Page 520 - Shigley's Mechanical Engineering Design

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494 Mechanical Engineering Design
These are the weld-bead lengths required by weld metal strength. The attachment shear
stress allowable in the base metal, from Table 9–4, is
τ all = 0.4S y = 0.4(36) = 14.4 kpsi
The shear stress τ in the base metal adjacent to the weld is

F F F
τ = = = = τ all = 14.4 kpsi
h(l 1 + l 2 ) h(l 1 + 1.4l 1 ) h(2.4l 1 )
from which

F 24
l 1 = = = 2.22 in
14.4h(2.4) 14.4(0.3125)2.4
l 2 = 1.4l 1 = 1.4(2.22) = 3.11 in
These are the weld-bead lengths required by base metal (attachment) strength. The base
metal controls the weld lengths. For the allowable tensile stress σ all in the shank of the
attachment, the AISC allowable for tension members is 0.6S y ; therefore,
σ all = 0.6S y = 0.6(36) = 21.6 kpsi
The nominal tensile stress σ is uniform across the attachment cross section because of
the load application at the centroid. The stress σ is
F 24
σ = = = 10.7 kpsi
A 0.75(2) + 2(0.375)
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Since σ ≤ σ all , the shank section is satisfactory. With l 1 set to a nominal 2 in, l 2 should
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be 1.4(2.25) = 3.15 in.
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Decision Set l 1 = 2 in, l 2 = 3 in. The small magnitude of the departure from l 2 /l 1 = 1.4 is not
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serious. The joint is essentially moment-free.

EXAMPLE 9–4 Perform an adequacy assessment of the statically loaded welded cantilever carrying
500 lbf depicted in Fig. 9–20. The cantilever is made of AISI 1018 HR steel and welded
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with a -in fillet weld as shown in the figure. An E6010 electrode was used, and the
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design factor was 3.0.
(a) Use the conventional method for the weld metal.
(b) Use the conventional method for the attachment (cantilever) metal.
(c) Use a welding code for the weld metal.

Solution (a) From Table 9–3, S y = 50 kpsi, S ut = 62 kpsi. From Table 9–2, second pattern, b =
0.375 in, d = 2 in, so
A = 1.414hd = 1.414(0.375)2 = 1.06 in 2
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3
I u = d /6 = 2 /6 = 1.33 in 3
I = 0.707hI u = 0.707(0.375)1.33 = 0.353 in 4