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Wednesday, 15 January 2020 09:59cat

Allowable shear force of the girder

Written by  TheStructuralEngineer.info

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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