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.

Question

[latex]\begin{array}{lrlll}\hline ext { Wall } & ext { Length }(\mathrm{mm}) & ext { Thickness }(\mathrm{mm}) & ext { Cell } & ext { Area }\left(\mathrm{mm}^2 ight) \\hline 12^{\mathrm{U}} & 1084 & 1.220 & ext { I } & 108400 \12^{\mathrm{L}} & 2160 & 1.625 & ext { II } & 202500 \14,23 & 127 & 0.915 & ext { III } & 528000 \34^{\mathrm{U}} & 797 & 0.915 & & \34^{\mathrm{L}} & 797 & 0.915 & & \\hline\end{array}[/latex]

Accepted Answer

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

[latex]\frac{\mathrm{d} heta}{\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} ight)[/latex] (23.6)

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 A_R G} \oint_R q \frac{\mathrm{d} s}{t}[/latex] (23.5)

[latex]\frac{\mathrm{d} heta}{\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} ight)-q_{\mathrm{II}} \delta_{34^ \mathrm{U}}-q_{\mathrm{III}}\left(\delta_{23}+\delta_{41} ight) ight][/latex] (i)

For Cell II

[latex]\frac{\mathrm{d} heta}{\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}}} ight)-q_{\mathrm{III}} \delta_{34^{\mathrm{L}}} ight][/latex] (ii)

For Cell III

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

In Eqs (i)–(iii)

[latex]\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}[/latex]

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

[latex]\begin{aligned}&\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 108400 G}\left(2037.1 q_{\mathrm{I}}-871.0 q_{\mathrm{II}}-277.6 q_{\mathrm{III}} ight) (iv) \&\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 202500 G}\left(-871.0 q_{\mathrm{I}}+1742.0 q_{\mathrm{II}}-871.0 q_{\mathrm{III}} ight) (v) \&\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 528000 G}\left(-277.6 q_{\mathrm{I}}-871.0 q_{\mathrm{II}}+2477.8 q_{\mathrm{III}} ight) (vi)\end{aligned}[/latex]

Also, from Eq. (23.4)

[latex]T=\sum_{R=1}^N 2 A_R q_R[/latex] (23.4)

[latex]565000 imes 10^3=2\left(108400 q_{\mathrm{I}}+202500 q_{\mathrm{II}}+528000 q_{\mathrm{III}} ight)[/latex] (vii)

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

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

[latex]\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}[/latex]

Question p.23.4: Determine the shear flow distribution for a torque of 56 500...

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