Principle Stress

DEN102Stress Analysis

Principal Stress

Dr P.H. Wen

Aims1. Recognise the stress and strain tensors ;2. Understand stress state of point and how to

calculate principal stresses and their directions;3. Recognise why principal stresses and their

directions are useful; 4. Understand what a yield criterion is and how it

can be used.

Structure Stress Analysis

• Analytical solutionHand calculation: beam, truss, shaft

• Numerical solutionANASYS, ABAQUS, DYNA, IDEAS

8.1. Normal and Shear Stresses in Solid• In general objects, i.e. multi-

dimensional objects, they are generated by exterior forces F, distributed loads or pressures p.

• Exterior loads result in interior forces, which are resultants of interior stresses.

• At an arbitrary cut at S-S (see figure), these stresses act on the surface.

• There are normal stresses σ(orthogonal to the surface) and tangential or shear stresses τ(parallel to the surface).

• On the opposite surface of the cut, stresses occur in the opposite direction.

• The stress vector t at a point P is defined by

• It is a vectorial sum of shear and normal stress:

AAA ddlim

0

FFtrr

r=

∆∆

=→∆

στ

σ

τt

t

Force diagram:

P P=τσFFFFFt +=+=

+==

AAAAnn

dd

dd

ddd

dd ττ

rrrrrr

8.1

8.2. General Stress State (Three Dimensional Element)

• ELEMENT

A three dimensional rectangularelement is of differential size:

dx×dy×dz=dV

Positive surfaces +veNegative surfaces -ve

Element is in a state of uniformstress.

xσ

xzτ

yxτxyτyzτ

yσ

zxτzσ

zyτ

x

y

zdx

dz

dy

• The stresses depend on the direction of the cut, i.e. the orientation of the surface area A.

• Generally, a Cartesian coordinate system is chosen (see Figure 8.1)

• The total stress state is determined by the normal and shear stresses at six surfaces of an infinitesiminal volume element (see Figure 8.1).

• This element is normally defined with surfaces orthogonal to the coordinate directions.

• The values of all stresses change if the volume element is cut in a different orientation (i.e. coordinate transformation 8.2).

• Positive direct stress are defined as tensile stresses ; negative as compressive stresses .

xσ

xzτ

yxτxyτ

yzτyσ

zxτzσ

zyτ

x

y

z

Figure 8.1

xσ

xzτ

yxτxyτyzτ

yσ

zxτ

zσ

zyτx

y

zFigure 8.2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

zzyzx

yzyyx

xzxyx

στττστττσ

σ

8.3. General Stress State (in Matrix Form)• The normal and shear

stresses are represented by the stress matrix also named stress tensor

• Stresses with non-identical indices are shear stresses, the other are normal stresses; the latter are on the diagonal of the matrix.

Two-dimension

(8.2); ⎥⎦

⎤⎢⎣

⎡=

yyx

xyx

σττσ

σ

Normal stresses

Shear stresses

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zzyzx

yzyyx

xzxyx

στττστττσ

σxσ

xzτ

yxτxyτyzτ

yσ

zxτzσ

zyτ

x

y

z

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zzyzx

yzyyx

xzxyx

στττστττσ

σ

Figure 8.3

Three-dimension

• The consequence is that the stress matrix is symmetric

• There are hence for each point if the structure 6 unknown stresses to be determined in stress analysis

8.4. Corresponding Shear Stresses

• This can be seen by taking the sum of the moments with respect to point O, i.e.

( )xyyx

xyyx

xyyxO

dV

M

ττ

ττ

ττ

=⇒

=−

=−=∑

0 therefore

0)dydz(dx)dxdz(dy :0

zxxzzyyzyxxy ττττττ === ;;

(8.3a)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zyzxz

yzyxy

xzxyx

στττστττσ

σ

xσ

yxxy ττ =

zxxz ττ =

zyyz ττ =

yσ

zσ x

y

z

xzτ

yxτxyτyzτ

zxτ

zyτ

dxdy

yxτxyτ

x

y

O

yσ

xσ

yxτ

xyτ

yσ

xσ

(8.3b)

Figure 8.4

• Shear stresses at two surfaces, which are perpendicular to one another are equal

Plane stress state:

• The stress tensor in matrix form is then

or

⎥⎦

⎤⎢⎣

⎡=

yxy

xyx

σττσ

σ

(8.4)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

00000

yxy

xyx

σττσ

σ

0=== zyzxz σττ

(8.5)

Figure 8.5

8.5. Two Dimensional Element (Plane Stress)

yσ

xσxyτ

dxdy

yxτxyτ

y yσ

xσxyτ

xσ

yxτyσ

x

Example 8.0:Illustrate stress state using stress tensor in matrix form (unit= N/mm2)

50

150

70

80

100

30

40

60

30

(a) (b) (c)

8.6. Stresses in Straight Bar• Normal exterior forces F leads to

normal stresses σ in the interior (constant over the cross section A)

• Interior normal force N is the resultant forces of normal stresses σ

• If the cut is not perpendicular to the axis of the beam, a normal force Fleads to normal stresses σ and shear stresses τ

]N/m[ 2

AF

=σ (8.6)

F Fcut S-S S

S

F F

σ σ

F F

N N]N[d AAN

A

⋅== ∫ σσ (8.7)

A

F FS

S

θ

F Fσθ

τθ

θθ cosAA =

Non-perpendicular cut S-S (glue)

θθσθθθ

θτ

θσθθθσ

θ

τθ

θθ

cossincossincos/sin

coscoscos/cos 22

====

====

AF

AF

AF

AF

AF

AFn

(8.8)

Fn Fτ

Fτ

F

Fn

θ

θ

Figure 8.6

Figure 8.7

]N/m[ 2yI

M

x

bz =σ (8.8)

Undeformed beam

Deformed beam

y

yy

zxy

x z

zx

8.7. Normal Stresses due to Bending Moments

• Bending moment leads to linear distributions of normal stresses

Mb Mbcut S-S S

S

equivalentto

Mbσ

σ

Figure 8.8

σmaxσmax

8.8. Shear Stress in Cylinder by Torque

• Normal exterior torque T0 leads to shear stresses τ in the interior

• Interior torque T is the resultant forces of shear stresses τ

T0 cut S-S S

S

JTR

JTr

== max2 ],N/m[ ττ (8.9)

TArJTArrT

AA

=== ∫∫ dd)( 2τ (8.10)

A

T0

T0T

Figure 8.9

τmax

τmax

Summary of Today

1. State of Stress at Point (Element)

2. State of Plane Stress

3. Stress Tensor

Example 1 (individual)

A hollow shaft, of external diameter D2 and internal diameter D1, where D1/D2=1/3, is required to transmit a torque of 100kNm and compressive axial load 1500kN as shown in Figure. If D2 is selected as 200mm, illustrate stress state on outer surface of the shaft using stress tensor in matrix form (unit= N/mm2)

D2D1

propeller shaft1500kN

100kNm

8.9. Coordinate Transformation• If stresses are known for one coordinate system and

stresses for another coordinate system are to be derived, one uses the following equations (balance of force in local coordinate ):

Therefore

0cos)sin(sin)sin(

sin)cos(cos)cos(:element) of thicknessis ,(direction in 0Fi

x

=−−

−−

×==∑

θθτθθσ

θθτθθσσ

dAdA

dAdAdAdAx

xyy

xyxx

ttc

0sin)sin(cos)sin(

cos)cos(sin)cos(

direction in 0Fiy

=+−

−+

=∑

θθτθθσ

θθτθθστ

dAdA

dAdAdA

y

xyy

xyxyx

θτθσστ

θτθσσσσσ

2cos2sin)(21

2sin2cos)(21)(

21

xyyxyx

xyyxyxx

+−−=

+−++=

dA

xyτxyτ

x

y

O

yσ

xσxyτ

xyτyσ

t

yox

(8.12)

c

xσ

yxτ

yσ

xyτxσ

xyτ

y

(normal) xy

x

θ θ

a

b

Figure 8.10

Example 8.1 At a point of steel plate, the state of plane stress (2D) is defined by σ,

where

with respect to the axes xoy. Fine the stresses (normal and shear) for the same point acting on the plane orientated at angle 450 to the x-axis.

50

10050

2N/mm 505050100

⎥⎦

⎤⎢⎣

⎡=σ

50

50

100

045x

Example 8.2 At a point of steel plate, the state of plane stress (2D) is defined by σ,

where

in the coordinate . Determine the stress matrix for the same point with respect to the axes , where x lies at 400 anticlockwise from x. Sketch the element in the axes and the stresses acting on its faces.

2N/mm 507575150

⎥⎦

⎤⎢⎣

⎡=σ

yoxxoy

150

5075

040

yox

σ

040x

x

Example 8.3Variation of stresses (normal and shear) with rotation angle of normal to the x-axis

xσ

1801651501351201059075604530150θ0

2

2

2

N/mm 20

N/mm 32

N/mm 64

=

−=

=

xy

y

x

τ

σ

σ

64

32

20θ

64

3220

θ

xσyxτ

yxτxσ

Example 8.3Variation of stresses (normal and shear) with rotation angle of normal to the x-axis

2

2

2

N/mm 20

N/mm 32

N/mm 64

=

−=

=

xy

y

x

τ

σ

σ

64

32

20θ

64

3220

θ

xσyxτ

θθτθθσ

2cos202sin482sin202cos4816

+−=++=

yx

x

Answer: Stresses acting on the face orientated with θ from (8.12) are

-60

-40

-20

0

20

40

60

80

0 15 30 45 60 75 90 105 120 135 150 165 180

xσ

yxτ

8.10. Principal Stresses (Method 1)• The value of the stresses depend on

the choice of the coordinate system: we do not know the highest normal or the highest shear stress.

• For evaluation of yield, failure or fatigue independent (invariant) quantities are required;

• The extreme normal stresses(maximum and minimum values) are obtained via

This leads to

• Two angles are determined by

0=θ

σd

d x

( ) 022cos22sin ** ==+−− yxxyyx τθτθσσyx

xy

σστ

θ−

=∴2

2tan *

*θ

1

2 ; then

2tan

21 **1* πθθθθ

σστ

θ ±==−

= −ppp

yx

xy (8.13)

Figure 8.11

• Inserting equation (8.13) into (8.14) leads to the principal stresses

)}(),(min{

)}(),(max{4

)(2

)(

2

1

22

2,1

ppxpx

ppxpx

xyyxyx

θσθσσ

θσθσσ

τσσσσ

σ

=

=

+−

±+

=

(14)

(8.16)

)2/(or ** πθθ ±

ppxyppyxyxppx

pxypyxyxpx

θτθσσσσθσ

θτθσσσσθσ

2sin2cos)(21)(

21)(

2sin2cos)(21)(

21)(

+−++=

+−++= (8.14)

• Substituting equation (8.13) into (8.12) leads to the principal stresses

• Shear stresses

0)()( == ppyxpyx θτθτ (8.15)

Figure 8.12

Example 8.4

In a concrete structure a two-dimensionalstress state was computed with

2

2

2

N/mm 20

N/mm 32

N/mm 64

=

−=

=

xy

y

x

τ

σ

σ

Determine:

1. The normal and shear stresses under an angle of 60°

2. The principle stresses and the principles directions

3. Sketch the element corresponding to principal stresses

4. In which direction do you expect fractures (concrete)?

64

32

20*θ

8.11. Principal Stresses (Method 2)• An easier way to compute principle

stresses is obtained by the mathematical scheme to compute Eigen value (natural values) of a matrix:

• Therefore:

22

2

2

12

2

1

4)(

2)(

4)(

2)(

στσσσσ

λ

στσσσσ

λ

=+−

−+

=

=+−

++

=

xyyxyx

xyyxyx

0detdet =−

−=−

λσττλσ

λyxy

xyxIσ

( )( )( ) ( ) 0

022

2

=−++−

=−−−

xyyxyx

xyyx

τσσσσλλ

τλσλσ

Example 8.5

Determine principal stresses by method 2.

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡10101010

yxy

xyx

σττσ

(8.17)

(8.18)

)(21)(

21)()( 21 σσσσθσθσ +=+== yxssxsx

8.13. Principal Shear Stress• The extreme shear stresses

(maximum and minimum values) are obtained via

This leads to

Two angles are determined by

0=θ

τd

d yx

( ) 02sin22cos **** =−−− θτθσσ xyyx

xy

yx

τσσ

θ2

2tan ** −−=

*θ

1

2 ; then

2tan

21 ****1** πθθθθ

τσσ

θ ±==−

−= −sss

xy

yx

(8.20)eq.(8.14)] [from 22

2122

(max)σστ

σστ −

±=+⎟⎟⎠

⎞⎜⎜⎝

⎛ −±= xy

yxyx

**θ

)( syx θτ )( sx θσ

(8.19)

Figure 8.15

• Maximal shear stress is obtained when an angle of 45° with respect to the principle directions is chosen

• Maximal shear stress is equal to

1σ

2σ

1σ

2σ

( ) ( )2121max 21

21 σσσσστ +=−±= (8.21)

450

maxτσ σ

maxτσ σ

Figure 8.16

8.14. Linear Elasticity (Hooke’s Law)• In one dimension we have

• E is Young’s modulus and describes the stiffness of the material (i.e. it depends only on the material) as long as it is elastic

• Steel: E = 210.109 N/m²• Aluminium E = 70.109 N/m²• Beyond the elastic limit (yield

stress), the material reacts plastically.

• In elastic state, material can be loaded and unloaded without remaining strains, i.e. the loading procedure is reversible

• Material may show linear or non-linear elastic behaviour

εσ ⋅= E (8.22)

Stre

ss σ

=F/A

200

400

600

500

10 15 20 25

L0

∆L ∆L ∆L

Strain: 0 / LL∆=ε

L0 = originallength

E

Yield stress: yieldσ

Figure 8.18

8.15. Multi-dimensional Linear Elasticity• In the multi-dimensional case,

stresses in one direction generate strains also in the other directions; we have (2D):

ν is the Poisson coefficient• In linear elasticity, it is sufficient to

describe the material by two parameters, e.g. Young’s modulus and Poisson coefficient; but you can use as well other parameters

( )

( )

G

E

E

xyxy

yxy

yxx

τγ

σνσε

νσσε

=

+−=

−=

1

1

(8.28) • Shear strains are related to shear stresses via (G is the shear modulus)

)1(2 ν+=

EG (8.29)

Figure 8.19

• Natural values:

• The natural values are the principal values

(2D)

(3D)

8.16. Strain Matrix and Strain Principle Values• In analogy to stresses, the strains

are grouped into a strain matrix(or strain tensor)

(2 Dimension) (3 Dimension)

• The strain matrix is as well symmetric

• Principle strains can be computed by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zzyzx

yzyyx

xzxyx

εγγγεγγγε

2/2/2/2/2/2/

ε (8.23) 2/

2/⎥⎦

⎤⎢⎣

⎡=

yyx

xyx

εγγε

ε

zyyzzxxzyxxy γγγγγγ === ;;

4)(

2)(

4)(

2)(

22

2

22

1

xyyxyx

xyyxyx

γεεεεε

γεεεεε

+−−

+=

+−+

+=

(8.25)

(8.24)

02/

2/det =

−−

λεγγλε

yxy

xyx

( )( )( ) ( ) 04/

04/22

2

=−++−

=−−−

xyyxyx

xyyx

γεεεελλ

γλελε

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

000000

εε

εε

(8.26);0

0

2

1⎥⎦

⎤⎢⎣

⎡=

εε

ε

2211 and λελε ==

(8.27)

Example 8.7

Principle directions of a beam subjected to a central unit force

Compression

Tension

2N/mm 5040

40100⎥⎦

⎤⎢⎣

⎡−

=σ

A stress (matrix) tensor for the plane stress elasticity is given by

The Young’s modulus and Poisson’s ratio of this material are E=200×103N/mm2 and ν=1/3 respectively.

(1) Draw a square element with the above stress components acting on its sides;

(2) Calculate the strain tensor in matrix form; (3) Determine the principal stresses and their directions;(4) Determine the principal strains.

8.17. Equivalent Stress and structure failure• To determine failure, the multi-

dimensional stress state should be taken into account;

• Although the one-dimensional stress limit is not reached, the structure may fail when subjected to multi-dimensional stresses;

• Multi-dimensionality of the stress state might improve or weaken.

• To obtain simple measures to evaluate multi-dimensional stress states, the concept of equivalent stress is used, i.e. a single value is obtained from the stress matrix, which is then compared to the one-dimensional limit stress (yield stress, ultimate stress, etc.)

n is the safety factor

• Simple approaches to determine the equivalent stress (the choice between them depends on the material):

– Normal stress hypothesis:Failure occurs when maximal normal stress is reached

– Shear stress hypothesis (2D):Failure occurs when the maximal shear stress is reached (Tresca)

– Energy hypothesis (2D):Failure occurs if the maximal elastic energy is reached (von Mises)

n/Limitequiv σσ ≤ (8.30)

1equiv σσ =

( ) ( ) 22max21equiv 42 xyyx τσστσσσ +−==−=

222

2122

21equiv

3 xyyxyx τσσσσ

σσσσσ

+−+=

−+=

(8.32)

(8.33)

(8.31)

Example (individual)

A hollow shaft, of external diameter D2 and internal diameter D1, where D1/D2=1/3, is required to transmit a torque of 100kNm and axial load 1500kN (compressive). If D2 is selected as 200mm, ultimate stress σLimit=300N/mm2, safety factor n=2.5 and von Mises criterion is considered, check whether the shaft is safe.

D2D1

propeller shaft1500kN

100kNm

Principle Stress

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