dubbel handbook of mechanical engineering volume 1 pdf

1.3 Combination and Resolution of Non-Concurrent Forces 1.3.1 Coplanar Forces Combination of Several" Forces to Form One Resultant Force Grapblcal Process wltb Polygon of Forces and Fu

DUB BEL

DUBBEL

Handbook of

MECHANICAL

ENGINEERING

With 1258 Figures

Springer-Verlag London Ltd

Wolfgang Beitz, Professor Dr.-Ing

Technische UniversWit Berlin,

Institut fur Maschinenkonstruktion, 10623 Berlin, Germany

Karl-Heinz Kiittner, Professor Dipl.-Ing

Formerly at Technische Fachhochschule Berlin

Address for correspondence: Miillerstrasse 120,

M.J Shields, FIInfSc, MITI

Literary and Technical Language Services,

Unit 10, Centenary Business Centre,

Attleborough Fields Industrial Estate,

Nuneaton, Warwickshire CVII 6RY, UK

DOl 10.1007/978-1-4471-3566-1

British library Cataloguing in Publication Data

Dubbel: Handbook of Mechanical Engineering

I Beitz, Wolfgang II Kiittner,

Karl·Heinz III Shields, Michael J

621

library of Congress Cataloging·in·Publication Data

Dubbel, Heinrich, 1873·1947

[Taschenbuch fur den Maschinenbau English 1

Handbook of mechanical engineering / Dubbe! ; [edited by 1 W Beitz and K.·H Kiittner

p cm

Includes bibliographical references and index

1 Mechanical engineering-Handbooks, manuals, etc I Beitz, Wolfgang

II Kiittner, Karl·Heinz III Title

to the publishers

© Springer-Verlag London 1994

Originally published by Springer-Verlag London Limited in 1994

Softcover reprint of the hardcover 1 st edition 1994

The publisher makes no representation, express or implied, with regard to the accuracy of the infor· mation contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made

Typeset by Photo· graphics, Honiton, Devon

69/3830-543210 Printed on acid-free paper

Dr J N Ashton, University of Manchester Institute of Science and Technology

Dr N C Baines, Imperial College of Science, Technology and Medicine, London

Professor C B Besant, Imperial College of Science, Technology and

Medi-cine, London

Dr B Lengyel, Imperial College of Science, Technology and Medicine, London

D A Robb, Imperial College of Science, Technology and Medicine, London

Dr C Ruiz, University of Oxford

Professor J E E Sharpe, Lancaster University

Contributors

B Behr, Rheinisch-Westfalische Technische Hochschule Aachen

Professor W Beitz, Technische Universitat Berlin

Professor A Burr, Fachhochschule Heilbronn

E Dannenmann, Universitat Stuttgart

Professor L Dom, Technische Universitat Berlin

Dr K.A Ebertt, Hattersheim

Professor K Ehrlenspiel, Technische Universitat Munchen

Professor D Fenler, Batelle-Institut e.Y., Frankfurt a.M

Professor H Gelbe, Technische Universitat Berlin

Professor K.-H Habig, Bundesanstalt fur Materialforschung und-prufung

(BAt\1.) , Berlin

Professor G Harsch, Fachhoschschu1e Heilbronn

Dr K Herfurth, Verein Deutscher GieBereifachleute VDG, Dusseldorf

Dr H Kerie, Technische Universitat Braunschweig

Professor K.H Kloos, Technische Hochschu1e Darmstadt

Professor K.-H Kuttner, Technische Fachhochschule Berlin

J Ladwig, Universitat Stuttgart

G Mauer, Rheinisch-Westfalische Technische Hochschu1e Aachen Professor H Mertens, Technische Universitat Berlin

Professor H.W Miiller, Technische Hochschu1e Darmstadt

Professor R Nordmann, Universitat Kaiserslautern

Professor G Pahi, Technische Hochschule Darmstadt

Professor H Peeken, Rheinisch-Westfalische Technische Hochschu1e

Aachen

Professor G Pritschow, UniversWit Stuttgart

W Reuter, Rheinisch-Westfalische Technische Hochschu1e Aachen Professor R Roper, Universitat Dortmund

Professor G Rumpel, Technische Fachhochschu1e Berlin

Professor G Seliger, Technische Universitat Berlin

Professor K Siegert, Universitat Stuttgart

viii Contributors

Professor H.D Sond.ershausen, Technische Fachhochschule Berlin Professor G Spur, Technische Universitat Berlin

Professor K Stephan, Universitat Stuttgart

Professor H.K Tonshoff, Universitat Hannover

Professor H.-J.Warneeke, Universitat Stuttgart

Professor M Week, Rheinisch-WestfaIische Technische Hochschule Aachen

T Werle, Universitat Stuttgart

Professor H Winter, Technische Universitat MOnchen

H Wosle, Technische Universitat Braunchschweig

Preface to the English Edition

It has been an education and a pleasure to assist in the preparation of this first English version of the widely used "DUBBEL: Taschenbuch fUr den Maschinen- bau", which has been a standard mechanical engineering reference book in German-speaking countries since 1914

All the chapters of primary interest to English-speaking mechanical engineers have been translated I trust that this "Pocket Book" will be a ready and authoritat- ive source of the best current practice in mechanical engineering It is up to date, having been revised regularly, with the last revision appearing in 1990

It provides an easily accessible theoretical and practical treatment of a wide range of mechanical engineering topics with comprehensive explanatory dia- grams, tables, formulae and worked examples

Much care has been given to ensuring a correct and easily understood larion of the German text For completeness, it was felt necessary to retain many German references and also DIN Standards Where possible, ISO equivalents have been given It is unlikely that this complex exercise is entirely error free but I believe that faith has been kept with the original text

trans-B John Davies

Emeritus Professor, Department of Mechanical Engineering, UMIST

May 1994

Introduction

Since 1914 the Dubbel Handbook of Mechanical Engineering has been the dard reference text used by generations of students and practising engineers in the German-speaking countries The book covers all fundamental Mechanical Engineering subjects

stan-Contributions are written by leading experts in their fields This handbook is not primarily intended for specialists in particular areas, but for students and practitioners, who, within the framework of their responsibilities, also need to know about the basics outside their own special area

The handbook deliberately focuses on fundamentals and on the solutions of problems, but it also covers a wide range of applications Charts and tables with general material values and specific parameters are included As a German hand- book, it relies more on the German Industrial Standards (DIN) and focuses on the components of German manufacturers This should not be a problem in this English-international edition owing to the exemplary character of these appli- cations and examples; and with the increasing referencing of EN- and ISO/IEC- standards, the national DIN standard becomes less significant

In parallel with the complete German edition, the selected subjects in this edition combine the fundamentals of theoretical sciences, materials and engineer- ing design with important mechanical engineering applications

I would like to thank all those involved in the production of this handbook for their enthusiastic co-operation, since this has made an important standard mechanical engineering text available to an international readership

w Beitz

Technische Universitat Berlin

November 1993

Contents

A Mechanics

1 1 Introduction

1.2 Combination and Resolution of Concurrent Forces

1.3 Combination and Resolution of Non-Concurrent Forces

AI9

A19 A22

Fluids) ASI

1.1 Stress and Strain

1.2 Strength and Properties of Materials

1.3 Failure Criteria, Equivalent Stresses

2.1 Tension and Compression

2.2 Transverse Shear Stresses

2.3 Contact Stresses and Bearing Pressures

4.3 Arbitrarily Curved Surfaces

7.2 Lateral Buckling of Beams

7.3 Buckling of Plates and Shells

A69

A69 A69

A72

Bl

B1 B4 B6

B7

B7 B7 B8 B8 B27 B31 B32

B36

B36 B36 B37

B~8

B38 B38 B38

B39

B39 B41 B41

B43

B43 B43 B43

B45

B45 B48 B48

8.1 Finite Elements

1.1 Systems, Boundaries of Systems, Surroundings

Processes

2.1 Adiabatic and Diathermal Walls

2.2 Zeroth Law and Empirical Temperature

3.1 General Formulation

4.1 The Principle of Irreversibility

4.2 General Formulation

4.3 Special Formulations

5.1 Exergy of a Closed System

5.3 Exergy of an Open System

5.4 Exergy and Heat

5 '5 Exergy Losses

6.1 Thermal State Variables of Gases and Vapours

6.2 Caloric Properties of Gases and Vapours

6.4 Mixing Temperature, Measurement of Specitk Heat

Capacities

7.1 Changes of State of Gases and Vapours at Rest

7.2 Changes of State of Gases and Vapours in Motion

B59 B76

C3

C3 C4 C4 C5

c6

C6 C6 C7

CI4

C14 CI5

CI6

C16

xvi Contents

8.2 Internal Combustion Engines

8.3 Cyclic Processes

8.4 Cooling and Heating

C18 C19 C22

1.4 Effect of Materials Structure, Manufacturing Process and

2.1 Fundamentals

2.2 Test Methods

3.1 Iron Base Materials

D26

D26 D43 D49 D53

D54

D54 D54 D54 D55 D57 D57 D58 D58 D59

E Fundamentals of Engineering Design

1.1 Energy, Material and Signal Transformation

1.2 Functional Interrelationship

1.3 Working Interrelationships

1.4 Constmctional Interrelationship

1.5 System Interrelationship

1.6 General Objectives and Constraints

2.1 General Working Method

2.2 General Problem-Solving

2.3 Abstracting to Identify Functions

2.4 Search for Solution Principles

3.5 Types of Engineering Design

4.1 Basic Rule of Embodiment Design

4.2 Principles of Embodiment Design

4.3 Guidelines for Embodiment Design

Contents

xvii

D62 D66 D67

D67

D67 D67 D68 D70 D71 D72

D76 D121

El

E1 E1 E2 E4 E4 E4

E4

E4 E4 E4 E5 E6

EI0

EIO

Ell E12 E12 E12

EU

E13 E14 El6

Modular Design E20

5.1 Similarity Laws E20 5.2 Decimal-Geometric Series of Preferred Numbers (Renard

1.7 Selecting Types of Connection

2.1 Uses, Characteristics, Properties

2.2 Metal Springs

2.3 Rubber Springs and Anti-vibration Mountings

2.4 Fibre Composite Springs

2 <; Gas Springs

3.1 Survey, Functions

3.2 Permanent Torsionally StifI Couplings

3.3 Permanent Elastic Couplings

3.4 Clutches

3.5 Automatic Clutches

4.1 Fundamentals

4.2 Types of Rolling Bearings

4.3 Load Capacity, Fatigue Life, Service Life

4.4 Lubrication of Rolling Bearings

4.5 Friction and Heating

4.6 Design of Rolling Bearing Assemblies

5 Plain Bearings

Fl

Fl F18 F21 F23 F28 F34 F47

F50

F50 F51 F59 F62 F63

F64

F64 F65 F66 F69 F74

F75

F75 F77 F79 F84 F87 F87

F89

5.2 Calculation of Plain Journal Bearings Under Steady Radial

Load F89 5.3 Calculation of Plain Journal Bearings Under Variable Radial

5.4 Turbulent Film Flow F92

5.5 Calculation of Plain Thrust Bearings

5.6 Form Design of Plain Bearings

5.7 Lobed and Multi-pad Plain Bearings

8.1 Spur and Helical Gears - Gear Tooth Geometry

8.2 Tooth Errors and Tolerances, Backlash

8.3 Lubrication and Cooling

8.4 Materials and Heat Treatment - Gear Manufacture

8.5 Load Capacity of Spur and Helical Gears

8.6 Bevel Gears

8.7 Crossed Helical Gears

8.8 Worm Gears

8.9 Epicyclic Gear Arrangements

8.10 Design of Geared Transmissions

10.3 Components of Crank Mechanism

11 Appendix F: Diagrams and Tables

12 References

Contents

G Hydraulic and Pneumatic Power Transmission

Systems

xix

F92 F96 F98 F98 F99 F99 F99

F101

FlOl FlOl F107 Fl09 F109

FllO

FllO FIll Fl12 F1l5

F1l6

F1l7 F124 F125 F126 F128 F137 Fl39 Fl39 Fl43 F153

F157

F157 F16l F166 F168

F168

F168 Fl70 Fl72

F176

F194

G1

4.1 Hydraulic Circuit Arrangements

4.2 Design of Hydraulic Circuits

5.1 Pneumatic Components

5.2 Circuits

H Components of Thermal Apparatus

1 Fundamentals

1.1 Heat Exchanger Characteristics

1.2 Thermodynamic and Fluid Dynamic Design

1.3 Heat Exchanger Flow Arrangements and Operating

Characteristics

1.4 Efficiency, Exergy Losses

2.1 Basis for Design Calculations

2.2 Cylindrical Shells and Tubes Under Internal Pressure

2.3 Cylindrical Shells Under External Pressure

2.4 Flat End Closures and Tube Plates

G15

GIS GIS

G16

G16 G17

G17 G18

H5

H5 H6 H6 H7 H7

HS

HS HlO H14 HIS

H21

2.4 Representation of Vibrations in the Time and Frequency

3.1 Basic Concepts

3.2 The Generation of Machine Noise

3.3 Methods for Reducing Machine Noise

Manufacturing Processes

J36

Kl

Kl K1 K2 K2 K3 K15

xx i i Contents

L

2.4 Forming of Metals and Ceramics by Powder Metallurgy

2.5 Other Methods of Primary Shaping

K17 K19

K96

K96

KlOO

K103 K105 K107

KI09

K118

L1 L1 L4 L21

125

L34

L34 L37 L41

2.4 Numerical Control (NC)

2.5 Equipment for Position Measurement at NC Machines

3.1 Shearing Machines

3.2 Blanking Machines

3.3 Nibbling Machines

3.4 Beam Cutting Machines

4.1 Characteristics of Presses and Hammers

6.1 Arc Welding Machines

6.2 Resistance Welding Machines

6.3 Soldering and Brazing Equipment

L52

L52 L53 L53 L54

L54

L54 L55 L60 L61 L65

L66

L66 L73 L79 L83 L83 L85 L86 L87 L89 L92 L94 L97

L97

L97 L99

LlOO

LI00

LlOO LlOI Ll02 Ll02 Ll02 Ll04 Ll07

LlO8

Index 1

Mechanics

G Rumpel and H D Sondershausen, Berlin

1.1 Introduction

Statics is the study of the equilibrium of solid bodies or

of systems of solid bodies Equilibrium prevails if a body

is at rest or is in uniform motion in a straight line Rigid

bodies as understood in statics are bodies of which the

deformations are so small that the points at which f()rce

is applied undergo negligible displacement

Forces These are vectors of varying direction and

dis-placeable in their lines of action, which cause changes in

the motions or shapes of bodies The determinant factors

of forces are magnitude direction and location (Fig Ia)

F = F, + Fy + F, = F,ex + Fye, + F,e,

= (F cos OI)ex + (F cos {3)ey + (F cos y)e, ' (1)

where

For the cosines of direction, cos 0: = Fxf F,

cos {3 = IV P, cos Y = F,I 1' and cos' 01 + cos' {3 +

cos' y = I

There are applied forces and reaction t<lrces as well as

external and internal t(lrces External forces are all the

for-ces that exert an effect on a body capable of free motion

from the outside (see AI.S) (loads and supporting

forces) Intenlal forces are all the cutting forces and

bind-ing forces occurrbind-ing inside a system

of a vector which is perpendicular to the plane of effect

In this situation r, F, M form a right helix (a right-handed system) Couples can be displaced at random in their planes of effect and perpendicular to it; in other words, the moment vector is a free vector, determined by the vector product

M = r X F = M, + My + M, = Mxex + M,ey + M,e,

= (M cos OI')e, + (M cos /3')ey + (M cos y)e, (3)

M = i~ = irl·IE1· sin cp = Fh = ,M~ + M; + M;(4)

M signifies the magnitude or amount of the moment, and provides a graphic representation of the area of the paral-

lelogram formed by rand F In this situation, h is the

moment ann perpendicular to F For the direction cosines

(Fig Ic), the following applies:

cos a' = M,/M cos /3' = M,/M cos Y = MjM

Moment of a Force Relative to a Point (Moment of Displacement) The effect of an individual force with a random point of contact in relation to a point 0 becomes clear if a null vector is added; i.e two mutually opposed forces of equal value, F and -F, at the point 0 (Fig 2a)

A single force F is derived at the point 0, as well as a pair of forces or moment M (displacement moment) the vector of which is perpendicular to the plane formed by

Mechanics • I Statics of Rigid Bodies

rand F If rand F are given in components x, y, z or FX'

F y , F, (Fig 2b), then the following applies:

The following applies to the components, the value of

the moment vector, and the direction cosines:

M, = Fyx - Fy:

M = IMi = Irl' F· sin cp = Ph = VM~ + M: + M;:

cos 1'* = MjM

If the force vector is located in the x, y plane, i.e if z

and P, are equal to zero, then it follows that (Fig 2c):

M = IMI = M, = Fyx _ FxY = Fr sin cp = Ph

Projection of a moment vector onto a given axis

(direction): if ijI is the angle between the vector and the

axis, and e, is the unit vector of the axis, then from the

scalar product we derive: M, = Me, = M cos ijI

1.2 Combination and Resolution of

Concurrent Forces

1.2.1 Systems of Coplanar Forces

CombinatioR of Forces to ORe Resultant

Force Forces are added together geometrically

(vectolially), for two forces with the parallelogram or

tri-angle of forces (Fig 3), and for several forcts with the

polygon of forcts (Fig 4: t')fce scale I em = K N)

The calculated solution is

(6)

where f~x = f~ cos (Xi' f~y = ~ sin Q'j Size and direction

of the resultant force are given by

a

y

b

Figure 3 Combination oJ two Figure 4 Combination of

sev-plane forces: a paraJlelognun of eral sev-plane forces: a location

(7)

ResolutioR of a Force in the Plane 'fbis is unique only in two directions, and the solution is multi-valued in three and more directions (statically indeterminate) For

a graphical solution, see Fig Sa, b

Calculated Solution (Fig Sc) F

components:

F sin a = F1 sin ill + FJ sin 0: 2 ; i.e l'~ = (F sin a - F, sin all/sin a, and therefore Feos a sin (}:2 - Fsin a cos ct 1

In other words, F, = P sin (a, - a)/sin (a, - al) and accordingly F, = F sin (a, - a)/sin (a, - a,) These for-mulae correspond to the sine law for a scalene triangle

(Fig Ga) In the x, y-plane (horizontal plane), vector addition gives the projection F~ of the resultant force FR

in the y,z-plane (vertical projection) F~ (Fig 6b) The projection of the F; in the z-direction produces the true component FR, of the resultant force FR, is combined with F~ in the extended plane of both components to form the resultant force FR (Fig Gc)

Calculated Solution This is

1.2 Combination and Resolution of Concurrent Forces 1.2.2 Forces in Space

rRl

I

I

I F!.'

b

F{'

Figure 6 Combination of forces in space: a location plan, b

poly-gon of forces, c overall resultant force

where F;x = 1'; cos iy;, F;y = F; cos 13;, r;, = 1'; cos y; The

size and direction of the resultant force are

cos "R = FRJF, cos f3R = FRY/!' cos YR = FR,/!'.(9)

The Resolution of a Force in a plane is uniquelv

poss-ible only in three directions; in four and more directions

the solution is multi-valued (statically indeterminate) In

the graphical solution (Fig 7a, b), the point Ii at which

force F (line of action 0) intersects the abscissa is initially

determined in the vertical and horizontal projection, and

therefore the plane A' D' £' extended from 0 and 1 is

deter-mined in the horizontal projection It meets the line of

intersection C' of plane B' (7' D' formed by 2 and 3 in the

horizontal projection at the projected intersection point

of line G' The projection of (;' into the vertical projection

provides G" and therefore efl Force P is now resolved in

the horizontal and vertical projections in directions J and

c, which are located in two different planes, namely ABD

and BCD F( is known as Culmann's auxiliary force, which

is subsequently resolved in directions 2 and 3 in the

hori-zontal and vertical projections The final value of PI is

obtained by combining the component F", obtained from

the horizontal projection, with the force F; obtained from

the horizontal projection in the plane extended from both

these forces The same applies correspondingly to F, and

As in Fig 8, the following applies to the direction cosines

of the three directions given:

cos 0', = xj ,x: +yi + zi,

cos 13, = yj Ixi + yi + zi,

Figure 7 Resolution of a force in space: a location plan, b gon of forces, c final forces

poly-cos Y; = zj yX~ + yi + zi·

It follows that

F, cos 13, + 1"2 cns 132 + 1", cos 13, = I' cos 13,

FJ cos YI +-FJ cos 1'2 + Pi cos y" :::: F cos "I

These three linear equations for the three unknown ces 1'" 1', and 1', have a unique solution only if their sys-tem determinant does not equal zero, i.e if the three direction vectors are not in one plane According to

for-Fig 8, FIe! + F2e 2 + F~e , = F, and, after multiplication

bye2 X e~,

1",e, (e, X e,) + 1",e, (e 2 X e,) + F,e,(e, X e,)

= F(e, X e,)

Because the vector (e, X e,) is perpendicular to both

e 2 as well as to e", the scale products are zero, and it

follows that

F,e, (e, X e,) = F(e 2 X e,)

and

(10)

1', = e,Fe,/ (e,e 2 e,), F, = e,e,F/ (e,eze,)

Fe,e" e,e,e, etc are parallelipipedal products, i.e scalar quantities, whose values determine the spatial volume of the parallelipiped formed by the three vectors The sol-ution is single-value if the parallelipiped product is e,e,e,

" 0, i.e the three vectors may not occur in one plane With e i = cos (tjex +-cos (3jey +-cos "tiez it follows that

IF cos ex cos 0'2 cos a:;,1

F, = F cos 13 l'(>S 13, cos 13.,

F cos Y cos Y2 cos Y-I

corresponding to F2 and F.:;,

I cos 0', : cos 13,

Mechanics • 1 Statics of Rigid Bodies

z

F

XI y

Figure 8e Calculated resolution of a force in space

1.3 Combination and Resolution of

Non-Concurrent Forces

1.3.1 Coplanar Forces

Combination of Several" Forces to Form One

Resultant Force

Grapblcal Process wltb Polygon of Forces and Funicular

Polygon The forces are added geometricaUy to the

result-ant force in the polygon of forces (Fig 9), a random

pole P is selected, and the radius vectors 1 to n are drawn

The paraUeis to these are transferred to the location plan

(Fig 9&) as rays of the funicular polygon l' to n', such

that the forces of a triangle of forces of the polar solid

angle intersects it in the ground plan at a point

(point-triangle rule) The point of intersection ofthe first and last

rays of the funicular polygon provides the contact point of

the resultant force, the value and direction of which are

derived from the polygon of forces

Calculation Procedure By reference to the zero point,

the plane group of forces provides a resultant force and

a resultant (displacement) moment (Fig lOa):

For a random point, the effect of the group of forces is

the same as that of their resultant force If the resultant

force is displaced parallel from the zero point so far that

M becomes zero, it fOllows for its position that MR = FRb

etc (Fig lOb)

bR = MRfFR or X R = MRfFRY or

y = -M.fFRx·

Resolution of a Force The resolution of force in a

plane is possible as a single value, in three given directions

that do not intersect at a point, and of which a maximum

of two may be paraUe! A force is graphicaUy resolved with

the aid of Culmann's auxiliary vectors (Fig 11a, b) In

addition to this, the force F is made to intersect one of

the three lines of application, and the other two lines of

application are made to intersect each other The line

join-ing the points of intersection A and B is the s<>-called

Cul-mann's auxiliary vector c After resolution of the force F

in the polygon of forces in the directions 3 and c, F., and

Fc are derived The force Fc is then again resolved in

direc-tions 1 and 2, giving F, and F z

a

c Figure 11 Resolution of a plane force: a location plan with Cul~

mann lines c, b polygon of forces, c calculated solution

The calculated solution follows from the condition that the application of force and moment of the individual for-ces F; and the force F must be equal in relation to the zero point (Fig 11c):

2: F; = F, 2: (ri X F;) = r X F,

PI COS C1, + F2 cos (X2 + F3 cos {1:\ = F cos a,

1', sin (x, + 1'2 sin l:r z + F3 sin C1 j ::::: F sin a;

F, (x, sin '" - y, cos "') + F2 (x 2 sin "2 - Y2 cos "2)

+ F,(x, sin ", - y, cos ",) = F(x sin" - y cos ,,);

or, instead of the last equation, F,b, + F 2 bz + F,b, = Fh,

where antidockwise moments are positive There are three equations for the three unknowns F" Fz, F, Their denominator determinants may not be zero; i.e the con-

1.4 Conditions of Equilibrium 1.4.1 System of Forces in Space

r,

Y2

Figure 12 Reduction of forces in space: a location plan, b force and moment resultant (: force and moment components

ditions specified for the graphical solution regarding the

position of the lines of action must be fulfilled, if it is

intended that the solution should be unique

1.3.2 Forces in Space

Force Combination (Reduction) A group of forces

in space, Fi = (Fix: Fi,; Fi')' the contact points of which

are given by the radius vectors r, = (Xi: Yi: Z,), can be

combined (reduced) in relation to a random point to give

a resultant force FR and a resultant moment MR' The

com-plex graphical solution is acquired in the projection

planes [I] The calculated solution (Fig 12), in relation

to the zero point, is

Parallelogram of Forces A further simplification of

the reduced system of forces is possible if there is an axis

that has a specific position in which the force vector and

the moment vector are parallel (Fig 13) This axis is

called the central axis It is derived by resolving MR in the

plane E formed by MR and FR into tbe components

MI' = M" cos cp (parallel to F R) and M, = M" sin cp

(perpendicular to FR ) Here, cp follows from the scalar

product MR' FR = ,tfHP R cos cpo i.e cos cp = MR FR

(MRFR) M, is then made equal to zero by displacing FR

perpendicular to the plane E by the amount a = MJF R

The vector that pertains to this is = (F R X M")/F~,

since its value is 1"1 = a = Fll.il1R sin !pI r~ = flrL) FR' The

vector equation for the central axis, in the direction of

which FR and Me take effect, then reads, with t as

para-meter:

r(t) = + FR t

Force Resolution in Space A force can be resolved

in space as a single value in six given directions If the directions are given by their direction cosines, and if the forces are designated F, ' F o, then

"

L F, cos <l<i = F cos <l<, L Fi cos f3i = F cos f3,

L Fi(Yi cos Yi - Z, cos f3,) = F(y cos Y - Z cos f3),

L F,(Zi cos (Xi - Xi cos y,) = F(z cos a - X cos Y),

L F,(Xi cos f3i - Yi cos ail = F(x cos f3 - y cos a)

From these six linear equations a unique solution may be derived, if the denominator determinant is not equal to zero

1.4 Conditions of Equilibrium

A body is in equilibrium if it is at rest or in uniform motion Since all acceleration values are zero, it follows from the basic principles of dynamics that no resultant force and no resultant moment are exerted on the body

1.4.1 System of Forces in Space The conditions of equilibrium are

FR = 'i.F, = 0 and M = lMi = 0:

for-Six unknown values (forces or moments) can be lated from the six conditions of equilibrium If more than six unknowns exist, the problem is described as statically indeterminate Its solution is possible only by invoking deformation values (see B2.7) If forces with common

calcu-Mechanics 1 Statics of Rigid Bodies

Figure 14 Graphical conditions of equilibrium

points of contact are present, then the moment

con-ditions of Eq (13) are fuIftlled with respect to the point

of intersection (and therefore also for all other points,

since MR is a free vector) Only the equilibrium conditions

of forces of Eq (13) then apply, from which three

unknown forces can be determined For the graphical

sol-ution, in this case, the polygon of force in space must be

closed due to FR = LF, = 0 (performed in horizontal and

vertical projection in accordance with Fig 7)

1.4.2 System of Coplanar Forces

The equation system (13) is reduced to three conditions

of equilibrium:

Both force equilibrium conditions can be replaced by

two further moment conditions The three points of

refer-ence for the moment equations lnust not lie on a straight

line From the three conditions of equilibrium of the

plane, three unknown values (forces or moments) can be

determined If more unknowns are present, the problent

is statically indeterminate

The graphical solution for equilibrium in the plane

fol-lows from the principle that polygons of force and

funicu-lar polygons must close (Fig 14) If the polygon offorce

closes, but the funicular polygon does not, then no

equi-librium prevails; a couple of forces remains (see Fig 14,

force F; and couple of forces, consisting of funicular poly~

gons 1 and 5') Special cases: Two forces are in

equilib-rium if they have the same lines of application are of

equal magnitude, and are in opposing directions Three

forces must intersect at a point, and the polygon of force

must close In the case of four forces, the polygon of force

must close, and the resultant force of each pair must lie

on the same line of action, be of equal magnitude, and be

in opposing directioos (Fig IS)

Forces with a Common Point of Contact in the

Plane For these, the moment condition in Eq (14) is

05 F,

F,

Figure IS Equilibrium of four Figure 16 Equilibrium of

for-plane forces ces with common points of

con-fuIftlled in an identical manner, and there remain only the two force conditions

(15)

The graphical solution follows from the vector equation

FR = 0, i.e the polygon of force must be closed (Fig 16)

1.4.3 Principle of Virtual Work

The principle replaces the equilibrium conditions, and states: If a rigid body is subjected to a minor (virtual) displacement that is compatible with its geometrical con-straints, and if the body is in eqUilibrium (Fig 17), then the total virtual work of all the external forces and moments affected - identified by superscript (e) - are equal to zero:

8W"" = LFf"' 8 , + LMf e ' 1i'P, = 0; (16)

or, in components,

8wCe) = L(Ff;' lix, + Pi;' liy, + Ff:' 8z,)

+ L(Mf2 1i'P" + Mf;' 1i'P,y + Mf:' 8'1',,) = 0; here ", = (x,; y,; z,) radius vectors to the points of contact

of the forces; 8 , = (Ox,; liy,; 8z,) variations (vector entials expressed mathematicalIy) of the radius vectors, which result from the formation of the first derivative; 1i'P I

differ-= angle of rotation differentials of the torsional motions

'P,

In natural coordinates, the principle takes the form

8 W Ce' = "iF f:' lis, + LM f:' 1i'P, = 0, (17)

where F i:) are the components of force in the direction

of the displacement, and M/~) are the components of the moments taking effect about the axis of rotation The prin-ciple is used, among other things, in statics to examine the equilibrium of displaceable systems and to calculate the effect of migrating loads on intersection and contact forces (lines of application)

Example Drafting Machine (Fig 18) A counterweight Po

and its lever arm I are to be determined in such a way that the dnlfting machine will be in eqUilibrium from its own weight r~ in any position

The system has two different degrees of freedom:

rC, = (-C sin 'P+ b sin 0/; bcoso/-C cos 'P),

rQ = (I sin 'P - a sin 0/: -a cos 0/ + I cos 'P), or" ~ (-c cos 'I' &'1' + b cos tjJ 01/1: -b sin tjJ &tjJ + c sin 'I' &rp),

&rv ~ (l cos 'I' &'1' -a cos tjJ &tjJ: a sin tjJ &tjJ - I sin 'I' 0'1')

With Fe; = (0; -l'~,) and P Q = (0; -1'Q)' it follows that

&W'eJ ~ 2Fi") or, ~ -F" (-b sin tjJ &tjJ + c sin rp &rp)

-F Q (a sin tjJ &tjJ - I sin rp &'1')

~ sin tjJ otjJ (F"b - FQa) + sin rp &'1' (-F,;c + FQI)

y

Figure 17 PrinCiple of virtual Figure 18 Dr.lfting machine

From aW le ) = 0, due to the random nature of r.p and t/J

F(,b - Fva = 0 and -F(,c + F(/ = 0

and therefore

fo = F(;b/a and 1= c "~JFq = calb

In addition,

fjlW(<') = cos t/J &1/1 (I\;b - Ftp) + cos r.p btpl (-F(,c + F(/)

It follows from this, wirh the solution values calculated that

?)2W(") = n, Le neutral equilibrium prevails (sec AlA 4)

ViA Types of Equilibrium

A distinction is drawn between stable unstable and

neu-tral equilibria (see Fig 19) Stable equilibrium prevails

when a body reverts to its initial position under a

displace-ment that is compatihle 'with its geometrical constraints;

unstable equilibrium wben tbe body seeks to leave its

initial position; neutral equilibrium prevails when any

adjacent position is a new position of equilihrium If, in

accordance with AIA.:\ the minor displacement is per·

ceived as virtual then, in accordance with the principk

of virtual work, 0 If.' ".) ~ 0 applies at the equilibrium

pos-ition If the hody is moved as in Fig 19a from a position

1 to a position 2, through the position of equilibrium O

then the work is 0 W u" ~ F, os > 0, in the range' to 0:

i.e positive, and in the range () to 2 o">-(/') < 0, i.e

nega-tive It follows from the function oW"" ~J(s) that the

increase of oU'7«(') js negative i.e ~)2W'(") < O if

equilib-rium is stable The general application fi)r tl1<" equilibequilib-rium

is: stable 8 l w<e) < O unstable 81 {f'U" > 0, neutral

If the problems involved are such that only weight

forces are acting, then with the potential U ~ F"z

or 0 U ~ F"oz the following applies:

oW'''' ~ F'e'or ,= (0:0: -1',,) (ox: 0,.: oz)

~ - f~l)z ~-'6U

and therefore the potential energy [1 is a minimum, while

in unstable equilihrium f/-(T < 0, and the potential energy

is a maximum

1.4.5 Stability

Bodies having supports t.hat can only absorb compressive

forces are in danger of overturning This is prevented if

1.6 Support Reactions 1.6.1 Plane Problems

the sum of the righting moments around the potential

tilt-ing edges A or B (Fig 20) is greater than the sum of the tilting moments, i.e if the resultant of the system of forces within the tilting edges intersects the standing surface Stability is the ratio of the total of all righting moments to

the total of all tilting moments hy reference to a tilting edge: S ~ LMsfLM, For S ?': l stability and equilihrium prevail

1.5 Types of Support; the 'Free Body'

Bodies are supported by hearings The supporting forces

function as reactive forces to the forces imposed on the body from outside Depending on the mode of constmc-tion of the bearings a maximum of three forces and a maximum of three moments can be transferred in space The reactive forces and fllotllents become external forces owing 10 the so-called 'freeing' of a body A body is freed

by being separated from its surrounding~ due to a dosed section through all the bearings, and hy all the bearing forces being applied as external forces (Fig 21, freeing

principle) Equal and opposite forces of action and tion (Newton's third law) then take effect on the bear-ings A distinction is dra.wn between a bearing with J sin-gle value and one with lip to six vailles deperitling on the bearing's stnlCturaJ nature and the number of the reaction values (Fig 22)

inde-tenninate in its bearings If the bearings are less than three-valued the system is statically underdetemlinated, i.e unstable and 1110vahle The calculation of the contact reactions is carried out by freeing the body and determin-ing the conditions of equilibrium

Figure 20 Stabilit) Figure 21 Freeing

principle: a ~upported

body with closed line of intersection, b free

om Mechanics 1 Statics of Rigid Bodies

Figure :J:J Types of bearings

Example Shaft (Fig l'a) The contact forces in A and B are

sought in terms of the forces P, and F}._

Calculated Solution The following applies to the freed shaft

(Fig.l'b):

IM" ~ 0 ~ -F,a + F,,! - 1',(1 + e), i.e

F ~ tF,a + F,(I + e)J/I;

IMiB = 0 = -FA,) + Fib - Fl.c, Le

F"y = (Fib - Flc)/l;

IF;, ~ 0 ~ FAx'

The condition of equilibrium F iy = 0 must likewise be fulfilled, and

can be used as a control equation

a

IF;y = FA)" - F, + Fa - F2

~ (F,b - F,c)/I - F, + [p,a + F,(I + e)lIl- F,

~ F,(a + b - f)/I + F,( -c + 1+ c - f)/I ~ O

c Fipre l, Shaft: a system, b freeing, c graphical solution

Graphical Solution (Fig.l'c) The polygon offorce and the polar solid angle, together with the funicular polygon pertaining to them,

are shown by the forces' and F'l.' The points of intersection of the two outcr radius vectors l' and 3' with the known lines of

action of both contact forces (since in this case FAX = 0) provides the clOSing side s', which 'closes' the radius vector Parallel transfer

to the polar solid angle provides the two contact forces FA and F

while maintaining the point-ttiangle rule (see AI.3.1) Both the funicular polygon and polygon of forces are then closed; i.e equilib- rium exists between the forces #\, F 2 , F B, FA'

Example Angled Carrier Element (Fig 14a) The angled

car-rier element is loaded by the two point forces'l and F2 and the

constant distributed load q The applied force in the fixed hearing

A and the force in the pendulum rod at B are to be determined

Calculated Solution With the resultant of the distributed load

F, ~ qc (Fig 14b):

IMiA = 0 = - FI sin u a - qc(a + b + e12)

- F~e + Fs cos aJ + F., sin ash,

and from this

From

IF l" = 0 = FA" + FI cos a l + F2 - Fs sin Us and

IF iy = 0 = FAy - F sin a - qc + Fs cos as

it follows that

in which the calculated value for Fs is to be used

Graphical Solution (Fig 14<:) After drawing the polygon offorce

from the given forces FI> FJ.> and Fq and the polar solid angle, the

funicular polygon pertaining to them is drawn, for which the first radius vector l' must be laid through the fixed bearing A; this is due to the fact that it is the only known point on the line of action

of FA' The point of intersection of the last vector radius 4' with the

(known) line of action of P s provides the closing side s', which

closes the funicular polygon The transfer of this to the polar solid angle, maintaining the point-triangle rule (see A1.3.1), provides first F and then, by closing the polygon of force, the force FA' Example Wagon on an Inclined Plane (Fig lSa) The wagon, loaded by the force of gravity FG and the trailer tractive force Pz, is held in equilibrium on the inclined plane by a cable

c Fipre 14 Angled bearing element: a system, b freeing, c graphi-

b

Fn1

Figure 25 Wagon on an inclined plane: a sy~tem, b freeing,

c graphical solution

winch The tensile force in the retaining cable and the supporting

force on the wheels are to be calculated, neglecting friction forces

Calculated Solution On the freed wagon (Fig 25b), equilibrium

conditions prevail:

":iF,,, = () = f'z F(, sin a -+- l'~ cos fl,

- P s (h!2) cos a F~«(J -+-2h) sin a;

~MiA = 0 = Fzh/4 - 2F"lh -+- f~_;(hI2) sin a -+- f~;b cos a

- F~(hI2) cos a - "~a sin a

from which follows

F", ~ - F z h/(8b) - Fd(h/2) sin a - b cos a]/(2b)

+ Fd(h/2) cos a + (a + 2b) sin a]/(2h) and

F", ~ Fzh/(8b) i-F,,](h/2) sin a + b cos a]/(2b)

- Fsl(h/2) cos a T a sin aj/(2b)

where the calculated value of F~ is to be used The condition

IF,v = 0 = Fnl -+- F"l - F(, cos a - Fs sin a can then be used as a

check equation

Graphical Solution (Fig 2.Sc) The imposed forces, F(i and F z

are combined to form the resultant imposed force F II , the position

of which is given by the poim of intersection of the lines of action

Fnl requires that the resultant of every two forces (e.g FR and Fnl

intersection of the lines of action of FI{ and Fnl or of Fnl and Fs

produces the Culmann's auxiliary vector c on which the two result~

ants must lie as opposing forces FII can be combined in the polygon

of forces with Fnl to F" and the opposed force -F, is then resolved

into Fnl and Fs

Example Carrier Element Under Vatiable Load: Line of Action

(Fig 2.6a) The contact force F" for a force F in a random load

a

~x)

1~

Figure 2.6 Load-bearing Figure 2.7 Plate supported in

element with variable load: space by six rods

1.6 Suppon Reactions 1.6.2 Body in Space

setting x derives from lM'1l = 0 = -F,J + F(l- x) as FA = F(I- x)/I ~ F'Y/(x) For F = 1, it follows that FA ~ 'l(xl ~

(/ - x)/I This function is a straight line (Fig 2.6b), the ordinates

of which represent the effect of the varying load F = 1 on the tact force FA' If, for example, a force FI = 300 N acts at the point

con-XI = 31/4, then FA = Fl 11(X l ) = 300 N (1/4) = 75 N For several single forces F; at the points Xi> it follows that FA = "2F i 11(X.) With

,

a distributed load q(x) in the range a ::s x::S b, FI\ = f Q(X)11(X)

dx It follows that, for a constant distributed load qo on the whole carrier element length,

dis-valued (n > 6), then the system bearing is (n - 6) times statically indeterminate If n < 6, it is statically underdeter-minate, in other words movable and unstable

Example A Plate Supported in SPace by Six Rods (Fig

27) The axial forces F t to F6 are to be determined by calculation

The conditions of equilibrium are presented in the form of force or moment equations as far as possible in such a way that only one unknown is contained

"2.F iy = 0 gives F~ "'" f~/cos a;

lM", ~ 0 gives F, ~ (F, - F,l/(2 cos a);

IF,, ~ 0 gives P, ~ (Fx + F,l/(2 cos a);

Example A Shaft with Helical Teeth (Fig 2.8), The contact

forces of the shaft are to be calculated The shaft can rotate about the x-axis, i.e lM,x = 0 does not apply The remaining five con- ditions of equilibrium are:

lFix = 0 gives FAx =: FIX - F 2 ,.;

IMmf = 0 gives FI\.} = -(Flxr! + Flyb + F1xr Z + Flyc)/l;

IMiBy = 0 gives F Az =: (F1zb + Flzc)/l;

IMl1" , = 0 gives F By =: [F]xr] - Flra + Flxrz + F 2y (l + c))!l;

lM", ~ 0 gives F., ~ 1F"a + F,,(/ + el]/1

The conditions IF,v = 0 and lFiz =: 0 can be used as a check

z

Mechanics • 1 Statics of Rigid Bodies

Figure :19 System of rigid bodies

1.7 Systems of Rigid Bodies

These consist of several bodies connected by elements,

such as (a) joints or (b) guides, or (c) guides with jointed

connections (Fig :19) A joint transfers forces in two

directions, but transfers no moment; a guide transfers one

force transverse to the guide and a moment, but transfers

no parallel force; a jointed guide transfers one force

trans-verse to the guide, but transfers no parallel force and no

moment Thus, we speak of two-value or single-value

con-necting elements If i is the sum of the values of the

sup-ports, and j the sum of the values of the connecting

elements, then in a system consisting of k bodies with 3k

conditions of plane equilibrium, the condition j + j = 3k

is fulfilled if it is intended that a stable system should be

statically detertninate

i.e if i + j = 3k + n, it is n times statically indetertninate

If i + j < 3k, the system is statically underdetertnined, and

is in any event unstable For the stable system in Fig :19,

j + j = 7 + 5 = 12 and 3k = 3·4 = 12, i.e the system is

statically detertninate In statically detertninate systems

the support reactions and reactions in the connecting

elements are ascertained, inasmuch as the conditions of

equilibrium for the freed individual bodies are fulfilled

Example 1bree10inted Frame or 1bree10lnted Arcb (Fig

~a)

Calculated Solution After freeing the two individual bodies (Fig

3Gb), conditions of equilibrium for body I are:

(l8a) (l8b)

and those for body II are:

IFly = 0 gives Fay = Fcy + fiy;

(l8d) (l8e)

+ F" (y, - (H - b)] + F" (I - x,) (l8f)

From Eqs (18c) and (l8f) are derived the pin forces Fe, and Fe"

used in Eqs (l8a), (ISb), (18d) and (l8e), and then the support

forces PM', FA'! F a,,-, Fay IM'L = 0 is used as a check for the

whole system

Graphical Solution (Fig~ 3Oc) The resultants FRI and 'R2 of the

imposed forces are formed for each body, and their effect on each

other are examined FRI must be in equilibrium with the forces FAI

and '81 in the bearings A and B The line of action of FBI must in

this instance pass through the points B and C, since moment

equilib-rium must prevail for random points in ihe body II, initially still to

be regarded as load-free However, the line of action of 'AI must also pass through the point of intersection D of this line of action

wiih F." if it is intended ihat equilibrium should prevail between

the three forces 'Rio 'AI and '01 (see A1.4.2) The values of 'AI and '01 are derived from the polygon of forces which is now ascer-tained From a similar construction for'R2 (where the polygon of forces'R2 usefully applies to 'RI), the forces 'A2 and '82 then fol-low The vectoral addition of 'AI and 'A2 gives 'A' and that of '01 and '02 gives 's Finally, the equilibrium condition in the polygon

of forces, 'I + '2 + F3 + '0 + 'A = 0, is fulfilled as reqUired

1.8 Pin·Jointed Frames 1.8.1 Plane Frames Pin-jointed frames consist of rods connected by freely rota-ting joints at nodal points The joints are assumed to be free of friction; i.e only forces in the direction of the rods are transferred The friction torque that actually exists at the nodal points, and the deflection-resistant connections, lead to secondary stresses, which as a rule are negligible The external forces act on the nodal points, or, in accord-ance with the lever ptinciple on the rod, are distributed over these points

If a pin-jointed frame has n nodes and s rods, and if it is externally statically determinate, beating on three support forces, then 2n = s + 3, s = 2n - 3 applies to a statically detertninate and stable pin-jointed frame (Fig ~la),

since two conditions of eqUilibrium exist for each node; i.e., of the 2n - 3 conditions of equilibrium, s unknown axial forces can be calculated A pin-jointed frame with

2 < 2n - 3 rods is statically underdetenninate and matically unstable (Fig ~ Ib), and a pin-jointed frame with s > 2n - 3 is internally statically indetertninate (Fig

stati-cally determinate and stable pin-jointed frames: Starting from a stable basic triangle, new nodal points are connected one after another by two rods, Figs ~la,

~:1a

A new pin-jointed frame is formed from two statically detertninate pin-jointed frames by means of three con-necting rods, the lines of action of which have no com-mon point of intersection (Fig ~:1b) Two rods may

in this situation be replaced by a node common to both frames (Fig ~:1b, right)

Any frame formed in accordance with these rules can be transformed into another statically detertninate and stable frame by transposing rods, provided that the transposed rod is fitted between two points that are capable of moving by reason of their distance from one another (Fig ~:ac)

Fa

c Figure 30 Three-jOinted frame: a system, b freeing c graphical solution

Figure 31 Framework: a statically determinate, b statically

underdeterminate, c statically indeterminate

New stable pin-frame systems can be formed from several

stable frames in accordance with the rules of rigid body

systems as per A1.7 (Fig ~2d)

Figure 32 Framework: a to d law.'; uf formation I to 4

Determining Axial Forces

Node intersection Procedure In general the axial forces

sand th<: three supporting forces for a statically

determi-nate pin-jointed frame are derived by setting up

equilib-rium conditions "iF,x ~ 0 and "iF" ~ () at all the n nodes

freed by circular cuts 2n linear equations are derived If

the denominator determinant of the equation system does

not equal zero, the frame is stable; if it equals zero, the

frame is unstable (displaceable) [I J It frequently occurs

(e.g after the support forces have been ascertained

beforehand from the conditions of equilibrium of the

entire system) that an initial node exists with only two

unknown axial forces, and further nodes connect to the

initial node with only two unknown axial forces in each

case, so that they can be calculated one after another from

the equilibrium conditions without an equation system

needing to be solved In the graphical solution, this leads

to what is known as the Cremona force diagram If there

are only two unknowns progressing from node to node,

they are graphically determined from the closed polygon

of forces The arrangement of the polygons of force next

to one another produces the Cremona t,)rce diagram,

where all the nodes follow one another in the same

direc-tion of rotadirec-tion Axial forces that impose on one node in

this situation provide a closed polygon of force in the

force diagram (see example)

Ritter's Method of Sections An analytical procedure in

which an entire pin-jointed frame is freed by cutting three

rods, and, after application of the three conditions of

equi-librium for this part, the three unknown axial forces are

calculated (see example)

Henneberg's Rod Transposition Process

Complex-stmc-tured frames can be referred back to simple stmctures

by the rod transposition process The axial force in the

equivalent member as a consequence of external load and

the force of the equivalent member must together total

zero; from this is derived the force in the equivalent

mem-I.S Pin-Jointed Frames • mem-I.S.1 Plane Frames

ber The method is also well suited for establishing the stability of a frame, since in the event of instability the force in the equivalent member approaches infinity

Influence Lines Resulting From Variable Loads

The calculation of an axial force PSi as a function of x resulting from a variable load F ~ 1 gives the application

function y/(x); the graphical representation of this is

called the influence line The evaluation of several

concen-trated loads Fi provides the axial force Fs, ~ "iFjy/(xj)

(see example)

Example Frame Supports (Fig ~~a) Given: PI = 5 kN, F2

= 10 kN, F.~ = 20 kN, a = 2 m, b = 3 rn, h = 2 m, a = 45''',

f3 ~ 3369°

ReqUired: axial forces

Node intersection Process The unknown axial forces FSi are

applied as positive tensile forces (Fig 'lob) For nodes E the

IF,x = 0 gives F.; =- F,) = + l-;.OO kJ'.:

(tensile force);

2.1'" = 0 gives Fs\ =- F\ = -20.00 kN

(pressure)

For nodes D, the following applies:

"'iF,} = 0 gives F~", = -(r~l sin a + Fs.\)/sin {3

'iF,x = 0 gives FI! = -F S6 = 55.00 kN

For nodes A the following applies'

"LF ix = 0 gives FA"", = FS4 + Fs,,; cos f3 = 60.00 kN;

2.F j \" = 0 gives FA~ = Fs'i sin {3 + F",~ = 30.00 kN The support forces also derive from the conditions of equilibrium

in the (non-intersected) total system

Cremona Force Diagram (Fig ~~c) Moving in a clockwise direction

Ritter's Method Of Sections The axial forces F S4 , Fs'"> and F~6 are determined by Ritter's method of sections (Fig ~~d)

2.MiD = 0 gives F~4 = (F 1 a + F)h)/h = + 15.00 kN;

lM" ~ 0 gives F", ~ -[F,(a + b) + F,b[/h ~ -55.00 kN;

IF,, ~ 0 gives F" ~ (F, + F,)/sin f3 ~ +54.08 kN

lnjluence J.ines for Axial Furce F s6 • The influence of a vertical able load Fv on the axial force FS(, is examined (in any random pos- ition x on the top chord) (Fig ~~e) From

vari-it follows wvari-ith Fl' = 1 Tj(X) ~ -1· (a + b - x)/h ~ -5/2 + x/(2 01)

in other words, a straight line (Fig ~~f) In view of the fact that

F J has no influence on P S( (see SM iA = 0), the evaluation for the

Mechanics • I Statics of Rigid Bodies

Fso = F1.TJ (x = 0) -r FjTJ (x = a)

~ 10 kN(-512) + 20 kN(-3/2) ~ -55 kN

1.8.2 Space Frames

In view of the fact that three conditions of equilibrium

pertain per node in space, and six bearing forces are

required for the stable, statically detertttined bearing of

the pin-frame as a whole, the numeric criterion

3n = s + 6 or s = 3n - 6 applies Otherwise, methods

apply to the plane frames which are analogous to the axial

force calculation, etc [2J

1.9 Cables and Chains

Cables and chains are regarded as flexible and bending;

i.e they can transfer only tensile forces If the elongation

of the individual elements is disregarded (first-order

theory), then the following apply for the plane problem,

as a result of a vertical uniformly distributed load, from

the conditions of equilibrium at the cable element (Fig

34a):

With a Given Load q(s) IF,x = 0, i.e dFH = 0, IF" = 0,

i.e dFy ~ q(s) d, i.e dFv = q(s) ds, in other words FIJ

= const and dFjds = qs According to Fig 34a, there

q(s) ds = q(x) dx, i.e

q(s) = q(x) dx/ds = q(x) cos", = q(x)/ ~I + y"

and therefore, according to Eq (19),

The solutions to these differential equations produce the

catenary curve y(x) The two integration constants that

occur in the process and the unknown (constant)

hori-zontal tensile force FH follow from the boundary

Figure 34 Cable; a dement, b cable under its own weight, C

cable under point load

ditionsy(x = Xl) = y, andy(x = x2 ) = y" as well as from

the given cable length L = J ds = J ~I + y'2 dx

1.9.1 The Catenary

For a cable with a constant cross-section, with q(s) =

const = q from Eq (19) with a = FH/q after separation

of the variables and integration, it follows that arcsinh y'

= (x - xu)/a or y = sinh[(x - xo)/a], and thus for the catenary

The extreme value of x0') follows from y' = 0 at the

point x = Xo to Ymin = Yn + a The unknown constants

xo, Yo and a = FH/ q are derived from the following three

= a sinh[(x, - xo)/a] + a sinh(xo/a)

From this are derived

Yo = -a cosh(x,,/a), x" = x,/2 - a arctanh(y,/L) and

sinh(x,/2a) = \L' - yl/(2a)

From the latter (transcendental) equation, a can be

cal-culated, and subsequently Xo and Yo The maximum sag I

against the chord follows at the point Xm = Xo +

a arcsinh(y,/x,) for 1= y,xm/x, - y(xm)

For the forces, the following applies:

FH = aq = const Fv(x) = FIIy' (x),

The maximum cable force occurs at the point at which

y' is a maximum, i.e in one of the ftxing pOints

Example Catenary', Fixing point PI (0; 0) and P2 (30001;

- SO m) Cable length L = 340 m, loading q(s) = 30 N/m From the

transcendental equation, by iteration, we derive a = 179.2 m and

therefore x() = 176." m and Yo = - 273.4 m, from which the

catenary is determined according to Eq (21), The maximum sag

against the chord occurs at the point xm - 146.8 m and has the

value J = 67.3 m The horizontal tensile force amounts to PH = aq

= 5.375 kN = const The greatest cable force occurs at the point PI:

Fv(x = 0) = FIJ lv'(x = 0)1 = 6.192 kN and therefore

Fs max = F.;(x = 0) = 8.2(} kN

1.9.2 Cable with Uttifonn Load over the Span

This includes not only cables with attached unifornl loads

over the span q(x) = const., but also cables with plane

sags under dead weight, since at q(s) = qo = const.,

because of q(s) -II + y" = q,,/cos 'P = q(x) with cos 'P

= COS" = const., q(x) = const = q also pertains Double

integration of Eq (20) provides y(x) = (q/F II )x'/2 +

C!x + C,; boundary conditions with given sagl in the

cen-tre; y(x, = 0) = 0, y(x = x,) = y" y(x = x,/2)

,Yz/2 -J

From this C, = 0, C, = (y, - 4j)!x2, I'll = qxl/(8f)

and therefore

y(x) = 0',/x,)x - (41/x1) (x2x - x') = (y,/x,) x - I(x) ,

where I(x) is the sag against the chord (Fig ~4b) In

addition, Fv(x) = FHY' (x) and Fs(x) = -IF~ + N (x);

Fs m= applies at the point of maximum slope

The length L of the cable follows from L =

x,

J <I + y" dY with a = FH/q to

x=o

L = (a/2) [(C, + x,/a) \1 + (C, + x,/a)

+ In(C, + x,/a + VI + (C, + x,/a)2)

- C! Vi + cf - In(C! + \(1 + CD]·

1.10 Centre of Gravity

For cables with plane sags, with the chord length

/ = -Ixl + yl, the approximation formula applies:

Vex) = -0.1667·x - 0.003 m-' (300 m·x - x')

= -1.064·x + 0.003 m-' .x'

At the point x = n y'm" = iy'(O)i = 1.064, i.e Fv m" = }~y~" =

5.408 kN and therefore F~.Olax = 7.42 kN

The approximation formula Eq (23) for the cable length then prOvides, with I = 304.1 m, the value L = 342.7 ffi The results show that the approximation solution does not deviate substantially from the exact values (see A1.9.1), although the 'plane' sag applies here only to a minor dc:gree

A cable with only plane sags with respect to the chord is considered (Fig ~4c, left) If x"y" x"y, are given, then with FH! = FHII = FII the following relationships apply:

I, = q, xU(8f;,), IH = q!! xi/(8FH), y(x) = (y,/x,)x - (qj/2FII )(x,x - x'),

y' (x) = (Y2/X2) - (q,/2FII )(x, - 2x), y' (x) = (ji,/x2) - (qn/2FH)(X, - 2X)

From the condition of equilibrium "iF;, = 0 = FYI +

F - F v , at the node P, (Fig ~4<:, right) it follows that

Fv = FII·ly'l taking into consideration thaty' is negative, and therefore Lv' I = -y

Flly,/x, + q,x2/2 + F + FHy,/x, + qllx,/2 = 0, i.e

FH = [-q!x, - q,.x2 - 2F]/[20',/x2 + y,/x,)]

This enables [, and [" to be calculated as shown, Fv(x)

and Fs(x) to be calculated by Eq (22), and L, and LII to

be calculated by Eq (23)

1.10 Centre of Gravity

The mass dements of a body of mass m are affected by the forces of gravity dF" = dmg, all of which are parallel

to one another The point of contact of their resultant

F" = J dF" is termed the centre of gravity (Fig ~Sa) Its position is determined by the condition that the moment

of the resultant must be equal to that of the individual forces, i.e

Mechanics • J Statics of Rigid Bodies

By analo!,'y, with a constant acceleration of graviry g for

the centre of gravity, a constant density p for the centre

of volume, and for the centroids of areas and centres of

lines in vector form, the following applies:

rs = (lim) f r dm; rs = (J/V) f r dV;

(25)

rs = (iIA) f r dA and rs = (lIs) f r ds,

If the bodies consist of an infinite number of parts with

known centres of gravity, then the following applies in

the components, for example for the centroid of area:

Centres of gravity determined by integration, of geneous bodies and of surfaces and lines, are shown in Tables 1 to~

homo-Example Centre of Gravity of a Cross-Section Through a

Bear-ing Element The centre of gravity for the combined bearing element cross-section is to be detennined (Fig ~Sb) The centre

of gravity lies on the axis of symmetry Ys is determined by tabular means, for which the hole is regarded as a 'negative surface'

Table 1 Centres of gravity of homogeneous bodies

Paraboloid

01 rotation

Truncated pyramid or cone

1.11 Friction • 1.11.1 Static and Sliding Friction

Table 2 Centres of gravity of surfắt's

em:!

2252.0

19398 1600.0 330.0 -90.0

L 6031.8

1.11.1 Static and Sliding Friction Static Friction (Friction at Rest) If a body remains under the influence of a resultant force F, which holds it against a supporting base, then static friction pertains

(Fig 36) The distribution of the surface pressure between the body and the supporting base is in most cases unknown, and is substituted by the reactive force

Fn- For reasons of eqUilibrium, Pn = Fs = F cos ex and

Fr = Ft = F sin 0', ịẹ Fr = 1: 1 tan Ct The body remains at rest until the reactive force J.~ reaches the limiting value

Fro = Fn tan Po = FnJ.Lo, Lẹ for as long as F, considered in

Mechanics 1 Statics of Rigid Bodies

Table~ Centres of gravity of lines

IlNI029

Figure JS Centre of gravity: a of a body, b of a bearing element

cross-section ¢ cross section of rectangle

Figure J6 Static friction Figure'7 Sliding friction

spatial tenns, lies within what is known as the cone of

friction with the angular aperture 2po The inequality

(27)

applies to the reactive force F,

The coefficient of static friction JLo depends on the

materials in contact, on the composition of their surfaces,

on the presence of a lubricant film, on the temperature

and the humidity, on the contact pressure, and on the

Random Hal arc

value of the nonnal pressure force; JLo accordingly ates between specific limits and may, if necessary, be detennined by expeliment [3] For reference values for

fluctu-JLo see Table 4

Sliding Friction (Friction of Motion) If the static friction is overcome, and the body is set in motion, then Coulomb's law of friction applies to the friction force

(Fig 37):

F,/ Fn = const = tan P = JL or F, = !LPn (28) The sliding friction force is an imposed force that is directed against the speed or displacement vector The sliding friction coefficient JL (or sliding friction angle p)

depends not only on the influences described under Al.1l.1 but particularly on the lubrication conditions (dry friction, semi-fluid friction, fluid friction; see D5.1), as well as, in part, on the velocity [4, 5] For reference values for JL, see Table 4

Example Stability of a Ladder (Fig, ~8) How far maya per·

son (gravitational force FQ) climb a ladder (length I, angle of nation {3, weight negligible), without the ladder slipping, ifthe static

incli-frictional angle between ladder and wall and ground is Po? How

great would the forces then be at the upper and lower points of contact? The cone of friction is drawn in at the head and foot points

As long as the force PI ) lies within the shaded area, it can be resolved

in an infinite number of ways so that the forces I'~} and Fu at the head and foot points lie within the cone of friction, and equilibrium prevails (i.e the problem is statically indeterminate, and the force are thus indeterminable) The limit value is attained when f'Q passes

through point A (Fig 3>8) The results follow from the illustration (owing to the right angle between the upper surface line and the lower one)

a = I cos({3 + Po) x = a cos Po = I cos Po cos(/3 l- Po) From the polygon of force, we read off:

1.11.2 Applications of Static and Sliding Friction

Friction on a Wedge The force F is being sought, which is required for the raising or lowering of a load at constant speed The solution is derived most simply from the sine law at the polygon of forces; e.g for raising the load according to Fig 39,

sin(90° + p,) F

sin [90° - + P2 + F,

sin(a + PI + P2)

sin (90° - PI) ,

1.11 Friction 1.11.2 Applications of Static and Sliding Friction

Table 4 Static and sliding friction coefficients

Substance pairs Coefficient of Coefficient of

static friction ~(l sliding friction I.l

Dry Lubricated Dry tubricated

1.0 Steel-steel 0.45 to 0.10 (lAO to 0.10

0.70 Aluminium-aluminium

0.94 to

1.35 51.37-St.37, polished O.IS

Sted-grey cast iron 0,18 to 0.10 0.17 to 0.0210

0.60 0.50 Cup leather metal 0.60 0.20 0.20 to 0.12

0.25 Steel-pol}1ctra-

The wedge must then be taken out or pushed out from

the other side The efficiency of the wedge system when

raiSing the load is 1) = /0;'; F; in this case, /0;, = FQ tan 0' is the force required without friction For PI = P2 = p, = p

there applies F = /0;, tan( 0' :+: 2p); self-locking for a "" 2p,

efficiency 1) = tan a/tan(O' + 2p) In the case of ing, 1) = tan 2p/tan 4p = 0.5 - 0.5 tan' 2p < 0.5

self-lock-Screws

Rectangular Thread (Square-Threaded Screw) (Fig

40a) The torque M for uniform raising or lowering of the load is required

'iF" = 0 = f dF cos(n + p) - FQ, F = FQ/cos(a + p), 'iM" = 0 = M - f dFsin(a + p)r m ,

M = FQrm tan(a + p)

Efficiency for raising 1) = Mo/ M = tan a/tanC a + p); Mo

is the required moment without friction During lowering,

-p occurs instead of p; M = FQrm tan(a - pl Self-locking

for M"" 0, i.e tan(a - p) "" 0 and therefore a"" p A negative moment is then required for lowering the load For a = p, it follows that

1) = tan p/tan 2p = 0.5 - 0.5 tan2 p < 0.5

Trapezoidal and Triangular Threads (Fig 40b) 'Ine

same equations apply as for the rectangular thread, if instead of ~ = tan p the friction coefficient ~'=

tan p' = ~/cos({3/2), is used, i.e instead of p the angle

of friction p' = arctan[~/cos({3/2)] Proof is according to Fig 4Ob, since instead of dF n the force

df~ = dFn/cos({3/2) obtains, and instead of df', = ~dFn

the force dF; = ~df'~ = [~/cos({3/2)]dFn = ~'df~ In this case, {3 is the flank angle of the thread Note: For screws

used for securing purposes, self-locking is required, i.e 0' s g~

Rope Friction (Fig 41) Sliding friction occurs when there is relative motion between rope and pulley (belt brake, bollard with running rope), static friction when a state of rest prevails between rope and pulley (belt drive, belt brake as retaining brake, bollard with

rope or cable at rest) Accordingly, J.L or }-4,) is taken as

the coefficient of friction Equilibrium in the normal and tangential direction at the rope element (Fig 41b) gives dF" = Fs dip, df~ = dF,: with dF, = ~dFn it follows that

con-tact a, there follows Euler's rope friction formula

III

Mechanics 1 Statics of Rigid Bodies

Figure 40 Friction on a flat- Figure 41 Cable

threaded and b sharp-threaded screw friction: a forces, b

Figure 42 Resistances: a rolling resistance, b tractive resistance,

c fIxed and d loose rope pulley, e pulley block

F,.,,2 = FSl exp J La (30)

The friction force is derived from F, = F s, - FSI and the

friction moment from M, = F,r

In cases in which the velocity of the rope cannot be

disregarded (e.g on belt drives), centrifugal forces

qF = m'v'lr occur on the rope (m' = mass per length

unit of the rope) In such cases, Fs is replaced by

Fs m*v 2

1.11.3 Rolling Resistance

If a cylindrical body or a body of similar shape rolls on a

supporting surface (Fig 42a), the deformation of the

supporting surface and of the body produces a resultant

lhat is directed at an oblique angle, the horizontal

compo-tllent of which is the resisting force F W' If the motion is

uniform, the driving force E, must retain the body in

equi-librium With En = Fo andf""" r, i.e tan 0' = sin 0' = fir,

it follows that

Fw = Foflr = Fo/L,

and Mw = Fwr = /L/'or = Fof as what is known as the moment of rolling friction, where /L, = fir is the coef-ficient of rolling friction The lever arm f of the rolling friction is determined empirically For steel wheels on rails, f = 0.05 cm, and for rolling bearings f = 0.0005 to 0.001 cm

The sum of the rolling resistance and the bearing

fric-tion resistance is designated the tractive resistance

simul-Fixed Pulley (Fig 42c) During lifting

l MA = 0 = F(r - a l ) - FQ (r + a,) - (F + Fo) r"

F = f'o (r + a, + r,)/(r - a l - r,) = FQI1)

1) is the efficiency of the fixed pulley during lifting (1) = 0.95) When lowering, 1) is replaced by 1 I 1) (r,

radius of the journal friction)

Free Pulley (Fig 42d) During lifting

l MA = 0 = F(2r + a, - a l) - FQ (r + a z + r z), i.e

F = (FQ/2) (r + a2 + r,)/(r + a,/2 - a 1/2)

= (FQ/2)11)

= (Fos!2)/(Fs) By way of approximation, it is likewise taken that 1) = 0.95 During lowering 1) is replaced by 1/1)

Pulley Tackle (Fig 42e) With the results for the fIXed and free pulley, FI = 1)F, F,=1)' FI = 1)2 F, etc Equilib-rium for the freed lower hlock leads to

With n load-bearing rope lines, the force and the overall

degree of efficiency for lifting become

F = f'o/[1)(1 - 1)n)/(I - 1))) and 1),", = Wn/W, = (FQsln)/(Fs) = 1) (1 - 1)n)/[(1 - 1))n)

For lowering, 1) is again to be substituted by 1/ 1)

Weight Force, Acceleration of Fall, Terms - DIN 1311: Oscillation Study - DIN 1342: Viscosity of Newtonian Fluids - DIN 5492: Formula Symbols for Fluid Mech-anics - DIN 5497: Mechanics; Rigid Bodies; Formula sym-bols

2.1 Motion of a Panicle 2.l.1 Introduction

Kinematics

Kinematics is the study of the geometrical and analytical

description of the states of motion of points and bodies

It does not take account of forces and moments as causes

of the motion

2.1 Motion of a Particle

2.1.1 Introduction

Trajectory A panicle moves as a function of time in

space along a trajectory The spatial coordinates of the

panicle are determined by the spatial vector (Fig la):

r(t) = x(t)e, + y(t)e, + z(l)e, = (x(t); y(t); z(t»

(1)

A panicle has three degrees of freedom in space, two in

guided motion along a surface and one degree of freedom

along a line

Velocity The velocity vector is acquired by

differen-tiating the spatial vector with respect to time:

vet) = dr/dl = r(t) = x(t)ex + y(t)e y + i(t)e, (2)

= (x(t); yet); i(I» = (v x ; v y ; v,)

The velocity vector is always tangential to the trajectory,

since in natural coordinates I, n, b (accompanying

tri-hedral, where I is the tangent direction in what is termed

the osculating plane, n is the nonnal direction in the

oscu-lating plane, and b is the binonnal direction venical to I

and n; see Fig la),

Acceleration The acceleration vector is ascenained by

differentiating the velocity vector with respect to time:

dv d2 r

,,(t) = dt = dl' = r(t) = x(t)ex + y(t)e y + i(t)e, (5)

= (x(I); yet); i(t» = (ax; a,; a,)

or, in natural coordinates,

dS v R env (see Fig Ib)

,,(t) = VI', + (v'/R)en = II, + lIn' (6) i.e the acceleration vector always lies in the osculating plane (Fig la) Its components in the tangential and nonnal direction are called tangential and normal acceler-

ation

a, = dv/dl = v(t) = set) (7) and

where R is the radius of curvature of the trajectory The nonnal acceleration is always directed to the mid-point of the curvature; in other words, it is always a centripetal acceleration Applicable to the value of the (resultant)

acceleration vector is

a = 1,,1 = Ja~ + a~ + a; = Ja; + a~ (9)

Uniform Motion This penains if v(t) = set) = Vo =

const By integration, it follows that

sCt) = f sCI) dl = vot + C,

or, with the initial condition s(t = (,) = s" from this C, =

s, - vol, and therefore

s(t) = vo(t - I,) + St·

Graphical representations of vet) and set) provide the velocity-time diagram and the distance-time diagram (Fig 2) From set), by differentiation, v(t) follows inversely

Uniformly Accelerated (and Retarded) Motion

(Fig ~a) This penains when

a,(1) = v(t) = s(t) = a", = const., i.e

vet) = a,o ( + C, and s(t) = a,o t' /2 + Cl + C 2·

From thiS, with the initial conditions vet = t I) = Vt and s(t = t,) = s" the constants follow:

v = Jvi + 2aw(s - s,), s = (v' - vD/(2a,o) + St·

For the special case of t, = 0, v, = 0, s, = 0 it follows that

vet) = awt, set) = a /,ot' /2, t = v / a,o,

a,o = v2/(2s), v = ~2a",s, s = v'/(2a,o)'

The mean velocity is derived by

IZImI Mechanics 2 Kinematics

In all the equations, a, can be positive or negative: Positive

a, signifies acceleration during the movement of a particle

in the positive s-direction, but retardation in motion in

the negative s-direction; negative at signifies retardation

in motion in the positive s-direction, but acceleration in

motion in the negative s-direction If sU) is given, vU)

and a,(t) are obtained by differentiation

Non-unHonnly Accelerated (and Retarded)

Motion 11J.is pertains when a,(t) = f,(t) (Fig 3b)

Integration leads to

vet) = J a,(1) dt = J f, (t) dt = f,(t) + C, and

set) = J vet) dt = J [f,(t) + C,l dt = /,(t) + C,t + C2·

The constants are determined from the initial conditions

vet = t,) = v, and set = t,) = s, or equivalent conditions

From v(t) = a,(t) it follows that, in cases in which vet)

assumes an extreme value (where v = 0), in the a"

t-dia-gram the function a,(t) passes through zero Similarly, it

follows from s(t) = vet) that set) has an extreme value

when v(t) in the v, t-diagram passes through zero The

mean velocity is given by vrn = (S2 - s,)/(t, - t,) In

accordance with the graphic significance of the integral

as the surface area, with a given a,(t) the values vU) and

s(t) are determined with the methods of graphical or

numerical integration

2.1.2 Plane Motion

Trajectory (Path), Velocity, Acceleration The

for-mulae of A2.1.1 apply, reduced to the two components x

and y (Fig 4a):

r(t) = x(t)e x + y(t)e y = (x(t);y(t»,

vet) = x(t)ex + y(t)e y = (xU); yU» = (vx ; v y ),

IIU) = x(t)ex + y(t)e y = (xU); yU» = (ax; a y)

or, in natural coordinates t and n:

x

Figure 4 Motion in a plane: a general, b circle

1I(t) = v(t)e, + (v2/R)en = (v(t); v2/R) = (a,; an)

If the trajectory is given with y(x) and the position of the particle with s(t), then a connection is derived berween

t and x across the arc length sex) = J ~l + y" dx from sex) = set) From this, t(x) or x(t) are explicitly calcu-lable only in simple cases (see next example)

Example Motion on a Trajectory y(x) (Fig 4b) The motion

of a particle on the orbit y(x) = -vr2 - r is examined in accordance with the distance-time law set) = Af According to Eqs (4), (7) and (8), we derive

vet) = s(t) = ZAt, a,(t) = set) = set) = ZA and