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,

Question

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: [latex]A_{34}=6450 \mathrm{~mm}^2, A_{16}=7750 \mathrm{~mm}^2[/latex]
Wall lengths: [latex]s_{34}=250 \mathrm{~mm}, \quad s_{16}=300 \mathrm{~mm}[/latex]
Wall thickness: [latex]t_{12}=1.63 \mathrm{~mm}, \quad t_{34}=0.56 \mathrm{~mm}[/latex]
[latex]\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}[/latex]

Accepted Answer

From Eq. (23.6) 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_{\mathrm{I}} G}\left[q_{\mathrm{I}}\left(\delta_{21}+\delta_{16}+\delta_{65}+\delta_{52} ight)-q_{\mathrm{II}} \delta_{52} ight][/latex] (i)
and for Cell II

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

In Eqs (i) and (ii)

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

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

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 220250 G}\left(1281.2 q_{\mathrm{I}}-236.2 q_{\mathrm{II}} ight)[/latex] (iii)

and for Cell II

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 351450 G}\left(-236.2 q_{\mathrm{I}}+2742.3 q_{\mathrm{II}} ight)[/latex] (iv)

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

[latex] au_{\max }= au_{65}=\frac{q_{\mathrm{I}}}{0.92}=1.087 q_{\mathrm{I}}[/latex]

and in Cell II

[latex] au_{\max }=\frac{q_{\mathrm{II}}}{0.56}=1.304 q_{\mathrm{I}}[/latex]

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

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

[latex]T=(2 imes 220250 imes 107.4+2 imes 351450 imes 78.4) imes 10^{-3}[/latex]

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

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 imes 220250 imes 26600}(1281.2 imes 107.4-236.2 imes 78.4)[/latex]

i.e.

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=1.02 imes 10^{-5} \mathrm{rad} / \mathrm{mm}[/latex]

Hence

[latex] heta=1.02 imes 10^{-5} imes 2500 imes\left(\frac{180}{\pi} ight)=1.46^{\circ}[/latex]

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

[latex]T=G J \frac{\mathrm{d} heta}{\mathrm{d} z}[/latex] (3.12)

[latex]G J=\frac{T}{(\mathrm{~d} heta / \mathrm{d} z)}=102417 imes 10^3 /\left(1.02 imes 10^{-5} ight)=10 imes 10^{12} \mathrm{Nmm}^2 / \mathrm{rad}[/latex]

Question p.23.2: Figure P.23.2 shows the cross-section of a two-cell torque b...

Related Answered Questions

A portion of a wing box is built-in at one end and carries a shear load of 2000 N through its shear centre and a torque of 1000 N m as shown in Fig. P.23.9. If the skin panel in the upper surface of the inboard bay is removed, calculate the shear flows in the spar webs and remaining skin panels, ...

Verified Answer:

Verified Answer:

Verified Answer:

Figure P.23.6 shows a singly symmetric, two-cell wing section in which all direct stresses are carried by the booms, shear stresses alone being carried by the walls. All walls are flat with the exception of the nose portion 45. Find the position of the shear centre S and the shear flow ...

Verified Answer:

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. ...

Verified Answer:

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. ...

Verified Answer:

Verified Answer:

Verified Answer:

A wing box has the skin panel on its undersurface removed between stations 2000 and 3000 and carries lift and drag loads which are constant between stations 1000 and 4000 as shown in Fig. 23.21(a). Determine the shear flows in the skin panels and spar webs and also the loads in the wing ribs at ...

Verified Answer:

Calculate the deflection at the free end of the two-cell beam shown in Fig. 23.15 allowing for both bending and shear effects. The booms carry all the direct stresses while the skin panels, of constant thickness throughout, are effective only in shear. ...

Verified Answer: