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ME 533 Unit 1 Static Force Analysis

This document outlines the course content for ME 533 Dynamics of Machines taught by Prof. P. Pal Pandian at Christ University. The course covers various topics in static and dynamic force analysis of machines including: static force analysis of mechanisms like four-bar linkages and slider-crank mechanisms, friction and belt drives, balancing of rotating and reciprocating masses, governors, gyroscopes, and analysis of cams. The first unit focuses on static force analysis and concepts like equilibrium conditions, free body diagrams, and static analysis of common machine components and linkages. Examples are provided for analyzing forces in four-bar linkages and slider-crank mechanisms subjected to different loading conditions.

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PalPandianP
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169 views35 pages

ME 533 Unit 1 Static Force Analysis

This document outlines the course content for ME 533 Dynamics of Machines taught by Prof. P. Pal Pandian at Christ University. The course covers various topics in static and dynamic force analysis of machines including: static force analysis of mechanisms like four-bar linkages and slider-crank mechanisms, friction and belt drives, balancing of rotating and reciprocating masses, governors, gyroscopes, and analysis of cams. The first unit focuses on static force analysis and concepts like equilibrium conditions, free body diagrams, and static analysis of common machine components and linkages. Examples are provided for analyzing forces in four-bar linkages and slider-crank mechanisms subjected to different loading conditions.

Uploaded by

PalPandianP
Copyright
© Attribution Non-Commercial (BY-NC)
Available Formats
Download as PDF, TXT or read online on Scribd
Download as pdf or txt
Download as pdf or txt
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ME 533 DYNAMICS OF MACHINES

Prof.P.PAL PANDIAN Department of Mechanical Engineering Christ University Faculty of Engineering

Abstract

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 Cams
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Unit - 1

Static Force Analysis

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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 mechanism and 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 the mechanism and their acceleration

Frictional forces:
Is the outcome of friction in the joints.

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Static force analysis


Assumption: Inertia forces & neglected gravity forces are

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Static equilibrium

A body is in static equilibrium if it remains in its 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. 2. 3.

4.

Two force member Three force member Two force and torque 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 of action of the forces intersect at the same point

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3.Two force and torque


The forces are equal in magnitude, parallel and 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 a link isolated from the mechanism showing all the active and reactive forces acting on the link in order to determine the nature of the forces acting in the link.

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Principle of Superposition

If number of forces act on a system, the net effect is equal to the sum of the individual effects of the forces taken one at a time. In a liner system, the output force is directly proportional to the input force.

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Force Convention

A force in unknown in Magnitude but Known in Direction is represented by a solid straight line A force unknown in Magnitude and Direction is represented by a wavy line.

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F12 Force exerted by the link-1 on link-2 F21 Force exerted by the link-2 on link-1 For Equilibrium of point @ B F12 = - F21 In general, Fij = - Fji
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Static force analysis of 4bar mechanism


One Known Force Two Known Force

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Static force analysis of 4bar mechanism :: One Known Force

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Static force analysis of 4bar mechanism


PROBLEM 1 A 4 bar mechanism is shown if fig., is acted upon by a force P=100N 120o on the link CD. The dimensions of various links are: AB=40mm; BC=60mm; CD=50mm; AD=30mm; DE=20mm. Determine the input torque on the link AB for the static equilibrium of the mechanism.

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Solution:

Scale : 1cm = 20N

F34 = -F43 = F23 = -F32 = ab = 1.4cm = 28N F14 = 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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Static force analysis of 4bar 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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Static force analysis of 4bar mechanism :: Two Known Force

The links 3 and 4 of a 4-bar mechanism are subjected to forces of Q=100N @600 and P=50N@450. The dimensions of various links are:

O2O4=800mm, BC=450mm, BD=200mm,

O2B=500mm, O4C=300mm, O4E=150mm.

Calculate the shaft torque T2 on the link 2 for static equilibrium of the mechanism. Also find the forces 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, the value 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. is subjected 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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