Mechanical Engineering Principles

8

Bending of beams

At the end of this chapter you should be

able to:

• define neutral layer


• define the neutral axis of a beam’s cross-
  section


• prove that

σ

y

=

M

I

=

E

R

• calculate the stresses in a beam due to
  bending


• calculate the radius of curvature of the
  neutral layer due to a pure bending
  momentM

8.1 Introduction

If a beam of symmetrical cross-section is subjected
to a bending momentM, then stresses due to bend-
ing action will occur. This can be illustrated by
the horizontal beam of Figure 8.1, which is of uni-
form cross-section. In pure or simple bending, the
beam will bend into an arc of a circle as shown in
Figure 8.2.

dx

MM

Figure 8.1

Now in Figure 8.2, it can be seen that due to
these couples M, the upper layers of the beam
will be in tension, because their lengths have been
increased, and the lower layers of the beam will
be in compression, because their lengths have been
decreased. Somewhere in between these two layers
lies a layer whose length has not changed, so that
its stress due to bending is zero. This layer is

(a) Beam (b) Cross-section

q

Neu

trallayer

M CA BD M

R

y

NA

Figure 8.2

called theneutral layerand its intersection with

the beam’s cross-section is called theneutral axis

(NA). Later on in this chapter it will be shown that

the neutral axis is also the centroidal axis described

in Chapter 7.

8.2 To prove that

σ

y

=

M

I

=

E

R

In the formula

σ

y

=

M

I

=

E

R

,

σ=the stress due to bending momentM,

occurring at a distanceyfrom the

neutral axisNA,

I=the second moment of area of the

beam’s cross-section aboutNA,

E =Young’s modulus of elasticity of the

beam’s material,

andR =radius of curvature of the neutral layer

of the beam due to the bending

momentM.

Now the original length of the beam element,

dx=Rθ ( 8. 1 )

At any distanceyfromNA, the lengthABincreases

its length to:

CD=(R+y)θ ( 8. 2 )