22-Mec-A4 Design and Manufacture of Machine Elements · May 2013

Question 8 of 8: Fillet-weld size for a lever welded to a tubular boss

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 2013 — 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. — grinding (Ch. 26), sheet-metal forming and the forming-limit diagram (Ch. 16), sand casting and casting defects (Ch. 11–12).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — combined stresses and static failure theories (Ch. 3, 5), fatigue (Ch. 6), shaft design (Ch. 7), bolted joints in shear (Ch. 8), welded joints (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — stress transformation and combined loading.
ASM Handbook Vol. 15, Casting — gas, penetration and mould-wall-movement defects in no-bake sand systems.

Check: Part B is figure-driven. Every dimension used below was read from the printed figures (Fig. S4–S7). Where the drawing dimensions a distance from a face rather than from a bolt centre (Q7), the reading is stated explicitly in Given so a grader can substitute a different interpretation without redoing the method.

Question 8: Fillet-weld size for a lever welded to a tubular boss (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.

Quantity	Symbol	Value
Hand force at the grip	$F$	100 lbf
Grip to the weld centre at A	$L$	16.0 in
Lever bar thickness × width	$t \times w$	0.500 × 2.00 in
Tubular boss at A: 1/2-in ID × 1-in OD × 2 in long	$d$	1.00 in OD
Number of circumferential fillet welds (one at each end of the boss)	$N$	2
Allowable shear stress in the weld throat	$\tau_{\text{allow}}$	3000 psi

Find. The fillet-weld leg size $h$ that keeps the shear stress in the weld throat at or below 3000 psi, then the nearest standard size to specify.

Fig. 8-1 — The lever of Fig. S7 viewed along the axis of the tubular boss. The hand force 16 in from A produces a moment whose vector lies along the tube axis, so the weld group is loaded in torsion about its own centroid plus the direct hand force.

Approach. Transfer the hand force to the weld group as a direct shear plus a torque about the tube axis, treat the two circumferential fillets as lines of unit throat, evaluate the torsional and direct shear stresses per unit weld size, superpose them at the worst point on the circle, and solve for the leg $h$.

Transfer the hand load to the weld group. The tube axis at A is perpendicular to the plane of the lever, and the 100-lbf hand force acts in that plane 16 in away. Its moment vector therefore lies along the tube axis, which means the weld sees a torque, not a bending moment: $$T = F L = 100\,(16.0) = 1600\ \text{lbf}\!\cdot\!\text{in}, \qquad V = F = 100\ \text{lbf}$$
Model the weld group as lines. The lever is secured by one circumferential fillet at each end of the 2-in boss, so the group is two circles of diameter $d = 1.00$ in. Treating each as a line of unit throat and using the standard circular-weld properties, $$A_u = N\pi d = 2\pi(1.00) = 6.283\ \text{in}, \qquad J_u = N\,\frac{\pi d^{3}}{4} = 2\,\frac{\pi(1.00)^{3}}{4} = 1.571\ \text{in}^{3}$$ The real throat area and polar second moment are these values multiplied by the throat $0.707h$.
Compute the torsional shear stress. The critical radius is the weld radius $r = d/2 = 0.500$ in, so $$\tau_{\text{tors}} = \frac{T r}{0.707h\,J_u} = \frac{1600\,(0.500)}{0.707h\,(1.571)} = \frac{720.3}{h}\ \text{psi (with }h\text{ in inches)}$$ This component is tangential to the weld circle at every point.
Compute the direct shear stress. The hand force is carried uniformly over the whole throat, $$\tau_{\text{dir}} = \frac{V}{0.707h\,A_u} = \frac{100}{0.707h\,(6.283)} = \frac{22.5}{h}\ \text{psi}$$ which is only about 3 % of the torsional term — the weld is essentially a torsion problem, as expected for a 16-in lever arm on a 1-in-diameter weld circle.
Superpose at the worst point. The two components are collinear at the point on the weld circle where the tangential torsional shear runs parallel to the direct shear, so they add arithmetically there: $$\tau_{\max} = \frac{720.3 + 22.5}{h} = \frac{742.8}{h}\ \text{psi}$$
Solve for the required leg size. Setting $\tau_{\max} = \tau_{\text{allow}} = 3000$ psi, $$\boxed{\,h = \frac{742.8}{3000} = 0.248\ \text{in}\,}$$
Specify a standard size and check it. Rounding up to the nearest commercially specified fillet gives $h = 1/4$ in (0.250 in), at which the throat stress is $$\tau = \frac{742.8}{0.250} = 2971\ \text{psi} < 3000\ \text{psi}\quad\checkmark$$ The specified weld is also compatible with the parts it joins: the tube wall is $(1.00-0.50)/2 = 0.250$ in thick and the lever is 0.500 in thick, so a 1/4-in leg neither exceeds the thinner member nor demands an unreasonable number of passes. It can be laid in a single pass with a 5/32-in electrode. A designer wanting a smaller weld would move the weld outward to a larger boss diameter, since the torsional capacity of a circular weld scales with $d^{2}$ — going from a 1-in to a 1.5-in weld circle would more than halve the required leg.

Quantity	Symbol	Result
Torque on the weld group	$T$	1600 lbf·in
Direct shear force	$V$	100 lbf
Unit weld area (2 circles, $d=1$ in)	$A_u$	6.283 in
Unit polar second moment	$J_u$	1.571 in³
Torsional shear (per unit $h$)	$\tau_{\text{tors}}h$	720 psi·in
Direct shear (per unit $h$)	$\tau_{\text{dir}}h$	22.5 psi·in
Required weld leg	$h$	0.248 in
Specified fillet weld	$h$	1/4 in (τ = 2971 psi)

Check: assumes the two fillet welds indicated in Fig. S7 run circumferentially around the 1-in-OD boss, one at each end of the 2-in-long tube, and that the lever is loaded only by the 100-lbf hand force. If instead a single circumferential weld is used, the required leg doubles to 0.50 in, which would be an unreasonable weld on a 0.25-in tube wall — confirming that the two-weld reading of the figure is the intended one.

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

Key relations: Throat $= 0.707h$; for a circular weld of diameter $d$, $A_u = \pi d$ and $J_u = \pi d^{3}/4$; torsional shear $\tau = Tr/(0.707h\,J_u)$; direct shear $\tau = V/(0.707h\,A_u)$; combine vectorially, or arithmetically where the components align.

Why this works: A fillet weld fails in shear on its throat plane regardless of the direction in which it is loaded, which is why every weld calculation reduces to one allowable shear stress. Because the throat dimension is small compared with the weld's plan dimensions, the weld can be idealised as a line of unit width: the geometric properties $A_u$ and $J_u$ then depend only on the weld pattern and are tabulated once and for all, and the unknown leg size $h$ factors cleanly out of the stress expression. That is what makes weld sizing a one-line calculation once the pattern is chosen.

Common pitfall: Treating the 1600 lbf·in as a bending moment on the weld. The moment vector here is parallel to the tube axis and therefore perpendicular to the plane of the weld circles, which makes it torsion; a bending treatment would use $I_u$ instead of $J_u$ and give the wrong answer by a factor of $\sqrt2$ or more. The other frequent error is omitting the 0.707 throat factor and sizing on the leg dimension directly, which under-sizes the weld by 41 %.

Source: Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §9-3 (stresses in fillet welds), §9-4 (torsion in welded joints) and Table 9-1 (torsional properties of common fillet-weld patterns).

← Question 7 Back to the paper →