22-Mec-A4 Design and Manufacture of Machine Elements · December 2018

Question 5 of 6: Welded Sheet-Steel Bracket — Allowable Load (25 marks)

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

Question 5: Welded Sheet-Steel Bracket — Allowable 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.

Find. The largest total load $W$ the bracket can carry without the combined stress in the weld metal exceeding 850 psi.

Approach. Use Shigley's weld-as-a-line method: reduce the weld group to a line of unit width, obtain the unit area and unit section modulus from Table 9-2, convert each to a throat property by multiplying by $0.707h$, then combine the primary (direct shear) and secondary (bending) stresses vectorially and set the resultant equal to the allowable.

1. Reduce the loading to the weld plane. The load is uniform over the 8-in projection of the top flange, so its resultant acts at mid-length: $$V = W \qquad M = W\!\left(\frac{8}{2}\right) = 4W \ \ \text{[lb}\cdot\text{in]}$$
2. Identify the weld pattern and its unit properties. Two parallel vertical fillet welds of length $d = 8$ in run up the sides of the mounting flange. From Shigley Table 9-2 (case 3), $$A_u = 2d = 16.0\ \text{in} \qquad I_u = \frac{d^{3}}{6} = 85.33\ \text{in}^{3} \qquad S_u = \frac{I_u}{d/2} = \frac{d^{2}}{3} = 21.33\ \text{in}^{2}$$ The 1-in flange width does not enter, because the bending axis is horizontal and both welds lie in the plane of the wall.
3. Convert to throat properties. A fillet weld of leg $h$ has throat $0.707h$, so $$A = 0.707hA_u = 0.707(0.0598)(16.0) = 0.6765\ \text{in}^{2}$$ $$S = 0.707hS_u = 0.707(0.0598)(21.33) = 0.9020\ \text{in}^{3}$$
4. Write the primary shear stress. The direct load is carried uniformly over the throat area: $$\tau' = \frac{V}{A} = \frac{W}{0.6765} = 1.478W$$
5. Write the secondary (bending) stress. The moment produces a normal stress on the throat, treated in this method as a stress component perpendicular to $\tau'$: $$\tau'' = \frac{M}{S} = \frac{4W}{0.9020} = 4.435W$$ Bending dominates by a factor of three, which is what one expects from an 8-in cantilever welded over only 8 in of height.
6. Combine the two components. They act at right angles on the throat, so $$\tau = \sqrt{(\tau')^{2}+(\tau'')^{2}} = W\sqrt{1.478^{2}+4.435^{2}} = 4.675W$$
7. Impose the allowable stress and solve. Setting $\tau = \tau_{\text{all}} = 850$ psi, $$W = \frac{850}{4.675}$$ $$\boxed{W \approx 182\ \text{lb}}$$

The result is modest because the whole assembly is 16-gauge sheet: the throat is only 0.042 in deep, so 16 in of weld provides just two-thirds of a square inch of shear area. If a larger capacity were needed, the effective move is to increase the welded height rather than the leg size, since the bending term varies as $d^{2}$ while the shear term varies only as $d$.