Determine the position of the shear center of the open section beam of Example 16.2.

Question

Accepted Answer

The shear centre S lies on the horizontal axis of symmetry a distance, [latex]\xi_{S}[/latex], say, from the centroid C. The shear flow distribution is given by Eq. (iii) of Ex. 16.2 , i.e., [latex]q_{ heta}=S_{y}(\cos heta-1) / \pi r[/latex]. Therefore, taking moments about C

[latex]q_{ heta}=\left(S_{y} / \pi r ight)(\cos heta-1)[/latex] (iii)

[latex]S_{y} \xi_{ S }=-\int_{0}^{2 \pi} q_{ heta} r d s=-\int_{0}^{2 \pi} q_{ heta} r^{2} d heta[/latex]

Then

[latex]S_{y} \xi_{ S }=-\left(S_{y} r / \pi ight) \int_{0}^{2 \pi}(\cos heta-1) d heta[/latex] (i)

Note that the assumed direction of [latex]q_{ heta}[/latex] in Ex. 16.2 is in the direction of increasing [latex] heta[/latex] so that we introduce a negative sign to allow for this when equating the moment of the external shear force to that of the internal shear flow. Integrating Eq. (i) we obtain

[latex]\xi_{ S }=-(r / \pi)[\sin heta- heta]_{0}^{2 \pi}[/latex]

which gives

[latex]\xi_{ S }=2 r[/latex]

Question 16.4: Determine the position of the shear center of the open secti...

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