Lecture 8 Stress–Strain Relation (1)

Stress–Strain Relation (1)

8.1 General Stress–Strain System

Objectives

- Understand tensor systems of stress and strain

- Study difference between displacement and deformation

Body force: gravitational force

Surface force: normal force, shear force

 

F ma d mV

   dt

g

x yx zx

x x

g a

x y z

     

 

   

  

xy y zy

y y

g a

x y x

  

 

  

   

  

xz yz z

z z

g a

x y z

     

 

   

  

Parallelepiped, cube → infinitesimal C.V.

8.1.1 Surface Stress

Surface stresses: normal stress – shear stress –

xx

xy

Surface force

lim0

( )

x

x

xx x

A x

x

A A y z

 

  



   

lim0

x

y

xy A

Ax

  

 lim⁰

x

z

xz A

x

F

 A

 

 



lim0

( )

y

x

yx A

y y

F A A x z

  

 



   

lim0

y

y

yy y

A y

F

  A



  

 ^{limy 0}

z

yz A

y

F

 A

 

 



lim0

( )

z

x

zx A

z z

F A A x y

  

 



   

lim0

z

y

zy A

z

F

 A

 

 

 lim⁰

z

z

zz z

A z

F

  A

 

  



, ,

x y z

F F F

  

F

where = component of force vector

Same direction but different surface

Same surface

but different direction

_x : subscript indicates the direction of stress

_xy : 1st - direction of the normal to the face on which  acts 2nd - direction in which  acts

• general stress system: stress tensor

~ 9 scalar components (Cauchy’s formula)

xx xy xz

yx yy yz

zx zy zz

  

  

  

 

 

 

 

 

~ an ordered array of entities which is invariant under coordinate transformation; includes scalars & vectors

~ 3ⁿ

0th order – 1 component, scalar (mass, length, pressure, energy) 1st order – 3 components, vector (velocity, force, acceleration) 2nd order – 9 components (stress, rate of strain, turbulent diffusion)

' ^y

y y y

y







 ^







' ^z

z z z

z







 ^







' ^xy

xy xy x

x







 ^







' ^yx

yx yx y

y







 ^







' ^zx

zx zx z

z







 ^







' ^x

x x

x

x

      



◈ Shear stress is symmetric. → moment is zero

→ Shear stress pairs with subscripts differing in order are equal.

→ [Proof]

In static equilibrium, sum of all moments and sum of all forces equal zero for the element.

First, apply Newton's 2nd law F m du

 dt



xx xy xz

yx yy yz

zx zy zz

  

  

  

 

 

 

 

 

xy yx







where I = moment of inertia = r²m r = radius of gyration

= angular acceleration

( ) ( 2 ) ( )

d d d d

T rmu r m I I

dt dt dt dt

  

   



d dt



Thus,

2 d T mr

dt







^(A)

Now, take a moment about a centroid axis in the z-direction

     

2 2 2

xy yx xy yx

x y x y z

y z x z

LHS   T                 

2

d

2

d

RHS dvolr x y z r

dt dt

 

 

    



^{xy yx}

 x y z

²

x y z r

²

d

dt

    

       

𝜏_𝑦𝑥

2

2

xy yx

r d

dt

     

2

,

lim

, 0

0

x y z

r

   



xy yx

0

   

xy yx

 

 

xx xy xz

xy yy yz

xz yz zz

  

  

  

 

 

 

 

 

 Components of Lame: 6 components

Motion and deformation of fluid element

translation (displacement) – 단순이동(변위) Motion

rotation - 회전

linear (normal) deformation – 선형변형(수축 및 팽창) Deformation

angular (shear) deformation – 각변형(전단변형)

(1) Motion: no change in shape 1) Translation: x, h

, d u dt u

dt

x

 

x

, d v dt v

dt

h

 

h

2) Rotation ← Shear flow

1

z 2

v u

x y



 ^ ^  ^ ^ tan

tan

v x t x

x u y t y

y

v t x

u t y

 



 



 

 









  



   







1) Linear deformation – normal strain (변형율)

y

y

   h



Non-dimensional

x

u u dx dt u dt x u

dx x dt x

 x

    

    

 

  

 

change in length original length

 

0

x x



 ^

x



0

y y



 ^

h

 change in volume.

- For incompressible fluid, if length in 2-D increases, then length in another 1-D decreases in order to make total volume unchanged.

x '

 x  dir

_.



_y

x ''

 x  dir

_.



_z

= elongation in the due to

= elongation in the due to

2) Angular deformation– shear strain

xy x y

h x



 ^  ^

 

○ Strain normal strain:  ← linear deformation shear strain:  ← angular deformation

ii) Deformation: due to system of external forces

1) Normal strain, 

 , ,  '  , , 

O x y z  O x  x y  h z  z

' ' ' '

OABC  O A B C

change in length original length

 

 , , 

' , ,

C x x y y z z

C x x x y x z x

x x x

x h z

x h z

      

  

            

    

 

~  is positive when element elongates under deformation

 

0 0

lim O'C' lim

x x x

x x

x x x

OC x

OC x x

x x

x x



   

            

  

     

  

 

 

0 0

lim O'A' lim

y y y

y y y y y

OA y

OA y y

h h h

 h

   

           

   

    

  

 

z z



 ^

z



OC

(8.1a)

(8.1b)

(8.1c)

~ change in angle between two originally perpendicular elements

For –plane

xy

   

, 0 , 0

lim lim tan tan

xy c A c A

x y x y

    

     

   

, 0

xlimy

x x

x x

x h

x

  

 

     



y y

y y

y x

h

 

 

   



x y

h x

 

   

   

   

 

 

 

x x

x x

      

  

 

C’D A’E

O’D O’E

(8.2a)

(8.2c) (8.2b)

xy x y

h x



 ^  ^

 

yz y z

z h



 ^  ^

 

zx z x

x z



 ^  ^

 

(2) displacement vector



i j k

 x

 

h



z

^(8.3)

change of volume of deformed element original volume

e 

  ^{x x y} y ^{y z z x y z}

d V x z

e V x y z

x  ^h  z

 

              

      

   

  

 

   

x y z

x y z

x h z   

  

     

  

x y z

e      

e x y z

x h z 

  

     

   --- divergence

(8.4) (8.5) (8.6)

O C  O A 

[Homework Assignment-Special work]

Due: 1 week from today

1. Fabricate your own “Stress Cube” using paper box.