Unit 1
1. Describe in details about Newton law of motion?( M/J 2014)
Newton’s law :
Since the muscle skeletal system is simply a series of objects in contact with each other , some of the basics physics principles developed by sir Isaac Newton.
Newton’s law are as follows:
Newton's First law : An object remains at rest (on continues moving at a constant velocity ) unless acted upon by an unbalance external force.
Newton's first law is called the Law of Inertia because it outlines a key property of matter related to motion.
Newton's second law : If there is an unbalanced force acting an a object it produces an acceleration in the direction of the force directly proportional to force (F=ma).
Newton's second law is arguably the most important law of motion because it shows how the forces that create motion (kinetics) are linked to the motion (kinematics). The second law is called the Law of Momentum or Law of Acceleration.
Newton's Third law: For every action (force) there is a reaction (opposite force)of equal magnitude but in the opposite direction.
From the first law It is clear that if a body is at rest there can be no unbalanced external force acting on it.In this situation ,termed static equilibrium ,all the external forces acting on a body must add to zero.
An extension of this law to objects larger than a particle is that the sum of the external moments acting on that body must also be equal to zero for the body to be rest.
Moments (m) is typically caused by the force(f) acting at a distance ® from the center of relation of a segment .A moment lends to cause a rotation and is defined by the cross product function M=r X f.
For 3D Analysis there are a total of six equations that must satisfied for static for static equilibrium.
F x = 0 F y= 0 F z = 0
M x = 0 M y= 0 M z = 0
For 2D analysis there are only two in plane force components & one perpendicular moments (torque) components:
F x = 0 F y= 0 M z = 0
When the body is not in static equilibrium Newton’s second law state that the unbalanced force and moments are proportional to the acceleration of the body.
2. Derive the Euler’s equation .(N/D2014)
A fluid in equilibrium is not affected by an shearing stress that is there is there are no forces acting in the horizontal direction .It the fluid is in motion these forces will be important as a result of the viscosity of fluid .let us consider initially an ideal fluid for which the viscosity forces the only stress that has to be considered in the fluid is the normal stress or pressure P.pressure is a scalar quantity and a function of a space coordinates and time,p(x,y,z,t).
In the simplest cases the fluid particle move in layer or sheet ,which constitute a laminar flow .If the particle trajectories are irregular the flow is turbulent.
Consider non rotational fluds,I n which the angular velocity fluids ,in which the deformation is negible.
Let us consider again a volume V in the fluid .the total force acting on this volume due to the interactions with the remaining fluid particles is
Where –v e sign is due to the fact that the Forces acts on the consider element ,to a scalar quality A, we have
Therefore the force by the remaining part of the fluid on the volume element dv is -p dv,and the force per unit volume is simply -p.
We can express the momentum conservation by.
ɐ X acceleration =-p.
where ɐ is again the fluid density .In the care of a particle of mass m underthe action of a force and moving with velocity v,the acceleration is simply a=F/M =dv/dt. for a fluids particle the velocity variation has two components :
dv1= __d v ___ d t
d t
which refer to the velocity variation at a fixed point in space that is having constant coordinates x,y,z in a time Val dt and
dv2= dx ∂v∂x + dy ∂v∂y + dz ∂v∂z
which refers to the velocity variation at a given time between two different points in space separated by
d r=dxi+dyj+d2k
which is the distance covered by the fluid particle in time dt(I,j,k are the unit vectors of the direction x,y,z)recalling tha
∇=ddx i+ddy j+ ddz kwe have
dv2= dx ddx + dy ddy + dz ddz
dv2=(dr-∇)v
the total variation dv is
dv =dv1+dv2 =dvdtdt+(dr.∇)v
the acceleration is them
acceleration =dvdt + (v. ∇) v
this derivation is called total derivation which is usually represented.
D Dt =ddt+(v. ∇)
The descrption of the fluid motion considering separately two terms,that is d/dt
In a fixed position and v, ∇ in given time corresponds to eulerian description on the other hand the description of the fluid motion in terms of the total derivative D/Dt corresponds to the lagrangian description .In this case we can imagine that we are following the motion of a fluid element .
Naleirally the lagrangian derivative D/Dt is related to eulerian derivatives
e X [dvdt + (v. ∇) v ] = -∇p
dvdt + (v. ∇) v -1e ∇p
D Dt =-1e ∇p
Different form of euler equation.
An important application of the equation of motion occurs when the fluid is in a gravitation fluid characterized by the acceleration (force per unit mass)g .In this case each volume unit is under the action of the force eg which is the gravitational force acting per unit volume so that the equation of motion is written as.
dvdt + (v. ∇) v -1e ∇p+g
More generally if the fluid is under the action of an external force field F(dyn/cm’)that is F is the force acting on a unit volume we have
dvdt + (v. ∇) v -1e ∇p+ 1e f
This is the Eulers equation.
3.Derive the Navier stroke equation of fluid mechanics.(M/J 2014)
Imagine a closed control volume .∀0 within the flow field is moving through it. The control volume occupies reasonably large ferrite region of the flow field .A control surface A0 is defined as the surface which bondsthe volume v0.
According to Reynolds transport theorem the rate of change of moments for asystem equals the sum of the rate of change of moments inside the control volume and the rate of efflux of momentum
Across the control surface.
The rate change of momentum for system is equal to the net external force acting on it
Rate of change of moments inside the control volume = ddt ᶴᶴᶴ ɐ d ∀
= ᶴᶴᶴddt (ɐ d ∀
∀o
Rate of efflux of moments through control surface
ᶴᶴA0 Pv→(→v.dA→)=ᶴᶴA0 Pvv→.n→dA
=ᶴᶴᶴ∀0(v→(∇.pv→)+pv→.∇.v→ )d ∀
Surface force acting on the control volume
=ᶴᶴA0 dA→σ
=ᶴᶴᶴ∀0(∇.σ)d∀
Body force acting on the control volume
=ᶴᶴᶴ∀0 ρ→dd∀
Intially we got
ᶴᶴᶴ∀0(ddt(ρv→)+(v→(∇.pv→)+pv→.∇.v→ ) d∀
=ᶴᶴᶴ∀0(∇.σ+ ρ→d)d∀
ρ dv→dt +v→de dt +pv→ . ∇v→+v→(∇.pv→)+pv→.∇.v→) =v.σ+ρ/b→
we known that
dedt+∇.pv→
=0 is the general from of mass conservation equation valid
for both compressible and in compressible flows.
ρ (dv→dt +v→. pv→)=∇.σ+ ρ→b
ρ Dv→Dt= ∇.σ+ ρ→b)
This equation is referred as Cauchy”& equation of motion .In this Equation,σ
is the stress tensor.
After having substituted σ We get .
∇.σ= -∇p+(μ'+μ)∇(∇.∇→)+μ∇2 v→
from stnoke’s hypothesis we get μ’+2/3 μ=0
Invoking above two relationship in to 1 we get
ρ Dv→Dt= -∇p+μ∇2 v→ +1/3 μ∇(∇. v→)+ρ→b
This the most general from of navier –stokes equation
4.Derive the Stress transformations of fluid mechanics.
Consider the rectangular bar subject to externally applied force that cause various mode of deformations which the bar .
Let p be a point within the structure .assume that a small cubical material element at point p with sides parallel to the sides of the sides of the bar is cut out and analyzed .
The material is subjected to a combination of normal (σx ∆ σy)and shear (τxy)stress in xy plane
Consider a second element at the same material point but with a different orientation than first element . Mechanical stress is symbolized with the Greek letter sigma and is defined as the force per unit area within a material ( = F/A). Mechanical stress is similarto the concept of pressure and has the same units (N/m2 and lbs/in2).
In the SI system one Newton per meter squared is one Pascal (Pa) of stress or pressure. Atmospheric gases that typically exert a pressure of 1 atm, 101.3 KPa (kilopascals), or 14.7 lbs/in2 on your body.
Note that mechanical stress is not vector quantity, but an even more complex quantity called a tensor.
Tensors are generalized vectors thathave multiple directions that must be accounted for, much like resolving a force into anatomically relevant axes like along a longitudinal axis and at right angles (shear).
The maximum force capacity of skeletal muscle is usually expressed as a maximum stress of about 25–40 N/cm2 or 36–57 lbs/in2.
This force potential per unit of cross-sectional area is the same across gender, with females tending to have about two-thirds of the muscular strength of males because they have about
two-thirds as much muscle mass a males.
One can assume that the second material element is obtained simply by rotating the first in the counter clockwise direction through an angle θ.
Let x1 and y1 be two mutually direction representing the normal’s to the surface of the transformed material element .
The stress distribution on the transformation material element would be different than of the first .in general the second element may be subjected to normal stresses (σx1 and σy1) and shear stress (τx1y1)
As well .if stress σx, σy and τxy1 and the angle of rotation θ given ,the stress σx1 , σy1 and τx1y1 can be calculate using the following formulas
σx1 = σx+σy2 + σx-σy2 cos(2θ)+ τxy sin(2θ)
σy1 = σy+σx2 + σy-σx2 cos(2θ) - τxy sin(2θ)
Τx1y1 = - σx-σy2 sin (2θ) + τxy cos (2θ)
There equations can be used for transforming stress from one set of coordinates (xy) another (x1 ,y1)
5.Derive the Strain energy function for fluid mechanics.
A material is said to be homogenous . when the distribution of the internal structure is such , that material point has same mechanical behavior. In heterogeneous material , the strain – energy function will additionally depend on the portion of the material point in the reference placement X .
The measure of the deformation of a material created by a load is called strain. This deformation is usually expressed as a ratio of the normal or resting length (L0) of the material. Strain can be calculated as a change in length divided by normal length (L – L0)/ L0.
Imagine stretching a rubber band between two fingers. If the band is elongated to 1.5 times its original length, you could say the band experiences 0.5 or50% tensile strain. This text will discuss the typical strains in musculoskeletal tissues in percentage units. Most engineers use muchmore rigid materials and typically talk in terms of units of microstrain.
For physical observation we conclude that the strain energy increase monotonically with the deformation ,
W(I)=0 and w(F)≥0
dw = Tij dij = r(dI-da)
w is the strain energy function , its derivation with respect to the strain is the stress .
6.Define Hooke’s law of elasticity.(M/J2013)
Hooke's law is a principle of physics that states that the force (F) needed to extend or compress a spring by some distance X scales linearly with respect to that distance.
F = kX,
where k is a constant factor characteristic of the spring: its stiffness, and X is small compared to the total possible deformation of the spring.
Soft tissues and cells exhibit several anelastic properties:
Hysteresis during loading & unloading of
Stress relaxation at constant strain
Creep at constant stress .
Strain rate dependence.
There properties can be modeled by theory of elasticity.
Visco elastic materials consists of polymers of variable chain length and filters resulting in to following stress strain diagram.
Slope of the major axis of the ellipse is a measure of material stiffness while the ratio of the minor stress strain relationship may be written as
a0 σ+i=1nai +ditdti =b0 +j=1nbj djtdtj
o is stress and ϵ is the strain In this relation of all the coefficients a0………an and bo…..bm are constants ,the material is referred to as linear elasticity.
7.Write short note on newtanion and non newtanion fluid ?(N/D2013)
Constitutive Equations :
Two types of fluids:
1. Newtonian
2. non-. Newtonian fluids
Newtonian Fluid
Newtonian fluid's viscosity remains constant, no matter the amount of shear applied for a constant temperature.. These fluids have a linear relationship between viscosity and shear stress.
A Newtonian fluid Is a viscous fluid for which stress is proportional to the velocity gradient .(i.e time –rate of strain )
τ= μ du/dy
τ-sharestress exeted by the fluid
μ-viscosity
du/dy-velocity gradient perpendicular to the direction of shear.
Non-. Newtonian fluids:
For a non- Newtonian fluids the viscosity change with the applied strain rate (velocity gradient ) As a result , non- Newtonian fluids may not have a well –defined viscosity.
Non-Newtonian fluids are the opposite of Newtonian fluids. When shear is applied to non-Newtonian fluids, the viscosity of the fluid changes.
Constitute relation for Newtonian fluids
The stress ten son can be decomposed into spherical and deviation parts
σ=τ-pI or σij= τij-pʆij
Where p=-1/3 τ σ =-1/3 σij
Is the mechanical pressure and τ is stress deviator shear stress tensor )
Constitutive relations for Newtonian fluid from three elementary hypotheses.
σ=μ(∇μ+∁∇u))T-pI
σij= μ(μij+μji)-p ʆij
This is a relation for in compressible fluid (ie when ∇, μ=0)
8.What is viscoelasticity? Explain its types. (N/D2013)
Viscoelasticity is the property of materials that exhibit both viscous and elastic characteristics when undergoing deformation. Viscous materials, like honey, resist shear flow and strain linearly with time when a stress is applied. Elastic materials strain when stretched and quickly return to their original state once the stress is removed.
Viscoelastic materials have elements of both of these properties and, as such, exhibit time-dependent strain. Whereas elasticity is usually the result of bond stretching along crystallographic planes in an ordered solid, viscosity is the result of the diffusion of atoms or molecules inside an amorphous material.
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Viscoelasticity was further examined in the late twentieth century when synthetic polymers were engineered and used in a variety of applications.
Viscoelasticity calculations depend heavily on the viscosity variable, η. The inverse of η is also known as fluidity, φ. The value of either can be derived as a function of temperature or as a given value .
Depending on the change of strain rate versus stress inside a material the viscosity can be categorized as having a linear, non-linear, or plastic response. When a material exhibits a linear response it is categorized as a Newtonian material. In this case the stress is linearly proportional to the strain rate. If the material exhibits a non-linear response to the strain rate, it is categorized
A viscoelastic material has the following properties:
Types of Viscoelasticity
Linear viscoelasticity is when the function is separable in both creep response and load. All linear viscoelastic models can be represented by a Volterra equationconnecting stress and strain:
Linear viscoelasticity is usually applicable only for small deformations.
Nonlinear viscoelasticity is when the function is not separable. It usually happens when the deformations are large or if the material changes its properties under deformations.
9.Write short note on the Kinetics and Kinematics of Motion ?(M/J 2014)
Kinetics is a term for the branch of classical mechanics that is concerned with the relationship between the motion of bodies and its causes, namely forces and torques.
In Kinetics, velocities, accelerations and the forces which creates the motion.
Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects) and systems of bodies (groups of objects) without consideration of the causes of motion.
In Kinematics, velocities and accelerations without the forces/torques which creates the motion. Kinematics & Kinetics are bound by Newton’s second law,which stales that the external force (f) on an object is proportional to product of that object mass (m) and linear acceleration(a)
F=ma
for conditions of state equilibrium ,there are no external force because there is no accelerations are the sum of the external forces can be set equal to zero .however when an object is accelerating ,the so called inertial force must be consider and the sum of the force is no longer equal to zero .
ex :static and dynamic Equilibrium .
If this performed very slowly so that the acceleration is negligible static equilibrium conditions can be applied and the force required is 200n .However ,if this same box is lifted with an accelerations of 5m/s2 ,then the sum of force is not equal to zero and the force required is 300N .
There is an analogous relationship for rotational motion ,in which the external moment (M) on an object proportional to that object moment of inertia (Z) and angular acceleration (α)
M=I α
Just as mass is a measure of a reuse trance to linear acceleration moment of inertia is a measure of resistance to angular acceleration .It is affected both by
total mass an the distance that mass is from the COR .π as follows
I=m π2
This is the kinetics and kinematics of motion.
10.Write short note on biomechanical principles and vector mechanics?
Stability
The lower the center of gravity the large the base of support the closer the line of gravity to the center of the base of support & the greater the mass the more stability increases.
Maximum effort
The production of maximum force requires the use of all the joint that can be used .(Ex foot ball player (kickers)
The production of maximum velocity requires the use of all the joints in order from largest to smallest.
(ex basket ball jump, short larger starts the motion & facter joint contributes)
Linear motion
The greater the applied impulse the greater the increase in velocity .Impulse =Force X time.
Movement usually occurs in the direction opposite that of the applied force.
Ex skiing, speed skating ,swimming.
Angular Motion :
Angular Motion is produced by the application of force acting at some distance from axis .(or a torque)
Angular movement is constant when an athlete or object is free in the air.
Ex: diver rotates in air momenlium constant while in air .
Vector Mechanics:
Biomechanical parameters can be represented as either scalar or vector quantities .A scalar is simply represented by its magnitude .
Ex :mass ,time & length A vector is generally describe as both magnituted any orientation Ex :forces & moments.
The mostcommon use of vectors in biomechanics is to represent force such as muscle and joint reactions and resistance graphically with the use of line with an arrow at one end.
Vector addition :
When study musculo skeletal biomechanics ,its common to have more then one force to consider .therefore it is important to understand how to work with more than one vector . when adding or subtracting two vectors , there some important properties to consider.
vector addition is cumulative.
A+B=B+A
A-B=A+(-B)
Vector addition is associative
A+(B+C)=(A+B)+c
Both magnitudes and direction on (orientation) vector must be taken into account .
It is called the principal of mechanics and vector mechanics.
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