Transformation of stress (2D only) (Ch. 7.5 in Shames)

445.204

Introduction to Mechanics of Materials (재료역학개론)

Chapter 11: Transformation of stress (2D only)

(Ch. 7.5 in Shames)

Myoung-Gyu Lee, 이명규

Tel. 880-1711; Email: myounglee@snu.ac.kr TA: Chanmi Moon, 문찬미

Lab: Materials Mechanics lab.(Office: 30-521) Email: chanmi0705@snu.ac.kr

Objectives of the chapter

- The most common problems in engineering mechanics involve

“transformation of axes”

- Question: stresses are known in x-y plane (or x-y coordinate).

Then, what is the stress in the coordinate rotated about θ degrees?

- Stress (and strain) is not coordinate dependent, but they have different components if the coordinates are different!

σ_y

σ_x τ_xy

τ_xy

x' y

x y'

θ

2

Objectives of the chapter

Wood

Wood Glue

P P

Two pieces of wood, cut at an angle, and glued together. The wood is being pulled apart by a tensile force P.

How do we know if the glued joint can sustain the resultant stress that this force produces?

(Assume that we know the tensile and shear properties of the glue)

Objectives of the chapter

Maximal principal stress distribution observed in three gorilla teeth of an unworn (left), a

lightly worn (middle) and a worn (right) condition

Researchers at the Max Planck Institute for Evolutionary Anthropology in Leipzig,

Germany, and the Senckenberg Research Institute in Frankfurt am Main, Germany, have conducted stress analyses on gorilla teeth of differing wear stages. Their

findings show that different features of the occlusal surface antagonize tensile stresses in the tooth to tooth contact during the chewing process. They further show that tooth wear with its loss of dental tissue and the reduction of the occlusal relief decreases tensile stresses in the tooth. The result, however, is that food processing becomes less effective. Thus, when the condition of the occlusal surface changes during an individual’s lifetime due to tooth wear, the biomechanical requirements on the existing dental material change as well – an evolutionary compromise for longer tooth preservation.

Transformation of Stresses: 2D

Consideration of static equilibrium

5

Transformation of Stresses: 2D

1D uniaxial tension σ_y

σ_y

θ σ_y

σ_y' τ_x'y'

A

A/cosθ

Force equilibrium in the y’ direction (normal to the y’ plane)

( ) σ

_y

A cos θ − σ

_y′

( A/cos θ ) = 0 θ

σ cos σ

_y′

=

_y ²

Force equilibrium in the tangential direction

cos θ sin θ

σ

_y

y

x' _′

= ⋅

τ

6

Transformation of Stresses:

2D, Direct approach

2D plane stress

Force equilibrium in the x’ direction

θ τ sin θ cos θ 2

σ sin θ

σ cos

σ ′

_x

=

_x ²

+

_y ²

+

_xy

cos2 θ τ

sin2 θ 2

sin2 θ σ 2

τ

_x′y′

= − σ

^x

+

^y

+

_xy

θ

σ_y τ_xy σ_x θ

τ_xy

τ_x'y’

σ_x' Acosθ

Asinθ

A

y' x'

Force equilibrium in the y’ direction

Transformation of Stresses:

2D, Direct approach

2D plane stress

Force equilibrium in the y’ direction

θ τ sin θ cos θ 2

σ cos θ

σ sin

σ ′

_y

=

_x ²

+

_y ²

−

_xy

θ σ_y

τ_xy

σ_x

τ_xy τ_x'y’ σ_y' Acosθ

Asinθ

A

y' x'

θ

Transformation of Stresses: 2D, Direct approach

2D plane stress: summary

( σ sin θ σ ) sin σ cos θ cos θ θ 2 ( sin cos θ cos sin θ )

σ σ

cos sin θ

θ 2 σ sin

θ σ cos

σ

2 2

xy x

y y

x

xy 2

y 2

x y

xy 2

y 2

x x

− +

−

=

− +

=

+ +

=

′

′

′

′

θ τ

τ

θ τ

θ τ

 







 







 







 







−

−

−

 =

 





 







′

′

′

′

xy y x

2 2

2 2

2 2

y x

y x

τ σ σ s

c sc

sc

2sc c

s

2sc s

c τ

σ σ

Matrix form

A σ σ ′ =

or

Transformation of Strain: 2D, Direct approach





















=





































−

−

−

=





















′

′

′

′

xy y x

xy y x

2 2

2 2

2 2

y x y x

2 γ 1 ε ε 2 γ

1 ε ε s

c sc sc

2sc c

s

2sc s

c

2 γ 1

ε ε

A

ε ε ′ = RAR

⁻¹

or

















=





















=





















=





































 =















−

′

′

′

′

′

′

′

′

′

′

′

′

xy y x

xy y x

y x y x

y x y x

y x

y x

γ ε ε 2 γ

1 ε ε 2 γ

1 ε ε 2 γ

1 ε ε 2

0 0

0 1 0

0 0 1 γ

ε ε

RAR 1

RA R

( ε ε ) sin θ cos θ γ ( cos θ sin θ )

γ

cos θ sin θ

γ θ

ε cos θ

ε sin ε

cos θ sin θ

γ θ

ε sin θ

ε cos ε

2 2

xy x

y y

x

xy 2

y 2

x y

xy 2

y 2

x x

− +

−

=

− +

=

+ +

=

′

′

′

′

σ −> ε

τ −> γ/2

Note:

Transformation of Stresses: 2D

( σ sin θ σ ) sin σ cos θ cos θ θ 2 ( sin cos θ cos sin θ )

σ σ

cos sin θ

θ 2 σ sin

θ σ cos

σ

2 2

xy x

y y

x

xy 2

y 2

x y

xy 2

y 2

x x

− +

−

=

− +

=

+ +

=

′

′

′

′

θ τ

τ

θ τ

θ τ

2

cos2θ θ 1

cos

2

cos2θ θ 1

sin

cosθ 2sinθ

sin2θ

2 2

= +

= −

=

Trigonometric Identities

cos2 θ τ

sin2 θ 2

σ τ σ

sin2 θ τ

θ 2 cos2

σ σ

2 σ σ σ

sin2 θ τ

θ 2 cos2

σ σ

2 σ σ σ

xy y

x y

x

xy y

x y

x y

xy y

x y

x x

− +

−

=

− − + −

=

− + + +

=

′

′

′

′

Transformation of Stresses: Example 7.1

Transformation of Stresses: 2D, Mohr Circle

2 xy 2

y 2 x

y x 2

y x

x

τ

2 σ τ σ

2 σ

σ σ  +



 



 −

=

 +



 



 +

−

_{′ ′}

′

y

τ

x_{′ ′}

x′

σ

y′

σ



 



 +

2 ,0 σ σx y

( σ

_x′

, τ

_x′_y′

)

R

2 xy 2

y

x τ

2 σ

R σ  +



 



 −

=

Transformation of Stresses: 2D, Mohr Circle

Steps to draw Mohr’s circle: illustration

σ_x= 5 τ_xy

τ_xy= 4

σ_y=-3 Step 1:

- Consider a shear stress acting in a clockwise-rotation sense as being positive (+), and counter-clockwise as negative (-)

- The shear stresses on the x and y faces have opposite signs

- The normal stresses are positive in tension and negative in

compression as usual

σ

x

σ

y

σ

x

σ

y

τ

xy

τ

xy

Positive Negative Positive Negative

Transformation of Stresses: 2D, Mohr Circle

Steps to draw Mohr’s circle: illustration Step 2:

- Construct a graph with t as the ordinate (y axis) and σ as abscissa (x axis).

- Plot the stresses on the x and y faces of the stress as two points on this

graph(follow the sign convention before)

Step 3:

- Connect these two points with a straight line

- Draw as circle with the line as a diameter

τ

( ) - 3,4

x ( + 5,-4 )

y

σ

Transformation of Stresses: 2D, Mohr Circle

Steps to draw Mohr’s circle: illustration

Step 4:

- Determine stresses on a square that has been

rotated through an angle θ with respect to the original square

- Rotate the line in the

same direction through 2θ.

This new end points of the line are labeled as x’ and y’.

τ

( ) - 3,4

x ( + 5,-4 )

y

σ

x′

y′

2 θ

Transformation of Stresses: Principal stress

- Normal stresses become maximum values and the shear stresses are zero

- These normal stresses are called “principal”

stresses, σ

_p1

and σ

_p2

τ

x y

σ

2 θ

p

σ

p1

σ

p2

2 σ σ_x − _y

τxy

2 σ σ_x + _y

Transformation of Stresses: Principal stress

( σ σ ) /2

tan2 θ τ

y x

xy

p

= −

(

^{x y}

)

^{x y 2}

( )

_xy ²

p2

p1,

τ

2 σ σ

2 σ

σ σ  +



 



 −

+ ±

=

( )

xy ² 2

y x

p2 p1

max

τ

2 σ σ

2 σ

τ σ  +



 



 −

− =

=

By Pythagorean construction

Principal stresses

Maximum shear stress

“The maximum shear are 90

^o

away from the

principal stress points on the Mohr’s circle”

Transformation of Stresses: 2D, Mohr Circle

Maximum shear stress and its plane

( )

xy ² 2

y x

p2 p1

max

τ

2 σ σ

2 σ

τ σ   +

 



 −

− =

=

Maximum shear stress

“ In the tensile specimen, the maximum shear are 45 ^o away from the loading

direction which is the direction of principal stress”

Transformation of Stresses: Invariant

cos2 θ τ

sin2 θ 2

σ τ σ

sin2 θ τ

θ 2 cos2

σ σ

2 σ σ σ

sin2 θ τ

θ 2 cos2

σ σ

2 σ σ σ

xy y

x y

x

xy y

x y

x y

xy y

x y

x x

− +

−

=

− − + −

=

− + + +

=

′

′

′

′

x y x y

σ ′ + σ ′ = σ + σ

Transformation of Stresses: 2D, Mohr Circle

Pure shear

= 0 +

2 σ

σ

_{x y}

When normal stresses vanish on the plane of maximum shear

Example: the stress state by the simple torsion

τ

x

σ σ

1

σ

2

y

These two stresses are equivalent

Transformation of Stresses: 2D, Mohr Circle

Under Pure shear

G γ τ =

2G ε

₁

= τ

(

_{1 2}

)

1

σ νσ

E

ε = 1 −

( ) ( τ ν - τ )

E 1 2G

τ = − G = 2 ( 1 E + ν )

γ/2

x

ε ε

1

y

Hooke’s law for shear

From the Mohr circle

Principal strain is related to the principal stress

From the Mohr’s circle