Referring to Fig. P.23.7 the horizontal x axis is an axis of symmetry so that I_{xy} = 0 and the shear centre lies on this axis. Further, applying an arbitrary shear load, S_y, through the shear centre then S_x = 0 and Eq. (20.6) simplifies to

\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}=2 \times 645 \times 102^2+2 \times 1290 \times 152^2+2 \times 1935 \times 150^2=162.4 \times 10^6 \mathrm{~mm}^4
Eq. (i) then becomes

q_{\mathrm{b}}=-6.16 \times 10^{-9} S_y \sum_{r=1}^n B_r y_r  (ii)

‘Cut’ the walls 34° and 23. Then, from Eq. (ii)

\begin{aligned}&q_{\mathrm{b}, 34^{\circ}}=q_{\mathrm{b}, 23}=0=q_{\mathrm{b}, 45} \quad(\text { from symmetry) } \\&q_{\mathrm{b}, 43^{\mathrm{i}}}=-6.16 \times 10^{-9} S_y \times 1935 \times(-152)=1.81 \times 10^{-3} S_y \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 65}=-6.16 \times 10^{-9} S_y \times 645 \times(-102)=0.41 \times 10^{-3} S_y \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 21} \\&\quad \text { (from symmetry) } \\&q_{\mathrm{b}, 52}=0.41 \times 10^{-3} S_y-6.16 \times 10^{-9} S_y \times 1290 \times(-152)=1.62 \times 10^{-3} S_y \mathrm{~N} / \mathrm{mm}\end{aligned}

Since the shear load, S_y, is applied through the shear centre of the section the rate of twist, dθ/dz, is zero. Thus, for Cell I, Eq. (23.10) reduces 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)

0=q_{s, 0, \mathrm{I}}\left(\delta_{34^{\circ}}+\delta_{34^{\mathrm{i}}}\right)-q_{s, 0, \mathrm{II}} \delta_{34^{\mathrm{i}}}+q_{\mathrm{b}, 43^{\mathrm{i}}} \delta_{34^{\mathrm{i}}}  (iii)

and for Cell II

0=-q_{s, 0, \mathrm{I}} \delta_{43^{\mathrm{i}}}+q_{s, 0,\mathrm{II}}\left(\delta_{23}+\delta_{34^{\mathrm{i}}}+\delta_{45}+\delta_{52}\right)+q_{\mathrm{b}, 52} \delta_{52}-q_{\mathrm{b}, 43^{\mathrm{i}}} \delta_{43^{\mathrm{i}}}  (iv)

in which

\begin{aligned}\delta_{34^{\circ}} &=1015 / 0.559=1815.7 \quad \delta_{34^{\mathrm{i}}}=304 / 2.030=149.8 \\\delta_{23} &=\delta_{45}=765 / 0.915=836.1 \\\delta_{25} &=304 / 1.625=187.1\end{aligned}

Thus Eq. (iii) becomes

1965.5 q_{s, 0, \mathrm{I}}-149.8 q_{s, 0, \mathrm{II}}+0.271 S_y=0

or

q_{s, 0, \mathrm{I}}-0.076 q_{s, 0, \mathrm{II}}+0.138 \times 10^{-3} S_y=0  (v)

and Eq. (iv) becomes
-149.8 q_{s, 0, \mathrm{I}}+2009.1 q_{s, 0, \mathrm{II}}+319.64 \times 10^{-4} S_y=0
or

q_{s, 0, \mathrm{I}}-13.411 q_{s, 0, \mathrm{II}}-0.213 \times 10^{-3} S_y=0  (vi)

Subtracting Eq. (vi) from (v)

13.335 q_{s, 0, \mathrm{II}}+0.351 \times 10^{-3} S_y=0

whence
q_{s, 0, \mathrm{II}}=-0.026 \times 10^{-3} S_y
Then from Eq. (vi)
q_{s .0 . \mathrm{I}}=-0.139 \times 10^{-3} S_y
Now taking moments about the mid-point of the web 43

\begin{aligned}S_y x_s=&-2 q_{\mathrm{b}, 21}(508 \times 152+50 \times 762)+q_{\mathrm{b}, 52} \times 304 \times 762+2 \times 258000 q_{s, 0, \mathrm{II}} \\&+2 \times 93000 q_{s, 0, \mathrm{I}}\end{aligned}

from which
x_s = 241.4 mm