Question 16.3: Calculate the position of the shear center of the thin-walle...

The shear center S lies on the horizontal axis of symmetry at some distance \xi_{S}, say, from the web. If we apply an arbitrary shear load S_{y} through the shear center, then the shear flow distribution is given by Eq. (16.14) and the moment about any point in the cross-section produced by these shear flows is equivalent to the moment of the applied shear load. S_{y} appears on both sides of the resulting equation and may therefore be eliminated to leave \xi_{S}.

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)

For the channel section, Cx is an axis of symmetry so that I_{x y}=0. Also S_{x}=0 and therefore Eq. (16.14) simplifies to

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

where

Substituting for I_{x x} in Eq. (i), we have

q_{s}=\frac{-12 S_{y}}{h^{3}(1+6 b / h)} \int_{0}^{s} y d s  (ii)

The amount of computation involved may be reduced by giving some thought to the requirements of the problem. In this case, we are asked to find the position of the shear center only, not a complete shear flow distribution. From symmetry, it is clear that the moments of the resultant shears on the top and bottom flanges about the midpoint of the web are numerically equal and act in the same rotational sense. Furthermore, the moment of the web shear about the same point is zero. We deduce that it is necessary to obtain the shear flow distribution only on either the top or bottom flange for a solution. Alternatively, choosing a web–flange junction as a moment center leads to the same conclusion. On the bottom flange, y=-h / 2, so that, from Eq. (ii), we have

q_{12}=\frac{6 S_{y}}{h^{2}(1+6 b / h)} s_{1}  (iii)

Equating the clockwise moments of the internal shears about the midpoint of the web to the clockwise moment of the applied shear load about the same point gives

or, by substitution from Eq. (iii),

from which

\xi_{ s }=\frac{3 b^{2}}{h(1+6 b / h)}  (iv)