BS02-09

Section 2-09 Normal Stresses in Bending

In deriving the relations for the normal stresses in beams we make the following idealizations:

1. The beam is subjected to pure bending; this means that 
   the shear force is zero, and that no torsion or axial 
   loads are present.
2. The material is isotropic and homogeneous.
3. The material obeys Hooke's law.
4. The beam is initially straight with a cross section
   that is constant throughout the beam length.
5. The beam has an axis of symmetry in the plane of 
   bending.
6. The proportions of the beam are such that it would
   fail by bending rather than by crushing, wrinkling, or
   sideways buckling.
7. Cross sections of the beam remain plane during bending.

In Figure 2-12a we visualize a portion of a beam acted upon by the positive bending moment M. The y axis is the axis of symmetry. The x axis is coincident with the neutral axis of the section, the xz plane, which contains the neutral axis of all sections, is called the neutral plane. Elements of the beam coincident with this plane have zero strain. The location of the neutral axis with respect to the cross section has not yet been defined.

Application of the positive moment will cause the upper surface of the beam to bend downward, and the neutral axis will then e curved, as shown in Figure 2-12b. Because of the curvature, a section AB originally parallel to CD, since the beam was straight, will rotate throuh the angle d phi to A'B'. Since AB and A'B' are both straight lines, we have utilized the assumption that plane sections remain plane during bending. If we now specify the radius of curvature of the neutral axis as rho, the length of a differential element of the neutral axis as ds, and the angle subtended by the two adjacent sides CD and A'B' as d phi, then, from the definition of curvature, we have


     1 / rho = d phi / ds             [a]

As shown in Figure 2-12b, the deformation of a "fiber" at distance y from the neutral axis is


     dx = y * d phi                   [b]

The strain is the deformation divided by the original length, or


     epsilon = - dx / ds              [c]

where the negative sign indicates compression. Solving equations [a], [b], and [c] simultaneously gives


     epsilon = - y / rho              [d]

Thus the strain is proportional to the distance y from the nuetral axis. Now, since sigma = E * epsilon, we have for the stress


     sigma = - E * y / rho            [e]

We are now dealing with pure bending, which means that there are no axial forces acting on the beam. We can state this in mathematical form by summing all the horizontal forces acting on the cross section and equating this sum to zero. The force acting on an element of area dA is sigma * dA; therefore


     INTEGRAL(sigma * dA) = - E / rho * INTEGRAL(y * dA) 
          = 0                                [f]

Equation [f] defines the location of the neutral axis. The moment of the area about the neutral axis is zero, and hence the neutral axis passes through the centroid of the cross-sectional area.

Next we observe that equilibrium requires that the internal bending moment created by the stress sigma must be the same as the external moment M. In other words,


     M = INTEGRAL(y * sigma * dA) =
             E/rho * INTEGRAL(y^2 * dA)      [g]

The second integral in equation [g] is the moment of inertia of the area about the z axis. That is,


     I = INTEGRAL(y^2 * dA)                 [2-28]

This is called the area moment of inertia; since area cannot truly have inertia, it should not be confused with the mass moment of inertia.

If we next solve equations [g] and [2-28] and rearrange them, we have


     1 / rho = M / (E*I)                   [2-29]

This is an important equation in the determination of the deflection of beams, and we shall employ it in Chapter 3. Finally, we eliminate rho from equations [e] and [2-29] and obtain


     sigma = - M * y / I                   [2-30]

Equation [2-30] states that the bending stress sigma is directly proportional to the distance y from the neutral axis and the bending moment M, as shown in Figure 2-13. It is customary to designate c = y_max, to omit the negative sign, and to write


     sigma = M * c / I                     [2-31]

where it is understood that equation [2-31] gives the maximum stress. Then tensile or compressive maximum stresses are determined by inspection when the sense of the moment is known.

Equation [2-31] is often written in the two alternative forms


     sigma = M / (I/c)
     sigma = M / Z                         [2-32]

where Z = I / c is called the section modulus.

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Mechanical Engineering Design Section 2-09 Normal Stresses in Bending