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## Q. 23.3

The wing section of Example 23.1 (Fig. 23.3) carries a vertically upward shear load of 86.8 kN in the plane of the web 572. The section has been idealized such that the booms resist all the direct stresses while the walls are effective only in shear. If the shear modulus of all walls is 27 600 N/mm² except for the wall 78 for which it is three times this value, calculate the shear flow distribution in the section and the rate of twist. Additional data are given below.

$\begin{array}{llll}\hline \text { Wall } & \text { Length }(\mathrm{mm}) & \text { Thickness }(\mathrm{mm}) & \text { Cell area }\left(\mathrm{mm}^2\right) \\\hline 12,56 & 1023 & 1.22 & A_{\mathrm{I}}=265000 \\23 & 1274 & 1.63 & A_{\mathrm{II}}=213000 \\34 & 2200 & 2.03 & A_{\mathrm{III}}=413000 \\483 & 400 & 2.64 & \\572 & 460 & 2.64 & \\61 & 330 & 1.63 & \\78 & 1270 & 1.22 & \\\hline\end{array}$

## Verified Solution

Choosing $G_{\text {REF }} \text { as } 27600 \mathrm{~N} / \mathrm{mm}^2$ then, from Eq. (23.9)

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

$t_{78}^*=\frac{3 \times 27600}{27600} \times 1.22=3.66 \mathrm{~mm}$

Hence

$\delta_{78}=\frac{1270}{3.66}=347$

Also

\begin{aligned}&\delta_{12}=\delta_{56}=840 \quad \delta_{23}=783 \quad \delta_{34}=1083 \quad \delta_{38}=57 \quad \delta_{84}=95 \quad \delta_{87}=347 \\&\delta_{27}=68 \quad \delta_{75}=106 \quad \delta_{16}=202\end{aligned}

We now ‘cut’ the top skin panels in each cell and calculate the ‘open section’ shear flows using Eq. (20.6) which, since the wing section is idealized, singly symmetrical (as far as the direct stress carrying area is concerned) and is subjected to a vertical shear load only, reduces 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)

where, from Example 23.1, $I_{x x}=809 \times 10^6 \mathrm{~mm}^4$. Thus, from Eq. (i)

$q_{\mathrm{b}}=-\frac{86.8 \times 10^3}{809 \times 10^6} \sum_{r=1}^n B_r y_r=-1.07 \times 10^{-4} \sum_{r=1}^n B_r y_r$  (ii)

Since $q_b$ = 0 at each ‘cut’, then $q_b$ = 0 for the skin panels 12, 23 and 34. The remaining $q_b$ shear flows are now calculated using Eq. (ii). Note that the order of the numerals in the subscript of $q_b$ indicates the direction of movement from boom to boom.

\begin{aligned}&q_{\mathrm{b}, 27}=-1.07 \times 10^{-4} \times 3880 \times 230=-95.5 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 16}=-1.07 \times 10^{-4} \times 2580 \times 165=-45.5 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 65}=-45.5-1.07 \times 10^{-4} \times 2580 \times(-165)=0 \\&q_{\mathrm{b}, 57}=-1.07 \times 10^{-4} \times 3880 \times(-230)=95.5 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 38}=-1.07 \times 10^{-4} \times 3230 \times 200=-69.0 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 48}=-1.07 \times 10^{-4} \times 3230 \times(-200)=69.0\mathrm{~N} / \mathrm{mm}\end{aligned}

Therefore, as $q_{\mathrm{b}, 83}=q_{\mathrm{b}, 48}\left(\text { or } q_{\mathrm{b}, 72}=q_{\mathrm{b}, 57}\right), q_{\mathrm{b}, 78}=0$. The distribution of the $q_b$ shear flows is shown in Fig. 23.11. The values of δ and $q_b$ are now substituted in Eq. (23.10) for each cell in turn.

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

• For cell I

$\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 265000 G_{\mathrm{REF}}}\left[q_{s, 0, \mathrm{I}}(1083+95+57)-57 q_{s, 0, \mathrm{II}}+69 \times 95+69 \times 57\right]$  (iii)

• For cell II

\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 213000 G_{\mathrm{REF}}} & {\left[-57 q_{s, 0, \mathrm{I}}+q_{s, 0, \mathrm{II}}(783+57+347+68)-68 q_{s, 0, \mathrm{III}}\right.} \\&+95.5 \times 68-69 \times 57]\end{aligned} (iv)

• For cell III

\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 413000 G_{\mathrm{REF}}} & {\left[-68 q_{s, 0, \mathrm{II}}+q_{s, 0, \mathrm{III}}(840+68+106+840+202)\right.} \\&+45.5 \times 202-95.5 \times 68-95.5 \times 106]\end{aligned}  (v)

The solely numerical terms in Eqs (iii)–(v) represent $\oint_R q_{\mathrm{b}}(\mathrm{d} s / t)$ for each cell. Care must be taken to ensure that the contribution of each $q_b$ value to this term is interpreted correctly. The path of the integration follows the positive direction of $q_{s,0}$ in each cell, i.e. anticlockwise. Thus, the positive contribution of $q_{\mathrm{b}, 83} \text { to } \oint_{\mathrm{I}} q_{\mathrm{b}}(\mathrm{d} s / t)$ becomes a negative contribution to $\oint_{\mathrm{II}} q_{\mathrm{b}}(\mathrm{d} s / t)$ and so on.

The fourth equation required for a solution is obtained from Eq. (23.12) by taking moments about the intersection of the x axis and the web 572. Thus

$0=\sum_{R=1}^N \oint_R q_{\mathrm{b}} p_0 \mathrm{~d} s+\sum_{R=1}^N 2 A_R q_{s, 0, R}$  (23.12)

\begin{aligned}0=&-69.0 \times 250 \times 1270-69.0 \times 150 \times 1270+45.5 \times 330 \times 1020 \\&+2 \times 265000 q_{s, 0, \mathrm{I}}+2 \times 213000 q_{s, 0, \mathrm{II}}+2 \times 413000 q_{s, 0, \mathrm{III}}\end{aligned}  (vi)

Simultaneous solution of Eqs (iii)–(vi) gives

$q_{s, 0, \mathrm{I}}=5.5 \mathrm{~N} / \mathrm{mm} \quad q_{s, 0, \mathrm{II}}=10.2 \mathrm{~N} / \mathrm{mm} \quad q_{s, 0, \mathrm{III}}=16.5 \mathrm{~N} / \mathrm{mm}$

Superimposing these shear flows on the $q_b$ distribution of Fig. 23.11, we obtain the final shear flow distribution. Thus

\begin{aligned}q_{34} &=5.5 \mathrm{~N} / \mathrm{mm} \quad q_{23}=q_{87}=10.2 \mathrm{~N} / \mathrm{mm} \quad q_{12}=q_{56}=16.5 \mathrm{~N} / \mathrm{mm} \\q_{61} &=62.0 \mathrm{~N} / \mathrm{mm} \quad q_{57}=79.0 \mathrm{~N} / \mathrm{mm} \quad q_{72}=89.2 \mathrm{~N} / \mathrm{mm} \\q_{48} &=74.5 \mathrm{~N} / \mathrm{mm} \quad q_{83}=64.3 \mathrm{~N} / \mathrm{mm}\end{aligned}

Finally, from any of Eqs (iii)–(v)

$\frac{\mathrm{d} \theta}{\mathrm{d} z}=1.16 \times 10^{-6} \mathrm{rad} / \mathrm{mm}$