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Chapter 23

Q. p.23.3

Determine the torsional stiffness of the four-cell wing section shown in Fig. P.23.3.

Data:
Wall                                   \begin{array}{lll}12 & 23 & 34\end{array}

                                 \begin{array}{llllllll}78 & 67 & 56 & 45^{\circ} & 45^{\mathrm{i}} & 36 & 27 & 18\end{array}

Peripheral length (mm) \begin{array}{llllllll}762  & 812 & 812 & 1525 & 356 & 406 & 356 & 254\end{array}
Thickness (mm)         \begin{array}{llllllll}0.915 & 0.915 & 0.915 & 0.711 & 1.220 & 1.625 & 1.220 & 0.915\end{array}

\begin{aligned}&\text { Cell areas }\left(\mathrm{mm}^2\right)\\&A_{\mathrm{I}}=161500 \quad A_{\mathrm{II}}=291000\\&A_{\mathrm{III}}=291000 \quad A_{\mathrm{IV}}=226000\end{aligned}
Screenshot 2022-10-10 102014

Step-by-Step

Verified Solution

From Eq. (23.6) for Cell I

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_R G}\left(-q_{R-1} \delta_{R-1, R}+q_R \delta_R-q_{R+1} \delta_{R+1, R}\right)  (23.6)

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{I}} G}\left[q_{\mathrm{I}}\left(\delta_{45^{\circ}}+\delta_{45^{\mathrm{i}}}\right)-q_{\mathrm{II}} \delta_{45^{\mathrm{i}}}\right]  (i)
For Cell II

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{II}} G}\left[-q_{\mathrm{I}} \delta_{45^{\mathrm{i}}}+q_{\mathrm{II}}\left(\delta_{34}+\delta_{45^{\mathrm{i}}}+\delta_{56}+\delta_{63}\right)-q_{\mathrm{III}} \delta_{63}\right]  (ii)

For Cell III

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{III}} G}\left[-q_{\mathrm{II}} \delta_{63}+q_{\mathrm{III}}\left(\delta_{23}+\delta_{36}+\delta_{67}+\delta_{72}\right)-q_{\mathrm{IV}} \delta_{72}\right]  (iii)

For Cell IV

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{IV}} G}\left[-q_{\mathrm{III}} \delta_{72}+q_{\mathrm{IV}}\left(\delta_{27}+\delta_{78}+\delta_{81}+\delta_{12}\right)\right]  (iv)

where

\begin{aligned}&\delta_{12}=\delta_{78}=762 / 0.915=832.8 \quad \delta_{23}=\delta_{67}=\delta_{34}=\delta_{56}=812 / 0.915=887.4\\&\delta_{45^{\mathrm{i}}}=356 / 1.220=291.8 \quad \delta_{45^{\circ}}=1525 / 0.711=2144.9\\&\delta_{36}=406 / 1.625=249.8 \quad \delta_{72}=356 / 1.22=291.8 \quad \delta_{81}=254 / 0.915=277.6\end{aligned}

Substituting these values in Eqs (i)–(iv)

\begin{aligned}\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 161500 G}\left(2436.7 q_{\mathrm{I}}-291.8 q_{\mathrm{II}}\right) (v) \\\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 291000 G}\left(-291.8 q_{\mathrm{I}}+2316.4 q_{\mathrm{II}}-249.8 q_{\mathrm{III}}\right)  (vi) \\\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 291000 G}\left(-249.8 q_{\mathrm{II}}+2316.4 q_{\mathrm{III}}-291.8 q_{\mathrm{IV}}\right)  (vii)\\\frac{\mathrm{d} \theta}{\mathrm{d} z} &=\frac{1}{2 \times 226000 G}\left(-291.8 q_{\mathrm{III}}+2235.0 q_{\mathrm{IV}}\right)  (viii)\end{aligned}

Also, from Eq. (23.4)

T=\sum_{R=1}^N 2 A_R q_R  (23.4)

T=2\left(161500 q_{\mathrm{I}}+291000 q_{\mathrm{II}}+291000 q_{\mathrm{III}}+226000 q_{\mathrm{IV}}\right)  (ix)

Equating Eqs (v) and (vi)
q_{\mathrm{I}}-0.607 q_{\mathrm{II}}+0.053 q_{\mathrm{III}}=0 (x)
Now equating Eqs (v) and (vii)
q_{\mathrm{I}}-0.063 q_{\mathrm{II}}-0.528 q_{\mathrm{III}}+0.066 q_{\mathrm{IV}}=0 (xi)

Equating Eqs (v) and (viii)
q_{\mathrm{I}}-0.120 q_{\mathrm{II}}+0.089q_{\mathrm{III}}-0.655 q_{\mathrm{IV}}=0 (xii)
From Eq. (ix)
q_{\mathrm{I}}+1.802 q_{\mathrm{II}}+1.802 q_{\mathrm{III}}+1.399 q_{\mathrm{IV}}=3.096 \times 10^{-6} T (xiii)
Subtracting Eq. (xi) from (x)
q_{\mathrm{II}}-1.068 q_{\mathrm{III}}+0.121 q_{\mathrm{IV}}=0 (xiv)
Subtracting Eq. (xii) from (x)
q_{\mathrm{II}}+0.074 q_{\mathrm{III}}-1.345 q_{\mathrm{IV}}=0 (xv)
Subtracting Eq. (xiii) from (x)
q_{\mathrm{II}}+0.726 q_{\mathrm{III}}+0.581 q_{\mathrm{IV}}=0 (xvi)
Now subtracting Eq. (xv) from (xiv)
q_{\mathrm{III}}-1.284 q_{\mathrm{IV}}=0 (xvii)
Subtracting Eq. (xvi) from (xiv)
q_{\mathrm{III}}+0.256 q_{\mathrm{IV}}=0.716 \times 10^{-6} T (xviii)
Finally, subtracting Eq. (xviii) from (xvii)
q_{\mathrm{IV}}=0.465 \times 10^{-6} T
and from Eq. (xvii)
q_{\mathrm{III}}=0.597 \times 10^{-6} T
Substituting for q_{\mathrm{III}} \text { and } q_{\mathrm{IV}} in Eq. (viii)

so that
T /(\mathrm{d} \theta / \mathrm{d} z)=522.5 \times 10^6 \mathrm{G} \mathrm{Nmm}^2 / \mathrm{rad}