Question 16.7: A thin-walled closed section beam has the singly symmetrical...

The shear center clearly lies on the horizontal axis of symmetry, so that it is necessary only to apply a shear load S_{y} through S and to determine \xi_{S}. If we take the x reference axis to coincide with the axis of symmetry, then I_{x y}=0, and since S_{x}=0, Eq. (16.15) simplifies to

q_{s}=-\left(\frac{S_{x} I_{x x}-S_{y} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}}\right) \int_{0}^{s} t x d s-\left(\frac{S_{y} I_{y y}-S_{x} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}}\right) \int_{0}^{s} t y d s+q_{s, 0}  (16.15)

q_{s}=-\frac{S_{y}}{I_{x x}} \int_{0}^{s} t y d s+q_{s, 0}  (i)

in which

Evaluating this expression gives I_{x x}=1,152 a^{3} t.

The basic shear flow distribution q_{ b } is obtained from the first term in Eq. (i). Then, for the wall 41,

q_{ b , 41}=\frac{-S_{y}}{1,152 a^{3} t} \int_{0}^{s_{1}} t\left(\frac{8}{10} s_{1}\right) d s_{1}=\frac{-S_{y}}{1,152 a^{3}}\left(\frac{2}{5} s_{1}^{2}\right)  (ii)

In the wall 12,

q_{ b , 12}=\frac{-S_{y}}{1,152 a^{3}}\left[\int_{0}^{s_{2}}\left(17 a-s_{2}\right) \frac{8}{17} d s_{2}+40 a^{2}\right]  (iii)

which gives

q_{ b , 12}=\frac{-S_{y}}{1,152 a^{3}}\left(-\frac{4}{17} s_{2}^{2}+8 a s_{2}+40 a^{2}\right)  (iv)

The q_{ b } distributions on the walls 23 and 34 follow from symmetry. Hence, from Eq. (16.28),

q_{s, 0}=-\frac{\oint q_{ b } d s}{\oint d s}  (16.28)

giving

q_{s, 0}=\frac{S_{y}}{1,152 a^{3}}\left(58.7 a^{2}\right)  (v)

Taking moments about the point 2, we have

or

S_{y}\left(\xi_{ S }+9 a\right)=\frac{S_{y} 34 a \sin \theta}{1,152 a^{3}} \int_{0}^{10 a}\left(-\frac{2}{5} s_{1}^{2}+58.7 a^{2}\right) d s_{1}  (vi)

We may replace \sin \theta by \sin \left(\theta_{1}-\theta_{2}\right)=\sin \theta_{1} \cos \theta_{2}-\cos \theta_{1} \sin \theta_{2}, where \sin \theta_{1}=15 / 17, \cos \theta_{2}=8 / 10, \cos \theta_{1}=8 / 17, and \sin \theta_{2}=6 / 10. Substituting these values and integrating Eq. (vi) gives

which means that the shear center is inside the beam section.