Question 9.3: A specially rolled A36 structural steel section for the atta...

The y coordinate of the section centroid of the attachment is

\overline {y} =  \frac {\sum{y_{i} A_{i}}}{\sum {A_{i}}} =\frac {1(0.75)2 + 3(0.375)2}{0.75(2) + 0.375(2)} = 1.67  in

Summing moments about point B to zero gives

\sum {M_{B}} = 0 = −F_{1}b + F  \overline {y} = −F_{1}(4) +24(1.67)

from which

F_{1} = 10  kip

It follows that

F_{2} = 24 − 10.0 = 14.0  kip

The weld throat areas have to be in the ratio 14/10 = 1.4, that is, l_{2} = 1.4l_{1}. The weld length design variables are coupled by this relation, so l_{1} is the weld length design variable. The other design variable is the fillet weld leg size h, which has been decided by the problem statement. From Table 9–4, the allowable shear stress on the throat τ_{all} is

τ_{all} = 0.3(70) = 21  kpsi

Table 9–4   Stresses Permitted by the AISC Code for Weld Metal

*The factor of safety n has been computed by using the distortion-energy theory.
†Shear stress on base metal should not exceed 0.40Sy of base metal.

The shear stress τ on the 45° throat is

τ =\frac{F}{(0.707)h(l_{1} + l_{2})} =\frac {F}{(0.707)h(l_{1} + 1.4l_{1})}

=\frac {F}{(0.707)h(2.4l_{1})} = τ_{all} = 21  kpsi

from which the weld length l_{1} is

l_{1} =\frac {24}{21(0.707)0.3125(2.4) }= 2.16  in

and

l_{2} = 1.4l_{1} = 1.4(2.16) = 3.02  in

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

τ =\frac {F}{h(l_{1} + l_{2})} =\frac {F}{h(l_{1} + 1.4l_{1})} =\frac {F}{h(2.4l_{1})} = τ_{all} = 14.4  kpsi

from which

l_{1} =\frac {F}{14.4h(2.4)} =\frac {24}{14.4(0.3125)2.4} = 2.22 in
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

σ =\frac {F}{A} =\frac {24}{0.75(2) + 2(0.375)} = 10.7  kpsi

Since σ_{all} ≥ σ , the shank section is satisfactory. With l_{1} set to a nominal 2\frac {1}{4} in, l_{2} should be 1.4(2.25) = 3.15 in.

Set l_{1} = 2\frac {1}{4}  in, l_{2} = 3\frac {1}{4}  in. The small magnitude of the departure from l_{2}/l_{1} = 1.4 is not serious. The joint is essentially moment-free.