An extruded beam has the cross section shown in Figure P9.70. The dimensions of this shape are b = 45 mm, h = 75 mm, and t = 4 mm. Assume that the thickness t is constant for all portions of the cross section. What is the distance e from the left-most element to the shear center O?

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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}I_{N A} &=\frac{(4 mm )(150 mm )^{3}}{12}+2\left[(45 mm )(4 mm )\left(\frac{75 mm }{2} ight)^{2} ight]+2\left[(45 mm )(4 mm )\left(\frac{150 mm }{2} ight)^{2} ight] \&=1,125,000 mm ^{4}+506,250 mm ^{4}+2,025,000 mm ^{4}=3,656,250 mm ^{4}\end{aligned}[/latex]

Shear Flow in Upper Flange: Derive an expression for Q for the upper flange as a function of a temporary variable s which originates at the free end of the flange.

[latex]Q_{ ext {upper }}=\bar{y}^{\prime} A^{\prime}=(75 mm )[(4 mm ) s]=\left(300 mm ^{2} ight) s[/latex]

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

[latex]q_{ upper }=\frac{V Q_{ ext {upper }}}{I}=\left(\frac{V\left(300 mm ^{2} ight)}{3,656,250 mm ^{4}} ight) s[/latex]

Integrate with respect to the temporary variable s to determine the resultant force in the upper stiffener.

[latex]\begin{aligned}F_{ ext {upper }} &=\int_{0}^{45 mm } q_{ upper } d s \&=\int_{0}^{45 mm }\left(\frac{V\left(300 mm ^{2} ight)}{3,656,250 mm ^{4}} ight) s d s \&=\frac{V\left(300 mm ^{2} ight)}{3,656,250 mm ^{4}}\left[\frac{1}{2} s^{2} ight]_{0}^{45 mm }=0.0830769 V\end{aligned}[/latex]

Shear Flow in Middle Flange: Derive an expression for Q for the middle flange as a function of a temporary variable s which originates at the free end of the flange.

[latex]Q_{ ext {middle }}=\bar{y}^{\prime} A^{\prime}=(37.5 mm )[(4 mm ) s]=\left(150 mm ^{2} ight) s[/latex]

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

[latex]q_{ ext {middle }}=\frac{V Q_{ ext {middle }}}{I}=\left(\frac{V\left(150 mm ^{2} ight)}{3,656,250 mm ^{4}} ight) s[/latex]

Integrate with respect to the temporary variable s to determine the resultant force in the middle flange.

[latex]\begin{aligned}F_{ ext {middle }} &=\int_{0}^{45 mm } q_{ ext {midde }} d s \&=\int_{0}^{45 mm }\left(\frac{V\left(150 mm ^{2} ight)}{3,656,250 mm ^{4}} ight) s d s \&=\frac{V\left(150 mm ^{2} ight)}{3,656,250 mm ^{4}}\left[\frac{1}{2} s^{2} ight]_{0}^{45 mm }=0.0415385 V\end{aligned}[/latex]

Shear center: To determine the shear center, sum moments about a central point in the middle of the web at the neutral axis location.

[latex]\begin{aligned}V e &=2 F_{ ext {upper }}(75 mm )+2 F_{ ext {middle }}(37.5 mm ) \e &=\frac{2 F_{ ext {upper }}(75 mm )+2 F_{ ext {middle }}(37.5 mm )}{V} \&=\frac{2(0.0830769 V )(75 mm )+2(0.0415385 V )(37.5 mm )}{V} \&=12.4615350 mm +3.1153846 mm =15.58 mm\end{aligned}[/latex]

Question 9.70: An extruded beam has the cross section shown in Figure P9.70...

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