1. simple stress and strains

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SIMPLE STRESS AND STRAINS
 Introduction:-
In earlier studies we have undergone a basic course in mechanics of Rigid Bodies,
more commonly referred to as Engineering Mechanics or Applied Mechanics.
Mechanics as such is subdivided into three branches; Mechanics of Rigid bodies,
Mechanics of Deformable Bodies and Mechanics of fluids.
Mechanics of Rigid Bodies assumed bodies to be perfectly rigid i.e. there is no
deformation of bodies under the action of loads to which they are subjected statics
and Dynamics are the two branches of Mechanics Of rigid Bodies involving stationary
and moving bodies respectively under the action of loads.
 Stress:-
i. The force of resistances per unit are offered by a body against deformation is known
as Stress. When a body is subjected to external loading the body undergoes some
deformation. At the same time internal force of resistance is due to the cohesion of
molecules inside the body. Thus stress is induced in the body upon external action
of load
ii. If the body is able to resist the external load , it is said to be stable , in equilibrium
and therefore for this condition the internal force of resistance should be equal to
the external load.
By Definition
Or
 Strain:-
As the body produce force of resistance to counter the external loading it
undergoes some deformation .The extent of deformation depend on the material
property like molecular cohesion .The ratio of change in dimension is known as
strain.
Changeindimension
Strain =
Originaldimension
Since Strain is ratio, it has no units. We shall denote strain by letter e. If L is the
original dimension and is change in dimension and then
δL
Strain = e =
L
 Types of Stress:-
1) Direct Stress and Direct Strain:-
When the force of resistance acts normal or
perpendicular to the area on which it acts, the
stress so produced is termed as Direct or Normal
Stress and corresponding strain is referred to as
Direct Strain.
forceof resistance
Stress =
Cross-sectionalarea
P
Stress =
A

forceof resistance toc/sarea
c/sarea on which forceof resistanceacts
Direct Stress
R
f
A
P
f
A



We shall denote direct stress by letter 'f'. Direct stress could be of tensile nature of
compressive nature.
2) Tensile stress:-
When the force of resistance acts away from the cross sectional area, the direct
stress is of tensile nature. Tensile stresses tend to cause an increase in the original
dimension.
3) Compressive Stress:-
When the force of resistance acts towards the cross sectional area, the direct stress is of
compressive in nature .Compressive stresses tend to cause a decrease in the original
dimension.
 Stress–Strain curve for Ductile Materials:-
1) Proportional Limit:-
It is the point (A) upto which stress is proportional
to strain and the material obeys Hooke’s Law.
2) Elastic Limit:-
It is a point (B) very close to A upto which the
material is said to be elastic.
3) Yield Point:-
It is a region from point C upto point D where the
curves is nearly flat or horizontal indicating no
increase in stress but appreciable increase in
strain.
4) Ultimate stress:-
It is the point E on the graph and is the maximum stress which the material can
with stand before it finally fails.
5) Breaking Point:-

Any ductile material does not fail at the ultimate stress point but before failure the
necking of the specimen takes place which leads to failure.
 Stress-Strain Curve for Brittle materials:-
i. Brittle materials undergo very little deformation on tensile loading and fail at low
tensile loads. Brittle materials absorb little energy on impact. Glass, cast-iron,
concrete etc. are examples of brittle materials.
ii. The stress-strain curve for a brittle material shows low proportional limit and also a
low ultimate stress value. The yield point (A) is not very well defined and
approximations are used to locate it. The Ultimate stress point and Breaking point
(B) are the same.
 Hooke’s Law:-
Stress is proportional to strain and the material is fully elastic in this region. This is
known as Hooke`s law.
f e
or f = E.e
E =
f
e

 Relation for Change in Length of a Material of a Member under Direct Stress:-

P
Wehave directstressf =
A
δL
also Strain e =
L
f
Also Modulus of ElectricityE =
e
Combining theabove equation weget
PL
Change in Lenght δL =
AE
The ratio of Ultimate stress to safe stress is known as the Factor of Safety. It works
as a protective shield against failure of material under stress.
Ultimatestress
Factory of Safety =
Safe stress
 Deformation of Uniformly Tapering Circular Section Bar:-
Consider a tapering rod with diameter D at one end which tapers uniformly to
diameter d at the other end over a length L. Let the rod be subjected to an axial tensile
load of P. Consider an element of length dx located at distance x from the smaller end.
The element would also be under the axial pull P.
Length of element eL dx
Diameter of the element e
D d
d d x
L
 
  
 
c/s area of the element
2
2
4 4
e e
D d
A d d x
L
     
    
  
Change in length of element e
e
P L
Le
A E
 
2
4
p dx
D d
d x E
L



   
   
  
Total change in length of the rod
2
0
4
L
Pdx
L
D d
E d x
L



   
  
  


0
4 1
L
P L
L
D dE D d
d x
L


 
 
 
       
4 1 1
( )
PL
L
E D d D d


 
    
4PL
L
EDd


 ……………… 1.7 Deformation of circular tapering rod
 Deformation of Uniformly Tapering Rectangular Section Bar:-
Consider a tapering bar of uniform thickness t with side a on one end which tapers
uniformly to side b at the other end over a length L. Let the rod be subjected to an axial
tensile load P. Consider an element of length dx located at distance x from the smaller
end. The element would also be under the axial pull P.
Length of element eL dx
Tapering side of the element =
a b
b x
L
 
 
 
c/s area of the element e
a b
A b x t
L
   
   
  
Change in length of element e
e
P L
Le
A E
 
P dx
a b
b x t E
L


   
   
  
Total change in length of the bar
0
L
Pdx
L
a b
tE b x
L
 
   
  
  

log
L
e
o
P L a b
L b x
tE a b L

    
        

log
( )
e
PL a
L
tE a b b

  
      
……….. Deformation of rectangular tapering bar
 Deformation due to Self Weight
Member when vertically suspended from one end, may
undergo elongation under its own weight. This deformation is
usually very small in magnitude as compared to other deformation.
However if the length of the suspended member is large, the
elongation due to self weight becomes appreciable.
 Uniform Rod Under Self Weight:-
Let us derive a relation for elongation of a uniform rod and
relation for direct stress when supported in a vertical
position by suspending if from the top-most point of the
rod.
Figure shows a rod of uniform c/s area A, modulus of
elasticity E and length L.
Let  be the unit weight of the material of the rod.
Consider a section x-x at distance y from the bottom.
The weight of the portion below the section xx ( )A y  
Stress at section x –x =
.
/
P wt of portionbelow x x
A c s area


 
. ..... stress at a distance y from the bottom
A y
f
A
f y


 

 
Consider an element of length ‘dy’ being elongated due to force P at
section x-x
Change in length of the element =
( )PL A y dy
AE AE
 

. y dy
E


 Total change in length of member =
0
.
L
y dy
E


2
2
L
L
E

  ….elongation of uniform rod under self weight

 Analysis of Members Made of Composite Materials:-
Member made up of more than one material are referred to as composite members.
Members are intentionally made of composite materials to increase mainly their load
carrying capacity and also other parameters effecting strength, stiffness and stability.
For solving problems on composite materials we develop two basic relations, these
are
(1) Load Shearing Relation :
This relation relates the individual loads carried by different materials to the
total load carried by the composite member.
(2) Strain Relation :
This relation relates the deformation of different materials of the composite
section.
 Shear Stress:-
When the load acts tangential to the cross-sectional area it tends to shear or cut the
member. When the blades of the scissor of the scissor cut a paper the force of the
blades act tangential to the cross-section of the
paper and the effect is shearing of the paper.
Shear stress is defined as the shear resistance per
unit shear area. We shall denote shear stress by
Greek letter  .
Shear Resistance
Share Stress =
Shear Area
R
=
b×d
Shear Force
ShearStress τ =
Shear Area
S.I. Unit of Shear Stress is N/m2
 Single Shear and Double Shear:-
1) Single Shear:-
When the shearing force causes the resistance shear to act at only one plane the
shear is referred to as single shear.
Refer to which shows two plates connected by a rivet. If force P acts in the two
plates as shown. The rivet is subjected to single shear. The shear resistance is
seen acting at one plane.
2) Double Shear:-
When the shearing force causes the resistance shear to act at two planes the shear
is referred to as double shear.

Refer to Fig. which shows a member connected by a bolt to a change. If force P acts
on the member as shown. The bolt is subjected to double shear. The shearing
resistance acts at two planes as seen in the figure
3) Temperature Stress:-
Materials tend to naturally expand or contract due to rise or fall in temperature.
This expansion or contraction which takes place is known as free expansion or free
contraction of the materials and is proportional to
A. the change in temperature ‘T’
B. coefficient of linear thermal expansion of the material ‘α’ and
C. the original length of the member ‘L’
Free expansion or contraction T L
No temperature stresses are generated
in the member if the free expansion or
contraction is allowed.
Whenever the free expansion or
contraction of a material is prevented,
temperature stresses which are
compressive or tensile in nature are
developed.
Temperature strain and temperature
stresses are calculated using the following
relation
PrExpansion or Contraction evented
Temperature Strain
Original Length

Temperature Stress E Temperature Strain 
Temperature stresses become important in situations where the temperature is
likely to change by large amount. For example, in a railway track.
4) Lateral Strain:-

When direct axial load is applied on the member the length of the member changes,
but this is accompanied by changes in lateral dimensions too. Lateral dimensions
are the dimensions at right angles to the axis of the member.
Fig. Shows a bar of length L, breadth b and depth d subjected to axial tensile load.
The length L increases by ,L but the lateral dimensions b and d decrease by
b and d  respectively. Fig shows axial compressive load and here the length
decreases the lateral dimensions increase.
We have seen Strain
L
L


Now we will be more specific, and call this strain which is along the direction of the
force as Longitudinal Strain.
δL
Longitudinal Strain, longitudinal=
L
The Lateral Strain is the ratio of change in lateral dimension to the original lateral
dimension.
Changein LaterealDimension
Letereal Strain =
OrigionalLeteraldimension
Lateral Strain, lateral =
b d
b d
 

In case of bars with circular cross-section
latereal= Where D is the rod`s diametere D
D

Lateral strains are always opposite in nature to the longitudinal strain. For tensile
stress, the length increases whereas the lateral dimension increases whereas the
length decreases. Lateral strains are smaller in magnitude as compared to
longitudinal strain.
5) Poisson’s Ratio:-
For a given material the ratio of lateral strain to the longitudinal strain is a
constant within certain limits. This ratio is termed as Poisson’s ratio.
1
It is denoted as
m
1 LeteralStrain
Poisson'sRatio =
m LongituinalStrain
1 elateral
=
m elogitudinal
Most of the materials used in engineering applications have Poisson`s Ratio
between 0.25 and 0.35.

6) Volumetric Strain:-
Volumetric strain is defined as the ratio change in volume to the original volume.
Volumetric strain is denoted as Ve
V
changein Volume
Volume Strain e =
OriginalVolume
Volumetric Strain for Uni-Axial Loading
A body subjected to axial force in one direction and hence the stress
developing along one single direction, is said to be under Uni-Axial loading.
Figure shows a bar subjected to stress in the x direction.
Volumetric Strain ve under Uni-Axial Loading,
log 2v itudinal laterale e e  …………………
or
1
2v
f f
e
E M E
 
    
2
1v
f
e
E M
 
    
……………….. Volumetric Strain in terms of stress
Note: In equation if the load is tensile longitudinale is positive and laterale is negative. For
compressive load, longitudinale is negative and laterale is positive. In equation stress f
is positive if the load is tensile and negative if load is compressive.
Volumetric Strain for Tri-Axial Loading
A body subjected to axial forces in all three directions and hence the
stresses developing along all the three directions, is said to be under Tri-Axial
loading.
Volumetric Strain ve under Tri-Axial Loading

v x y ze e e e  
Here ,xe ye and ze are the strains in the ,x y and z directions respectively.
yx z
x
ff f
e
E mE mE
  
y x z
y
f f f
e
E mE mE
  
yxz
z
fff
e
E mE mE
  
2x yf f f f   …….Volumetric Strain in terms of stress
We shall use +ve values of stress if they are tensile and -ve values of stress if they
are compressive.
7) Bulk Modulus:-
The ratio of direct stress to the volumetric strain produced in the material within
certain limits is a constant and is referred to as Bulk Modulus. It is a material
properly like E and G.
Bulk Modulus is denoted by letter K and its S.I. unit is 2
N / m .
Hence Bulk Modulus
v
f
K
e

Relation between E and G
Consider a solid cube subjected to shear force F at the top face. This produces
shear stress  on faces AB and CD and complementary shear stress on face BC and AD.
The distortion due to shear stress is shown by dotted lines. The diagonal AC also distorts
to AC.
Now, Shear strain
'CC
BC
 
Modulus of Rigidity
Shear Stress
G
Shear Strain

Shear Stress
Shear strain
G
 
'CC
BC G


…………… (1)
From C drop a  on the distorted diagonal 'AC
' ' 'cos 45
( / cos45)
EC EC CC
strain of the diagonal
AE AC BC
   

1
2 1E G
m
 
   
 
………………………. (2)
Substituting from (1) and (2) we get
Strain of the diagonal
2G


Let f be the direct stress induced in the diagonal AC due to the shear stress 
2 2
f
strain of the diagonal
G G

   ……………………………. (3)
The diagonal AC is subjected to direct tensile stress while the diagonal BD is
subjected to direct compressive stress.
The total strain of the diagonal AC would therefore be
1
.
f f
E M E
 
1
1
f
E m
 
  
 
………….. (4)
Comparing equations 3 and 4, we get
1
1
2
f f
G E m
 
  
 
1
2 1E G
m
 
   
 
Relation between E and K
Instead of shear stress  , Now let the same solid cube be subjected to direct
stress f on all the faces.
From equation
3 2
1v
f
e
E m
 
   
Since 2x yf f f f  
We have
3 2
1v
f
e
E m
 
   
………………. (1)
From equation
v
f
e
k
 ……………….(2)
Equating equation (1) and (2)

3 2
1
f f
k E m
 
   
9
3
KG
E
K G
 

Relation between E, G and K
The relation between the three moduli can be obtained by eliminating m from the
equations and
From equation
1 2
2
E G
m G


From equation
1 3
6
K E
m K


Equation we get
2 3
2 6
E G K E
G K
 

6 12 6 2EK KG KG EG  
 18 6 2KG E K G 
9
3
KG
E
K G
 


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