Question 16.2: Calculate the shear flow distribution in the thin-walled ope...

The centroid of the section coincides with the center of the circle. Also the Cx axis is an axis of symmetry so that I_{x y}=0 and since S_{x}=0 Eq. (16.14) reduces 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  (16.14)

q_{s}=-\left(S_{y} / I_{x x}\right) \int_{0}^{s} t y d s  (i)

The second moment of area, I_{x x}, of the section about Cx may be deduced from the second moment of area of the semi-circular section shown in Fig. 15.33 and is \pi r^{3} t. Then, at any point a distance s from one edge of the narrow slit

q_{s}=-\left(S_{y} / \pi r^{3} t\right) \int_{0}^{s} t y d s  (ii)

Working with angular coordinates for convenience Eq. (ii) becomes

i.e.,

which gives

Then

q_{\theta}=\left(S_{y} / \pi r\right)(\cos \theta-1)  (iii)

From Eq. (iii), when \theta=0, q_{\theta}=0 (as expected at an open edge) and when \theta=\pi, q_{\theta}=-2 S_{y} / \pi r. Further analysis of Eq. (iii) shows that q_{\theta} is also zero at \theta=2 \pi and that q_{\theta} is a maximum when \theta=\pi. Also q_{\theta} is negative, i.e., in the opposite sense to increasing values of \theta, for all values of \theta The complete shear flow distribution is shown in Fig. 16.9.