1. ELASTIC CONSTANTS IN ISOTROPIC MATERIALS
1. Elasticity Modulus (E)
2. Poisson’s Ratio (n)
3. Shear Modulus (G)
4. Bulk Modulus (K)

2. 1. Modulus of Elasticity, E (Young’s Modulus)
simple
tension
test
1. Modulus of Elasticity, E (Young’s Modulus) s
Linear-
elastic E e
s = E e
Units:
E: [GPa]

3. 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.

4. 2. Poisson's ratio, n F eT eL n eT n = - eL eT : Transverse Strain
simple
tension
test
2. Poisson's ratio, n eT eL n
“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 eT n = - eL
eT : Transverse Strain
eL : Longitudional Strain
Units:
n: dimensionless

5. 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

6. 3. Shear Modulus, G M t G t = G g g M Units: G: [GPa] simple torsion
test
t = G g M
Units:
G: [GPa]

7. 4. Bulk Modulus, K P Initial Volume = V0 Volume Change = DV P P savg =
σavg is the average of three stresses applied along three principal directions. D V V o K
Units:
K: [GPa]

8. Elastic Constants Stresses Strains s = E e t = G g savg = K D V o
Normal
Shear
Volumetric
Stresses
Strains

9. Example: Uniaxial Loading of a Prismatic Specimen
P=1000 kgf
9.9 cm
Before
10 cm
Determine E and n
10.4 cm
9.9 cm
After

10. 1000 P=1000kgf → σ= = 10kgf/cm2 10*10 E= σ ε = 10 0.04 = 250 kgf/cm2
Δd/2=0.05cm
Δl/2=0.2cm E= σ ε = 10
0.04
= 250 kgf/cm2
10cm
εlong= Δl l0 =
=0.04
0.4 10
εlat= Δd d0 =
= -0.01
-0.1 10
10cm
ν = -
-0.01
0.04
= 0.25

11. For an isotropic material the stress-strain relations are as follows:

12. RELATION B/W K & E Consider a cube with a unit volumeσ
σ causes an elongation in the direction CD and contraction in the directions AB & BC. 1 σ D C B A
The new dimensions of the cube is :
CD direction is 1+ε
BC direction is 1-νε
AB direction is 1-νε

13. ε is small, ε2 & ε3 are smaller and can be neglected.
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)

14. Ξ + SΔV = 3ε (1-2ν) = ε (1-2ν) + ε (1-2ν) + ε (1-2ν) savg (σ+σ+σ)/3 σE
3 (1-2ν)
K = = = =
DV/V0
3ε (1-2ν)
3ε (1-2ν)
K = E
3 (1-2ν)

15. K = E
3 (1-2ν)
The relation between K and E is :
Moreover the relation between G and E is :
G = E
2 (1+ν)
The relation between G, E and K is : E 1 = + 9K 3G
Therefore, out of the four elastic constants only two of them are independent.

16. 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.

17. 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.

18. Elastic Constants of Some Materials
E(psi)x106 (GPa)
G(psi)x106 (GPa)
ν (-)
Cast Iron
0.17
Steel
0.26
Aluminum
0.33
Concrete
0.2
Wood
Long
Tang ?