Download - Mechanical and Metal Trade Handbook 01
8102019 Mechanical and Metal Trade Handbook 01
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8102019 Mechanical and Metal Trade Handbook 01
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Original title
Tabellenbuch Metall 45th edition 2011
Authors
Ulrich Fischer Dipl-Ing (FH) Reutlingen
Roland Gomeringer Dipl-Gwl Meszligstetten
Max Heinzler Dipl-Ing (FH) Wangen im Allgaumlu
Roland Kilgus Dipl-Gwl Neckartenzlingen
Friedrich Naumlher Dipl-Ing (FH) Balingen
Stefan Oesterle Dipl-Ing Amtzell
Heinz Paetzold Dipl-Ing (FH) Muumlhlacker
Andreas Stephan Dipl-Ing (FH) Marktoberdorf
Editor
Ulrich Fischer Reutlingen
Graphic design
Design office of Verlag Europa-Lehrmittel Ostfildern Germany
The publisher and its affiliates have taken care to collect the information given in this book to the best of their
ability However no responsibility is accepted by the publisher or any of its affiliates regarding its content or
any statement herein or omission there from which may result in any loss or damage to any party using the
data shown above Warranty claims against the authors or the publisher are excluded
Most recent editions of standards and other regulations govern their use
They can be ordered from Beuth Verlag GmbH Burggrafenstr 6 10787 Berlin Germany
The content of the chapter bdquoProgram structure of CNC machines according to PALldquo (page 412 to 424) complies
with the publications of the PAL Pruumlfungs- und Lehrmittelentwicklungsstelle (Institute for the development of
training and testing material) of the IHK Region Stuttgart (Chamber of Commerce and Industry of the Stuttgartregion)
English edition Mechanical and Metal Trades Handbook
3rd edition 2012
6 5 4 3 2 1
All printings of this edition may be used concurrently in the classroom since they are unchanged except for somecorrections to typographical errors and slight changes in standards
ISBN 13 978-3-8085-1914-1
Cover design includes a photograph from TESABrown amp Sharpe Renens Switzerland
All rights reserved This publication is protected under copyright law Any use other than those permitted by law
must be approved in writing by the publisher
copy 2012 by Verlag Europa-Lehrmittel Nourney Vollmer GmbH amp Co KG 42781 Haan-Gruiten Germanyhttpwwweuropa-lehrmittelde
Translation Eva Schwarz 76879 Ottersheim Germany wwwtechnische-uebersetzungen-eva-schwarzde
Typesetting Satz+Layout Werkstatt Kluth GmbH D-50374 Erftstadt Germany wwwslw-kluthde
Printed by M P Media-Print Informationstechnologie GmbH D-33100 Paderborn Germany
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3
Preface
The Mechanical and Metal Trades Handbook is well-suited
for shop reference tooling machine building maintenance
and as a general book of knowledge It is also useful for ed-
ucational purposes especially in practical work or curricula
and continuing education programsTarget Groups
bull Industrial and trade mechanics
bull Tool amp die makers
bull Machinists
bull Millwrights
bull Draftspersons
bull Technical Instructors
bull Apprentices in above trade areas
bull Practitioners in trades and industry
bull Mechanical Engineering students
Notes for the user
The contents of this book include tables and formulae in
eight chapters including Tables of Contents Subject Index
and Standards Index
The tables contain the most important guidelines designs
types dimensions and standard values for their subject
areas
Units are not specified in the legends for the formulae if sev-
eral units are possible However the calculation examples
for each formula use those units normally applied in practice
The Table of Contents in the front of the book is expandedfurther at the beginning of each chapter in form of a partial
Table of Contents
The Subject Index at the end of the book (pages 435 ndash 444) is
extensive
The Standards Index (pages 425 ndash 434) lists all the current
standards and regulations cited in the book In many cases
previous standards are also listed to ease the transition from
older more familiar standards to new ones
Changes in the 3rd edition
In the present edition we have updated the cited standards
and restructured updated enhanced or added the follow-ing chapters in line with new developments in engineering
Acknowledgement
Special thanks to Alexander Huter Vocational TrainingSpecialist ndash Tool and Die Ontario for his input into the
English translation of this book His assistance has been
extremely valuable
November 2012 Authors and publisher
M1 Mathematics
9 ndash 28
P2 Physics
29 ndash 50
TD3 Technical
Drawing
51 ndash 110
MS4 Material
Science
111 ndash 200
ME5 Machine
Elements
201 ndash 268
PE6 Production
Engineering
269 ndash 366
A7 Automation and
Information Tech-
nology 367 ndash 424
S8 International Material
Comparison ChartStandards 425ndash 434
ndash Fundamentals of technical
mathematics
ndash Strength of materials
ndash Plastics
ndash Production management
ndash Forming
ndash Welding
ndash PAL programming system
for NC turning and NC
milling
ndash Steel types
ndash Material testing
ndash Machining processes
ndash Injection molding (new)
ndash GRAFCET
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5Table of Contents
4 Materials Science (MS) 11141 Materials Material characteristics of solids 112 Material characteristics of liquids
and gases 113 Periodic table of the elements 114
42 Steels designation system Definition and classification of steel 116 Standardization of steel products 117 Designation system for steels 118
43 Steels steel types Overwiew of steel products 123 Unalloyed steels 126 Case hardened steels quenched amp
tempered steels 129 Nitriding steels free cutting steels 131 Tool steels stainless steels 132 Steels for bright steel products 137
44 Steels finished products Sheet and strip metal tubes 138
Profiles 142 Linear mass density and area mass
d e n s i t y 1 5 1
45 Heat treatment Iron-carbon phase diagram 152 Heat treatement of steels 153 Hardening of aluminium alloys 156
46 Cast iron materials Designation and material codes 157 Classification 158 Cast iron 159 Malleable cast iron cast steel 160
47 Foundry technology 1 6 148 Light alloys Overview of aluminum alloys 163 Wrought aluminum alloys 165 Aluminum profiles 168 Magnesium and titanium alloys 171
49 Heavy non-ferrous metals Overview and designation system 172 Copper and refined zinc alloys 174
410 Other metallic materials 176
411 Plastics Overview and designation 178 Thermoset plastics 181 Thermoplastics elastomers 182
Plastics processing testing of plastics 186
3 Technical Drawing (TD) 51
31 Graphs Cartesian coordinate system 52
Graph types 53
32 Basic geometric constructions Lines and angles 54
Tangents circular arcs 55 Inscribed circles ellipses 56
Cycloids involute curves parabolas 57
33 Elements of drawingFonts 58
Preferred numbers radii scales 59
Drawing layout bills of materials 60 Line types 62
34 Representation Projection methods 64 Views 66
Sectional views 68 Hatching 70
35 Dimensioning drawings Dimensioning lines dimension values 71
Dimensioning rules 72 Elements of drawing 73 Tolerance specifications 75 Types of dimensioning 76 Simplified presentation in drawings 78
36 Machine elements representation Gear types 79 Roller bearings 80 Seals 81 Retaining rings springs 82
37 Object elements Bosses object edges 83 Thread runouts thread undercuts 84 Threads screw joints 85 Center holes knurls undercuts 86
38 Welding and soldering Graphical symbols 88 Dimensioning examples 91
39 Surfaces Hardness specifications in drawings 92 Form deviations roughness 93 Surface testing surface indications 94
310 ISO tolerances and fits Fundamentals 98 Basic hole system basic shaft system 102 General tolerances 106 Roller bearing fits 106 Fit recommendations 107 Geometric tolerancing 108
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
8102019 Mechanical and Metal Trade Handbook 01
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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Original title
Tabellenbuch Metall 45th edition 2011
Authors
Ulrich Fischer Dipl-Ing (FH) Reutlingen
Roland Gomeringer Dipl-Gwl Meszligstetten
Max Heinzler Dipl-Ing (FH) Wangen im Allgaumlu
Roland Kilgus Dipl-Gwl Neckartenzlingen
Friedrich Naumlher Dipl-Ing (FH) Balingen
Stefan Oesterle Dipl-Ing Amtzell
Heinz Paetzold Dipl-Ing (FH) Muumlhlacker
Andreas Stephan Dipl-Ing (FH) Marktoberdorf
Editor
Ulrich Fischer Reutlingen
Graphic design
Design office of Verlag Europa-Lehrmittel Ostfildern Germany
The publisher and its affiliates have taken care to collect the information given in this book to the best of their
ability However no responsibility is accepted by the publisher or any of its affiliates regarding its content or
any statement herein or omission there from which may result in any loss or damage to any party using the
data shown above Warranty claims against the authors or the publisher are excluded
Most recent editions of standards and other regulations govern their use
They can be ordered from Beuth Verlag GmbH Burggrafenstr 6 10787 Berlin Germany
The content of the chapter bdquoProgram structure of CNC machines according to PALldquo (page 412 to 424) complies
with the publications of the PAL Pruumlfungs- und Lehrmittelentwicklungsstelle (Institute for the development of
training and testing material) of the IHK Region Stuttgart (Chamber of Commerce and Industry of the Stuttgartregion)
English edition Mechanical and Metal Trades Handbook
3rd edition 2012
6 5 4 3 2 1
All printings of this edition may be used concurrently in the classroom since they are unchanged except for somecorrections to typographical errors and slight changes in standards
ISBN 13 978-3-8085-1914-1
Cover design includes a photograph from TESABrown amp Sharpe Renens Switzerland
All rights reserved This publication is protected under copyright law Any use other than those permitted by law
must be approved in writing by the publisher
copy 2012 by Verlag Europa-Lehrmittel Nourney Vollmer GmbH amp Co KG 42781 Haan-Gruiten Germanyhttpwwweuropa-lehrmittelde
Translation Eva Schwarz 76879 Ottersheim Germany wwwtechnische-uebersetzungen-eva-schwarzde
Typesetting Satz+Layout Werkstatt Kluth GmbH D-50374 Erftstadt Germany wwwslw-kluthde
Printed by M P Media-Print Informationstechnologie GmbH D-33100 Paderborn Germany
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3
Preface
The Mechanical and Metal Trades Handbook is well-suited
for shop reference tooling machine building maintenance
and as a general book of knowledge It is also useful for ed-
ucational purposes especially in practical work or curricula
and continuing education programsTarget Groups
bull Industrial and trade mechanics
bull Tool amp die makers
bull Machinists
bull Millwrights
bull Draftspersons
bull Technical Instructors
bull Apprentices in above trade areas
bull Practitioners in trades and industry
bull Mechanical Engineering students
Notes for the user
The contents of this book include tables and formulae in
eight chapters including Tables of Contents Subject Index
and Standards Index
The tables contain the most important guidelines designs
types dimensions and standard values for their subject
areas
Units are not specified in the legends for the formulae if sev-
eral units are possible However the calculation examples
for each formula use those units normally applied in practice
The Table of Contents in the front of the book is expandedfurther at the beginning of each chapter in form of a partial
Table of Contents
The Subject Index at the end of the book (pages 435 ndash 444) is
extensive
The Standards Index (pages 425 ndash 434) lists all the current
standards and regulations cited in the book In many cases
previous standards are also listed to ease the transition from
older more familiar standards to new ones
Changes in the 3rd edition
In the present edition we have updated the cited standards
and restructured updated enhanced or added the follow-ing chapters in line with new developments in engineering
Acknowledgement
Special thanks to Alexander Huter Vocational TrainingSpecialist ndash Tool and Die Ontario for his input into the
English translation of this book His assistance has been
extremely valuable
November 2012 Authors and publisher
M1 Mathematics
9 ndash 28
P2 Physics
29 ndash 50
TD3 Technical
Drawing
51 ndash 110
MS4 Material
Science
111 ndash 200
ME5 Machine
Elements
201 ndash 268
PE6 Production
Engineering
269 ndash 366
A7 Automation and
Information Tech-
nology 367 ndash 424
S8 International Material
Comparison ChartStandards 425ndash 434
ndash Fundamentals of technical
mathematics
ndash Strength of materials
ndash Plastics
ndash Production management
ndash Forming
ndash Welding
ndash PAL programming system
for NC turning and NC
milling
ndash Steel types
ndash Material testing
ndash Machining processes
ndash Injection molding (new)
ndash GRAFCET
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5Table of Contents
4 Materials Science (MS) 11141 Materials Material characteristics of solids 112 Material characteristics of liquids
and gases 113 Periodic table of the elements 114
42 Steels designation system Definition and classification of steel 116 Standardization of steel products 117 Designation system for steels 118
43 Steels steel types Overwiew of steel products 123 Unalloyed steels 126 Case hardened steels quenched amp
tempered steels 129 Nitriding steels free cutting steels 131 Tool steels stainless steels 132 Steels for bright steel products 137
44 Steels finished products Sheet and strip metal tubes 138
Profiles 142 Linear mass density and area mass
d e n s i t y 1 5 1
45 Heat treatment Iron-carbon phase diagram 152 Heat treatement of steels 153 Hardening of aluminium alloys 156
46 Cast iron materials Designation and material codes 157 Classification 158 Cast iron 159 Malleable cast iron cast steel 160
47 Foundry technology 1 6 148 Light alloys Overview of aluminum alloys 163 Wrought aluminum alloys 165 Aluminum profiles 168 Magnesium and titanium alloys 171
49 Heavy non-ferrous metals Overview and designation system 172 Copper and refined zinc alloys 174
410 Other metallic materials 176
411 Plastics Overview and designation 178 Thermoset plastics 181 Thermoplastics elastomers 182
Plastics processing testing of plastics 186
3 Technical Drawing (TD) 51
31 Graphs Cartesian coordinate system 52
Graph types 53
32 Basic geometric constructions Lines and angles 54
Tangents circular arcs 55 Inscribed circles ellipses 56
Cycloids involute curves parabolas 57
33 Elements of drawingFonts 58
Preferred numbers radii scales 59
Drawing layout bills of materials 60 Line types 62
34 Representation Projection methods 64 Views 66
Sectional views 68 Hatching 70
35 Dimensioning drawings Dimensioning lines dimension values 71
Dimensioning rules 72 Elements of drawing 73 Tolerance specifications 75 Types of dimensioning 76 Simplified presentation in drawings 78
36 Machine elements representation Gear types 79 Roller bearings 80 Seals 81 Retaining rings springs 82
37 Object elements Bosses object edges 83 Thread runouts thread undercuts 84 Threads screw joints 85 Center holes knurls undercuts 86
38 Welding and soldering Graphical symbols 88 Dimensioning examples 91
39 Surfaces Hardness specifications in drawings 92 Form deviations roughness 93 Surface testing surface indications 94
310 ISO tolerances and fits Fundamentals 98 Basic hole system basic shaft system 102 General tolerances 106 Roller bearing fits 106 Fit recommendations 107 Geometric tolerancing 108
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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3
Preface
The Mechanical and Metal Trades Handbook is well-suited
for shop reference tooling machine building maintenance
and as a general book of knowledge It is also useful for ed-
ucational purposes especially in practical work or curricula
and continuing education programsTarget Groups
bull Industrial and trade mechanics
bull Tool amp die makers
bull Machinists
bull Millwrights
bull Draftspersons
bull Technical Instructors
bull Apprentices in above trade areas
bull Practitioners in trades and industry
bull Mechanical Engineering students
Notes for the user
The contents of this book include tables and formulae in
eight chapters including Tables of Contents Subject Index
and Standards Index
The tables contain the most important guidelines designs
types dimensions and standard values for their subject
areas
Units are not specified in the legends for the formulae if sev-
eral units are possible However the calculation examples
for each formula use those units normally applied in practice
The Table of Contents in the front of the book is expandedfurther at the beginning of each chapter in form of a partial
Table of Contents
The Subject Index at the end of the book (pages 435 ndash 444) is
extensive
The Standards Index (pages 425 ndash 434) lists all the current
standards and regulations cited in the book In many cases
previous standards are also listed to ease the transition from
older more familiar standards to new ones
Changes in the 3rd edition
In the present edition we have updated the cited standards
and restructured updated enhanced or added the follow-ing chapters in line with new developments in engineering
Acknowledgement
Special thanks to Alexander Huter Vocational TrainingSpecialist ndash Tool and Die Ontario for his input into the
English translation of this book His assistance has been
extremely valuable
November 2012 Authors and publisher
M1 Mathematics
9 ndash 28
P2 Physics
29 ndash 50
TD3 Technical
Drawing
51 ndash 110
MS4 Material
Science
111 ndash 200
ME5 Machine
Elements
201 ndash 268
PE6 Production
Engineering
269 ndash 366
A7 Automation and
Information Tech-
nology 367 ndash 424
S8 International Material
Comparison ChartStandards 425ndash 434
ndash Fundamentals of technical
mathematics
ndash Strength of materials
ndash Plastics
ndash Production management
ndash Forming
ndash Welding
ndash PAL programming system
for NC turning and NC
milling
ndash Steel types
ndash Material testing
ndash Machining processes
ndash Injection molding (new)
ndash GRAFCET
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5Table of Contents
4 Materials Science (MS) 11141 Materials Material characteristics of solids 112 Material characteristics of liquids
and gases 113 Periodic table of the elements 114
42 Steels designation system Definition and classification of steel 116 Standardization of steel products 117 Designation system for steels 118
43 Steels steel types Overwiew of steel products 123 Unalloyed steels 126 Case hardened steels quenched amp
tempered steels 129 Nitriding steels free cutting steels 131 Tool steels stainless steels 132 Steels for bright steel products 137
44 Steels finished products Sheet and strip metal tubes 138
Profiles 142 Linear mass density and area mass
d e n s i t y 1 5 1
45 Heat treatment Iron-carbon phase diagram 152 Heat treatement of steels 153 Hardening of aluminium alloys 156
46 Cast iron materials Designation and material codes 157 Classification 158 Cast iron 159 Malleable cast iron cast steel 160
47 Foundry technology 1 6 148 Light alloys Overview of aluminum alloys 163 Wrought aluminum alloys 165 Aluminum profiles 168 Magnesium and titanium alloys 171
49 Heavy non-ferrous metals Overview and designation system 172 Copper and refined zinc alloys 174
410 Other metallic materials 176
411 Plastics Overview and designation 178 Thermoset plastics 181 Thermoplastics elastomers 182
Plastics processing testing of plastics 186
3 Technical Drawing (TD) 51
31 Graphs Cartesian coordinate system 52
Graph types 53
32 Basic geometric constructions Lines and angles 54
Tangents circular arcs 55 Inscribed circles ellipses 56
Cycloids involute curves parabolas 57
33 Elements of drawingFonts 58
Preferred numbers radii scales 59
Drawing layout bills of materials 60 Line types 62
34 Representation Projection methods 64 Views 66
Sectional views 68 Hatching 70
35 Dimensioning drawings Dimensioning lines dimension values 71
Dimensioning rules 72 Elements of drawing 73 Tolerance specifications 75 Types of dimensioning 76 Simplified presentation in drawings 78
36 Machine elements representation Gear types 79 Roller bearings 80 Seals 81 Retaining rings springs 82
37 Object elements Bosses object edges 83 Thread runouts thread undercuts 84 Threads screw joints 85 Center holes knurls undercuts 86
38 Welding and soldering Graphical symbols 88 Dimensioning examples 91
39 Surfaces Hardness specifications in drawings 92 Form deviations roughness 93 Surface testing surface indications 94
310 ISO tolerances and fits Fundamentals 98 Basic hole system basic shaft system 102 General tolerances 106 Roller bearing fits 106 Fit recommendations 107 Geometric tolerancing 108
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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5Table of Contents
4 Materials Science (MS) 11141 Materials Material characteristics of solids 112 Material characteristics of liquids
and gases 113 Periodic table of the elements 114
42 Steels designation system Definition and classification of steel 116 Standardization of steel products 117 Designation system for steels 118
43 Steels steel types Overwiew of steel products 123 Unalloyed steels 126 Case hardened steels quenched amp
tempered steels 129 Nitriding steels free cutting steels 131 Tool steels stainless steels 132 Steels for bright steel products 137
44 Steels finished products Sheet and strip metal tubes 138
Profiles 142 Linear mass density and area mass
d e n s i t y 1 5 1
45 Heat treatment Iron-carbon phase diagram 152 Heat treatement of steels 153 Hardening of aluminium alloys 156
46 Cast iron materials Designation and material codes 157 Classification 158 Cast iron 159 Malleable cast iron cast steel 160
47 Foundry technology 1 6 148 Light alloys Overview of aluminum alloys 163 Wrought aluminum alloys 165 Aluminum profiles 168 Magnesium and titanium alloys 171
49 Heavy non-ferrous metals Overview and designation system 172 Copper and refined zinc alloys 174
410 Other metallic materials 176
411 Plastics Overview and designation 178 Thermoset plastics 181 Thermoplastics elastomers 182
Plastics processing testing of plastics 186
3 Technical Drawing (TD) 51
31 Graphs Cartesian coordinate system 52
Graph types 53
32 Basic geometric constructions Lines and angles 54
Tangents circular arcs 55 Inscribed circles ellipses 56
Cycloids involute curves parabolas 57
33 Elements of drawingFonts 58
Preferred numbers radii scales 59
Drawing layout bills of materials 60 Line types 62
34 Representation Projection methods 64 Views 66
Sectional views 68 Hatching 70
35 Dimensioning drawings Dimensioning lines dimension values 71
Dimensioning rules 72 Elements of drawing 73 Tolerance specifications 75 Types of dimensioning 76 Simplified presentation in drawings 78
36 Machine elements representation Gear types 79 Roller bearings 80 Seals 81 Retaining rings springs 82
37 Object elements Bosses object edges 83 Thread runouts thread undercuts 84 Threads screw joints 85 Center holes knurls undercuts 86
38 Welding and soldering Graphical symbols 88 Dimensioning examples 91
39 Surfaces Hardness specifications in drawings 92 Form deviations roughness 93 Surface testing surface indications 94
310 ISO tolerances and fits Fundamentals 98 Basic hole system basic shaft system 102 General tolerances 106 Roller bearing fits 106 Fit recommendations 107 Geometric tolerancing 108
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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5Table of Contents
4 Materials Science (MS) 11141 Materials Material characteristics of solids 112 Material characteristics of liquids
and gases 113 Periodic table of the elements 114
42 Steels designation system Definition and classification of steel 116 Standardization of steel products 117 Designation system for steels 118
43 Steels steel types Overwiew of steel products 123 Unalloyed steels 126 Case hardened steels quenched amp
tempered steels 129 Nitriding steels free cutting steels 131 Tool steels stainless steels 132 Steels for bright steel products 137
44 Steels finished products Sheet and strip metal tubes 138
Profiles 142 Linear mass density and area mass
d e n s i t y 1 5 1
45 Heat treatment Iron-carbon phase diagram 152 Heat treatement of steels 153 Hardening of aluminium alloys 156
46 Cast iron materials Designation and material codes 157 Classification 158 Cast iron 159 Malleable cast iron cast steel 160
47 Foundry technology 1 6 148 Light alloys Overview of aluminum alloys 163 Wrought aluminum alloys 165 Aluminum profiles 168 Magnesium and titanium alloys 171
49 Heavy non-ferrous metals Overview and designation system 172 Copper and refined zinc alloys 174
410 Other metallic materials 176
411 Plastics Overview and designation 178 Thermoset plastics 181 Thermoplastics elastomers 182
Plastics processing testing of plastics 186
3 Technical Drawing (TD) 51
31 Graphs Cartesian coordinate system 52
Graph types 53
32 Basic geometric constructions Lines and angles 54
Tangents circular arcs 55 Inscribed circles ellipses 56
Cycloids involute curves parabolas 57
33 Elements of drawingFonts 58
Preferred numbers radii scales 59
Drawing layout bills of materials 60 Line types 62
34 Representation Projection methods 64 Views 66
Sectional views 68 Hatching 70
35 Dimensioning drawings Dimensioning lines dimension values 71
Dimensioning rules 72 Elements of drawing 73 Tolerance specifications 75 Types of dimensioning 76 Simplified presentation in drawings 78
36 Machine elements representation Gear types 79 Roller bearings 80 Seals 81 Retaining rings springs 82
37 Object elements Bosses object edges 83 Thread runouts thread undercuts 84 Threads screw joints 85 Center holes knurls undercuts 86
38 Welding and soldering Graphical symbols 88 Dimensioning examples 91
39 Surfaces Hardness specifications in drawings 92 Form deviations roughness 93 Surface testing surface indications 94
310 ISO tolerances and fits Fundamentals 98 Basic hole system basic shaft system 102 General tolerances 106 Roller bearing fits 106 Fit recommendations 107 Geometric tolerancing 108
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
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Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
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Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
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Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
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Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
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Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
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FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
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A
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Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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7Table of Contents
81 International material comparisonchart 425
82 Index of cited standards and otherregulations 430
8 Material Chart Standards (S) 425 ndash 434
Subject Index 435 ndash 444
7 Automation and Information Technology (A) 367
71 Control engineering basic terminology
Basic terminology code letters symbols 368 Analog controllers 370 Discontinuous and digital controllers 371
Binary logic 372 Numbering systems 373 Information processing 374
72 Electrical circuits Circuit symbols 375 Designations in circuit plans 377 Circuit diagrams 378 Sensors 379 Safety precautions 380
73 GRAFCET Important basic terms 382 Steps transitions 383 Actions 384 Branchings 386
74 Programmable logical controllers PLC PLC programming languages overview 388 Ladder diagram (LD) 389 Instruction list (IL) 390 PLC programming languages
comparison 391 Programming example 392
75 Hydraulics pneumatics
Circuit symbols 393 Proportional valves 395 Circuit diagrams 396 Pneumatic control 397 Electro-pneumatic control 398 Electro-hydraulic control 399 Hydraulic fluids 400 Pneumatic cylinders 401
Hydraulic pumps 402 Tubes 403
76 Handling and robot systems
Coordinate systems and axes 404 Robot designs 405 Grippers job safety 406
77 CNC technology Coordinate axes 407 Program structure 408 Tool offset and cutter compensation 409 Program structure according to DIN 410 Program structure according to PAL 412 PAL functions for lathes 413 PAL cycles for lathes 414
PAL functions for milling machines 417 PAL cycles for milling machines 418
67 Separation by cutting Cutting force 325 Cutting tool 326
68 Forming Bending 330 Deep drawing 334
69 Injection molding Injection molding tools 338 Shrinkage cooling batching 341
610 Joining Welding processes 343 Weld preparation 345 Gas-shielded welding 346 Arc welding 348 Beam cutting 350 Gas cylinders identification 352
Brazing bonded joints 354
611 Workplace and environmental protection Safety colors 359 Warning signs safety signs 360 Sound and noise 366
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
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MS
ME
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A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
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A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
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Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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8
Standards and other Regulations
Standardization and standards terms
Standardization is the systematic achievement of uniformity of material and non-material objects such as compo-nents calculation methods process flows and services for the benefit of the general public
Types of standards and regulations (selection)
Standards term Example Explanation
Standard DIN 7157 A standard is the published work of standardization e g the selection of particularfits in DIN 7157
Part DIN 30910-2 Standards can comprise several parts associated with each other The part num-bers are appended to the main standard number with hyphens DIN 30910-2describes sintered materials for filters for example whereas Part 3 and 4 deal withsintered materials for bearings and formed parts
Supplement DIN 743Suppl 1
A supplement contains information for a standard however no additional specifi-cations The supplement DIN 743 Suppl 1 for example contains applicationexamples of load capacity calculations for shafts and axles described in DIN 743
Draft E DIN 743(2008-10)
Draft standards are made available to the public for examination and comment-ing The planned new version of DIN 743 on load-bearing calculations of shaftsand axes for example has been published since October 2008 as Draft E DIN 743
Preliminary
standard
DIN V 66304
(1991-04)
A preliminary standard contains the results of standardization which have not
been released as a standard because of certain provisos DIN V 66304 for examplediscusses a format for exchange of standard part data for computer-aided design
Issue date DIN 76-1(2004-06)
Date of publication which is made public in the DIN publication guide this is thedate at which time the standard becomes valid DIN 76-1 which sets undercutsfor metric ISO threads has been valid since June 2004 for example
Type Abbreviation Explanation Purpose and contents
InternationalStandards(ISO standards)
ISO International Organization forStandardization Geneva (O and Sare reversed in the abbreviation)
Simplifies the international exchange ofgoods and services as well as cooperationin scientific technical and economic areas
EuropeanStandards
(EN standards)
EN European Committee for Standardi-zation (Comiteacute Europeacuteen de
Normalisation) Brussels
Technical harmonization and the associatedreduction of trade barriers for the advance-
ment of the European market and the coa-lescence of Europe
GermanStandards(DIN standards)
DIN Deutsches Institut fuumlr Normung eVBerlin (German Institute forStandardization)
National standardization facilitates rational-ization quality assurance environmentalprotection and common understanding ineconomics technology science manage-ment and public relations
DIN EN European standard for which theGerman version has attained thestatus of a German standard
DIN ISO German standard for which an inter-national standard has been adoptedwithout change
DIN EN ISO European standard for which aninternational standard has been
adopted unchanged and the Ger-man version has the status of a Ger-man standard
DIN VDE Printed publication of the VDEwhich has the status of a Germanstandard
VDI Guidelines VDI Verein Deutscher Ingenieure eVDuumlsseldorf (Association of GermanEngineers)
These guidelines give an account of the cur-rent state of the art in specific subject areasand contain for example concrete procedu-ral guidelines for the performing calculationsor designing processes in mechanical orelectrical engineering
VDE printedpublications
VDE Verband Deutscher ElektrotechnikereV Frankfurt (Association forElectrical Electronic amp InformationTechnologies)
DGQpublications DGQ Deutsche Gesellschaft fuumlr QualitaumlteV Frankfurt (German Society forQuality)
Recommendations in the area of qualitytechnology
REFA sheets REFA Association for Work Design Indus-trial Organization and CorporateDevelopment REFA eV Darmstadt
Recommendations in the area of produc-tion and work planning
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9Table of Contents
1 Mathematics
M
P
TD
MS
ME
PE
A
S
Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
P
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MS
ME
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A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
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9Table of Contents
1 Mathematics
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A
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Quantity SymbolUnit
Name Symbol
Lengths Œ meter m
11 Units of measurement
SI base quantities and base units 10 Derived quantities and their units 11
Non-SI units 12
12 Formulas
Formula symbols mathematical symbols 13 Formulas equations graphs 14
Transformation of formulas 15 Quantities and units 16 Calculation with quantities 17 Percentage and interest calculation 17
13 Angels and triangels
Types of angels sum of angels in a triangle 18 Theorem of intersecting lines 18 Functions of right triangles 19 Functions of oblique triangles 19
14 Lengths
Division of lengths 20Spring wire lengths 21
Rough lengths 21
15 Areas
Angular areas 22 Triangle polygon circle 23 Circular sector circular segment 24 Ellipse 24
16 Volume and surface area
Cube cylinder pyramid 25
Truncated pyramid cone truncated cone sphere 26 Volumes of composite solids 27
17 Mass
General calculations 27 Linear mass density 27 Area mass density 27
18 Centroids
Centroids of lines 28 Centroids of plane areas 28
sine = opposite side hypotenuse
cosine = adjacent side
hypotenuse
tangent = opposite side
adjacent side
xsx
y s
y
S1
S2
S
m inkg
m 1 m
d
A d h d
s = +p
p
middot middot middot middot
24
2
Surface area
AM = p middot d middot h
Lateral surface area
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10 11 Units of measurement
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Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
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Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
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Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
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FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
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Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
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Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
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S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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10 11 Units of measurement
M
P
D
S
E
E
A
Units of measurement
SI1) Base quantities and base units cf DIN 1301-1 (2010-10) -2 (1978-02) -3 (1979-10)
Base quantities derived quantities and their units
Length Area Volume Angle
Mechanics
Basequantity
Length Mass TimeElectriccurrent
Thermo-dynamic
temperature
Amount ofsubstance
Luminousintensity
Base
units meter
kilo-
gram second ampere kelvin mole candela
Unitsymbol
m
1) The units for measurement are defined in the International System of Units SI (Systegraveme International drsquoUniteacutes) Itis based on the seven basic units (SI units) from which other units are derived
kg s A K mol cd
Quantity
Symbol
Œ meter m 1 m = 10 dm = 100 cm = 1000 mm1 mm = 1000 microm1 km = 1000 m
1 inch = 254 mmIn aviation and nautical applicationsthe following applies1 international nautical mile = 1852 m
Unit
Name SymbolRelationship
RemarksExamples of application
Length
A S square meter
arehectare
m2
aha
1 m2 = 10 000 cm2
= 1 000 000 mm2
1 a = 100 m2
1 ha = 100 a = 10 000 m2
100 ha = 1 km2
Symbol S only for cross-sectionalareas
Are and hectare only for land
Area
V cubic meter
liter
m3
mdash L
1 m3 = 1000 dm3
= 1 000 000 cm3
1 mdash = 1 L = 1 dm3 = 10 d mdash =0001 m3
1 m mdash = 1 cm
3
Mostly for fluids and gases
Volume
a b g hellip radian
degrees
minutes
seconds
rad
deg
+
1 rad = 1 mm = 572957hellipdeg= 180deg p
1deg = p
rad = 60 180
1 = 1deg60 = 60+1+ = 1 60 = 1deg3600
1 rad is the angle formed by theintersection of a circle around thecenter of 1 m radius with an arc of 1m lengthIn technical calculations instead ofa = 33deg 17 276+ better use is a =33291deg
Planeangle(angle)
asymp steradian sr 1 sr = 1 m2 m2 An object whose extension measures1 rad in one direction and perpendicu-larly to this also 1 rad covers a solidangle of 1 sr
Solid angle
m kilogramgram
megagrammetric ton
kgg
Mgt
1 kg = 1000 g1 g = 1000 mg
1 metric t = 1000 kg = 1 Mg02 g = 1 ct
Mass in the sense of a scale result or aweight is a quantity of the type of mass(unit kg)
Mass for precious stones in carat (ct)
Mass
m kilogramper meter
kgm 1 kgm = 1 gmm For calculating the mass of bars pro-files pipes
Linear massdensity
m+ kilogramper squaremeter
kgm2 1 kgm2 = 01 gcm2 To calculate the mass of sheet metalArea massdensity
r kilogramper cubicmeter
kgm3 1000 kgm3 = 1 metric tm3 = 1 kgdm3
= 1 gcm3
= 1 gml
= 1 mgmm3
The density is a quantity independentof location
Density
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1111 Units of measurement
M
P
TD
MS
ME
PE
A
S
Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
M
P
D
S
E
E
A
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
M
P
TD
MS
ME
PE
A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
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Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
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1111 Units of measurement
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Units of measurement
J kilogram xsquaremeter
kg middot m2 The following applies for ahomogenous bodyJ = r middot r 2 middot V
The moment of inertia (2nd momentof mass) is dependent upon the totalmass of the body as well as its formand the position of the axis of rotation
Momentof inertia 2ndMoment ofmass
Quantities and Units (continued)
Mechanics
F
F G G
newton N1 N = 1
kg middot m= 1
J s2 m
1 MN = 103 kN = 1 000 000 N
The force 1 N effects a change invelocity of 1 ms in 1 s in a 1 kg mass
Force
Weight
M M bT
newton xmeter
N middot m1 N middot m = 1
kg middot m2
s21 N middot m is the moment that a force of1 N effects with a lever arm of 1 m
TorqueBending momTorsionalmom
p kilogram xmeter
per second
kgmiddot ms 1 kg middot ms = 1 N middot s The momentum is the product of themass times velocity It has the direc-
tion of the velocity
Momentum
p
s t
pascal
newtonper squaremillimeter
Pa
Nmm2
1 Pa = 1 Nm2 = 001 mbar1 bar = 100 000 Nm2
= 10 Ncm2 = 105 Pa1 mbar = 1 hPa1 Nmm2 = 10 bar = 1 MNm2
= 1 MPa1 daNcm2 = 01 Nmm2
Pressure refers to the force per unitarea For gage pressure the symbol p g is used (DIN 1314)
1 bar = 145 psi (pounds per squareinch )
Pressure
Mechanicalstress
I meter to thefourth powercentimeterto the fourthpower
m4
cm4
1 m4 = 100 000 000 cm4 Previously Geometrical moment ofinertia
Secondmoment ofarea
E W joule J 1 J = 1 Nmiddot m = 1 Wmiddot s = 1 kgmiddot m2 s2
Joule for all forms of energy kWmiddot hpreferred for electrical energy
Energy WorkQuantity ofheat
PG
watt W 1 W = 1 Js = 1 N middot ms = 1 V middot A = 1 m2 middot kgs3
Power describes the work which isachieved within a specific time
PowerHeat flux
Time
t seconds minuteshoursdayyear
sminhda
1 min = 60 s1 h = 60 min = 3600 s1 d = 24 h = 86 400 s
3 h means a time span (3 hrs)3h means a point in time (3 orsquoclock)If points in time are written in mixedform eg 3h24m10s the symbol mincan be shortened to m
Time Time spanDuration
f v hertz Hz 1 Hz = 1s 1 Hz Dagger 1 cycle in 1 secondFrequency
n 1 per second
1 per minute
1s
1min
1s = 60min = 60 minndash1
1min = 1 minndash1 = 1
60 s
The number of revolutions per unit oftime gives the revolution frequencyalso called rpm
RotationalspeedRotationalfrequency
v meters persecond
meters perminute
kilometers perhour
ms
mmin
kmh
1 ms = 60 mmin= 36 kmh
1 mmin =1 m
60 s
1 kmh =1 m
36 s
Nautical velocity in knots (kn)
1 kn = 1852 kmh
miles per hour = 1 mileh = 1 mph1 mph = 160934 kmh
Velocity
w 1 per second
radians persecond
1s
rads
w = 2 p middot n For a rpm of n = 2s the angular
velocity w = 4 p s
Angular-
velocity
a g meters persecondsquared
ms21 ms2 =
1 ms 1 s
Symbol g only for acceleration due togravity
g = 981 ms2 fi 10 ms2
Acceleration
QuantitySym-bol
Unit
Name SymbolRelationship
RemarksExamples of application
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12 11 Units of measurement
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Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
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FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
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A
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Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
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Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
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Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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12 11 Units of measurement
M
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Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Units of measurement
I
E
R
G
amperevolt
ohm
siemens
AV
O
S
1 V = 1 W1 A = 1 JC
1 O = 1 V1 A
1 S = 1 A1 V = 1 O
The motion of an electrical charge iscalled current The electromotive forceis equal to the potential difference be-tween two points in an electric field Thereciprocal of the electrical resistance iscalled the electrical conductivity
Frequency of public electric utilityEU 50 Hz USACanada 60 Hz
Electric currentElectromotiveforceElectricalresistance
Electricalconductance
r
g k
ohm xmeter
siemensper meter
O middotm
Sm
10ndash6 O middot m = 1 O middot mm2 m r
k= 1 2
in mm
m
Ω middot
kr
= 1
2in
m
mmΩ middot
Specificresistance
Conductivity
f hertz Hz 1 Hz = 1s1000 Hz = 1 kHz
Frequency
In atomic and nuclear physics the uniteV (electron volt) is used
W joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 36 MJ1 W middot h = 36 kJ
Electrical energy
The angle between current and voltagein inductive or capacitive load
j ndash ndash for alternating current
cos j = P
U middot I
Phasedifference
In electrical power engineeringApparent power S in V middot A
P watt W 1 W = 1 Js = 1 Nmiddotms = 1 V middot A
Power Effective power
E F
Q C
Q
U Q t = = = middotI
E Q C L
volts per metercoulombfarad
henry
VmCFH
1 C = 1 A middot 1 s 1 A middot h = 36 kC1 F = 1 CV1 H = 1 V middot sA
Elect field strength
Elect chargeElect capacitanceinductance
Quantities and units (continued)
Electricity and Magnetism
QuantitySym-bol
Unit
NameRelationship
RemarksExamples of application
Sym-bol
Non-SI units
T Q
t h
kelvin
degreesCelsius
K
degC
0 K = ndash 27315 degC
0 degC = 27315 K0 degC = 32 degF0 degF = ndash 1777 degC
Kelvin (K) and degrees Celsius (degC) areused for temperatures and tempera-ture differencest = T ndash T 0 T 0 = 27315 Kdegrees Fahrenheit (degF) 18 degF = 1 degC
Thermo-
dynamic
temperature
Celsiustemperature
Q joule J 1 J = 1 W middot s = 1 N middot m1 kW middot h = 3 600 000 J = 36 MJ
1 kcal Dagger 41868 kJQuantity ofheat
H u joule per
kilogramJoule percubic meter
Jkg
Jm3
1 MJkg = 1 000 000 Jkg
1 MJm3 = 1 000 000 Jm3
Thermal energy released per kg fuel
minus the heat of vaporization of thewater vapor contained in the exhaustgases
Net calorific
value
Length Area Volume Mass Energy Power
1 inch (in) = 254 mm
1 foot (ft) = 03048 m
1 yard(yd) = 09144 m
1 nautical
mile = 1852 km1 mile = 16093 km
1 sqin = 6452 cm2
1 sqft = 929 dm2
1 sqyd = 08361 m2
1 acre = 4046856m2
Pressure
1 bar = 145poundin2
1 Nmm2 = 145038poundin2
1 cuin = 1639 cm3
1 cuft = 2832 dm3
1 cuyd = 7646 dm3
1 gallon1 (US) = 3785 mdash
1 gallon1 (UK) = 4546 mdash
1 barrel = 1588 mdash
1 oz = 2835 g
1 lb = 4536 g
1 t = 1000 kg
1 shortton = 9072 kg
1 Karat = 02 g
1 poundin3 = 2768gcm3
1 PSh = 0735 kWh
1 PS = 735 W
1 kcal = 41868 Ws
1 kcal = 1166 Wh
1 kpms = 9807 W
1 Btu = 1055 Ws
1 hp = 7457 W
Sym-bol
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1312 Formulas
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A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
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A
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Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
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Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
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1913 Angels and Triangels
M
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Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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1312 Formulas
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A
S
FormulaMeaning
symbol
Œ Length
w Width
h Height
s Linear distance
r R Radius
d D Diameter
A S Area Cross-sectional area
V Volume
a b g Planar angle
W Solid angle
l Wave length
FormulaMeaning
symbol
FormulaMeaning
symbol
Formula symbols Mathematical symbols
Formula symbols cf DIN 1304-1 (1994-03)
Length Area Volume Angle
t Time Duration
T Cycle duration
n Revolution frequency
Speed
f v Frequency
v u Velocity
w Angular velocity
a Acceleration
g Gravitational acceleration
a Angular acceleration
Qmiddot
V q v Volumetric flow rate
Time
Q Electric charge Quantity ofelectricity
E Electromotive force C Capacitance I Electric current
L Inductance R Resistance r Specific resistance g k Electrical conductivity
X Reactance Z Impedance j Phase difference N Number of turns
Electricity
T Q Thermodynamictemperature
DT Dt Dh Temperature difference t h Celsius temperature
a mdash a Coefficient of linearexpansion
Q Heat Quantity of heat l Thermal conductivity a Heat transition coefficient k Heat transmission
coefficient
G middotQ Heat flow a Thermal diffusivity c Specific heat H net Net calorific value
Heat
E Illuminance f Focal length n Refractive index
I Luminous intensity Q W Radiant energy
Light Electromagnetic radiation
p Acoustic pressurec Acoustic velocity
fi approx equals aroundabout
Dagger equivalent to hellip and so on etc 6 infinity
= equal to Iuml not equal to ==
def is equal to by definition lt less than
permil less than or equal to gt greater than
rsaquo greater than or equal to + plus
ndash minus middot times multiplied by ndash divide over divided by per to S sigma (summation)
proportional an
a to the n-th power the n-th
power of a 0 3
square root of
n0 3 n-th root of
aeligxaelig absolute value of x o perpendicular to oslash is parallel to parallel in the same direction
ordm parallel in the opposite direction angle
trade triangle 9 congruent to
Dx delta x (difference betweentwo values)
percent of a hundred permil per mil of a thousand
log logarithm (general)
lg common logarithm
ln natural logarithm
e Euler number (e = 2718281hellip)
sin sine cos cosine tan tangent cot cotangent
() [] parentheses brackets open and closed
p pi (circle constant =314159 hellip)
AB line segment AB
AB pound arc AB a a+ a prime a double prime a1 a2 a sub 1 a sub 2
LP Acoustic pressure level I Sound intensity
N Loudness LN Loudness level
Acoustics
m Mass
m Linear mass density
m+ Area mass density
r Density
J Moment of inertia
p Pressure
p abs Absolute pressure
p amb Ambient pressure
p g Gage pressure
F Force
F W W Gravitational force WeightM Torque
T Torsional moment
M b Bending moment
s Normal stress
t Shear stress
e Normal strain
E Modulus of elasticity
G Shear modulus
micro f Coefficient of friction
W Section modulus
I Second moment of an area
W E Work Energy
W p E p Potential energy
W k E k Kinetic energy
P Power
n Efficiency
Mechanics
Mathematical symbols cf DIN 1302 (1999-12)
MathSpoken
symbol
MathSpoken
symbol
MathSpoken
symbol
ordm
ordmordm
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
8102019 Mechanical and Metal Trade Handbook 01
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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1512 Formulas
M
P
TD
MS
ME
PE
A
S
Example formula L = Œ1 + Œ2 transformation to find Œ2
Example formula A = Œ middot b transformation to find Œ
Example formula n =
ŒŒ1 + s
transformation to find s
Example formula c =
a2 + b2 transformation to find a
Transformation of formulas
Transformation of formulas
Transformations of sums
Transformations of products
Transformations of fractions
Transformations of roots
Formulas and numerical equations are transformed so that the quantity to be obtainedstands alone on the left side of the equation The value of the left side and right side ofthe formula must not change during the transformation The following rule applies to allsteps of the formula transformation
Changes applied to theleft formula side
=Changes applied to theright formula side
To be able to trace each step of the transformation it is useful to mark it to the right nextto the formula
aeligmiddot t int both sides of the formula are multiplied by t
aelig F int both sides of the formula are divided by F
Formula
P =
F middot s
t
left side right sideof the = of theformula formula
1222222
122222
1
1
1
1
3
3
4
4
2
2
2
2
3
3
4
4
5
5
6
6
L = Œ1 + Œ2 aeligndash Œ1
A = Œ middot b aelig b
n =
ŒŒ1 + s
aeligmiddot (Œ1 + s )
c = a 2 + b 2 aelig( )2
L ndash Œ1 = Œ2
A
b = Œ
n middot Œ1 ndash n middot Œ1 + n middot s = Œ ndash n middot Œ1 aelig n
a 2 = c 2 ndash b 2 aelig12
L ndash Œ1 = Œ1+ Œ2 ndash Œ1
A
b = Œ middot b
b
n middot (Œ1 + s ) = Œ middot (Œ1 + s )
(Œ1 + s )
c 2 = a 2 + b 2 aeligndash b 2
n middot Œ1 + n middot s = Œ aeligndash n middot Œ1
c 2 ndash b 2 = a 2 + b 2 ndash b 2
Œ2 = L ndash Œ1
Œ = A
b
s middot n
n
= Œ ndash n middot Œ1
n
a 2 = c 2 ndash b 2
s =
Œ ndash n middot Œ1n
a = c 2 ndash b 2
subtract Œ1
divide by b
multiply by (Œ1 + s )
square equation
invert both sides
invert both sides
subtractdivide by n
extract the root
performsubtraction
cancel b
cancel (Œ1 + s ) on theright side
solve the term inbrackets
subtract b 2
subtract ndash n middot Œ1
subtractinvert both sides
transformedformula
transformedformula
cancel n
simplify theexpression
transformedformula
transformedformula
122 1222222
1222222
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
8102019 Mechanical and Metal Trade Handbook 01
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20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
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16 12 Formulas
M
P
D
S
E
E
A
Conversion of units
Quantities and units
Decimal multiples or factors of units cf DIN 1301-1 (2004-10)
Conversion factors for units (excerpt)
Numerical values and units
Prefix Powerof ten
Mathematicaldesignation
ExamplesSymbol Name
T tera 1012 trillion 12 000 000 000 000 N = 12 middot 1012 N = 12 TN (teranewtons)
G giga 109 billion 45 000 000 000 W = 45 middot 109 W = 45 GW (gigawatts)
M mega 106 million 8 500 000 V = 85 middot 106 V = 85 MV (megavolts)
k kilo 103 thousand 12 600 W = 126 middot 103 W = 126 kW (kilowatts)
h hecto 102 hundred 500 mdash
= 5 middot 102 mdash
= 5 h mdash
(hectoliters)
da deca 101 ten 32 m = 32 middot 101 m = 32 dam (decameters)
ndash ndash 100 one 15 m = 15 middot 100 m
d deci 10ndash1 tenth 05 mdash = 5 middot 10ndash1 mdash = 5 d mdash (deciliters)
c centi 10ndash2 hundredth 025 m = 25 middot 10ndash2 m = 25 cm (centimeters)
m milli 10ndash3 thousandth 0375 A = 375 middot 10ndash3 A = 375 mA (milliamperes)
micro micro 10ndash6 millionth 0000 052 m = 52 middot 10ndash6 m = 52 microm (micrometers)
n nano 10ndash9 billionth 0000 000 075 m = 75 middot 10ndash9 m = 75 nm (nanometers)
p pico 10ndash12 trillionth 0000 000 000 006 F = 6 middot 10ndash12 F = 6 pF (picofarads)
Quantity Conversion factors e g Quantity Conversion factors e g
Length 1 =
10 mm1 cm
=
1000 mm1 m
=
1 m1000 mm
=
1 km1000 m
Time 1 =
60 min1 h
=
3600 s1 h
=
60 s1 min
=
1 min60 s
Area 1 =
100 mm2
1 cm2 =
100 cm2
1 dm2 =
Angle 1 =
60rsquo1deg
=
60rsquorsquo1rsquo
=
3600rsquorsquo1deg
=
1deg60 s
Volume 1 =
1000 mm3
1 cm3 =
1000 cm3
1 dm3 = Inch 1 inch = 254 mm 1 mm =
1
254 inches
Calculations with physical units are only possible if these units refer to the same base in this calculation Whensolving mathematical problems units often must be converted to basic units e g mm to m s to h mm2 to m2 Thisis done with the help of conversion factors that represent the value 1 (coherent units)
Physical quantities e g 125 mm consist of a
bull numerical value which is determined by measurement or calculation and abull unit e g m kg
Units are standardized in accordance with DIN 1301-1 (page 10)Very large or very small numerical values may be represented in a simplified wayas decimal multiples or factors with the help of prefixes e g 0004 mm = 4 microm
Physical quantity
10 mm
Numerical Unitvalue
Convert volume V = 3416 mm3 to cm3
Volume V is multiplied by a conversion factor Its numerator has the unit cm3 and its denominator the unitmm3
V = 3416 mm3 =1 cm3 middot 3416 mm3
1000 mm3 =
3416 cm3
1000 = 3416 cm 3
The angle size specification a = 42deg 16rsquo is to be expressed in degrees (deg)
The partial angle 16rsquo must be converted to degrees (deg) The value is multiplied by a conversion factor thenumerator of which has the unit degree (deg) and the denominator the unit minute (rsquo)
983137 = 42deg + 16rsquo middot1deg60rsquo
= 42deg +16 middot 1deg
60 = 42deg + 0267deg = 42267deg
1st example
2nd example
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8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1820
aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
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1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 2020
20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1720
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1820
aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1920
1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 2020
20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1820
aring iquest
b a
c
copy
Sum of angles in a triangle
iquest
para
aring copy
g
g2
g1
18 13 Angels and Triangels
M
P
D
S
E
E
A
b2
a2
c2
abc
G 0 3
P1 X
Z
P2
RK I
Types of angles Theorem of intersecting lines Angles in a triangle Pythagorean theorem
Types of angles
Theorem of intersecting lines
Pythagorean theorem
D d
a2
daggerto
daggerti
a1
b 1
b 2
g straight line
g 1 g 2 parallel straight lines
a b corresponding angles
b d opposite anglesa d alternate angles
a g adjacent angles
If two parallels are intersected by a straight line thereare geometrical interrelationships between the resultingangles
If two intersecting lines are intercepted by a pair ofparallels the resulting segments form equal ratios
a = b
Corresponding angles
b = d
Opposite angles
a = d
Alternate angles
a + g = 180deg
Adjacent angles
a
b
a
b
1
1
2
2
=
b
d
b
D
1 2=
a
a
b
b
d
D
1
2
1
2
2
2
= =
Theorem ofintersecting lines
a + b + g = 180deg
Sum of angles ina triangle
In every triangle thesum of the interiorangles equals 180deg
D = 40 mm d = 30 mmtta = 135 Nmm2 tti =
Example
a = 21deg b = 95deg g =
983143 = 180deg ndash a ndash b = 180deg ndash 21deg ndash 95deg = 64deg
Example
t
t
t
tti
to
tito
= rArr =
=
d
D
d
D
middot
135 Nmm middot 30 mm
4
2
00 mm= 101 Nmm225
tto outer torsional stress
tti inner torsional stress
a b c sides of the triangle
a b g angles in the triangle
c a b
c a
=
=
35 mm = 21 mm =
= mm)2 2b ndash ( ndash2 35 ((21 mm)2= 28mm
1st example
c 2 = a 2 + b 2
Length of thehypotenuse
In a right triangle the square on the hypotenuse is equalto the sum of the squares on the sides meeting the rightangle
a sideb sidec hypotenuse
CNC programm with R = 50 mm and I = 25 mmK =
2nd example
c a b
R K
K R
2 2 2
2 2 2
2 2 2 250 25
= +
= +
= =
=
I
I ndash ndashmm mm2 2
K 433mm
c a b = +2 2
Square on thehypotenuse
a c b = 2 2ndash
b c a = 2 2ndash
Length of the sides
meeting the right angle
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1920
1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 2020
20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 1920
1913 Angels and Triangels
M
P
TD
MS
ME
PE
A
S
Functions of triangles
c hypotenuse
b adjacent side of aring
a adjacentside of aringaring
b opposite side of iquest
aadjacentside of iquest
iquestc hypotenuse
F
L3=140mm
L
L
1 = 1 5 0 m
m
L
2 = 3 0 m m
aring
szlig
copy
aring iquest
b a
c
Fd
Fz
F
4 0
1 2
szlig = 3 8
aring = 4 0
1 2
copy = 1 0
2
Fd
Fz
F
Diagram of forces
Functions of right triangles (trigonometric functions)
Functions of oblique triangles (law of sines law of cosines)
sine = opposite side
hypotenuse
cosine = adjacent side hypotenuse
tangent = opposite side
adjacent side
Trigonometric functionsc hypotenuse (longest side)
a b sidesndash b is the adjacent side of a ndash a is the opposite side of a
a b g angles in the triangle g = 90deg
sin notation of sine
cos notation of cosine
tan notation of tangent
sina sine of angle a
According to the law of sines the ratios of
the sides correspond to the sine of theiropposite angles in the triangle If one sideand two angles are known the other valuescan be calculated with the help of thisfunction
Side a int opposite angle sinaSide b int opposite angle sinbHypothenuse c int opposite angle sing
The calculation of an angle in degrees (deg)
or as a circular measure (rad) is done withthe help of inverse trigonometric functionse g arcsine
L1 = 150 mm L2 = 30 mm L3 = 140 mmangle a =
Angle 983137 = 52deg
1st example
tan a =+
= =L L
L
1 2
3
1 286180 mm
140 mm
L1 = 150 mm L2 = 30 mm a = 52degLength of the shock absorber L =
2nd example
L =+
= =L L1 2
sina
180 mm
sin 52deg22842mm
F = 800 N a = 40deg b = 38deg F z = F d =
The forces are calculated with the help
of the forces diagram
Example
F F F
F
sin sin
middot sin
sina b
b
a= rArr =
=
zz
800 N middot sin38degF z
ssin40deg= 76624 N
F F F
F
sin sin
middot sin
sina g
g
a= rArr =
=
dd
800 N middot sin102F d
degdeg
sin40deg= 121738 N
Relations applying to angle a
sina =
a
c cosa =
b
c tana =
a
b
Relations applying to angle b
sinb =
b
c cosb =
a
c tanb =
b
a
There are many transformationoptions
a =
b middot sinasinb
=
c middot sinasing
b =
a middot sinbsina
=
c middot sinbsing
c =
a middot singsina
=
b middot singsinb
Law of sines
a b c = sina sinb sing
a
sina =
b
sinb =
c
sing
Law of cosines
a 2 = b 2 + c 2 ndash 2 middot b middot c middot cosab 2 = a 2 + c 2 ndash 2 middot a middot c middot cosbc 2 = a 2 + b 2 ndash 2 middot a middot b middot cosg
The calculation of an angle indegrees (deg) or as a circularmeasure (rad) is done with thehelp of inverse trigonometricfunctions e g arcsine
Transformation e g
cosa =
b 2 + c 2 ndash a 2
2 middot b middot c
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 2020
20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A
8102019 Mechanical and Metal Trade Handbook 01
httpslidepdfcomreaderfullmechanical-and-metal-trade-handbook-01 2020
20 14 Lengths
Division of lengths Arc length Composite length
Œ total length n number of holesp spacing
Spacing
Spacing
Sub-dividing lengths
Arc length
L = Œ1 + Œ2 + hellip
Composite length
D outside diameter d inside diameterd m mean diameter t thicknessŒ1 Œ2 section lengths L composite lengtha angle at center
Composite length
D = 360 mm t = 5 mm a = 270deg Œ2 = 70 mmd m = L =
Example (composite length picture left)
Edge distance = spacing
Edge distance) spacing
Example
Œ total length n number of holesp spacing a b edge distances
Example
Number of pieces
Subdividing into pieces Œ bar length s saw cutting widthz number of pieces Œr remaining lengthŒs piece length
Example
Œr = Œ ndash z middot (Œs + s )
Remaining length
Example Torsion spring Œa arc length a angle at centerr radius d diameter
Example
Arc length
M
P
D
S
E
E
A