Consider a prismatic member AB possessing a plane of symmetry and subjected to equal and opposite couples M and M' acting in that plane as shown:
It is observed that if a section is passed through the member AB at some arbitrary point C, the conditions of equilibrium of the portion AC of the member require that the internal forces in the section be equivalent to the couple M as shown:
Thus, the internal forces in any cross section of a symmetric member in pure bending are equivalent to a couple. The moment M of that couple is referred to as the bending moment in the section. Following the usual convention, a positive sign will be assigned to M when the member is bent as shown in the first figure, i.e., when the concavity of the beam faces upward, and a negative sign otherwise.
Denoting by σx the normal stress at a given point of the cross section and by τxy and τxz the components of the shearing stress, the system of the elementary internal forcs exerted on the section is expressed as being equivalent to the couple M.
Recall from statics that a couple M actually consists of two equal and opposite forces. The sum of the components of these forces in any direction is therefore equal to zero. Moreover, the moment of the couple is the same about any axis perpendicular to its plane, and is zero about any axis contained in that plane. Selecting arbitrarily the z-axis as shown:
The equivalence of the elementary internal forces and of the couple M is expressed by writing the sums of the components and of the moments of the elementary forces as being equal to the corresponding components and moments of the couple M:
x-components:
∫σx dA = 0
moments about y-axis:
∫zσx dA = 0
moments about z-axis:
∫(−yσx dA) = M
Three additional equations could be obtained by setting equal to zero the sums of the y-components, z-components, and moments about the x-axis, but these equations would involve only the components of the shearing stress and, as in the next lesson, the components of the shearing stress are both equal to zero.
Two remarks should be made at this point: (1) The minus sign in Eq3 is due to the fact that a tensile stress (σx > 0) leads to a negative moment (clockwise) of the normal force σxdA about the z-axis. (2) Eq2 could have been anticipated, since the application of couples in the plane of symmetry of member AB will result in a distribution of normal stresses that is symmetric about the y-axis.
Once more, note that the actual distribution of stresses in a given cross section cannot be determined from statics alone. It is statically indeterminate and may be obtained only by analyzing the deformations produced in the member.