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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 mm^{4}

P_{max} = 2*2000 KN/m

Q = Q_{flange}=A_{F}*d_{F} = 500*50*(250+50/2) = 6.88E+06 mm^{3}

P_{max} = V_{max}*Q/I_{x} = >V_{max} = P_{max}*I_{x}/Q = > V_{max} = 2.51E+06

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