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Allowable shear force of the girder
This calculation is an example problem in structural engineering. This content is sponsored by PPI. For similar problems, see the list of review books by PPI
Calculate the maximum allowable shear force Vmax for the girder. The welded steel girder is having the cross section shown in the figure. It is fabricated of two 500 mm x 50 mm flange plates and a 500 mm x 50 mm web plate. The plates are joined by four fillet welds that run continuously at the length of the girder. Each weld has an allowable load in shear of 2000 kN/m.
Solution
Ix = (1/12)*50*5003+2*((1/12)*500*503+(500*50)*(250+50/2)2) = 4.31E+09 mm4
Pmax = 2*2000 KN/m
Q = Qflange=AF*dF = 500*50*(250+50/2) = 6.88E+06 mm3
Pmax = Vmax*Q/Ix = >Vmax = Pmax*Ix/Q = > Vmax = 2.51E+06
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