Referring to Fig. P.23.6, the horizontal x axis is an axis of symmetry so that I_{xy} = 0. Further, S_x = 0 so that, from Eq. (20.6)

\begin{aligned}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_{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{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r\right)\end{aligned}  (20.6)

q_{\mathrm{b}}=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r  (i)

in which
I_{x x}=4 \times 1290 \times 153^2+4 \times 645 \times 153^2=181.2 \times 10^6 \mathrm{~mm}^4
Eq. (i) then becomes

q_{\mathrm{b}}=-\frac{66750}{181.2 \times 10^6} \sum_{r=1}^n B_r y_r=-3.68 \times 10^{-4} \sum_{r=1}^n B_r y_r

Now, ‘cutting’ Cell I in the wall 45 and Cell II in the wall 12

\begin{aligned}&q_{\mathrm{b}, 45}=0=q_{\mathrm{b}, 12} \\&q_{\mathrm{b}, 43}=-3.68 \times 10^{-4} \times 645 \times 153=-36.3 \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 65} \quad \text { (from symmetry) } \\&q_{\mathrm{b}, 18}=-3.68 \times 10^{-4} \times 1290 \times 153=-72.6 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 78}=0 \quad(\text { from symmetry) } \\&q_{\mathrm{b}, 76}=-3.68 \times 10^{-4} \times 645 \times(-153)=36.3 \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 32} \quad \quad \quad \text { (from symmetry) } \\&q_{\mathrm{b}, 63}=36.3+36.3-3.68 \times 10^{-4} \times 1290 \times(-153)=145.2 \mathrm{~N} /\mathrm{mm}\end{aligned}

The shear load is applied through the shear centre of the section so that the rate of twist of the section, dθ/dz, is zero and Eq. (23.10) for Cell I simplifies to

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_R G}\left(-q_{s, 0, R-1} \delta_{R-1, R}+q_{s, 0, R} \delta_R-q_{s, 0, R+1} \delta_{R+1, R}+\oint_R q_{\mathrm{b}} \frac{\mathrm{d} s}{t}\right)  (23.10)

\begin{aligned}0=\frac{1}{2 A_{\mathrm{I}} G_{\mathrm{REF}}}[& q_{s, 0, \mathrm{I}}\left(\delta_{34}+\delta_{45}+\delta_{56}+\delta_{63}\right) \\&\left.-q_{s, 0, \mathrm{II}} \delta_{63}+q_{\mathrm{b}, 63} \delta_{63}+q_{\mathrm{b}, 34} \delta_{34}+q_{\mathrm{b}, 56} \delta_{56}\right]\end{aligned}  (ii)

and for Cell II

\begin{aligned}&0=\frac{1}{2 A_{\mathrm{II}} G_{\mathrm{REF}}}\left[-q_{s, 0, \mathrm{I}} \delta_{63}+q_{s, 0, \mathrm{II}}\left(\delta_{12}+\delta_{23}+\delta_{36}+\delta_{67}+\delta_{78}+\delta_{81}\right)+q_{\mathrm{b}, 81} \delta_{81}\right.\\&\left.+q_{\mathrm{b}, 23} \delta_{23}+q_{\mathrm{b}, 36} \delta_{36}+q_{\mathrm{b}, 67} \delta_{67}\right]\end{aligned}  (iii)

in which G_{\mathrm{REF}}=24200 \mathrm{~N} / \mathrm{mm}^2. Then, from Eq. (23.9)

t^*=\frac{G}{G_{\mathrm{REF}}} t  (23.9)

\begin{aligned}&t_{34}^*=t_{56}^*=\frac{20700}{24200} \times 0.915=0.783 \mathrm{~mm} \\&t_{36}^*=t_{81}^*=t_{45}^*=\frac{24800}{24200} \times 1.220=1.250 \mathrm{~mm}\end{aligned}

Thus

\begin{aligned}&\delta_{34}=\delta_{56}=380 / 0.783=485.3 \\&\delta_{12}=\delta_{23}=\delta_{67}=\delta_{78}=356 / 0.915=389.1 \\&\delta_{36}=\delta_{81}=306 / 1.250=244.8 \\&\delta_{45}=610 / 1.250=488.0\end{aligned}

Eq. (ii) then becomes

1703.4 q_{s, 0, \mathrm{I}}-244.8 q_{s, 0, \mathrm{II}}+70777.7=0

or

q_{s, 0, \mathrm{I}}-0.144 q_{s, 0, \mathrm{II}}+41.55=0  (iv)

and Eq. (iii) becomes

-244.8 q_{s, 0, \mathrm{I}}+2046 q_{s, 0, \mathrm{II}}-46021.1=0

or
q_{s, 0, \mathrm{I}}-8.358 q_{s, 0, \mathrm{II}}+188.0=0 (v)
Subtracting Eq. (v) from (iv) gives
q_{s, 0, \mathrm{II}}=17.8 \mathrm{~N} / \mathrm{mm}
Then, from Eq. (v)
q_{s, 0, \mathrm{I}}=-39.2 \mathrm{~N} / \mathrm{mm}

The resulting shear flows are then

\begin{aligned}&q_{12}=q_{78}=17.8 \mathrm{~N} / \mathrm{mm} \quad q_{32}=q_{76}=36.3-17.8=18.5 \mathrm{~N} / \mathrm{mm} \\&q_{63}=145.2-17.8-39.2=88.2 \mathrm{~N} / \mathrm{mm} \\&q_{43}=q_{65}=39.2-36.3=2.9 \mathrm{~N} / \mathrm{mm} \quad q_{54}=39.2 \mathrm{~N} / \mathrm{mm} \\&q_{81}=72.6+17.8=90.4 \mathrm{~N} / \mathrm{mm}\end{aligned}

Now taking moments about the mid-point of the web 63

\begin{aligned}66750 x_{\mathrm{S}}=&-2 \times q_{76} \times 356 \times 153+2 \times q_{78} \times 356 \times 153+q_{81} \times 306 \times 712 \\&-2 \times q_{43} \times 380 \times 153-q_{54} \times 2(51500+153 \times 380)(\text { see Eq. (20.10)) }\end{aligned}

M_q=2 A q_{12}  (20.10)

from which
x_S = 160.1 mm