Page 24 - Handbook of Properties of Textile and Technical Fibres

Introduction to the science of fibers 5

0
From the above two equations we can see that the imposed strain in D D is then the
increase in length divided by the original, unstrained, length of the beam.
Induced strain is

0
0
D D C C rq þ Sq rq Sq
0
D D ¼ ¼ ¼ (1.1)
C C rq r
0
0
The line D D experiences a tensile stress. If we accept that the section of the beam
denoted by the line D D has a very small thickness, the stress is given by the force, dF,
0
on this elementary part of the beam divided by its cross-section dA. For an elastic
material, Hooke’s law relates stress, s, strain, ε, and stiffness, E. The latter is called
the Young’s modulus and we can write s ¼ E$ε.
So that
dF S
¼ E$ (1.2)
dA r
As D D is a distance S from the neutral line C C, the force dF produces a turning
0
0
moment dF$S in the beam so that from Eq. (1.2) we obtain
S 2
dF$S ¼ E$ dA
r
This means that the total bending moment M S is given by

S E 2 E
Z 2 Z
M S ¼ E$ dA ¼ S dA h I A (1.3)
r r r
where I A is known as the second moment of inertia.
It should be noted that this is to do with bending and nothing to do with movement,
as in inertia defined by Newton’s first law.
If we consider that our fiber is circular in cross-section we can work out the second
moment of inertia for a circular beam. Fig. 1.2 shows the cross-section of the circular
beam.
We must write a relationship for the cross-section of the elementary section at a dis-
tance S from the neutral access, which runs through the centre of the fiber. We see,
from Fig. 1.2, that, in polar coordinates, dA can be written as r$dr$da and also that
S ¼ r sin a.
From Eq. (1.3) we can now write

Z 2p Z R
2
2
I A ¼ r sin aðr$dr$daÞ:
0 0
R 2p 2 R R 3
So that I A ¼ sin a$da r dr
0 0