From equilibrium principles:τxy = τyx , τxz = τzx , τzy = τyz

3D Stress Components

Normal Stresses

Shear Stresses

Normal stress (σ) : the subscript identifies the face on which the stress acts. Tension is positive and compression is negative.Shear stress (τ) : it has two subscripts. The first subscript denotes the face on which the stress acts. The second subscript denotes the direction on that face. A shear stress is positive if it acts on a positive face and positive direction or if it acts in a negative face and negative direction.

The most general state of stress at a point may be represented by 6 components

zyx σσσ

xzyzxy τττ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

==

zyzxz

zyyxy

zxyxx

ij

στττστττσ

σσ

For static equilibrium τxy = τyx , τxz = τzx , τzy = τyz resulting in six independent scalar quantities. These six scalars can be arranged in a 3x3 matrix, giving us a stress tensor.

The sign convention for the stress elements is that a positive force on a positive face or a negative force on a negative face is positive. All others are negative.

The stress state is a second order tensor since it is a quantity associated with two directions (two subscripts direction of the surface normal and direction of the stress).Same state of stress is represented by a different set of components if axes are rotated. There is a special set of components (when axes are rotated) where all the shear components are zero (principal stresses).

A property of a symmetric tensor is that there exists an orthogonal set of axes 1, 2 and 3 (called principal axes) with respect to which the tensor elements are all zero except for those in the diagonal.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

==

zyzxz

zyyxy

zxyxx

ij

στττστττσ

σσ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

3

2

1'

000000

'σ

σσ

σ ijσEigen values

In matrix notation the transformation is known as the Eigen-values.The principal stresses are the “new-axes” coordinate system. The angles between the “old-axes” and the “new-axes” are known as the Eigen-vectors.

principal stress

Cosine of angle between X and the

principal stress

Cosine of angle between Y and the

principal stress

Cosine of angle between Z and the

principal stress

σ1 k1 l1 m1σ2 k2 l2 m2σ3 k3 l3 m3

State of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by

.0and,, xy === zyzxzyx ττστσσ

Plane Stress

Sign Conventions for Shear Stress and Strain

The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element produce a counterclockwise (ccw) torque (couple).

A shear strain in an element is positive when the angle between two positive faces (or two negative faces) is reduced, and is negative if the angle is increased.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00000

yyxy

yxxx

σττσ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00000

1111

1111

yyyx

xyxx

σττσ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0000000

2

1

σσ

1

2σ2 σ1

Stresses on Inclined SectionsKnowing the normal and shear stresses acting in the element denoted by the xy axes, we will calculate the normal and shear stresses acting in the element denoted by the axis x1y1.

x1y1

x

y

θ

θ

σy

σx

σx1

τx1y

1

τyx

τxy

Equilibrium of forces: Acting in x1

θθθτ

θθθσθτθσ

θσ

CosSinA

CosSinASinAAA

OYXOYOXYOXO

X ⋅+⋅+⋅+⋅=⋅ cossincoscos1

Eliminating Ao , secθ = 1/cosθ and τxy=τyx

θθτθσθσσ cossin2sincos 221 XYYXX ++=

Acting in y1

τx1y1Aosecθ = − σxAosinθ + τxyAocosθ + σyAotanθcosθ − τyxAotanθsinθ

Eliminating Ao , secθ = 1/cosθ and τxy=τyx

θθτθσθσσ cossin2cossin 221 XYYXY −+=

( )θθτθθσθθστ 2211 sincoscossincossin −+⋅⋅+⋅⋅−= xyyxyx

Using the following trigonometric identities:Cos2θ = ½ (1+ cos 2θ) Sin2θ = ½ (1- cos 2θ) Sin θ cos θ = ½ sin 2θ

These equations are known as the transformation equations for plane stress.

θτθσσ

τ

θτθσσσσ

σ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

2sin2cos22

11

1

1

xyyx

yx

xyyxyx

y

xyyxyx

x

+−

−=

−−

−+

=

+−

++

=

Transformation Equations for Plane Stress

Case 1: Uniaxial stress

Special Cases

⎟⎠⎞

⎜⎝⎛⋅−=

⎟⎠⎞

⎜⎝⎛ +

⋅=

===

22

221

0 0

1,1

1

xy

θστ

θσσ

ττσ

Sin

Cos

xyx

xx

yxy

Case 2 : Pure Shear

θττ

θτσ

σσ

2

2

0

1,1

1

Cos

Sin

xyyx

xyx

yx

⋅=

⋅=

==

θτθσσ

τ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

11

1

xyyx

yx

xyyxyx

x

+−

−=

+−

++

=

Case 3: Biaxial stress

θσσ

τ

θσσσσ

σ

τ

22

222

0

1,1

1

Sin

Cos

yxyx

yxyxx

xy

⋅−

−=

⋅−

++

=

=

An element in plane stress is subjected to stresses σx=16000psi, σy=6000psi, and τxy=τyx= 4000psi (as shown in figure below). Determine the stresses acting on anelement inclined at an angle θ=45o (counterclockwise - ccw).

Solution: We will use the following transformation equations:

θτθσσ

τ

θτθσσσσ

σ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

2sin2cos22

11

1

1

xyyx

yx

xyyxyx

y

xyyxyx

x

+−

−=

−−

−+

=

+−

++

=

Numerical substitution ( )

( )

( ) ( )( ) ( )

( ) ( ) psi

psipsi

psi

psi

psi

yx

y

x

xy

yx

yx

50000400015000

700014000050001100015000140000500011000

4000

50002

6000160002

110002

6000160002

11

1

1

−=+−=

=−−==++=

=

=−

=−

=+

=+

τ

σσ

τ

σσ

σσ

( ) ( )( ) ( ) 0902

190245

0

0

0

==

==

+=

CosCosSinSin

(ccw) θFor x-axis

θ

θ

( ) ( )( ) ( ) 02702

127029045

0

0

00

==

−==

++=

CosCosSinSin

(ccw) θFor y-axis

θ

θ

Note: 11 yxyx σσσσ +=+

A plane stress condition exists at a point on the surface of a loaded structure such as shown below. Determine the stresses acting on an element that is oriented at a clockwise (cw) angle of 15o with respect to the original element, the principal stresses, the maximum shear stress and the angle of inclination for the principal stresses

Solution: We will use the following transformation equations:

θτθσσ

τ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

11

1

xyyx

yx

xyyxyx

x

+−

−=

+−

++

=

( )

( )

( ) ( )( ) ( )( )( )( ) ( )( ) MPa

MPa

MPa

MPa

MPa

yx

x

xy

yx

yx

31866.0195.0296.325.019866.02917

19

292

12462

172

12462

11

1

−=−+−−=−=−−+−+−=

−=

−=−−

=−

−=+−

=+

τσ

τ

σσ

σσ

( ) ( )( ) ( ) 866.0302

5.030215

0

0

0

=−=

−=−=

−=

CosCosSinSin

(cw) θFor x-axis

θ

θ

MPay

yxyx

4.11

11

−=

+=+

σ

σσσσ

Determine the normal stress sw acting perpendicular to the line or the weld and the shear stress tw acting parallel to the weld. (Assume sw is positive when it acts in tension and tw is positive when it acts counterclockwise against the weld).

Solution:

psi

psipsi

psipsipsi

y

yxx

xyyxyx

25

375 375

0 4252

1752

1

111

−=

−==

==−

=+

σ

τσ

τσσσσ

00 92612963053tan

.θ.θ

θ For x-axis

=⇒=

⇒=

psiy

yxyx

251

11

−=

+=+

σ

σσσσ -375psi

Θ = 30.96o

375psi25psi

x

y

Stresses acting on the weldσw

θτw

25psi

375psiθ = 30.96o

A rectangular plate of dimensions 3.0 in x 5.0 in is formed by welding two triangular plates (see figure). The plate is subjected to a tensile stress of 600psi in the long direction and a compressive stress of 250psi in the short direction.

σw = -25psi and τw = 375psi

Principal Stresses and Maximum Shear StressesThe sum of the normal stresses acting on perpendicular faces of plane stress elements is constant and independent of the angle θ.

YXYX σσσσ +=+ 11

As we change the angle θ there will be maximum and minimum normal and shear stresses that are needed for design purposes.

The maximum and minimum normal stresses are known as the principal stresses. These stresses are found by taking the derivative of σx1 with respect to θ and setting equal to zero.

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

=+−−=

+−

++

=

2

2tan

02cos22sin)(

2sin2cos22

1

1

yx

xyP

xyyxx

xyyxyx

x

σστ

θ

θτθσσδθ

δσ

θτθσσσσ

σ

θθτθσθσσ cossin2sincos 221 XYYXX ++=

θθτθσθσσ cossin2cossin 221 XYYXY −+=

The subscript p indicates that the angle θp defines the orientation of the principal planes. The angle θp has two values that differ by 90o. They are known as the principal angles. For one of these angles σx1 is a maximum principal stress and for the other a minimum. The principal stresses occur in mutually perpendicular planes.

( )2

2tanyx

xyP σσ

τθ

−=

2

)(2cos 2sin

RRyx

Pxy

P

σσθ

τθ

−==

P

1

stress maximum for the

2sin2cos22

θθ

θτθσσσσ

σ

=

+−

++

= xyyxyx

x

( )

( )

( ) ( )22

12

2

22

1

22

1

22

2 But

21

2

1122222

xyyxyx

xyyx

xyyxyx

xyyxyxxy

xyyxyxyx

R

R

RRRR

τσσσσ

στσσ

τσσσσ

σ

τσσσστ

τσσσσσσ

σ

+⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

+=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛+

+=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−+

+=

The plus sign gives the algebraically larger principal stress and the minus sign the algebraically smaller principal stress.

This are the in-plane principal stresses. The third stress is zero in plane stress conditions

( )

( ) R

R

Averagexyyxyx

Averagexyyxyx

−=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

+=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

στσσσσ

σ

στσσσσ

σ

22

2

22

1

22

22

Principal Stresses

The location of the angle for the maximum shear stress is obtained by taking the derivative of τx1y1 with respect to θ and setting it equal to zero.

θτθσσ

τ 2cos2sin211 xy

yxyx +

−−=

( )

xy

yx

S

xyyxyx

τ

σσθ

θτθσσδθ

δτ

22tan

02sin22cos)(11

−−=

=−−−=

Maximum Shear Stresses

( )( )

( )( )o

P

oP

Po

Po

P

P

S

S

PPxy

yxS

902cos902sin

290cos290sin

2sin2cos

2cos2sin

2cot2tan1

2)(

2tan

−−

=−−−

=−=

−=−=−

−=

θθ

θθ

θθ

θθ

θθτ

σσθ

0P1s111 45 and

2)(

2sin 2cos −=−

−== θθσσ

θτ

θRR

yxs

xys

The planes for maximum shear stress occurs at 45o to the principal planes. The plane of the maximum positive shear stress τmax is defined by the angle θS1 for which the following equations apply:

The corresponding maximum shear is given by the equation

Another expression for the maximum shear stress

The normal stresses associated with the maximum shear stress are equal to

( ) Rxyyx

MAX =+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= 2

2

2τ

σστ

( )2

21 σστ

−=MAX

Therefore, 2θs-2θp=-90o or θs= θp +/- 45o

( )2

yxAVER

σσσ

+=

Equations of a Circle

θτθσσ

σσ

θτθσσσσ

σ

2sin2cos2

2sin2cos22

1

1

xyyx

AVERx

xyyxyx

x

+−

=−

+−

++

=General equation

Consider

Equation (1)

Equation (2)

θτθσσ

τ 2cos2sin211 xy

yxyx +

−−=

Equation (1) + Equation (2)

( )2

21 2sin2cos

2 ⎥⎦

⎤⎢⎣

⎡+

−=− θτθ

σσσσ xy

yxAVERx

( )2

211 2cos2sin

2 ⎥⎦

⎤⎢⎣

⎡+

−−= θτθ

σστ xy

yxyx

( ) ( )22

211

21 2cos2sin

22sin2cos

2 ⎥⎦

⎤⎢⎣

⎡+

−−+⎥

⎦

⎤⎢⎣

⎡+

−=+− θτθ

σσθτθ

σστσσ xy

yxxy

yxyxAVERx

( ) ( )

( ) ( )

( ) 222

22222

22222

2

2cos2sin2

22cos2sin2

2cos2sin2

2cos2sin2

22sin2cos2

2sin2cos2

RSUM xyyx

xyyx

xyyx

xyyx

xyyx

xyyx

xyyx

=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−=⎥

⎦

⎤⎢⎣

⎡+

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛ −++⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=⎥

⎦

⎤⎢⎣

⎡+

−

τσσ

θθτσσ

θτθσσ

θτθσσ

θθτσσ

θτθσσ

θτθσσ

( ) ( ) 2211

21 RyxAVERx =+− τσσ

τx

y

2θP

σ

τ

C

( )22

2 xyyxR τ

σσ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +

2yx σσ

The radius of the Mohr circle is the magnitude R.

Mohr Circle

( )22

2 xyyxR τ

σσ+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

The center of the Mohr circle is the magnitude

( )2

yxAVER

σσσ

+=

( ) ( ) ( )22

211

21 2

xyyx

yxAVERx τσσ

τσσ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=+−State of Stresses

Alternative sign conversion for shear stresses: (a)clockwise shear stress,(b)counterclockwise shear stress, and (c) axes for Mohr’s circle.

Note that clockwise shear stresses are plotted upward and counterclockwise shear stresses are plotted downward.

a) We can plot the normal stress σx1 positive to the right and the shear stress τx1y1 positive downwards, i.e. the angle 2θ will be positive when counterclockwise or

b) We can plot the normal stress σx1 positive to the right and the shear stress τx1y1positive upwards, i.e. the angle 2θ will be positive when clockwise.

Both forms are mathematically correct. We use (a)

Forms of Mohr’s Circle

Two forms of Mohr’s circle:

τx1y1 is positive downward and the angle 2θ is positive counterclockwise, and

τx1y1 is positive upward and the angle 2θ is positive clockwise. (Note: The first form is used here)

Construction of Mohr’s circle for plane stress.

At a point on the surface of a pressurized cylinder, the material is subjected to biaxial stresses σx = 90MPa and σy = 20MPa as shown in the element below.Using the Mohr circle, determine the stresses acting on an element inclined at an angle θ = 30o (Sketch a properly oriented element).

Because the shear stress is zero, these are the principal stresses.Construction of the Mohr’s circle

(σx = 90MPa, σy = 20MPa and τxy = 0MPa)

( ) ( ) MPayxAverage 55

22090

2=

+=

+=

σσσ

The center of the circle is

The radius of the circle is ( ) ( ) MPaR xy

yx 3502

20902

22

22

=+⎟⎠⎞

⎜⎝⎛ −

=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= τ

σσ

Stresses on an element inclined at θ = 30o

By inspection of the circle, the coordinates of point D are

( )MPaCosCosR

MPaSinSinR

MPaCosCosR

Averagey

ooyx

Averagex

5.3760355560

3.30603560

5.7260355560

01

11

01

=⋅−=⋅−=

−=−=⋅−=

=⋅+=⋅+=

σσ

τ

σσ

An element in plane stress at the surface of a large machine is subjected to stresses σx = 15000psi, σy = 5000psi and τxy = 4000psi.

Using the Mohr’s circle determine the following:a) The stresses acting on an element inclined at an

angle θ = 40o

b) The principal stresses andc) The maximum shear stresses.

Construction of Mohr’s circle:Center of the circle (Point C):

Radius of the circle:

Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = 15000psi and τx1y1 = 4000psi

Point B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 5000psi and τy1x1 = - 4000psiThe circle is now drawn through points A and B with center C and radius R

( ) ( ) psiyxAverage 10000

2500015000

2=

+=

+=

σσσ

( ) ( ) psiR xyyx 64034000

2500015000

22

22

2

=+⎟⎠⎞

⎜⎝⎛ −

=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= τ

σσ

By inspection the angle ACP1 for the principal stresses (point P1) is :

Then, the angle P1CD is 80o –38.66o = 41.34o

( ) oACPACP 66.3850004000tan 11 =⇒=

Stresses on an element inclined at θ= 40o

These are given by the coordinates of point D which is at an angle 2θ = 80o

from point A

Knowing this angle, we can calculate the coordinates of point D (by inspection)

( )psiCosCosR

psiSinSinR

psiCosCosR

ooAveragey

ooyx

ooAveragex

519034.4164031000034.41

423034.41640334.41

1481034.4164031000034.41

1

11

1

=⋅−=⋅−=

−=−=⋅−=

=⋅+=⋅+=

σσ

τ

σσ

Principal StressesThe principal stresses are represented by points P1 and P2 on Mohr’s circle.The angle it was found to be 2θ = 38.66o

or θ = 19.3o

psiR

psiR

Average

Average

3597640310000

16403640310000

2

1

=−=−=

=+=+=

σσ

σσ

And of course, the sum of the normal stresses is 14810psi + 5190psi = 15000psi + 5000psi

Maximum Shear StressesThese are represented by point S1 and S2 in Mohr’s circle. Algebraically the maximum shear stress is given by the radius of the circle.The angle ACS1 from point A to point S1 is 2 θS1 = 51.34o. This angle is negative because is measured clockwise on the circle. Then the corresponding θS1 value is –25.7o.

Psi⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0000500040000400015000

Psi⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0000359700016403

3-D stress stateTransform to

In matrix notation the transformation is known as the Eigen-values.

The principal stresses are the “new-axes” coordinate system. The angles between the “old-axes” and the “new-axes” are known as the Eigen-vectors.

principal stress

Cosine of angle between X and the

principal stress

Cosine of angle between Y and the

principal stress

Cosine of angle between Z and the

principal stress16403.1242 0.94362832 0.331006939 03596.876 -0.33101 0.943628 0

0 0 0 1

At a point on the surface of a generator shaft the stresses are σx = -50MPa, σy = 10MPa and τxy = - 40MPa as shown in the figure. Using Mohr’s circle determine the following: (a)Stresses acting on an element inclined at an angle θ = 45o,(b)The principal stresses and(c)The maximum shear stresses

Construction of Mohr’s circleCenter of the circle (Point C):

Radius of the circle:.

Point A, representing the stresses on the x face of the element (θ = 0o) has the coordinates σx1 = -50MPa and τx1y1 = - 40MPaPoint B, representing the stresses on the y face of the element (θ = 90o) has the coordinates σy1 = 10MPa and τy1x1 = 40MPaThe circle is now drawn through points A and B with center C and radius R.

( ) ( ) ( ) MPayxAverage 20

21050

2−=

+−=

+=

σσσ

( ) ( ) ( ) ( ) MPaR xyyx 5040

21050

22

22

2

=−+⎟⎠⎞

⎜⎝⎛ −−

=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −= τ

σσ

Stresses on an element inclined at θ = 45o

These stresses are given by the coordinates of point D (2θ = 90o of point A). To calculate its magnitude we need to determine the angles ACP2 and P2CD.

Then, the coordinates of point D are

And of course, the sum of the normal stresses is -50MPa+10MPa = -60MPa +20MPa

( )( )

( ) MPaCosCosR

MPaSinSinR

MPaCosCosR

ooAveragey

ooyx

ooAveragex

2087.36502087.36

3087.365087.36

6087.36502087.36

1

11

1

=⋅+−=⋅−=

==⋅=

−=⋅+−=⋅+=

σσ

τ

σσ

tan ACP2=40/30=4/3 ACP2=53.13o

P2CD = 90o – 53.13o = 36.87o

Principal StressesThey are represented by points P1 and P2 on Mohr’s circle.

The angle ACP1 is 2θP1 = 180o + 53.13o

= 233.13o or θP1 = 116.6o

The angle ACP2 is 2θP2 = 53.13o or θP2= 26.6o

Maximum Shear StressesThese are represented by point S1 and S2in Mohr’s circle. The angle ACS1 is 2θS1 = 90o + 53.13o = 143.13o or θ = 71.6o . The magnitude of the maximum shear stress is 50MPa and the normal stresses corresponding to point S1 is the average stress -20MPa.

MPaR

MPaR

Average

Average

705020

305020

2

1

−=−−=−=

=+−=+=

σσ

σσ

MPa⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−−

0000104004050

MPa⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

00007000030

3-D stress state Transform to

In matrix notation the transformation is known as the Eigen-values.

The principal stresses are the “new-axes” coordinate system. The angles between the “old-axes” and the “new-axes” are known as the Eigen-vectors.

principal stress

Cosine of angle between X and the

principal stress

Cosine of angle between Y and the

principal stress

Cosine of angle between Z and the

principal stress

30 -0.44721359 0.894427193 0-70 0.894427 0.447214 00 0 0 1

The stress transformations equations were derived solely from equilibrium conditions and they are material independent.Here the material properties will be considered (strain) taking into account the following:a)The material is uniform throughout the body (homogeneous)b)The material has the same properties in all directions (isotropic)c)The material follows Hooke’s law (linearly elastic material)Hooke’s law: Linear relationship between stress and strainFor uniaxial stress: (E = modulus of elasticity or Young’s modulus)

Poisson’s ratio:

For pure shear : (G = Shear modulus of elasticity)

εσ E=

allongitudin

transverse

strainastrainlateral

εευ −=−=

xial

γτ G=

Hooke’s Law for Plane Stress

Element of material in plane stress (σz = 0).

Consider the normal strains εx, εy, εz in plane stress.All are shown with positive elongation.

The strains can be expressed in terms of the stresses by superimposing the effect of the individual stresses.For instance the strain εx in the x direction:a)Due to the stress σx is equal to σx/E.b)Due to the stress σy is equal to –νσy/E.The resulting strain is:

EE

EE

EE

yxz

yxy

yxx

σνσνε

σσνε

σνσε

−−=

+−=

−=

The shear stress causes a distortion of the element such that each z face becomes a rhombus.

GXY

XYτ

γ =

The normal stresses σx and σy have no effect on the shear strain γxy

or rearranging the equations:GXY

XYτ

γ =

( ) ( ) ( ) ( ) XYyx

Yyx

x GEEEE γτ

νε

νενσ

νεν

νεσ =+

−+

−=

−+

−= XY2222

11

11

EEEEEEyx

zyx

yyx

x

σνσνε

σσνεσ

νσε −−=+−=−=

These equations are known collectively as Hooke’s Law for plane stressThese equations contain three material constants (E, G and ν) but only two are independent because of the relationship:

)1(2 ν+=

EG

Special cases of Hooke’s law (σz = 0)

EEx

zx

x

xyy

συεεσε

τσ

−===

==

y

0 0Uniaxial stress :

Gxy

xyzyx

yx

τγεεε

σσ

====

==

0

0Pure Shear :

Biaxial stress :

Eσ

- EEE

εEσ

- Eσ yxyx

yyx

x

xy

υσυεσσυυε

τ

−=+−==

=

z

0

When a solid object undergoes strains, both its dimensions and its volume will change.

Consider an object of dimensions a, b, c. The original volume is Vo = abc and its final volume is

V1 = (a + aεx) (b + bεy) (c + cεz) V1= abc (1+εx) (1+εy) (1+εz)

Volume Change

Upon expanding the terms:V1 = Vo (1 + εx + εy + εz + εxεy + εxεz + εyεz + εxεyεz)

For small strains:V1 = Vo (1 + εx + εy + εz )

The volume change is ΔV = V1 – Vo = Vo ( εx + εy + εz )

The unit volume change e, also known as dilatation is defines as: e = ΔV / Vo = εx + εy + εz

Positive strains are considered for elongations and negative strains for shortening, i.e. positive values of e for an increase in volume.

EEEEEEyx

zyx

yyx

x

σνσνε

σσνεσ

νσε −−=+−=−=

( )( )EV

Ve yxυσσ 21−

+=Δ

= For uniaxial stress σy = 0 ( )E

e xυσ 21−

=

We can notice that the maximum possible value of Poisson’s ratio is 0.5, because a larger value means that the volume decreases when the material is in tension (contrary to physical behavior).

The strain energy density u is the strain energy stored in a unit volume of the material.Because the normal and shear strains occur independently, we can add the strain energy of these two elements to obtain the total energy.

Work done = Force x distance

STRAIN ENERGY DENSITY IN PLANE STRESS

( )( ) ( )

( )( ) ( )y

x

bac

abc

εσ

εσ

2direction-y in the doneWork

2direction- xin the doneWork

y

x

=

=

The sum of the energies due to normal stresses:( )yyxx

abcU εσεσ +=2

Then the strain energy density (strain per unit volume) ( )yyxxu εσεσ +=21

1

Similarly, the strain energy density associated with the shear strain:

By combining the strain energy densities for the normal and shear strains:

xyxyu γτ21

2 =

( )xyxyyyxxu γτεσεσ ++=21

The strain energy density in terms of stress alone: GEEE

u XYYXYX

222

222 τσσυσσ+−+=

The strain energy density in terms of strain alone: ( )( ) 222

2 22

12 XYYXYXGEu γευεεε

υ+−+

−=

For the special case of uniaxial stress: For the special case of pure shear:

2

or 2

0 0 022xx

xyxyxyy

EuE

u

ε

εσ

γυετσ

==

=−===

2

or 2

0 0 022xyxy

xyyx

Gu

Gu

ε

γτ

εσσ

==

====

An element of the material subjected to normal stresses σx, σy and σz acting in three mutually perpendicular directions is said to be in a state of triaxial stress. Since there is no shear in x, yor z faces then the stresses σx, σy and σz are the principal stresses in the material.

TRIAXIAL STRESS

If an inclined plane parallel to the z-axis is cut through the element, the only stress of the inclined face are the normal stress σ and the shear stress τ, both of which act parallel to the xy plane.

The same general conclusion hold for normal and shear stresses acting on inclined planes cut through the element parallel to the x and y axes.

Maximum Shear Stress For a material in triaxial stress, the maximum shear stresses occur on elements oriented at angles of 45o to the x, y and z axes.

( )

( )

( )2

2

2

ZXYMAX

ZYXMAX

YXZMAX

σστ

σστ

σστ

−±=

−±=

−±=

Because these stresses are independent of the σz, we can use the transformation equations of plane stress, as well as the Mohr’s circle for plane stress, when determining the stresses σ and τ in triaxial stress.

for the inclined plane // z-axis

for the inclined plane // x-axis

for the inclined plane // y-axis

The absolute maximum of the shear stress is the numerically largest of the above.

The stresses acting on elements oriented at various angles to the x, y and z axes can be visualized with the aid of the Mohr’s circle.

In this case σx > σy > σz

Hooke’s Law for Triaxial Stress

If Hooke’s law is obeyed, it is possible to obtain the relationship between normal stresses and normal strains using the same procedure as for plane stress.

EEE

EEE

EEE

zyxz

zyxy

zyxx

σσνσνε

σνσσνε

σνσ

νσε

+−−=

−+−=

−−=

Or

( )( ) ( )[ ]

( )( ) ( )[ ]

( )( ) ( )[ ]zyxz

zyxy

zyxx

E

E

E

ευυευευυ

σ

υεευυευυ

σ

υευεευυυ

σ

−++−+

=

+−+−+

=

++−−+

=

1211

1211

1211

They are known as the Hooke’s law for triaxialstress.

Unit Volume ChangeThe unit volume change (or dilatation) for an element in triaxial stress is obtained in the same manner as for plane stress.

( ) ( )ZYXO

ZYXO

EVVe

VVe

σσσυ

εεε

++−

=Δ

=

++=Δ

=

21 thenapply, laws sHooke' If

Strain Energy Density

The strain energy density for an element in triaxial stress is obtained by the same method used for plane stress.

( ) ( )zyzxyxzyx

zzyyxx

EEu

u

σσσσσσυσσσ

εσεσεσ

++++=

++=

-21

:stresses of In terms

222

222

( )( ) ( )( ) ( )[ ] 21 2112

222zyzxyxzyx

Eu εεεεεευεεευυυ

+++++−−+

=

In terms of the strains:

Spherical StressA special case of triaxial stress, called spherical stress, occurs whenever all three normal stresses are equal:

0σσσσ === zyx

The Mohr’s circle is reduced to a single point. Any plane cut through the element will be free of shear stress and will be subjected to the same normal stress so and it is a principal plane.The normal strains in spherical stress are also the same in all directions, provided the material is isotropic and if Hooke’s law applies:

( )

( )E

E

O

OO

υσε

υσε

2133e

change volumeThe 21

O−

==

−=

Element in spherical stress.

K = bulk or volume modulus of elasticity

( ) eKEK 0

213σ

υ=

−=

If ν = 1/3 then K = EIf ν = 0 then K = E/3If ν = 1/2 then K = infinite (rigid material having no change in volume)These formulas also apply to and element in uniform compression,for example rock deep within the earth or material submerged in water (hydrostatic stress).

PLAIN STRAIN

Strains are measured by strain gages.A material is said to be in a state of plain strain if the only deformations are those in the xy plane, i.e. it has only three strain components εx, εy and γxy.

Plain stress is analogous to plane stress, but under ordinary conditions they do not occur simultaneouslyException when σx = -σy and when ν = 0

Strain components εx, εy, and γxy in the xy plane (plane strain).

Comparison of plane stress and plane strain.

The transformation equations for plane stress are valid even when σz is not zero, because σz does not enter the equations of equilibrium.Therefore, the transformations equations for plane stress can also be used for stresses in plane strain.Similarly, the strain transformation equations that will be derived for the case of plain strain in the xy plane are valid even when εz is not zero, because the strain εzdoes not affect the geometric relationship used for the derivation.Therefore, the transformations equations for plane strain can also be used for strains in plane stress.

APPLICATION OF THE TRANSFORMATION EQUATIONS

Transformation Equations for Plain Strain

We will assume that the strain εx, εy and γxyassociated with the xy plane are known.We need to determine the normal and shear strains (εx1 and γx1y1) associated with the x1y1 axis. εy1 can be obtained from the equation of εx1 by substituting θ +90 for θ.

For an element of size dx, dyIn the x direction: the strain εx produces an elongation εx dx.The diagonal increases in length by εx dx cos θ.

In the y direction: the strain εy produces an elongation εy dy.The diagonal increases in length by εy dy sin θ.

The shear strain γxy in the plane xy produces a distortion of the element such that the angle at the lower left corner decreases by an amount equal to the shear strain. Consequently, the upper face moves to the right by an amount γxy dy. This deformation results in an increase in the length of the diagonal equal to: γxy dy cos θ

The total increase Δδ of the diagonal is the sum of the preceding three expressions, thus: ( ) ( ) ( )θγθεθε CosdySindyCosdxd xyyx ++=Δ

But ( ) ( ) ( )

θθ

θγθεθεε

SindsdyCos

dsdx

CosdsdySin

dsdyCos

dsdx

dsd

xyyxx

==

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

Δ=

1

θθγθεθεε CosSinSinCos xyyxx ++= 221

Shear Strain γx1y1 associated with x1y1 axes.

γ

This strain is equal to the decrease in angle between lines in the material that were initially along the x1 and y1 axes.Oa and Ob were the lines initially along the x1 and y1 axis respectively. The deformation caused by the strains εx, εy and γxy caused the Oa and Ob lines to rotate and angle α and β from the x1 and y1 axis respectively. The shear strain γx1y1 is the decrease in angle between the two lines that originally were at right angles, therefore, γx1y1=α+β.

The angle α can be found from the deformations produced by the strains εx, εy and γxy . The strains εx and γxy produce a clockwise rotation, while the strain εy produces a counterclockwise rotation.

Let us denote the angle of rotation produced by εx , εy and γxy as α1 , α2 and α3 respectively.

dsdySin

dsdyCos

dsdxSin

xy

y

x

θγα

θεα

θεα

=

=

=

1

2

1 θθ SindsdyCos

dsdx

==

( ) θγθθεεαααα 2321 SinCosSin xyyx −−−=−+−=

The rotation of line Ob which initially was at 90o to the line Oa can be found by substituting θ +90 for θ in the expression for α. Because β is positive when clockwise. Thus:

( ) ( ) ( ) ( )( ) θγθθεεβ

θγθθεεβ2

2 909090

CosCosSin

SinCosSin

xyyx

xyyx

+−−=

++++−=

( ) [ ]θθγ

θθεεβαγ 2211 2

SinCosCosSin xyyxyx −+−−=+=

Transformation Equations for Plain StrainUsing the following trigonometric identities:Cos2θ = ½ (1+ Cos 2θ) Sin2θ = ½ (1- Cos 2θ) Sin θ cos θ = ½ Sin 2θ

( ) ( )

( )θ

γθ

εεγ

θγ

θεεεε

ε

2cos2

2sin22

2sin2

2cos22

11

1

xyyxyx

xyyxyxx

+−

−=

+−

++

=

11var yxyxiantIn εεεε +=+= ( )2

yxAverage

εεε

+=

PRINCIPAL STRAINS

The angle for the principal strains is :

The value for the principal strains are( )

yx

xy

yx

xy

P εεγ

εε

γθ

−=

−==

2

22tan

( )

( ) 22

2

22

1

222

222

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+

+=

xyyxyx

xyyxyx

γεεεεε

γεεεεε

Maximum Shear

The maximum shear strains in the xy plane are associated with axes at 45o to the directions of the principal strains

( )21

22

or 222

εεγγεεγ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+= Max

xyyxMax

For isotropic materials, at a given point in an stressed body, the principal strains and principal stresses occur in the same directions.

MOHR’S CIRCLE FOR PLANE STRAIN

It is constructed in the same manner as the Mohr’s circle for plane stress with the following similarities:

2

2

11

1

11

1

yx

x

xy

y

x

yx

x

xy

y

x

γε

γεε

τστσσ

Strain MeasurementsAn electrical-resistance strain gage is a device for measuring normal strains (ε) on the surface of a stressed object.The gages are small (less than ½ inch) made of wires that are bonded to the surface of the object. Each gage that is stretched or shortened when the object is strained at the point, changes its electrical resistance. This change in resistance is converted into a measurement of strain.From three measurements it is possible to calculate the strains in any direction. A group of three gages arranged in a particular pattern is called a strain rosette.Because the rosette is mounted in the surface of the body, where the material is in plane stress, we can use the transformation equations for plane strain to calculate the strains in various directions.

45° strain rosette, and element oriented at an angle θ to the xy axes.

General Equations

Other Strain Rosette

An element of material in plane strain undergoes the following strains: εx=340x10-6

; εy = 110x10-6 ; γxy = 180x10-6 . Determine the following quantities: (a) the strains of an element oriented at an angle θ = 30o ; (b) the principal strains and (c) the maximum shear strains.

(a) Element oriented at an angle θ = 30o (2θ = 60o)

( ) ( )

61

61

1

10360

10602

180602

1103402

110340

2sin2

2cos22

−

−

×=

×⎥⎦⎤

⎢⎣⎡ +

−+

+=

+−

++

=

x

x

xyyxyxx

SinCos

ε

ε

θγ

θεεεε

ε

με

εμμμ

εεεε

90

360110340

1

1

11

=

+=+

+=+

y

y

yxyx

με 225=Average

( )

6611

11

105510602

180602

1103402

2cos2

2sin22

−− ×−=×⎥⎦⎤

⎢⎣⎡ +

−−=

+−

−=

CosSinyx

xyyxyx

γ

θγ

θεεγ

(b) Principal Strains and Angle of Rotation

( )

μμμμε

μμμμε

γεεεεε

802

1802

110340225

3702

1802

110340225

222

22

2

22

1

22

2,1

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −

−=

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±

+= xyyxyx

019

7826.0110340

552tan

=

=−

−=

−=

P

yx

xyP

θ

μμμ

εεγ

θ

(c) In-Plane Maximum Shear Strain

( ) μμμεεγ

μμμμγεεγ

29080370

1452

1802110-340

222

21

2222

=−=−=

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=

Max

xyyxMax

με 225=Average

(d) Out-of-Plane Maximum Shear Strain

( ) μμμεεγ 3700370 31 =−=−=Max

( )θθγθθεθθεγ 2211 sincos2

cossincossin2

−++−= XYYX

YX

θθγθεθεε cossin22

sincos 221 ⎟

⎠⎞

⎜⎝⎛++= XY

YXX

θθγθεθεε cossin22

cossin 221 ⎟

⎠⎞

⎜⎝⎛−+= XY

YXY

Transformation Equations

[ ]

[ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

×=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

×=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

22

22

11

1

11

11

1

1

YX

Y

X

XY

Y

X

XY

Y

X

YX

Y

X

T

T

γεε

γεε

γεε

γεε

[ ]( )⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

θθθθθθθθθθ

θθθθ

22

22

22

sincoscossincossincossin2cossin

cossin2sincosT

( )⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

2sincoscossincossincossin2cossin

cossin2sincos

222

22

22

11

1

1

XY

Y

X

YX

Y

X

γεε

θθθθθθθθθθ

θθθθ

γεε

( )

μμγεε

γεε

γεε

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

8.556.883.361

90110340

5.0438.0438.0876.075.025.0

876.025.075.0

2

230sin30cos30cos30sin30cos30sin30cos30sin230cos30sin

30cos30sin230sin30cos

2

11

1

1

22

22

22

11

1

1

YX

Y

X

XY

Y

X

YX

Y

X

For Θ=30 degrees

[ ] TensorStrain

zzyzxz

zyyyxy

zxyxxx

_

21

21

21

21

21

21

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

εγγ

γεγ

γγε

ε

[ ]

μ

μ

εγγ

γεγ

γγε

ε

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⇒

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

000079000371

_

00001102

18002

180340

21

21

21

21

21

21

ValuesEigen

zzyzxz

zyyyxy

zxyxxx

Example

A 45o strain rosette (rectangular rosette) consists of three electrical-resistance strain gages, arranged to measure strains in two perpendicular directions and also at a 45o angle (as shown below). The rosette is bonded to the surface of the structure before it is loaded. Gages A, B and C measure the normal strains εa, εb and εc in the directions of the lines Oa, Ob and Oc, respectively.Explain how to obtain the strains εx1, εy1 and γx1y1, associated with an element oriented at an angle θ to the xy axes.

θθγ

θεθεε cossin22

sincos 22⎟⎟⎠

⎞⎜⎜⎝

⎛++= xy

yxa

Angles with respect to x-axis: (a) is zero ; (b) is 45 degrees CCW ; (c) 90 degrees CCW

0cos0sin22

0sin0cos 22⎟⎟⎠

⎞⎜⎜⎝

⎛++= xy

yxa

γεεε

( )45cos45sin22

45sin45cos 22⎟⎟⎠

⎞⎜⎜⎝

⎛++= xy

yxb

γεεε

( )90cos90sin22

90sin90cos 22⎟⎟⎠

⎞⎜⎜⎝

⎛++= xy

yxb

γεεε

cabxy

by

ax

εεεγ

εεεε

−−=

==

2

The following results are obtained from a 600 strain gauge rosette:Strain in direction of strain gauge A = 750μ;Strain in direction of SG B, 600 to A = 350μ;Strain in direction of SG C, 1200 to A = 100 μ.Determine the principal strains and their directions.

μεε 750== xa

( )

( ) ( ) ( )433.02

75.025.0

60cos60sin22

60sin60cos 22

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

xyyxb

xyyxb

γεεε

γεεε

( )

( ) ( ) ( )433.02

75.025.0

120cos120sin22

120sin120cos 22

−⎟⎟⎠

⎞⎜⎜⎝

⎛++=

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

xyyxb

xyyxb

γεεε

γεεε

μγμε

28950

==

⇒xy

y

[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⇒

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⇒

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

1000981.01945.001945.0981.0

_

000021000779

_

0000502

28902

289750

21

21

21

21

21

21

VectorsEigen

ValuesEigen

zzyzxz

zyyyxy

zxyxxx

μ

μ

εγγ

γεγ

γγε

ε

ArcCos(angle)=0.981

Angle =11.2degress

(A) Using the transformation equations define the maximum and minimum principal strains, maximum shearing strain and principal angles given εX = 3500μ ; εY = 700μ and γXY = -1050μ

(B) Repeat using the Mohr’s circle.

[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −⇒

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⇒

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

1000984.0178.00178.0984.0

_

00008.6040002.3595

_

00007002

105002

10503500

VectorsEigen

ValuesEigen μ

με

ArcCos(angle)=0.984

Angle =10.28degress

(c) In-Plane Maximum Shear Strain( ) μμμεεγ 4.29908.6042.3595 21 =−=−=Max

(d) Out-of-Plane Maximum Shear Strain

( ) μμμεεγ 2.359502.3595 31 =−=−=Max

The state of stress at a point in a structural member is determined to be as shown. Knowing that for this material E=210GPa and ν=0.3, use the Mohr’s circle to determine: (1) the principal stresses; (2) the in-plane maximum shear stress; (3) the absolute maximum shear stress; (4) principal angles; (5) the strains and the principal strains; (6) the maximum shear strain; (7) the principal angles.

14MPa

56MPa11.2MPa

x

y ( ) RAveragexyyxyx ±=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±⎟⎟

⎠

⎞⎜⎜⎝

⎛ += στ

σσσσσ 2

2

2,1 22

( ) ( )

( ) MPa

MPaR

Average 352

1456

8.232.112

1456 22

−=⎟⎠⎞

⎜⎝⎛ −+−

=

=+⎟⎠⎞

⎜⎝⎛ −−−

=

σ

MPaMPa

8.588.23352.118.2335

2

1

−=−−=−=+−=

σσ

1. Principal Stresses

[ ] MPaTensorStress⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−==

0000142.1102.1156

_σ MPaValuesEigen⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−⇒

8.580002.110000

_

MPaMPa

MPa

8.582.11

0

3

2

1

−=−=

=

σσσ

2. In-Plane Maximum Shear Stress MPaRMax 8.23==τ

3. The Absolute Maximum Shear Stress (Out of plane)

( ) MPaMax 4.292

8.5802

31 =−−

=−

=σστ

4. Angle between the x-axis and the Principal Stresses

( ) ( ) ( )( )

( )

deg07.282

533.0212.112tan

21456

2.11

2

2tan)_(

−=

−=−

=

−−−=

−==

P

P

yx

xyPPlaneIn

θ

θ

σστ

θ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⇒

12425.09701.009701.02425.0000

_VectorsEigen

For σ1=0ArcCos(angle)=0.0Angle =90degress

For σ2=-11.2MPaArcCos(angle)=0.2425Angle =76degress

For σ3=-58.8MPaArcCos(angle)=0.9701Angle =14degress

5. Strains and Principal Strains

EEE

EEE

EEE

zyxz

zyxy

zyxx

σσνσνε

σνσσνε

σνσ

νσε

+−−=

−+−=

−−=

GXY

XYτ

γ =

( ) ( )

( ) ( )

( ) ( ) ( ) μμε

μμε

μμε

100 2080 210000

143.0210000

563.0

4.136.6680210000

14210000

563.0

6.246206.266 210000

143.0210000

56

=+=−

−−

−=

=−=−

+−

−=

−=+−=−

−−

=

z

y

x

μτ

γ 13980770

2.11===

Gxy

xy

( ) GPaEG 77.803.012

210)1(2

=+

=+

=ν

[ ]

μ

μ

εγγ

γεγ

γγε

ε

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⇒

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.2640008.30000100

_

1000004.132

13902

1396.246

21

21

21

21

21

21

ValuesEigen

zzyzxz

zyyyxy

zxyxxx

( )

RAverage

xyyxyx

±=

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±

+=

εε

γεεεεε

2,1

22

2,1 222

( )

μ

με

4.1472

1392

4.136.246

6.1162

4.136.246

22

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −−

=

−=+−

=

R

Average

μεμε

2648.30

2

1

−==

6. Maximum Shear Strain and Absolute Maximum Shear Strain

( )

μγ

μγ

8.294

4.1472

1392

4.136.246_2

22

=

=⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −−

==

Max

Max Rplanein

( ) ( )

μγ

μεεγ

364

1822

2641002

__2

31

=

=−−

=−

=

Max

Max planeofout

( ) ( ) 534.04.136.246

139

2

22tan −=−−

=−

=−

==yx

xy

yx

xy

P εεγ

εε

γθ

7. In-plane and Out-of-plane angles

( )deg07.282533.02tan

−=−=

P

P

θθ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⇒

12425.09701.009701.02425.0000

_VectorsEigen

For ε1=100μArcCos(angle)=0.0Angle =90degress

For ε2=30.8μArcCos(angle)=0.2425Angle =76degress

For ε3=-264μArcCos(angle)=0.9701Angle =14degress