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Chapter 23

Q. p.23.2

Figure P.23.2 shows the cross-section of a two-cell torque box. If the shear stress in any wall must not exceed 140 N/mm², find the maximum torque which can be applied to the box.

If this torque were applied at one end and resisted at the other end of such a box of span 2500 mm, find the twist in degrees of one end relative to the other and the torsional rigidity of the box. The shear modulus G = 26 600 N/mm² for all walls. Data:

Shaded areas: A_{34}=6450 \mathrm{~mm}^2, A_{16}=7750 \mathrm{~mm}^2
Wall lengths: s_{34}=250 \mathrm{~mm}, \quad s_{16}=300 \mathrm{~mm}
Wall thickness: t_{12}=1.63 \mathrm{~mm}, \quad t_{34}=0.56 \mathrm{~mm}
\begin{aligned}&t_{23}=t_{45}=t_{56}=0.92 \mathrm{~mm} \\&t_{61}=2.03 \mathrm{~mm} \\&t_{25}=2.54 \mathrm{~mm}\end{aligned}

Screenshot 2022-10-10 100607

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_{21}+\delta_{16}+\delta_{65}+\delta_{52}\right)-q_{\mathrm{II}} \delta_{52}\right]  (i)
and for Cell II

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_{\mathrm{II}} G}\left[-q_{\mathrm{I}} \delta_{52}+q_{\mathrm{II}}\left(\delta_{32}+\delta_{25}+\delta_{54}+\delta_{43}\right)\right]

In Eqs (i) and (ii)

\begin{aligned}&A_{\mathrm{I}}=7750+(250+600) \times 500 / 2=220250 \mathrm{~mm}^2 \\&A_{\mathrm{II}}=6450+(150+600) \times 920 / 2=351450 \mathrm{~mm}^2 \\&\delta_{21}=\left(\sqrt{250^2+500^2}\right) / 1.63=343.0 \quad \delta_{16}=300 / 2.03=147.8 \\&\delta_{65}=\left(\sqrt{100^2+500^2}\right) / 0.92=554.2 \quad \delta_{52}=600 / 2.54=236.2 \\&\delta_{54}=\left(\sqrt{250^2+920^2}\right) / 0.92=1036.3 \quad \delta_{43}=250 / 0.56=446.4 \\&\delta_{32}=\left(\sqrt{200^2+920^2}\right) / 0.92=1023.4\end{aligned}

Substituting these values in Eqs (i) and (ii) gives, for Cell I

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 220250 G}\left(1281.2 q_{\mathrm{I}}-236.2 q_{\mathrm{II}}\right)  (iii)

and for Cell II

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 351450 G}\left(-236.2 q_{\mathrm{I}}+2742.3 q_{\mathrm{II}}\right)  (iv)

Equating Eqs (iii) and (iv) gives
q_{\mathrm{II}}=0.73 q_{\mathrm{I}} (v)
Then, in Cell I

\tau_{\max }=\tau_{65}=\frac{q_{\mathrm{I}}}{0.92}=1.087 q_{\mathrm{I}}

and in Cell II

\tau_{\max }=\frac{q_{\mathrm{II}}}{0.56}=1.304 q_{\mathrm{I}}

In the wall 52
\tau_{52}=\frac{q_{\mathrm{I}}-q_{\mathrm{II}}}{2.54}=0.106 q_{\mathrm{I}}
Therefore
\tau_{\max }=1.304 q_{\mathrm{I}}=140 \mathrm{~N} / \mathrm{mm}^2
which gives
q_{\mathrm{I}}=107.4 \mathrm{~N} / \mathrm{mm}
and, from Eq. (v)
q_{\mathrm{II}}=78.4 \mathrm{~N} / \mathrm{mm}
Substituting for q_{\mathrm{I}} \text { and } q_{\mathrm{II}} in Eq. (23.4)

T=\sum_{R=1}^N 2 A_R q_R  (23.4)

T=(2 \times 220250 \times 107.4+2 \times 351450 \times 78.4) \times 10^{-3}

i.e.
T = 102 417 Nm
From Eq. (iii) (or Eq. (iv))

\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 220250 \times 26600}(1281.2 \times 107.4-236.2 \times 78.4)

i.e.

\frac{\mathrm{d} \theta}{\mathrm{d} z}=1.02 \times 10^{-5} \mathrm{rad} / \mathrm{mm}

Hence

\theta=1.02 \times 10^{-5} \times 2500 \times\left(\frac{180}{\pi}\right)=1.46^{\circ}

The torsional stiffness is obtained from Eq. (3.12), thus

T=G J \frac{\mathrm{d} \theta}{\mathrm{d} z}  (3.12)

G J=\frac{T}{(\mathrm{~d} \theta / \mathrm{d} z)}=102417 \times 10^3 /\left(1.02 \times 10^{-5}\right)=10 \times 10^{12} \mathrm{Nmm}^2 / \mathrm{rad}