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. (Cambridge)

Question

Accepted Answer

The second moment of area of the web about the centroidal axis is

[latex] \frac{1}{2} (0.010) (0.45)^{3} = 0.0760 imes 10^{-3} m^{4}[/latex]

The second moment of are of each flange about the centroidal axis is

[latex](0.225) (0.0125)(0.231)^{2} = 0.150 imes 10^{-3} m^{4}[/latex]

The total second moment of area is then

[latex]I_{x} = [0.076 + 2(0.150)] 10^{-3}=0.376 imes 10^{-3} m^{4}[/latex]

At a distance y above the neutral axis, the shearing stress from equation (10.9) is

[latex] au =\frac{q}{t}=\frac{F}{2I_{x}}[bh+s_{2}(h-s_{2})][/latex] (10.9)

[latex] au =\frac{F}{2I_{x}}\left[\left\lgroup bh+\frac{1}{4}h^{2} ight group -y^{2} ight][/latex]

[latex]=\frac{200 imes 10^{3}}{2 imes 0.376 imes 10^{-3}} \left[(0.225)(0.4625)+\frac{1}{4}(0.4625)^{2}-y^{2} ight][/latex]

where [latex]s_{2}=h / 2 - y[/latex]

At the top of the web, we have y =0.231 m, and

[latex] au =27.7 MN/m^{2}[/latex]

While at the middle of the web, where y = 0, we have

[latex] au =41.9 MN/m^{2}[/latex]

Question 10.1: The web of a girder of I-section is 45 cm deep and 1 cm thic...

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