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## Q. p.23.6

Figure P.23.6 shows a singly symmetric, two-cell wing section in which all direct stresses are carried by the booms, shear stresses alone being carried by the walls. All walls are flat with the exception of the nose portion 45. Find the position of the shear centre S and the shear flow distribution for a load of $S_y$ = 66 750 N through S. Tabulated below are lengths, thicknesses and shear moduli of the shear carrying walls. Note that dotted line 45 is not a wall.

$\begin{array}{lclllc}\hline \text { Wall } & \text { Length }(\mathrm{mm}) & \text { Thickness }(\mathrm{mm}) & G\left(\mathrm{~N} / \mathrm{mm}^2\right) & \text { Boom } & \text { Area }\left(\mathrm{mm}^2\right) \\\hline 34,56 & 380 & 0.915 & 20700 & 1,3,6,8 & 1290 \\12,23,67,78 & 356 & 0.915 & 24200 & 2,4,5,7 & 645 \\36,81 & 306 & 1.220 & 24800 & & \\45 & 610 & 1.220 & 24800 & & \\\text { Nose area } N_1=51500 \mathrm{~mm}^2 & & & & \\\hline\end{array}$

## Verified Solution

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