Transcript Forces

Forces
Normal Stress
 A stress measures the surface
force per unit area.

F
 Elastic for small changes
A
 A normal stress acts normal to
a surface.
 Compression or tension
Dx
A

F

 F
t 
A
Strain
 Deformation is relative to the
size of an object.
 The displacement compared to
the length is the strain e.
DL
L
e
DL
L
Shear Stress
 A shear stress acts parallel to
a surface.

F
 Also elastic for small changes
A
 Ideal fluids at rest have no
shear stress.

F
Dx
 Solids
 Viscous fluids
L
A (goes into screen)

 F
t 
A
Volume Stress
 Fluids exert a force in all
directions.

F
 Same force in all directions
A

F

F
 The force compared to the
area is the pressure.
P
DV
V
A (surface area)
F
P
A
Surface Force
 Any area in the fluid
experiences equal forces from
each direction.
 Law of inertia
 All forces balanced
 Any arbitrary volume in the
fluid has balanced forces.
Force Prism
 
t (dS1 )

dS1
 
t ( dS 2 )

dS 2
 Consider a small prism of fluid
in a continuous fluid.
 Stress vector t at any point
 Normal area vectors S form a
triangle
 The stress function is linear.



t (dS1  dS 2 )


 dS1  dS 2
 
 
t (cdS )  ct (dS )


 
t (dS )  t (dS )

 
 
 
t (dS1 )  t (dS 2 )  t (dS1  dS 2 )
Stress Function
 The stress function is symmetric with 6 components.
 To represent the stress function requires something
more than a vector.

 
 Define a tensor
P(dS )  P  dS
 If the only stress is pressure the
tensor is diagonal.

 
P(dS )  P  dS  p1 dS
 The total force is found by integration.


F   P  dS
S
Transformation Matrix
 A Cartesian vector can be
defined by its transformation
rule.
 Another transformation matrix
T transforms similarly.
x3
x3
x2
x2
x j l ij xi
x1
x1
xi  lij x j
Tpq l ipl jqTij
Tij l ipl jqTpq
Order and Rank
 For a Cartesian coordinate
system a tensor is defined by
its transformation rule.
 The order or rank of a tensor
determines the number of
separate transformations.
 Rank 0: scalar
 Rank 1: vector
 Rank 2 and up: Tensor
 The Kronecker delta is the unit
rank-2 tensor.
ss
Scalars are independent
of coordinate system.
x j l ij xi
Tpq l ipl jqTij
Tp1pn l i1 p1 l in pnTi1in
 pq l ip l jq  ij
Direct Product
 A rank 2 tensor can be
represented as a matrix.
 Two vectors can be combined
into a matrix.
 Vector direct product
 Old name dyad
 Indices transform as separate
vectors
C11 C12 C13 
C  C21 C22 C23 
C31 C32 C33 
   T
C  AB  A  B
 a1b1
C  a2b1
 a3b1
a1b2
a2b2
a3b2
a1b3 
a2b3 
a3b3 
Tensor Algebra
1T  T
f ( gT)  ( fg )T
( f  g )T  fT  gT
f ( T  U )  fT  fU
TU  UT
 Tensors form a linear vector
space.
 Tensors T, U
 Scalars f, g
 Tensor algebra includes
addition and scalar
multiplication.
 Operations by component
 Usual rules of algebra
T  U  Tij  U ij
Contraction
 The summation rule applies to
tensors of different ranks.
 Dot product
 Sum of ranks reduce by 2
 A tensor can be contracted by
summing over a pair of
indices.
 Reduces rank by 2
 Rank 2 tensor contracts to the
trace
ci  Aijb j
Tik  e ijk v j
s  ai bi
Tij  Tii
3
tr T   ijTij   Tii
i 1
Symmetric Tensor
 The transpose of a rank-2
tensor reverses the indices.
 Transposed products and
products transposed
 A symmetric tensor is its own
transpose.
 Antisymmetric is negative
transpose
 All tensors are the sums of
symmetric and antisymmetric
parts.
~
T T ij  Tij  T ji
(TU ) T  U T TT
Sij  S ji
Aij   A ji

 
~
~
T  12 T  T  12 T  T
TSA

Stress Tensor
 Represent the stress function
by a tensor.
 Normal vector n = dS
 Tij component acts on surface
element
 The components transform like
a tensor.
 Transformation l
 Dummy subscript changes



t j ( x , t , dS )  niTij ( x , t )
t j  niTij
l pjt p  lqi nqTij
l pj niTip  lqi nqTij
lrjl pj niTip  lrjlqi nqTij
 rpniTip  niTir  lrjlqi nqTij
nqTqr  nqlqilrjTij
Tqr  lqilrjTij
Symmetric Form
 The stress tensor includes normal
and shear stresses.
 Diagonal normal
 Off-diagonal shear
 An ideal fluid has only pressure.
 Normal stress
 Isotropic
 A viscous fluid includes shear.
 1  12  13 
T   21  2  23 
 31  32  3 
Tij  P ij
 Symmetric
 6 component tensor
Tij  T ji
 1  12  13 
T   12  2  23 
 13  23  3 
Force Density


F   T  dS
S

F    T  nˆdS
 The total force is found by
integration.
 Closed volume with Gauss’
law
 Outward unit vectors
S

F      TdV
V

f S    P
 A force density due to stress
can be defined from the tensor.
 Due to differences in stress as
a function of position