22-Mec-A4 Design and Manufacture of Machine Elements · December 2019

Question 6 of 6: 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 Examinations, December 2019 — 16-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions divided into Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element design); candidates answer two from Part A and two from Part B, and all questions carry equal value (25 % each). All six questions are solved here so the paper can be used as a complete study resource.

Reference texts. R. G. Budynas & J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. (Ch. 3 stress, Ch. 5 static failure, Ch. 6–7 fatigue and shafts, Ch. 9 welded joints); S. Kalpakjian & S. Schmid, Manufacturing Engineering and Technology, 8th ed. (Ch. 15 bulk deformation, Ch. 16 sheet-metal forming); M. P. Groover, Fundamentals of Modern Manufacturing, 7th ed. (Ch. 19–20); R. C. Hibbeler, Mechanics of Materials, 10th ed.

Question 6: Fillet-Weld Size for a Lever Welded to a Tubular Boss (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
Hand force at the grip	$F$	100 lbf
Moment arm, grip to the tube axis at A	$L$	16 in
Tube outside diameter (= weld circle diameter)	$d$	1.0 in
Tube inside diameter, length	—	0.5 in ID, 2 in long, two required
Number of circumferential fillet welds	$n_w$	2 (one at each end of the boss)
Lever bar thickness, width	$t,\ w$	0.5 in, 2 in
Allowable shear stress in the weld throat	$\tau_{all}$	3000 psi

Find. A safe fillet-weld leg size $h$ such that the maximum shear stress in the weld throat, from the combination of the in-plane moment and the direct shear, does not exceed 3000 psi.

Figure 6 — The hand force acting 16 in from the boss twists the weld group about the tube axis; the weld is treated as a line of unit throat.

Approach. Treat the two circumferential fillets as a weld group loaded in its own plane by a direct shear $F$ and an in-plane moment $T = FL$; use the weld-as-a-line method to get the unit area and unit polar second moment, superpose the two shear components at the worst point on the circle, and solve for the leg size that brings the resultant to the allowable.

Reduce the hand load to the weld group. The 100 lbf hand force acts 16 in from the axis of the tubular boss, so at the weld group it becomes a direct shear force plus an in-plane moment about the tube axis:

$$V = F = 100\ \text{lbf}, \qquad T = FL = 100 \times 16 = 1600\ \text{lbf}\cdot\text{in}$$

The moment is what dominates; the direct shear will turn out to be a 3 % correction.
Set up the weld group as a line. Each fillet runs right around the 1-in outside diameter of the boss, and there are two of them, one at each end of the 2-in-long tube. Treating the weld as a line of unit throat, the unit area and the unit polar second moment of area are

$$A_u = n_w \pi d = 2\pi(1.0) = 6.283\ \text{in}, \qquad J_u = n_w\frac{\pi d^3}{4} = \frac{2\pi(1.0)^3}{4} = 1.571\ \text{in}^3$$

The actual throat area and polar second moment follow by multiplying by the throat $0.707h$, which is why $h$ can be carried symbolically to the end.
Compute the torsional shear component. The moment produces a shear that is largest at the greatest radius from the group centroid, i.e. at the tube surface $r = d/2 = 0.5$ in, directed tangentially:

$$\tau'' = \frac{T r}{0.707 h\, J_u} = \frac{1600(0.5)}{0.707h(1.571)} = \frac{720.4}{h}\ \text{psi (with } h \text{ in inches)}$$
Compute the direct shear component. The direct shear is taken as uniform over the throat area:

$$\tau' = \frac{V}{0.707h\,A_u} = \frac{100}{0.707h(6.283)} = \frac{22.5}{h}\ \text{psi}$$
Combine at the critical point. The two components are vectors. At the two points on the weld circle where the torsional shear is parallel to the direct shear, they add arithmetically, and that is the governing point:

$$\tau_{max} = \frac{\tau''_{unit} + \tau'_{unit}}{h} = \frac{720.4 + 22.5}{h} = \frac{742.9}{h}\ \text{psi}$$
Solve for the required leg size. Setting $\tau_{max} = \tau_{all} = 3000$ psi,

$$h = \frac{742.9}{3000} = 0.248\ \text{in}$$

$$\boxed{\,h_{req} = 0.248\ \text{in} \ \Rightarrow\ \text{specify a } \tfrac{1}{4}\text{-in fillet weld}\,}$$
Check the specified size. At $h = 0.250$ in the actual maximum throat shear is

$$\tau = \frac{742.9}{0.250} = 2971\ \text{psi} \le 3000\ \text{psi} \quad\checkmark$$

so the ¼-in fillet satisfies the requirement with 1 % to spare. A ¼-in leg is also compatible with the 0.5-in bar thickness — the usual guidance being that the leg should not exceed the thickness of the thinner part joined, and preferably sits between about ¾ t and t for a lightly loaded attachment like this.

The answer is dominated by the moment: had the two welds been placed only at one end of the boss, $J_u$ would have halved and the required leg would have grown to nearly ½ in. Welding both ends of the 2-in boss is therefore not merely good practice for sealing the joint, it halves the weld metal actually required.

Check: This treatment assumes the weld group carries the load purely in its own plane about the tube axis — the standard reading of "the lever is welded to the tubular part at A", and the reading implied by the two circumferential welds. If the tool is instead used to pry (loading the lever out of plane), the weld group would see a bending moment about a diameter, for which $I_u = \pi d^3/4$ per weld and the resulting leg size would be similar but the stress state normal rather than shear. The 3000 psi allowable given in the question is deliberately low and already embeds a factor of safety for a hand tool, so no further design factor is applied.

Quantity	Symbol	Result
Moment on the weld group	$T$	1600 lbf·in
Unit weld area	$A_u$	6.283 in
Unit polar second moment	$J_u$	1.571 in³
Torsional shear (× h)	$\tau''h$	720.4 psi·in
Direct shear (× h)	$\tau' h$	22.5 psi·in
Required weld leg	$h$	0.248 in
Specified weld	—	¼-in fillet, all round, both ends of the boss ($\tau = 2971$ psi)

Topic. Fillet-weld groups loaded by an in-plane moment; the weld-as-a-line method.

Key relations. Throat $= 0.707h$ for an equal-leg fillet; unit properties for a circular weld $A_u = \pi d$, $J_u = \pi d^3/4$, $I_u = \pi d^3/8$ about a diameter; direct shear $\tau' = V/(0.707hA_u)$; torsional shear $\tau'' = Tr/(0.707hJ_u)$; resultant $\tau = \sqrt{\tau'^2 + \tau''^2 + 2\tau'\tau''\cos\theta}$, which degenerates to the arithmetic sum where the two are collinear.

Why this works. A fillet weld fails in shear on its throat plane whatever the direction of the applied load, so the whole design reduces to finding the largest throat shear anywhere in the group. The weld-as-a-line trick is to compute the group's geometric properties for a line of unit width and carry the throat dimension $0.707h$ as a symbolic multiplier; the leg size then falls out linearly at the end, instead of appearing inside a second moment that must be re-derived for each trial size. For a circular group, symmetry places the centroid at the tube axis, the torsional shear is constant in magnitude around the circle, and the direct shear is uniform, so the critical point is simply wherever the two happen to be parallel.

Common pitfall. Combining the two components as $\sqrt{\tau'^2+\tau''^2}$ everywhere. That is the resultant only where they are perpendicular; at the governing point they are collinear and must be added directly, and using the root-sum-square there under-predicts the stress by about 3 % in this problem and much more when the direct shear is a larger fraction. The other frequent slip is using the tube inside diameter, or the mean diameter, as the weld circle — the fillet sits on the outside of the boss, so $d$ is the OD.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §9-1 to §9-4 (butt and fillet welds, stresses in welded joints in torsion, Table 9-1 torsional properties of fillet welds treated as lines).

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