on its throat plane regardless of how the joint is loaded, so all weld stresses are reduced to a shear on that plane. Because the throat thickness is the only \(h\)-dependent factor and it appears linearly, the weld properties can be computed once for a line of unit throat — the "unit weld" — and then scaled, which is why \(h\) drops out of the geometry and reappears as a single divisor at the end. Recognising the load as torsion rather than bending is the crux: the moment vector here is parallel to the tube axis and therefore normal to the plane of the circular weld, which is exactly the condition for \(Tr/J\) rather than \(Mc/I\). If the lever were loaded out of plane instead, the same welds would be in bending and \(I_u = \pi d^{3}/8\) would be used.

Common pitfall. Using \(J\) for a solid 1-in shaft (\(\pi d^4/32\)) instead of the unit-weld \(J_u\) for a thin circular weld line; the two differ by a factor of \(d/8\) and the resulting weld size is wrong by an order of magnitude. A second pitfall is combining \(\tau'\) and \(\tau''\) as a root-sum-square everywhere on the circle; on a circular weld they are genuinely collinear at one point, so the direct sum is the correct governing value.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §9-3 (Stresses in welded joints in torsion) and Table 9-1 (torsional properties of fillet welds); AWS D1.1 / CSA W59 for minimum fillet sizes.

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