Locate the shear center for the cross section shown in Figure P9.72. Assume that the web thickness is the same as the flange thickness.

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Accepted Answer

Section Properties:

Horizontal distance to centroid (measured from center of left-side flange).

[latex]\begin{aligned}\bar{z} &=\frac{(5 in .)(10 in .)(0.375 in .)+(10 in .)(0.375 in .)(5 in .)}{(0.375 in .)(3 in .)+(10 in .)(0.375 in .)+(0.375 in .)(5 in .)} \&=\frac{18.75 in .{ }^{3}+18.75 in .{ }^{3}}{1.125 in .{ }^{2}+3.75 in .{ }^{2}+1.875 in .{ }^{2}}=\frac{37.5 in .^{3}}{6.75 in .^{2}}=5.5555556 in .\end{aligned}[/latex]

Moment of inertia about z axis.

[latex]I_{z}=\frac{1}{12}(0.375 in .)(3 in .)^{3}+\frac{1}{12}(0.375 in .)(5 in .)^{3}=4.75 in. ^{4}[/latex]

Shear Flow Resultant: Consider the right-side flange. Q for the 5-in. deep flange can be expressed as:

[latex]\begin{aligned}Q=&\left(\frac{2.5 in .+y}{2} ight)((0.375 in .)(2.5 in .-y)) \&=\left(1.25 in .+\frac{y}{2} ight)\left[0.9375 in .^{2}-(0.375 in .) y ight] \&=1.1718750 in .^{3}+\left(0.46875 in .^{2} ight) y \&-\left(0.46875 in .^{2} ight) y-(0.1875 in .) y^{2} \&=1.1718750 in .^{3}-(0.1875 in .) y^{2} \q=& \frac{V Q}{I}=\frac{V}{4.75 in .^{4}}\left[1.1718750 in .^{3}-(0.1875 in .) y^{2} ight]\end{aligned}[/latex]

[latex]\begin{aligned}F_{ ext {right }} &=\int q d y \&=\int_{-2.5 ext { in. }}^{2.5 ext { in. }} \frac{V}{4.75 ext { in. }^{4}}\left[1.1718750 ext { in. }^{3}-(0.1875 ext { in. }) y^{2} ight] d y \&=\frac{V}{4.75 ext { in. }^{4}}\left[\left(1.1718750 ext { in. }^{3} ight) y-\frac{(0.1875 ext { in. })}{3} y^{3} ight]_{-2.5 ext { in. }}^{2.5 ext { in. }} \&=\frac{V}{4.75 ext { in. }^{4}}\left[2.9296875 in .^{4}-0.9765625 in .^{4}-\left(-2.9296875 in .^{4} ight)-0.9765625 in. ^{4} ight] \&=0.8223684 V\end{aligned}[/latex]

Shear Center: Sum moments about the center of the left-side flange to find

[latex]\begin{aligned}&V(\bar{z}+e)=F_{ ext {right }}(10 ext { in. }) \&\bar{z}+e=\frac{F_{ ext {right }}(10 ext { in. })}{V}=\frac{(0.8223684 V)(10 in .)}{V}=8.223684 ext { in. } \&e=8.223684 ext { in. }-5.5555556 in .=2.67 in .\end{aligned}[/latex]

Question 9.72: Locate the shear center for the cross section shown in Figur...

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