Mechanical Behavior Notes-2009d

Module Module #3#3

Transformation of stresses in 3-D

READING LISTDIETER: Ch. 2, pp. 27-36

Ch. 3 in RoeslerCh. 2 in McClintock and Argon

Ch. 7 in Edelglass

HOMEWORKFrom Dieter2-3, 2-4, 3-7

The Stress TensorThe Stress Tensor

x

y

z

xx

xy

xz

zz

zyzx

yy

yx

yz

xy

zO

xx xy xz

yy yzij yx

zx z zzy

In three dimensions the state of stress is described by the stress tensor.

We can transform from one coordinate system to another in the same way that we did for two dimensions.

• Lets resolve an arbitrary 3D state of stress onto an oblique plane ABC (area A).

• To make the problem easier, let S be parallel to the plane normal (meaning that it is a principal stress acting on a principal plane (i.e., the plane w/o shear).

MethodMethod

DIRECTION COSINESl = cos θxm = cos θyn = cos θz

x

y

z

xz

y

normal

yx

S =

x

y

z

A

C

Byy

yz

xy

xz

zy zxO

xx

zz

A

yx

S =

x

y

z

A

C

Byy

yz

xy

xz

zy zxO

xx

zz

yx

S =

x

y

z

A

C

Byy

yz

xy

xz

zy zxO

xx

zz

x

y

z

xz

y

normal

x

y

z

xz

y

normal

DIRECTION COSINESl = cos θxm = cos θyn = cos θz

L*

A

• The components of S parallel to the original x-, y-, and z-axes (i.e., Sx, Sy, Sz) are:

– Sx = Sl = σl Area COB = Al.– Sy = Sm = σm Area AOC = Am.– Sz = Sn = σn Area AOB = An.

2 2 2 2x y zS S S S

Recall:l = cos θxm = cos θyn = cos θz

To balance force we need the areas that each stress

components acts on

yxS

x

y

z

A

C

Byy

yz

xy

xz

zy zxO

xx

zz

AxS

zS

yS

x

y

z

xz

y

normal

( ) 0( ) 0

( ) 0

x xx yx zx

y yy xy zy

z zz xz yz

x xx yx zx

x

y x

y

z

y yy zy

z xz yz zz

xx yx zx

xy yy zy

xz yz zz

F A Al Al Am AnF A Am Am Al AnF A An An Al Am

F l m nF l m nF l

S SS SS S

SS

S

S

m

S

n

S

000

lmn

All forces must balance to meet the conditions for static equilibrium (i.e., ∑F=0):

When written in matrix

form

• The solution of the determinant of the matrix on the left yields a cubic equation in terms of S.

• In this problem, S =.

• Thus, the three roots of this cubic equation represent the principal stresses, 1, 2, and 3.

2 2 2

2 2 2

1 2 3

3 2

3 2

( ) ( )

( 2 ) 0

or0

xx yy zz xx yy yy zz xx zz xy yz xz

xx yy zz xy yz xz xx yz yy xz zz xy

S S S

S IS SI I

Method Method –– contcont’’dd

Method Method –– contcont’’dd• The directions in which the principal stresses act are

determined by substituting 1, 2, and 3, each for S in:

Then the resulting equations must be solved simultaneously for l, m, and n (using the relationship l2+m2+n2 = 1).

(a) Substitute 1 for S ; solve for l, m, and n;

(b) Substitute 2 for S ; solve for l, m, and n;

(c) Substitute 3 for S ; solve for l, m, and n.

( ) 0( ) 0

( ) 0

xx yx zx

xy yy zy

xz yz zz

SS

l m nl ml nS

nm

Invariants of the Stress TensorInvariants of the Stress Tensor

Whenever stresses are transformed from one coordinate system to another, these three quantities remain constant.

1

2 2 22

2 2 23 2

xx yy zz

xx xy yy yz xx xzxx yy yy zz xx zz xy yz xz

yx yy zy zz zx zz

xx xy xz

yx yy yz xx yy zz xy yz xz xx yz yy xz zz xy

zx zy zz

I

I

I

• Determine the principal normal stresses for the following state of stress:

Example Problem #1Example Problem #1

0, 10, 75,

50, 0xx yy zz

xy yz xz

0 50 050 10 0 MPa

0 0 75

or

• This problem can be solved by substituting the known state of stress into the cubic equation:

where S = σ.

• This is detailed on the next page.

Example Problem #1 Example Problem #1 –– solutionsolution

23

32

1 0I IS S S I

2 2 2

2

3 2

2

22

2

3 2 20 10

0 50 050 10 0 MPa

0 0 75

( ) ( )

( 2 ) 0

( ) [(0 10) (10 75) (0 75) ( 50) (0)7(

5 (0) ][ 0

xx yy yy zz xx zz xy yz xz

xx yy zz xy yz xz xx yz yy xz zz x

xx yy z

y

z

2

2

3 2

2 2

3

10 75) 2( 50 0 0) (0 0 ) (10 0 ) ( 75 50 )] 0

( ) [ 3250] [187500] 065

6 3250 187500 05

I1 I2 I3

-100 -50 0 50 100-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1 = 55

2 = -45

3 + 6

52 -

3250

- 1

8750

0 (1

06 MPa

)

(MPa)

3 = -75

You can determine the principal stresses by plotting this equatiYou can determine the principal stresses by plotting this equation on OROR you can solve it using more traditional means. you can solve it using more traditional means.

• This problem is easier than most because there are no shear stresses along the z-axis.

• It should have been obvious toyou that one of the principal stresses is = -75 MPa (since zx = xz = 0 and zy = yz = 0).

• Can you determine the directions in which the principal stresses act?

Example Problem #1 Example Problem #1 –– solutionsolution

(I RECOMMEND THAT YOU TRY IT)

Resources on the WebResources on the Web

• There are many useful eigenvalue calculators on the world wide web. Here are a few:

• http://portal.cs.umass.edu/projects/mohr/

• http://www.engapplets.vt.edu/Mohr/java/nsfapplets/MohrCircles2-3D/Applets/applet.htm

H*

• Determine (a) the principal stresses, (b) maximum shear stress, and (c) the orientations of the principal planes for the state of stress provided below:

Example Problem #2Example Problem #2

2 2 2

2 2 2

2

3 2

3 2 28

80 20 5020 40 30 MPa50 30 60

- ( ) ( )

( 2 ) 0

( ) [(80 40) ( 40 60) (80 60) (20) (30) (500 4 6 )0 0

xx yy yy zz xx zz xy yz xz

xx yy zz xy yz xz xx yz yy xz zz xy

xx yy zz

2

2

3

3 2

2

2

2

][(80 40 60) 2(20 30 60) (80 30 ) ( 40 50 ) (60 20 )] 0

( ) [ 4600] [ 248000] 0

4600 248000 0

100

100

1

p. 1/11

-100 -50 0 50 100 150-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1 = 1212 = 36

3 - 10

02 -

4600

+

2480

00 (

106 M

Pa)

(MPa)

3 = -57

1 = 121 MPa; 2 = 36 MPa; 3 =-57 MPa1

3 2 4600 248001 0 000

p. 2/11

1 2 3

( ) 0( ) 0

( ) 0

(80 ) 20 50 020 ( 40 ) 30 0

50 30 (60 ) 0

substitute , , and in place of and solve simultaneous equations

xx yx zx

xy yy zy

xz yz zz

l m nl m nl m n

l m nl m n

l m n

1 121(80 ) 20 50 0 41 20 50 0 1

20 ( 40 ) 30 0 20 161 30 0 250 30 (60 ) 0 50 30 61 0 3

l m n l m nl m n l m n

l m n l m n

80 20 5020 40 30 MPa50 30 60

123 60 150 0100 805 150 0

23 745 0 0; 0.0309

l m nl m n

l m n m l

(3 [1]) + (5 [2]) yields:2

p. 3/11

1 2 3substitute , , and in place of and solve simultaneous equations

123 60 150 0100 60 122 0

23 0 28 0; 0.821

l m nl m n

l m n n l

(3 [1]) + (-2 [3]) yields:3

Sample Problem #2cont’d

p. 4/11

1 121(80 ) 20 50 0 41 20 50 0 1

20 ( 40 ) 30 0 20 161 30 0 250 30 (60 ) 0 50 30 61 0 3


l m n l m n

2 2 2

2 22 2

1

0.031 0.821 1.673 1

1 0.7731.6730.031 0.024

0.821 0.635

substitute expresions for & l m n

m n

l l l l

l

m l

n l

4


Orientations of principal planes associated with

1

p. 5/11

2 36(80 ) 20 50 0 44 20 50 0 [4]

20 ( 40 ) 30 0 20 76 30 0 [5]50 30 (60 ) 0 50 30 24 0 [6]


l m n l m n

132 60 150 0100 380 150 0

232 320 0 0; 0.725

l m nl m n

l m n m l

(3 [4]) + (5 [5]) yields:5


p. 6/11


132 60 150 0100 60 48 0232 0 198 0; 1.172

l m nl m nl m n n l

(3 [4]) + (-2 [6]) yields:6


p. 7/11


2 36(80 ) 20 50 0 44 20 50 0 [4]

20 ( 40 ) 30 0 20 76 30 0 [5]50 30 (60 ) 0 50 30 24 0 [6]


l m n l m n

2 2 2

2 22 2

1

0.725 1.172 2.899

1 0.5872.899

0.725 0.426

1.172 0.688

l m nm n

l l l l

l

m l

n l

substitute expresions for &

7



2

p. 8/11

3 57(80 ) 20 50 0 137 20 50 0 7

20 ( 40 ) 30 0 20 17 30 0 850 30 (60 ) 0 50 30 117 0 9


l m n l m n

411 60 150 0100 85 150 0

511 145 0 0; 3.524

l m nl m nl m n m l

(3 [7]) + (5 [8]) yields:8


p. 9/11


(3 [7]) + (-2 [6]) yields:9

411 60 150 0100 60 234 0

511 0 384 0; 1.331

l m nl m n

l m n n l


p. 10/11


3 57(80 ) 20 50 0 137 20 50 0 7

20 ( 40 ) 30 0 20 17 30 0 850 30 (60 ) 0 50 30 117 0 9


l m n l m n

2 2 2

2 22 2

1

3.524 1.331 15.189

1 0.25715.1893.524 0.904

1.331 0.342

substitute expresions for & l m n

m n

l l l l

l

m l

n l

10

10 1 3max

121 ( 57) 892 2

MPa



3

p. 11/11

5 minute break

• In previous example/method, we assumed that the stress on the inclined plane was a principal stress.

• What if the stress on the new plane is not a principal stress?

• The math is nearlythe same.

General Method for Triaxial States of StressGeneral Method for Triaxial States of Stress[p. 29-30 in Dieter]

2 2 22 22x ysn zS S S S

yx

S

x

y

z

A

C

Byy

yz

xy

xz

zy zxO

xx

zz

s

n

x

y

z

xz

y

normal

Triaxial Stress States Triaxial Stress States –– contcont’’ddFrom summation of forces parallel to the x, y, z axes, we find

the components Sx, Sy, Sz:

The normal stress on the oblique plane equals the sum of the components Sx, Sy, Sz parallel to the plane normal.

x xx yx zx

y xy yy zy

z xz yz zz

S l m n

S l m n

S l m n

2 2 2 2 2 2n x y z

xx yy zz xy yz zx

S l S m S n

l m n lm mn nl

Triaxial Stress States Triaxial Stress States –– contcont’’ddFrom the expression S2 = σ2 + τ2 , the shear stress can be obtained.

When written in terms of principal axes, it becomes:

The maximum shear stress occurs when:

2 22 2 2 2 2 2 21 2 1 3 2 3

2s l m l n m n

1 3 max minmax 2 2

2 2 2snS

MohrMohr’’s Circle in 3s Circle in 3--DD• We can use a 3-D Mohr’s circle to visualize the state of

stress and to determine principal stresses.

• Essentially three 2-D Mohr’scircles corresponding to the x-y, x-z, and y-z faces of the elemental cubic element.

x

y

z

xx

xy

xz

zz

zyzx

yy

yx

yz

13

2

123

Adapted from G.E. Dieter, Mechanical Metallurgy, 3rd ed., McGraw-Hill (1986) p. 37

3 = 2 = 0

max

1

3 = 0

max = 2

1

2

1 3

max

1 = 2 = 03

2 = 3

max = 2= 3

1

2=3

max = 2= 3

1

1

3

2

1

σ3

2

1

σ3

2

1 = -22 = -23

1 = 22 = 23

1

3

2

UniaxialTension

3

1

2

UniaxialCompression

BiaxialTension

Triaxial Tension(unequal)

Uniaxial Tension plusBiaxial Compression

1 0 00 0 00 0 0

1

2

0 00 00 0 0

1

2

3

0 00 00 0

3

0 0 00 0 00 0

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Mechanical Behavior Notes-2009d