2.4.1: Transformation of Stress• Direction cosines

• Orthogonality properties and unit length

• Is there a less redundant description?

xZθ

xYθ

xXθ

y

x

zX

Y

Z

x y z

X l1 m1 n1

Y l2 m2 n2

Z l3 m3 n3

000

1 2 1 2 1 2

1 3 1 3 1 3

2 3 2 3 2 3

l l m m n nl l m m n nl l m m n n

+ + =+ + =

+ + =

00

0

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

l m l m l ml n l n l nm n m n m n

+ + =

+ + =

+ + =

1

1

1

2 2 21 2 3

2 2 21 2 32 2 21 2 3

l l l

m m m

n n n

+ + =

+ + =

+ + =

1

1

1

2 2 21 1 12 2 22 2 22 2 23 3 3

l m n

l m n

l m n

+ + =

+ + =

+ + =

Stresses on an inclined plane• Normal vector to plane

y

x

zyσ−

zσ−

xσ−

N

Pσ

O

C

AP

ˆˆ ˆ l m n N i j k= + +

l; m; n

Stress vector on plane P:

l m n ˆˆ ˆσ σ σ

x y z

OBC OAC OBA

ABC ABC ABC

P x y z

P P P P

A A AA A A

σ σ σ σ

σ i j k

= = =

= + +

= + +

Components of the stress vector σ : σ l σ m σ n σ

σ l σ m σ n σ

x

y

P P xx yx zx

P xy yy zy

= + +

= + +

σ l σ m σ n σzP xz yz zz= + +

Normal and shear stresses

• Normal and shear components

• We often need the maximal and minimal normal stresses and maximum shear stress. Why?

( ) ( ) ( )N

N

S N x y z N

P P

P xx yy zz yz xz xy

2 2 2 2 2 2P P P P P P P

σ N σ

σ σ σ σ 2 σ 2 σ 2 σ

σ σ σ σ σ σ σ

2 2 2l m n m n n l l m

= •

= + + + + +

= − = + + −

Stress transformation• Using the equation for normal stresses

• Similarly, obtain

• How do we reduce these to a single short line?

σ σ σ σ 2

σ σ σ σ 2

σ 2 σ

σ σ σ σ 2 σ 2 σ 2 σ

2

σ σ 2 σ

σ

2

2 2 2ZZ 3 xx 3 yy 3 zz 3 3

2 2 2XX 1

2 2 2YY 2 xx 2 yy 2 zz 2 2 y

xx 1 yy 1 zz 1 1 yz 1 1 xz 1 1 x

z 2 2 xz 2 2 x

yz 3 3 xz 3 3

y

y

y

x

l m n m n

l

l m

m n m n n l l m

n m n n l l m

n l l m

= + + + + +

= +

=

+ + +

+

+

+ + + +

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )σ σ σ σ σ σ

σ σ σ σ

σ σ σ σ σ σ m

σ σ m σ

σ

m σ

XY 1 2 xx 1 2 yy 1 2 zz 1 2 2 1 yz 1 2 2 1 xz 1 2 2 1 x

XZ 1 3 xx 1 3 yy 1 3 z

YZ 2 3 xx 2 3

z 1 3 3 1 yz

yy 2 3 zz 2 3 3 2 yz 2 3 3

1 3 3

2 xz

y

2 3 3 2 xy

1 xz 1 3 3 1 xyl l m m n n m n m

l l m m n n m n m n l n l n

n l n

l

l l m m n n m n m n l n l

m l

l n

l

m

n m

l

l

l

= + + + + +

= + + + + + +

= + + + + + + +

+

+

+

+

+

+

More compact notation

• Stress tensor transformation

• Matrix notation

xx xy xz1 1 1 1 2 3

new 2 2 2 xy yy yz 1 2 3

3 3 3 1 2 3xz yz zz

l m n l l lT l m n m m m

l m n n n n

σ σ σ = σ σ σ σ σ σ

1 2 3

1 2 3

1 2 3

Tnew old

l l lr m m m

n n n

T r T r

rotation matrix:measured from old system = =

=

Principal stresses

• Can obtain principal stresses by posing two seemingly unrelated questions– What plane will give us maximum or minimal

normal stresses?– What plane will give us zero shear stresses?– Both approaches give the same answer: A

coordinate system with zero shear stresses and extreme normal stresses

• Textbook derives equations based on zero shear stresses

Derivation• Stress vector normal to plane, hence

• That is, [l,m,n] is eigenvector of stress matrix

• Principal stresses are the eigenvalues.• What properties eigenvalues of symmetric matrix?

σ l σ m σ n σ l σ

σ l σ m σ n σ m σ

σ l σ m σ n σ n σ

x

y

z

P xx yx zx

P xy yy zy

P xz yz zz

= + + =

= + + =

= + + =

xx xy xz xx xy xz

xy yy yz xy yy yz

xz yz yy xz yz zz

σ σ σ σ σ σ σ σ0σ σ σ σ 0 σ σ σ σ 0

0σ σ σ σ σ σ σ σ

lm =m

− − − → − =

− −

Stress invariants• Setting determinant to zero gives

• Why are the I’s invariant to coordinate system?

3 21 2 3

1 xx yy zz

xx xy yy yzxx xz 2 2 22 xy xz yz xx yy xx zz yy zz

xz zzxy yy yz zz

xx xy xz2 2 2

3 xy yy yz xx yy zz xy yz xz xx yz yy xz zz xy

xz yz zz

σ I σ I σ I 0I σ σ σ

σ σ σ σσ σI σ σ σ σ σ σ σ σ σ

σ σσ σ σ σ

σ σ σ

I σ σ σ σ σ σ 2σ σ σ σ σ σ σ σ σ

σ σ σ

− − − =

= + +

= − − − = + + − − −

= = + − − −

Example

s=[1,1,1;1,1,1;1,1,1]s = 1 1 1

1 1 11 1 1

>> [l,sig]=eig(s)

l =0.4082 0.7071 0.57740.4082 -0.7071 0.5774-0.8165 0 0.5774

sig = 0 0 00 0 00 0 3

What are easy checks that the answersobtained from Matlab are correct?

Another example

s = 1 1 01 1 10 1 1

>> [l,sig]=eig(s)

l =0.5000 -0.7071 0.5000-0.7071 0.0000 0.70710.5000 0.7071 0.5000

sig =-0.4142 0 00 1.0000 00 0 2.4142

If we swapped the solutions of the two examples. How could we tell that something is wrong?

Mean and deviator stresses

• Mean normal stress• We divide stress tensor

as

• Mean stress responsible for volume change

• Deviator for yielding

113 3

xx yy zzm I

σ σ σσ

+ += =

m dT T T= +0 0

0 00 0

m xx m xy xz

m m d xy yy m yz

m xz yz m

T Tσ σ σ σ σ

σ σ σ σ σσ σ σ σ

− = = −

Plane stress

• 2-D state of stress

• Direction cosines

xx xyxx xy

zz xz yz yx yyyx yy

σ σ 0σ σ

σ σ σ 0 T σ σ 0σ σ

0 0 0

= = = → = =

x

y

X

Y

θ

θ

2π θ−

2π θ+

x y z

X n1=0

Y n2=0

Z l3=0 m3=0 n3=1

1l cosθ=

2l sinθ= −

1m sinθ=

2m cosθ=

Stress transformation

• Matrix version

• Or

• Review Mohr’s circle in textbook

xx xyXX XY

YX YY yx yy

σ σcos sin cos sinT

sin cos σ σ sin cos σ σ θ θ θ − θ

= = σ σ − θ θ θ θ

( ) ( )

2 2

2 2

2 2

σ σ cos θ σ sin θ 2σ cosθsin θ

σ σ sin θ σ cos θ 2σ cosθsin θ

σ σ σ sin θ cosθ σ cos θ sin θ

XX xx yy xy

YY xx yy xy

XY xx yy xy

= + +

= + −

= − − + −

( ) ( )( ) ( )( )

1 12 2

1 12 2

12

σ σ σ σ σ cos 2θ σ sin 2θ

σ σ σ σ σ cos 2θ σ sin 2θ

σ σ σ sin 2θ σ cos 2θ

XX xx yy xx yy xy

YY xx yy xx yy xy

XY xx yy xy

= + + − +

= + − − −

= − − +

Reading assignmentSections 2.5-6: Question: Where do we use the first of

equations 2.45 in elementary beam theory?

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