BME18R311

BIOFLUIDS AND DYNAMICS

GEORGE EBENEZER ARUL SAMUVEL

BIO MECHANICS
• Biological fluid mechanics (or biofluid mechanics) is the study of the
motion of biological fluids in any possible context.
1. blood circulation

2. hearth pumping

3. flow in the systemic arteries

4. flow in the pulmonary arteries

5. flow in the microcirculation

Use of biological fluid mechanics
• Pure physiology: understanding how animals, and in particular
humans, work.
• Pathophysiology: understanding why they might go wrong. In other
words, understanding
• the origins and development of diseases.
• Diagnosis: recognizing diseases from possibly non-traumatic
measurements.
• Cure: providing support to surgery and to the design of prosthetic
devices.
NEWTON LAWS
NEWTON LAWS
NEWTON'S FIRST LAW

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted
upon by an external force. It may be seen as a statement about inertia, that objects will remain in their state of
motion unless a force acts to change the motion.
NEWTON’S SECOND LAW

Newton's Second Law as stated below applies to a wide range of physical phenomena, but it is not a
fundamental principle like the Conservation Laws. It is applicable only if the force is the net external force.
NEWTON’S THIRD LAW

Newton’s Third Law All forces in the universe occur in equal but oppositely directed pairs. There are no isolated
forces; for every external force that acts on an object there is a force of equal magnitude but opposite direction
which acts back on the object which exerted that external force.
Stress
• When solid bodies are deformed, internal forces get distributed in the
material. These are called stresses. Stress has the unit of force per area.
Stress
• Lets assume,
Before the deformation, the surface is characterized by the area dA and
the normal vector N. After the deformation, these become da and n.
Stress
• The equation can written as

An axially loaded bar.

An infinitesimal surface in the original and deformed configurations.

 Stress

where F is the deformation gradient tensor and

where
 Stress

will give a closed system of equations for the displacement vector.

Here

Stress Components on a Rotated Plane

is called the second Piola-Kirchhoff stress tensor. This is a symmetric tensor that is For the axially loaded bar, it is easy to think about the stress as a scalar number

energy conjugate to the Green-Lagrange strain. and state that on this bar, only a normal stress exists. The full stress tensor is

The first and second Piola-Kirchhoff stress tensors are related via:

This formula makes it possible to rewrite the momentum balance equation as:

which together with a constitutive relation of the form where is the angle between the axis of the bar and the normal to the surface.
 1D Elasticity (axially loaded bar)
y A(x) = cross section at x
b(x) = body force distribution
F (force per unit length)
x E(x) = Young’s modulus
x u(x) = displacement of the bar
x=0 x=L at x

1. Strong formulation: Equilibrium equation + boundary

conditions
d
Equilibrium equation  b  0; 0 xL
dx
Boundary conditions u  0 at x  0
du
EA  F at x  L
dx
 du
2. Strain-displacement relationship: ε(x) 
dx

3. Stress-strain (constitutive) relation :  (x)  E ε(x)

E: Elastic (Young’s) modulus of bar
 3D Elasticity
Problem definition

Surface (S) V: Volume of body

S: Total surface of the body
Volume (V) The deformation at point
w x =[x,y,z]T
v is given by the 3
u components of its u 
z  
x displacement u  v
w 
NOTE: u= u(x,y,z), i.e., each  
displacement component is a function
y
x of position
 3D Elasticity:
EXTERNAL FORCES ACTING ON THE BODY

Two basic types of external forces act on a body

1. Body force (force per unit volume) e.g., weight, inertia, etc
2. Surface traction (force per unit surface area) e.g., friction
 BODY FORCE

Volume
Xc dV
element dV Body force: distributed
Xb dV force per unit volume (e.g.,
Xa dV weight, inertia, etc)
w
Volume (V) X a 
v  
u X  X b 
z Surface (S) X 
x  c
NOTE: If the body is accelerating,  u 
 
then the inertia force  u
   v 
y  w 
x   
may be considered as part of X
~
X  X   u
 SURFACE TRACTION
Volume pz
Xc dV Traction: Distributed
element dV
Xb dV p py force per unit surface
x area
Xa dV
w
Volume (V)
v ST
u p x 
z  
x T S  p y 
p 
 z
y
x
 3D Elasticity:
INTERNAL FORCES
Volume sz
element dV
tzy
tzx ty
w txz t sy
z
Volume (V) xy
v tyx
u
z sx
x

y
x If I take out a chunk of material from the body, I will see that,
due to the external forces applied to it, there are reaction
forces (e.g., due to the loads applied to a truss structure, internal
forces develop in each truss member). For the cube in the figure,
the internal reaction forces per unit area(red arrows) , on each
surface, may be decomposed into three orthogonal components.
 3D Elasticity
sz
sx, sy and sz are normal stresses.
tzy
tzx ty The rest 6 are the shear stresses
txz s Convention
y
z t xy
z
txy is the stress on the face
tyx perpendicular to the x-axis and points
sx in the +ve y direction
Total of 9 stress components of which
y  xy   yx
x only 6 are independent since
 x   yz   zy
   zx   xz
The stress vector is therefore  y

 z 
  
 xy 
 yz 
 
 zx 
  x 
 
Strains: 6 independent strain components  y
  z 
  
 xy 
 yz 
 
 zx 

Consider the equilibrium of a differential volume element to

obtain the 3 equilibrium equations of elasticity
 x  xy  xz
   Xa  0
x y z
 xy  y  yz
   Xb  0
x y z
 xz  yz  z
   Xc  0
x y z
 Compactly;
EQUILIBRIUM (1)
  X 0
T
EQUATIONS

where  
 x 0 0
  
0 0
 y 
 
0 0
z 
 
 0
 y x 
  
0 
 z y 
 0

 z x 
 TS
nz pz Traction: Distributed
force per unit area
n
py p x 
ny  
T S  p y 
nx px p 
ST  z
n x 
 
If the unit outward normal to ST : n  n y 
n 
 z
Then p x   x nx   xy n y   xz nz
p y   xy nx   y n y   yz nz
p z   xz nx   zy n y   z nz
 In 2D n py TS
txy
ny
q q q
dy ds nx sx dy ds px
txy dx
dx
y ST
sy
x Consider the equilibrium of the wedge
in x-direction
dx p x ds   x dy   xy dx
sin    ny
ds dy dx
dy  px   x   xy
cos    nx ds ds
ds  p x   x n x   xy n y
Similarly
p y   xy n x   y n y
2. Strain-displacement relationships:
u
x 
x
v
y 
y
w
z 
z
u v
 xy  
y x
v w
 yz  
z y
u w
 zx  
z x
Compactly;  u (2)

 
 x 0 0
  
 x  0 0
   y 
 y   u 
  z  0 0  
    z  u  v
 xy   
0
w 
 
 yz   y x 
    
 zx 
0 
 z y 
 0

 z x 
 u
dy
y
C’
In 2D v y
v dy
y
C 2
B’
u A’ 1 v
dy dx
x
v
A dx B
x
u
  u   u dx

 dx   u  dx   u 
  dx x
A' B' AB   x   u
x   
AB dx x
  v  
 dy  
 v  dy 
  v   dy
A' C' AC  y  v
   
y   
AC dy y
π
 xy   angle (C' A' B' )  β 1  β 2  tanβ 1  tanβ 2
2
v u
 
x x
3. Stress-Strain relationship:

Linear elastic material (Hooke’s Law)

  D (3)

Linear elastic isotropic material

1     0 0 0 
  1   0 0 0 

   1  0 0 0 
E  1  2 
D  0 0 0
2
0 0 
(1   )(1  2 )  
1  2
 0 0 0 0 0 
 2 
 1  2 
 0 0 0 0 0
2 
PLANE STRESS: Only the in-plane stress components are nonzero
Area
Nonzero stress components  x ,  y , xy
element dA h
 xy y
 xy
x
D
y Assumptions:
1. h<<D
2. Top and bottom surfaces are free from
traction
x 3. Xc=0 and pz=0
PLANE STRESS Examples:
1. Thin plate with a hole

 xy y
 xy
x

2. Thin cantilever plate

 PLANE STRESS
Nonzero stresses:  x ,  y , xy
Nonzero strains:  x ,  y ,  z ,  xy

Isotropic linear elastic stress-strain law   D 

 x    
 1  0   x  
 
 y  
E
 1 0   y  z    x   y 
  1   2
 1    1 
 xy  0 0   xy 
 2 

Hence, the D matrix for the plane stress case is

 
1  0
E  
D  1 0 
1  
2
1  
0 0 
 2 
PLANE STRAIN: Only the in-plane strain components are nonzero

Nonzero strain components  x ,  y ,  xy

Area
 xy y
element dA
 xy
x

Assumptions:
y 1. Displacement components u,v functions
of (x,y) only and w=0
2. Top and bottom surfaces are fixed
x 3. Xc=0
4. px and py do not vary with z
z
PLANE STRAIN Examples:
1. Dam Slice of unit
thickness
1

 xy y
y  xy
x
z

x
z
2. Long cylindrical pressure vessel subjected to internal/external
pressure and constrained at the ends
 PLANE STRAIN
Nonzero stress:  x ,  y ,  z , xy
Nonzero strain components:  x ,  y ,  xy
Isotropic linear elastic stress-strain law   D 

 x   
1   0   x 
  E
      z    x   y 
 y   1  0  y 
  1   1  2   0 1  2   
 xy   0   xy 
 2 

Hence, the D matrix for the plane strain case is

 
1   0 
E
D   1  0 
1   1  2   1  2 
 0 0 
 2 