Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.75.

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Section Properties: For the arc

[latex]\begin{aligned}y &=r \cos heta \quad d A=t d s=t r d heta \d I_{ arc } &=y^{2} d A=r^{2} \cos ^{2} heta(t r d heta)=r^{3} t \sin ^{2} heta d heta \I_{ arc } &=r^{3} t \int_{0}^{\pi} \cos ^{2} heta d heta \&=r^{3} t \int_{0}^{\pi}\left(\frac{\cos 2 heta+1}{2} ight) d heta \&=\frac{1}{2} \pi r^{3} t\end{aligned}[/latex]

The moment of inertia for the complete cross section is:

[latex]I=I_{ arc }+2\left[r t(r)^{2} ight]=\frac{1}{2} \pi r^{3} t+2 r^{3} t=\frac{r^{3} t}{2}(\pi+4)[/latex]

Derive an expression for the value of Q for the upper horizontal portion in terms of a temporary variable u:

[latex]Q_{ ext {upper }}=r(u imes t)=r^{2} t[/latex]

For locations in the arc, Q can be expressed as:

[latex]Q_{ ext {arc }}=Q_{ ext {upper }}+\int d Q[/latex]

where [latex]d Q=y d A=r \cos heta(t r d heta)=r^{2} t \cos heta d heta[/latex]

[latex]Q_{ arc }=r^{2} t+r^{2} t \int_{0}^{ heta} \cos heta d heta=r^{2} t(1+\sin heta)[/latex]

Shear Flow Resultant (for the upper portion):

[latex]\begin{aligned}q_{ ext {upper }} &=\frac{V Q_{u}}{I}=\frac{V r t}{\frac{r^{3} t}{2}(\pi+4)} u=\frac{2 V}{r^{2}(\pi+4)} u \F_{ ext {upper }} &=\int q_{ ext {upper }} d u \&=\int_{0}^{r} \frac{2 V}{r^{2}(\pi+4)} u d u=\frac{V}{\pi+4}\end{aligned}[/latex]

Shear Flow Resultant (for the arc):

[latex]\begin{aligned}q_{ arc } &=\frac{V Q_{ arc }}{I}=\frac{V r^{2} t(1+\sin heta)}{\frac{r^{3} t}{2}(\pi+4)}=\frac{2 V}{r(\pi+4)}(1+\sin heta) \F_{ arc } &=\int q_{ arc } d s \&=\int_{0}^{\pi} \frac{2 V}{r(\pi+4)}(1+\sin heta) r d heta \&=\frac{2 V}{\pi+4} \int_{0}^{\pi}(1+\sin heta) d heta=\frac{2 V(\pi+2)}{\pi+4}\end{aligned}[/latex]

Shear Center: Sum moments about the center of the arc to find

[latex]\begin{aligned}V e &=2 F_{ ext {upper }} r+F_{ ext {arc }} r \V e &=2\left(\frac{V}{\pi+4} ight) r+\left(\frac{2 V(\pi+2)}{\pi+4} ight) r \&=\frac{2 V r}{\pi+4}(\pi+3) \& herefore e=\left(\frac{\pi+3}{\pi+4} ight) 2 r\end{aligned}[/latex]

Question 9.75: Determine the location of the shear center O of a thin-walle...

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