2)Stress Analysis 2

7/27/2019 2)Stress Analysis 2

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2 STRESS ANALYSIS

2.1 stress state of a point

2.1.1 I nternal forces and external forces

Questions:

• What is stress?

• What is the cause of stress?• How many stress components need we know to specify

the unique state of a stressed point in a continuous body?

2.1.2 Contact force and body force

2.1.3 Stress at a point in a continuous body is in equi l ibrium

Stress at a point in the considered plane can be defined as A F

A

0lim

Note: 1) Residual stress are not considered in this course;

2) Stress at a point varies with the plane it acts upon. Therefore,

whenever we refer to a stress, we must specify the plane.



2 STRESS ANALYSIS

The stress state of the point depends on the stresses on

all the surfaces of arbitrarily selected polyhedron surrounding the point.When the volume of the polyhedron is contracted around the point,

stress on each face of the polyhedron will approach a constant value. In

the most commonly used Cartesian system, hexahedron can be used.

The total number of stress components on six faces is 6 3 = 18 .But what is the least number of independent stress components?

2.1.5 Stress state as represented by a stress tensor

2.1.4 Resolution of stress into normal stress and shear stress

Stress can be conveniently resolved intonormal stress and shear stress and A

F t A

0lim

A

F n A

0lim

Stress state* - The overall stress situationQuestion: How many independent stress components do we need to

describe a stress state at a point?

Reasoning:



2 STRESS ANALYSIS

Z

X Yo

y

z

x

xy yx

xz

zx yz

zy

xz

designating the direction

of the normal to the the

plane on which the stress

acts.

Indicating the direction

of the stress component

A double subscript notationis often used to define eachstress component

For simplicity, it is customary

to delete the second subscript

for normal stress component

P



2 STRESS ANALYSIS

• For each stress component in a positive surface, there should be a

stress component in a corresponding opposite surface with the samemagnitude but in the opposite direction.

• From couple equilibrium ( the components of the stresses are

balanced ), we may find that

xy= yx; yz = zy; zx= xz

Or ij = ji for short

The nine components, of which 6 are independent, make up a new

physical quantity defining the stress state at a point ( if we know this

set of stress components, we will be sure to find the stress acting on any

plane passing through the point by force equilibrium). This quantity is

called stress tensor and can be written in the form as shown below:



2 STRESS ANALYSIS

Sometimes. it is more convenient to employ other

coordinates systems, such as cylindrical and spherical polar coordinate systems, to specify the stress components at a point.

In those cases, the subscripts may change accordingly, but the

number of independent stress components remain the same.

In general, the stress state of a point can be specified by a

group of components acting on any three mutually orthogonal

surfaces intersecting at the point.

Why?

z zy zx

yz y yx

xz xy x

ij

where

ij is the short form for thetensor while i and j are iterated

over x, y and z respectively.

Why?



2 STRESS ANALYSIS

Question: If given the stress tensor of a point ( ij , which is

composed of 9 components), are we able to find

the stress at the same point but on a oblique plane

arbitrarily inclined to the three Cartesian axes? If

so, how do we do it? With this question in mind,

we come to the following section.

Question: • Will the stress state of a point vary with thechange of coordinate system?

• Will the value of each stress component of a

stress tensor vary with the change of

coordinate system?



2 STRESS ANALYSIS

S

2.2 Stress on a oblique plane

z

A

C

B

O

Z

Y

X

o

y

x

S z

yx

xz

zx

zy

xy

yz

S yS x

N(l,m,n)

Given a stress state ij of

a point “O”, the relatedstress components areshown in the figure. Letthe area of the oblique

plane ABC be unity, then

area OBC=l; OAC=mOAB=n

where l , m and n are thedirection cosines of the

normal to the oblique plane. The resultant stresson the oblique plane isdenoted S which can beresolved into S x , S y and S z

respectively in OX, OYa n d O Z d i r e c t i o n s .



2 STRESS ANALYSIS

Since the tetrahedron is in equilibrium, the resultant force in all the

three coordinates direction should be zero,

The resultant stress S can be resolved into normal stress S n and

shear stress S s . Resolving forces normal to the oblique plane produces

0

0

0

nml S

nml S

nml S

z yz xz z

zy y xy y

zx yx x x

2.1

nS mS l S S z y xn 2.3

Substituting for S x , S y and S z from equation (2.1) gives normal stress

)(2222 nl mnlmnml S zx yz xy z y xn 2.4

222

z y x S S S S The resultant stress: 2.2



2 STRESS ANALYSIS

The shear stress S s can be determined by

22

n s

S S S

Homework:

2

31

2

2

1

According to the stressed element in the

figure ,

1) write down the values of each stresscomponent ;

2) write down the stress state of the stressed

element in tensor format and

3) determine the normal stress and shear

stress in a plane with its normal in equal

angle to the three coordinate axes.

Z

YX

o



2 STRESS ANALYSIS

2.3 Principal stresses

If the direction of the vector S and S n coincide, then the shear stress

on the oblique plane is zero. The oblique plane is then a principal plane,

S n is then the principal stress and its direction is a principal direction.

Hence,

If these values are substituted into equation (2.1), we then have

nS

mS l S

z

y

x

2.5

0)(

0)(

0)(

nml nml S

nml nml S

nml nml S

z yz xz z yz xz z

zy y xy zy y xy y

zx yx x zx yx x x

2.6



2 STRESS ANALYSIS

Equations (2.6) are consistent in l , m and n . To find nontrival

solution for l , m and n , the determinant of the coefficients must

vanish, resulting in

0

z zy zx

yz y yx

xz xy x

2.7

This determinant can be expanded as

where

032

2

1

3 J J J 2.8

)(2

)(

222

3

222

2

1

xy z zx y yz x zx yz xy z y x

zx yz xy x z z y y x

z y x

J

J

J

2.9



2 STRESS ANALYSIS

The cubic equation (2.8) has three real roots which will be designated 1,

2, and 3, respectively. Once the principal stresses are known, we may

put them back in equation (2.6), with the constraint condition

we are able to find the solution for l , m and n for each principal stress.

1222 nml 2.10

J1, J2, J3 are called invariants of the stress tensor because J1, J2, J3

are independent of the orientation of the axes. Those coefficients are

known as the first, second and third invariants of the stress tensor.Although stress components change with the coordinate system, J1, J2,

J3 remain constants. They are related to the metal properties closely.



2 STRESS ANALYSIS

Some typical stress states

Spherical state

Cylindrical state

3=22

1

1

Uniaxial tension

Triaxial state

1

3

2

3

2

Plane stress

xy

xy

Pure shear



2 STRESS ANALYSIS

2.4 Principal shear stress

Similar to principal stress, for any given stress state, we can alwaysfind three principal shear stresses by finding the extremum values of

shear stress with respect to direction cosines. The orientations of

principal shear stress planes and the values of the principal shear stress in

terms of principal stresses are give in the figure bellow.

1

3

2

2

2112

2

3223

2

1331



2 STRESS ANALYSIS

2.5 Equilibrium equations

X

Y

xy xz

zx zy

P

Z

P’

dx

x

x x

dx x xz

xz

dz z

z z

dy

y

y

y

dx x

xy

xy

dy y

yz yz

dy y

yx

yx

dz z

zx zx

dz

z

zy

zy

x

y

z

yz

yx

dy

dz

dx



2 STRESS ANALYSIS

The element is in equilibrium. Resolving forces in OX direction gives

0

dxdydxdydz z

dxdz

dxdz dy y

dydz dydz dx x

zx zx

zx yx

yx

yx x x

x

After simplifying and dividing by dxdydz : 0

z y x

zx yx x

If the forces are resolved in the OY

and OZ direction, the following

equations are obtained. It is called

equil ibrium equations .

0

0

0

z y x

z y x

z y x

z yz xz

zy y xy

zx yx x

2.11



2 STRESS ANALYSIS

2.6 Spherical and deviator stress tensors

In general, the deformation due to the stress state consists of twocomponents: a) a volumetric or hydrostatic component , and b) adistortional component which produce a change in geometry of the

body. The corresponding stress tensor are called spherical stresstensor and deviator stress tensor .

'

'

'

00

00

00

z zy zx

yz y yx

xz xy x

m

m

m

z zy zx

yz y yx

xz xy x

Where m=(

x+ y+

z ) / 3 is average stress, i j is the Kronecker delta.

'ijmijij or 2.12



2 STRESS ANALYSIS

It is now evident that the deviator stress is composed only of the deviator

stress components, x’ =

x - m ,

y’ = y -

m and z =

z - m and the

shear stress components yx , yx and yx . Spherical stress tensor

produces only volumetric change while deviator stress tensor produces

only change in geometry.

We can see that spherical stress tensor is in effect a special hydrostatic

pressure state.

Exercise:

Calculate the sphericalstress components and the

deviator stress components

in uniaxial compression

and equibiaxial stretching.Uniaxial

compression

Equibiaxial

stretching



2 STRESS ANALYSIS

2.7 Representative or equivalent(effective) stress

Another important invariant of the state of stress is known asRepresentative or equivalent(effective) stress which is defined by

Exercise:

1) Calculate the equivalent stress of a

uniaxial stressed state.2) Calculate the shear stress and normal

stress in a octahedral plane with itstensor (u, v, w)

1

2

13

2

32

2

21

222222

)()()(21

)(6)()()(2

1

zx yz xy x z z y y x

2.13

Important: the equivalent stress of a uniaxial stress state ( 1 0, 2 = 3

=0 ) is equal to the non vanishing principal stress( 1 ).



2 STRESS ANALYSIS

2.8 Mohr’s circle for Stress transformation

11 AB 12 = 12OA+ 22OB

where 1 2 : the cosine angle between 1 and

2 axis 11 12 = 1211 + 2212 (1)

11 AB 11 = 11OA+ 21OB

11 11 = 1111 + 2112 (2)

and

(1) 12+(2) 11, since

11 2 + 12

2 = 1,

11 = 11 11 11+ 1112 12 + 12 1121 + 12 12 22

then

A

B

O

2

1

11

12

1

22

21

1 1 = ?

2

Let’s first see simple stress transformation between two

axes system:12→1 2 , with ,

From the balance of the force,



2 STRESS ANALYSIS

Express 1 1 as Generally in three axes system, m n =mi nj i j

Using above equation, we can now perform any stress

transformation between two axis system,

For example, x,y,z x , y , z ,

x y = x i y j i j = x x y x xx + x x y y xy + x x y z xz + x y y x yx + x y y y yy + x y y z yz + x z y x zx + x z y y zy + x z y z zz

= x x y x x + x y y y y + x z y z z + (x x y y + x y y x ) xy + (x x

y z + x z y x ) xz + (x y y z + x z y y ) zy

x x = x x x x xx + x x x y xy + x x x z xz + x y x x yx + x y x y yy + x y x z yz + x z x x zx + x z x y zy + x z x z zz

= x x 2

x + x y

2y + x z 2

z + 2

x x

x y

xy + 2

x x

x z

xz + 2

x y x z

yz

11 = 1i 1 jij

2.8 Mohr’s circle for Stress transformation

2 S SS S S



2 STRESS ANALYSIS

Mohr’s circles

Use the stress transformation equations, associatedwith the rotation angle between two axes,

1 1 = 1i 1 j ij

= 111111+11 12 12+12 11 21+121222

=cos2 11 +sin2 22 +2sincos 12

=(1+cos2)/211 +(1-cos2)/222 +sin2 12

=(11+22)/2+(11-22)/2cos2 + 12 sin2

2 2 = 2i 2 j ij

=(11+22)/2-(11-22)/2cos2 + 12 sin2

If the plane the stress which acts normally rotates an

angle , the line 11-22 will rotate an angle 2 to

11-22

112 2

11, 12

2

22 , 21

1

2

2 STRESS ANALYSIS



2 STRESS ANALYSIS

2/122

21 }]2/){[(2/ xy y x

2sin

]2cos))[(2/1(2/2,1

xy

y x y x

Mohr’s circle

A Mohr’s circle diagram is a graphic

representation of the above equations. It

form a circle of the radium ½(1- 2) with

the center at ½(1+ 2). The equationdescribing the angle follows

y x xy /22tan

Stress transformation from x,y to 1,2,

which principal stress acts.

2 STRESS ANALYSIS



2 STRESS ANALYSIS

2

212

2

21

2

132

2

13

2

322

2

32

22

22

22

2.19

1 32

o )2

( 21

)2

( 13

)2

( 32

The above equations

represent three circles as

shown in the figure.They are called Mohr

circle. Mohr circles are

useful tools in stress

analysis

2 STRESS ANALYSIS



2 STRESS ANALYSIS

2) When the oblique plane rotates from one principal plane to the other with its normal perpendicular to the direction of the third principal

stress, the - stress pair on that plane will movealong the correspondingcircle.

1 32

o

)2

( 13

Important points:

1) The radius of the circles are the principal shear stresses.

3) The correspondingangle in Mohr circle is

twice those in physical plane.

2

3

1

2 STRESS ANALYSIS



2 STRESS ANALYSIS

The conventions for constructing Mohr circles is the same as in the

course of Materials Strength:

o

x

y

xy

yx

1 22

y, yx

x , xy

Exercise:

Find the direction of the

principal stresses according

the Mohr circles.

To make a complete view of a

stress state of a point, two

additional circles have to be

added. This should be born inmind when giving a Mohr circle

representation of a stress state.

2 STRESS ANALYSIS



2 STRESS ANALYSIS

You may wonder how the stresses on a arbitrary plane should look

like in a Mohr circle representation. When1 >=

2 >=

3 , fromequation 2.18, we find

0))((

0))((

0))((

23132

1232

2

3121

2

n

m

l

This means that the point

representing the stress pair

in the plane should belocated inside the largest

circle and outside the two

smaller circles.

o

2 STRESS ANALYSIS



2 STRESS ANALYSIS

Exercises:

Give a full Mohr circle representation of the following stress state of a

stressed element.

1) Uniaxial tension;

2) Pure shear

2 STRESS ANALYSIS



2 STRESS ANALYSIS

2) The extent of plastic deformation is profoundly influenced by stress

state. When the stress in a spherical state of stress is compressive,the stress state is commonly referred to as “hydrostatic pressure” .

Hydrostatic pressure is beneficial to plastic deformation.

Important points:

1) Plastic deformation in macro scale is dislocation movement in microscale and is caused by shear stress. Normal stress alone can not

produce plastic deformation.

3) For any stress state, we can always find three principal stresses and at

least three principal directions.

Note: Hydrostatic pressure does not cause plastic deformation, but it

may help improve formability.

2 STRESS ANALYSIS



2 STRESS ANALYSIS

Plane stress component with σx=50 kg/cm2，σy=10 kg/cm2，τxy=40kg/cm2 , as shown in the figure. Try to answer the followingquestions by using the methods of (a) force balance and (b) Mohr’scircle：

(1)When θ=30,

the normal stress σn

and the shear stress τ ?

(2)The maximum shear

Stress, τ max

and θ

(3)σ1 σ2？

10 (kg/cm2)

50 (kg/cm2)

10 (kg/cm2)

50(kg/cm2)

40 (kg/cm2)

x

y

30

Homework

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2)Stress Analysis 2