Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.77.

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Accepted Answer

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}b &=\frac{6 mm }{\sin 30^{\circ}}=12 mm \h &=35 mm \|A B| &=\frac{35 mm }{\sin 30^{\circ}}=70 mm\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(\frac{1}{2} s \sin 30^{\circ} ight)[(6 mm ) s]=(1.5 mm ) 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(1.5 mm )}{514,500 mm ^{4}} ight) s^{2}[/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}^{70 mm } q_{A B} d s \&=\int_{0}^{70 mm }\left(\frac{V(1.5 mm )}{514,500 mm ^{4}} ight) s^{2} d s \&=\frac{V(1.5 mm )}{514,500 mm ^{4}}\left[\frac{1}{3} s^{3} ight]_{0}^{70 mm }=0.3333333 V\end{aligned}[/latex]

Shear Center: Sum moments about E to find

[latex]\begin{aligned}&V e=(70 mm ) F_{A B} \cos 30^{\circ} \&e=\frac{(70 mm )(0.3333333 V) \cos 30^{\circ}}{V}=20.2 mm\end{aligned}[/latex]

Question 9.77: Determine the location of the shear center O of a thin-walle...

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