Stress transformation, Study notes of Linear Algebra

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Stress transformation:

tractors, linear algebra, and circles within circles.

Henri P. Gavin

EGR 201L.

Intro Solid Mechanics

Department of Civil & Environmental Engineering

Duke University

Spring 2017

traction is stress at a point on a plane

traction forces in equilibrium

Stress Transformation tractors EGR 201L. Duke H.P.G Spring 2017 2 / 36

nonuniform distribution of traction stress components

σ(x)

τ(x)

Stress Transformation tractors EGR 201L. Duke H.P.G Spring 2017 3 / 36

the state of stress is coordinate invariant

the state of stress in 2D

I The state of stress on an infinitesimal cubic volume is completely

described by the traction vectors acting on the faces of the cube.

I In 2D (plane stress), the complete set of traction stresses is

S =

[

t x

t y

]

[

σ xx

τ xy

τ yx σ yy

]

I S is called the stress tensor

I τ xy

= τ yx

... S is a symmetric tensor

σ

σ

σ

σ

dl

dl

P

xx

xx

yy

yy

xy

τ

yx

τ

τ xy

τ yx

M P

+σ xx

(∆A )(dl/ 2 ) − σ xx

(∆A )(dl/ 2 )

+σ yy

(∆A )(dl/ 2 ) − σ yy

(∆A )(dl/ 2 )

+τ xy

(∆A )(dl) − τ yx

(∆A )(dl) = 0

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 5 / 36

traction stress vectors on any plane:

t

n

n · S

x

y

σ

σ

A sin

A cos

θ

θ

θ

θ

A

x

t

t

n

n

t

y

t

x

t

y

yx

τ

xx

xy

τ

yy

equilibrium on the wedge:

A

t n = (A cos θ)

t x +(A sin θ)

t y

t n

= [cos θ sin θ] ·

[

t x

t y

]

~n = [cos θ sin θ]

S =

[

t x

t y

]

t n

= ~n · S

The traction

t n

n · S is expressed in the xy coordinate system.

note: (A cos θ) = An x and (A sin θ) = An y

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 7 / 36

traction stress vectors in other coordinates:

t

′

n

t

n

· T

I Let’s express this traction

t n

in a new coordinate system x

′

y

′

,

where x

′

is along the unit normal

n and y

′

is perpendicular to x

′

.

I These rotated components of

t n are the normal stress and shear

stress on the plane with normal

n: t

′

n

= [σ x

′ x

′ , τ x

′ y

′ ]

y

x

σ

n

t

t

t

n

nx

ny

x’x’

y’

x’

x’y’

τ

c = cos θ

s = sin θ

σ x

′ x

′ = t nx

c + t ny

s

τ x

′ y

′ = −t nx s + t ny c

[

σ x

′ x

′ , τ x

′ y

′

]

[

t nx , t ny

]

[

c −s

s c

]

t

′

n

t n

· T ; T

− 1

= T

T

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 8 / 36

summary

x

y

A sin

A cos

θ

θ

θ

θ

A

x

t

t n

n

t y

t x

t y

yx

xx

xy

yy

y

x

σ

n

t

t t n

nx

ny

x’x’

y’

x’

x’y’ τ

I In the xy coordinate system, the

traction on a plane with unit

normal n is

t n

= ~n · S

= [cos θ , sin θ] · S

I The coordinate transformation

from

t n (in xy) to

t

′

n

(in x

′

y

′

) is

t

′

n

t n

· T

I Unit vectors can be similarly

transformed, ~n

′

= ~n · T ... and ...

n =

n

′

· T

T

[cos θ , sin θ] = [ 1 , 0 ] · T

T

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 10 / 36

putting it all together ... stress transformation

I Given:

• stress in an xy coordinate system, S.
• a unit vector ~ n along x

′

in an x

′

y

′

coordinate system.

I Find: The stress S

′

in the x

′

y

′

coordinate system.

I Substitute ...

t

′

n

t n

· T

t

′

n

n · S · T

t

′

n

= ~n

′

· T

T

· S · T

t

′

n

n

′

· S

′

S

′

= T

T

· S · T

In 2D, the stress transformation formula for a CCW rotation θ is:

[

σ x

′ x

′ τ x

′ y

′

τ y

′ x

′ σ y

′ y

′

]

[

c s

−s c

] [

σ xx

τ xy

τ yx σ yy

] [

c −s

s c

]

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 11 / 36

For what angle θ is τ

x

′ y

′

1. On what planes is the shear stress zero?
2. What is the transformation matrix T that diagonalizes S

′

?

3. What are the normal stresses on planes with no shear stress?

[

c s

−s c

] [

σ xx τ xy

τ yx

σ yy

] [

c −s

s c

]

[

σ 1 0

0 σ 2

]

[

c s

−s c

] [

σ xx

τ xy

τ yx

σ yy

]

[

σ 1

0

0 σ 2

] [

c s

−s c

]

[

c s

]

[

σ xx

τ xy

τ yx σ yy

]

= σ 1

[

c s

]

[

−s c

]

[

σ xx

τ xy

τ yx σ yy

]

= σ 2

[

−s c

]

1. Planes with no shear stress are normal to eigenvectors of S.
2. The transformation matrix T of eigenvectors of S diagonalizes S.
3. The normal stresses on planes with no shear stress are the

eigenvalues of S, σ 1

and σ 2

. These are called principle stresses.

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 13 / 36

principle stresses are the eigenvalues of the stress tensor

That sounds impressive.

But what does it mean?

Let’s look at an example in 2D.

S =

[

t x

t y

]

[

σ xx τ xy

τ yx

σ yy

]

[

]

kPa

Here’s a plot of

t n (θ) =

n(θ) · S for all values of

n(θ) = [cos θ , sin θ]

0

5

10

-10 -5 0 5 10

n

y

, t

xy

n x , t xx

n(θ)

t n

(θ)

σ 1 =-2.10 kPa

σ 2

= 8.10 kPa

t n (0)

Stress Transformation 2D stress EGR 201L. Duke H.P.G Spring 2017 14 / 36

Mohr’s Circle (for plane stress)

trigonometry and algebra lead to Mohr’s circle

Substitute:

cos

2

θ = ( 1 + cos 2θ)/ 2

sin

2

θ = ( 1 − cos 2θ)/ 2

2 sin θ cos θ = sin 2θ

into:

σ x

′ x

′ = σ xx cos

2

θ + σ yy sin

2

θ + 2 τ xy sin θ cos θ

τ x

′ y

′ = (σ yy − σ xx ) sin θ cos θ + τ xy (cos

2

θ − sin

2

θ)

square equations, add them, and do some algebra to obtain:

σ x

′ x

′ −

σ xx

• σ yy

2

2

• τ

2

x

′ y

σ xx − σ yy

2

2

• τ

2

xy

and compare to:

(x − c)

2

• y

2

= r

2

to recognize the equation for a circle with center (c, 0 ) and radius r!

Stress Transformation 2D Mohr’s circle EGR 201L. Duke H.P.G Spring 2017 16 / 36