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Stress-Strain Relationships  Basics of Stress and Strain ME 445 Mechanical Properties 1

# Stress Strain PartI

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Stress-Strain Relationships

Basics of Stress and Strain

ME 445 Mechanical Properties 1 8/2/2019 Stress Strain PartI

General Stress State (3D)6 independent stress components

3 normal stresses:

3 shear stresses:

zzyyxx ,,

xzyzxy ,,

yx xy    yy

yxyz

Symmetry:  zx xz

y xx

xy

zy

xz xy xx

x

zz

zz yz xz

yz yy xy

ME 445 Mechanical Properties 2 8/2/2019 Stress Strain PartI

Strain in Three Dimensions

• Strain is derived from displacements•

dyvv,dyuu

y

dy

x

(u,v)

dxx

v,dxx

u

ME 445 Mechanical Properties 3 8/2/2019 Stress Strain PartI

Strain in Three Dimensions

• Normal strains are due to change in dimensions (dilatation)

uudxx

u

u

xdx

xx

v

y

v

dy

vyyvyy

w Similarly (in 3D):

z

w

dzz

zz

ME 445 Mechanical Properties 4 8/2/2019 Stress Strain PartI

Strain in Three Dimensions

• Shear strains arise from changes in angles (rotation)

vdy

y11

x

22yutan

2y

Total change in angle:

vu dx

x1dx

x x

xy x y

2

21

ME 445 Mechanical Properties 5 8/2/2019 Stress Strain PartI

Strain in Three Dimensions

• Similar to stress, strain is also a symmetric second-order tensor.

Six inde endent strain com onents ε  ε  ε

γxy/2, γyz/2, γzx/2

• Need a constitutive law relating stress to strain

xxxx

zz

yy

zz

yy

matrix6x6

xz

yz

xy

xz

yz

xy

36 constants in general form!

ME 445 Mechanical Properties 6 8/2/2019 Stress Strain PartI

General Stress-Strain Relationships• In case of uniaxial stress (only σx component as shown in figure a),the stress element extends in the longitudinal (x) direction and

.produces the following normal strains for homogeneous, isotropic

material:

longitudinal strain

x x

xv

xv E

lateral strain

x

E  x

E vv x

x y

x

x

E

x

v  E

x

E  x z

xv E

(a) (b)

ME 445 Mechanical Properties 77 8/2/2019 Stress Strain PartI

General Stress-Strain Relationships• For an element subjected to triaxial stress (σx , σy , and σz), the totalstrain in x direction will be due to longitudinal strain produced by σx

an a era s ra ns pro uce y σy an σz .

• Using the principal of superposition, the total strain in x-direction is:  y

• Similarly, total strains in y- and z-direction are:

z y x z x

x v E  E

v E

v E

x

z

z x y y v E

1  x

y

• In the case of shear strain, there is no lateral strain, hence the τ-

y x z z v E

y

z

xz

relationship is the same for both uniaxial and complex strain systems.

xy

xy

yz

yz

zx

zx

Where:

EG

ME 445 Mechanical Properties 8 8/2/2019 Stress Strain PartI

General Stress-Strain Relationshi s

’

zzyyxxxxE

G

xyxy

zzxxyyyy

1

xzxz

xxyyzzzz

E

Gzy

ME 445 Mechanical Properties 9 8/2/2019 Stress Strain PartI

General Stress-Strain Relationshi s

Generalized Hooke’s Law

This simplified constitutive law for homogeneous,

isotro ic s stems can be written in a tensor form

xxxx000

EEE1

zz

yy

zz

yy

000E

1EE

000EEE

yz

xy

yz

xy

010000

00G

1000

xzxzG

100000

ME 445 Mechanical Properties 10 8/2/2019 Stress Strain PartI

Example: Generalized Hooke’s Law• A material subjected to a compressive stress σz isconfined so that it cannot deform in y-dir, butdeformation is ermitted in the x-dir.• Assuming, linear-elastic behavior, determine thefollowing in terms of σz and elastic constants E & v :

. -

b.The strain in the z-direction

c.The strain in the x-direction

d.The stiffness E ́= σz /εz in the z-direction.

Solution: (Plain Strain Conditions)Given: No deformation in y-dir: εy = 0 , Free to deform in x-dir: σx = 0

a) Use Hooke’s law for 3D case. Write equation for εy

z x y y v

E

v    01

0 z y v

ME 445 Mechanical Properties 1111

E 8/2/2019 Stress Strain PartI

Example: Generalized Hooke’s Law

Solution:b) The strain in the z-direction:

y x z z v

E

1

v21

c) The strain in the x-direction:

z z z vv E

0 z z E

z y x x v E

1

1 vv 1

d) The apparent stiffness in z-direction:

z z x vv E

z x E

- Using equation for εz (E ́ is larger than E )

΄

21 v

E  E

z

z

ME 445 Mechanical Properties 1212

- or v = . = . 8/2/2019 Stress Strain PartI

Volumetric Strain and Hydrostatic Stress

• Hydrostatic stress σh refers to a normal (tensile or compressive) stress

applied equally (internally or externally) to a body in all dimensions.

• Defined as the average normal stress:

• H drostatic stress results in chan e in volume of the material

3

z y x

h

associated with normal strains

• Note: shear strains case no change in volume, or distortion).

, , . , ,

increase in length in all (x, y, z) directions. Then:

Original volume = V = L·W·H

New volume = V1= (W + dW)(L + dL)(H + dH)

Neglecting products of small quantities:

V1 = WLH + WH dL + L H dW + LW dH= V + WH dL + L H dW + LW dH

ME 445 Mechanical Properties 131313

, 1 - 8/2/2019 Stress Strain PartI

Volumetric Strain and Hydrostatic Stress

• Divide dV by V (where V=WLH):

dH dW dLdV

• The three terms on the right hand side are

the normal strains in x, y, and z directions:

L

dL x

dH

W  y

• A volumetric strainε

v is defined as:

H  z

(sum of normal strains) z y xv

dV

ME 445 Mechanical Properties 1414 8/2/2019 Stress Strain PartI

Volumetric Strain and Hydrostatic Stress

• Using the normal strain expressions defined by generalized Hooke’s

law, the volumetric strain can be expressed in terms of stresses:

21 v

• Recall the definition of hydrostatic stress:

z y xv E

3

z y x

h

,

hv

E

v

213

• One can then define a material constant known as bulk modulus B :

E

Bh

• From the above equation, if v = 0.5, then value of B becomes infinitely

large. Hence, the body (material) is said to be incompressible.

v

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