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ELASTIC CONSTANTS IN 
ISOTROPIC MATERIALS 
1. Elasticity Modulus (E) 
2. Poisson’s Ratio (n) 
3. Shear Modulus (G) 
4. Bulk Modulus (K) 
ebooks.edhole.com
1. Modulus of Elasticity, E 
(Young’s Modulus) 
s = E e 
s 
Linear-elastic 
E 
e 
F 
F 
simple 
tension 
test 
Units: 
E: [GPa] ebooks.edhole.com
Slope of stress strain plot (which is 
proportional to the elastic modulus) 
depends on bond strength of metal 
Adapted from Fig. 6.7, 
Callister 7e. 
E= 
ebooks.edhole.com
2. Poisson's ratio, n 
Units: 
n: dimensionless 
eT 
eL 
n 
F 
F 
simple 
tension 
test 
 “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 
ebooks.edhole.com
 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 
ebooks.edhole.com
3. Shear Modulus, G 
t 
G 
g 
t = G g 
M 
simple 
torsion 
test 
M 
Units: 
G: [GPa] ebooks.edhole.com
4. Bulk Modulus, K 
savg = K 
DV 
Vo savg 
DV 
K 
Vo 
P 
P 
P 
Initial Volume = V0 
Volume Change = DV 
Units: 
K: [GPa] 
σavg is the average of 
three stresses applied 
along three principal 
directions. 
ebooks.edhole.com
Elastic Constants 
s = E e 
t = G g 
savg = K 
DV 
Vo 
Stresses Strains 
Normal 
Shear 
Volumetric 
ebooks.edhole.com
Example: 
Uniaxial Loading of a Prismatic Specimen 
9.9 cm 
Before After 
10 cm 
10 cm 
10 cm 
10.4 cm 
9.9 cm 
Determine 
E and n 
P=1000 kgf 
ebooks.edhole.com
P=1000 kgf 
10cm 
10cm 
Δl/2=0.2cm 
Δd/2=0.05cm 
1000 kgf 
P=1000kgf → σ= 
1000 
10*10 
= 10kgf/cm2 
E= 
σ 
ε = 
10 
0.04 
= 250 kgf/cm2 
εlong= 
Δl 
l0 
0.4 
10 
= =0.04 
εlat= 
Δd 
d0 
-0.1 
10 
= = -0.01 
ν = - 
-0.01 
0.04 
= 0.25 
ebooks.edhole.com
For an isotropic material the stress-strain 
relations are as follows: 
ebooks.edhole.com
RELATION B/W K & E 
 Consider a cube with a unit volume 
σ 
1 
1 
1 
σ 
D 
C 
A B 
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 
ebooks.edhole.com
 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 : 
ebooks.ΔedVh=ole3.cεom(1-2ν)
σ 
σ 
σ 
σ 
σ 
σ 
Ξ + + 
SΔV = 3ε (1-2ν) = ε (1-2ν) + ε (1-2ν) + ε (1-2ν) 
K = 
(σ+σ+σ)/3 σ 
= 
3ε (1-2ν) 
= 
3ε (1-2ν) 
E 
3 (1-2ν) 
K = 
= 
E 
savg 
DV/V0 
3 (1-2ν) 
ebooks.edhole.com
 Moreover the relation 
between G and E is : 
G = 
E 
2 (1+ν) 
 The relation between 
G, E and K is : 
1 1 
E 
= 
1 
+ 
9K 3G 
K = 
E 
3 (1-2ν) 
The relation between 
K and E is : 
Therefore, out of the four elastic 
constants only two of them are 
indepeebnoodkesn.etd.hole.com
 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. 
ebooks.edhole.com
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. 
ebooks.edhole.com
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 
Concrete 1.5-5.5 10-40 0.62-2.30 4-15 0.2 
Wood Long 1.81 12 
Tang 0.10 0.7 
0.11 0.7 
0.03 0.2 
? 
ebooks.edhole.com

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  • 2. ELASTIC CONSTANTS IN ISOTROPIC MATERIALS 1. Elasticity Modulus (E) 2. Poisson’s Ratio (n) 3. Shear Modulus (G) 4. Bulk Modulus (K) ebooks.edhole.com
  • 3. 1. Modulus of Elasticity, E (Young’s Modulus) s = E e s Linear-elastic E e F F simple tension test Units: E: [GPa] ebooks.edhole.com
  • 4. Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal Adapted from Fig. 6.7, Callister 7e. E= ebooks.edhole.com
  • 5. 2. Poisson's ratio, n Units: n: dimensionless eT eL n F F simple tension test  “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 ebooks.edhole.com
  • 6.  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 ebooks.edhole.com
  • 7. 3. Shear Modulus, G t G g t = G g M simple torsion test M Units: G: [GPa] ebooks.edhole.com
  • 8. 4. Bulk Modulus, K savg = K DV Vo savg DV K Vo P P P Initial Volume = V0 Volume Change = DV Units: K: [GPa] σavg is the average of three stresses applied along three principal directions. ebooks.edhole.com
  • 9. Elastic Constants s = E e t = G g savg = K DV Vo Stresses Strains Normal Shear Volumetric ebooks.edhole.com
  • 10. Example: Uniaxial Loading of a Prismatic Specimen 9.9 cm Before After 10 cm 10 cm 10 cm 10.4 cm 9.9 cm Determine E and n P=1000 kgf ebooks.edhole.com
  • 11. P=1000 kgf 10cm 10cm Δl/2=0.2cm Δd/2=0.05cm 1000 kgf P=1000kgf → σ= 1000 10*10 = 10kgf/cm2 E= σ ε = 10 0.04 = 250 kgf/cm2 εlong= Δl l0 0.4 10 = =0.04 εlat= Δd d0 -0.1 10 = = -0.01 ν = - -0.01 0.04 = 0.25 ebooks.edhole.com
  • 12. For an isotropic material the stress-strain relations are as follows: ebooks.edhole.com
  • 13. RELATION B/W K & E  Consider a cube with a unit volume σ 1 1 1 σ D C A B 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 ebooks.edhole.com
  • 14.  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 : ebooks.ΔedVh=ole3.cεom(1-2ν)
  • 15. σ σ σ σ σ σ Ξ + + SΔV = 3ε (1-2ν) = ε (1-2ν) + ε (1-2ν) + ε (1-2ν) K = (σ+σ+σ)/3 σ = 3ε (1-2ν) = 3ε (1-2ν) E 3 (1-2ν) K = = E savg DV/V0 3 (1-2ν) ebooks.edhole.com
  • 16.  Moreover the relation between G and E is : G = E 2 (1+ν)  The relation between G, E and K is : 1 1 E = 1 + 9K 3G K = E 3 (1-2ν) The relation between K and E is : Therefore, out of the four elastic constants only two of them are indepeebnoodkesn.etd.hole.com
  • 17.  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. ebooks.edhole.com
  • 18. 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. ebooks.edhole.com
  • 19. 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 Concrete 1.5-5.5 10-40 0.62-2.30 4-15 0.2 Wood Long 1.81 12 Tang 0.10 0.7 0.11 0.7 0.03 0.2 ? ebooks.edhole.com