02_Stress and Strain Relationships for Elastic Behaviour

Stress and strain relationships for elastic behaviour

Chapter 2

Subjects of interest • Introduction/Objectives • Description of stress at a point • State of stresses in two dimensions (Plane stress) stress) • Mohr’ Mohr’s s circle of of stress stress – two dimensio dimensions ns • State of stress in three dimensions • Strain at a point • Hydrostatic and deviator components of stress • Elastic stress-strain relations • Strain energy • Stress concentratio concentration n • Finite element method Suranaree University of Technology

Tapan Tapany y Udomp Udompho holl

May-Aug 2007

Objectives

• This chapter chapter provides mathema mathematical tical relationships relationships to understand relationships relationships between stress and strain in a solid which obeys Hook’s law. • Stress and strain at a point and in three dimensions will be understood.

Suranaree University of Technology


May-Aug 2007

Objectives

• This chapter chapter provides mathema mathematical tical relationships relationships to understand relationships relationships between stress and strain in a solid which obeys Hook’s law. • Stress and strain at a point and in three dimensions will be understood.



May-Aug 2007

Introduction Force

Elastic behaviour Solid Force

Stress-strain relationship

Plastic behaviour

Two dimensional Three dimensional

Note: if solid is metal, we need to include metallurgical factors. Suranaree University of Technology


May-Aug 2007

Description of stress at a point • Stress at a point is resolved into normal and shear components. • Shear components are at arbitrary angles to the coordinate axes. • The normal stress σ σ x acting on the plane perpendicular to the x the x direction. direction. (this also applies to σ σy and σ σz .)

z σ σz τ τzx

τ τzy

τ τyx τ τ xz

σ σy τ τyz

τ τyx τ τ xy

σ σ x

τ τyz ∆z ∆z σ σy y ∆ x

∆y

x Stress acting on an element cube.


• The shearing stress has two components component s and need two subscripts; - The first subscript is subscript is the plane in which the stress acts. - The second subscript is subscript is the direction in which the stress acts. Ex: τ τyz is the shear stress in the plane perpendicular to the y y axis axis in the z direction. Tapan Tapany y Udomp Udompho holl

May-Aug 2007

Sign convention for shear stress • A shear stress is is positive positive if it points in the positive the positive direction on the positive the positive face of a unit cube. (and negative direction on the negative face). • A shear stress is negative if it points in the negative direction of a positive face of a unit cube. (and positive direction on the negative face). +y

-x

+y

+x

-y (a) Positive shear stresses. Suranaree University of Technology

-x

+x

-y (b) Negative shear stresses. Tapan Tapany y Udomp Udompho holl

May-Aug 2007

Stress components • In order to establish state of stress at a point, nine quantities must be defined; σ σ x , σ σy , σ σz , τ τ xy , τ τ xz , τ τyx , τ τyx , τ τzx and τ τzy . • If stress are slowly varying across the infinitesimal cube, moment equilibrium about the centroid of the cube requires that

τ xy

= τ yx

, τ xz

= τ zx

, τ yz

= τ zy

…Eq. 1

• Nine stress components can now reduce to six independent quantities σ σ x , σ σy , σ σz , τ τ xy , τ τ xz , and τ τzy , which can be written as

σ ij


 σ x  = τ xy τ  xz

τ xy σ y τ yz

τ xz 

 τ yz   σ z 

Tapany Udomphol

…Eq. 2

May-Aug 2007

State of stress in two dimensions (Plane stress) Definition: Plane stress is a stress condition in which the stresses are zero in one of the primary directions. • In a thin plate where load will be on the plane of the plate and there will be no stress acting perpendicular to the surface of the plate. • The stress system consists of two normal stresses σ σ x and σ σy and a shear stress τ τ xy .


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Stress on oblique plane • Consider an oblique plane normal to the plane of the paper crossing x and y axis. • The direction x’ is normal to the oblique plane and y’ direction is lying in the oblique plane. • The normal stress σ σ and shear stress ττ are acting on this plane and A is the area on the oblique plane. • S x and S y denote the x and y components of the total stress acting on the inclined face. The direction cosines between x’ and the x and y axes are l and m, hence l = cosθ θ and m = sinθ θ. Summation of the forces x direction

S x A = σ x Al + τ xy Am

y direction

S y A = σ y Am + τ xy Al

or

S x S y

Al = area Am = area

to x to y

= σ x cos θ + τ xy sin θ = σ y sin θ + τ xy cos θ Stress on oblique plane (two dimensions)


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May-Aug 2007

Stress on oblique plane • The components of S x and S y in the direction of the normal stress σ σ are • The normal stress acting on the oblique plane is given by

S xN

= S x cos θ

and S yn

= S y sin θ

σ x '

= S x cos θ + S y sin θ

σ x '

= σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ …Eq. 3

• The shearing stress on the oblique plane is given by

τ x ' y '

= S y cosθ − S x sin θ

τ x ' y '

= τ xy (cos 2 θ − sin 2 θ ) + (σ y − σ x )sin θ cos θ

• The stress σ σy’ can be found by θ substituting θ + π π /2 for θ, we then have

…Eq. 4

…Eq. 5

σ y '

= σ x sin 2 θ + σ y cos 2 θ − 2τ xy sin θ cosθ If stresses in an xy coordinate system and the angle θ θ are known, we will get the stresses in any x’y’ coordinate. Suranaree University of Technology

Stress on oblique plane (two dimensions) May-Aug 2007

Stress on oblique plane • Equations 3-5 can be expressed in terms of double angle 2 θ θ. σ x ' σ y ' τ x ' y '

= = =

σ x

+ σ y

+

…Eq. 6

σ + σ σ + σ σ x+ σ − −σ = y y cos 2θ + τ sin 2θ 2 −2 cos 2θ − τ xy sin 2θ

…Eq. 7

x

σ x '

2

σ y

− σ y

cos 2θ + τ xy sin 2θ

2 σ x

σ x y

x

y

xy

σ y '

=

τ x ' y '

=

σ x

− σ xσ 2

2

y

+ σ y 2

−

2 − σ

σ x

y

2

cos 2θ − τ xy sin 2θ

− σ sin 2sinθ 2θ ++τ cos τ xycos 2θ 2θ x

2

xy

…Eq. 8

Note: σ σ x’ + σ σy’ = σ σ x + σ σy  Thus the sum of the normal stresses on two perpendicular plan is an invariant quantity and independent of orientation or angle θ θ. Eq. 3-8 describe the normal stress and shear stress on any plane through a point in a body subjected to a plane-stress situation. Suranaree University of Technology

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May-Aug 2007

Variation of normal stress and shear stress with θ for the biaxial - - plane stress situation. 1) When τ τ is zero give max and min values of σ σ. 2) The max and min values of σ σ and τ τ occur at angles which are 90 o apart. 3) The max τ τ occurs at an angle halfway between the max and min σ σ. σ and ττ occurs in 4) The variation of σ the form of a sine wave, with a period of θ θ = 180o.

The relationships are valid for any state of stress.


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May-Aug 2007

Principal stresses • When there is no shear stresses acting on the planes  giving the maximum normal stress acting on the planes. • These planes are called the principal planes, and stresses normal to these planes are the principal stresses σ σ1 , σ σ2 and σ σ3 which in general do not coincide with the cartesian- coordinate axes x, y, z . Directions of principal stresses are 1, 2 and 3. Biaxial-plane stress condition • Two principal stresses, σ σ1 and σ σ2 . Triaxial-plane strain condition • Three principal stresses, σ σ1 , σ σ2 and σ σ3 , where σ σ1 > σ σ2 > σ σ3 . Biaxial stresses Suranaree University of Technology

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Triaxial stresses May-Aug 2007

Maximum and minimum principal stresses in biaxial state of stress • On a principal plane there is no shear stress; thereby, τ τ x’y’ =0 From Eq. 5

τ xy cos 2 θ − sin 2 θ τ xy σ x

− σ y

tan 2θ =

=

+

σ y

− σ x )sin θ cos θ = 0

sin θ cos θ cos θ − sin θ 2

2

=

1 2

(sin 2θ ) cos 2θ

=

1 2

tan 2θ

2τ xy σ x

…Eq. 9

− σ y

2 θ θ ), Eq.9 has two roots, θ θ 2 • Since tan2 θ θ = tan( π π+ θ1 and θ = θ θ1 + nπ π /2 . These roots define two mutually perpendicular planes which are free from shear .

• The maximum and minimum principal stresses for biaxial state of stress are given by 12 2    σ max = σ 1  σ x + σ y  σ x − σ y   + τ xy2  ±  = σ min = σ 2  2  2   


Tapany Udomphol

…Eq. 10

May-Aug 2007

Directions of the maximum principal stress

• The largest principal stress σ σ1 will lie between the largest normal stress and the shear diagonal . σ x σ 1 , and if • (If there is no shear, σ =σ there is only shear the principal stress σ σ1 would exist along the shear diagonal. If both normal and shear stresses act on the element, then σ σ1 lies between the influence of these two effects.


σ σy τ τ xy σ σ x σ σ x > σ σy

Shear diagonal

σ σ1

Method of establishing direction of σ σ1

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May-Aug 2007

The maximum shear stress in biaxial stress σ σy

• The maximum shear stress can be found by differentiating Eq. 8 and set to zero.

d τ x ' y ' d θ

τ τ xy σ σ x

= (σ y − σ x )cos 2θ − 2τ xy sin 2θ = 0

tan 2θ s

=

σ y

− σ x

2τ xy

=−

σ x

σ σ x > σ σy

− σ y

2τ xy

…Eq. 11

Note: tan2 θs is the negative reciprocal of tan2 θ n. θ s and 2 θ This means that 2 θ θn are orthogonal and that θ θs and θ θn are separated in space by 45 o.


σ σ1

Shear diagonal

Method of establishing direction of σ σ1 .

• The magnitude of the maximum shear stress is given by τ max

 σ x − σ y  2 2   + τ xy  = ±   2  

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12

May-Aug 2007

Example: The state of stress is given by σ σ x = 25p and σ σy = 5p plus shearing stresses τ τ xy . On a plane at 45 o counterclockwise to the plane on which σ σ x acts on the state of stress is 50 MPa tension and 5 MPa shear. Determine the values of σ σ x , σ σy , τ τ xy . From Eq.6 σ x '

=

σ x

+ σ y 2

+

σ x

− σ y 2

25 p + 5 p

50 × 10 6

=

50 × 10 6

= 15 p + τ xy

2

+

cos 2θ + τ xy sin 2θ

25 p − 5 p 2

cos 90 o

+ τ xy sin 90 o

τ x ' y '

=

5 × 10 6

τ xy

= 50 × 10 6 − 15(−5 × 10 5 )

τ xy

= 57.5 MPa

Since σ σ x + σ σy = σ σ x’ + σ σy’

From Eq.8 σ y

σ y

= 25(−5 × 10 6 ) = −12.5 MPa = 5( p) = 2.5 MPa

σ x

− σ x

sin 2θ + τ xy cos 2θ

2  5 p − 25 p 

=  2 p = −5 × 10 6 Pa

σ y '

= σ x + σ y − σ x

σ y '

= −12.5 − 2.5 − 50 = −65 MPa

'

 sin 90 o + τ xy cos 90 o 


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May-Aug 2007

Mohr ’s circle of stress two dimensions O. Mohr used a graphical method to represent the state of stress at a point on an oblique plane through the point. σ x + σ y σ x − σ y From Eq. 6 = σ x − cos 2θ + τ xy sin 2θ and Eq. 8 2 2 '

τ x ' y '

=

σ y

− σ x 2

sin 2θ + τ xy cos 2θ

…Eq. 6 …Eq. 8

By squaring each of these equations and adding, we have 2

2

σ + σ y    σ − σ y  σ x − x  + τ x2 y =  x  + τ xy2 2    2  '

…Eq. 12

' '

Eq. 12 is in the equation of a circle of the form (x-h)2 + y 2 = r 2 . Therefore Mohr’s circle is a circle in σ σ x’ and τ τ x’y’ coordinates +σ which r = τ τmax and the centre is displaced ( σ σ x σy )/2 to the right of the origin. Suranaree University of Technology

Tapany Udomphol

May-Aug 2007

Mohr ’s circle Conventions • A shear stress causing a clockwise rotation about any point in the physical element is plotted above the horizontal axis of the Mohr’s circle. • A point on Mohr’s circle gives the magnitude and direction of the normal and shear stresses on any plane in the physical element. • Normal stresses are plotted along the x axis, shear stresses along the y axis. • The stresses on the planes normal to the x and y axes are plotted as points A and B.

B D

E

A

Mohr’s circle for two-dimensional state of stress Suranaree University of Technology

• The shear stress is zero at points D and E , representing the values of the principal stresses σ σ1 and σ σ2 respectively. • The Angle between σ σ x and σ σ1 on May-Aug Mohr’s circle is 2 θ θ.

2007

State of stress in three dimensions

• In general three dimensional state of stress consists of three unequal principal stresses acting at a point, which is called a triaxial state of stress. • If two of the three principal stresses are equal  cylindrical . • If σ σ1 = σ σ2 = σ σ3  hydrostatic or spherical .


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May-Aug 2007

Stress in three dimensions • Considering an elemental free body with diagonal plane JKL of area A, which is assumed to be a principal plane cutting through the unit cube.

K

σ is acting normal to • The principal stress σ the plane JKL. The direction cosines of σ σ and x , y and z axes is l , m and n respectively.

In equilibrium, the forces acting on each of its face must balance. S x , S y and S z are the components of σ σ along the axes. σ l S x = σ Area KOL =Al

S y = σ σm Area JOK = Am

σ n S z = σ Area JOL = An

σ σ

o

L

J

Stresses acting on elemental free body

Taking summation of the forces in the x direction results in

σ Al − σ x Al − τ yx Am − τ zx An = 0 Which reduces to

(σ − σ x )l − τ yx m − τ zx n = 0


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…Eq. 13 May-Aug 2007

(σ − σ x )l − τ yx m − τ zx n = 0

…Eq. 13(a)

K

Summation of the forces in the other two directions results in

σ σ

− τ xy l + (σ − σ y )m − τ zy n = 0 − τ xz l − τ yz m + (σ − σ z )n = 0 By setting the σ − σ x determinant of the coefficients of l , m and − τ xy − τ xz n=0

…Eq. 13(b) L

…Eq. 13(c)

− τ yx σ − σ y − τ yx

− τ zx − τ zy = 0 σ − σ z

J

Stresses acting on elemental free body

Will give the solution 3 2 2 2 2 ( ) − + + + + + − − − σ σ σ σ σ σ σ σ σ σ σ τ τ τ ( x y z x y y z x z xy yz xz )σ of the determinant which results in a − (σ xσ yσ z + 2τ xyτ yz τ xz − σ xτ yz 2 − σ yτ xz 2 − σ z τ xy2 ) = 0 cubic equation in σ σ I 1 = σ x + σ y + σ z And invariant coefficients

Note:

I 2

2 = σ xσ y + σ yσ z + σ xσ z − τ xy2 − τ yz − τ xz 2

I 3

2 = σ xσ yσ y + 2τ xyτ yz τ xz − σ xτ yz − σ yτ xz 2 − σ z τ xy2

σ x

+ σ y + σ z = σ x + σ y + σ z = σ 1 + σ 2 + σ 3


'

'

'

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May-Aug 2007

Determination of the normal stress on any oblique plane K • On any oblique plane whose normal has the direction cosines l, m, n with the x, y, z axes, The total stress on the plane S will not now be coaxial with the normal stress, and that S 2 = σ σ2 + τ τ2 .

σ σ L J

• The total stress can be resolved into components S x , S y , S z , so that

S 2

• Taking the summation of the forces in the x, y, z directions, the expressions for the orthogonal components of the total stress are given by;

S x

• The normal stress σ σ on the oblique plane;

σ = S x l + S y m + S z n

= S x2 + S y2 + S z 2

…Eq. 15

= σ x l + τ yx m +τ zxn S y = τ xy l + σ y m + τ zy n …Eq. 16 S z = τ xz l + τ yz m + σ z n

Substituting Eq and simplifying τ τ xy = τ τyx Etc.

σ = σ x l 2

+ σ y m 2 + σ z n 2 + 2τ xy lm + 2τ yz mn + 2τ zx nl


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The maximum or principal shear stress • Since plastic flow involves shearing stresses, it is important to identify the planes on which the maximum or principal shear stresses occur. • The principal shear planes can be defined in terms of the three principal axes 1, 2, 3.

τ 2

= (σ 1 − σ 2 )2 l 2 m 2 + (σ 1 − σ 3 )2 l 2 n 2 + (σ 2 − σ 3 )2 m 2 n 2

…Eq. 14

Where l , m and n are the direction cosines between the normal to the oblique plane and the principal axes. The principal shear stresses occur for the following combination of direction cosines that bisect the angle between two of the three principal axes:


l

m

± ±

1

±

0 1 2 1 2

n 2

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2 1

±

0

±

1

±

1 2

2 0

τ σ 2

τ 1

=

τ 2

=

σ 1

τ 3

=

σ 1

− σ 3 2

− σ 3 2

− σ 2 2 May-Aug 2007

The maximum or principal shear stress • Principal shear stresses for a cube whose faces are the principal planes. • For each pair of principal stresses, there are two planes of principal shear stress, which bisect the directions of the principal stresses. According to convention σ σ1 is the greatest principal normal stress and σ σ3 is the smallest principal stress, τ τ2 therefore has the largest value


τ 3

=

σ 1

− σ 2

τ max

2

τ 1

= τ 2 =

=

σ 2

σ 1

− σ 3 2

− σ 3 2

Planes of principal shear stresses.

• The maximum principal shear stress τ τmax is given by τ max Tapany Udomphol

=

σ 1

− σ 3 2


Stress tensor Stress tensor is used to simplify the equations for the transformation of the stress components from one set of coordinate axes to another coordinate system. • First, we consider the transformation of a vector (first-rank tensor) from one coordinate system to another.

x 3

S = S 1i 1 + S 2 i 2 + S 3i 3 when the unit vectors i 1 , i 2 , i 3 are in the direction x 1 , x 2 , x 3.

S 3

S 2

Where S 1 , S 2 , S 3 are the components of S referred to the axes x 1, x 2 , x 3. • The components of S referred to the x ’ 1 , x ’ 2 , x ’ 3 is obtained by resolving S 1 , S 2 , S 3 along the new direction x ’ 1.

S 1'

= S 1 cos( x1 x1' ) + S 2 cos( x 2 x1' ) + S 3 cos( x3 x1' )

Where

S ’ 1 x ’ 1

x 2

S ’ 2 S 1

x ’ 2

x 1

Transformation of axes for a vector.

= a11 S 1 + a12 S 2 + a13 S 3 S 2' = a 21 S 1 + a 22 S 2 + a 23 S 3 S 3' = a31 S 1 + a32 S 2 + a33 S 3

' or S 1

a11 is the direction cosine between x ’ 1 and x 1 a12 is the direction cosine between x ’ 1 and x 2 , etc.. Suranaree University of Technology

S

Tapany Udomphol

…Eq. 19

May-Aug 2007

• We could write the equations (from last slide) as 3

' 1

S

= ∑ a1 j S j

3

' 2

S

j =1

= ∑ a 2 j S j

3

' 3

S

j =1

∑a

3 j

S j

j =1

• These three equations could be combined by writing; 3

' i

S

= ∑ aij S j (i = 1,2,3) = ai1 S 1 + ai 2 S 2 + ai 3 S 3 j =1

…Eq. 20

• In greater brevity, the equation is obtained by writing in the Einstein suffix notation.

S i'

= aij S j

…Eq. 21

• The suffix notation (j) indicates summation when a suffix occurs twice in the same term. The summation will take place over j . • The transformation of the stress tensor σ σij from the x 1, x 2 , x 3 system of axes to the x’ 1 , x’ 2 , x’ 3 axes is given by

σ kl Suranaree University of Technology

= a ki alj σ ij Tapany Udomphol

May-Aug 2007

Rank of tensors • Scalars are tensors of rank zero, in which their quantity remains unchanged with the transformation of axes. • Vectors are tensors of first rank . • Physical quantities such as stress, strain and many other quantities are second-rank tensors which transform with coordinate axes.

The number of components required to specify a quantity is 3n, when n is the rank of the tensor.


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May-Aug 2007

Mohr ’s circle – three dimensions Mohr’s circle in three dimensions show how a triaxial state of stress is presented.

• All the possible stress conditions within the body fall within the shaded area between the circles. • The y axis is the shear stress τ τ and the x axis is the normal stress σ σ. • σ σ1 , σ σ2 , σ σ3 are shown on the x axis and τ τ1 , τ τ2 , τ τ3 are shown on the y axis.


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May-Aug 2007

Mohr ’ ’s circle for various states of stress • Mohr’s circle give a geometrical representation of the equations that express the transformation of stress components to different sets of axes. • The introduction of σ σ2 at a right angle to σ σ1 results in a reduction in the principal shear stress, but τ τ2 remains similar see fig (c). • The maximum shear stress is reduced appreciably when the third principal stress is introduced, see fig (d). Suranaree University of Technology

σ 1 = σ σ 2 σ 3 , • Hydrostatic tension σ =σ Mohr’s circle reduces to a point with no shear stress. Tapany Udomphol

May-Aug 2007

Mohr ’ ’s circle for various states of stress Biaxial and triaxial tension stresses effectively reduce the shear stresses and this results in a considerable decrease in ductility of the material, because plastic deformation is produced by shear stresses. Triaxial state of stress

Reduction in ductility Brittle fracture

Notch

Stress raiser

Uniaxial tension plus biaxial compression stresses produce high value of shear stress and contribute to an excellent opportunity to deform plastically without fracturing. Ex: forming by metal extrusion gives better ductility than simple uniaxial tension. Suranaree University of Technology

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May-Aug 2007

Description of strain at a point • Deformation of a solid may be made up of dilatation (change in volume), or distortion (change in shape). This results in displacement of points in a continuum body. y • Consider a solid body in fixed coordinates x , y, z with a displacement from point Q to Q’. • The components of displacement are u, v, w . • The displacement of Q is the vector uQ = f(u,v,w) • Displacement is a function of distance, ui = f(x ) i and for elastic solid and small displacement, ui is a linear function of x i .

Q Q ’ y

o

z

y+v x

x

z+w x+u

z

Displacement of point Q.

Note: in other materials the displacement may not be linear with distance, which leads to cumbersome mathematical relationships. Suranaree University of Technology

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May-Aug 2007

One- dimensional dimensional strain • Consider a simple one-dimensional strain, which has been deform from the original distance AB to A’B’. • Displacement u is in one dimension (as a function of x ).

dx

x A

• The normal strain is given by

B

u

∆ L A B − AB = = '

e x

=

L

'

B

dx +

∂u dx − dx ∂u ∂ x = ∂ x dx

u+ A’

∂u dx ∂ x

B’

dx +

P

∂u dx ∂ x

…Eq. 22

The displacement in one dimensional case is given by

u Suranaree University of Technology

= e x x Tapany Udomphol

One-dimensional strain

…Eq. 23

May-Aug 2007

Three - dimensional strain • In three dimensional strain, each of the component u = e xx x + e xy y + e xz z of the displacement will be linearly related to each of v = e yx x + e yy y + e yz z the three initial coordinates of the point. w = e zx x + e zy y + e zz z • Three coefficients for ∂u ∂v ∂w = = e e e xx or u i = eij x j the normal strains, ∂ x yy ∂ y zz ∂ z …Eq. 24

• Consider an angular distortion of an element in the xy plane by shearing stresses, we have y

∂u BB ' ∂v e xy = = = e yx = DA ∂ y AB ∂ x ∂u ∂u ∂u ∂ x ∂ y ∂ z e xx e xy e xz ∂v ∂v ∂v eij = e yx e yy e yz = ∂ x ∂ y ∂ z e zx e zy e zz ∂w ∂w ∂w ∂ x ∂ y ∂ z

C’

DD '


D

D’

C

u e xy

eyx A

B’

x

B

Angular distortion of an element

…Eq. 25

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May-Aug 2007

Strain tensor and rotation tensor • In general, displacement components such as e xy , eyx .. produce both shear strain and rigid-body rotation. • From tensor theory, any second-rank tensor can be decomposed into a symmetric tensor and an anisymmetric tensor. y

= (eij + e ji ) + eij = ε ij + ω ij eij

1 2

1 2

(eij

− e ji )

…Eq. 26

where

y

e xy = eyx

y

e xy = -eyx

x

e xy = γγ eyx = 0

x

x

(a) Pure shear (b) Pure rotation (c) Simple shear. without rotation. without shear.

∂u  1  ∂u =  i + i  Strain tensor 2  ∂ x j ∂ xi  ∂u  1  ∂u ω ij =  i − i  Rotation tensor 2  ∂ x j ∂ x i  ε ij


The general displacement equations

ui

Tapany Udomphol

= ε ij x j + ω ij x j


Shear strain The shear strain γ was defined as the total angular change from a right angle.

γ = e xy y

+ e yx = ε xy + ε yx = 2ε xy

And the definition of shear strain, γ γij = ε εij is called the engineering shear strain.

…Eq. 28

e xy = eyx

x

(a) Pure shear without rotation.

=

…Eq. 29

Since strain is a second-rank tensor , following the transformation of stress previously done, the strain tensor may be transformed one set of coordinate axes of coordinate axes to a new system of axes by

ε kl Suranaree University of Technology

∂u ∂v + ∂ y ∂ x ∂w ∂u γ xz = + ∂ x ∂ z ∂w ∂v + γ yz = ∂ y ∂ z γ xy

= a ki alj ε ij

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Principal shear strain τ Following Eq.17 , substituting ε ε for σ σ and γ γ /2 for τ. The normal strain on an oblique plane is given by ε = ε x l 2

+ ε y m 2 + ε z n 2 + γ xy lm + γ yz mn + γ xz nl

…Eq. 31

Similar to stress, the direction of the principal strains coincide with the principal stress directions. The three principal strains are the roots of the cubic equation. ε 3

− I 1ε 2 + I 2ε − I 3 = 0

…Eq. 32

Where

The maximum shearing strains can be obtained from


I 1

= ε x + ε y + ε z

I 2

= ε x ε y + ε y ε z + ε z ε x − 14 (γ xy2 + γ zx2 + γ yz 2 )

I 3

= ε x ε y ε z + 14 (ε x γ yz 2 + ε y γ zx2 + ε z γ xy2 ) = ε 2 − ε 3 = γ 2 = ε 1 − ε 3 γ 3 = ε 1 − ε 2

…Eq. 33

γ 1

γ max

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…Eq. 34

May-Aug 2007

Strain tensor Strain tensor can be divided into a hydrostatic or mean strain and a strain deviator . 1) Hydrostatic or mean strain (volume change) …Eq. 35 ε m

=

ε x

+ ε y + ε z 3

=

ε kk 3

=

∆ 3

Where ∆ is the volume strain (change in volume).

2) Strain deviator (shape change) We can do by subtracting ε εm from the normal strain components, thus ε x …Eq. 36 ε ' ij

=

− ε m ε yx ε zx

ε xy ε y

− ε m ε zy


ε xz ε yz ε z − ε m

These strains represent elongations or contractions along the principal axes that change the shape of the body at constant volume. Tapany Udomphol

May-Aug 2007

Strain measurement • Strain can be measured by using a bonded-wire resistance gauge or strain gauge. • When the body is deformed, the wires in the strain gauge are strained and their electrical resistance is altered. • The change in resistance, which is proportional to strain can therefore be determined. • Strain gauges can make only direct readings of linear strain, while shear strains must be determined indirectly. c b b

a

Rectangular Suranaree University of Technology

60 o

c

a

60 o

Typical strain gauge rosettes.

Delta Tapany Udomphol

May-Aug 2007

Hydrostatic and deviator components of stress Similar to strain tensor, the total stress tensor can be divided into 1) Hydrostatic or mean stress tensor , σ σm , which involves only pure tension or compression.  Produce elastic volume changes. σ m

=

σ kk 3

=

σ x

+ σ y + σ z 3

=

σ 1

+ σ 2 + σ 3 3

…Eq. 37

2) Deviator stress tensor σ σ’ ij . Which represents the shear stress in the total state of stress.  important in causing plastic deformation.


Tapany Udomphol

May-Aug 2007

Deviator of stress tensor Since the decomposition of the stress tensor is given by σ ij

= σ ij' + 13 δ ijσ kk

…Eq. 38

σ ’ ij Stress deviator involves shear stress. For example, referring σ to a system of principal axes. 2σ 1

− σ 2 − σ 3

− σ 2 ) + (σ 1 − σ 3 )

=

σ 1'

σ − σ  2 2  σ − σ =  1 2 + 1 3  = (τ 3 + τ 2 ) 3  2 2  3

3

=

(σ 1

σ 1'

3

…Eq. 39

Where τ τ3 and τ τ2 are principal shearing stresses.


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May-Aug 2007

Deviator of stress tensor • The principal values of the stress deviator are the roots of the cubic equation: (σ ' ) 3

− J 1 (σ ' ) 2 − J 2σ ' − J 3 = 0

Where J 1 , J 2 , J 3 are the invariants of the deviator stress tensor. J 1

= (σ x − σ m ) + (σ y − σ m ) + (σ z − σ m ) = 0

J 2

= τ xy2 + τ yz 2 + τ xz 2 − σ x' σ y' − σ y' σ z ' − σ x' σ z '

J 2

=

1 6

[(σ

x

− σ y ) 2 + (σ y − σ z ) 2 + (σ x − σ z ) 2 + 6(τ xy2 + τ yz 2 + τ xz 2 )]


Tapany Udomphol

May-Aug 2007

Elastic stress – strain relations The elastic stress is linearly related to elastic strain following Hooke’s law. Where E is the modulus of elasticity in tension or compression.

σ x

= E ε x

…Eq. 40

However, during linear extension, i.e., in x axis, the contraction in the transverse y and z direction causes a constant fraction of the strain in the longitudinal direction known as Poisson’s ratio v .

ε y

= ε z = −vε x = −

vσ x E

…Eq. 41

Note: for most metals v ~ 0.33 Suranaree University of Technology

Tapany Udomphol

May-Aug 2007

Strain energy

σ σ

When a material is deformed by an external loading, work done during elastic deformation is stored as elastic energy and will be recovered when the load is released.  Strain energy .

∆z ∆ x

∆y σ σ

Work is force x distance over it acts, therefore

P

U = 12 P δ U = 12 P δ

…Eq. 42

Where P/2 is the average force from zero. δ δ is the extension.

extension

For linear elastic (i.e., x axis) then Hooke’s law is applied ( σ σ = E ε ε ) U o


=

1 2

2

σ x ε x

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=

1 σ x

2 E

=

1 2

ε x2 E …Eq. 43 May-Aug 2007

Stress concentration • Discontinuity such as a hole or a notch results in non-uniform stress distribution at the vicinity of the discontinuity. stress concentration or stress raiser . • The distribution of the axial stress reaches a high value at the edges of the hole and drops of rapidly with distance away from the hole.

• The stress concentration is expressed by a theoretical stress-concentration factor K t .

K t

=

σ max σ nom

…Eq. 44

Stress distributions due to (a) circular hole and (b) elliptical hole. Suranaree University of Technology

Tapany Udomphol

May-Aug 2007

Stress concentration at a circular hole in a plate • For the circular hole in a plate subjected to an axial load, a radial stress is produced as well as a longitudinal stress. • From elastic analysis, the stresses can be expressed as: a4 σ  a 2  σ  σ r = 1 − 2  + 1 + 3 4 2  r  2  r

…Eq. 45

−4

a 2 

 cos 2θ

r 2 

σ  a 2  σ  a 4  σ θ = 1 + 2  − 1 + 3 4  cos 2θ 2  r  2  r  a σ  τ = − 1 − 3 4 2  r

4

a 2 

A

+ 2 2  sin 2θ r 

The maximum stress occurs at point A when θ r=a θ = π π /2 and σ θ

= 3σ = σ max

Stress distribution at a circular hole

…Eq. 46

The theoretical stress-concentration factor = 3 Suranaree University of Technology

Tapany Udomphol

May-Aug 2007

Stress concentration at an elliptical hole in a plate In the case of an elliptical hole in a plate, the maximum stress at the ends of the hole is given by the equation

σ max

a/b

a  = σ  1 + 2  b  

…Eq. 47

stress

There for a very sharp crack normal to the tensile direction will result in a very high stress concentration. Stress distribution at an elliptical hole


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May-Aug 2007

Stress concentrations for different geometrical shapes


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May-Aug 2007

Stress raiser in brittle and ductile materials • The effect of stress raiser is much more pronounced in brittle materials than in ductile materials.

Ductile materials • Plastic deformation occurs when the yield stress is exceeded at the point of maximum stress. • In ductile materials, further increase in load produce strain hardening (work hardening).  redistribution of stress  the materials will not develop the full theoretically stressconcentration factor.


Brittle materials • Stress redistribution will not occur in to any extent in brittle materials.  stress concentration of close to the theoretical value will result.

Tapany Udomphol

May-Aug 2007

Finite element method • Finite element method (FEM) is a very powerful technique for determining stresses and deflections in structures too complex to analyse by strictly analytical methods. • The structure is divided into a network of small elements connected to each other at node points. One node has one degree of freedom (see fig showing two and three dimensional element)

(a) Simple rectangular element

(b) Two elements joined to model a structure Some common elements used in FEM analysis


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May-Aug 2007

Finite element method

• A finite element solution involves calculating the stiffness matrices for every element in the structure. • A cumbersome part of the finite element solution is the preparation of the input data. Required data such as topology of the element mesh, node numbers, coordinates of the node points. Propeller designed by means of FEM analysis


Tapany Udomphol

May-Aug 2007

02_Stress and Strain Relationships for Elastic Behaviour

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