An extruded beam has the cross section shown in Figure P9.69. For this shape, use dimensions of b = 50 mm, h = 40 mm, and t = 3 mm. What is the distance e to the shear center O?

Question

Accepted Answer

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

[latex]\begin{aligned}I_{N A} &=2 \frac{(3 mm )(40 mm )^{3}}{12}+2\left[(50 mm )(3 mm )\left(\frac{40 mm }{2} ight)^{2} ight] \&=32,000 mm ^{4}+120,000 mm ^{4}=152,000 mm ^{4}\end{aligned}[/latex]

Shear Flow in Upper Stiffener: Derive an expression for Q for the upper stiffener as a function of a temporary variable v which originates at the tip of the upper stiffener.

[latex]Q_{s}=\bar{y}^{\prime} A^{\prime}=\left(\frac{v}{2} ight)[(3 mm ) v]=(1.5 mm ) v^{2}[/latex]

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

[latex]q=\frac{V Q}{I}=\frac{V}{152,000 mm ^{4}}\left[(1.5 mm ) v^{2} ight][/latex]

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

[latex]\begin{aligned}F_{s} &=\int_{0}^{20 mm } q d v \&=\int_{0}^{20 mm } \frac{V}{152,000 mm ^{4}}\left[(1.5 mm ) v^{2} ight] d v \&=\frac{V}{152,000 mm ^{4}}\left[\frac{1.5 mm }{3} v^{3} ight]_{0}^{20 mm }=\frac{V}{38}=0.0263158 V\end{aligned}[/latex]

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

[latex]\begin{aligned}Q_{f} &=\bar{y}^{\prime} A^{\prime}=(10 mm )(3 mm )(20 mm )+(20 mm )[(3 mm ) u] \&=600 mm ^{3}+\left(60 mm ^{2} ight) u\end{aligned}[/latex]

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

[latex]q=\frac{V Q}{I}=\frac{V}{152,000 mm ^{4}}\left[600 mm ^{3}+\left(60 mm ^{2} ight) u ight][/latex]

Integrate with respect to the temporary variable u to determine the resultant force in the upper flange.

[latex]\begin{aligned}F_{f} &=\int_{0}^{50 mm } q d u \&=\int_{0}^{50 mm } \frac{V}{152,000 mm ^{4}}\left[600 mm ^{3}+\left(60 mm ^{2} ight) u ight] d u \&=\frac{V}{152,000 mm ^{4}}\left[\left(600 mm ^{3} ight) u+\left(30 mm ^{2} ight) u^{2} ight]_{0}^{50 mm } \&=\frac{V}{152,000 mm ^{4}}\left[30,000 mm ^{4}+75,000 mm ^{4} ight]=0.6907895 V\end{aligned}[/latex]

Shear center: To determine the shear center, sum moments about the lower left corner of the channel web.

[latex]\begin{aligned}V e &=F_{f}(40 mm )+2 F_{s}(50 mm ) \e &=\frac{F_{f}(40 mm )+2 F_{s}(50 mm )}{V} \&=\frac{(0.6907895 V )(40 mm )+2(0.0263158 V )(50 mm )}{V} \&=27.6315800 mm +2.6315800 mm =30.3 mm\end{aligned}[/latex]

Question 9.69: An extruded beam has the cross section shown in Figure P9.69...

Related Answered Questions