mechanical behavior notes-2009c [recovered] · 2018-10-08 · mohr’s circle for strain reading...

Module #4Module #4

Fundamentals of strainThe strain deviator

Mohr’s circle for strain

READING LISTDIETER: Ch. 2, Pages 38-46

Pages 11-12 in HosfordCh. 6 in Nye

HOMEWORKFrom Dieter

2-7

StrainStrain• When a solid is subjected to a load, parts of the solid are

displaced from their original positions.• Think of it like this; the atoms making up the solid are displaced

from their original positions.

• This displacement of points or particles under an applied stressis termed strain.

OO´

Load

Load

AA´

BB´ C

C´

Rotation

x

y

v

-v

Pure Shear

x

y

v

u

+

TotalDisplacement

= x

y

x

y

Translation

+

Where to beginWhere to begin• Consider a point A in a solid located at position x,y,z.

• Apply force to the body and point A (x,y,z) is displaced to A′ (x+u,y+v,z+w).

A (x,y,z)

x

y

z

A′ (x+u,y+v,z+w)

u

v

w

uA

Displacement of pointsDisplacement of points• Displacement vector:

uA = f(u,v,w)where u, v, and w are units of translation along the x, y, zaxes.

• Solids are composed of many particles.

• If uA is constant for all particles, no deformation occurs (only translation).

• If uA varies from particle to particle, i.e., ui = f(xi), the solid deforms.

Displacement = Translation + Rotation + Shear

uAA′ (x+u,y+v,z+w)

x

y

z

A (x,y,z)

11--D Linear StrainD Linear Strain

• Points A and B are displaced from their original positions

• The amount of displacement is a function of x. Point B moves farther than Point A.• Let distance A A = u.

• Thus, distance B B = u+(Δu/Δx)dx = u+(u/x)dx.

A B

A´ B´F

dx

0 1 2 3 4 5

x

• Strain is defined by the following relationship:

• Integrating yields the displacement.

• uo ≈ rigid body translation, which we can subtract, yielding:xxu e x

11--D Linear Strain D Linear Strain –– contcont’’d.d.

xx

udx dx dxL A B AB u uxeL AB dx x x

o xxu u e x

A B

A´ B´F

dx

x

Generalization to 3Generalization to 3--DD• Displacement is related to the initial coordinates of the

point.

• Normal / Linear Strains:

xy xz

yx yz

xx

yy

zzzx zy

i ij j

u x e y e z

e x y e z

e x e y z

u e

e

e

e

xv

w

, ,

, ,

xx yy zz

xx yy zz

u v we e ex y zu v we e ex y z

A (x,y,z)A′ (x+u,y+v,z+w)

x

y

z

uA

If we orient the system such that the load is applied parallel to the x-axis. The variables u, v, and w are displacements parallel to the x, y, and z axes.

u

v

w

Shear Strains in 2Shear Strains in 2--D and 3D and 3--DD• Consider a square or cubic element that is distorted by shear.

Incremental displacement in x-direction = u.

Incremental displacement in y-direction = v.

Incremental displacement in z-direction = w.

x

y

A

D

B

C

C´

D´

B´

u

v

• Displacement of AD increases with distance along the y-axis resulting in an angular distortion of y-axis.

2D2D

shear distortion of they-axis in the x-direc

or

n

tio

xyu uDD

h DA y ye

shear distortion of thex-axis in the y-direc

or

n

tio

yxv vBB

h AB x xe

x

y

A

D

B

C

C´

D´

B´

u

v

• An analogous event occurs along the x-axis.

Strain in 3Strain in 3--DD• The displacement strain is defined by nine strain components:

– exx, exy, exz, eyy, eyx, eyz, ezz, ezx, ezy

– The strains on the negative faces are equal to satisfy the requirements for equilibrium.

• Notation is similar to stress; subscripts reversed:

– eij: i = direction of displacementj = plane on which strain acts

• Convention– (+)ive when both i & j are (+)ive– (+)ive when both i & j are (-)ive– (-)ive when both i & j are opposite

• Tension: eij = positive• Compression: eij = negative

x

y

z

exx

eyx

ezx

ezz

eyzexz

eyy

exy

ezy

3D Displacement Strain Matrix

xx xy xz

ij yx yy yz

zx zy zz

u u u u u ux y z x y ze e e

e e e ex y z x y z

e e e

x y z x

v v v v v v

w w w w w wy z

• The displacement strain matrix.

• Can produce pure shear strain and rigid-body rotation.

(2)Rotation

x

y

-v

u

exy = -eyx

(1)Pure Shear

w/o Rotation

x

y

v

u

exy = eyx

(3)Simple Shear

x

yu

exy = u/y eyx = 0

• We need to break the displacement matrix into strain and rotational components.

• We can decompose the total strain matrix into symmetric and anti-symmetric components.

y

x

Decomposition of StrainDecomposition of Strain

12

12

12

12

ij

ij ji

ji

j i

iij j

ij ji

ji

j i

e e

uu

e e

x

e

xuuxx

Symmetric

Shear

Anti-symmetric

Rotation

Displacement strain[matrix]

Shear strain[tensor]

Rotation[tensor]

ij

1 12 2

1 1ε2 21 12 2

xx xy yx xz zx

xx xy xz

yx yy yz xy yx yy yz zy

zx zy zz

xz zx yz zy zz

e e e e e

e e e e e

e e e e e

ij

1 102 2

1 102 21 1 02 2

xy yx xz zx

xx xy xz

xy yy yz yx xy yz zy

xz yz zz

zx xz zy yz

e e e e

e e e e

e e e e

xx xy xz

yx yy yz

zx zy zz

e e ee e ee e e

=

+

=

+

Shear StrainShear Strain• Total angular change from a right angle.

(1)Pure Shear

w/o Rotation

x

y

v

u

exy = eyx

2 ( 0)

2 (engineering shear strain)

xy yx xy ij

ij ij

xy

xz

yz

e e

u vy xw ux zw vy z

Transformation of StrainsTransformation of Strains• Equations for strain, analogous to those for stress, can be

written by substituting for and /2 for .

• We can also define a coordinate system where there will be no shear strains. These will be principal axes

2 2 2

2 2

normal

normal2

2 2 2xx yy zz xy yz zx

xx yy zz xy yz zx

l m n lm mn nl

l m n lm mn nl

3 2 2 2 2

2 2 2

1 33 2

2

1- - - -4

1 1- - 04 4

or- - 0

xx yy yy zz xx zz xy yz xz

xx yy zz xy yz xz xx

xx yy zz

yz yy xz zz xy

II I

• The directions in which the principal strains act are determined by substituting 1, 2, and 3, each for in:

and then solving the resulting equations simultaneously for l, m, and n(using the relationship l2 + m2 + n2 = 1).

(a) Substitute 1 for in & solve; (b) Substitute 2 for in & solve; (c) Substitute 3 for in & solve.

( - )2 0( - )2 0

( - )2 0

xx yx zx

xy yy zy

xz yz zz

l m nl m nl m n

Equations for Principal Shearing StrainsEquations for Principal Shearing Strains

• Deformation of a solid involves a combination of volumechange and shape change.

• We can separate strain into hydrostatic (volume change) and deviatoric (shape change) components.

1 2 3

max 2 1 3

3 1 2

xy

z

dxdy

dz

Hydrostatic ComponentHydrostatic Component

• The volume strain is:

• If we neglect the products of strains (i.e., εii×εjj), this becomes:

which is equal to the first invariant of the strain tensor

• Volume = dxdydz

• Volume of strained element = 1 1 1xx yy zz dxdydz

1 1 1

1 1 1 1

xx yy zz

xx yy zz

dxdydz dxdydzdxdydz

xx yy zz

• The hydrostatic component of strain, i.e., the mean strain, is:

The mean strain does not induce shape change. It causes volume change. It is the hydrostatic component.

• The part that causes shape change is called the strain deviator.We get the strain deviator by subtracting the mean strain from the normal strain components.

3 3 3xx yy zz ii

mean

xx mean xy xz

ij yx yy mean yz

zx zy zz mean

23

23

23

xx mean xy xz

ij yx yy mean yz

zx zy zz mean

xx yy zzxy xz

yy zz xxyx yz

zz xx yyzx zy

The Strain DeviatorThe Strain Deviator

3 3ij ij m ij ij ij

MohrMohr’’s Circle for Strains Circle for Strain

Intersection with the -axis is min=2

Intersection withthe -axis ismax = 1

ε

max/2

C(average, 0)

+γ/2

2

CW

CCW

V(xx, -xy/2)

H(yy, yx/2)

(1, )(2, )

Allows us to determine the magnitude and directions of the principal strains.

tan 2

tan 2

xynormal

xx yy

xx yyshear

xy

MAXIMUM & MINUMUM PRINCIPAL STRAINS IN 2-D STATE

2 2max 1

min 2 2 2 2xx yy xx yy xy

2 2max 3 xx yy xy

Strain MeasurementStrain Measurement• Strain can be measured using a strain gauge.

• When an object is deformed, the wires in the strain gate are strained which changes their electrical resistance, which is proportional to strain.

• Strain gauges make only direct readings of linear strain. Shear strain must be determined indirectly.

45°

45°

60° 60°

60°

Rectangular Delta

y

x

45

x

60 120

State of stress at a point:

State of strain at a point:

There are many different systems of notation.BE WARY!

11 12 13

21 22 23

31 32 33

11 12 13

12 22 23

13 23 33

xx xy xz

yx yy yz

zx zy zz

11 12 13 11 12 13

21 22 23 12 22 23

31 32 33 13 23 33

Matrix NotationMatrix Notation

• We often replace the indices with matrix notation for simplicity

• This will be particularly important when we discuss higher order tensors and tensor relationships (i.e., elastic properties)

11 1 11 2 33 323 4 13 5 12 6

11 12 13 1 6 522 23 2 4

33 3

xx yy zzyz xz xy

General forms for stress and strain in matrix notationGeneral forms for stress and strain in matrix notation

1 6 5

6 2 4

5 4 3

6 51

6 42

5 53

2 2

2 2

2 2

1 11 2 22 3 33

4 23 23

5 13 13

6 12 12

NOTE; ;

222

Special definitions