Minimum fillet-weld size for an eccentrically loaded bracket: 22-Mec-A4 Design and Manufacture of Machine Elements, May 2017, Q5

Question 5: Minimum fillet-weld size for an eccentrically loaded bracket

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	Value
Applied load, V (vertical, through the hole centre)	4000 lb
Eccentricity from the weld plane to the load line, e	3 in
Weld length, d (each side)	4 in, vertical
Number of welds	2 (both sides of the bracket plate)
Electrode yield strength, S_y	48 ksi
Factor of safety, n	3

Find. The minimum fillet-weld leg size h, rounded up to a standard size.

Elevation of the bracket. The two vertical fillet welds form the weld group; the 4000 lb load acts 3 in. out from their plane, so the group carries direct shear plus a bending moment about its horizontal centroidal axis.

Approach. Treat the welds as lines of unit throat (the standard weld-as-a-line method), superpose the direct shear per unit length with the bending shear per unit length, take the resultant, and divide by the allowable shear stress to obtain the throat — then convert throat to leg.

Weld-group properties as a line. Two vertical welds each of length d = 4 in, both at essentially the same station (they are separated only by the plate thickness): $$A_u = 2d = 2(4) = 8\ \text{in},\qquad I_u = \frac{2d^{3}}{12} = \frac{2(4)^{3}}{12} = 10.67\ \text{in}^{3},\qquad c=\frac{d}{2}=2\ \text{in}$$ These are per unit throat: the true area is A = 0.707h A_u and the true second moment is I = 0.707h I_u, so the throat can be carried through symbolically and solved for at the end.
Primary (direct) shear. The whole load is carried in shear along the weld: $$f' = \frac{V}{A_u} = \frac{4000}{8} = 500\ \text{lb/in}$$
Secondary (bending) shear. Moving the 4000 lb load back to the weld plane leaves a moment $$M = V e = 4000(3) = 12{,}000\ \text{lb}\cdot\text{in}$$ about the horizontal centroidal axis of the weld group, which produces a vertical-plane stress that is largest at the top and bottom ends of the welds: $$f'' = \frac{Mc}{I_u} = \frac{12{,}000(2)}{10.67} = 2250\ \text{lb/in}$$
Resultant at the critical point. The direct shear acts downward along the weld and the bending term acts normal to it, so the two are perpendicular and combine vectorially at the weld ends: $$f = \sqrt{(f')^{2}+(f'')^{2}} = \sqrt{500^{2}+2250^{2}} = \boxed{2305\ \text{lb/in}}$$ The bending term dominates by a factor of 4.5, so the answer is governed almost entirely by the eccentricity, not by the magnitude of the load in direct shear.
Allowable shear stress in the weld metal. Using the distortion-energy relation between shear yield and tensile yield, S_sy = 0.577 S_y, and dividing by the factor of safety: $$\tau_{\text{all}}=\frac{0.577\,S_y}{n}=\frac{0.577(48{,}000)}{3}= 9232\ \text{psi}$$
Solve for the throat, then the leg. The stress on the throat is the force per unit length divided by the throat dimension: $$t_{\text{throat}} = \frac{f}{\tau_{\text{all}}} = \frac{2305}{9232}=0.2497\ \text{in}$$ $$h = \frac{t_{\text{throat}}}{0.707} = \frac{0.2497}{0.707}=\boxed{0.353\ \text{in}}$$
Specify a standard weld size. Fillet welds are called out in sixteenths; rounding up from 0.353 in gives $$h = \tfrac{3}{8}\ \text{in}\ (0.375\ \text{in})$$ which provides a realised factor of safety of 3×(0.375/0.353) = 3.19. A 3/8 in fillet on each side over the full 4 in length is the specification.

Check: the allowable stress basis is the one judgement call in this problem, because the question supplies a yield strength and a bare factor of safety without naming a code. The distortion-energy basis used above (τ_all = 0.577S_y/n = 9.23 ksi) is Shigley’s. A maximum-shear-stress basis (τ_all = 0.5S_y/n = 8.00 ksi) is the conservative alternative and gives h = 0.408 in → 7/16 in. Either is defensible provided the basis is stated; the difference is one standard weld size. The weld is also assumed to be an equal-leg fillet loaded on its throat, with the base metal and the plate itself not governing.

Quantity	Value
Weld-group length as a line, A_u	8 in
Weld-group second moment as a line, I_u	10.67 in³
Primary (direct) shear, f′	500 lb/in
Bending moment on the group, M	12,000 lb·in
Secondary (bending) shear, f″	2250 lb/in
Resultant force per unit length, f	2305 lb/in
Allowable shear stress, τ_all	9232 psi
Required throat	0.250 in
Minimum weld leg required	0.353 in
Specify	3/8 in fillet, both sides, full 4 in length

Topic. Eccentrically loaded fillet welds analysed by the weld-as-a-line method (out-of-plane moment).

Key relations. f′ = V/A_u; f″ = Mc/I_u; f = √(f′² + f″²); throat = f/τ_all; h = throat/0.707; τ_all = 0.577S_y/n.

Why this works. Because every fillet weld is checked on its throat, the throat dimension appears as a single common factor in every stress term. Setting it to unity turns the weld group into a line whose “area” and “second moment” are pure geometry, the load effects are computed once, and the real throat drops out at the end by a single division. The same tabulated A_u, I_u and J_u then serve every weld pattern.

Common pitfall. Dividing the load by the full weld area and stopping there. Direct shear alone would give a 0.05 in throat — a seventh of what is needed. The eccentricity is the whole problem, and its moment must be taken about the weld-group centroid, not about the outer edge of the bracket.

Source. Budynas & Nisbett, Shigley’s Mechanical Engineering Design, 10th ed., Ch. 9.4–9.5 (“Stresses in Welded Joints in Torsion / in Bending”), Tables 9-2 and 9-3.

Question 5 of 6: Minimum fillet-weld size for an eccentrically loaded bracket

Question 5: Minimum fillet-weld size for an eccentrically loaded bracket