Behaviour
of Materials
1. Introduction
When
a force is applied on a body it suffers a change in shape, that is, it deforms.
A force to resist the deformation is also set up simultaneously within the body
and it increases as the deformation continues. The process of deformation stops
when the internal resisting force equals the externally applied force. If the
body is unable to put up full resistance to external action, the process of
deformation continues until failure takes place. The deformation of a body
under external action and accompanying resistance to deform are referred to by
the terms strain and stress respectively.
2. Stresses
Stress
is defined as the internal resistance set up by a body when it is deformed. It is measured in N/m2
and this unit is specifically called Pascal (Pa). A bigger unit of stress is
the mega Pascal (MPa).
1 Pa = 1N/m2,
1MPa = 106 N/m2 =1N/mm2.
2.1. Three Basic Types of Stresses
Basically
three different types of stresses can be identified. These are related to the
nature of the deforming force applied on the body. That is, whether they are
tensile, compressive or shearing.
2.1.1. Tensile Stress
Consider
a uniform bar of cross sectional area A subjected to an axial tensile force P.
The stress at any section x-x normal to the line of action of the tensile force
P is specifically called tensile stress pt . Since internal
resistance R at x-x is equal to the applied force P, we have,
pt
= (internal resistance at
x-x)/(resisting area at x-x)
=R/A
=P/A.
Under
tensile stress the bar suffers stretching or elongation.
2.1.2. Compressive Stress
If
the bar is subjected to axial compression instead of axial tension, the stress
developed at x-x is specifically called compressive stress pc.
pc
=R/A
= P/A.
Under
compressive stress the bar suffers shortening.
2.1.3. Shear Stress
Consider
the section x-x of the rivet forming joint between two plates subjected to a
tensile force P as shown in figure.
The
stresses set up at the section x-x acts along the surface of the section, that
is, along a direction tangential to the section. It is specifically called
shear or tangential stress at the section and is denoted by q.
q =R/A
=P/A.
2.2. Normal or Direct Stresses
When
the stress acts at a section or normal to the plane of the section, it is
called a normal stress or a direct stress. It is a term used to mean both the
tensile stress and the compressive stress.
2.3. Simple and Pure Stresses
The
three basic types of stresses are tensile, compressive and shear stresses. The
stress developed in a body is said to be simple tension, simple compression and
simple shear when the stress induced in the body is (a) single and (b) uniform.
If the condition (a) alone is satisfied, the stress is called pure tension or
pure compression or pure shear, as the case may be.
2.4. Volumetric Stress
Three
mutually perpendicular like direct stresses of same intensity produced in a
body constitute a volumetric stress. For example consider a body in the shape
of a cube subjected equal normal pushes on all its six faces. It is now
subjected to equal compressive stresses p in all the three mutually
perpendicular directions. The body is now said to be subjected to a volumetric
compressive stress p.
Volumetric
stress produces a change in volume of the body without producing any distortion
to the shape of the body.
3. Strains
Strain
is defined a the ratio of change in dimension to original dimension of a body
when it is deformed. It is a dimensionless quantity as it is a ratio between
two quantities of same dimension.
3.1. Linear Strain
Linear
strain of a deformed body is defined as the ratio of the change in length of
the body due to the deformation to its original length in the direction of the
force. If l is the original length and δl the change in length occurred due to
the deformation, the linear strain e induced is given by
e=δl/l.
Linear
strain may be a tensile strain , et or a compressive strain ec according
as δl refers to an increase in length or a decrease in length of the body. If
we consider one of these as +ve then the other should be considered as –ve, as
these are opposite in nature.
3.2.Lateral Strain
Lateral
strain of a deformed body is defined as the ratio of the change in length
(breadth of a rectangular bar or diameter of a circular bar) of the body due to
the deformation to its original length (breadth of a rectangular bar or
diameter of a circular bar) in the direction perpendicular to the force.
3.3. Volumetric Strain
Volumetric
strain of a deformed body is defined as the ratio of the change in volume of the
body to the deformation to its original volume. If V is the original volume and
δV the change in volume occurred due to the deformation, the volumetric strain
ev induced is given by ev
=δV/V
Consider
a uniform rectangular bar of length l, breadth b and depth d as shown in
figure. Its volume V is given by,
V =
lbd
δV = δl bd + δb ld + δd lb
δV /V = (δl / l) + (δb / b) + (δd / d)
ev = ex +ey +ez
This
means that volumetric strain of a deformed body is the sum of the linear
strains in three mutually perpendicular directions.
3.4. Shear Strain
Shear
strain is defined as the strain accompanying a shearing action. It is the angle
in radian measure through which the body gets distorted when subjected to an
external shearing action. It is denoted by Ф.
Consider
a cube ABCD subjected to equal and opposite forces Q across the top and bottom
forces AB and CD. If the bottom face is taken fixed, the cube gets distorted
through angle f to the shape ABC’D’. Now strain or deformation per
unit length is
Shear
strain of cube = CC’ / CD = CC’ / BC = f
radian
4. Relationship between Stress and Strain
Relationship
between Stress and Strain are derived on the basis of the elastic behaviour of
material bodies.
A standard mild steel specimen is
subjected to a gradually increasing pull by Universal Testing Machine. The
stress-strain curve obtained is as shown below.
A -Elastic Limit
B - Upper Yield Stress
C - Lower Yield Stress
D -Ultimate Stress
E -Breaking Stress
4.1. Elasticity and Elastic Limit
Elasticity
of a body is the property of the body by virtue of which the body regains its
original size and shape when the deformation force is removed. Most materials
are elastic in nature to a lesser or greater extend, even though perfectly elastic
materials are very rare.
The
maximum stress upto which a material can exhibit the property of elasticity is
called the elastic limit. If the deformation forces applied causes the stress
in the material to exceed the elastic limit, there will be a permanent set in
it. That is the body will not regain its original shape and size even after the
removal of the deforming force completely. There will be some residual strain
left in it.
Yield stress
When
a specimen is loaded beyond the elastic limit the stress increases and reach a
point at which the material starts yielding this stress is called yield stress.
Ultimate stress
Ultimate
load is defined as maximum load which can be placed prior to the breaking of
the specimen. Stress corresponding to the ultimate load is known as ultimate
stress.
Working stress
Working
stress= Yield stress/Factor of safety.
4.2. Hooke’s Law
Hooke’s
law states that stress is proportional to strain upto elastic limit. If I is
the stress induced in a material and e the corresponding strain, then according
to Hooke’s law,
p
/ e = E, a constant.
This
constant E is called the modulus of elasticity or Young’s Modulus, (named after
the English scientist Thomas Young).
It
has later been established that Hooke’s law is valid only upto a stress called
the limit of proportionality which is slightly less than the elastic limit.
4.3. Elastic Constants
Elastic
constants are used to express the relationship between stresses and strains.
Hooke’s law , is stress/strain = a constant, within a certain limit. This means
that any stress/corresponding strain = a constant, within certain limit. It follows
that there can be three different types of such constants. (which we may call
the elastic constants or elastic modulae) corresponding to three distinct types
of stresses and strains. These are given below.
(i)Modulus of Elasticity or Young’s Modulus (E)
Modulus
of Elasticity is the ratio of direct stress to corresponding linear strain
within elastic limit. If p is any direct stress below the elastic limit and e
the corresponding linear strain, then E
= p / e.
(ii)Modulus of Rigidity or Shear Modulus (G)
Modulus of
Rigidity is the ratio of shear stress to shear strain within elastic limit. It
is denoted by N,C or G. if q is the shear stress within elastic limit and f
the corresponding shear strain, then G =
q / f.
(iii) Bulk Modulus (K)
Bulk
Modulus is the ratio of volumetric stress to volumetric strain within the
elastic limit. If pv is the volumetric stress within elastic limit
and ev the corresponding volumetric strain, we have K = pv /
ev.
5. Poisson’s Ratio
Any
direct stress is accompanied by a strain in its own direction and called linear
strain and an opposite kind strain in every direction at right angles to it,
lateral strain. This lateral strain bears a constant ratio with the linear
strain. This ratio is called the Poisson’s ratio and is denoted by (1/m) or µ.
Poisson’s
Ratio = Lateral Strain / Linear Strain.
Value
of the Poisson’s ratio for most materials lies between 0.25 and 0.33.
6. Complementary Strain
Consider
a rectangular element ABCD of a body subjected to simple shear of intensity q
as shown. Let t be the thickness of the element.
Total
force on face AB is , FAB = stress X area = q X AB X t.
Total
force on face CD is, FCD = q X CD Xt = q XAB Xt.
FAB
and FCD being equal and opposite, constitute a couple whose moment is given by,
M =FAB X BC = q XAB X
BC X t
Since
the element is in equilibrium within the body, there must be a balancing couple
which can be formed only by another shear stress of some intensity q’ on the
faces BC and DA. This shear stress is called the complementary stress.
FBC = q’ X BC X t
FDA = q’ X DA X t = q’
X BC Xt
The
couple formed by these two forces is M’ = FBC X AB = q’ X BC X t
For
equilibrium, M’ = M.
Therefore
q’ = q
This
enables us to make the following statement.
In
a state of simple shear a shear stress
of any intensity along a plane is always accompanied by a complementary shear
stress of same intensity along a plane at right angles to the plane.
7. Direct Stresses Developed Due to Simple Shear.
Consider
a square element of side a and thickness t in a state of simple shear as shown
in figure. It is clear that the shear stress on the forces of element causes it
to elongate in the direction of the diagonal BD. Therefore a tensile stress of
same intensity pt is induced in the elements along BD. ie, across the plane of
the diagonal AC. The triangular portion ABC of the element is in equilibrium
under the action of the following.
(1)FAC
= Normal force on face AC = pt X AC X t = pt X √2 aXt
(2)FAB
= Tangential force on face AB = q X BC X t = q aXt
(3)FBC
= Tangential force on face BC = q X BC X t = q aXt
For
equilibrium in the direction normal to AC,
FAC
– FAB cos45 – FBC cos45 = 0
Pt
X √2 at – q at X 1/√2 - q at X 1/√2 = 0
√2
pt – 2 q /√2 = 0
pt
= q
It
can also be seen that the shear stress on the faces of the element causes it to
foreshorten in the direction of the diagonal BD. Therefore a compressive stress
pc is induced in the element in the direction AC, ie across the plane of the
diagonal BD. It can also be shown that pc = q.
It
can thus be concluded that simple shear of any intensity gives rise to direct
stresses of same intensity along the two planes inclined at 45º to the shearing
plane. The stress along one of these planes being tensile and that along the other being
compressive.
8.
Relationship among the elastic constants
8.1. Relationship between modulus of elasticity and
modulus of rigidity
Consider
a square element ABCD of side ‘a’ subjected to simple shear of intensity q as
shown in figure.
It
is deformed to the shape AB’C’D under the shear stress. Drop perpendicular BE
to the diagonal DB’. Let Ф be the shear strain induced and let N be the modulus
of rigidity.
The
diagonal DB gets elongated to DB’. Hence there is tensile strain et in the
diagonal.
et = (DB’ – DB) / DB = EB’ / DB
since
this deformation is very small we can take L BB’E = 45º
EB’
= BB’ / √2 = AB tan Ф / √2 = a tan Ф / √2
DB
= √2 a
et
= (a tan Ф / √2)/ √2 a = tan Ф /2 = = Ф / 2 since Ф is small
ie
et = ½ X q/N ------------------ (1)
We
know that stress along the diagonal DB is a pure tensile stress pt = q and that
along the diagonal AC is a pure compressive stress pc also equal to q. hence
the strain along the diagonal DB is et = q/E + 1/m X q/E
Ie
et = q/E (1+1/m) ------------------ (2)
From
(1) and (2) we have,
E
= 2N(1+1/m)
This
is the required relationship between E and N.
8.2. Relationship between Modulus of Elasticity E and
Bulk Modulus K
Consider
a cube element subjected to volumetric tensile stress pυ in X,Y and Z
directions. Stress in each direction is equal to pυ. ie px = py = pz =pυ
Consider
strains induced in X-direction by these stresses. px induces tensile strain ,
while py and pz induces compressive strains. Therefore,
ex
= px/E – 1/m[py/E + pz/E] = pυ/E[1-2/m]
due
to the perfect symmetry in geometry and stresses
ey = pυ/E[1-2/m]
ez = pυ/E[1-2/m]
K = p υ / e υ =p υ/(ex+ey+ez) = pυ/[3pυ/E(1-2/m)]
ie
E = 3K(1-2/m ) is the required relationship.
8.3. Relationship among the constants
From above,
E = 2N[1+(1/m)] and E = 3K[1-(2/m
)]
E = 3K[1-2(E/2N -1)] = 3K[1-E/N
+2]
9K = E[1+(3K/N)] = E[(N+3K)/N]
E = 9NK/(N+3K)
9. Bars of uniform section
Consider a bar of length l and
Cross sectional area A. Let P be the axial pull on the bar,p the stresss
induced ,e the strain in the bar and δl is the elongation.
Then p= P/A
e=
p/E =P/(AE) -------------(1)
e= δl/l -------------(2)
equating (1)
and (2)
δl = Pl / (AE)
