Week 13 - Stress Transformation

1

1

CVEN2301 Mechanics of Solids

STRESS TRANSFORMATION

Chongmin Song

School of Civil and Environmental EngineeringThe University of New South Wales

Stress and Strain at a Point

zzyzx

yzyyx

xzxyx

στττστττσ

zzyzx

yzyyx

xzxyx

στττστττσ

σσσσ zz

σσσσ xx

σσσσ yy

ττττzy

ττττzx

ττττxz

ττττxy

ττττyxττττyz

Plane xPlane z

Plane y

y

xz

3D Stresses at a Point

Stress tensor

),,,( zyxji

jiij

=

=ττ

Complementary property of shear

2

Plane stresses 0zz zx zyσ τ τ= = =

Sign convention: Positive normal stress acts outward fromall faces and positive shear stress acts upward on the right-hand face of the element.

Plane StressesA structural member may be subjected to both normal

stress and shear stress .

Which stress do we use for design purposes?

3

Design assumption: material behaviour (elastic/plastic/in failure) is identical along the bar.

Normal stress:

Shear stress:

0=τ

Along Section Plane a-a

4

Along Section Plane b-b

N40030sin800

N8.69230cos800

=×=

=×=

�

�

V

N

: force Shear

:force Normal

Along Section Plane b-b

kPa375101.8475

8.692

kPa217101.8475

400

m101.8475)60sin/04.0(04.0

3-

3-

23-

=×

==

=×

==

×=×=

A

N

A

V

A

σ

τ

:stress Normal

:stress Shear

:area Sectional�

5

Equivalent Stress States

Stress Transformation

At a specified location under a given loading:• The material behaviour (elastic/plastic/in failure) is

uniquely determined.• The stress states of an element will change with its

orientation but are equivalent.

The equations for stress transformation provides us with unique values to be used in structural designs.

6

EXAMPLE 1

Equilibrium (of force) in x’, y’ coordinates

The answer is independent of ∆A

7

Equilibrium (of force) in x’, y’ coordinates

Stresses on plane b-b

8

Stresses on plane b-b

Equivalent stress states

9

• Stress transformation from x, y axes to x’, y’ axes• The orientation of an inclined plane will be defined using

the angle θ. • Angle θ is measured from x-axis to x’-axis in the

counterclockwise direction


σσσσ x'x'

ττττ y'x'

ττττ x'y'

σσσσ y'y'

x'

y'

System 2

θθθθ xσσσσxx

ττττyx

ττττxy

σσσσyy

x

y

System 1

θ

σσσσxx

ττττyx

ττττxy

σσσσyy

x

y

System 1

• Consider the equilibrium of a triangular segment


θ

10


θτθσσσσ

σ

θθθθ

θθθ

θθτθσθσσ

θθσθθτθθσθθτσ

2sin2cos22

2/)2cos1(cos

;2/)2cos1(sin

;cossin22sin

)cossin2(sincos

0cos)cos(sin)cos(

sin)sin(cos)sin(

'

2

2

22'

'

xyyxyx

x

xyyxx

xxy

yxyx

AA

AAA

+−

++

=

+=

−==

++=

=∆−∆−

∆−∆−∆

:stress Normal

:identitiesric trigonomet Using

ion x'-directin mEquilibriu


θτθσσ

τ

θθτθθσστ

θθσθθτθθσθθττ

2cos2sin2

)sin(coscossin)(

0sin)cos(cos)cos(

cos)sin(sin)sin(

''

22''

''

xyyx

yx

xyxyyx

xxy

yxyyx

AA

AAA

+−

−=

−+−=

=∆−∆−

∆−∆+∆

:stress Shear

ion y'-directin mEquilibriu

11


θτθσσσσ

σ

θθ

θτθσσσσ

σ

2sin2cos22

)

2sin2cos22

'

'

xyyxyx

y

xyyxyx

x

−−

−+

=

+←

+−

++

=

:90( ion y'-directin stress Normal

:ion x'-directin stress Normal

�

x

y’

x’

EXAMPLE 2

12

EXAMPLE 2

MPa15.4

)30sin()30(2cos2

5080

2

5080

2sin2cos22

MPa8.25

)30sin()30(2cos2

5080

2

5080

2sin2cos22

'

'

−=

−−−−−−+−=

−−

−+

=

−=

−+−−−++−=

+−

++

=

��

��

xy

xyyxyx

y

xy

xyyxyx

x

τ

θτθσσσσ

σ

τ

θτθσσσσ

σ

:system coordinate y', x'in Stresses

Stresses in x, y coordinate system

EXAMPLE 2

MPa8.68

)30(2cos)25(2

5080

2cos2sin2''

−=

−−+−−−=

+−

−=

�

θτθσσ

τ xyyx

yx

Note the signs and directions of stresses

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Principal Stresses

( )

( )

121

'

90

:2180tan(2tan

2/2tan

02cos22sin22

ppp

ppp

yx

xyp

pxypyx

p

x

d

d

θθθ

θθθ

σστ

θ

θτθσσ

θθθσ

+=

+=

−=

=+−

−==

�

�

and

for exist solutions two ), As

The principal stresses represent the maximum and minimum normal stress at the point.

θτθσσσσ

σ 2sin2cos22' xy

yxyxx +

−+

+=

• Orientation of the planes of maximum and minimum normal stress (principal planes)

• Normal stress

Principal Stresses

( )

2

2

1

2

2

1

2/

22cos

2/2sin

0 0

xyyxyx

p

xyyx

xyp

xyyx

τσσσσ

θ

τσσ

τθ

τσσ

+

−

−=

+

−=

>>− and case the gConsiderin

pxyyxyx

x θθθτθσσσσ

σ =+−

++

= with ;2sin2cos22'

Normal stress acting on principal planes

2

2

2

2

2

2

2/

22cos

;2

/2sin

xyyxyx

p

xyyx

xyp

τσσσσ

θ

τσσ

τθ

+

−

−−=

+

−−=

( ) 2/2tan

yx

xyp σσ

τθ

−=

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Principal Stresses

2

2

2

2

2

2

111

22

2/

22

2sin2cos22

xyyxyx

xyyx

xyyxyx

pxypyxyx

τσσσσ

τσσ

τσσσσ

θτθσσσσ

σ

+

−+

+=

+

−

+

−+

+=

+−

++

=

Maximum in-plane normal stress

Minimum in-plane normal stress

2

2

222

22

2sin2cos22

xyyxyx

pxypyxyx

τσσσσ

θτθσσσσ

σ

+

−−

+=

+−

++

=

Principal Stresses

2

2

2,1 22 xyyxyx τ

σσσσσ +

−±

+=

Maximum/ Minimum in-plane normal stress (σ1 ≥ σ2)

Shear stress on the principal planes

02cos2sin2

2cos2sin2

;2

2cos

2sin2tan

'' =+−

−=

=−

⇒−

==

pxypyx

yx

pxypyx

yx

xy

p

pp

θτθσσ

τ

θτθσσ

σστ

θθ

θ As

No shear stress acts on the principal planes!

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σσσσxx = 3

τ = 2τ = 2τ = 2τ = 2

τ = 2τ = 2τ = 2τ = 2

σσσσyy =1

x

y

Find the principal stresses and show their sense on a properly oriented element.

Unit: MPa

EXAMPLE 3

MPa

MPa

MPa

xy

y

x

2

1

3

=

==

τσσ

EXAMPLE 3Principal Direction:

x'

y'

θ = 31.72 θ = 31.72 θ = 31.72 θ = 31.72 deg x

4.24 MPa

-0.24 MPa

( ) 22

422tan ==

−=

yx

xyp σσ

τθ

MPa24.4

)72.312sin(2)72.312cos(2

13

2

13

2sin2cos22'

=

×+×−++=

+−

++

=

��

θτθσσσσ

σ xyyxyx

x

MPa24.0

)72.312sin(2)72.312cos(2

13

2

13

2sin2cos22'

−=

×−×−−+=

−−

−+

=

��

θτθσσσσ

σ xyyxyx

y

MPa

MPa

MPa

xy

y

x

2

1

3

=

==

τσσ

MPa24.0

MPa24.4

2

1

−==

σσ

�72.31=pθ

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Maximum Shear Stress

( )( )

psps

yx

xyp

pxy

yxs

sxysyx

s

yx

xyyx

yx

d

d

θθθθ

σστ

θθτ

σσθ

θτθσσ

θθθτ

θτθσσ

τ

+=⇒+=

−=−=

−−=

=−−

−==

+−

−=

�� 45;2902

2/2tan ;

2tan

12/2tan

02sin22cos22

2cos2sin2

''

''

stress shear maximum of plane the of nOrientatio

:stress Shear

The plane for maximum shear stress is orientated 45°from principal planes.

2

2

2cos2sin2

2cos2sin

avg''

22

''

yxyx

xyyx

xyyx

yx

ss

σσσσσ

τσσ

τ

θτθσσ

τ

θθ

+===

+

−=

+−

−=

stress shear maximum of

planes the on stress normal the Similarly,

stress shear Maximum

:into and for values the ngSubstituti

max

Maximum Shear Stresses

( )xy

yxs τ

σσθ

2/2tan

−−=

17

EXAMPLE 4

EXAMPLE 4

02

02

00

avg

222

=+

=

±=+=+

−=

−===

yx

xyyx

xyyx

σσσ

τττσσ

τ

ττσσ

stress normal Average

stress shear Maximum

Stresses

max

Ductile materials (e.g. mild steel) often fail due to shear stress

18

EXAMPLE 4

θ θ

EXAMPLE 4

Brittle materials (e.g. cast iron) often fail due to normal stress

Stress element

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Principal stresses

Which of the following statement is incorrect?

A)The principal stresses represent the maximum and minimum normal stress at the point

B)When the state of stress is represented by the principal stresses, no shear stress will act on the element

C)When the state of stress is represented in terms of the maximum in-plane shear stress, no normal stress will act on the element

D)For the state of stress at a point, the maximum in-plane shear stress usually associated with the average normal stress.

Quiz

21

Mohr’s Circle of StressA graphical solution for plane stress transformation Often convenient and easy to use. There are several ways of drawing a Mohr’s circle

)3( 2cos2sin2

)2(2sin2cos22

)1(2sin2cos22

''

'

'

θτθσσ

τ

θτθσσσσ

σ

θτθσσσσ

σ

xyyx

yx

xyyxyx

y

xyyxyx

x

+−

−=

−−

−+

=

+−

++

=

Equation for stress transformation:

Mohr’s Circle of Stress

( )

( ) ( )

2

2

avg

22''

2avg'

2

22

''

2

'

2222

''

'

2;

2

22

12sin2cos)()(

)3( 2cos2sin2

)4(2sin2cos22

xyyxyx

xyyx

yxx

xyyx

yxyx

x

xyyx

yx

xyyxyx

x

R

R

τσσσσ

σ

τσστσσ

τσσ

τσσ

σ

θθ

θτθσσ

τ

θτθσσσσ

σ

+

−=

+=

=+−

+

−=+

+−

=++

+−

−=

+−

=+

−

constants) known are , ,( with

as Rewritten

using and 3 Eq.4 Eq.

Rewrite Eq. 1

22


( ) ( ) 22''

2avg'

avg

'''

)0,

R

R

yxx

yxx

=+− τσσ

στσ

( at centre and

radius having circle Mohr' a on is ) ,(

Sign Convention:• σ is positive to

the right, and • τ is positive

downward.


xyyxyx

xyyxxx

ττσσθττσσθ

−======

'''

'''

90

0

and , :G

and , : Apoints Reference

�

�

A rotation of θ of the x’ axis correspondsto a rotation 2θ on the circle in the same direction

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Mohr’s Circle of StressDraw a Mohr’s circle• Coordinates σ, τ (positive downward)• Centre of circle C (σavg, 0)• Reference point A (σx, τxy)• Connect CA (θ=0o) and determine R• Sketch the circle

σ

τ

C

Aθ=0

o

Mohr’s Circle of StressPrincipal stress• σ1 at point B on circle• Angle 2θp1 measured from reference line CA to CB• σ2 at point D on circle; θp1 and θp2 are 90o apart• Sketch a stress element with the principal axis (+x’) rotate

by θp1 from the reference axis (+x)

σ

τ

C

A

2θθθθp1

B

D

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Mohr’s Circle of StressMaximum shear• τmax at point E or F• Angle 2θs1 measured from reference line CA to CE or CF• Sketch a stress element

σ

τ

C

A2θθθθs1

B

D

E

F

Mohr’s Circle of StressStresses on Arbitrary Plane• Measure an angle 2θ on the circle from the reference line

CA in the same direction as angle θ• Determine the coordinates of P

σ

τ

C

A

B

DP

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MPa60

MPa12

−==

−=

xy

y

x

τσσ

( )

�

�

5.22

45612

6tan2

plane principal ofn Orientatio

MPa49.1449.86

MPa49.2649.8

stresses Principle

MPa49.86)612(

circle theof Radius

(-12,-6)A point initial The

0) (-6, Ccenter The

MPa6

2/)012(2/

2

12

2

1

22

avg

=

=−

=

−=−−==−=

=+−=

−=

+−=+=

−

p

p

yx

R

θ

θ

σσ

σσσ

(MPa)σ

26

MPa6

MPa12

MPa8

−=

=−=

xy

y

x

τσσ

( )

)0(CA from measured

clockwisecounter 60)2(30at

CP line radial a Draw

MPa66.116)10(

circle theof Radius

(-8,-6)A point initial The

0) (2, Ccenter The

MPa22/)128(2/

22

avg

�

��

=

=

=+=

=+−=+=

θ

σσσ

R

yx

27

MPa66.504.29sin66.11

MPa20.804.29cos66.112

04.2996.3060

96.3010

6tan

circle theofgeometry theFrom

''

'

1-

==

−=−=

=−=

==

�

�

��

�

yx

x

τ

σ

ψ

φ

Determine the coordinate of point P(σx’, τx’y’)

• Face DE is 60o clockwise from the x-axis.

• Stresses represented by point Q

MPa66.504.29sin66.11

MPa2.1204.29cos66.112

circle theofgeometry theFrom

''

'

−=−=

=+=�

�

yx

x

τ

σ

Week 13 - Stress Transformation

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Transcript of Week 13 - Stress Transformation