me 533 unit 1 static force analysis

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ME 533

DYNAMICS OF MACHINES

Prof.P.PAL PANDIANDepartment of Mechanical Engineering

Christ University Faculty of Engineering


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Unit 1

Static Force Analysis

Unit 2

Dynamic Force Analysis

Unit 3

Friction and Belt Drives

Balancing of Rotating Masses

Unit 4

Balancing of Reciprocating Masses

Governors

Unit 5

Gyroscope

Analysis of CamsProf.P.PAL PANDIAN 2

Abstract


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Unit - 1

Static Force Analysis

Prof.P.PAL PANDIAN 3


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Unit I Abstract

Introduction Static equilibrium. Equilibrium of two and three force members. Members with two forces and torque. Free body diagrams. Principle of virtual work / Principle of

Superposition

Static force analysis of four bar mechanismand slider-crank mechanism with and without

friction.



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Introduction

Sources of Forces Forces of gravity Forces of assembly Forces from applied load Forces from energy transmission Frictional forces Spring forces

Impact forces Forces due to change of temperature Inertia Forces Applied Forces



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Introduction

Applied forces:

Acts from outside on the mechanism

Inertia forces:

Arises due to the mass of the links of themechanism and their acceleration

Frictional forces:

Is the outcome of friction in the joints.



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Static force analysis

Assumption:

Inertia forces & gravity forces areneglected



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Static equilibrium

A body is in static equilibrium if it remains inits state of rest or of motion.

Conditions: The vector sum of all the forces acting on the

body is zero.F=0 The vector sum of all the moments about any

arbitrary point is zero.M=0 In a planar mechanism, forces can be described by

2D vectors. Fx=0 Fy=0 Mx=0



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Equilibrium of members

1. Two force member

2. Three force member

3. Two force and torque

4. Four force member



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1.Two force member

The forces are of the same magnitude

The forces are collinear

The forces are in opposite directions



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2.Three force member

The resultant force is zero

The forces are concurrent, i.e., line ofaction of the forces intersect at the same

point



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3.Two force and torque

The forces are equal in magnitude, paralleland opposite in direction.

Two forces form a couple, which is equal

and opposite to the applied torque.



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4. Four force members



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Free body diagram

A free body diagram is the diagram of alinkisolated from the mechanism showingall the active and reactive forces acting on

the link in order to determine the natureof the forces acting in the link.



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Principle of Superposition

If number of forces act on a system, thenet effect is equal to the sum of theindividual effects of the forces taken one

at a time.

In a liner system, the output force isdirectly proportional to the input force.



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Force Convention

A force in unknown in Magnitude butKnown in Direction is represented by asolid straight line

A force unknown in Magnitude andDirection is represented by a wavy line.



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F12Force exerted by the link-1 on link-2F21Force exerted by the link-2 on link-1

For Equilibrium of point @ BF12 = - F21

In general, Fij = - Fji


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Static force analysis of 4bar

mechanism One Known Force

Two Known Force


S i f i f 4


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mechanism :: One Known Force



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mechanismPROBLEM 1

A 4 bar mechanism is shown if fig., is acted upon by aforce P=100N120o on the link CD. The dimensions ofvarious links are: AB=40mm; BC=60mm; CD=50mm;

AD=30mm; DE=20mm. Determine the input torque onthe link AB for the static equilibrium of the mechanism.



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Solution:


Scale :

1cm = 20N

F34 = -F43 = F23 = -F32 = ab = 1.4cm = 28NF14 = bo = 4.9cm = 98N

h = 39mm

T= -F32 x h = -28 x 39 = -1092Nmm [CCW]

Input Torque = T2= -T = 1092Nmm [CW]


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mechanism :: Two Known Force


UNKNOWNS:

Link-2: mag.& dir of F12 & F32 and T2

Link-3: mag. & dir of F23 & F43

Link-4: mag.& dir of F34 and F14


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The mag. of Ft34 is found by taking moments about O4:

Mag of Ft43= -Ft34

UNKNOWN: mag.&dir of F23 and mag of

Fn34

The mag of Fn34 can be found by taking

moments about point B


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UNKNOWN:

mag.&dir of F23


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mechanism :: Two Known Force The links 3 and 4 of a 4-bar

mechanism are subjected toforces of Q=100N @600 andP=50N@450. The dimensions ofvarious links are:

O2O4=800mm, O2B=500mm,

BC=450mm, O4C=300mm,

BD=200mm, O4E=150mm.

Calculate the shaft torque T2 onthe link 2 for static equilibrium ofthe mechanism. Also find theforces in the joints.



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Static force analysis of Slider Crank

mechanism :: One Known Force

In a slider crank mechanism shown in fig, thevalue of force applied to slider 4 is 2kN. The

dimensions of the various links are: AB=80mm,

BC=240mm. Determine the forces on various

links and the driving torque T2



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F14=ab=1.5cm=600N

F34=ob=5.3cm=2120N

F34=-F43=F32=-F23=F21=-F12


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Couple = T = F32 x h = -2120 x 79.2 = -167.9Ncm [ccw]

Torque on link AB = T2 = -T = 167.9Ncm [cw]


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Static force analysis of Slider Crank

mechanism :: Two Known Force

A slider crank mechanism shown in fig. issubjected to two forces: P=3kN and Q=1000N @

600. The dimensions of various links are:

AB=250mm, BC=600mm, BD=250mm.

Determine the torque T2 applied on the link 2



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Qsin60 x BD = Ft43 x BC

Ft43 = 360.8N


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F34 = ca

ca = 6.7cm = 3350N

To find : F23 = -F32

F23 = -F32 = 7.7cm = 3850N

T = F32 X h = -3850 x 0.21

= -808.5Nm[ccw]

Torque on link AB = T2 = -T

=808.5Nm[cw]


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me 533 unit 1 static force analysis

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