ELASTIC CONSTANTS IN
ISOTROPIC MATERIALS
1. Elasticity Modulus (E)
2. Poisson’s Ratio (n)
3. Shear Modulus (G)
4. Bulk Modulus (K)
1. Modulus of Elasticity, E
(Young’s Modulus)
F
F
simple
tension
test
s
E
e
s=Ee
Linearelastic
Units:
E: [GPa]
Slope of stress strain plot (which is
proportional to the elastic modulus)
depends on bond strength of metal
E=
Adapted from Fig. 6.7,
Callister 7e.
F
2. Poisson's ratio, n
eT
 “n” is the ratio of
transverse contraction strain
to longitudinal extension
strain in the direction of
stretching force.
Either transverse strain or
longitudional strain is
negative,  ν is positive
F
simple
eL tension
test
n
eT
n=eL
eT : Transverse Strain
eL : Longitudional Strain
Units:
n: dimensionless


Virtually all common materials undergo a
transverse contraction when stretched in one
direction and a transverse expansion when
compressed.
In an isotropic material the allowable (theoretical)
range of Poisson's ratio is from -1.0 to +0.5,
based on the theory of elasticity.
metals: n ~ 0.33
ceramics: n ~ 0.25
polymers: n ~ 0.40
3. Shear Modulus, G
t
M
simple
torsion
test
G
g
t=Gg
M
Units:
G: [GPa]
4. Bulk Modulus, K
P
Initial Volume = V0
Volume Change = DV
savg = K
P
DV
Vo
σavg is the average of
three stresses applied
along three principal
directions.
P
savg
DV
K
Vo
Units:
K: [GPa]
Elastic Constants
s= E
e
t= G
g
DV
savg = K
Stresses
Vo
Strains
Normal
Shear
Volumetric
Example:
Uniaxial Loading of a Prismatic Specimen
P=1000 kgf
9.9 cm
10 cm
Determine
E and n
10.4 cm
10 cm
10 cm
Before
9.9 cm
After
1000
= 10kgf/cm2
P=1000kgf → σ=
10*10
P=1000 kgf
Δd/2=0.05cm
Δl/2=0.2cm
σ
E=
ε
10cm
εlong=
10cm
1000 kgf
εlat=
10
= 250 kgf/cm2
=
0.04
Δl
l0
Δd
d0
ν=-
=
0.4
10
=0.04
=
-0.1
10
= -0.01
-0.01
0.04
= 0.25
 For an isotropic material the stress-strain
relations are as follows:
RELATION B/W K & E

Consider a cube with a unit volume
σ
D
1
C
1
A
1
B
σ
σ causes an elongation in the direction
CD and contraction in the directions AB
& BC.
The new dimensions of the cube is :
• CD direction is 1+ε
• BC direction is 1-νε
• AB direction is 1-νε
 V0 = 1
 Final volume Vf of the cube is now:
(1+ε) (1-νε) (1-νε) = (1+ε) (1-2νε+μ2ε2)
= 1 - 2νε + μ2ε2 + ε-2νε2 + μ2ε3
= 1 + ε - 2νε - 2νε2 + μ2ε2 + μ2ε3
ε is small, ε2 & ε3 are smaller and can be neglected.
 Vf = 1+ ε - 2νε → ΔV = Vf - V0 = ε (1-2ν)
 If equal tensile stresses are applied to each
of the other two pairs of faces of the cube
than the total change in volume will be :
ΔV = 3ε (1-2ν)
σ
σ
σ
σ
σ
Ξ
K=
+
σ
SΔV = 3ε (1-2ν) =
K=
+
savg
DV/V0
=
ε (1-2ν)
(σ+σ+σ)/3
3ε (1-2ν)
E
3 (1-2ν)
+
=
ε (1-2ν)
σ
3ε (1-2ν)
+
=
ε (1-2ν)
E
3 (1-2ν)
 The relation between
K and E is :
K=
E
3 (1-2ν)
 Moreover the relation
between G and E is :
 The relation between
G, E and K is :
G=
1
E
E
2 (1+ν)
=
1
1
+
9K
3G
Therefore, out of the four elastic
constants only two of them are
independent.
 For very soft materials such as pastes, gels,
putties, K is very large
 Note that as K → ∞ → ν → 0.5 & E ≈ 3G
 If K is very large → ΔV/V0 ≈ 0 *No volume
change
 For materials like metals, fibers & certain
plastics K must be considered.
 Modulus of Elasticity :
• High in covalent compounds such as diamond
• Lower in metallic and ionic crystals
• Lowest in molecular amorphous solids such
as plastics and rubber.
Elastic Constants of Some Materials
E(psi)x106 (GPa) G(psi)x106 (GPa)
ν (-)
Cast Iron
16
110
7.4
50
0.17
Steel
30
205
11.8
80
0.26
Aluminum
10
70
3.6
25
0.33
4-15
0.2
0.7
0.2
?
Concrete
1.5-5.5
Wood
Long 1.81
Tang 0.10
10-40 0.62-2.30
12
0.7
0.11
0.03