1.1 Introduction
When an external load acts on a beam, shear forces and bending moments are set up at all sections of the
beam. Due to the shear forces and bending moments, the beam undergoes certain deformations. The
material of the beam will offer resistance or stresses against these deformations. These stresses, with
certain assumptions, can be calculated. The stresses introduced by bending moment are known as
bending stresses.
1.2 Pure Bending/Simple Bending
If a length of beam is subjected to a constant bending moment and no shear force (i.e. 0 shear force), then
the stresses will be set up in that length of beam due to bending moments only and that length of beam
is said to be in pure bending or simple bending. The stresses set up in that length of beam are known as
bending stresses.
Figure 7.1
In Figure 7.1 (a), a beam is simply supported at A and B and overhanging by the same length at each
support. A point load P, is applied at each end of the overhanging portion. The shear force diagram and
bending moment diagram for the beam are drawn in Figure 7.1 (b) and (c) respectively. From these
diagrams, it is clear that there is no shear force between A and B but the bending moment between A and
B is constant. This implies that between A and B. the bean is subjected to a constant bending moment
only. This condition for the beam between A and B is known as pure bending or simple bending.
, 1.3 Assumptions made in the Theory of Simple Bending
The following assumptions are made in the theory of simple bending:
(i) The material of the beam is homogeneous and isotropic. Homogeneous meaning that it is of the
same kind of material throughout and isotropic meaning that it has equal elastic properties in all
directions.
(ii) The value of Young’s Modulus of elasticity, E, is the same in tension and compression.
(iii) The transverse sections which were plane before bending remain plane after bending.
(iv) The beam is initially straight and all longitudinal filaments bend into circular arcs with a common
centre of curvature.
(v) The radius of curvature is very large compared to the dimensions of the cross-section.
(vi) Each layer of the beam is free to expand or contract, independently of the layer above or below
it.
Figure 7.2
Figure 7.2 (a) shows a part of a beam subjected to simple bending. Consider a small length 𝛿𝑥 of this part
of beam. Consider two sections AB and CD which are normal to the axis of the beam N-N. Due to the
action of bending moment, the part of length 𝛿𝑥 will be deformed as shown in Figure 7.2 (b). From this
figure, it is clear that all layers of the beam, which were originally of the same length, do not remain of
the same length any more. The top layer AC has deformed into A’C’. This layer has been shortened in its
length. The bottom layer BD has been deformed to B’D’. This layer has been elongated.
From Figure 7.2 (b), it is clear that some of the layers have been shortened and others elongated. At a
level between the top and bottom of the beam, there will be a layer which is neither shortened nor
elongated. This layer is known as neutral layer or neutral surface. This layer in Figure 7.2 (a) is shown by
N-N and in Figure 7.2 (b) by N’-N’. The line of intersection of the neutral layer on a cross-section of a beam
is known as neutral axis written as N.A.
The layers above N-N (or N’-N’) have been shortened and those below have been elongated. Due to
decrease in lengths of the layers above N-N, these layers will be subjected to compressive stresses. Due
to increase in the lengths of layers below N-N, these layers will be subjected to tensile stresses. Also, note
that the top layer has been shortened maximum. As we proceed towards the layer N-N, the decrease in
length of the layers decreases. At layer N-N, there is no change in length. This means that the compressive
stresses will be maximum at the top layer.
Similarly, the increases in length will be maximum in the bottom layer. As we proceed from the bottom
layer towards N-N, the increase in length of the layers decreases. Hence, the amount by which a layer