Question 6 of 6: Fillet-weld size for the lever at A

Question 6: Fillet-weld size for the lever at A

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.

Find. A safe fillet-weld leg size $h$ for the joint between the lever bar and the tube at A.

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 by the weld-as-a-line method to get the throat area and polar second moment per unit leg, then size the leg so that the largest combined throat shear equals the allowable.

1. Reduce the load to the weld group. Moving the 150 lbf force to the tube axis leaves a direct shear and a couple:$$V = 150\ \text{lbf}, \qquad T = F\ell = 150(16) = 2400\ \text{lbf}\cdot\text{in}$$Because the force acts in the plane of the lever, the moment vector is parallel to the tube axis - the weld ring is in torsion, not bending. This is the single decision that shapes the whole solution; a load perpendicular to the lever plane would put the same weld in bending and require $I_{u}$ rather than $J_{u}$.
2. Weld-as-a-line properties of one circumferential fillet. For a weld all round a circle of diameter $d = 1$ in (Shigley Table 9-1):$$A_{u} = \pi d = 3.1416\ \text{in}, \qquad J_{u} = \frac{\pi d^{3}}{4} = 0.7854\ \text{in}^{3}$$These are the properties of the line; multiplying by the throat $0.707h$ converts them to true area and polar second moment.
3. Group properties for the two welds. The bar is welded to the tube on both faces, so two identical rings act:$$A = 2(0.707h)A_{u} = 4.442h\ \text{in}^{2}, \qquad J = 2(0.707h)J_{u} = 1.111h\ \text{in}^{4}$$
4. Primary and secondary shear stresses. The direct shear is spread uniformly over the throat area, and the torsional shear is greatest at the outside of the ring, $r = d/2 = 0.5$ in:$$\tau' = \frac{V}{A} = \frac{150}{4.442h} = \frac{33.8}{h}\ \text{psi}, \qquad \tau'' = \frac{Tr}{J} = \frac{2400(0.5)}{1.111h} = \frac{1080.5}{h}\ \text{psi}$$The torsional component is 32 times the direct shear, which is the usual state of affairs on a lever: the weld is sized by the moment, and the direct shear is a small correction that should still be carried.
5. Combine at the critical point. Around the ring the torsional shear is everywhere tangential while the direct shear is everywhere parallel to $F$; at the two points where they are collinear they add arithmetically, and that point governs:$$\tau_{\max} = \frac{\tau' + \tau''}{1} = \frac{33.8 + 1080.5}{h} = \frac{1114.3}{h}\ \text{psi}$$(At the points where they are perpendicular the resultant is only $1081/h$ psi, so using the root-sum-square everywhere would be unconservative by 3 %.)
6. Size the weld. Setting $\tau_{\max} = \tau_{\text{all}}$,$$h \geq \frac{1114.3}{3000} = 0.371\ \text{in}$$Rounding up to the next standard fillet leg,$$\boxed{h = \tfrac{3}{8}\ \text{in fillet weld, all round, both sides}}$$At 3/8 in the actual throat shear is $1114.3/0.375 = 2971$ psi, just inside the 3000 psi allowable.

A check on proportion is worthwhile before the size is issued. A 3/8 in fillet is large against a 0.6 in bar and a 0.25 in tube wall - the general rule is that the leg should not exceed the thickness of the thinner part joined, which here is the 0.25 in tube wall. The honest engineering answer is therefore to specify 3/8 in as calculated but to note that the joint is weld-limited by the allowable stress imposed in the question: at a more usual allowable for E70 electrodes the required leg would fall to about 3/16 in. If the 3000 psi figure is to stand, the practical remedy is to weld all round both faces as assumed and to increase the tube wall, or to extend the weld onto a saddle so that the effective $J_{u}$ rises.