Determine the location e of the shear center, point O, for the thin-walled member having the cross section shown. The member segments have the same thickness t.

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Accepted Answer

Summing moments about A,

[latex]P e=F(4)+2 V_{1}(1.8)[/latex] [latex] ext{(1)}[/latex]

[latex]I=2\left[\frac{1}{12} t\left(4^{3} ight) ight]-\frac{1}{12} t\left(2^{3} ight)+2\left[(1.80)(t)\left(2^{2} ight) ight]=24.4 t \mathrm{in}^{4}[/latex]

[latex]Q_{1}=\bar{y}_{1}^{\prime} A^{\prime}=\left(1+\frac{y}{2} ight)(y t)=t\left(y+\frac{y^{2}}{2} ight)[/latex]

[latex]Q_{2}=\Sigma \bar{y}^{\prime} A^{\prime}=1.5(1)(t)+2(x)(t)=t(1.5+2 x)[/latex]

[latex]q_{1}=\frac{V Q_{1}}{I}=\frac{P t\left(y+\frac{y^{2}}{2} ight)}{24.4 t}=\frac{P\left(y+\frac{y^{2}}{2} ight)}{24.4}[/latex]

[latex]q_{2}=\frac{V Q_{2}}{I}=\frac{P t(1.5+2 x)}{24.4 t}=\frac{P(1.5+2 x)}{24.4}[/latex]

[latex]V_{1}=\int q_{1} d y=\frac{P}{24.4} \int_{0}^{1}\left(y+\frac{y^{2}}{2} ight) d y=0.0273 P[/latex]

[latex]F=\int q_{2} d y=\frac{P}{24.4} \int_{0}^{1.80}(1.5+2 x) d x=0.2434 P[/latex]

From Eq.(1),

[latex]P e=0.2434 P(4)+2(0.0273) P(1.8)[/latex]

[latex]e=1.07 in.[/latex]

Question 7.63: Determine the location e of the shear center, point O, for t...

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