B.tech admission in india

H. Garmestani, Professor
School of Materials Science and Engineering
Georgia Institute of Technology
 Outline:
 Materials Behavior
Tensile behavior
…
admission.edhole.com

 Constitutive equation is the relation between kinetics (stress, stress-rate)
quantities and kinematics (strain, strain-rate) quantities for a
specific material. It is a mathematical description of the actual
behavior of a material. The same material may exhibit different
behavior at different temperatures, rates of loading and duration of
loading time.). Though researchers always attempt to widen the
range of temperature, strain rate and time, every model has a given
range of applicability.
 Constitutive equations distinguish between solids and liquids; and
between different solids.
 In solids, we have: Metals, polymers, wood, ceramics, composites,
concrete, soils…
 In fluids we have: Water, oil air, reactive and inert gases

/ axial strain
diametral strain
/ stress
l e
d
a
e
a
e
d
l l
= D =
P A
e
s
= = -
=
= =
0
Poisson's Ratio
Load-displacement response
is Young's modulus (or modulus of elasticity)
E E
Y
k e
is bulk modulus, is dilatation (for an elastic material)
shear modulus (for a cylindrical bar of circular corss
s
e
a
=
e
M l
t
q
Y
k
=
m
s
p
I
=
section of radius r to a torsional moment along the cylinder axis)

Examples of Materials
Behavior
Uniaxial loading-unloading stress-strain curves for
(a) linear elastic;
(b) nonlinear elastic; and
(c) inelastic behavior.

 Elastic behavior is characterized by the
following two conditions:
 (1) where the stress in a material (s) is a
unique function of the strain (e),
 (2) where the material has the property for
complete recovery to a “natural” shape upon
removal of the applied forces
 Elastic behavior may be Linear or non-linear

 The constitutive equation for
elastic behavior in its most
general form as

s=Ce
where
C is a symmetric tensor-valued
function and e is a strain tensor we
introduced earlier.
Linear elastic s = Ce
Nonlinear-elastic s
= C(e) e

Boundary Value Problems
we assume that the strain is small and there is no rigid body rotation.
Further we assume that the material is governed by linear elastic isotropic
material model.
Field Equations
(1)
Eij = 1
( ui, j + u ) j.i 
(1)
2
(2) Stress Strain Relations
(3)Cauchy Traction Conditions (Cauchy Formula)
(4)

sij=lEkkdij+2mEij (2)

ti=sjinj
sji,j+Xj=0
sji,j+rBi
=0®For Statics
sji,j+rBi
=rai
® For Dynamics

In general, We know that
For small displacement
Thus
¶sij
¶xj
+rBi =rai
Bi is the body force/mass
rBi is the body force/volume=Xi
ai is the acceleration
i i x = X
v u
i
j
+ ¶
j
admission.edhole.com x
fixed
u
= = ¶
i i
v Dx
i Dt
t
x
i
¶
¶

Assume v << 1, then
For small displacement,
fixed
u
v u
t
= ¶
i i
= ¶
¶
a v
¶
dV dV E
Since 1
o kk
= ¶
r r o kk
r
kk
» -
Thus for small displacement/rotation problem
( )
( )
( kk ) o
i
x
i
i
E
E
E
t
t
i
r
= +
+
=
= +
¶
-
1
1
1
1
0
1
2
2
r » ro
¶sij
¶xj
+rBi
=r¶2ui
admission.edhole.com ¶t2

Consider a Hookean elastic solid, then
Thus, equation of equilibrium becomes
¶ 2
= + + ¶

sij =lEkkdij +2mEij
=luk,kdij +m ui, j +uj,i ( )
sij, j =luk,kjdij +m ui,ij +uj,ij ( )
B E
( )
i
u
i j
kk
i
o i
i
u
+ ¶
o ¶
t
2
¶
x
¶ x ¶
x
2
r r l m m

=
ui
0 2
2
¶
t
For static Equilibrium Then
¶
E
+ ¶
( )
kk
l m m r
x x x x
¶
E
+ ¶
( )
0
0
2
ö
u + o
B
= ÷ ø
÷
2 1 1
3
2
ö
÷ ÷
u + B
= ø
l m m r
o
2 2 2
3
2
2
2
2
2
2
2
kk
x x x x
2
2
2
1
2
2
1
2
1
2
ö
æ
+ ¶
ç ç
è
æ
¶
+ ¶
ç ç
è
æ
+ ¶
¶
+ ¶
¶
+ ¶
+ ¶
¶
+ ¶
¶
+ ¶
¶
+ ¶
¶
E
+ ¶
( ) 0
÷ ÷
u + B
= ø
l m m r
o
2 3 3
3
2
2
2
1
kk
x x x x
3
ç ç
è
¶
¶
¶
¶
The above equations are called Navier's equations of motion.
In terms of displacement components
2
E div u B u kk o o ¶
admission.eld+hmoleÑ.com+ m Ñ + r = r ¶
( ) 2
1 t

In a number of engineering applications, the geometry of
the body and loading allow us to model the problem using
2-D approximation. Such a study is called ''Plane
elasticity''. There are two categories of plane elasticity,
plane stress and plane strain. After these, we will study
two special case: simple extension and torsion of a circular
cylinder.

For plane stress,
(a) Thus equilibrium equation reduces to
s ij =s ij ( x1, x2 ) (i, j = 1,2)
b
s s r
+ + =
11,1 12,2 1
s s r
+ + =
21,1 22,2 2
s s s
= = =
(b) Strain-displacement relations are
(c) With the compatibility conditions,
0
0
0
13 23 33
b
11 1,1 22 2,2 12 1,2 2,1 E = u E = u 2E = u + u
E E E
11,22 22,11 12,12 2
E
12
2
1 2
1
E
2
22
2
2
E
2
11
2
= ¶
x x
x
x
¶ ¶
+ ¶
¶
¶
¶
+ =

(d) Constitutive law becomes, Inverting the left relations,
( )
( )
1
v
E
= -
s s ( )
11 11 22
E
Y
1
v
= -
s s
22 22 11
= + = =
s s g
12 12
E
E
Y
E v
1 2
12 12
v
E G G
Y
E v
33 ( 11 22 ) ( 11 22 )
1
E E
v
= - + = -
E
Y
+
-
s s
s
Y
11 2 11 22
1
Y
22 2 22 11
1
s g g
Thus the equations in the matrix form become:
Note that
ü
ì
ù
é
ù
s
é
E
11
22
11
s
22
1 0
1 0
E
EY
(e) In terms of displacements (Navier's equation)
( )
12 1 12 2 ( 1
) 12 12
s
= ×
+
=
+
=
+
-
=
+
-
=
G
v
E E
v
E
E vE
v
E
E vE
v
E
Y Y
ïþ
ïý
ïî
ïí
ú ú ú
û
ê ê ê
ë
-
-
=
ú ú ú
û
ê ê ê
ë
12
2
12
0 0 1
1
E
v
v
v
v
s
( ) ( ) 0 ( , 1,2)
admission.edhoY le.com r
+
+ = =
2 1 +
, 2 1 +
, u b i j
v
u E
v
E
i ji i
Y
i jj

(b) Inverting the relations, can e -s be written as:
E v
= + - -
1 1
[( ) s s
]
[( ) ]
( )
E G
11 11 22
Y
E v
= + - -
1 1
s s
22 22 11
E v
v v
E
v v
E
= + =
Y
Y
2 2
2 1
12 12
12
s s
(c) Navier's equation for displacement can be written as:
E
E
Y ( ) u +
Y
u + r
b = ( i j
=
)
2 1 +
v
i , jj
( 0 , 1,22 1 + v )( 1 -
2 v
) j , ji i

Relationship between kinetics (stress, stress rate) and kinematics (strain, strain-rate)
determines constitutive properties of materials.
Internal constitution describes the material's response to external thermo-mechanical
conditions. This is what distinguishes between fluids and solids, and
between solids wood from platinum and plastics from ceramics.
Elastic solid
Uniaxial test:
The test often used to get the mechanical properties
s = P
A0
=engineering stress
e = Dl
l0
=engineering strain
E =s
e

If is Cauchy tensor and is small strain tensor, then in general,

s ij Eij

sij=CijklEkl
Cijkl
where is a fourth order tensor, since T and E are second order
tensors. is called elasticity tensor. The values of these components
with respect to the primed basis ei’ and the unprimed basis ei are related by
the transformation law
ijkl mi ni rk sl mnrs C¢ = Q Q Q Q C
However, we know that E k l = E l k and then
Cijkl = C jikl = Ciklk [C] 4´4

sij=sji
We have symmetric matrix with 36 constants, If
elasticity is a unique scalar function of stress and strain, strain energy is given by
dU= sijdEkl or U= sijEij
Then sij = ¶U
¶Eij
ÞCijkl =Cklij
ÞNumber of independent constants=21
Cijkl

Show that if for a linearly elastic solid, then
Solution:
Since for linearly elastic solid , therefore
Thus
from , we have
Now, since
Therefore,
sij = ¶U
¶Eij
ijkl klij C = C

sij=CijklEkl

¶sij
¶Ers
=Cijrs

sij = ¶U
2
C = ¶
U
¶Eijrs ij ¶ E ¶
E
rs ij
¶2 2
U
U
E E
= ¶
¶ E ¶
E
rs ij ij rs ¶ ¶
ijkl klij C = C

Now consider that there is one plane of symmetry (monoclinic) material, then
One plane of symmetry => 13
If there are 3 planes of symmetry, it is called an ORTHOTROPIC material, then
orthortropy => 3 planes of symmetry => 9
Where there is isotropy in a single plane, then
Planar isotropy => 5
When the material is completely isotropic (no dependence on orientation)
Isotropic => 2

Crystal structure Rotational symmetry
Number of
independent
elastic
constants
Triclinic
Monoclinic
Orthorhombic
Tetragonal
Hexagonal
Cubic
Isotropic
None
1 twofold rotation
2 perpendicular twofold rotations
1 fourfold rotation
1 six fold rotation
4 threefold rotations
21
13
9
6
5
3
2

A material is isotropic if its mechanical properties are
independent of direction
Isotropy means
Note that the isotropy of a tensor is equivalent to the isotropy of
a material defined by the tensor.
Most general form of (Fourth order) is a function
ijkl C
ijkl ijkl ijkl ijkl C A B H
= g + a +
b
= + +
gd d ad d bd d
ij kl ik jl il jk

sij = CijklEkl
¢ s ij = C ¢ ijkl E ¢ kl
Cijkl =C ¢ ijkl

 Thus for isotropic material
 and are called Lame's constants.
 is also the shear modulus of the material (sometimes designated as G).

sij =CijklEkl
= (g dijdkl +adikdjl +bdildjk)Ekl
=g dijdklEkl +adikdjlEkl +bdildjkEkl
=g dijEkk +aEij +bEji
=g dije+ (a +b)Eij
=ledij +2mEij
when i¹j sij =2mEij
when i=j sij =le+2mEij
l
m
m
admissions.e=dlheoIl+e2.mcoEm

(3l+2m)skk
We know that
So we have
ù
úû
skkdij
3l+2m
sij- l
é
êë
2m
Also, w e have
sij=ledij+2mEij

skk=(3l+2m)e or e= 1

Eij=1

E v v E k v
l m m m
, , , , ,
Y Y
v
vE
l l 2
m m m
( ) ( ) ( )
( )
v v
E
m
m m Y
m m
( )
kv
v
3
( )
( )
k E
v
l m m m
( )
-
E
E
v
( )
( ) ( )
k -
v
3 1 2
k
E
E
v
v
v
v
2 1
+
m
3 1 2 3 3
2
3 3 1 2
m ( 3 l +
2 m
) 2 ( 1 ) 3 ( 1 2
)
v ( ) v v E v
E E m
v E k v
Y Y Y
Y
Y
Y Y
Y
Y Y
1
l m
l
2 2
2 1
2 1
1
3
2
1 2
1 1 2
-
+
+ -
+
- -
-
+
+
+
- +
+ - -
l m m
Note: Lame’s constants, the Young’s modulus, the shear modulus, the Poisson’s
ratio and the bulk modulus are all interrelated. Only two of them are independent
for a linear, elastic isotropic materials,

B.tech admission in india

More Related Content

What's hot

Similar to B.tech admission in india

More from Edhole.com

Recently uploaded

B.tech admission in india