Question

The web of a girder of i-section is 45 cm deep and 1 cm thick; the flanges are each 22.5 cm wide by 1.25 cm thick. The girder at some particular section has to withstand a total shearing force of 200 kn. Calculate the shearing stresses at the top and middle of the web.

Answer 1

The 2nd moment of area about its center axis is

1/12 (0.010)(0.45)^3= 0.0760 x 10^-3 m4

The 2nd moment of area of flange around its center axis

(0.225)(0.0125)(0.231)^2 = 0.150 x 10^-3 m4

Total 2nd moment

Ix= [0.0760 + 2(0.150)] x 10^-3= 0.376 x 10^-3 m4

At y distance, the shearing stress will be

tau= F/2Ix [ (bh+1/4h2) - y2]

     = 200 x 10^3/ 2 x 0.376 x  10^-3[(0.225)(0.4625) + 1/4(0.4625)2 - y2]

y= 0.231

tau= 34.6 MN/m2

when y=0

tau = 52.2 MN/m2