Plane Strain
Each strain is acting independent of one another
Due to normal strain x
Due to normal strain y
y
dv3
dv2
dy
dv1
x
x
dx
du1
x =
du3
du2
du
dx
y 
 du   x dx
dv
dy
dv   y dy
Due to shear strain  xy
y
 xy
Negligible
2
 xy
2
1
x
General Equations of Plane Strain Transformation
y
dv = y dy
B
 xy
2
dy
O
x
A
dx
 xy
du = x dx
2
Sign Conventions
(1) Normal strains are + ve, if they cause elongations along x and y axes, respectivelty.
(2) Normal strains are - ve, if they cause shortening along x and y axes, respectively
(3) Shear strains are + ve, if the interior angle AOB becomes less than 900.
(4) Shear strains are – ve, if the interior angle AOB becomes greater than 900.
y
y
x


dy
dx
dy
dy

O
x
dx
2
From the figure
dy
 cos
dy 
dx dx 

cos
dy  dy 
 (tan  ) cos
 sin 
dx
 cos 
dx 
dy
 sin 
dx 
Problem
Using the above orientations of axes, determine the strains along xoy axes due to
 x ,  y and xy , defined w.r.t. xoy axes.
Effect of normal strain x, along x axis
y
y
x
du


dy
dx
dy
dy
du1

dv1
x
du
=xdx
dx
Effect of normal strain y, along x axis
y
dv1 = y dycos
y
x

du1 =dv sin = x dysin
dv = ydy


dy
dx
dy

x
dx
3
Effect of shear strain xy, along x - axis
y
xy dy
x
dv1=xydysin
du1=xydycos

xy
dx
dy

x
dx
[Assume that dx remains fixed in position, and the shear strain xy is represented by the
change in angle of dy]


du   du1  du1 +du1
=  x dx cos     y dy sin     xy dy cos  
 dx 
 dy 
 dy 
 du 
 x 
  x cos 
   y sin  
   xy cos 

dx 
 dx  
 dx  
 dx  
  x cos cos    y sin  sin     xy cos sin  
  x cos 2    y sin 2    xy sin  cos


dv   dv1  dv1 +dv1
=   x dxsin  
 dycos  
y
xy
dysin 
dv 
 dx 
 dy 
 dy 
  x sin  
   y cos 
   xy sin  

dx 
 dx  
 dx  
 dx  
  x sin  cos     y cos  sin     xy sin  sin  
4
(I)
 ( x   y ) cos sin    xy sin 2 
=  (the angle of shear distortion along x axis)
By rotating the angle through 900, in the clockwise direction, the rotation  of elemental
length dy can be obtained.
  ( x   y ) cos(90   ) sin( 90   )   xy sin 2 (90   )
   x   y  sin  cos    xy cos 
2
   x   y sin  cos   xy cos 2 
(-) = rotation of the right angle xoy

 2  x   y sin  cos   xy cos 2   sin 2 

   x   y sin 2   xy cos 2
  xy
(II)
From Equation I,

 x   x cos2    y sin 2    xy sin  cos

 1  cos 2 
 1  cos 2    xy
 x
sin 2 
y

2
2



  2

x y
 
 2
 x y
  
  2


 cos 2  xy sin 2
2

(III)
From Equation II,
  xy     x   y

  
2
 2  

 
 sin 2   xy  cos 2

 2 
5
(IV)
 y  can be obtained by introducing (90+) for  in  x 
 y    x cos 2 90      sin 2 90      xy sin 90    cos90   
=  x ( sin  ) 2   y cos    xy  cos sin  
2
  x sin 2    y cos 2    xy sin  cos
 1  cos 2 
 1  cos 2   xy
= x 
sin 2
y

2
2



 2
x y
= 
 2
 x y
  
  2

 
 cos 2   xy  sin 2

 2 
To find the principle strain

d x
0
dx
i.e.,   x   y sin 2   xy cos 2  0
 tan 2  tan 2 p 
x y

 2
   xy 
  

2
 

2
 xy
x y
2
  xy

 2
2p
x y

 2



 xy



sin 2 p 
6
2
x y

 2
   xy
  
  2
2



2

cos 2 p 
x
y
2
x y

 2
   xy 
  

  2 
2
  x   y   x   y 
  xy   xy 



  
  x   y   2  2 

 2  2 
 
 x   1  

2
2
2
 2 
  x   y    xy 
  x   y    xy
 = p

   

  
2
2
2

  

  2
x y
= 
 2

 

x y

 2
   xy 
  

  2 
2
2
  x   y   x   y 
  xy   xy 






2  2 
2  2 
x y 



 
 y   2  

2
2
2
 2 




 x
 xy 
  x   y    xy
y 

  


  
 = p
 2   2 
 2   2
x y
= 
 2

 

x y

 2
   xy 
  

  2 
2
  max 
 2 
 1



 2  in plane  2 
2
x y

 2
7
   xy 
  

2
 

2
2



2
2



2
Similarity Between Stress and Strain Transformation Equations
Stresses at a point
Strains at a point
8
Principle Stresses
Principle Strains
 x  y
d x
2 sin 2   2 xy cos 2  0

d
2
tan 2 p 
 xy
 x   y / 2
for p1, sin 2 p1 
 1, 2 
 x  y

2

 x  y
2
tan 2 p 
 xy
  x  y 


cos 2 p1
d x
  x   y sin 2   xy cos 2  0
dx
2
 1, 2 
2
   xy2




  x  y

2

 x  y
 
2

2

   xy2

2

   xy2

9
x y
2
 xy
 x   y / 2
x y
 
 2
2

   xy2

Maximum in-plane shear stress
Maximum in-plane Shear Strain
Principal stress plane and maximum
Principal strain plane and maximum (in-
shear stress planes are inclined at 450.
plane) shear stress planes are inclined at
Consequently,
450 to one another.
twice the values of these
Consequently, twice the values of these
angles will be inclined at 900.
angles will be inclined at 900.
 tan 2 p tan 2 s   1
 tan 2 p tan 2 s   1
As a result
As a result
tan 2 s 
 max
in plane
 avg 
  x   y  / 2
tan 2 s 
 xy
 x  y
 
2

 max
2

   xy2

in plane
2
 x  y
 avg 
2
10
  x   y  / 2
 xy
x y
 
 2
x y
2
  xy
 
2

2
Mohr’s Circle
For Plane Stress
For Plane Strain
2
1
11
Material Property Relationships
1
[ x   ( x   y )]
E
1
 y  [ y   ( z   x )]
E
1
 z  [ z   ( x   y )]
E
x 
12
When only shear stresses are acting
 1 ,  2    xy 2
  xy
  xy
 1 ,  2  
 2


  xy
2

2
  xy 
   xy
1
 2  

E  E
E
E
 1 max  
  xy 
1   
 xy
 xy

1    
i.e.,
E
2

E
2G
1
2G
i.e., E  21   G
(A)
When a body is subjected to normal stresses  x   y   z  p, the body under goes
only change in volume.
Volume change = ( x   y   z )V
1  2  p
1
[ p   ( p  p)] 
E
E

1
1  2  p
 y  [ p   ( p  p)] 
E
E

1
1  2  p
 z  [ p   ( p  p)] 
E
E
x 
p

x
 y  z 
3

( p  p  p)
p
3
(volume change)/unit volume = ( x   y   z )
13
 1  2
=
 E
p/e = bulk modulus = K =
=
3(1  2 ) p

e
3 p 
E

p E
3(1  2 ) p
E
3(1  2 )
Theories of Failure
In the design of structural members, it becomes important to place an upper limit on the
state of stress that defines the material's failure.
Ductile Materials
Brittle Materials
Stress
Stress

Y - Yield stress (steel)

ult – Ultimate stress
-

ult – Ultimate stress
Not used since strain is very
high at this stress level

0.1% Proof stress
(stress at o.1% elongation)
(Aluminum)
Strain
Strain


y – Yield strain (0.15 to 0.2
for mild steel)
ult – Ultimate strain (0.2% to
0.3% for CAST IRON)
 ult – Ultimate strain (20 to 25%
for mild steel)
14
The material behaviour – either ductile or brittle – does not remain a constant one for
any material. It is dependent on:
Temperature
Rate of loading
Chemical environment
Forming/shaping methods
In order to apply the theories of failure:
(i)
The state of stress in a structure, at a point where the maximum stresses
are expected -  x ,  y ,  xy ,  z ,  yz ,  zx - are determined first.
(ii)
Thereafter, the principal stresses and maximum shear stresses are
determined -
1   2   3
 
12 max
,
  ,  
23 max
31 max
Failure Theories
For ductile materials
For brittle materials
1. Maximum shear stress theory
1. Maximum normal stress theory
Proposed by Tresca
2. Mohr’s failure criterion
2. Maximum distortion energy
theory – Proposed first by Huber
- Proposed by Otto Mohr
and refined later by Von Mises and
Hencky
-
Huber–Mises–Hencky theory
15
1.
Maximum Shear Stress Theory
“Failure (by yielding) will occur in a material (at a point) when the maximum shear
stress in the material is equal to the maximum shear stress that will occur when the
material is subjected to an axial tensile test’.
P = (Y) A
450
Thin mild
steel strip
Luder’s lines
For a two-dimensional stress system,
 max 
 max   min
2
(I)
Under simple tension test,
 max   Y
 min  0
 0
  max  Y
2
 
i.e.  Y    max
 2 
Using in Eqn. I
16
(II)
Y
2

 max   min
2
 max   min   Y
Governing criteria
(III)
Considering a three-dimensional element (with two-dimensional stress state)
3 = 0
3
1
2
1
2
Arranging the stresses in the order of decreasing magnitudes,
(i)
Case (a):
 1   2   3 ( 0)
Hence  max   1
 min  0 (  3 )
Failure will occur first in the  1   3 plane.
 1  0   Y
i.e.,  1   Y
(IV)
Failure in shear will occur, when the maximum principal stress is equal to Y.
(ii)
Case (b):
 1   2 ( 0)   3 (ve)
 max   1
17
 min   3 (ve)
 1   3   Y
V (a)
Failure will occur in the plane containing  1 ,  3 stresses
Generalizing this for a plane-stress failure wherein  1 ,  2 act along x-y axes and 3 acts
along z-axis, (zero stress), one can rewrite Equation (V (a)) as
1   2   Y
V (b)
Failure occurs in
(1, 2) plane
1  
Y
2  
(-ve)
2 = Y, 3 =0
-Failure in this
plane (1, 3)
2 (+ve)
2
Y
Y
-1, +2
2
-ve
1 = Y (3 =0)
-Failure in this
plane (1 , 3)
+ve
Y
Y
1, -2
-Failure occurs in
plane (1, 3)
+ve
Y
(-ve)
Failure envelope or Yield loci
18
1 (+ve)
1  
-ve
2  
Failure occurs in
(1, 2) plane
Y
2
Y
2
2.
Maximum Distortion Energy Theory
“ Failure (by yielding) will occur when the shear or distortion energy in the material
(at a point) reaches the equivalent value that will occur when a material is subjected to
uniaxial tensile test”.
Let us say that the principal stresses in an element, at a point, is given by  1 ,  2 ,  3
Total strain energy stored in the given system = Total volumetric strain energy + Total
distortion strain energy
ut
=
uv
+
ud
(VI)
3
3
3

=
2

2
2
1
1
3
1

31
+
21
2
11
1
19
 1     11
 2     21
 3     31
 
1   2   3
(VII)
3
1   2   3 
   11
3


 1     11  
1
2 1   2   3 
3
  2 3 
    21   1
   21
3


1
 21  2 2   3   1 
3
  2 3 
    31   1
   11
3


1
 31  2 3   1   2 
3
  11 
2

3

1
Also  11   21   31  [( 2 1   2   3 )  (2 2   3   1 )  (2 3   1   2 )]
3
=0
(VIII)
Using the earlier stress-strain relationships






1 1
[ 1    21   31 ]
E
1
 21  [ 21    31   11 ]
E
1
 31  [ 31    11   21 ]
E
 11 
Considering the volumetric strains due to  11 ,  21
20
and  31
 v1   11   21   31

 


1 1
 1   21   31    21   31   31   11   11   21
E
1 1

 1   21   31  2  11   21   31
E
(1  2 ) 1

 1   21   31
E






=0
(IX)
[Since  11   21   31  0 according to Eqn. (VIII)]
Equation (IX) states that no volumetric change occurs in the material due to the stresses
 11 ,  21
and  31 (but it does produce a change of shape). Due to the three stresses
 11 ,  21
and  31 ,
1
( 1 1   2  2   3 3 )dv
2
1 3
   i  i dv
2 i 1
ut 
(Total strain energy)
Hence strain energy per unit volume
3
=  i i
i 1
Considering only 
and
 [the mean stresses and strains due to (  1 ,  2 and  3 )
and (  1 ,  2 and  3 )],
1
u v  [        ]
2
 1  1  2   
1

 3     3  
 
2

2  E  

3 (1  2 ) 2

2 E
21
 
1   2   3
3
1
   (   )
E
(1  2 )


E

31  2   ( 1   2   3 ) 


2 E 
3


2
(1  2 )
[ 1   2   3 ] 2
6
(X)
Considering the normal (or principal) stresses and strains,
1
 1 1   2 2   3 3 
2
1   1
   1
   1
 
  1   1    2   3    2   2    3   1    3   3    1   2 
2   E
   E
   E
 
1


 12   22   32     1 2   1 3   2 3   2 1   3 1   3 2 
2E
ut 






1
 12   22   32  2  1 2   2 3   3 1 
2E
(XI)
Since u t  u v  u s ,
u s  ut  uv
1
 1  2 
2
 12   22   32  2  1 2   2 3   3 1   
 1   2   3 
2E
 6E 
2
2
2
1  3  1   2   3  6  1 2   2 3   3 1   (1  2 )( 12  2 1 2   22  2 2 3   32  2 3 1 ) 



2E 
3








1
[(3 12  3 22  3 32   12   22   32 )  2 ( 12   22   32 )
6E
 (6 1 2  6 2 3  6 3 1  4 1 2  4 2 3  4 3 1 )  2( 1 2   2 3   3 1 )]

1
[2( 12   22   32 )  2 ( 12   22   32 )  2 ( 1 2   2 3   3 1 )  2( 1 2   2 3   3 1 )]
6E

(1   )
 1   2 2   2   3 2   3   1 2
6E

When the specimen is under uniaxial tension,
22

(XII)
1   y
2 3  0
From eqn. (XII),
(1   )
1   2
1  
2
2
us  
(2 Y2 )
Y  0 Y  
 2 Y 
6{2(1   )G}
 6E 
 6E 

 u s 

 
 Y2
(XIII)
6G
For a general state of stress,
us 

(1   )
 1   2 2   2   3 2   3   1 2
6{2(1   )G}
=

1
 1   2 2   2   3 2   3   1 2
12G


(XIV)
From Eqns. (XIII) and (XIV), equating the distortional energies due to an uniaxial state
of stress and that due to a multiaxial state of stresses,
 1   2 2   2   3 2   3   1 2  2 Y2
(XV)
For a two-dimensional state of stresses,
3  0
Hence equation reduce to

  2    22   1  2 Y2
2
1
i.e.,
2
 12   1 2   22   Y 2
This is an equation to an inclined ellipse.
23
(XVI)
2
Y
 Y Y 
 
,

3
3

450
Y
Y
1
450
450
 
Y

,  Y 
3
 3
Y
Maximum
distortionenergy theory
Maximum
shear stress
theory
Plot of Eqn. (XVI) gives the failure envelope or yield loci for a system subjected to a
two-dimensional state of stress.
24
Brittle Materials
Applicable to cast iron that tends to fail suddenly by fracture, without any warning.
1. Maximum normal stress Theory:
In a tension (or compression) test, brittle fracture occurs when the normal stress reaches
the ultimate stress ult.
In a torsion test, brittle fracture occurs due to a maximum tensile stress (in a plane 450 to
the shear direction) when it reaches the ultimate stress ult.
Tension
Torsional
shear
Compression
2
Failure criteria or failure loci:
ult
ult
25
1
Statement
When the maximum principal stress  1 (or  2 ) in the material reaches a limiting value
that is equal to the ultimate normal stress the material can sustain, failure by fracture
occurs.
 1   ult
 2   ult
-Eg. Chalk: under tension, under bending and under torsion.
2. Mohr’s Failure Criterion
For materials (brittle) that have different fracture properties in tension and compression,
this criterion holds good.
-Specially for metals
-For nonmetals like concrete (Rock, concrete, soils) another theory
is applicable (we will briefly deal with this later)
Three tests done to determine failure criteria –
- Tension test that gives (ult)t
-
Compression test that gives (ult)c
-
Torsion test that gives ult
26
Circle B
1 = ult
2 = 0
3 = -ult
A
Circle C
1 = (ult)t
2 = 0
3 = 0
(ult)c

B
C
ult

Circle A
1 = 0
2 = 0
3 = -(ult)c
Failure
envelope
Mohr’s circle for each test
Material is
under limiting
condition
2
(11, 21)
(ult)t
(ult)c
(ult)t
1
x (1, 2) material
has failed
(ult)c
Mohr’s failure criteria
Failure occurs when the absolute value of either one of the principal stresses reaches a
value greater than (ult)t or (ult)c or in general, if the stress at a point is defined by the
stress coordinate (1, 2), which is plotted on the boundary or outside the shaded area.
27