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Chapter 23

Q. p.23.4

Determine the shear flow distribution for a torque of 56 500 N m for the three cell section shown in Fig. P.23.4. The section has a constant shear modulus throughout.

\begin{array}{lrlll}\hline \text { Wall } & \text { Length }(\mathrm{mm}) & \text { Thickness }(\mathrm{mm}) & \text { Cell } & \text { Area }\left(\mathrm{mm}^2\right) \\\hline 12^{\mathrm{U}} & 1084 & 1.220 & \text { I } & 108400 \\12^{\mathrm{L}} & 2160 & 1.625 & \text { II } & 202500 \\14,23 & 127 & 0.915 & \text { III } & 528000 \\34^{\mathrm{U}} & 797 & 0.915 & & \\34^{\mathrm{L}} & 797 & 0.915 & & \\\hline\end{array}
Screenshot 2022-10-10 104312

Step-by-Step

Verified Solution

In this problem the cells are not connected consecutively so that Eq. (23.6) does not apply. Therefore, from Eq. (23.5) 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_R G} \oint_R q \frac{\mathrm{d} s}{t}  (23.5)

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{I}} G}\left[q_{\mathrm{I}}\left(\delta_{12^ \mathrm{U}}+\delta_{23}+\delta_{34^ \mathrm{U}}+\delta_{41}\right)-q_{\mathrm{II}} \delta_{34^ \mathrm{U}}-q_{\mathrm{III}}\left(\delta_{23}+\delta_{41}\right)\right]  (i)

For Cell II

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

For Cell III

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{III}} G}\left[-q_{\mathrm{I}}\left(\delta_{23}+\delta_{41}\right)\right.\left.-q_{\mathrm{II}} \delta_{34^ \mathrm{~L}}+q_{\mathrm{III}}\left(\delta_{14}+\delta_{43^\mathrm{~L}} +\delta_{32}+\delta_{21^{\mathrm{L}}}\right)\right]  (iii)

In Eqs (i)–(iii)

\begin{aligned}\delta_{12^ \mathrm{U}} &=1084 / 1.220=888.5 \quad \delta_{12^ \mathrm{~L}}=2160 / 1.625=1329.2 \\\delta_{14} &=\delta_{23}=127 / 0.915=138.8 \quad \delta_{34^ \mathrm{U}}=\delta_{34 ^\mathrm{~L}}=797 / 0.915=871.0\end{aligned}

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

\begin{aligned}&\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 108400 G}\left(2037.1 q_{\mathrm{I}}-871.0 q_{\mathrm{II}}-277.6 q_{\mathrm{III}}\right)  (iv) \\&\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 202500 G}\left(-871.0 q_{\mathrm{I}}+1742.0 q_{\mathrm{II}}-871.0 q_{\mathrm{III}}\right)  (v) \\&\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 528000 G}\left(-277.6 q_{\mathrm{I}}-871.0 q_{\mathrm{II}}+2477.8 q_{\mathrm{III}}\right)  (vi)\end{aligned}

Also, from Eq. (23.4)

T=\sum_{R=1}^N 2 A_R q_R  (23.4)

565000 \times 10^3=2\left(108400 q_{\mathrm{I}}+202500 q_{\mathrm{II}}+528000 q_{\mathrm{III}}\right)  (vii)

Equating Eqs (iv) and (v)
q_{\mathrm{I}}-0.720 q_{\mathrm{II}}+0.075 q_{\mathrm{III}}=0 (viii)
Equating Eqs (iv) and (vi)
q_{\mathrm{I}}-0.331 q_{\mathrm{II}}-0.375 q_{\mathrm{III}}=0 (ix)
From Eq. (vii)
q_{\mathrm{I}}+1.868 q_{\mathrm{II}}+4.871 q_{\mathrm{III}}=260.61 (x)
Now subtracting Eq. (ix) from (viii)
q_{\mathrm{II}}-1.157 q_{\mathrm{III}}=0 (xi)
Subtracting Eq. (x) from (viii)
q_{\mathrm{II}}+1.853 q_{\mathrm{III}}=100.70 (xii)
Finally, subtracting Eq. (xii) from (xi)
q_{\mathrm{III}}=33.5 \mathrm{~N} / \mathrm{mm}

Then, from Eq. (xi)
q_{\mathrm{II}}=38.8 \mathrm{~N} / \mathrm{mm}
and from Eq. (ix)
q_{\mathrm{I}}=25.4 \mathrm{~N} / \mathrm{mm}
Thus

\begin{aligned}&q_{12^\mathrm{U}} =25.4 \mathrm{~N} / \mathrm{mm} \quad q_{21^\mathrm{~L}} =33.5 \mathrm{~N} / \mathrm{mm} \quad q_{14}=q_{32}=33.5-25.4=8.1 \mathrm{~N} / \mathrm{mm} \\&q_{43^\mathrm{U}} =38.8-25.4=13.4 \mathrm{~N} /\mathrm{mm} \quad q_{34^\mathrm{~L}} =38.8-33.5=5.3 \mathrm{~N} /\mathrm{mm}\end{aligned}