Question 24.3: A three-flange wing section is stiffened by the wing rib sho......

A three-flange wing section is statically determinate (see Section 23.1), so that the shear flows applied to the wing rib may be found by considering the equilibrium of the wing rib. From Fig. S.24.3(a) and resolving forces horizontally,

600q_{12}-600q_{34}-1200=0

whence,

q_{12}-q_{34}=20        (i)

Now resolving vertically and noting that q_{51}=q_{45},

400q_{45}-400q_{23}+8000=0

i.e.,

q_{45}-q_{23}=-20        (ii)

Taking moments about 4,

q_{12} \times 600 \times 400+2\left(\frac{\pi\times 200^2}{2}+\frac{1}{2} \times 400 \times 600\right) q_{23}-12000 \times 200-8000\times 600=0

so that

q_{12}+1.52q_{23}=30        (iii)

Subtracting Eq. (iii) from (i) and noting that q_{34}=q_{23},

-2.52q_{23}=-10

or

q_{23}=4.0\mathrm{N/mm}=q_{34}

Then from Eq. (i),

q_{12}=24.0\mathrm{N/mm}

and from Eq. (ii),

q_{45}=-16.0\mathrm{N/mm}=q_{51}

Consider the nose portion of the wing rib in Fig. S.24.3(b). Taking moments about 3,

P_{2}\times400-2\times{\frac{\pi\times200^{2}}{2}}\times4.0=0

from which

P_{2}=1256.6\mathrm{N}(\mathrm{tension})

From horizontal equilibrium,

P_{3}+P_{2}=0

whence,

P_{3}=-1256.6N{\mathrm{(compression)}}\,

and from vertical equilibrium,

q_{1}=4.0\mathrm{N/mm}

From the vertical equilibrium of the stiffener 154 in Fig. S.24.3(c),

q_{2}\times200+q_{3}\times200-16\times400=0

i.e.,

q_{2}+q_{3}=32        (iv)

Also, in 15 at any distance h from 1,

P_{15}+16h-q_{2}h=0

i.e.,

P_{15}=(q_{2}-16)h        (v)

and in 54,

P_{54}+16h-q_{2}\times200-q_{3}(h-200)=0

whence,

P_{S4}=200(q_{2}-q_{3})+(q_{3}-16)h          (vi)

Fig. S.24.3(d) shows the stiffener 56. From horizontal equilibrium,

600q_{2}-600q_{3}-12000=0

or

q_{2}-q_{3}=20          (vii)

Adding Eqs (iv) and (vii),

2q_{2}=52

i.e.,

q_{2}=26{\mathrm{N}}/{\mathrm{mm}}

and from Eq. (iv),

q_{3}=6\mathrm{N}/\mathrm{mm}

Then, from Eq. (v),

P_{15}=10h          (viii)

and P_{15} varies linearly from zero at 1 to 2000 N (tension) at 5. From Eq. (vi),

P _{54}=200(26-6)+(6-16)h

i.e.,

P_{54}=4000-10h        (ix)

so that P_{54} varies linearly from 2000 N (tension) at 5 to zero at 4. Now from Fig. S.24.3(d) at any
section z,

P_{56}+q_2 z-q_3 z-12000=0

i.e.,

P_{56}=-20z+12000        (x)

Thus P_{56} varies linearly from 12,000 N (tension) at 5 to zero at 6.
Consider the flange 12 in Fig. S.24.3(e). At any section a distance z from 1,

P_{12}+24z-26z=0

i.e.,

P_{12}=2z          (xi)

Hence, P_{12} varies linearly from zero at 1 to 1200 N (tension) at 2.
Now consider the bottom flange in Fig. S.24.3(f). At any section a distance z from 4,

\ P_{43}+6z-4z=0

i.e.,

P_{43}=-2z        (xii)

Thus P_{43} varies linearly from zero at 4 to –1200 N (compression) at 3. (The discrepancy between P_{2} in 12 and P_{2} in 23 and between P_{3} in 43 and P_{3} in 23 is due to the rounding-off error in the shear flow q_{1}.)

In Fig. S.24.3(g), the load in the stiffener at any section a distance h from 2 is given by

P_{26}+26h+4h=0

i.e.,

P_{26}=-30h          (xiii)

Therefore P_{26} varies linearly from zero at 2 to –6000 N (compression) at 6. In 63,

P_{63}+26\times200+4h+6(h-200)=0

i.e.,

P_{63}=-4000-10h      (xiv)

Thus P_{63} varies linearly from –6000 N (compression) at 6 to –8000 N (compression) at 3.