Deformations in a Symmetric Member in Pure Bending
Details
In this lesson the deformations of a prismatic member possessing a plane of symmetry and subjected at its ends to equal and opposite couples M and M' acting in the plane of symmetry will be analyzed. The member will bend under the action of the couples, but will remain symmetric with respect to that plane as shown:
Moreover, since the bending moment M is the same in any cross section, the member will bend uniformly. Thus, the line AB along which the upper face of the member intersects the plane of the couples will have a constant curvature. In other words, the line AB, which was originally a straight line, will be transformed into a circle of center C, and so will the line A'B' (not shown in the figure) along which the lower face of the member intersects the plane of symmetry. Also note that the line AB will decrease in length when the member is bent as shown above, i.e., when M > 0, while A'B' will become longer.
Next it will be proved that any cross section perpendicular to the axis of the member remains plane, and that the plane of the section passes through C. If this were not the case, a point E of the original section through D as shown:
could be found which, after the member has been bent, would not lie in the plane perpendicular to the plane of symmetry that contains line CD as shown above. But, because of the symmetry of the member, there would be another point E' that would be transformed exactly in the same way. Assume that, after the beam has been bent, both points would be located to the left of the plane defined by CD, as shown above. Since the bending moment M is the same throughout the member, a similar situation would prevail in any other cross section, and the points corresponding to E and E' would also move to the left. Thus, an observer at A would conclude that the loading causes the points E and E' in the various cross sections to move forward (toward the observer). But an observer at B, to whom the loading looks the same, and who observes the points E and E' in the same positions (except that they are now inverted) would reach the opposite conclusion. This inconsistency leads to the conclusion that E and E' will lie in the plane defined by CD and, therefore, that the section remains plane and passes through C. It should be noted, however, that thislesson does not rule out the possibility of deformations within the plane of the section, as in the lesson Deformations in a Transverse Cross Section.
Suppose that the member is divided into a large number of small cubic elements with faces respectively parallel to the three coordinate planes. The property that has been established requires that these elements be transformed as shown in Fig4 when the member is subjected to the couples M and M'. Since all the faces represented in the two projections of Fig4 are at 90° to each other, it is concluded that γxy = γzx = 0 and, thus, that τxy = τxz = 0. Regarding the three stress components that have not yet been discussed, namely, σy, σz, and τyz, it is noted that they must be zero on the surface of the member. Since, on the other hand, the deformations involved do not require any interaction between the elements of a given transverse cross section, it can be assumed that these three stress components are equal to zero throughout the member. This assumption is verified, both from experimental evidence and from the theory of elasticity, for slender members undergoing small deformations. It is concluded that the only nonzero stress component exerted on any of the small cubic elements considered here is the normal component σx. Thus, at any point of a slender member in pure bending, there is a state of uniaxial stress. Recalling that, for M > 0, lines AB and A'B' are observed, respectively, to decrease and increase in length, it is noted that the strain εx and the stress σx are negative in the upper portion of the member (compression) and positive in the lower portion (tension).
It follows from the above that there must exist a surface parallel to the upper and lower faces of the member, where εx and σx are zero. This surface is called the neutral surface. The neutral surface intersects the plane of symmetry along an arc of circle DE as shown in Fig6, and it intersects a transverse section along a straight line called the neutral axis of the section as shown in Fig7. The origin of coordinates will now be selected on the neutral surface, rather than on the lower face of the member as done earlier, so that the distance from any point to the neutral surface will be measured by its coordinate y.
Denoting by ρ the radius of arc DE as in Fig6, by θ the central angle corresponding to DE, and observing that the length of DE is equal to the length L of the undeformed member:
(Eq1)
L = ρθ
Considering now the arc JK located at a distance y above the neutral surface, its length L' is:
(Eq2)
L' = (ρ − y)θ
Since the original length of arc JK was equal to L, the deformation of JK is:
(Eq3)
δ = L' − L
or, if Eq1 and Eq2 are substituted into Eq3:
(Eq4)
δ = (ρ − y)θ − ρθ
The longitudinal strain εx in the elements of JK is obtained by dividing δ by the original length L of JK:
(Eq5)
εx =
δ
L
=
−yθ
ρθ
or:
(Eq6)
εx = −
y
ρ
The minus sign is due to the fact that it has been assumed the bending moment is positive and, thus, the beam is concave upward.
Because of the requirement that transverse sections remain plane, identical deformations will occur in all planes parallel to the plane of symmetry. Thus the value of the strain given by Eq6 is valid anywhere, and it is concluded that the longitudinal normal strain εx varies linearly with the distance y from the neutral surface.
The strain εx reaches its maximum absolute value when y itself is largest. Denoting by c the largest distance from the neutral surface (which corresponds to either the upper or the lower surface of the member), and by εm the maximum absolute value of the strain:
(Eq7)
εm =
c
ρ
Solving Eq7 for ρ and substituting the value obtained into Eq6:
(Eq8)
εx = −
y
c
εm
This analysis of the deformations of a member in pure bending is concluded by observing that the strain or stress at a given point of the member still can not be computed, since the neutral surface of the member has not yet been located. In order to locate this surface, the stress-strain relation of the material to be used must first be specified. Note, however, that if the member possesses both a vertical and a horizontal plane of symmetry (e.g., a member with a rectangular cross section), and if the stress-strain curve is the same in tension and compression, the neutral surface will coincide with the plane of symmetry.