Determine the location of the shear center for the cross section shown in Figure P9.71. Use dimensions of a = 50 mm, b = 100 mm, h = 300 mm, and t = 5 mm. Assume that the thickness t is constant for all portions of the cross section.

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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}I_{N A} &=\frac{(5 mm )(300 mm )^{3}}{12}+2\left[(50 mm +100 mm )(5 mm )\left(\frac{300 mm }{2} ight)^{2} ight] \&=11,250,000 mm ^{4}+33,750,000 mm ^{4}=45,000,000 mm ^{4}\end{aligned}[/latex]

Shear Flow in Right-side Flange: Derive an expression for Q for the right-side flange as a function of a temporary variable u which originates at the right end of the right-side flange.

[latex]Q_{ ext {right }}=\bar{y}^{\prime} A^{\prime}=(150 mm )[(5 mm ) u]=\left(750 mm ^{2} ight) u[/latex]

Express the shear flow q in the right-side flange using [latex]Q_{ ext {right }}[/latex].

[latex]q_{ right }=\frac{V Q_{ ext {right }}}{I}=\left(\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}} ight) u[/latex]

Integrate with respect to the temporary variable u to determine the resultant force in the right-side flange.

[latex]\begin{aligned}F_{ ext {right }} &=\int_{0}^{100 mm } q_{ ext {right }} d u \&=\int_{0}^{100 mm }\left(\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}} ight) u d u \&=\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}}\left[\frac{1}{2} u^{2} ight]_{0}^{100 mm }=0.0833333 V\end{aligned}[/latex]

Shear Flow in Left-side Flange: Derive an expression for Q for the leftside flange as a function of a temporary variable v which originates at the left end of the left-side flange.

[latex]Q_{ ext {left }}=\bar{y}^{\prime} A^{\prime}=(150 mm )[(5 mm ) v]=\left(750 mm ^{2} ight) v[/latex]

Express the shear flow q in the left-side flange using [latex]Q_{ ext {left }}[/latex].

[latex]q_{ ext {left }}=\frac{V Q_{ ext {left }}}{I}=\left(\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}} ight) v[/latex]

Integrate with respect to the temporary variable v to determine the resultant force in the left-side flange.

[latex]\begin{aligned}F_{ ext {left }} &=\int_{0}^{50 mm } q_{ ext {left }} d v \&=\int_{0}^{50 mm }\left(\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}} ight) v d v \&=\frac{V\left(750 mm ^{2} ight)}{45,000,000 mm ^{4}}\left[\frac{1}{2} v^{2} ight]_{0}^{50 mm }=0.0208333 V\end{aligned}[/latex]

Shear center: To determine the shear center, sum moments about the point where the web connects to the lower flange.

[latex]\begin{aligned}V e &=F_{ ext {right }}(300 mm )-F_{ ext {left }}(300 mm ) \e &=\frac{F_{ ext {right }}(300 mm )-F_{ ext {left }}(300 mm )}{V} \&=\frac{(0.0833333 V )(300 mm )-(0.0208333 V )(300 mm )}{V} \&=25 mm -6.25 mm =18.75 mm\end{aligned}[/latex]

Question 9.71: Determine the location of the shear center for the cross sec...

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