9. force analysis

7/29/2019 9. Force Analysis

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9. FORCE ANALYSIS OF MACHINERY

In the design of machine parts, the forces and torques acting onindividual elements should be analyzed carefully.

Bearings, pins, and other fasteners are usually critical elementsin machinery.

that is because of the concentration of forces acting on them.

9.1. INERTIA FORCE AND INERTIA TORQUE Consider a rigid body in planemotion subjected to the actionof external forces F 1, F 2 ,, F n as shown in fig 1


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The resultant of these forces is

produces the linear acceleration of the mass center a G and angularacceleration of the rigid body.

The angular acceleration vector need not be located at themass center G.

Let P be any element of the body, the mass of P being dm . The acceleration of P is

where a G is the acceleration of the mass center G , anda P/G is the acceleration of P relative to G .

)1(F R

)3()()(

)2(

/ /

/

t GPnGPGP

GPGP

aaaa

or

aaa


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Multiplying a P by the mass dm ,

Where and are the angular velocity and angularacceleration of the rigid body.

The resultant of a Pdm is made up of:i. The resultant of all forces like a G dm ,ii. The resultant of all forces like ( a P/G )n dm , andiii. The resultant of all forces like ( a P/G )t dm

All forces like a Gdm are all in the direction of the acceleration ofthe mass center G .

The resultant of these forces is

,)(,)(

)4()()(

/ 2

/

/ /

r aand r abut

dmadmadmadma

t GPnGP

t GPnGPGP

)5( GGG madmadma


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Forces like ( a P/G )ndm pass through the center of mass G andtheir resultant is given by

being the same for all lines on the rigid body,

But, rdm is the definition of the centroid and rdm = 0.

Forces like (a P/G )t dm have their resultant to be equal to

For the same reason as above the resultant of these forces is

)6()()(2

/ dmr dma nGP

)7()()( 22 / dmr dmr dma nGP

)8(0)( 2 / dmr dma nGP

)9()()( / dmr dmr dma t GP

)10(0)( / dmr dma t GP


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Hence we conclude that the resultant of all forces acting on abody producing acceleration of the body is

The resultant of all forces a Gdm and ( a P/G )ndm passes through themass center and does not create any moment.

But the force ( a P/G )t dm do not pass through the mass center and

act at a radial distance of r ; hence they produce a torque givenby

Knowing that the mass moment of inertia about the mass center I G is defined as

)11(aFR Gm

)12(

)(2

/

dmr

dmr r r dmaT t GP

)13(2 G I dmr


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The torque due to the resultant of the forces like ( a P/G )t dm isgiven by

Hence, the effect of the external forces acting on a rigid body inmotion is to produce

linear acceleration of the mass center G, and

Angular acceleration of the rigid body

The resultant force R and the torque T can be replaced by asingle force R acting at a distance of e from the mass centerwhere e is given by

)14(T G I

)15( R

I

R

T e G


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9.2. DYNAMIC EQUILIBRIUM In dynamic analysis of machines, accelerations are usually

known from kinematic analysis. Thus the forces required to produce these accelerations can be

determined from Newtons law of motion

Equation (16) and (17) can be written as

From equation (18) we have that the sum of all external forcesacting upon a body plus the fictitious force ma G is equal tozero.

The fictitious force ma G is called inertia force , and has thesame line of action as the acceleration of the mass center a G,but opposite in sense.

)17(T

)16(aFR

G

G

I

m

)19(0

)18(0

G

G

I T

maF


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Similarly, from equation (19) the sum of moments of externalforces about an axis through the mass center G and the externaltorques acting upon the body plus the fictitious couple -I G iszero.

The fictitious couple -I G is known as the inertia torque and itis opposite in sense to the angular acceleration .

The addition of the inertia force and the inertia torque to adynamic system transforms it into an artificial state of equilibriumknown as dynamic equilibrium .

Which can then be treated as if the system were static. This transformation of a dynamic system into an artificial static

system is known as DAlemberts Principle .

Equation (18) and (19) ca be written as

where F e and T e are the inertia force and inertia torque, respectively.

)21(

)20(

T I T

F maF

Ge

Ge


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9.3. LINKAGE FORCE ANALYSIS There are three methods of linkage force analysis:

a) Superposition Methodb) Use of Transverse and Radial Components, andc) Virtual Work Method.

These methods which can be solved either graphically oranalytically, have their relative advantage and disadvantage.

9.4. FORCE DETERMINATION In considering equilibrium of forces, the following points should

be considered carefully.i. A rigid body acted upon by two forces is in equilibrium only if the

two forces are collinear, equal in magnitude and opposite in sense.ii. A rigid body acted upon by three forces is in static equilibrium if the

lines of action of these forces are concurrent through a point.iii. A rigid body acted upon by a couple is in static equilibrium if only it

is acted upon by another coplanar couple equal in magnitude andopposite in sense.


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9.5. LINKAGE FORCE ANALYSIS BY SUPERPOSITION METHOD In the superposition method, a linkage with several forces acting

on it can easily be analyzed by determining the effect of each

force separately, one at a time, and then combining the results toget the cumulative effect. The disadvantage of this method are:

i. It is tedious because each link has to be analyzed several times, andii. It does not give accurate solution where friction has to be

considered. This method is best discussed by analyzing some examples.

9.6. RADIAL AND TRANSVERSE COMPONENTS In determining the forces between two links subjected to inertia

forces, it is possible to work simultaneously with the forces indetermining their effect on the joint connecting the link.

In this case, the forces are resolved into componentsperpendicular and parallel to the link to obtain the transverse andradial components respectively.


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Hence, treating all the joints that connect two links, we can findthe effect of all forces and torques.

In force analysis of linkages it is necessary to deal with a singleforce that acts upon a two-joint link with a line of action crossingthe link at some point between the two joints.

fig 2 shows four-bar linkage with force F 3 acting at E b/n A and B . F 3 is resolved into its radial and transverse components with

respect to A.


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By taking moments about A, the effect of the transversecomponent acting on link 3 is to produce a transverse force

on link 4 at point B. The moment of these forces about A are equal.

From this equation can be obtained as

By considering similar triangles, due to can be foundgraphically as shown in fig 2

TAF 3

TAF 34

)22(334 AE F ABF TATA

TAF 34

)23(334 AB AE

F F TATA

TAF 34

TAF 3


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If the particle moves from A to A by the action of the force, thework done during the displacement s is given by the dotproduct of vectors F and s .

For a virtual displacement s , the virtual work done is zero; i.e.

If the system is in equilibrium under the action of all externalforces and torques as well as the inertia forces and torques,then

Dividing each term by dt , we obtain

)24(cos

sF

sF U

)25(0U

)26(0neGennnn nnn T SF T SF U

)27(0neGennnn nnn T V F T V F U


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where

In using the virtual work method, the required velocity andacceleration terms are obtained from kinematic analysis of themechanism and by substituting in the virtual work equation therequired quantities can be obtained.

)32(

)31(

)30(

)29(

)28(

nnnne

GGnGe

n

G

G

n

I T

V amV F

and dt

dt

sV

dt

sV

n

nnnn

n

n


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9. force analysis

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