## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. p.23.7

A singly symmetric wing section consists of two closed cells and one open cell (see Fig. P.23.7). The webs 25, 34 and the walls 12, 56 are straight, while all other walls are curved. All walls of the section are assumed to be effective in carrying shear stresses only, direct stresses being carried by booms 1–6. Calculate the distance xS of the shear centre S aft of the web 34. The shear modulus G is the same for all walls.

$\begin{array}{llllllc}\hline \text { Wall } & \text { Length }(\mathrm{mm}) & \text { Thickness }(\mathrm{mm}) & \text { Boom } & \text { Area }\left(\mathrm{mm}^2\right) & \text { Cell } & \text { Area }\left(\mathrm{mm}^2\right) \\\hline 12,56 & 510 & 0.559 & 1,6 & 645 & \text { I } & 93000 \\23,45 & 765 & 0.915 & 2,5 & 1290 & \text { II } & 258000 \\34^{\circ} & 1015 & 0.559 & 3,4 & 1935 & & \\34^{\mathrm{i}} & 304 & 2.030 & & & & \\25 & 304 & 1.625 & & & & \\\hline\end{array}$

## Verified Solution

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