Question 20.3: Figure P.20.3 shows the cross-section of a single-cell, thin......

The shear center, S, lies on the horizontal axis of symmetry, the x axis. Therefore apply an arbitrary shear load, $S_{y},$ through S (Fig. S.20.3(a)). The internal shear flow distribution is given by Eq. (20.11), which, since $I_{x y}=0,\,S_{x}=0,\,{\mathrm{and}}\,\,t_{\mathrm{D}}=0,$ simplifies to

$q_{s}=-\left({\frac{S_{x}I_{x x}-S_{y}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}}\right) \left(\int_{0}^{s}t_{\mathrm{D}}x\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}x_{r}\right) -\left(\frac{S_{y}I_{y y}-S_{x}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}\right) \left(\int_{0}^{s}t_{\mathrm{DJ}}\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}y_{r}\right)+q_{s,0}$           (20.11)

$q_{s}=-{\frac{S_{y}}{I_{x x}}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0}$          (i)

in which

Equation (i) then becomes

$q_{s}=-0.5\times10^{-7}S_{y}\sum_{r=1}^{n}B_{r}y_{r}+q_{s,0}$          (ii)

The first term on the right-hand side of Eq. (ii) is the $q_{b}$ distribution (see Eq. (17.16)). To determine $q_{b}$, ‘cut’ the section in the wall 23. Then

$q_{s}=q_{b}+q_{s,0}$              (17.16)

The value of shear flow at the ‘cut’ is obtained using Eq. (17.28), which, since G= constant, becomes

$q_{s,0}=-{\frac{\oint q_{b}\;\mathrm{d}s}{\oint\mathrm{d}s}}$                (17.28)

$q_{s,0}=-{\frac{\oint(q_{\mathrm{b}}/t)\mathrm{d}s}{\oint\mathrm{d}s/t}}$         (iii)

In Eq. (iii),

Then, from Eq. (iii) and the above $q_{b}$ distribution,

i.e.,

The complete shear flow distribution is shown in Fig. S.20.3(b).
Now taking moments about O in Fig. S.20.3(b) and using the result of Eq. (20.10),

$M_{q}=2A q_{12}$            (20.10)

which gives