Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.76.

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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 follows. For segment AB,

[latex]I_{A B}=b t\left(b \sin 60^{\circ} ight)^{2}=t b^{3} \sin ^{2} 60^{\circ}[/latex]

For segment BC,

[latex]\begin{aligned}d A &=t d s \quad y=s \sin 60^{\circ} \d I &=y^{2} d A=\left(s \sin 60^{\circ} ight)^{2} t d s=t \sin ^{2} 60^{\circ} s^{2} d s \I_{B C} &=t \sin ^{2} 60^{\circ} \int_{0}^{b} s^{2} d s \&=\frac{t b^{3} \sin ^{2} 60^{\circ}}{3}\end{aligned}[/latex]

Therefore, the moment of inertia about the horizontal centroidal axis is

[latex]\begin{aligned}I &=2\left[t b^{3} \sin ^{2} 60^{\circ}+\frac{t b^{3} \sin ^{2} 60^{\circ}}{3} ight] \&=2 t b^{3}\end{aligned}[/latex]

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

[latex]Q=\bar{y}^{\prime} A^{\prime}=b \sin 60^{\circ}[s t]=\left(t b \sin 60^{\circ} ight) s[/latex]

Express the shear flow q in the flange using Q.

[latex]q=\frac{V Q}{I}=\frac{V\left(t b \sin 60^{\circ} ight) s}{2 t b^{3}}=\frac{\sqrt{3} V}{4 b^{2}} s[/latex]

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

[latex]F_{f}=\int_{0}^{b} q d s=\int_{0}^{b} \frac{\sqrt{3} V}{4 b^{2}} s d s=\frac{\sqrt{3} V}{8 b^{2}}\left[s^{2} ight]_{0}^{b}=\frac{\sqrt{3}}{8} V[/latex]

Shear Center: Sum moments about C to find

[latex]\begin{gathered}V e=2 F_{f}\left(b \sin 60^{\circ} ight)=2\left(\frac{\sqrt{3}}{8} V ight)\left(\frac{\sqrt{3}}{2} ight) b \ herefore e=\frac{3}{8} b=0.375 b\end{gathered}[/latex]

Question 9.76: Determine the location of the shear center O of a thin-walle...

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