Home > Articles

  • Print
  • + Share This
This chapter is from the book

5.3 Pure Bending of Beams of Asymmetrical Cross Section

In this section, we extend the discussion in Section 5.2 to the more general case in which a beam of arbitrary cross section is subjected to end couples My and Mz about the y and z axes, respectively (Fig. 5.3). Following a procedure similar to that described in Section 5.2, plane sections are again taken to remain plane. Assume that the normal stress σx acting at a point within dA is a linear function of the y and z coordinates of the point; assume further that the remaining stresses are zero. Then the stress field is

FIGURE 5.3

Figure 5.3. Pure bending of beams of asymmetrical cross section.

where c1, c2, c3 are constants to be evaluated.

The equilibrium conditions at the beam ends, as before, relate to the force and bending moment:

Carrying σx, as given by Eq. (5.11), into Eqs. (a), (b), and (c) results in the following expressions:

For the origin of the y and z axes to be coincident with the centroid of the section, it is required that

Based on Eq. (d), we conclude that c1 = 0; based on Eqs. (5.11), we conclude that σx = 0 at the origin. Thus, the neutral axis passes through the centroid, as in the beam of symmetrical section. In addition, the field of stress described by Eqs. (5.11) satisfies the equations of equilibrium and compatibility and the lateral surfaces are free of stress. Now consider the defining relationships

The quantities Iy and Iz are the moments of inertia about the y and z axes, respectively, and Iyz is the product of inertia about the y and z axes. From Eqs. (e) and (f), together with Eqs. (5.12), we obtain expressions for c2 and c3.

5.3.1 Stress Distribution

Substitution of the constants into Eqs. (5.11) results in the following generalized flexure formula:

The equation of the neutral axis is found by equating this expression to zero:

This is an inclined line through the centroid C. The angle ϕ between the neutral axis and the z axis is determined as follows:

The angle ϕ (measured from the z axis) is positive in the clockwise direction, as shown in Fig. 5.3. The highest bending stress occurs at a point located farthest from the neutral axis.

There is a specific orientation of the y and z axes for which the product of inertia Iyz vanishes. Labeling the axes so oriented as y′ and z′, we have Iy′z′ = 0. The flexure formula under these circumstances becomes

The y′ and z′ axes now coincide with the principal axes of inertia of the cross section, and we can find the stresses at any point by applying Eq. (5.13) or Eq. (5.16).

The kinematic relationships discussed in Section 5.2 are valid for beams of asymmetrical section provided that y and z represent the principal axes.

5.3.2 Transformation of Inertia Moments

Recall that the two-dimensional stress (or strain) and the moment of inertia of an area are second-order tensors (Section 1.17). Thus, the transformation equations for stress and moment of inertia are analogous (Section C.2.2). In turn, the Mohr′s circle analysis and all conclusions drawn for stress apply to the moment of inertia. With reference to the coordinate axes shown in Fig. 5.3, applying Eq. (C.12a), the moment of inertia about the y′ axis is

From Eq. (C.13), the orientation of the principal axes is given by

The principal moments of inertia, I1 and I2 from Eq. (C.14) are

where the subscripts 1 and 2 refer to the maximum and minimum values, respectively.

Determination of the moments of inertia and stresses in an asymmetrical section is illustrated in Example 5.1.

  • + Share This
  • 🔖 Save To Your Account

Related Resources

There are currently no related titles. Please check back later.