3D Stress Components

8/10/2019 3D Stress Components

1/71

From equilibrium principles:

xy

=

yx

,

xz

=

zx

,

zy

=

yz

3D Stress Components

Normal Stresses

Shear Stresses

Normal stress () : the subscript identifies the face on which thestress acts. Tension is positive and compression is negative.

Shear stress () : 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 itacts 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

xzyzxy


2/71

==

zyzxz

zyyxy

zxyxx

ij

For static equilibrium xy

=

yx

,

xz

=

zx

,

zy

=

yz

resulting insix independent

scalar quantities. These six scalars can be arranged in a 3x3 matrix, giving us astress

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).


3/71

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

==

3

2

1

'

00

00

00

'

ij

Eigen values

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

The principal stresses are the new-axescoordinate system. The angles between the

old-axesand the new-axesare known as theEigen-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

1 k1 l1 m1

2 k2 l2 m2

3 k3 l3 m3


4/71

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

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).


5/71

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.

000

0

0

yyxy

yxxx

000

0

0

1111

1111

yyyx

xyxx

000

00

00

2

1

1

22

1


6/71

Stresses on Inclined SectionsKnowing the normal and shear stresses acting in the element denoted by thexy axes,

we will calculate the normal and shear stresses acting in the element denoted by the

axisx1y1.

x1y1

x

y

y

x

x1

x1y1

yxxy

Equilibrium of forces:

Acting inx1

Cos

SinA

Cos

SinASinAA

AOYXOYOXYOX

OX +++= cossincos

cos1


7/71

EliminatingAo ,sec = 1/cos and

xy=yx

cossin2sincos 221 XYYXX ++=

Acting in y1

x1y1Aosec = xAosin + xyAocos + yAotancos yxAotansin

EliminatingAo ,sec = 1/cos and xy=yx

cossin2cossin 221 XYYXY +=

22

11 sincoscossincossin ++= xyyxyx


8/71


9/71

Case 1: Uniaxial stress

Special Cases

=

+=

===

2

2

2

21

00

1,1

1

xy

Sin

Cos

xyx

xx

yxy

Case 2 : Pure Shear

2

2

0

1,1

1

Cos

Sin

xyyx

xyx

yx

=

=

==

2cos2sin2

2sin2cos22

11

1

xy

yx

yx

xy

yxyx

x

+

=

+

++

=

Case 3: Biaxial stress

22

222

0

1,1

1

Sin

Cos

yx

yx

yxyx

x

xy

=

++

=

=


10/71

An element in plane stress is subjected to stresses x=16000psi, y=6000psi, and

xy=yx= 4000psi (as shown in figure below). Determine the stresses acting on an

element inclined at an angle =45o (counterclockwise - ccw).

Solution: We will use the following transformation equations:

2cos2sin

2

2sin2cos22

2sin2cos22

11

1

1

xy

yx

yx

xyyxyxy

xy

yxyx

x

+

=

+

=

+++=

( )600016000 ++


11/71

Numerical substitution

( )

( )

( ) ( )

( ) ( )( ) ( ) psi

psi

psi

psi

psi

psi

yx

y

x

xy

yx

yx

50000400015000

7000140000500011000

15000140000500011000

4000

50002

600016000

2

110002

600016000

2

11

1

1

=+===

=++=

=

=

=

=+

=+

( ) ( )( ) ( ) 0902

1902

45

0

0

0

==

==

+=

CosCos

SinSin

(ccw)For x-axis

( ) ( )( ) ( ) 02702

12702

9045

0

0

00

====

++=

CosCos

SinSin

(ccw)For y-axis

Note:11 yxyx

+=+


12/71

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 principalstresses, the maximum shear stress and the angle of inclination for the principal

stressesSolution: We will use the following transformation equations:

2cos2sin2

2sin2cos22

11

1

xy

yx

yx

xyyxyx

x

+

=

+++=

( )

( )

( ) ( )( ) ( )( )

( )( ) ( )( ) MPa

MPaMPa

MPa

MPa

yx

x

xy

yx

yx

31866.0195.029

6.325.019866.0291719

292

1246

2

172

1246

2

11

1

=+=

=++==

=

=

=+

=+

( ) ( )( ) ( ) 866.0302

5.0302

15

0

0

0

==

==

=

CosCos

SinSin

(cw)For x-axis

MPay

yxyx

4.11

11

=

+=+

A rectangular plate of dimensions 3 0 in x 5 0 in is formed by welding two


13/71

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 positivewhen it acts counterclockwise against the weld).

Solution:

psi

psipsi

psipsipsi

y

yxx

xy

yxyx

25

375375

04252

1752

1

111

===

==

=+

00 926129630

5

3tan

..

For x-axis

==

=

psiy

yxyx

251

11

=

+=+

-375psi

= 30.96o

375psi25psi

x

y

Stresses acting on the weld

w

w

25psi

375psi = 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.

w

= -25psi and

w

= 375psi


14/71

Principal Stresses and Maximum Shear Stresses

The sum of the normal stresses acting on perpendicular faces of plane stress

elements is constant and independent of the angle .

YXYX +=+ 11As we change the angle there will bemaximum 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 bytaking the derivative of x1 with respect

to and setting equal to zero.

=

=+=

+

++

=

2

2tan

02cos22sin)(

2sin2cos22

1

1

yx

xy

P

xyyxx

xy

yxyx

x

cossin2sincos 221 XYYXX ++=

cossin2cossin 221 XYYXY +=


15/71

The subscriptp indicates that the anglep defines the orientation of the principal

planes. The angle p has two values that differ by90o. They are known as the principal

angles. For one of these angles x1 is amaximumprincipal stress and for the other a

minimum. The principal stresses occur in mutually perpendicular planes.

( )2

2tanyx

xy

P

=

2

)(2cos2sin

RR

yx

P

xy

P

==

P

1

stressmaximumfor the

2sin2cos22

=

+

++

= xyyxyx

x

( )

( )

( ) ( )22

12

2

2

2

1

2

2

1

22

2But

2

1

2

11

22222

xyyxyx

xyyx

xy

yxyx

xy

yxyxxy

xy

yxyxyx

R

R

RRRR

+

++=+

=

+

++

=

+

+

+=

+

+

+=

P i i l St


16/71

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. Thethird stress is zero in

plane stress conditions

( )

( ) R

R

Averagexy

yxyx

Averagexy

yxyx

=+

+=

+=+

+

+=

2

2

2

2

2

1

22

22

Principal Stresses

The location of the angle for the maximum shear stress is obtained by taking thederivative of x1y1 with respect to and setting it equal to zero.

2cos2sin

211 xy

yx

yx +

=

( )

xy

yx

S

xyyx

yx

22tan

02sin22cos)(11

=

==

Maximum Shear Stresses

045d)(

2i2

yxxy


17/71

( )

( )

( )

( )o

P

o

P

P

o

P

o

P

P

S

S

P

Pxy

yx

S

902cos

902sin

290cos

290sin

2sin

2cos

2cos

2sin

2cot2tan

1

2

)(2tan

=

==

==

=

0

P1s111 45and2

)(2sin2cos ===

RR

yx

s

xy

s

The planes for maximum shearstress occurs at45o to the principal

planes. The plane of the maximum

positive shear stress max is defined

by the angle S1 for which thefollowing 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

2

( )

2

21 =

MAX

Therefore,2s-2p=-90o or

s= p +/- 45o

2

yx

AVER

+

=


18/71

Equations of a Circle

2sin2cos2

2sin2cos22

1

1

xy

yx

AVERx

xy

yxyx

x

+

=

+

++

=General equation

Consider

Equation (1)

Equation (2)

2cos2sin211 xy

yx

yx +

=

Equation (1) + Equation (2)

( )2

2

1 2sin2cos2

+= xy

yx

AVERx

( )2

2

11 2cos2sin2

+

=

xy

yx

yx

( )22


19/71

( ) ( )2112

1 2cos2sin2

2sin2cos2

+

+

+

=+

xy

yx

xy

yx

yxAVERx

( ) ( )

( ) ( )

( ) 222

222

22

222

22

2

2cos2sin2

22cos2sin2

2cos2sin2

2cos2sin2

22sin2cos2

2sin2cos2

RSUM xyyx

xy

yx

xy

yx

xy

yx

xyyx

xyyx

xyyx

=+

=

+

=

+

++

=

+

( ) ( )22

11

2

1 RyxAVERx =+

Mohr Circle


20/71

x

y

P

C

( )22

2 xy

yxR

+

=

+

2

yx

The radius of the Mohr circle is the

magnitudeR.

Mohr Circle

( )22

2 xy

yxR

+

=

The center of the Mohr circle is the

magnitude

2

yx

AVER

+=

( ) ( ) ( )22

211

21

2 xyyxyxAVERx

+

=+State of Stresses


21/71

Alternative sign conversion for shear

stresses:

(a)clockwise shear stress,

(b)counterclockwise shear stress, and

(c) axes for Mohrs circle.

Note that clockwise shear stresses are

plotted upward and counterclockwise

shear stresses are plotted downward.

a) We can plot the normal stress x1 positive to the right and the shear stress

x1y1 positive downwards, i.e. the angle2 will be positive when

counterclockwise or

b) We can plot the normal stress x1positive to the right and the shear stress x1y1positive upwards, i.e. the angle2 will be positive when clockwise.

Both forms are mathematically correct. We use (a)

Forms of Mohrs Circle

T f f M h i l


22/71

Two forms of Mohrs circle:

x1y1 is positive downward and the angle2 is

positive counterclockwise, and

x1y1 is positive upward and the angle2 is positive

clockwise. (Note: The first form is used here)

Construction of Mohrs circle for plane stress.


23/71

Construction of Mohr s circle for plane stress.


24/71

At a point on the surface of a pressurized cylinder, the

material is subjected to biaxial stresses x = 90MPa and y =

20MPa as shown in the element below.

Using the Mohr circle, determine the stresses acting on an

element inclined at an angle = 30o (Sketch a properly

oriented element).

Because the shear stress is zero,

these are the principal stresses.

Construction of the Mohrs circle

(x = 90MPa, y = 20MPa and xy = 0MPa)

( )MPa

yx

Average 552

2090

2=

+=

+=

The center of the circle is

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

yx350

2

2090

2

2

2

2

2

=+

=+

=

Stresses on an element inclined at = 30o


25/71

Stresses on an element inclined at = 30

By inspection of the circle, the coordinates of pointD are

( )MPaCosCosR

MPaSinSinR

MPaCosCosR

Averagey

oo

yx

Averagex

5.3760355560

3.30603560

5.7260355560

0

1

11

0

1

===

===

=+=+=

An element in plane stress at the surface of a large


26/71

machine is subjected to stresses x = 15000psi, y =

5000psi and xy = 4000psi.

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

angle = 40o

b) The principal stresses and

c) The maximum shear stresses.

Construction of Mohrs circle:

Center of the circle (Point C):

Radius of the circle:

Point A, representing the stresses on thex face of the element ( = 0o

) has thecoordinates x1 = 15000psi and x1y1 = 4000psi

Point B, representing the stresses on they face of the element ( = 90o) has the

coordinates y1 = 5000psi and y1x1 = - 4000psiThe circle is now drawn through points A and B with center C and radius R

( )psi

yx

Average 100002

500015000

2=

+=

+=

( ) ( ) psiR xyyx

640340002

500015000

2

2

2

2

2

=+

=+

=

Stresses on an element inclined at


27/71

By inspection the angle ACP1 for

the principal stresses (point P1) is :

Then, the angle P1CD is 80o

38.66o = 41.34o

( ) oACPACP 66.385000

4000tan 11 ==

= 40o

These are given by the coordinates of

pointD which is at an angle2 = 80o

from point A

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

( )

psiCosCosR

psiSinSinR

psiCosCosR

ooAveragey

oo

yx

oo

Averagex

519034.4164031000034.41

423034.41640334.41

1481034.4164031000034.41

1

11

1

===

===

=+=+=

And of course, the sum of the normal stresses is


28/71

Principal Stresses

The principal stresses are represented by

points P1 and P2 on Mohrs circle.

The angle it was found to be 2 = 38.66o

or = 19.3o

psiR

psiR

Average

Average

3597640310000

16403640310000

2

1

===

=+=+=

14810psi + 5190psi = 15000psi + 5000psi

Maximum Shear Stresses


29/71

These are represented by point S1 and S2 in Mohrs circle. Algebraically the

maximum shear stress is given by theradius of the circle.

The angle ACS1 from point A to point S1 is 2 S1 = 51.34o. This angle is negativebecause is measured clockwise on the circle. Then the corresponding S1 value is 25.7o.

0400015000 0016403


30/71

Psi

000

050004000

0400015000

Psi

000

035970

00164033-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 betweenthe old-axes and the new-axes are known as the Eigen-vectors.

principal stress

Cosine of angle

between X and theprincipal stress

Cosine of angle

between Y and theprincipal stress

Cosine of angle

between Z and theprincipal stress

16403.1242 0.94362832 0.331006939 0

3596.876 -0.33101 0.943628 0

0 0 0 1


31/71

At a point on the surface of a generator shaft the stresses are

x = -50MPa, y = 10MPa and xy = - 40MPa as shown in the

figure. Using Mohrs circle determine the following:(a)Stresses acting on an element inclined at an angle = 45o,(b)The principal stresses and

(c)The maximum shear stresses

Construction of Mohrs circle

Center of the circle (Point C):

Radius of the circle:.

Point A, representing the stresses on thex face of the element ( = 0o) has the

coordinates x1 = -50MPa and x1y1 = - 40MPaPoint B, representing the stresses on they face of the element ( = 90o) has thecoordinates y1 = 10MPa and y1x1 = 40MPaThe circle is now drawn through points A and B with center C and radius R.

( ) ( )MPa

yx

Average 202

1050

2=

+=

+=

( ) ( ) ( ) ( ) MPaR xyyx 504021050

22

2

2

2

=+

=+

=

Stresses on an element


32/71

inclined at = 45o

These stresses are given by

the coordinates of point D(2 = 90o of point A). Tocalculate 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

oo

Averagey

oo

yx

oo

Averagex

2087.36502087.36

3087.365087.36

6087.36502087.36

1

11

1

=+==

===

=+=+=

tan ACP2=40/30=4/3

ACP2=53.13o

P2CD = 90o 53.13o = 36.87o

Principal Stresses

Th d b i P dMaximum Shear Stresses


33/71

They are represented by points P1 and

P2 on Mohrs circle.

The angle ACP1 is 2P1 = 180o + 53.13o

= 233.13o or P1 = 116.6o

The angle ACP2 is 2P2 = 53.13o or P2= 26.6o

These are represented by point S1 and S2in Mohrs circle.

The angle ACS1 is 2S1 = 90o + 53.13o =143.13o or = 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

====+=+=

04050 0030


34/71

MPa

000

01040

04050

MPa

000

0700

00303-D stress state Transform to

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

The principal stresses are the new-axes coordinate system. The angles betweenthe old-axes and the new-axes are known as theEigen-vectors.

principal stress

Cosine of angle

between X and theprincipal stress

Cosine of angle

between Y and theprincipal stress

Cosine of angle

between Z and theprincipal stress

30 -0.44721359 0.894427193 0

-70 0.894427 0.447214 0

0 0 0 1

Hookes Law for Plane Stress


35/71

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 Hookes law (linearly elastic material)

Hookes law: Linear relationship between stress and strain

For uniaxial stress:

(E = modulus of elasticity or Youngs modulus)

Poissons ratio:

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

E=allongitudin

transverse

straina

strainlateral

==

xial

G=

Element of material in plane stress (z

= 0).


36/71

Consider the normal strains x

,

y

,

z

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 x

in the x direction:a)Due to the stress

x

is equal to x

/E .

b)Due to the stress y

is equal to y

/E .

The resulting strain is: EE

EE

EE

yx

z

yxy

yxx

=

+=

=

The shear stress causes a distortion of the element

h th t h f b h b


37/71

such that each z face becomes a rhombus.

GXYXY

=

The normal stressesx

and y

have no

effect on the shear strain xy

or rearranging the equations:

G

XYXY

=

( ) ( ) ( ) ( ) XYyx

Y

yxx G

EEEE

=+

+

=

+

= XY2222 11

11

EEEEEE

yxz

yxy

yxx

=+==

These equations are known collectively as Hookes Law for plane stress

These equations contain three material constants (E, G and

) but only two are

independent because of the relationship:

)1(2 += EG

Special cases of Hookes law

z

= 0)


38/71

EEx

zx

x

xyy

===

==

y

00

Uniaxial stress :

G

xy

xyzyx

yx

====

==

0

0

Pure Shear :

Biaxial stress :

-

- yxyx

y

yxx

xy

=+==

=

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

V

1

= a + a

x

) b + b

y

) c + c

z

)

V

1

= abc 1+

x

) 1+

y

) 1+

z

)

Volume Change

Upon expanding the terms:


39/71

V

1

= V

o

1 +

x

+

y

+

z

+

x

y

+

x

z

+

y

z

+

x

y

z

)

For small strains:

V

1

= V

o

1 +

x

+

y

+

z

)

The volume change is

V = V

1

V

o

= V

o

x

+

y

+

z

)

The unit volume change e, also known as dilatation is defines as:e = V / V

o

=

x

+

y

+

z

Positive strains are considered for elongations and negative strains for

shortening, i.e. positive values of efor an increase in volume.

EEEEEE

yxz

yxy

yxx

=+==

( )( )EV

Ve yx 21+== For uniaxial stress y = 0 ( )

Ee x 21=

We can notice that the maximum possible value of Poissons ratio is 0.5, because a

larger value means that the volume decreases when the material is in tension(contrary to physical behavior).

STRAIN ENERGY DENSITY IN PLANE STRESS


40/71

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

( )( )( )

( )( ) ( )y

x

bac

abc

2direction-yin thedoneWork

2direction-xin thedoneWork

y

x

=

=

The sum of the energies due to normal stresses:

( )yyxxabcU += 2

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

2

11

Similarly, the strain energy density associated with the

shear strain:

By combining the strain energy densities for the normaland shear strains:

xyxyu 2

12=

( )xyxyyyxxu ++=21

The strain energy density in terms

f t l GEEEu XYYXYX

222

222

++=


41/71

of stress alone: GEEE 222

The strain energy density in termsof strain alone: ( )( )

2222 2

212

XYYXYX GEu ++=

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

2

or2

000

22

xx

xyxyxyy

Euu

==

====

2

or2

000

22

xyxy

xyyx

Gu

Gu

==

====

An element of the material subjected to normal

stresses x

,

y

and z

acting in three mutually

perpendicular directions is said to be in a stateof triaxial stress. Since there is no shear in x, y

or z faces then the stresses x

, y

and z

are

the principal stresses in the material.

TRIAXIAL STRESS

If an inclined plane parallel to thez-axis is cut through the element, the only stress

of the inclined face are the normal stress and the shear stress both of which act


42/71

of the inclined face are the normal stressand the shear stress,both of which act

parallel to thexyplane.

The same general conclusion hold for normal and shear stresses

acting on inclined planes cut through the element parallel to the

x andy axes.

Maximum Shear Stress For a material in triaxial stress, the maximum shear stressesoccur on elements oriented at angles of45o to thex,y andz

axes.

( )

( )

( ) 2

2

2

ZX

YMAX

ZY

XMAX

YX

ZMAX

=

=

=

Because these stresses are independent of the z, we can usethe transformation equations of plane stress, as well as the

Mohrs circle for plane stress, when determining the stresses

and 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


43/71

various angles to thex,y andz axes can be

visualized with the aid of the Mohrs

circle.

In this case x

>

y

>

z

Hookes Law for Triaxial Stress

If H k l i b d i i ibl b i h


44/71

If Hookes 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

zyxz

zyxy

zyxx

+=

+=

=

Or

( )( )( )[ ]

( )( ) ( )[ ]

( )( ) ( )[ ]zyxz

zyxy

zyxx

E

E

E

+++

=

+++

=

+++=

1211

1211

1211

They are known as the

Hookes 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

ZYX

O

EV

Ve

VVe

++==

++==

21

thenapply,lawssHooke'If

Strain Energy Density


45/71

The strain energy density for an

element in triaxial stress is

obtained by the same method

used for plane stress.

( ) ( )zyzxyxzyx

zzyyxx

EEu

u

++++=

++=

-2

1

:stressesofIn terms

222

222

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

2112

222

zyzxyxzyx

Eu

+++++

+=

In terms of the strains:

Spherical StressA special case of triaxial stress, called spherical stress, occurs

whenever all three normal stresses are equal:0 === zyx

The Mohrs 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 stressso andit is a principal plane.

The normal strains in spherical stress are also

the same in all directions, provided the material

is isotropic and if Hookes law applies:

( )

( )

E

O

OO

2133e

changevolumeThe21

O

==

=

K= bulk or volume modulus of elasticity

E


46/71

Element in spherical stress.

( ) eK

EK 0

213

=

=

If = 1/3 thenK = E

If = 0 thenK = E/3

If = 1/2 thenK= 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


47/71

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 thexyplane, i.e. it has only three strain components x, y and xy.

Plain stress is analogous to plane stress, but under ordinary conditions they do

not occur simultaneously

Exception when x = -y and when = 0

Strain components x, y, and xy in the

xyplane (plane strain).


48/71

Comparison of plane stress and plane strain.

Th f i i f l lid h i

APPLICATION OF THE TRANSFORMATION EQUATIONS


49/71

The transformation equations for plane stress are valid even when z is not zero,

because z

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 thexyplane are valid even when z is not zero, because the strain zdoes 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.

Transformation Equations for Plain Strain

We will assume that the strain x, y and xyassociated with thexyplane are known.

We need to determine the normal and shear strains(x1 and x1y1) associated with thex1y1 axis. y1 can

be obtained from the equation of x1by substituting

+90 for .

For an element of sizedx,dy

In thex direction:


50/71

the strain xproduces an elongation xdx.

The diagonal increases in length by xdx cos.

In they direction:

the strain yproduces an elongation ydy.

The diagonal increases in length by ydy sin .

The shear strain xy in the planexyproduces a

distortion of the element such that the angle at thelower left corner decreases by an amount equal to

the shear strain. Consequently, the upper face

moves to the right by an amount xydy. This

deformation results in an increase in the length ofthe diagonal equal to: xydy cos

The total increase of the diagonal is the sum of the preceding three expressions,

thus: ( ) ( ) ( ) CosdySindyCosdxd xyyx ++=


51/71

( ) ( ) ( ) yy xyyx

But ( ) ( ) ( )

Sinds

dyCos

ds

dx

Cosds

dySinds

dyCosds

dx

ds

dxyyxx

==

++=

=

1

CosSinSinCos xyyxx ++=22

1

Shear Strain x1y1 associated withx1y1 axes.

This strain is equal to the decrease in angle

between lines in the material that were initially

along thex1 andy1 axes.

Oa and Ob were the lines initially along thex1 and

y1 axis respectively. The deformation caused by

the strains x, y and xy caused the Oa and Ob lines

to rotate and angle and from thex1 andy1 axis

respectively. The shear strain x1y1

is the decrease

in angle between the two lines that originally were

at right angles, therefore, x1y1=+.

The angle can be found from the deformations

produced by the strains x,y and xy . The strains


52/71

p y x y xyx andxyproduce a clockwise rotation, while the

strain y produces a counterclockwise rotation.

Let us denote the angle of rotation produced by

x , y and xy as 1 , 2 and 3 respectively.

dsdySin

ds

dyCos

ds

dx

Sin

xy

y

x

=

=

=

1

2

1 Sinds

dy

Cosds

dx

==

( ) 2321 SinCosSin xyyx =+=


53/71

PRINCIPAL STRAINS

Th l f th i i l t i ixy


54/71

The angle for the principal strains is :

The value for the principal strains are ( ) yxxy

yx

xy

P

=

==

2

22tan

( )

( ) 222

22

1

222

222

+

+=

+

+

+=

xyyxyx

xyyxyx

Maximum Shear

The maximum shear strains in

thexyplane are associated withaxes at45o to the directions of

the principal strains

( )21

22

or222

=

+

+= Max

xyyxMax

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

MOHRS CIRCLE FOR PLANE STRAIN

It is constructed in the same mannerx

x


55/71

as the Mohrs circle for plane stress

with the following similarities:

2

2

11

1

11

1

yx

x

xy

y

yx

x

xy

y

Strain Measurements

An electrical-resistancestrain gage is a device for measuring normal strains () on

the surface of a stressed object


56/71

the surface of a stressed object.The gages are small (less than 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 astrain rosette.

Because the rosette is mounted in the surface of the body, where the material is inplane 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

to thexy axes.

General Equations


57/71

Other Strain Rosette

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

; = 110x10-6 ; = 180x10-6 Determine the following quantities:


58/71

;y = 110x106 ; xy = 180x10

6 . Determine the following quantities:

(a) the strains of an element oriented at an angle = 30o

;(b) the principal strains and

(c) the maximum shear strains.

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

6

1

6

1

1

10360

10602

18060

2

110340

2

110340

2sin2

2cos22

=

+

++

=

+

++

=

x

x

xyyxyx

x

SinCos

90

360110340

1

1

11

=

+=+

+=+

y

y

yxyx

225=Average

6611

11

105510602

18060

2

110340

2

2cos2

2sin22

=

+

=

+=

CosSinyx

xyyxyx

(b)Principal Strains and Angle of Rotation

( ) 22

+ 78260

552tan =

== xy


59/71

( )

802

180

2

110340225

3702

180

2

110340225

222

22

2

22

1

2,1

=

+

=

=

+

+=

+

+= xyyxyx

019

7826.0110340

2tan

=

=

=

=

P

yx

P

(c) In-Plane Maximum Shear Strain

180110-3402222

xyyxM


60/71

( )

29080370

145

2

180

2

110340

222

21 ===

=

+

=

+

+=

Max

xyyxMax

225=Average

(d) Out-of-Plane Maximum Shear Strain

( ) 370037031 ===Max

cossin22

sincos 221

++= XYYXX

Transformation Equations


61/71

( ) 2211 sincos2

cossincossin

2

++= XYYXYX

21

YXX

cossin22

cossin 221

+= XYYXY

[ ]

[ ]

=

=

22

22

11

1

1

1

11

1

1

YX

Y

X

XY

Y

X

XY

Y

X

YX

Y

X

T

T

[ ]

( )

=

22

22

22

sincoscossincossin

cossin2cossin

cossin2sincos

T

( )

=

2

sincoscossincossin

cossin2cossincossin2sincos

2

22

22

22

11

1

1

XY

Y

X

YX

Y

X

30cos30sin230sin30cos22

1 XX

For =30 degrees


62/71

( )

=

=

=

8.55

6.883.361

90

110340

5.0438.0438.0

876.075.025.0876.025.075.0

2

230sin30cos30cos30sin30cos30sin30cos30sin230cos30sin

2

11

1

1

22

22

11

1

YX

Y

X

XY

Y

YX

Y

zxyxxx

112

1

2

1


63/71

[ ] TensorStrain

zzyzxz

zyyyxy _

2

1

2

12

1

2

1

=

=

[ ]

=

=

000

0790

00371

_

000

01102

180

02

180340

2

1

2

121

21

2

1

2

1

ValuesEigen

zzyzxz

zyyyxy

zxyxxx

Example

A45o strain rosette (rectangular rosette) consists of three electrical-resistance strain gages,

arranged to measure strains in two perpendicular directions and also at a45o angle (as shown

b l ) Th i b d d h f f h b f i i l d d G A B d


64/71

below). The rosette is bonded to the surface of the structure before it is loaded. GagesA,B and

Cmeasure the normal strainsa, b and c in the directions of the lines Oa, Ob and Oc,respectively.

Explain how to obtain the strainsx1, y1 and x1y1, associated with an element oriented at an

angle

to thexy axes.

cossin22

sincos22

++= xyyxa

Angles with respect to x-axis: (a) is zero ; (b) is

45 degrees CCW ; (c) 90 degrees CCW

0cos0sin220sin0cos

22

++=

xy

yxa

( )45cos45sin22

45sin45cos 22

++= xyyxb

( )90cos90sin22

90sin90cos 22

++= xyyxb

cabxy

by

ax

=

==

2

The following results are obtained from a 600 strain gauge

rosette:


65/71

Strain in direction of strain gauge A = 750;

Strain in direction of SG B, 600 to A = 350;

Strain in direction of SG C, 1200 to A = 100 .

Determine the principal strains and their directions.

750== xa

( )

( ) ( ) ( )433.0275.025.0

60cos60sin22

60sin60cos 22

++=

++=

xy

yxb

xy

yxb

( )

( ) ( ) ( )433.02

75.025.0

120cos120sin22

120sin120cos 22

++=

++=

xy

yxb

xy

yxb

289

50

=

=

xy

y

[ ]

=

= 050289

02

289750112

1

2

1zxyxxx


66/71

[ ]

=

=

100

0981.01945.0

01945.0981.0

_

000

0210

00779

_

000

0502

2

1

2

1 22

VectorsEigen

ValuesEigen

zzyzxz

zyyyxy

ArcCos(angle)=0.981Angle =11.2degress

(A) Using the transformation equations define the maximum and minimum

principal strains, maximum shearing strain and principal angles given


67/71

p p , g p p g g

X = 3500 ; Y = 700 and XY = -1050

(B) Repeat using the Mohrs circle.

[ ]

=

100

0984.0178.0

0178.0984.0

_

000

08.6040

002.3595

_

000

07002

1050

02

10503500

VectorsEigen

ValuesEigen

ArcCos(angle)=0.984

Angle =10.28degress

(c) In-Plane Maximum Shear Strain

( ) 4.29908.6042.359521 ===Max

(d) Out-of-Plane Maximum Shear Strain

( ) 2.359502.359531 ===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

=0.3, use the Mohrs circle to

determine: (1) the principal stresses; (2) the in-plane maximum shear stress; (3) the


68/71

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 principalstrains; (6) the maximum shear strain; (7) the principal angles.

14MPa

56MPa

11.2MPa

x

y ( ) RAveragexy

yxyx =+

+= 2

2

2,122

( )

( )( )

MPa

MPaR

Average 352

1456

8.232.112

1456 22

=

+=

=+

=

MPaMPa8.588.23352.118.2335

2

1

===+=

1. Principal Stresses

[ ] MPaTensorStress

==000

0142.11

02.1156

_ MPaValuesEigen

8.5800

02.110

000

_

MPaMPa

MPa

8.582.11

0

3

2

1

==

=

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

3 The Absolute Maximum Shear Stress (Out of plane)


69/71

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

( )MPaMax 4.29

2

8.580

231 =

=

=

4. Angle between the x-axis and the

Principal Stresses

( ) ( ) ( )( )

( )

deg07.282

533.021

2.112tan

21456

2.11

2

2tan)_(

=

=

=

=

==

P

P

yx

xy

PPlaneIn

12425.09701.0

09701.02425.0

000

_VectorsEigen

For 1=0ArcCos(angle)=0.0

Angle =90degress

For 2=-11.2MPaArcCos(angle)=0.2425

Angle =76degress

For 3=-58.8MPaArcCos(angle)=0.9701

Angle =14degress

5. Strains and Principal Strains

EEEzyx

x

= ( ) ( ) 6.246206.266

210000

143.0

210000

56=+=

=x


70/71

EEE

zyxz

zyxy

+=

+=

G

XYXY

=

( ) ( )

( ) ( ) ( )

1002080210000

143.0

210000

563.0

4.136.6680210000

14210000

563.0

=+=

=

==+=

z

y

13980770

2.11===

G

xy

xy

( ) GPaE

G 77.803.012

210

)1(2 =+=+=

[ ]

=

=

0.26400

08.300

00100

_

1000004.132139

02

1396.246

2

1

2

12

1

2

12

1

2

1

ValuesEigen

zzyzxz

zyyyxy

zxyxxx

( )

RAverage

xyyxyx

=

+

+=

2,1

22

2,1222

( )

4.1472

139

2

4.136.246

6.1162

4.136.246

22

=

+

=

=+

=

R

Average

264

8.30

2

1

=

=

6. Maximum Shear Strain and Absolute Maximum Shear Strain

( )

4.1472

139

2

4.136.246_

2

22

=

+

==Max Rplanein


71/71

8.294=Max

( ) ( )

364

1822

264100

2__

2

31

=

=

=

=

Max

Max planeofout

( ) ( ) 534.04.136.246139

2

22tan ===== yx

xy

yx

xy

P

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

( )

deg07.282

533.02tan

=

=

P

P

12425.09701.009701.02425.0

000

_VectorsEigen

For 1=100

ArcCos(angle)=0.0Angle =90degress

For 2=30.8


For 3=-264


Documents

3D Stress Components