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

Question

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 distribution for a load of [latex]S_y[/latex] = 66 750 N through S. Tabulated below are lengths, thicknesses and shear moduli of the shear carrying walls. Note that dotted line 45 is not a wall.

[latex]\begin{array}{lclllc}\hline ext { Wall } & ext { Length }(\mathrm{mm}) & ext { Thickness }(\mathrm{mm}) & G\left(\mathrm{~N} / \mathrm{mm}^2 ight) & ext { Boom } & ext { Area }\left(\mathrm{mm}^2 ight) \\hline 34,56 & 380 & 0.915 & 20700 & 1,3,6,8 & 1290 \12,23,67,78 & 356 & 0.915 & 24200 & 2,4,5,7 & 645 \36,81 & 306 & 1.220 & 24800 & & \45 & 610 & 1.220 & 24800 & & \ ext { Nose area } N_1=51500 \mathrm{~mm}^2 & & & & \\hline\end{array}[/latex]

Accepted Answer

Referring to Fig. P.23.6, the horizontal x axis is an axis of symmetry so that [latex]I_{xy}[/latex] = 0. Further, [latex]S_x[/latex] = 0 so that, from Eq. (20.6)

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)\end{aligned}[/latex] (20.6)

[latex]q_{\mathrm{b}}=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r[/latex] (i)

in which
[latex]I_{x x}=4 imes 1290 imes 153^2+4 imes 645 imes 153^2=181.2 imes 10^6 \mathrm{~mm}^4[/latex]
Eq. (i) then becomes

[latex]q_{\mathrm{b}}=-\frac{66750}{181.2 imes 10^6} \sum_{r=1}^n B_r y_r=-3.68 imes 10^{-4} \sum_{r=1}^n B_r y_r[/latex]

Now, ‘cutting’ Cell I in the wall 45 and Cell II in the wall 12

[latex]\begin{aligned}&q_{\mathrm{b}, 45}=0=q_{\mathrm{b}, 12} \&q_{\mathrm{b}, 43}=-3.68 imes 10^{-4} imes 645 imes 153=-36.3 \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 65} \quad ext { (from symmetry) } \&q_{\mathrm{b}, 18}=-3.68 imes 10^{-4} imes 1290 imes 153=-72.6 \mathrm{~N} / \mathrm{mm} \&q_{\mathrm{b}, 78}=0 \quad( ext { from symmetry) } \&q_{\mathrm{b}, 76}=-3.68 imes 10^{-4} imes 645 imes(-153)=36.3 \mathrm{~N} / \mathrm{mm}=q_{\mathrm{b}, 32} \quad \quad \quad ext { (from symmetry) } \&q_{\mathrm{b}, 63}=36.3+36.3-3.68 imes 10^{-4} imes 1290 imes(-153)=145.2 \mathrm{~N} /\mathrm{mm}\end{aligned}[/latex]

The shear load is applied through the shear centre of the section so that the rate of twist of the section, dθ/dz, is zero and Eq. (23.10) for Cell I simplifies to

[latex]\frac{\mathrm{d} heta}{\mathrm{d} z}=\frac{1}{2 A_R G}\left(-q_{s, 0, R-1} \delta_{R-1, R}+q_{s, 0, R} \delta_R-q_{s, 0, R+1} \delta_{R+1, R}+\oint_R q_{\mathrm{b}} \frac{\mathrm{d} s}{t} ight)[/latex] (23.10)

[latex]\begin{aligned}0=\frac{1}{2 A_{\mathrm{I}} G_{\mathrm{REF}}}[& q_{s, 0, \mathrm{I}}\left(\delta_{34}+\delta_{45}+\delta_{56}+\delta_{63} ight) \&\left.-q_{s, 0, \mathrm{II}} \delta_{63}+q_{\mathrm{b}, 63} \delta_{63}+q_{\mathrm{b}, 34} \delta_{34}+q_{\mathrm{b}, 56} \delta_{56} ight]\end{aligned}[/latex] (ii)

and for Cell II

[latex]\begin{aligned}&0=\frac{1}{2 A_{\mathrm{II}} G_{\mathrm{REF}}}\left[-q_{s, 0, \mathrm{I}} \delta_{63}+q_{s, 0, \mathrm{II}}\left(\delta_{12}+\delta_{23}+\delta_{36}+\delta_{67}+\delta_{78}+\delta_{81} ight)+q_{\mathrm{b}, 81} \delta_{81} ight.\&\left.+q_{\mathrm{b}, 23} \delta_{23}+q_{\mathrm{b}, 36} \delta_{36}+q_{\mathrm{b}, 67} \delta_{67} ight]\end{aligned}[/latex] (iii)

in which [latex]G_{\mathrm{REF}}=24200 \mathrm{~N} / \mathrm{mm}^2[/latex]. Then, from Eq. (23.9)

[latex]t^*=\frac{G}{G_{\mathrm{REF}}} t[/latex] (23.9)

[latex]\begin{aligned}&t_{34}^*=t_{56}^*=\frac{20700}{24200} imes 0.915=0.783 \mathrm{~mm} \&t_{36}^*=t_{81}^*=t_{45}^*=\frac{24800}{24200} imes 1.220=1.250 \mathrm{~mm}\end{aligned}[/latex]

Thus

[latex]\begin{aligned}&\delta_{34}=\delta_{56}=380 / 0.783=485.3 \&\delta_{12}=\delta_{23}=\delta_{67}=\delta_{78}=356 / 0.915=389.1 \&\delta_{36}=\delta_{81}=306 / 1.250=244.8 \&\delta_{45}=610 / 1.250=488.0\end{aligned}[/latex]

Eq. (ii) then becomes

[latex]1703.4 q_{s, 0, \mathrm{I}}-244.8 q_{s, 0, \mathrm{II}}+70777.7=0[/latex]

or

[latex]q_{s, 0, \mathrm{I}}-0.144 q_{s, 0, \mathrm{II}}+41.55=0[/latex] (iv)

and Eq. (iii) becomes

[latex]-244.8 q_{s, 0, \mathrm{I}}+2046 q_{s, 0, \mathrm{II}}-46021.1=0[/latex]

or
[latex]q_{s, 0, \mathrm{I}}-8.358 q_{s, 0, \mathrm{II}}+188.0=0[/latex] (v)
Subtracting Eq. (v) from (iv) gives
[latex]q_{s, 0, \mathrm{II}}=17.8 \mathrm{~N} / \mathrm{mm}[/latex]
Then, from Eq. (v)
[latex]q_{s, 0, \mathrm{I}}=-39.2 \mathrm{~N} / \mathrm{mm}[/latex]

The resulting shear flows are then

[latex]\begin{aligned}&q_{12}=q_{78}=17.8 \mathrm{~N} / \mathrm{mm} \quad q_{32}=q_{76}=36.3-17.8=18.5 \mathrm{~N} / \mathrm{mm} \&q_{63}=145.2-17.8-39.2=88.2 \mathrm{~N} / \mathrm{mm} \&q_{43}=q_{65}=39.2-36.3=2.9 \mathrm{~N} / \mathrm{mm} \quad q_{54}=39.2 \mathrm{~N} / \mathrm{mm} \&q_{81}=72.6+17.8=90.4 \mathrm{~N} / \mathrm{mm}\end{aligned}[/latex]

Now taking moments about the mid-point of the web 63

[latex]\begin{aligned}66750 x_{\mathrm{S}}=&-2 imes q_{76} imes 356 imes 153+2 imes q_{78} imes 356 imes 153+q_{81} imes 306 imes 712 \&-2 imes q_{43} imes 380 imes 153-q_{54} imes 2(51500+153 imes 380)( ext { see Eq. (20.10)) }\end{aligned}[/latex]

[latex]M_q=2 A q_{12}[/latex] (20.10)

from which
[latex]x_S[/latex] = 160.1 mm

Question p.23.6: Figure P.23.6 shows a singly symmetric, two-cell wing sectio...

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:

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:

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, ...

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: