Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.78.

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Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as:

[latex]\begin{aligned} an heta &=\frac{3 ext { in. }}{4 ext { in. }}=0.75 \quad herefore heta=36.8699^{\circ} \b &=\frac{0.25 in .}{\sin 36.8699^{\circ}}=0.41667 in. \h &=3 in . \|B D| &=\frac{3 in .}{\sin 36.8699^{\circ}}=5 in.\end{aligned}[/latex]

[latex]\begin{aligned}I_{N A}=& 2 I_{A B}+2 I_{B D} \=& 2\left[\frac{(0.25 in .)(2 in .)^{3}}{12}+(0.25 in .)(2 in .)(4 in .)^{2} ight] \ &+2\left[\frac{(0.41667 in .)(3 in .)^{3}}{3} ight] \=& 16.3333 in .^{4}+7.5 in .^{4}=23.8333 in .{ }^{4}\end{aligned}[/latex]

Shear Flow in Element AB: Derive an expression for Q for element AB as a function of a temporary variable s which originates at A and extends toward B.

[latex]Q_{A B}=\bar{y}^{\prime} A^{\prime}=\left(5 ext { in. }-\frac{s}{2} ight)[(0.25 in .) s]=\left(1.25 in. ^{2} ight) s-(0.125 in .) s^{2}[/latex]

Express the shear flow q in element AB using [latex]Q_{A B}[/latex].

[latex]q_{A B}=\frac{V Q_{A B}}{I}=\left(\frac{V}{23.8333 in .^{4}} ight)\left[\left(1.25 in .{ }^{2} ight) s-(0.125 in .) s ^{2} ight][/latex]

Integrate with respect to the temporary variable s to determine the resultant force in element AB.

[latex]\begin{aligned}F_{A B} &=\int_{0}^{2 ext { in. }} q_{A B} d s \&=\int_{0}^{2 ext { in. }}\left(\frac{V}{23.8333 in. ^{4}} ight)\left[\left(1.25 in .{ }^{2} ight) s-(0.125 in .) s^{2} ight] d s \&=\frac{V}{23.8333 in .^{4}}\left[\frac{1.25 in. ^{2}}{2} s^{2}-\frac{0.125 in .}{3} s^{3} ight]_{0}^{2 in .}=0.0909092 V\end{aligned}[/latex]

Shear Center: Sum moments about D to find

[latex]\begin{aligned}&V e=2(4 ext { in. }) F_{A B} \&e=\frac{2(4 ext { in. })(0.0909092 V)}{V}=0.727 ext { in. }\end{aligned}[/latex]

Question 9.78: Determine the location of the shear center O of a thin-walle...

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