재료의 기계적 거동

(Mechanical Behavior of Materials)

Lecture 7 – Elastic Behavior Heung Nam Han

Professor

Department of Materials Science & Engineering College of Engineering

Seoul National University Seoul 151-744, Korea

Tel : +82-2-880-9240 Fax : +82-2-885-9647

- Elasticity are extremely important because engineering design is done in the elastic region.

- Material fracture is related to elastic properties because the elastic energy release is one of driving force for

fracture.

- Elastic behavior is inherently anisotropic for individual grains. However, most polycrystalline materials are

elastically isotropic. Polycrystalline materials can be anisotropic if they are textured.

Elasticity

Basis for linear elasticity



Consider two atoms

F

is a force that should be applied to separate the atom from r

position ; external force F

F = 𝑭_𝒂𝒕𝒕+𝑭_𝒓𝒆𝒑

𝑭_𝒓𝒆𝒑

𝑭_𝒆𝒙𝒕 r

-𝝏𝑼/𝝏𝒓

𝒓_𝟎 Δr



Consider cubic crystal material



Slope : modulus

Potential energy increase

Basis for linear elasticity

(Young’s modulus)

F = 𝑭_𝒂𝒕𝒕+𝑭_𝒓𝒆𝒑

𝑭_𝒆𝒙𝒕 r

F = -𝝏𝑼/𝝏𝑹

𝒓_𝟎 Δr

a [100]

[010]

[001]

Applied force: increase atomic distance

decrease atomic distance

decrease atomic distance

𝑼_𝟎

U

r

𝒓_𝟎

0

Basis for linear elasticity

(Bulk modulus)



Relate elastic modulus to volume change

U area

F  



 









 





 









 

 U





Bulk modulus

2

U

K  

  



 

 

  







U ^Ω𝟎 Ω _atomic

volume 2Ω_𝟎

Ω

σ= (−𝝏𝑼 𝝏Ω)

Ω_𝟎

2Ω_𝟎 σ_𝒂𝒑𝒑 Ω_𝟎+ 𝚫𝛀

Basis for linear elasticity

(Temperature effect)



Bulk (Young’s) moduli relates to



Curvature of bonding energy



Bonding energy correlates with the melting temperature



Temperature (heat) increases atomic vibration



Thermal energy added



Potential increased



Curvature of bonding energy decreases

kT

U 

 kT 

E

Basis for linear elasticity

(anisotropy)



The forces between atoms, molecules, or ions in crystals depends on the distances between them. Thus, they also vary with crystallographic direction so it should

not be surprising that crystalline moduli

are anisotropic.

Hooke's Law in One Dimension

Hooke's Law in Three Dimensions

kl ijkl

ij

C 

  and 

 ^S



kl ijkl

ij

S 

  ₌ 

 ^S



Hooke's Law in Three Dimensions

Hooke's Law in Three Dimensions

















3 4

5

4 2

6

5 6

1



































33 32

31

23 22

21

13 12

12



















=































3 4

5

4 2

6

5 6

1

2 / 2

/

2 / 2

/

2 / 2

/



= C





= C





= S





= S



Hooke's Law in Three Dimensions

Elastic Strain Energy

   

.

Elastic Strain Energy

If the straining is carried out isothermally and reversibly, the energy expended is equal to the change in free energy (d) of the body.

Since the free energy is a state property, this is a perfect differential and the order of differentiation is immaterial.

The matrix array of the components of stiffness is symmetrical. There can be no more than twenty-one independent components of stiffness.

φ = w = (1/2) C





= (1/2) 



= (1/2) S





ij ji

C  C

Effect of Materials Symmetry on Elastic Constants (Cubic System)

If the crystal is rotated through π/2 about a fourfold axis,

Effect of Materials Symmetry on

Elastic Constants (Cubic System)

Effect of Materials Symmetry on Elastic Constants (Isotropic System)

Obviously, this includes cubic symmetry as a special case. Accordingly, let us

transform the stiffness tensor of cubic material for a rotation of  about x-axis,

A rotation of  about x-axis

x1 x₂ x₃ x1

x2

x3

0

0 0

0 1

 cos

sin  sin

cos













































0 0 0 ) 2 (

0 0 0 )

2 (

0 0 0 )

2 (

x y z

x ’ y ’ z ’

Effect of Materials Symmetry on Elastic Constants (Isotropic System)

We can determine the compliances simply by taking the inverse of the matrix of stiffness components,































) (

2 0

0 0

0 0

0 )

( 2 0

0 0 0

0 0

) (

2 0 0 0

0 0

0

0 0

0

0 0

0

12 11 12

11 12

11 11

12 12

12 11 12

12 12 11

S S S

S S

S S

S S

S S S

S S S

) 2 3 S₁₁ (













 

) 2 3 ( S₁₂ 2











 

Suppose that an elastically isotropic sample is acted on solely uniaxial stress along x-axis,

Young's modulus, E=1/S₁₁ Poisson's ratio, =-S₁₂/S₁₁

Effect of Materials Symmetry on Elastic Constants (Isotropic System)

Suppose now that the sole applied stress is a shear stress ₄ ,

Shear modulus, G=

2(1 ) E





Let us consider the effect of a hydrostatic stress _m ,

Bulk modulus, B=

3(1 2 ) E





) 2 1 ( 3

) 1

( 2







  G

B

2 ) / ( 6

2 ) / ( 3



 

G B

G



Compressive load (uniaxial)

Positive Poisson's ratio Negative Poisson's ratio

Bottle stopper (Cork)

Negative Poisson's ratio

Reentrant honeycomb structure

α

Mechanism: Negative Poisson's ratio

Twisted kagome lattice Isotropic elasticity

with vanishing bulk modulus (no deformation on lattice)

Regarding photonic feature, (a) Periodic structure

(b) Small applied force - Huge conformation change : Negative Poisson's ratio

: Extremely low bulk modulus

Twisted kagome lattice

Isotropy considerations

For these systems, anisotropy is defined by the Zener ratio:

When the Zener ratio = 1, the material is isotropic.

11 12

44 2

C C

C 

 S₄₄  2(S₁₁  S₁₂)

Elastic Moduli in Cubic Materials

We can use the different relations among elastic constants to ascertain elastic moduli along any orientation,

where 𝑙

, 𝑙

, 𝑙

equal the direction cosines between the [ijk]

direction and the [100], [010], and [001] directions.

(i.e., axes x, y, and z)