## Chapter 23

## Q. p.23.9

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.

## Step-by-Step

## Verified Solution

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 q_1 in the spar webs. Then

q_1=\frac{2000}{2 \times 200}=5 \mathrm{~N} / \mathrm{mm}Similarly in bay ②

q_2=\frac{2000}{2 \times 200}=5 \mathrm{~N} / \mathrm{mm}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

\pm \frac{2000 \times 1000}{2 \times 200}=\pm 5000 \mathrm{~N}at the built-in end. Then, at the built-in end

P_1=P_4=-P_2=-P_3=5000 \mathrm{~N}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

which gives

q_1 = 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

2 P_1=500 q_1=500 \times 12.5

so that

P_1 = 3125 N

Now considering the equilibrium of flange 1 in bay ②

P_1+q_2 \times 500-q_3 \times 500=0

whence

q_2-q_3=-6.25 (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

2 \times 2 \times \frac{1}{2} \times 200 \times 200 q_2+2 \times 2 \times \frac{1}{2} \times 400 \times 100 q_3=1000 \times 10^3i.e.

q_2+q_3=12.5 (ii)

Adding Eqs (i) and (ii) gives

q_2 = 3.125 N/mm

whence

q_3 = 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: P_1 = 5000 − 3125 = 1875 N (tension)

At the central rib: P_1 = 2500 + 3125 = 5625 N (tension)

Bay ②

At the central rib: P_1 = 3625 N (tension)

At the free end: P_1 = 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