1. 6.5 UNSYMMETRIC BENDING
When developing the flexure formula we imposed a condition that the cross-sectional area be symmetric about an axis perpendicular to the neutral axis; furthermore, the resultant internal moment M acts along the neutral axis. Such is the case for the T or channel sections shown in the figures below.
Moment Arbitrarily Applied.
Sometimes a member may be loaded such that the resultant internal moment does not act about one of the principal axes of the cross section. When this occurs, the moment should first be resolved into components directed along the principal axes. The flexure formula can then be used to determine the normal stress cause by each moment component. Finally, using the principle of superposition, the resultant normal stress at the point can be determined.
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2. 6.5 UNSYMMETRIC BENDING
Applying the flexure formula to each moment component in figure (b) and figure(c), we can express the resultant normal stress at any point on the cross section, figure(d), in general form as
................(1)
Where
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3. 6.5 UNSYMMETRIC BENDING
Orientation of the Neutral Axis. The angle α of the neutral axis in figure (d) can be determined by applying equation (1) with 휎=0, since by definition no normal stress acts on the neutral axis, we have
This is the equation of the line that defines neutral axis for the cross section. Since the slope of this line is tanα=y/z, then
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4. 4
Example 6.5 (Hibbeler)
The rectangular cross section shown in figure is subjected to a bending moment of M = 12 kN.m. Determine the normal stress developed at each corner of the section, and specify the orientation of the neutral axis.
5. 5
Example 6.6 (Hibbeler)
A T-beam is subjected to the bending moment of 15 kN.m. as shown in figure. Determine the maximum normal stress in the beam and the orientation of the neutral axis.
6. 6
Example 6.7(Hibbeler)
The cantilevered wide-flange steel beam is subjected to the concentrated force of at its end. Determine the maximum bending stress developed in the beam at section A.
