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,

Question

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, the distribution of load in the spar flanges and the loading on the central rib. Assume that the spar webs and skin panels are effective in resisting shear stresses only.

Accepted Answer

Consider first the flange loads and shear flows produced by the shear load acting through the shear centre of the wing box. Referring to Fig. S.23.9(a), in bay ① the shear load is resisted by the shear flows [latex]q_1[/latex] in the spar webs. Then

[latex]q_1=\frac{2000}{2 imes 200}=5 \mathrm{~N} / \mathrm{mm}[/latex]

Similarly in bay ②

[latex]q_2=\frac{2000}{2 imes 200}=5 \mathrm{~N} / \mathrm{mm}[/latex]

From symmetry the bending moment produced by the shear load will produce equal but opposite loads in the top and bottom flanges. These flange loads will increase with bending moment, i.e. linearly, from zero at the free end to

[latex]\pm \frac{2000 imes 1000}{2 imes 200}=\pm 5000 \mathrm{~N}[/latex]

at the built-in end. Then, at the built-in end

[latex]P_1=P_4=-P_2=-P_3=5000 \mathrm{~N}[/latex]

Alternatively, the flange loads may be determined by considering the equilibrium of a single flange subjected to the flange load and the shear flows in the adjacent spar webs.
Now consider the action of the applied torque in Fig. S.23.9(b). In bay ① the torque is resisted by differential bending of the spar webs. Thus

[latex]q_1 imes 200 imes 400=1000 imes 10^3[/latex]

which gives
[latex]q_1[/latex] = 12.5 N/mm
The differential bending of the spar webs in bay ① induces flange loads as shown in Fig. S.23.9(c). For equilibrium of flange 1

[latex]2 P_1=500 q_1=500 imes 12.5[/latex]
so that
[latex]P_1[/latex] = 3125 N
Now considering the equilibrium of flange 1 in bay ②
[latex]P_1+q_2 imes 500-q_3 imes 500=0[/latex]
whence
[latex]q_2-q_3=-6.25[/latex] (i)

Also, the resultant of the shear flows in the spar webs and skin panels in bay ② is equivalent to the applied torque. Thus

[latex]2 imes 2 imes \frac{1}{2} imes 200 imes 200 q_2+2 imes 2 imes \frac{1}{2} imes 400 imes 100 q_3=1000 imes 10^3[/latex]

i.e.
[latex]q_2+q_3=12.5[/latex] (ii)
Adding Eqs (i) and (ii) gives
[latex]q_2[/latex] = 3.125 N/mm
whence
[latex]q_3[/latex] = 9.375 N/mm

The shear flows due to the combined action of the shear and torsional loads are then as follows:
Bay ①

Spar webs: q = 12.5 − 5 = 7.5 N/mm
Bay ②
Spar webs: q = 5 − 3.125 = 1.875 N/mm
Skin panels: q = 9.375 N/mm
The flange loads are:
Bay ①
At the built-in end: [latex]P_1[/latex] = 5000 − 3125 = 1875 N (tension)
At the central rib: [latex]P_1[/latex] = 2500 + 3125 = 5625 N (tension)
Bay ②

At the central rib: [latex]P_1[/latex] = 3625 N (tension)
At the free end: [latex]P_1[/latex] = 0
Finally the shear flows on the central rib are:
On the horizontal edges: q = 9.375 N/mm
On the vertical edges: q = 7.5 + 1.875 = 9.375 N/mm

Question p.23.9: A portion of a wing box is built-in at one end and carries a...

Related Answered Questions

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:

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: