Textbooks & Solution Manuals

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

Your Ultimate AI Essay Writer & Assistant.

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

Need Help? We got you covered.

Products

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

Your Ultimate AI Essay Writer & Assistant.

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

Need Help? We got you covered.

Chapter 23

Q. p.23.5

The idealized cross-section of a two-cell thin-walled wing box is shown in Fig. P.23.5. If the wing box supports a load of 44 500 N acting along the web 25, calculate the shear flow distribution. The shear modulus G is the same for all walls of the wing box.

\begin{array}{llllc}\hline \text { Wall } & \text { Length }(\mathrm{mm}) & \text { Thickness }(\mathrm{mm}) & \text { Boom } & \text { Area }\left(\mathrm{mm}^2\right) \\\hline 16 & 254 & 1.625 & 1,6 & 1290 \\25 & 406 & 2.032 & 2,5 & 1936 \\34 & 202 & 1.220 & 3,4 & 645 \\12,56 & 647 & 0.915 & & \\23,45 & 775 & 0.559 & & \\\text { Cell areas: } A_{\mathrm{I}}=232000 \mathrm{~mm}^2, A_{\mathrm{II}}=258000 \mathrm{~mm}^2 & & \\\hline\end{array}
Screenshot 2022-10-10 111645

Step-by-Step

Verified Solution

In Eq. (23.10) the q_b shear flow distribution is given by Eq. (20.6) in which, since the x axis is an axis of symmetry (Fig. S.23.5), I_{xy} = 0; also S_x = 0. Thus

\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)

\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)

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 1290 \times 127^2+2 \times 1936 \times 203^2+2 \times 645 \times 101^2=214.3 \times 10^6 \mathrm{~mm}^4

Then Eq. (i) becomes

q_{\mathrm{b}}=-\frac{44500}{214.3 \times 10^6} \sum_{r=1}^n B_r y_r=-2.08 \times 10^{-4} \sum_{r=1}^n B_r y_r

‘Cut’ the walls 65 and 54. Then

\begin{aligned}&q_{\mathrm{b}, 65}=q_{\mathrm{b}, 54}=0 \\&q_{\mathrm{b}, 61}=-2.08 \times 10^{-4} \times 1290 \times 127=-32.8 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 12}=q_{\mathrm{b}, 23}=0 \quad(\text { from symmetry }) \\&q_{\mathrm{b}, 25}=-2.08 \times 10^{-4} \times 1936(-203)=81.7 \mathrm{~N} / \mathrm{mm} \\&q_{\mathrm{b}, 34}=-2.08 \times 10^{-4} \times 645(-101)=13.6 \mathrm{~N} / \mathrm{mm}\end{aligned}

From Eq. (23.10) for Cell I

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{J}} G}\left[q_{s, 0, \mathrm{I}}\left(\delta_{56}+\delta_{61}+\delta_{12}+\delta_{24}\right)-q_{s, 0, \mathrm{I}} \delta_{25}+q_{\mathrm{b}, 25} \delta_{25}+q_{\mathrm{b}, 61} \delta_{61}\right]  (ii)

For Cell II

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{II}} G}\left[-q_{s, 0, \mathrm{I}} \delta_{25}+q_{s, 0,\mathrm{II}}\left(\delta_{45}+\delta_{52}+\delta_{23}+\delta_{34}\right)+q_{\mathrm{b}, 34} \delta_{34}+q_{\mathrm{b}, 52} \delta_{25}\right]  (iii)

in which

\begin{aligned}&\delta_{56}=\delta_{12}=647 / 0.915=707.1 \quad \delta_{45}=\delta_{23}=775 / 0.559=1386.4 \\&\delta_{61}=254 / 1.625=156.3 \quad \delta_{52}=406 / 2.032=199.8 \quad \delta_{34}=202 / 1.220=165.6\end{aligned}

Substituting these values in Eqs (ii) and (iii)

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 232000 G}\left(1770.3 q_{s, 0, \mathrm{I}}-199.8 q_{s, 0, \mathrm{II}}+11197.0\right)  (iv)

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 258000 G}\left(-199.8 q_{s, 0, \mathrm{I}}+3138.2 q_{s, 0, \mathrm{II}}-14071.5\right)  (v)

Also, taking moments about the mid-point of the web 25 and from Eq. (23.11) (or Eq. (23.12))

S_x \eta_0-S_y \xi_0=\sum_{R=1}^N M_{q, R}=\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}  (23.11)

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)

0=13.6 \times 202 \times 763-32.8 \times 254 \times 635+2 A_{\mathrm{I}} q_{s, 0, \mathrm{I}}+2 A_{\mathrm{II}} q_{s, 0, \mathrm{II}}  (vi)

Equating Eqs (iv) and (v)
q_{s, 0, \mathrm{I}}-1.55 q_{s, 0, \mathrm{II}}+12.23=0 (vii)
From Eq. (vi)
q_{s, 0, \mathrm{I}}+1.11 q_{s, 0, \mathrm{II}}-6.88=0 (viii)
Subtracting Eq. (viii) from (vii) gives
q_{s, 0, \mathrm{II}}=7.2 \mathrm{~N} / \mathrm{mm}

Then, from Eq. (vii)
q_{s, 0, \mathrm{I}}=-1.1 \mathrm{~N} / \mathrm{mm}

Thus

\begin{aligned}&q_{16}=32.8+1.1=33.9 \mathrm{~N} / \mathrm{mm} \quad q_{65}=q_{21}=1.1 \mathrm{~N} / \mathrm{mm} \\&q_{45}=q_{23}=7.2 \mathrm{~N} / \mathrm{mm} \quad q_{34}=13.6+7.2=20.8 \mathrm{~N} / \mathrm{mm} \\&q_{25}=81.7-1.1-7.2=73.4 \mathrm{~N} / \mathrm{mm}\end{aligned}
Screenshot 2022-10-10 112551