22-Mec-A4 Design and Manufacture of Machine Elements · May 2015

Question 5 of 6: Load Capacity of a Welded Sheet-Steel Bracket

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

Notes on this paper

Paper format. National Examinations, May 2015 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element design). The rubric asks for two questions from each part, and all questions carry equal value (25 % each). All six are solved here, because this set is a study resource rather than an examination script.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. — sheet-metal forming, deep drawing, abrasive machining.
M. P. Groover, Fundamentals of Modern Manufacturing, 7th ed. — stretch forming and the flow-curve treatment of forming loads.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. — bolted and welded joints (Ch. 8, 9) and brakes (Ch. 16).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — net-section bending and bearing stress.

Units. The paper mixes systems deliberately: Q1, Q5 and Q6 are in US customary units (inch, pound, psi) and Q4 is metric. Each question is solved in the units in which it is set, as the exam intends.

Question 5: Load Capacity of a Welded Sheet-Steel Bracket (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
Sheet gauge (#16 ga)	$h$	0.0598 in
Mounting flange (welded to the support)	$b \times d$	1 in × 8 in
Top flange (same dimensions), load-carrying reach	—	8 in from the support
Load	$W$	uniformly distributed along the top flange
Allowable combined stress in the weld metal	$\tau_{allow}$	900 psi

Find. The total load W the welded bracket can support without the combined stress on the weld throat exceeding 900 psi.

Figure 4 — the bracket in elevation. The 1 in by 8 in mounting flange is fillet welded all round to the support; W is uniformly distributed along the 8 in top flange, so its resultant acts 4 in out from the weld plane and the weld group carries direct shear plus an out-of-plane couple.

Check: two modelling choices. First, the fillet leg is taken equal to the sheet thickness, $h = 0.0598$ in, which is standard practice for welding sheet to a heavier support — a larger leg cannot be deposited without burning through the sheet. Second, the weld is assumed to run all round the perimeter of the 1 in by 8 in mounting flange (18 in of weld); welding only the two 8 in edges would reduce the throat area to 0.676 in² and $I_u$ to 85.3 in³, lowering the answer to about 219 lb. Both assumptions are stated because the drawing shows the flange but not a weld symbol.

Approach. Treat the fillet weld as a line, compute the unit area and unit second moment of the rectangular weld pattern from Shigley's table, convert both to throat properties, then superpose the uniform primary shear from the vertical load with the normal stress from the couple and set the resultant equal to the allowable.

Resolve the load at the weld plane. W is distributed uniformly along the 8 in top flange, so its resultant acts at the mid-length of that flange: $$V = W, \qquad M = W(4.0) = 4W\ \text{lbf}\!\cdot\!\text{in}$$ The couple bends the bracket away from the support, so it stresses the weld throat in tension at the top and compression at the bottom.
Unit properties of the weld pattern. For a rectangle welded all round, with $b = 1$ in across and $d = 8$ in deep, Shigley Table 9–2 gives $$A_u = 2(b + d) = 2(1 + 8) = 18.0\ \text{in}, \qquad I_u = \frac{d^2(3b + d)}{6} = \frac{64\,(3 + 8)}{6} = 117.3\ \text{in}^3$$
Convert to throat properties. With a throat of $0.707h = 0.707(0.0598) = 0.04228$ in, $$A = 0.707h\,A_u = 0.04228(18.0) = 0.7610\ \text{in}^2$$ $$I = 0.707h\,I_u = 0.04228(117.3) = 4.961\ \text{in}^4$$
Primary shear on the throat. The vertical load is carried uniformly by the whole weld group: $$\tau' = \frac{V}{A} = \frac{W}{0.7610} = 1.314W\ \text{psi}$$
Normal stress from the couple. The extreme fibre of the weld group is $c = d/2 = 4.0$ in from its centroid: $$\sigma = \frac{Mc}{I} = \frac{4W(4.0)}{4.961} = 3.225W\ \text{psi}$$ Bending dominates the shear by a factor of nearly two and a half, as it must for a bracket this slender.
Combine and solve for W. The two components act on perpendicular axes of the same throat area, so $$\tau = \sqrt{(\tau')^2 + \sigma^2} = W\sqrt{(1.314)^2 + (3.225)^2} = 3.483W\ \text{psi}$$ Setting this equal to the 900 psi allowable, $$W = \frac{900}{3.483}$$ $$\boxed{W \approx 258\ \text{lb}}$$

Two hundred and fifty-eight pounds is a modest capacity, and correctly so: the limiting element is 0.06 in sheet, and the 900 psi allowable is a small fraction of what weld metal itself can carry. The number is really telling us that the sheet, not the weld metal, sets the strength of this joint, and that the bracket would be considerably stronger if the mounting flange were made deeper, since the capacity scales with $I_u$ and hence roughly with the square of the flange depth.

Quantity	Result
Weld throat	0.0423 in
Throat area A	0.761 in²
Throat second moment I	4.96 in⁴
Primary shear	1.314 W psi
Bending stress at the extreme fibre	3.225 W psi
Combined stress	3.483 W psi
Total load W	258 lb

Topic. Fillet welds loaded by an out-of-plane bending moment, and the weld-as-a-line method.

Key relations. Throat $= 0.707h$; $A = 0.707h\,A_u$ and $I = 0.707h\,I_u$ with $A_u$ and $I_u$ from Shigley Table 9–2; for a rectangle welded all round $A_u = 2(b+d)$ and $I_u = d^2(3b+d)/6$; primary shear $\tau' = V/A$; bending $\sigma = Mc/I$; combined $\tau = \sqrt{(\tau')^2 + \sigma^2}$.

Why this works. A fillet weld fails in shear on its throat regardless of how the joint is loaded, so the design method deliberately refers every stress component to that one plane and then combines them as if they were all shear. Treating the weld as a line of unit width is what makes this tractable: the throat dimension factors out of every property, so a single table of $A_u$ and $I_u$ covers every leg size, and the designer multiplies by $0.707h$ at the end. The method also makes the design lever obvious — $I_u$ grows as the square of the flange depth but only linearly with its width, so deepening a bracket is far more effective than widening it.

Common pitfall. Taking the moment arm to the far end of the top flange rather than to the resultant of a uniformly distributed load, which doubles the moment and halves the answer. The other frequent error is using the leg $h$ instead of the throat $0.707h$ when converting the unit properties, which overstates the capacity by about 41 %.

Source. Budynas and Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §9–3 and §9–4 with Table 9–2 (torsional and bending properties of fillet-weld groups).

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