PART SEVENMULTICELL BEAMS
because for bending only the boom areas are used. So in order to do this calculation just use the equations derived in Chapter 3.
Look at a wing section made of 'N' cells carrying a torque 'T'.
TORSION
(5.1), which is: Substituting this gives: However since we have 'N' cells this equation is statically indeterminate, so we need a compatibility relationship.
Look at an elemental section of wing of length z, where three skin sections are in contact withEffect of Booms on Shear Flow Distribution a boom of area B, Figure 81.
the boom experiences forces only produced by shear flows because no direct stresses are present. Therefore the booms don't affect the analysis in pure torsion.
Summing the forces along boom gives:
which simplifies to:
If no booms are present a similar relationship is obtained.
This equation means that the sum of the shear flows into a junction must equal the sum of the
Due to the ribs in aircraft wings, the rate of twist is constant for all cells in the section. So lookCompatibility Equation for Each Cell at the rate of twist of cell in the wing section of Figure 80 . Equation (4.21) then becomes foricell equal to:
In order to derive the rate of twist for cell i, it is necessary to look at the three cells: i, i - 1andi , and determine the bound integral at cell i + 1 of equation (4.21).
i
giving:
Rearranging this gives:
and in general terms for the configuration of Figure 82, the rate of twist equation for cell i ibecomes: where: D _{ ( i - 1 , i )} = length of wall common with the cell and i cell divided by thei - 1multiple of its thickness times its shear modulus D _{ i } = Sum of all Ds for cell
Di_{ ( i + 1 , i )} = length of wall common with the and i cell divided by the multiplei + 1of its thickness times its shear modulus With the rate of twist equation for all cells in the structure + the torque equation, you can solve them simultaneously to determine the value of the shear flows in the structure. Example 11: Calculate shear stress distribution in the walls of the 3 cell wing section subjected to a positive torque of 12000 kNmm
_{I} = 258000 mm^{2}, A_{II} = 355000 mm^{2}, A_{III} = 161000 mm^{2}
the following way: and Solving iv, v & vi simultaneously gives: q
To determine the shear flow distribution in a multicell beam we use the same analysis as for aSHEAR LOADS single close cell beam.
to determine the shear flow due to an open beam section, then determine the cut skin sections, shear flow , add this to the open beam shear flow and determine the true shear flowq_{s,0}experienced by the section. A multicell wing section can be made statically determinate by cutting a skin panel in each cell. The best place to cut the cells is always at the centre of the top or bottom skin panels. The reason for this is that when a cell is loaded vertically, the shear flows are zero at these positions, so that when determining the cut section shear flow its value will be close to zero,minimising numerical errors. If we do this to our idealised wing section, then that panel's shear flow would be zero. If that section was symmetrical, the bottom panel will also have a zero open cell shear flow. The section will look like this:
in Chapter 6 and was given by equations (6.12), (6.13) and (6.15). where q _{b} is the cut section shear flow and q_{bi} is the open beam shear flow.
By using equation (6.13) the open section shear flow distribution can be found in the multicellwing section. However there remain number of unknown values of shear flows at each of theNcuts, ie: , which are constant for each cell, shown in Figure 88.
q_{s,0,I}, q_{s,0,II}, q_{s,0,III}, . . . . , q_{s,0,N}
^{th} cell:
Substituting equation (6.15) gives:
but because q is a constant, then this term in the integral is similar to that of equation (7.3),_{s,0,i}thus giving: where as before: where the bound integral term of equation (7.4) can be represented in summation of the open beam shear flow multiplied by D for all the skin sections surrounding cell , equation (7.5).
where:
k = Skin segment around celli
and the compatibility equation is then that as for section 7.2:
Using (7.4) and (7.6) for all cells, (N - 1) equations can be generated, requiring 1 more to make iN equations to solve for the unknown . The extra equation is found by considering theq_{s,0,i's}moment equilibrium of all cells. Look at the i ^{th} cell, the moment M_{q,i} produced by the total shear flow about some point '0' is as follows:
loads so: And if moments are taken about the point where the loads are applied then: Example 12: The following wing section carries a vertical load of 80 kN in the plane of web 34, determine the shear flow distribution and the rate of twist. Boom areas, B _{1,6} = 2580 mm ^{2}, B_{2,5} = 3880 mm^{2}, B_{3,4} = 3250 mm^{2}.
_{I} = 256000 mm^{2}, A_{II} = 560000 mm^{2}, A_{III} = 413000 mm^{2}
_{xx} = 810.99x10^{6} mm^{4}
2) Determine the open section shear flow by cutting the top skin panels, like this:
Figure 93.
The position of the shear centre is found in an identical manner to that of section 4.5. ArbitrarySHEAR CENTRE loads S _{y} and S_{x} are applied in turn through the shear centre s.c., the corresponding shear flowdistributions determined and moments taken about some convenient point. The shear flow distributions are found as described in the previous section, except that the N equations derived using equation (7.4) are enough to determine the unknown shear flows. This is because the rate of twist of a beam loaded at the shear centre is zero, (). |