When determining the bending stresses, there is no difference if the section is multicell or not
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'.

Figure 80: Multicell wing structure with applied torque T

The torque T, generates individual torques in each cell, the sum of which must equal T, so:

But a torque in a closed cell (beam section) produces a constant shear flow, given by equation
(5.1), which is:

Substituting this gives:

However since we have 'N' cells this equation is statically indeterminate, so we need a
compatibility relationship.

Effect of Booms on Shear Flow Distribution

Look at an elemental section of wing of length z, where three skin sections are in contact with
a boom of area B, Figure 81.

Figure 81: Equilibrium of boom/skin/web junction

At this boom/skin/web junction, the shear flows must all be in equilibrium. The element dz of
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
shear flows out of the junction. This is similar to fluid flow in a multi pipe system. Which is why
we use the term shear flow.

Compatibility Equation for Each Cell

Due to the ribs in aircraft wings, the rate of twist is constant for all cells in the section. So look
at the rate of twist of cell i in the wing section of Figure 80 . Equation (4.21) then becomes for
cell i equal to:

In order to derive the rate of twist for cell i, it is necessary to look at the three cells: i - 1, i and
i + 1, and determine the bound integral at cell i of equation (4.21).

Figure 82: Shear flow distribution in the ith cell of an 'N' cell wing beam section

Let :

It is now necessary to determine the rate of twist for cell i giving:

Rearranging this gives:

and in general terms for the configuration of Figure 82, the rate of twist equation for cell i


      D ( i - 1 , i )  =      length of wall common with the cell i and i - 1 cell divided by the
                              multiple of its thickness times its shear modulus

      D i             =     Sum of all Ds for cell i

      D ( i + 1 , i ) =     length of wall common with the i and i + 1 cell divided by the multiple
                              of 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

Figure 83: Three cell wing section loaded by a pure torque

Where the cell areas are : AI = 258000 mm2, AII = 355000 mm2, AIII = 161000 mm2

Figure 84: Table where D is calculated

Using equation 7.3, determine the rate of twist of all cells:

For cell I

For cell II

For cell III

And from equation 7.1:

These four equations can now be solved simultaneously to give the shear flows. This is done in
the following way:


Solving iv, v & vi simultaneously gives:

qI = 8.6923 N/mm, qII = 8.4515 N/mm, qIII = 4.7021 N/mm

Figure 85: Shear flow distribution in three cell wing section with applied torque

Note:     By dividing the above values of shear flow by the respective wall thickness, gives the
              Shear Stresses in the walls in MPa.


To determine the shear flow distribution in a multicell beam we use the same analysis as for a
single close cell beam.

Figure 86: Multicell wing section with applied shear loads

Because a single closed beam is a statically indeterminate structure, we need to cut it in order
to determine the shear flow due to an open beam section, then determine the cut skin sections
shear flow
, qs,0, add this to the open beam shear flow and determine the true shear flow
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:

Figure 87: Multicell wing structure loaded by shear loads, indicating the best location for cutting the cell.

The shear flow equation for an idealised beam close section subjected to shear loads was found
in Chapter 6 and was given by equations (6.12), (6.13) and (6.15).

where qb is the cut section shear flow and qbi is the open beam shear flow.

By using equation (6.13) the open section shear flow distribution can be found in the multicell
wing section. However there remain N number of unknown values of shear flows at each of the
cuts, ie: qs,0,I, qs,0,II, qs,0,III, . . . . , qs,0,N, which are constant for each cell, shown in Figure 88.

Figure 88: Redundant shear flow in the ith cell of N cell wing structure loaded in shear

From equation (4.21), the rate of twist for ith cell:

Substituting equation (6.15) gives:

but because qs,0,i is a constant, then this term in the integral is similar to that of equation (7.3),
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 i, equation (7.5).


         k = Skin segment around cell i

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
N equations to solve for the unknown qs,0,i's. The extra equation is found by considering the
moment equilibrium of all cells.

Look at the ith cell, the moment Mq,i produced by the total shear flow about some point '0' is as

Figure 89: Moment equilibrium of ith cell

The moment about point 0 due to cell i is given by:

substituting equation (6.15) gives:

which simplifies to:

The sum of the moment of the individual cells is equal to the moment produced by the applied
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, B1,6
                       = 2580 mm2, B2,5 = 3880 mm2, B3,4 = 3250 mm2.

Figure 90: Idealised wing section with applied vertical shear load

Where the cell areas are AI = 256000 mm2, AII = 560000 mm2, AIII = 413000 mm2

Figure 91: Table with calculated D

1)      Determine sectional properties:

Because the section is symmetrical we only need Ixx = 810.99x106 mm4

2)      Determine the open section shear flow by cutting the top skin panels, like this:

Figure 92: Three cell wing section cut to determine open beam shear flows

Using equations (6.12) and (6.13) the open beam shear flows were calculated in the table of
Figure 93.

Figure 93: Excel table where the open beam shear flows were calculated

The shear flow distribution in the open beam section looks like this:

Figure 94: Idealised wing section showing open beam shear flows

3)      Determine rate of twist for each cell using equation (7.4):

For Cell I

For Cell II

For Cell III

Because all rates of twist are the same we can now equate them together:

4)     Equate moments about the point of application of load gives:

which simplifies to:

Solving these 3 equations simultaneously gives:

Superimposing these shear flows with the open beam values gives:

Figure 95: Wing section with final shear flows

5)     Determine rate of twist:

By substituting these values into any of the rate of twist equations we can find it to be :


The position of the shear centre is found in an identical manner to that of section 4.5. Arbitrary
loads Sy and Sx are applied in turn through the shear centre s.c., the corresponding shear flow
distributions 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, ().

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