BENDING OF BEAMS WITH NON-SYMMETRICAL CROSS SECTIONS

The majority of aircraft structural components consist of beams with non-symmetrical cross section acting in bending . For this reason an expression needs to be derived to allow for the determination of the stresses induced by bending moments to such sections.

Look at the Oxyz system of axis, with an arbitrary beam parallel to the z-axis:

Where :

T = Torque

M = Bending Moment

S = Shear Force

w = Distributed load

P = Axial or direct load

u,v,w = Axial displacements

All of these externally applied loads are positive in the direction indicated in the figure. Internal moment and forces applied to face A are in the same direction and sense as externally applied loads. However on face B, the positive internal moments and forces are in the opposite sense.

A bending moment M applied in any plane parallel to the z-axis can be resolved into M_x and M_y components by normal vector rules.

By doing it in a visual way it will be easier to see:

From Figure 9, the following relationships can be obtained:

M_x = M sin θ

M_y = M cos θ

and that these moments can have different sign depending on the value of θ. For example if θ > π/2, M_x is positive and M_y is negative.

Consider a beam of arbitrary non-symmetrical cross section, which supports bending moments Mx and My, bending about some axis in the cross section. This is the plane of no bending stress, called Neutral Axis (N.A.).

Let the axis origin coincide with the centroid G of the cross section, and that the neutral axis is a distance p from G.

The direct stress s_z on element d_A at point (x,y) and distance x from the neutral axis is:

Look at the beam in a plane parallel to the neutral axis with two segments ij and kl which are of equal length when the beam is undeflected:

Once the beam has been deflected this section will look like this:

where:

        R   =  the radius of curvature

        dq  =  angle between planes ik and jl

The strain in plane kl can be defined as:

with

and

giving

By substituting back into the stress equation it gives:

(3.2)

Now that a stress equation has been obtained, it is necessary to satisfy both rotational and linear equilibrium at the ends of the beam. That is, the summation of the forces through the depth of the beam is equal to '0', and the summation of the moments through the depth of the beam is equal the applied moments Mx and My.

As the beam supports pure bending, the resultant load on the end section must be zero. Hence

Substituting equation 3.2 gives:

This equation defines the location of the centroid of the section, it follows that the neutral axis must pass through the centroid.

Rather than use this equation to find the location of the centroid, it is much easier to locate the centroid about the xy-axis by using equation 3.3.

In order to see this in more detail, Figure 14 shown a close up of the axis and the area dA.

If the inclination of the Neutral Axis (N.A.) is at an angle from the x-axis then :

x = x sin α + y cos α (3.4)

and substituting into 3.2 gives:

(3.5)

The moment resultants have the same sense as the applied moments, hence:

, (3.6)

and substituting equation 3.4 into equations 3.5 gives:

and

Both the sin α and cos α terms are not a function of dA, so they can be removed from the integration. What remain are terms which only have to do with the characteristics of the cross sectional shapes of the beam, and these are just the 2^nd moments of area of the beam.

The second moments of area about the xy axes are:

which gives:

solving simultaneously gives,

Substituting into Equation (3.5) gives:

By defining the terms Effective Bending Moment, as:

and

Equation 3.8 can be re-written as follows:

and the x & y axes are the principal axes.

The location of the Neutral Axis was defined by Equation (3.3).

For a beam with a symmetrical cross section, the centroid is the point defined by equation 3.3 and the Neutral Axis is parallel to the x and y axis. For a non-symmetrical beam cross section however, the Neutral Axis passes at some angle a with respect to the x-axis. What needs to be done define the angle of the neutral axis.

At the N.A. the normal bending stresses are equal to s_z = 0, giving that:

where:

        x_N.A. and y_N.A. are coordinates of points along the neutral axis, giving:

By taking the inverse tan of the angle 'a', the angle of the Neutral Axis with respect to the x-axis can be found, given by equation 3.12.

and

b)   Shift axis to centroid

c)   Determine other sectional properties

d)   Calculate effective bending moments

Since M_x = 1500 and M_y = 0, substituting these values into Equations (3.9) and (3.10) gives:

e)   Determine bending stress equation and cross section stresses

Substitute the values calculated above into Equation (3.11).

Substituting the coordinates of all the corner points from Figure 16 give that:

Due to the thin nature of aircraft structures, the assumption can be made that stresses are constant through the thickness 't' of the skin. By saying this, the square and higher powers of 't' can be neglected from the computation of sectional properties.

To see this, look at a channel section:

This indicates that the sectional properties may be calculated as if the section was represented by a thin line, as shown in Figure 18, disregarding any t² or higher terms.

For this discrete thin walled section, the sectional properties would be found as follows:

1)    Determine location of centroid

2)    Shift axis to centroid

3)    Determining moments of inertia

These results are exactly the same as for the section considering the skin material thickness and then disregarding all t² and higher terms.

Since not all skin sections will lie parallel to either the x or y axis, the local moments or area for a section of skin at an angle q with respect to the x-axis are given by the following equations.

Consider an element of length dz from a beam with an unsymmetrical cross section with all types of loads applied in the y-z plane, Figure 21.

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