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—
Mechanical and Metal
Trades Handbook
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Europa No 1910X 2/431
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uj /-?J LEHRM ITTEL EUROPA-TECHNICAL BOOK SERIES
for the M etalworking Trades
Ulrich Fischer Ma x Heinzler Friedrich Naher Heinz Paetzold
Roland Gom eringer Roland Kilgus Stefan Oesterle Andreas Stephan
Mechanical and
Metal Trades
Handbook
2nd English edition
Eur op a-No.: 1910X
VERLAG EUROPA LEHRMITTEL • Nou rney, Vollmer G mbH & Co. KG 3/431
Dusselberger StraBe 23 • 42781 Haan-Gruiten • Germany
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Original title:
Tabellenbuch Metall, 44th edition, 2008
Authors: Dipl.-lng. (FH) Reutlingen
Dipl.-Gwl. MeBstetten
Ulrich Fischer Dipl.-lng. (FH) Wangen im Allgau
Roland Gomeringer
Max Heinzler Dipl.-Gwl. Neckartenzlingen
Roland Kilgus Dipl.-lng. (FH) Balingen
Friedrich Naher Dipl.-lng. Amtzell
Stefan Oesterle Dipl.-lng. (FH) Muhlacker
Heinz Paetzold Dipl.-lng. (FH) Kressbronn
Andreas Stephan
Editor:
Ulrich Fischer, Reutlingen
Graphic design:
Design office of Verlag E uropa-Lehrm ittel, Leinfelden-Echterdingen, 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 p ublisher 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 Gm bH, Burggrafenstr. 6, 10787 Berlin, Germany.
The content of the chapter Program structure of CNC m achines according to PAL (page 386 to 400) complies with
the publications of the PAL Priifungs- und Lehrmittelentwicklungsstelle (Institute for the development of training and
testing material) of the IHK Region Stuttgart (Chamber of Commerce and Industry of the Stuttgart region).
English edition: Mechanical and Metal Trades Handbook
2nd edition, 2010
654321
All printings of this edition may be used concurrently in the classroom since they are unchanged, except for some
corrections to typographical errors and slight changes in standards.
ISBN 13 978-3-8085-1913-4 4/431
Cover design includes a photograph from TESA/Brown & Sharpe, Renens, Sw itzerland
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.
© 2010 by Verlag Europa-Lehrmittel, Nourney, Vollmer GmbH & Co. KG, 42781 Haan-Gruiten, Germany
http://www.europa-lehrmittel.de
Translation: Techni-Translate, 72667 Schlaitdorf, Germany; www.techni-translate.com
Eva Schwarz, 76879 Ottersheim, G ermany; www.technische-uebersetzun gen-eva-schwarz.de
Typesetting: YellowHand GbR, 73257 Kongen, Germany; www.yellowhand.de
Printed by: Media Print Informationstechnologie, D -33100 , Paderborn, Germany
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3
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 programs. 2 Physics
Target Groups p
• Industrial and trade m echanics 33-56
• Tool & Die makers
3 Technical TD
• Machinists drawing
• Millwrights 57-114
• Draftspersons
• Technical Instructors 4 M aterial science
• Apprentices in above trade areas
• Practitioners in trades and ind ustry MS
• Mechanical Engineering students 115-200
Notes for th e user 5 Machine ME
elements
The contents of this book include tables and formulae in 201-272
eight chapters, including Tables of Contents, Subject Index
and Standards Index. 6 Production PE
Engineering
The tables contain the most important guidelines, designs,
types, dimensions and standard values for their subject 273-344
areas.
Units are not specified in the legend s for the formulae if sev-
eral units are possible. However, the calculation examples
for each formula use those u nits norm ally a pplied in practice.
Designation examples, which are included for all standard
parts, materials and drawing designations, are highlighted
by a red arrow (=>).
The Table of Contents in the front of the book is expanded
further at the beginning of each chapter in form of a partial
Table of Contents.
The Subject Index at the end of the book (pages 417-428) is
extensive.
The Standards Index (pages 407-416) lists all the current
standards and regulations cited in the book. In many cases
previou s stand ards are also listed to ease the transition from
older, more familiar standards to new ones.
W e have thoro ugh ly revised the 2 n d e d i t i o n of the Mechan- 7 A u t o m a t i o n an d
ical and Metal Trades Handb ook in line with the 44th ed ition
of the German version Tabellenbuch Metall . The section Information Tech- A
dealing with PAL programming of CNC machine tools was
updated (to the state of 2008) and considerably enhanced. nology 345-406
Special thanks to the Magna Technical Training Centre for
their input into the English translation of this book. Their
assistance has been extremely valuable.
The authors and the publisher will be grateful for any sug- 8 International material S
gestions and constructive comments. comparison ch art,
Standards 407-416
Spring 2010 Authors and publisher
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Table of Contents
1 Mathematics 9
1.1 Numerical tables 10 1.5 Lengths 23
Square root, Area of a circle 11 Calculations in a right triangle 24
Sine, Cosine 12 Sub-dividing lengths, Arc length 25
Tangent, Cotangent Flat lengths, Rough lengths
26
1.2 Trigonometric Functions 1.6 Areas
Angu lar areas 27
Definitions 13 Equilateral triangle, Polygons, 28
Circle
Sine, Cosine, Tangent, Cotangent . . . . 13 Circular areas 29
Laws of sines and cosines 14 1.7 Volu m e and Surface area 30
Cube, Cylinder, Pyramid 31
Angles, Theorem of intersecting Truncated p yram id, Cone,
Truncated cone, Sphere 3311
lines 14 Com posite solids 31
1.3 F u n d a m e n t a l s o f M a t h e m a t i cs 1.8 Mass 32
32
Using brackets, pow ers, roots 15 GLienneearraml caaslscudleantisointys
Area mass density
Equations 16
1.9 Centroids
Powers often, Interest calculation .... 17 Centroids of lines
Centroids of plane areas
Percentage and proportion
1 .4 cSaylmcublaotliso,nsU n i t s 18
Formula symbols, Mathematical
symbols 19
SI quantities and units of 20
measurement 22
Non-SI units
2 Physics 33
2.1 Motion Bending, Torsion 47
SUpneifeodrms oaf nmdaacchcineelesrated mo tion 3354 SShtaatpice mfaocmtoersnti,nSsetrcetniognthmodulus, 48
Mo men t of inertia 49
2.2 Forces Comparison of various
cross-sectional shapes 50
Adding and resolving force vectors ... 36
We ight, Spring force 36
Lever principle, Bearing forces 37 2.7 Thermodynamics
Temperatures, Linear
Torques, Centrifugal force 37 expansion, Shrinkage
Quantity of heat
2.3 Wo rk , Power, Efficiency 38 Heat flux, Heat of com bustion 51
Mechanical work 39 51
Simple machines 40 52
Power and Efficiency
2.8 Electricity
Ohm's Law, Conductor resistance . . . . 53
2.4 FFrriiccttioionnforce 41 Resistor circuits 54
Coefficients of friction 41
Friction in bearings 41 Types of current 55
2.5 Pressure in liquids and gases 42 Electrical wo rk and pow er 56
Pressure, definition and types 42
Buoyancy 42
Pressure changes in gases
43
2 .6 S t re n g t h o f m a t e ri a l s
Load cases, Load types 44
Safety factors, Mechanical
strength properties 45
Tension, Compression, 46
Surface pressure
Shear, Buckling
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Table of Contents 5
3 Technical drawing 57
3 .1 B as ic g e o m e t ri c c o n s t ru c t i o n s 3 .6 M a c h i n e e l e m e n t s
Gear types
Lines and angles 58 Roller bearings 84
Seals 85
Tangents, Circular arcs, Polygons . . . . 59 Retaining rings, Springs 86
87
Inscribed circles, Ellipses, Spirals 60
Cycloids, Involute curves, Parabolas .. 61
3.2 Graphs 3 .7 W o rk p i e c e e l e m e n t s
Cartesian coord inate system
Graph types 62 Bosses, Workpiece edges 88
63
Thread runouts, Thread undercuts . . . 89
3 .3 D ra w i n g e l e m e n t s Threads, Screw joints 90
Fonts
Preferred num bers, Radii, Scales 64 Center holes, Knurls, Undercuts 91
Drawing layout 65
Line types 66 3 .8 W e l d i n g a n d S o l d e ri n g 93
67 Graphical sym bols 95
Dimensioning examples
3.4 Representation 3.9 Surfaces
Projection methods
Views 69 Hardness specifications in drawings .. 97
71
Form deviations, Roughness 98
Sectional views 73 Surface testing , Surface indications .. 99
Hatching 75 3.10 ISO Tolerances an d Fits
3 .5 E n t e rin g d i m e n s i o n s Fundamentals 102
Dimen sioning rules 76 Basic hole and basic shaft systems .. 106
Diameters, Radii, Spheres, Chamfers, General Tolerances, Roller
Inclines, Tapers, Arc dime nsion s 78 bearing fits 110
Tolerance specifications 80 Fit recomm endations 111
Types of dime nsioning 81 Geometric tolerancing 112
Simp lified presentation in drawings .. 83 GD&T (Geometric
Dimensioning & Tolerancing) 113
4 Materials science 116 4 .7 F o u n d ry t e c h n o l o g y 115
Patterns, Pattern equ ipm ent
4.1 Materials 117 Shrinkage allowances, 162
Material characteristics of solids 118 Dimensional tolerances
Material characteristics of liquids 163
and gases
Periodic table of the elemen ts 4.8 Light alloys , Overview of Al alloys .. 164
4 .2 D e s i g n a t i o n s y s t e m f o r s t e el s Wrought aluminum alloys 166
Definition a nd classification of steel . 120 Alu min um casting alloys 168
Material codes, Designation 121 Alu m inum profiles 169
4.3 Steel typ es, Over view 126 Magnesium and titanium alloys 172
Structural steels 128
4.9 Heavy non -ferrous m etals,
Case hardened, quenched and tem- ODevesirgvnieawtion system 117734
pered, nitrided, free c utting steels . . . 132 Copper alloys 175
Tool steels 135 4 .1 0 O t h e r m e t a l l i c m a t e ri al s 177
Com posite m aterials, 178
Stainless steels, Spring steels 136 Ceramic materials 1 79
Sintered metals 182
4.4 Finished steel prod ucts 139 184
Sheet, strip, pipes 143 4.11 Plastics, Ov erview 186
Profiles Thermoplastics
Thermoset plastics, Elastomers
4 .5 H e a t t re a t m e n t 153 Plastics proce ssing
Iron-Carbon phase diagram 154
Processes
4.6 Cast iron m aterials
CDleassisginfiacatitoionn, Material codes 115589 4 .1 2 MOvaet ervriiaelwt e st i n g m e t h o d s, 188
Cast iron 160 Tensile tes ting 190
Malleable cast iron, Cast steel 161 Hardness test 192
4.13 Corrosion, Corrosion protection . . 196
4.14 Hazardo us m aterials 197
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6 Table of Contents
5 Machine elements 201
5.1 Threads ov erview ) 2 02 Grooved pins, Grooved drive studs,
Me tric ISO threads 204
W hitw orth threads, Pipe threads 206 Clevis pins 238
Trapezoidal and buttress threads 207
Thread tolerances 208 5 .7 S h a f t -h u b c o n n e c t i o n s 239
Tapered and feather keys 240
5 .2 B o l t s a n d s c re w s o v e rv i e w ) 209 Parallel and woo druff keys
Designations, strength 210
Hexagon head bolts & screws 212 Splined shafts, Blind rivets 241
Other bolts & screws 215 Tool tapers 242
Screw join t calculations 221
Locking fasteners 222 5 .8 S p ri n g s , c o m p o n e n t s o f j i g s 244
W idths across flats, Bolt and and tools 247
screw drive systems 223 Springs 251
2 24 Drill bushings
5.3 Coun tersink s Standard stamp ing parts
Countersinks for countersunk 224
head screws 5.9 Drive elemen ts 253
Belts 256
Gears 259
Transm ission ratios
5.4 CNouutsnteorvboerrevsiefwor) cap screws 222256 5.10 SBpeeaerdinggrsaph 260
Designations, Strength 227 Plain bearings (overview)
Hexagon nuts 228 Plain bearing bush ings 261
Other nuts 231 Antifriction bearings (overview) 262
Types of roller bearings 263
5.5 Washers ov erview ) 2 33 Retaining rings 265
Sealing elements 269
Flat wash ers 234 Lubricating oils 270
Lubricating greases 271
HV Clevis pin, Conical spring washers . 235 272
5.6 Pins and clevis pins ov erview ) . . . 236
Dow el pins, Taper pins, Spring pins . 237
6 Production Engineering 274 Shearing 273
276 Location of punch holder shank
6 .1 Q u a l i t y m a n a g e m e n t 277 316
279 317
Standards, Terminology 281
Quality planning, Quality testing 6.6 Forming 318
Statistical analysis Bending 320
Statistical process control Deep drawing
Process capa bility
6.7 Joining
6 .2 P ro d u c t i o n p l a n n i n g We lding processes 322
Weld preparation 323
Time accou nting according to REFA . 282 Gas we lding 324
Gas shielded metal arc we lding 325
Cost accounting 284
Mac hine hou rly rates 285
6 .3 M a c h i n i n g p ro c es s e s Arc welding 327
Thermal cutting 329
Productive time 287 Identification of gas cylinders 331
Soldering and brazing 333
Machining coolants 292 Adhesive bonding 336
Cutting tool materials, Inserts,
Tool holders 294
Forces and pow er 298 6 .8 W o rk p l a c e s a f et y a n d e n v i ro n m e n t a l
Cutting data: Drilling, Reaming, protection
Turning 301 Proh ibitive signs 338
Cutting data: Taper turn ing 304 Warning signs 339
Cutting data: Milling 305 Mandatory signs,
Indexing 307 Escape routes and rescue signs 340
Cutting data: Grinding and honing .. 308 Information signs 341
6 .4 M a t e ri a l re m o v a l Danger symbols 342
Cutting data
Processes 313 Identification of pipe lines 343
314
Sound and noise 344
6 .5 S e p a ra t io n b y c u t t i n g 315
Cutting forces
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Table of Contents
7 Automation and Information 345
7 .1 B as ic t e rm i n o l o g y f o r c o n t ro l Structured text (ST) 374
Instruction list 375
engineering Simple functions 376
Basic terminology, Code letters, 7 .6 H a n d l i n g a n d ro b o t s y s t e m s 378
Coordinate systems and axes
Symbols 346
Analog controllers 348
Discontinuous and digital controllers .. 349
Binary logic 350 GRoribpoptedrse,sjiogbnssafety 337890
7.2 Electrical circ uits 7 .7 N u m e ri c a l C o n t ro l N C) t e c h n o l o g y
Circuit sym bols
Designations in circuit diagrams 351 Coordinate systems 381
Circuit diagram s 353
Sensors 354 Program structure according to DIN .. 382
Protective precautions 355
356 Tool offset and Cutter compe nsation . 383
Machining mo tions as per DIN 384
Mach ining mo tions as per PAL
7 .3 F u n c t i o n c h a rt s a n d f u n c t i o n d i a g ra m s (German association) 386
Function charts 358 PAL programm ing system for turning . 388
Function diagrams 361 PAL program ming system for milling . 392
7 .4 P n e u m a t i c s a n d h y d ra u l ic s 7 .8 In f o r m a t i o n t e c h n o l o g y
Circuit sym bols 363 Num bering systems 401
Layout of circuit diagrams 365 ASCII code 402
Controllers 366
Hydraulic fluids 368 Program flow chart, Structograms .. 403
Pneumatic cylinders 369
Forces, Speeds, Power 370 WO RD-and EXEL comm ands 4 05
Precision steel tube 372
7 .5 P ro g ra m m a b l e l o g i c c o n t ro l 373
PLC progra mm ing languages 374
Ladder diagra m (LD) 374
Function block language (FBL)
8 Mate rial c hart. Standards 407
8 .1 I n t e rn a t i o n a l m a t e ri a l 407
comparison chart .412
8.2 DIN, DIN EN, ISO etc. stand ards
Subject index 417
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8
Standards and o ther Regulation s
Standardization and Standards term s
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.
Standards term Ex ample Explanation
Standard DIN 7157 Ain sDtaINnd7a1r5d7.is the pub lished result of standardization, e.g. the selection of ce rtain fits
Part DIN 30910-2
The part of a standard associated with other parts wit h the same main num ber. DIN
30910-2 for example describes sintered materials for filters, while Part 3 and 4
describe sintered materials for bearings and formed parts.
Supplement DIN 743 A supplem ent contains infor mation for a standard, how e ver no additional specifi-
Suppl. 1 cations. The supplem ent DIN 743 Supp l. 1, for exam ple, contains a pplication
examples of load capacity calculations for shafts and axles described in DIN 743.
Draft E DIN 6316 A draft standard contains the preliminary finished results of a standardization;
(2007-02) this version of the intended standard is made available to the public for com -
ments. For example, the planned new ver sion of D IN 6316 for goose- neck
clam ps has been available to the pu blic since February 2007 as Draft E
Preliminary DIN V 66304 DIN 6316.
standard (1991-12) A preliminary standard contains the results of standardization which are not released
by DIN as a standard, because of certain provisos. DIN V 66304, for e xam ple, discuss-
es a format for exchange of standard part data for com puter-aided design.
Issue date DIN 76-1 Date of publication which is made public in the DIN publication guide; this is the
(2004-06) date at which time the standard becomes valid. DIN 76-1, which sets undercuts
for metric ISO threads has been valid since June 2004 for example.
Types of Standards and Regulations selection)
Type Abbreviation Ex planation P u rp o s e a n d c o n t e n t s
International ISO International O rganization for Simplifies the international exchange of
Standards Standardization, Geneva (O and S goods and services, as well as cooperation
(ISO standards) are reversed in the abbreviation) in scientific, technical and economic areas.
European EN Eur opean C om mittee for Standar di- Technical harmonization and the associated
Standards zation (Comite Europeen de reduction of trade barriers for the advance-
(EN standards) Normalisation), Brussels ment of the European market and the coa-
lescence of Europe.
DIN Deutsches Institut fur Normung e.V.,
Berlin (Germ an Institute for National standardization facilitates rational-
DIN EN Standardization) ization, quality assurance, environmental
protection and common understanding in
European standard for which the economics, technology, science, manage-
German version has attained the sta- ment and public relations.
tus of a German standard.
German DIN ISO German standard for which an inter-
Standards national standard has been adopted
(DIN standards) without change.
DIN EN ISO European standard for which an
international standard has been
adopted unchanged and the German
version has the status of a German
standard.
DIN VDE Printed publication of the VDE, which
has the s tatus of a German standard.
VDI Guidelines VDI Verein Deutscher Ingenieure e.V., These guidelines give an account of the cur-
Dusseldorf (Society of German rent state of the art in specific subject areas
VDE printed VDE Engineers) and contain, for example, concrete procedu-
publications ral guidelines for the performing calculations
Verband Deutscher Elektrotechniker or designing processes in mechanical or
e.V., Frankfurt (Organization of Ger- electrical engineering.
man Electrical Engineers)
DGQ publica- DGQ DFreauntskfcuhret G(GeseermllsacnhaAfst sfuorc iQautiaolnitafot re.V., tReecchonmolmogeyn.dations in the ar ea of quality
tions Quality)
REFA she ets REFA Association for W ork Design/Work Recommendations in the area of produc-
Structure, Industrial Organization and tion and work planning.
Corporate Development REFA e.V.,
Darmstadt
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d id A ~ 4 Table of Contents 9
1 1.0000 0.7854 1 Mathematics 10
2 1.4142 3.1416 11
1.1 Num erical tables 12
3 1.7321 7.0686 Square root, Area of a circle
Sine, Cosine 13
sine oppo site side Tangent, Cotangent 13
cosine hypotenuse 14
tangent 1.2 Trigono metric Functions 14
cotangent = adjacent side Definitions
hypotenuse Sine, Cosine, Tangent, Cotangent 1156
Laws of sines and cosines 17
oppo site side Angles, Theorem of intersecting lines 18
adjacent side
1.3 Fundamentals of Mathem atics 19
adjacent side 20
opposite side EUqsuinagtiobnrasckets, pow ers, roots 22
Powers of ten, Interest calculation
- + - = - • ( 3 + 5) Percentage and proportion calculations 23
24
XXX 1.4 Sym bols , Units 25
Formula symbo ls, Mathematical sym bols
SI quan tities and units of me asurement 26
Non-SI units 27
28
1.5 Lengths
Calculations in a right triangle
Sub-dividing lengths, Arc length
Flat lengths, Rough lengths
1.6 Areas
Angu lar areas
Equilateral triang le, Polygo ns, Circle
Circular areas
1.7 VCoulbuem, Ce yalinnddeSru, rPfaycreamaridea 29
Truncated pyramid, Cone, Truncated cone, Sphere 30
Composite solids 31
1.8 Mass 31
General calculations 31
Linear mass den sity 31
Area mass density
1.9 Centroids 32
Centroids of lines 32
Centroids of plane areas
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10 Mathematics: 1.1 Numerical tables
Square root, Area o f a circle
ri i/T r lId A~ 4 ri lId rj lId A ~ 4
A~ 4 Li 1a
u U 4u
1 1.0000 0.7854 51 7.1414 2042.82 101 10.049 9 8011.85 151 12.2882 17907.9
2 1.4142 3.1416 52 7.2111 2123.72 102 10.0995 8171.28 152 12.3288 18145.8
3 1.7321 7.0686 53 7.2801 2206.18 103 10.1489 8332.29 153 12.3693 18385.4
4 2.0000 12.5664 54 7.3485 2290.22 104 10.1980 8494.87 154 12.4097 18626.5
5 2.236 1 19.6350 55 7.4162 2375.83 105 10.2470 8659.01 155 12.4499 18869.2
6 2.4495 28.2743 56 7.4833 2463.01 106 10.2956 8824.73 156 12.4900 19113.4
7 2.6458 38.4845 57 7.5498 2551.76 107 10.3441 8992.02 157 12.5300 19359.3
8 2.8284 50.2655 58 7.6158 2642.08 108 10.3923 9160.88 158 12.5698 19606.7
9 3.0000 63.6173 59 7.6811 2733.97 109 10.4403 9331.32 159 12.6095 19855.7
10 3.1623 78.5398 60 7.7460 2827.43 110 10.4881 9503.32 160 12.6491 20106.2
11 3.3166 95.0332 61 7.8102 2922.47 111 10.5357 9676.89 161 12.6886 20358.3
12 3.4641 113.097 62 7.8740 3019.07 112 10.5830 9852.03 162 12.7279 20612.0
13 3.6056 132.732 63 7.9373 3117.25 113 10.6301 10028.7 163 12.7671 20867.2
14 3.7417 153.938 64 8.0000 3216.99 114 10.6771 10207.0 164 12.8062 21124.1
15 3.8730 176.715 65 8.0623 3318.31 115 10.7238 10386.9 165 12.8452 21382.5
16 4.0000 201.062 66 8.1240 3421.19 116 10.7703 10568.3 166 12.8841 21642.4
17 4.1231 226.980 67 8.1854 3525.65 117 10.8167 10751.3 167 12.9228 21904.0
18 4.2426 254.469 68 8.2462 3631.68 118 10.8628 10935.9 168 12.9615 22167.1
19 4.3589 283.529 69 8.3066 3739.28 119 10.9087 11122.0 169 13.0000 22431.8
20 4.4721 314.159 70 8.3666 3848.45 120 10.9545 11309.7 170 13.0384 22698.0
21 4.5826 346.361 71 8.4261 3959.19 121 11.0000 11499.0 171 13.0767 22965.8
22 4.6904 380.133 72 8.4853 4071.50 122 11.0454 11689.9 172 13.1149 23235.2
23 4.7958 415.476 73 8.5440 4185.39 123 11.0905 11882.3 173 13.1529 23506.2
24 4.8990 452.389 74 8.6023 4300.84 124 11.1355 12076.3 174 13.1909 23778.7
25 5.0000 490.874 75 8.6603 4417.86 125 11.1803 12271.8 175 13.2288 24052.8
26 5.0990 530.929 76 8.7178 4536.46 126 11.2250 12469.0 176 13.2665 24328.5
177 13.3041 24605.7
27 5.1962 572.555 77 8.7750 4656.63 127 11.2694 12667.7 178 13.3417 24884.6
28 5.2915 615.752 78 8.8318 4778.36 128 11.3137 12868.0 179 13.3791 25164.9
29 5.3852 660.520 79 8.8882 4901.67 129 11.3578 13069.8 180 13.4164 25446.9
30 5.4772 706.858 80 8.9443 5026.55 130 11.4018 13273.2 181 13.4536 25730.4
182 13.4907 26015.5
31 5.5678 754.768 81 9.0000 5153.00 131 11.4455 13478.2 183 13.5277 26302.2
32 5.6569 804.248 82 9.0554 5281.02 132 11.4891 13684.8 184 13.5647 26590.4
33 5.7446 855.299 83 9.1104 5410.61 133 11.5326 13892.9 185 13.6015 26880.3
34 5.8310 907.920 84 9.1652 5541.77 134 11.5758 14102.6
35 5.9161 962.113 85 9.2195 5674.50 135 11.6190 14313.9 186 13.6382 27171.6
187 13.6748 27464.6
36 6.0000 1017.88 86 9.2736 5808.80 136 11.6619 14526.7 188 13.7113 27759.1
37 6.0828 1075.21 87 9.3274 5944.68 137 11.7047 14741.1 189 13.7477 28055.2
38 6.1644 1134.11 88 9.3808 6082.12 138 11.7473 14957.1 190 13.7840 28352.9
39 6.2450 1194.59 89 9.4340 6221.14 139 11.7898 15174.7
40 6.3246 1256.64 90 9.4868 6361.73 140 11.8322 15393.8 191 13.8203 28652.1
192 13.8564 28952.9
41 6.4031 1320.25 91 9.5394 6503.88 141 11.8743 15614.5 193 13.8924 29255.3
42 6.4807 1385.44 92 9.5917 6647.61 142 11.9164 15836.8 194 13.9284 29559.2
43 6.5574 1452.20 93 9.6437 6792.91 143 11.9583 16060.6 195 13.9642 29864.8
44 6.6332 1520.53 94 9.6954 6939.78 144 12.0000 16286.0
45 6.7082 1590.43 95 9.7468 7088.22 145 12.0416 16513.0 196 14.0000 30171.9
197 14.0357 30480.5
46 6.7823 1661.90 96 9.7980 7238.23 146 12.0830 16741.5 198 14.0712 30790.7
199 14.1067 31102.6
47 6.8557 1734.94 97 9.8489 7389.81 147 12.1244 16971.7 200 14.1421 31415.9
48 6.9282 1809.56 98 9.8995 7542.96 148 12.1655 17203.4
49 7.0000 1885.74 99 9.9499 7697.69 149 12.2066 17436.6
50 7.0711 1963.50 100 10.0000 7853.98 150 12.2474 17671.5
Table values of id and A are rounded off.
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Mathematics: 1.1 Numerical tables
Values of Sine and Cosine Trigonometric Functions
de- sine 0° to 45° de- sine 45° to 90°
grees
grees Co
I 0' 15' 30' 45' 60' 0' 15' 30' 45' 60'
0° 0.0000 0.0044 0.0087 0.0131 0.0175 89° 45° 0.7071 0.7102 0.7133 0.7163 0.7193 44°
21°° 00..00137459 00..00329138 00..00423662 00..00340850 00..00354239 8878°° 4476°° 00..77139134 00..77324234 00..77237534 00..77420824 00..77433114 44 23 °°
3° 0.0523 0.0567 0.0610 0.0654 0.0698 86° 48° 0.7431 0.7461 0.7490 0.7518 0.7547 41°
4° 0.0698 0.0741 0.0785 0.0828 0.0872 85° 49° 0.7547 0.7576 0.7604 0.7632 0.7660 40°
5° 0.0872 0.0915 0.0958 0.1002 0.1045 84° 50° 0.7660 0.7688 0.7716 0.7744 0.7771 39°
6° 0.1045 0.1089 0.1132 0.1175 0.1219 83° 51° 0.7771 0.7799 0.7826 0.7853 0.7880 38°
7° 0.1219 0.1262 0.1305 0.1349 0.1392 82° 52° 0.7880 0.7907 0.7934 0.7960 0.7986 37°
8° 0.1392 0.1435 0.1478 0.1521 0.1564 81° 53° 0.7986 0.8013 0.8039 0.8064 0.8090 36°
9° 0.1564 0.1607 0.1650 0.1693 0.1736 80° 54° 0.8090 0.8116 0.8141 0.8166 0.8192 35°
10° 0.1736 0.1779 0.1822 0.1865 0.1908 79° 55° 0.8192 0.8216 0.8241 0.8266 0.8290 34°
11° 0.1908 0.1951 0.1994 0.2036 0.2079 78° 56° 0.8290 0.8315 0.8339 0.8363 0.8387 33°
12° 0.2079 0.2122 0.2164 0.2207 0.2250 77° 57° 0.8387 0.8410 0.8434 0.8457 0.8480 32°
13° 0.2250 0.2292 0.2334 0.2377 0.2419 76° 58° 0.8480 0.8504 0.8526 0.8549 0.8572 31°
14° 0.2419 0.2462 0.2504 0.2546 0.2588 75° 59° 0.8572 0.8594 0.8616 0.8638 0.8660 30°
15° 0.2588 0.2630 0.2672 0.2714 0.2756 74° 60° 0.8660 0.8682 0.8704 0.8725 0.8746 29°
16° 0.2756 0.2798 0.2840 0.2882 0.2924 73° 61° 0.8746 0.8767 0.8788 0.8809 0.8829 28°
17° 0.2924 0.2965 0.3007 0.3049 0.3090 72° 62° 0.8829 0.8850 0.8870 0.8890 0.8910 27°
18° 0.3090 0.3132 0.3173 0.3214 0.3256 71° 63° 0.8910 0.8930 0.8949 0.8969 0.8988 26°
19° 0.3256 0.3297 0.3338 0.3379 0.3420 70° 64° 0.8988 0.9007 0.9026 0.9045 0.9063 25°
20° 0.3420 0.3461 0.3502 0.3543 0.3584 69° 65° 0.9063 0.9081 0.9100 0.9118 0.9135 24°
21° 0.3584 0.3624 0.3665 0.3706 0.3746 68° 66° 0.9135 0.9153 0.9171 0.9188 0.9205 23°
22° 0.3746 0.3786 0.3827 0.3867 0.3907 67° 67° 0.9205 0.9222 0.9239 0.9255 0.9272 22°
23° 0.3907 0.3947 0.3987 0.4027 0.4067 66° 68° 0.9272 0.9288 0.9304 0.9320 0.9336 21°
24° 0.4067 0.4107 0.4147 0.4187 0.4226 65° 69° 0.9336 0.9351 0.9367 0.9382 0.9397 20°
25° 0.4226 0.4266 0.4305 0.4344 0.4384 64° 70° 0.9397 0.9412 0.9426 0.9441 0.9455 19°
26° 0.4384 0.4423 0.4462 0.4501 0.4540 63° 71° 0.9455 0.9469 0.9483 0.9497 0.9511 18°
27° 0.4540 0.4579 0.4617 0.4656 0.4695 62° 72° 0.9511 0.9524 0.9537 0.9550 0.9563 17°
28° 0.4695 0.4733 0.4772 0.4810 0.4848 61° 73° 0.9563 0.9576 0.9588 0.9600 0.9613 16°
29° 0.4848 0.4886 0.4924 0.4962 0.5000 60° 74° 0.9613 0.9625 0.9636 0.9648 0.9659 15°
30° 0.5000 0.5038 0.5075 0.5113 0.5150 59° 75° 0.9659 0.9670 0.9681 0.9692 0.9703 14°
31° 0.5150 0.5188 0.5225 0.5262 0.5299 58° 76° 0.9703 0.9713 0.9724 0.9734 0.9744 13°
32° 0.5299 0.5336 0.5373 0.5410 0.5446 57° 77° 0.9744 0.9753 0.9763 0.9772 0.9781 12°
33° 0.5446 0.5483 0.5519 0.5556 0.5592 56° 78° 0.9781 0.9790 0.9799 0.9808 0.9816 11°
34° 0.5592 0.5628 0.5664 0.5700 0.5736 55° 79° 0.9816 0.9825 0.9833 0.9840 0.9848 10°
35° 0.5736 0.5771 0.5807 0.5842 0.5878 54° 80° 0.9848 0.9856 0.9863 0.9870 0.9877 9°
36° 0.5878 0.5913 0.5948 0.5983 0.6018 53° 81° 0.9877 0.9884 0.9890 0.9897 0.9903 8°
37° 0.6018 0.6053 0.6088 0.6122 0.6157 52° 82° 0.9903 0.9909 0.9914 0.9920 0.9925 7°
38° 0.6157 0.6191 0.6225 0.6259 0.6293 51° 83° 0.9925 0.9931 0.9936 0.9941 0.9945 6°
39° 0.6293 0.6327 0.6361 0.6394 0.6428 50° 84° 0.9945 0.9950 0.9954 0.9958 0.9962 5°
40° 0.6428 0.6461 0.6494 0.6528 0.6561 49° 85° 0.9962 0.9966 0.9969 0.9973 0.9976 4°
41° 0.6561 0.6593 0.6626 0.6659 0.6691 48° 86° 0.9976 0.9979 0.9981 0.9984 0.9986 3°
42° 0.6691 0.6724 0.6756 0.6788 0.6820 47° 87° 0.9986 0.9988 0.9990 0.9992 0.9994 2°
43° 0.6820 0.6852 0.6884 0.6915 0.6947 46° 88° 0.9994 0.9995 0.9997 0.9998 0.99985 1°
44° 0.6947 0.6978 0.7009 0.7040 0.7071 45° 89° 0.99985 0.99991 0.99996 0.99999 1.0000 0°
t
60' 45' 30' 15' 0' t 60' 45' 30' 15' 0'
minuies gdree-es minuies gdere-es
cosine 45° to 90° cosine 0° to 45°
Table values of the trigonometric functions are rounded o ff to four decimal places.
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12 Mathematics: 1.1 Numerical tables
Values of Tangent and Cotangent Trigonometric Functions
de- tangent 0° to 45° de- tangent 45° to 90°
grees
grees — 111111 u ic; 60' — 1 1 III IUICC
0' 15' 30'
1 0' 15' 30' 45' 0.0175 89° 45° 45' 60'
1.0355 4 4 °
0° 0.0000 0.0044 0.0087 0.0131 1.0000 1.0088 1.0176 1.0265
21°° 00..00134795 00..00329183 00..00423672 00..00438006 00..00354294 88 87 °° 46° 11..00375254 11..00844168 11..00953138 11..10063009 11..10172046 4432°°
47° 1.1106 1.1204 1.1303 1.1403 1.1504 41°
1.1504 1.1606 1.1708 1.1812 1.1918 40°
3° 0.0524 0.0568 0.0612 0.0655 0.0699 86° 48°
1.2131
4° 0.0699 0.0743 0.0787 0.0831 0.0875 85° 49° 1.2572
1.3032
5° 0.0875 0.0919 0.0963 0.1007 0.1051 84° 50° 1.1918 1.2024 1.3514 1.2239 1.2349 39°
6° 0.1051 0.1095 0.1139 0.1184 0.1228 83° 51° 1.2349 1.2460 1.4019 1.2685 1.2799 38°
7° 0.1228 0.1272 0.1317 0.1361 0.1405 82° 52° 1.2799 1.2915 1.3151 1.3270 37°
8° 0.1405 0.1450 0.1495 0.1539 0.1584 81° 53° 1.3270 1.3392 1.4550 1.3638 1.3764 36°
9° 0.1584 0.1629 0.1673 0.1718 0.1763 80° 54° 1.3764 1.3891 1.5108 1.4150 1.4281 35°
10° 0.1763 0.1808 0.1853 0.1899 0.1944 79° 55° 1.4281 1.4415 1.5697 1.4687 1.4826 34°
11° 0.1944 0.1989 0.2035 0.2080 0.2126 78° 56° 1.4826 1.4966 1.6319 1.5253 1.5399 33°
1.6977
12° 0.2126 0.2171 0.2217 0.2263 0.2309 77° 57° 1.5399 1.5547 1.5849 1.6003 32°
13° 0.2309 0.2355 0.2401 0.2447 0.2493 76° 58° 1.6003 1.6160 1.6479 1.6643 31°
14° 0.2493 0.2540 0.2586 0.2633 0.2679 75° 59° 1.6643 1.6808 1.7147 1.7321 30°
15° 0.2679 0.2726 0.2773 0.2820 0.2867 7 4 ° 60° 1.7321 1.7496 1.7675 1.7856 1.8040 29°
16° 0.2867 0.2915 0.2962 0.3010 0.3057 73° 61° 1.8040 1.8228 1.8418 1.8611 1.8807 28°
17° 0.3057 0.3105 0.3153 0.3201 0.3249 72° 62° 1.8807 1.9007 1.9210 1.9416 1.9626 27°
18° 0.3249 0.3298 0.3346 0.3395 0.3443 71° 63° 1.9626 1.9840 2.0057 2.0278 2.0503 26°
19° 0.3443 0.3492 0.3541 0.3590 0.3640 70° 64° 2.0503 2.0732 2.0965 2.1203 2.1445 25°
20° 0.3640 0.3689 0.3739 0.3789 0.3839 69° 65° 2.1445 2.1692 2.1943 2.2199 2.2460 24°
21° 0.3839 0.3889 0.3939 0.3990 0.4040 68° 66° 2.2460 2.2727 2.2998 2.3276 2.3559
22° 0.4040 0.4091 0.4142 0.4193 0.4245 67° 67° 2.3559 2.3847 2.4142 2.4443 2.4751 23°
23° 0.4245 0.4296 0.4348 0.4400 0.4452 66° 68° 2.4751 2.5065 2.5386 2.5715 2.6051 22°
24° 0.4452 0.4505 0.4557 0.4610 0.4663 65° 69° 2.6051 2.6395 2.6746 2.7106 2.7475 21°
20°
25° 0.4663 0.4716 0.4770 0.4823 0.4877 64° 70° 2.8239
26° 0.4877 0.4931 0.4986 0.5040 0.5095 63° 71° 2.7475 2.7852 2.9887 2.8636 2.9042 19°
27° 0.5095 0.5150 0.5206 0.5261 0.5317 62° 72° 2.9042 2.9459 3.1716 3.0326 3.0777 18°
28° 0.5317 0.5373 0.5430 0.5486 0.5543 61° 73° 3.0777 3.1240 3.3759 3.2205 3.2709 17°
29° 0.5543 0.5600 0.5658 0.5715 0.5774 60° 74° 3.2709 3.3226 3.6059 3.4308 3.4874 16°
3.4874 3.5457 3.6680 3.7321 15°
30° 0.5774 0.5832 0.5890 0.5949 0.6009 59° 75° 3.8667
31° 0.6009 0.6068 0.6128 0.6188 0.6249 58° 76° 3.7321 3.7983 4.1653 3.9375 4.0108 14°
32° 0.6249 0.6310 0.6371 0.6432 0.6494 57° 77° 4.0108 4.0876 4.5107 4.2468 4.3315
33° 0.6494 0.6556 0.6619 0.6682 0.6745 56° 78° 4.3315 4.4194 4.9152 4.6057 4.7046 13°
34° 0.6745 0.6809 0.6873 0.6937 0.7002 55° 79° 4.7046 4.8077 5.3955 5.0273 5.1446 12°
5.1446 5.2672 5.5301 5.6713 11°
10°
35° 0.7002 0.7067 0.7133 0.7199 0.7265 54° 80° 5.6713 5.8197 5.9758 6.1402 6.3138 9°
36° 0.7265 0.7332 0.7400 0.7467 0.7536 53° 81° 6.3138 6.4971 6.6912 6.8969 7.1154 8°
37° 0.7536 0.7604 0.7673 0.7743 0.7813 52° 82° 7.115 4 7.3479 7.5958 7.8606 8.1443 7°
38° 0.7813 0.7883 0.7954 0.8026 0.8098 51° 83° 8.1443 8.4490 8.7769 9.1309 9.5144 6°
39° 0.8098 0.8170 0.8243 0.8317 0.8391 50° 84° 9.5144 9.9310 10.3854 10.8829 11.4301 5°
40° 0.8391 0.8466 0.8541 0.8617 0.8693 49° 85° 11.4301 12.0346 12.7062 13.4566 14.3007 4°
41° 0.8693 0.8770 0.8847 0.8925 0.9004 48° 86° 14.3007 15.2571 16.3499 17.6106 19.0811 3°
42° 0.9004 0.9083 0.9163 0.9244 0.9325 4 7 ° 87° 19.0811 20.8188 22.9038 25.4517 28.6363
43° 0.9325 0.9407 0.9490 0.9573 0.9657 46° 88° 28.6363 32.7303 38.1885 45.8294 57.2900 2°
1°
4 4 ° 0.9657 0.9742 0.9827 0.9913 1.0000 45° 89° 57.2900 76.3900 114.5887 229.1817 00 0°
t
60' 45' 30' 15' 0' t 60' 45' 30' 15' 0' de-
grees
minuies de- minuies
cotangent 45° to 90° grees cotangent 0° to 45°
Table values of the trigonometric functions are rounded off to four decimal places.
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M a t h e m a t i c s : 1.2 T r i g o n o m e t r i c F u n c t i o n s 13
Trigonometric functions of right triangles
Definitions Definitions of th e Application
ratios of the sides
Designations in a for < a for < 0
right triangle
hypotenuse s<3idoeppoof saite sine opposite side sin a = — sin (i = -
cosine co s a = cos/3 = -
7 £ a adjacent hypotenuse tan a = -=- tan/8 = -
b adjacent side of side of ft tangent adjacent side cot a = — cot/? = 4
c hypotenuse hypotenuse
opposite side
adjacent side
opposite side of fi cotangent = adjacent side
opposite side
Graph of the trigonometric functions between 0° and 360c
Representation on a unit circle Graph of the trigonometric functions
The values of the trigonometric functions of angles > 90° can be derived from the values of the angles b etween 0° and
90° and then read from the tables (pages 11 and 12). Refer to the graphed curves of the trigono me tric functions for
the correct sign. Calculators wi th trigonom etric functions d isplay both the value and sign for the desired angle.
Example: Relationships for Quadrant II
Relationships Exam ple: Function va lues for the an gle 120° (a = 30° in the formulae)
sin (90° + a) = +cos a sin (90° + 30°) = sin 120° =+ 0.8660 cos 30° =+0.8660
cos (90° + a) =- s in a cos (90° + 30°) = cos 120° = -0.5 000 -s in 30° = -0.5000
tan (90° + a) = - c o t a tan (90° + 30°) = tan 120° = -1.7 32 1 -c ot 30°= -1.7321
Function values for selected angles
Function 0° 90° 180° 270° 360° Func tion 0° 90° 180° 270° 360°
sin 0 + 1 0 - 1 0 tan 0 0 0
cos + 1 0 -1 0 + 1 cot 00
Relationships betw een th e functions of an angle
s in 2 a + c o s 2 a = 1 ta n a • cot a = 1
sin or ta n a = sin a cot a = co s a
co s a sin a
cos or Example: Calculation of tana from sina and cosa for a = 30°:
tan a = s ina /co sa = 0.5000/0.8660 = 0.5774
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14 Ma them atics: 1.2 Trigon om etric Fu nctions
Trigonometric functions of oblique triangles, Angles, Theorem of intersecting lines
Law of sines and L aw of cosines
Law of sines Law of cosines
a: b: c = sin a : sin/3 : siny a2 = b2 + c2 - 2 •b •c •co s a
b2 = a2 + c2 - 2 • a • c • cos/3
a _ b _ c c2 = a2 + b2 - 2 •a -b •co s y
sina sin/3 sin/
App lication in calculating sides and angles
Calculation of sides Calculation of angles
using the Law of sines using the Law of cosines using the Law of sines using the Law of cosines
b-sina _ csina a = jb2 + c2 - 2 •b •c c• osa a sin /3 _ a - si n / b2+c 2 -a 2
a = sin/3 b = yja2 + c2 - 2 •a c• c•os /3 sina = b " c cos a = 2-b- c
sin/ c = yja2 + b2 - 2 •a b• •cos /
b = a- sin/3 _ c-sin/3 sin/3 = 6-sina _ b-siny cos/3 =a 2 +c2-b 2
sina siny a c 2 a- c
asiny _ bsiny c-sina c-sin^ a2 +b2-c 2
c = sina sin/ = cos y = 2-ab
sin/3
Types of angles If tw o parallels g- \and g2 are intersected Corresponding angles
Sum of angles in a triangle by a straight line g, there are geo metrical a=P
interrelationships between the corre-
sponding, opposite, alternate and adja- Opposite angles
p=d
cent angles.
Alternate angles
a=d
Adjacent angles
a + y = 180c
In every triangle the sum of the interior Sum of angles
angles equals 180°. in a triangle
a + /3 + y = 1 8 0 c
Theorem of intersecting lines
If two lines extending from Point A are Theorem of intersecting
intersected by two parallel lines BC and lines
B-|C1( the segments of the parallel lines
and the corresponding ray segments of a b_ c
the lines extending from A form equal
ratios. by Cl
ab
b~ c Cl
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Mathematics: 1. nts
Using b rackets, powers and roots
I Calculations wit h b rackets
Type Explanation Example
3 x + 5 x = x ( 3 + 5) = 8 x
Factoring out Common factors (divisors) in addition and subtraction are
placed before a bracket.
+I- .<3+5)
XXX
A fraction bar combines terms in the same manner as —a 2+-• /b? , =(.a +.b. 2h-
brackets.
Expanding A bracketed term is multiplied by a value (number, varia- 5 • (b + c )= 5b +5 c
bracketed terms ble, another bracketed term), by multiplying each term (a + b ) •(c - d) = ac - ad +be - bd
inside the brackets by this value.
(a +b):c =a:c +b:c
A bracketed term is divided by a value (num ber, variable, a-b a b
another bracketed term), by dividing each term inside the
bracket by this value. 5 ~ 5~ 5
Binomial A binom ial formula is a formula in wh ich the term (a + b) (a + b )2 = a 2 + la b +b2
formulae or (a - b) is multiplied by itself. (a - b)2 = a2 - 2ab +b2
(a + b)(a-b) = a2 - b2
Multiplication/divi- In mixed equations, the bracketed terms must be solved
sion and first. Then multiplication and division calculations are per- a • (3x- 5x) - b •( My- 2 y)
addition/subtracti- formed, and finally ad dition and subtraction. = a • (-2 x) - b • 10 y
on calculations = -lax - 10 by
Powers
Definitions a base; x exponent; y exponential value a* = y
Product of identical factors a •a •a • a = a 4
4 • 4 • 4 • 4 = 4 4 = 256
Addition Powers with the same base and the same exponents are
Subtraction treated like equal numbers. 3 a 3 + 5 a 3 - 4 a 3
= a 3 • ( 3 + 5 - 4 ) = 4a3
Multiplication Powers with the same base are multiplied (divided) by
Division adding (subtracting) the exponents and keeping the base. a4 a2 = a a-a-a a a =a6
I 4 • I2 = 2 ( 4 + 2 ) = 2 6 = 64
3 2 -r 3 3 = 3 ( 2 - 3 ) = 3 _ 1 = 1/3
Negative Num bers w ith negative exponents can also be written as 1 1 1
exponent fractions. The base is then given a positive exponent and
is placed in the denominator. m =—7 = —
Fractions in
exponents Powers with fractional exponents can also be written as rrr m
Zero in roots.
exponents Every power with a zero exponent has the value of one. a - 33 1
Roots x root's expo nent; a radicand; y root value 4 = a—3
Definitions Even number exponents of the root give positive and a3=fc
negative values, if the radicand is positive. A negative radi-
Signs cand results in an imaginary number. (m + n)° = 1
Odd number exponents of the root give positive values if
the radicand is positive and negative values if the radicand a4 + a4 = a ( 4 _ 4 ) = a 0 = 1
is negative. 2°= 1
y/a = yo r a 1 / x = y
\/9 = ±3
= + 3i
\/8 = l
yf-8=-l
AS ud bd ti rt iaocnt i o n Identical root expressions can be added and subtracted. \la+3\la-l\[a=l\[a
Multiplication Roots with the same exponents are multiplied (divided) by
Division taking the root of the product (quotient) of the radicands.
ra J i "
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16 Mathematics: 1. nts
Types of equations, Rules of transformation
Equations
Type Explanation Example
Variable v = ji • d •n
equation Equivalent terms (formula terms of equal value ) form rela- ( a+ b )2 = a 2 + 2ab+ b2
tionships between variables (see also, Rules of transfor-
Compatible units mation). p_ M-n p j n k jf
equation
Immediate conversion of units and constants to an SI unit 9550
in the result. n in 1/min and M in Nm
Only used in special cases, e.g. if engineering parameters
are specified or for simplification. x+3=8
x=8-3=5
Single variable Calculation of the value of a variable.
equation y = f (x)
* real n umb ers
Function Assigned function equation: y is a function of x w i t h x as
equation the independent variable; ya s the dependent variable. y = f (x )= b
The number pair (x,y ) of a value table form the graph of
the function in the (x,y) coordinate system. y = f (x) = mx
y= 2x
Constant function y = f (x) = m x +b
The graph is a line parallel to the x-axis. y= 0 . 5 x + 1
Proportional function y = f (x) = X 2
The graph is a straight line through the origin. y = a 2 x 2 + a-|X+ a 0
Linear func tion
The graph is a straight line w ith slope m an d y intercept b
(example below).
Quadratic func tion
Every quadratic function graphs as a parabola
(example below).
linear example: quadratic example:
function fyu=nxc2ti o n ~y= 0 . 5 - x 2 /
y= m x + b
\\ tI 2 " I/
2 -1 f ^ H 1 1
-1 - 1x 2 •3
Rules of transform ation
Equations are usually transformed to obtain an equation in which the unknown variable stands alone on the left side
of the equation.
Addition The same number can be added or subtracted from both x + 5 = 1 5 |- 5
Subtraction sides.
x + 5 - 5 = 1 5 - 5
In the equation s X+ 5 = 1 5 and x + 5 - 5 = 15 - 5, x has the
same value, i.e. the equations are equivalent. x = 10
y-c - d | + c
y-c + c=d + c
y = d + c
Multiplication It is possible to m ultiply or divide each side of the e quation ax = b | r a
Division by the same number.
ax b
Powers The expressions on both sides of the equations can be
raised to the same exp onential power. aa
b
x = —
a
s/x=a + b | () 2
22
(Vx) =(a +b)
x =a 2 +2ab + b2
Roots The root of the expressions on both sides of the equation x2=a + b \yf
can be taken using the same root exponent. (\[x)2 = -J a+ b
x -±\ja +b
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M a t h e m a t ic s : 1. nts
Decimal m ultiples and factors of units. Interest calculation
Decimal m ultiples and factors of u nits cf. DIN 1301-1 (2002-10)
Mathematics SI units
Unit
Power of Name Multipliiccaattiioonn ffaaccttor Prefix Examples
ten Nam e Character Mea ning
1 0 1 8 quintillion 1 000 000 000 000 000 000 exa E Em 10 1 85 meters
1 0 1 5 quadrillion 1 000 000 000 000 000 peta P Pm 10 1 meters
trilli on 1 000 000 000 000 tera T
1012 billio n 1 000 000 000 G TV 1 0 1 2 volts
million 1 000 000 giga M GW 109 watts
10 9 thous and 1 000 mega k MW 106 watts
hundred 100 h kN 103 newtons
106 ten 10 kilo da hi 102 liters
103 one 1 hecto dam 10 1 meters
102 deca
101 tenth
hundredth
10° thous andth m 10° meter
10" 1 mb i il ll il oi onnt ht h 0.1 deci d d m 10~1 meters
10-2 trillion th 0.01 centi c cm 10"2 meters
10"3 quadrillionth 0 . 00 1 milli m mV 10"3 volts
quin tillionth
6 0.0000.000000 000011 mnaicnroo Mn- 6
0.000 000 000 001 pico
11 00 "" 9 0.000 000 000 000 001 fem to P nHmA 1100""9 ammepteerrse
10"12 0.000 000 000 000 000 001 atto f pF 1 0 - 12 farad
1 0 -15 a fF 10"15 farads
am 10" 18 meters
1 0 -1 8
values Num bers greater than 1 are expressed wit h positive exponents and num-
< -f- > bers less than 1 are expressed w ith negative exponents.
11 I 10 100 1000 Examples: 4300 = 4.3 •1 000 = 4.3 • 1 03
14638= 1.4638 • 1 04
1000 100 I • • ••
-H—h-
007= 7 10 -2
1 0 " 3 1 0 - 2 1 0 " 1 10° 101 102 103 i^o = -
Simple interest
P principle I interest time in days, Interest
interest period
A amo unt accumulated r interest rate per year 1 interest year (1 a) = 360 days (360 d)
360 d = 12 months
1st example:
1 interest month = 30 days
P = $2800.00; r = 6 - ; t= 1/2a; / = ?
a*
$2800.00-6- -0.5a =$84.00
I= -t— -
100%
2nd example:
P = $4800.00; r = 5 . 1 ,0 -\ t = 50d; / = ?
$ 4 8 0 0 .0 0 - 5 .1 % • 5 0 d
I = ^ = $ 34.00
100%-360 d
a
Compound interest calculation for one-time paymen t
P principle I interest n ti m e Amount accumulated
A am ount accum ulated r interest rate per year q com poun ding factor A=P-<7 n
Example:
P = $8000.00; n = 7 years; r = 6.5 ;A = ? Compound ing factor
q = 1 + 6.5% = 1.065
100%
A = P- qn = $ 8000.00 • 1.0657 = $ 8000.00 • 1.553986
= $12431.89
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Mathematics: 1. nts
18
Percentage calculation, Proportion calculations
Percentage calculation
Th e percentage rate gives the fraction of the base value in h undred ths. Percent value
Th e base value is the value from wh ich the percentage is to be calculated. p _gVPr
Th e percent value is the amount representing the percentage of the base value.
v 100%
P r percentage rate, in percent P v percent value £ v base value.
Percentage rate
1st exam ple:
Workpiece rough part weight 250 kg (base value); material loss 2%
(percentage rate); material loss in kg = ? (percent value)
% _ 250 kg • 2 %
100% 100%
2nd example:
Rough we ight of a casting 150 kg; we ight after m achining 126 kg;
we ight percent rate (%) of material loss?
P„ r = BP,m • *1r0\r\0n%/ = 1 5 0kg- 12 6 k g • 100% = 1 6 %
150 kg
Proportion calculations
Three steps for calculating direct prop ortional ratios
Example: 6 Q e | b o w p j p e s w e j g h 330 kg. W hat is the weig ht of
35 elbow pipes?
1 elbow pipe weighs 330 kg
60
100 200 kg 300 3rd step: Calculate the total by multiplying
weight
35 elbow pipes we igh 330 kg • 35 = 1 9 2 5 R g
bO
Three steps for c alculating inverse prop ortional ratios
Example: It takes 3 workers 170 hours to process one order. How man y
hours do 12 worke rs need to process the same order?
It takes 1 worker 3 • 170 hrs
2 U 6 8 10 12 14 3rd step: Calculate the total by dividing
workers »
Ittakes12 workers 3- 170 hrs = 42.5 hrs
12
Using the three steps for calculating direct and inverse proportions
Example: 1st application of 3 steps:
5 machines produce 660 workpieces in 24 days
660 workpieces are manufactu- 1 machine produces 660 workpieces in 24 • 5 days
red by 5 machines in 24 days.
9 machines produce 660 workpieces in 24-5 days
How much time does it take for 2nd applicati9onmaocf h3insetespsp:roduce 660 workpieces in — -— days
9 machines to produce 9 machines produce 1 workp iece in 24 • 5 d a y s
312 workpieces of the same 9 • 660
type?
9 machines produce 312 workpieces in 24 5 • 312
— ' — = 6.3 days
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Formula symbols, Mathem atical symbols
Formula sym bols cf. DIN 1304-1 (1994-03)
Meaning
Formula Meaning Formula Meaning Formula
symbol symbol symbol
Length, Area, Volum e, Angle
/ Length r, R Radius Planar angle
QA WSoalvide alennggleth
wh WHeidigthht Ad,,DS ADriaema,eCterross-sectional area
s Linear distance V Volume
Mechanics
m Mass F Force G Shear m odulus
Gravitational force, W eight Coefficient of friction
rri Linear mass density F\n, IN Torque M,f Section m odulus
M Torsional moment W Second moment of an area
rri' Area mass density T Bending m oment I Work, Energy
Mb Normal stress Potential energy
Q Density a Shear stress W,E Kinetic energy
X Normal strain W p,E p Power
J Moment of inertia £ Modulus of elasticity W k,E k Efficiency
Pressure E
P Absolute pressure P
Ambient pressure
Pabs Gage pressure
Pamb
Pg
Time Time, Duration f,v Frequency a Acceleration
Cycle duration V, u Velocity Gravitational acceleration
t Revolution frequency, 0) Angular velocity 9 Angular acceleration
T Speed a Volumetric flow rate
n a V ,qy
Electricity L Inductance X Reactance
Q Electric charge, Quantity of R Resistance z Impedance
electricity Q Specific resistance Phase difference
E Electromotive force y,x Electrical condu ctivity <p Number of turns
C Capacitance N
I Electric current
Heat
T,Q Ttehmeprmeroadtuyrneamic Q THheeartm, Qaluacnotnitdyuocftihveitayt 0,Q Heat flow
AT, At, Ad Temperature difference a Thermal diffusivity
A c Specific heat
Celsius temperature a Heat transition coefficient Hiet Net calorific value
k Heat transm ission
Coefficient of linear coefficient
<*\,a expansion
Light, Electromagnetic radiation f Focal length / Luminous intensity
E Illuminance n Refractive index Q, W Radiant energy
Acoustics LP Acoustic pressure level N Loudness
P Acoustic pressure I Sound intensity Ln Loudness level
c Acoustic velocity
Mathematical symbols cf. DIN 1302 (1999-12)
Math, Spoken Math, Spoken Math, Spoken
symbol symbol symbol
proportional logarithm (general)
approx. equals, around, a~n a to the n-th power, the n-th log common logarithm
about power of a ig natural logarithm
equivalent to ff square root of In Euler number (e = 2.718281...)
and so on, etc. n-th root of e
00 infinity I x I absolute value of x sin sine
_L perpendicular to cos cosine
equal to is parallel to tan tangent
* not equal to Ut t parallel in the same d irection cot cotangent
is equal to by definition parallel in the opposite direction parentheses, brackets
def less than < angle (),[],{} open and closed
< tcroiannggruleent to 3p.i1(4c1ir5c9le..c.)onstant =
A delta x (difference between K
V less than or equal to two values) line segment AB
A greater than Ax percent, of a hundred AB arc AB
per m il, of a thousand AB a prime, a double prime
% a', a" a sub 1, a sub 2
ava2
L gprluesater than or equal to %0
minus
times, multiplied by
over, divided by, per, to
I sigma (summation)
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20 M a t h e m a t i c s : 1 .4 S y m b o l s , U n i t s
SI quantities and units of measurement
S l 1) Base quantities and base units cf. DIN 1301 1 (2002-10), -2 (1978-02), -3 (1979-10)
Base Length Mass Time Electric Thermo- Amount of Luminous
quantity current dynamic substance intensity
temperature
uBnaistse meter gkrialom- second ampere kelvin mole candela
Unit m kg s A K mol cd
symbol
1 ) The units for me asurem ent are defined in the International System of Un its SI (System e International d'Un ites). It
is based on the seven basic units (SI units), from wh ich other u nits are derived.
Base quantities, derived q uantities and their units
Quantity Unit Relationship Remarks
Examples of application
Symbol Name Symbol
Length, Area, Volum e, Ang le
Length / meter m 1m =10 dm = 100 cm 1 inch = 25.4 mm
= 1000 mm In aviation and nautical applications
the following applies:
1m m = 1000 (jm 1 international nautical mile = 1852 m
1km = 1000 m
Area A S square meter m 2 1 m 2 = 10000 cm 2 Symbol S only for cross-sectional
Volume
1a = 1000000 mm 2 areas
are 1 ha =100 m 2
hectare a 100 ha = 100 a = 10000 m 2 Are and hectare only for land
ha = 1 k m 2
1/ cubic meter m 3 1 m 3 = 1000 d m 3 Mostly for fluids and gases
= 1000000 cm 3
liter I, L
11 = 1 L = 1 d m 3 = 10 dl =
0.001 m 3
1 ml =1 cm 3
Plane radian rad 1 rad = 1 m/m = 57.2957...° 1 rad is the angle formed by the inter-
angle 0
(angle) degrees = 180%t section of a circle around the center of
minutes sr
Solid angle seconds 1° = 180 rad = 60' 1 m radius w ith an arc of 1 m length.
Q steradian 1' =1 76 0 = 60" In technical calculations instead of
a = 33° 17' 27.6", better use is a =
1" = 1760 = 173600 33.291°.
1 sr =1 m 2/m 2 An object whose extension measures
1 rad in one direction and perpendicu-
larly to this also 1 rad, covers a solid
angle of 1 sr.
Mechanics m kilogram gkg 1k g = 1000 g Mass in the sense of a scale result or a
Mass gram 1 g = 1000 mg weig ht is a quantity of the type of m ass
Mg (unit kg).
megagram t
metric ton 1 metric t = 1000 kg = 1 M g
kg/m
0.2 g = 1 ct Mass for precious s tones in carat (ct).
Linear mass m' kilogram 1 kg/m = 1 g/mm For calculating the mass of bars, pro-
files, pipes.
density per meter
Area mass m kilogram kg / m 2 1 kg/m 2 = 0.1 g / cm 2 To calculate the mass of sheet metal.
density
per square
meter
Density e kilogram kg/m3 1000 kg /m 3 = 1 metric t /m 3 The density is a quantity independent
per cubic = 1 kg / d m 3 of location.
meter = 1 g / cm 3
= 1 g/ml
= 1 m g / m m 3
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M a t h e m a t i c s : 1 .4 S y m b o l s , U n i t s 21
SI quantities and units of m easurement
I Quantities and Units (continued)
Quantity Sym- Unit Relationship Remarks
bol Examples of application
Name Symbol
I Mechanics
oMfoinmeertniat , 2nd J skqiluoagrream x kg • m 2 Thhoemofoglelonwoiunsg baopdpyli:es for a Tmhaessm) omis endteopfenindeerntita (u2pnodn mtohme entottoafl
Moment of meter J =e - r 2 . V mass of the body as well as its form
mass and the position of the axis of rotation.
F newton
Force N 1 N _ i M _ m _ i J The force 1 N effects a change in vel-
N • m s^ m ocity of 1 m/s in 1 s in a 1 kg ma ss.
Weight F g ,G 1 MN = 103 kN = 1000000 N
Torque M newton x 1 N - m is the moment that a force of
Bending mom. meter
Torsional Mb s^ 1 N effects with a lever arm of 1 m .
T kilogram x
meter
Momentum P per second kg • m/s 1 kg • m/s = 1 N • s The momentum is the product of the
pascal mass times v elocity. It has the direction
of the velocity.
Pressure P Pa 1 Pa = 1 N/ m 2 = 0.01 mbar Pressure refers to the force per unit
N/mm2 1 bar = 100000 N/m 2 area. For gage pressure the sym bol p g
Mechanical O, T n e wto n = 10 N/cm 2 = 105 Pa is used (DIN 1314).
stress per square
millimeter 1 mbar = 1 hPa 1 bar = 14.5 psi (pounds per square
1 N/m m 2 = 10 bar = 1 M N / m 2 inch )
= 1 M P a
1 d a N/cm 2 = 0.1 N / m m 2
Second I meter to the m 4 1 m 4 = 100000000 cm 4 Previously: Geometrical moment of
moment of fourth power cm4 inertia
area centimeter
to the fourth J 1 J = 1 N • m = 1 W- s Joule for all forms of energy, kW- h
Energy, Work, power
Quantity of = 1 kg • m 2 / s 2
heat E, W joule
preferred for electrical energy.
Power P watt W 1 W = 1 J/s = 1 N • m/s Power describes the work which is
Heat flux <P = 1 V • A = 1 m 2 • kg/s3 achieved within a specific time.
I Time t seconds s 3 h means a time span (3 hrs.),
Time, minutes min 1 m in = 60 s 3 h means a point in time (3 o'clock).
Time span, hours h 1 h =6 0 min = 3600 s If points in time are written in mixed
Duration day d 1 d = 24 h = 86400 s form, e.g. 3h24m10s, the symbol min
year a can be shortened to m .
Frequency
Rotational f,v hertz Hz 1 Hz = 1/s 1 Hz = 1 cycle in 1 second.
speed,
Rotational n 1 per second 1/s 1/s = 60/min = 60 m in - 1 The number of revolutions per unit of
frequency time gives the revolution frequency,
Velocity 1 per minute 1/min 1/min = 1 m i n - 1 =6701^—s also called rpm.
Av enlgoucliat yr - V meters per m/s 1 m/s = 60 m/m in Nautical veloc ity in knots (kn):
= 3.6 km/h 1 kn = 1.852 km/h
Acceleration second miles per hour = 1 mile/h = 1 m p h
1 mp h = 1.60934 km/h
meters per m/m in 1 m / m i n = 1 m
minute 60s For a rpm of n = 2/s the angular veloci-
ty a )= 4 JI / S.
kilometers per km/h 1 km/h = 3.16™s
hour Symbol g only for acceleration due to
gravity.
CD r1adpiearnssepceornd r1a/sd/s a) = 2 t c• n g = 9.81 m /s 2 » 10 m/s 2
second m/s2
1 m/s2 = \1 m / s
3,9 meters per
second s
squared
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22 M a t h e m a t i c s : 1 .4 S y m b o l s , U n i t s
SI quantities and units of m easurement
Quantities and units (continued)
Quantity Sym- Unit Sym- Relationship Remarks
bol Examples of application
Name bol
Electricity and Magnetism
EE lleeccttrri co mc uortri ev ne t EI ampere A 1 V = 1 W/1 A = 1 J/C The m ovem ent of an electrical charge is
force volt V called current. The electromotive force
Electrical is equal to the poten tial difference bet-
resistance R ohm Q 1 Q = 1 V/1 A ween tw o points in an electric field. The
Electrical reciprocal of the electrical resistance is
conductance G siemens S 1 S = 1 A/1 V = 1/Q called the electrical conductivity.
Specific e ohm x Q • m 10 "6 Q • m = 1 Q • m m 2 / m 1 . Q • m m 2
resistance
Conductivity meter S/m Q = — in
Y, sie me n s
Frequency Hz 1 Hz = 1/s xm
per meter 1000 Hz = 1 kHz
1. m
f hertz
x = — in -
FreqQ uencQy o• fmpmub2lic electric utility:
EU 50 Hz, USA/Canada 60 Hz
Electrical energy W joule J 1 J = 1 W • s = 1 N • m In atomic and nuclear physics the unit
1 kW • h = 3.6 MJ eV (electron volt) is used.
1 W • h = 3.6 kJ
Phase <P for alternating current: The angle between current and voltage
difference in inductive or capacitive load.
C0S(p=ih
Elect, field strength
Elect, charge E volts per meter V/m
Elect, capacitance Q coulomb
inductance C farad C 1 C = 1 A • 1 s; 1 A • h = 3.6 kC Q =U Q = 1 •t
L henry F 1 F = 1 C/V
H 1 H = 1 V • s/A
Power P watt W 1 W = 1 J/s = 1 N • m/s In electrical power engineering:
Effective powe r = 1 V - A Apparent power S in V • A
Thermod ynamics and Heat transfer
Thermo- T,e kelvin K OK = -273.15°C Kelvin (K) and degrees Celsius (°C) are
dynamic used for temperatures and tempera-
temperature degrees °C 0°C =273.15 K ture differences.
Celsius Celsius 0°C = 32 °F t = T - T0; 0T = 273.15 K
temperature 0°F = -17. 77 °C degrees Fahrenhe it (°F): 1.8 °F = 1 °C
Quantity of Q joule J 1J = 1 W • s = 1 N • m 1 kcal s 4.1868 kJ
heat 1 kW -h = 3600000 J = 3.6 MJ
Net calorific joule per J/kg 1 MJ/kg = 1000000 J/kg Thermal energy released per kg fuel
value Kiet kilogram J/m 3 1 MJ /m 3= 1000000 J/m 3 minus the heat of vaporization of the
water vapor contained in the exhaust
Non-SI units Joule per
gases.
cubic meter
Length Area Volume Mass Energy, Power
1 inch = 25.4 m m 1 sq.in = 6.452 cm 2 1 cu.in =16.39 cm 3 1 oz = 28.35 g 1 PSh = 0.735 kWh
1 foot = 0.3048 m 1 sq.ft = 9.29 dm 2 1 cu.ft = 28.32 d m 3 1 lb = 453.6 g 1 PS = 735 W
1 yard = 0.9144 m 1 sq.yd = 0.8361 m 2 1 cu.yd = 764.6 d m 3 1 me tric t = 1000 kg 1 kcal = 4186.8 Ws
1 nautical 1 US gallon = 3.785 dm 3 1 short ton = 907.2 kg 1 kcal = 1.166 W h
1 Imp. gallon = 4.536 d m 3 1 carat = 0.2 g 1 kpm/s = 9.807 W
mile = 1.852 km Pressure =158.8 dm 3 1 Btu = 1055 W s
1 mile = 1.609 km 1 hp = 745.7 W
1 barrel
1 bar = 14.5 psi
Prefixes of decimal factors and mu ltiples
Prefix pico nano micro milli centi deci deca hecto kilo mega giga tera
Prefix sym bol P n M m c d da h kMG T
Power of ten
1 0 -1 2 10-9 10-6 10 "3 10" 2 io-1 10 1 102 10 3 1 06 109 1012
Factor Multiple
1 mm = 10"3 m = 1/1000 m, 1 km = 1000 m, 1 kg = 1000 g, 1 GB (Gigabyte) = 1000000000 bytes
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Ma them atics: 1. en 23
Calculations in a right triangle
The Pythagorean Theorem
In a right triangle the square of the hypotenuse is equal Square of the
to the sum of the squares of the two sides. hypotenuse
a side
c 2 = a2 + b2
cb shiydpeotenuse
1st example: Length of the
hypotenuse
c = 35 m m ; a = 21 m m ; b = ?
b = > / c 2 - a 2 = 7(35 m m ) 2 - ( 2 1 m m )2 = 2 8 mm c = \la2 +b2
2nd example: Length of th e sides
CNC program w ith R = 50 mm and 1= 25 mm . a = yjc2 -b2
b = \lc2 -a2
K=?
c2 =a2+b 2
22 2
P2 R =I f l 2 +-K1 2 = V50 2 m m 2 - 252 mm2
K = V
X£ 3
K = 43.3 mm
Euclidean Theo rem (Theore m of sides)
The square ov er one side is equal in area to a rectangle Square over the side
formed by the hypotenuse and the adjacent hy potenuse b2 = c• q
segment.
a, b sides
c hypotenuse
p, q hypotenuse segments
b/ a Example: a2 = c • p
fq c
A rectangle with c = 6 cm and p = 3 cm should be
c-q C'P changed into a square with the same area.
How long is the side of the square a?
a2 =c •p
a = yjc •p = V 6 cm - 3 cm = 4.24 cm
P y t h ag o r ea n t h e o r e m o f h e i g h t
/ h The square of height h is equal in area to the rectangle of Square of the height
the hypotenuse sections p an d q. h2 = p •q
/q h height
p, q hypote nuse sections
P
Example:
p.q P
Right triangle
p = 6 c m ; q = 2c m ; h = ?
h2 = p •q
h = V p • Q = 76 cm • 2 cm = Vl 2 cm 2 = 3.46 cm
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24 M a t h em a t ic s : 1. nts
Division of lengths. Arc length. Composite length
Sub-dividing lengths
Edge distance = spacing I total length n numbe r of holes Spacing
p spacing
P PPP P = n +1
Example:
-A
i 1 = 2m; n = 24 holes; p = ?
/ 2000 m m
P = n 1+ 2 4 + 1 = 8 0 mm
Edge distance ^ spacing / total length n numb er of holes Spacing
P PPP p spacing a, b edge distances
l-(a +b)
eeeee Example: P n-1
/ = 1950 mm ; a = 100 mm; b = 50 mm ;
Subdividing into pieces
n = 25 holes; p = ?
/ l-(a + b ) 1950 mm -1 50 mm _
Ir
p = n-1 :— = 2—5 - 1: 75 m m
Arc length
/ bar length s saw cutting width Num ber of pieces
Example: Torsion spring
z num ber of pieces /r remaining length Remaining length
/s piece length
| /r = / - z - ( / s ~
Example:
/ = 6 0 0 0 m m ; l s = 230 mm ; s = 1.2 mm ; z = ?; /r = ?
z= / = 6000 m m = _25.9_5 = 25 p.ieces
l r = l/ s- z+ s• (/s2 +30s)m=m60+001.m2 mmm-2 5 • (230 mm + 1.2 mm)
= 220 m m
la arc length a angle at center Arc length
r radius d diameter
Example:
r = 36 mm; a = 120°; L = ?
n - r- a x- 36 mm -120° = 75.36 mm
'a = 180° 180°
Composite length
D outside diameter d inside diameter
t thickness
dm mean diameter L comp osite length
/•i, l 2 section lengths
a angle at center
Example (composite length, picture left):
D = 36 0 m m ; t =5 m m ; a = 270°; 12 = 7 0 m m ; Composite length
dm = ?;L = ?
L = lA + l 2 + ...
d m = D-t =360 mm - 5 mm = 355 m m
L + = 360 + /o
ji • 355 mm • 270°+ 70 m m = 906.45 mm
360°
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Mathematics: 1.5 Lengths 25
Effective length, Spring w ire length. Rough length
Effective lengths
Circular ring D outside diameter Effective length
d inside diameter of a circular ring
dm mean diameter / = n • d
m
It tehfifcekcntievses length
Effective lengt h of a
a angle at center circular ring sector
l=ji-dm-a
360°
Circular ring sector Example (circular ring sector):
I D = 36 m m ; t = 4 mm ; a = 240°; dm = 1) 1= 1 Mean diameter
drn = D
d m = D - f = 36 mm - 4 mm = 32 m m t
/ = —7x-d—m-a = jt-32 m m -240°= 67.02 m m
dm = d+t
360° 360c
Spring wire length
Example: Compression spring / effective length of the helix Effective length
D m mean coil diameter of the helix
/ num ber of active coils
I = JT • D M • / +
2 • n • Dm,
1=71- Dm. (1 + 2)
Example:
D m = 16 m m ; /'= 8.5; / = ?
I = ji • D m • /' + 2 • it • D m
D , = 71-16 mm • 8.5 + 2 • Jt • 16 mm = 52 8 mm
Rough length of forged parts and pressed parts
When forming without scaling loss the volume of the rough Volume with out sca-
part is the same as the volum e of the finished part. If there ling loss
is scaling loss or burr formation, this is compensated by a
factor that is applied to the volum e of the finished piece. Va-V e
Va vvoolulummeeoof fththeefirnoiusghhedpapratrt V o l u me w i th s c al in g
addition factor for scaling loss or loss due to burrs loss
cross-sectional area of the rou gh part
cross-sectional area of the finished part Va=Ve + q
initial length of the addition
length of the solid forged part v a = v e (1 + q)
Example: A| • /< | =^ 2 • •H + q)
A cylindrical peg d = 24 mm and I2 = 60 mm is pressed
onto a flat steel workpiece 50 x 30 mm. The scaling
loss is 10 %. What is the initial length ^ of the forged
scaling loss addition?
+ q)
Av I, = A 2 • / 2 - f l + g)
1= • / 2 - d + g)
A
jt • (24 mm)2 60 mm • (1 +0.1) = 20 m m
4 • 50 mm • 30 mm
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26 Mathem atics: 1. s
Angular areas
Square
A area d length of diagonal Area
I length of side
/ / Example: A = l2
/ /' — / = 14 mm ; A = 7 ;d=? Length of diagonal
A =l 2 = (1 4 m m ) 2 = 196 m m 2
/' d=fZ •/
d = / 2 • / = { • 14 mm = 19.8 m m
/ Area
t A = l •w
Rhombus (lozenge)
A area w width
/ length of side
Example:
/ = 9 m m ; w= 8.5 mm; A = ?
2
A = I •w= 9 m m • 8.5 mm = 76.5 m m
Rectangle
A area w width Area
I length d length of diagona l
A=I•w
Example:
/= 12 m m ; w = 11 m m ; A = ? ;d=? Length of diagonal
A =1 - w = 12 m m • 11 m m = 132 m m 2 \~l~7i 2 + w2
I d = v / 2 + w2 = V(12 m m )2 + (11 m m ) 2
= 16.28 m m
Rhomboid (parallelogram)
A area w width Area
/ length
A = I •w
Example:
/ = 36 mm ; w= 15 m m ; A = ?
/ A =1 -w= 36 mm • 15 mm = 54 0 m m 2
Trapezoid
A area / m average length Area
/•I longer length w width
l 2 shorter length >1 Z1 /2
A = -—- •w
Example: 2
/-i = 23 m m ; l2 = 2 0 m m ; w= 17 m m ; A = 7 Average length
A.b±k 23 m m + 20 m m • 17 mm 'm ; 1 2+ / 2
w =
= 365.5 m m 2
Triangle
>4 area w width Area
I length of side
Example:
ly = 6 2 m m ; w = 2 9 m m ; A = ?
AM U -w = 6 2m m • 29 m m „„„ m m _2
= J 899
22
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7/18/2019 Mechanical and Metal Trades Handbook 27
Equilateral triangle M athe m atics : 1. rea
Regular polygon s Triangle, Polygon, Circle
A area Diameter of
d diameter of inscribed circle circumsc ribed circle Area
/ length of side
h height D = — • yj3 •I = 2- d 4
D diameter of circumscribed 3
circle
Example: Diameter of
I = 4 2 m m ; A = 7; inscribed circle Triangle height
A =--V3-/2= -V 3-(42mm)2 3 2 2
44
- 763.9 m m 2
A area Diameter of Area
inscribed circle
>4 = n-l-d
/D ldeinagmthet eorf soidf ec i rc ums c ri bed d = ^ D2-1 2
circle Length of side
Diameter of
d diam eter of inscribed circle circumscribed circle
n no. of vertices
a angle at center
vertex angle
Example: Angle at center
H e x a g o n w i t h D = 80 m m ; / = ?; d = ?; A = 7 36 0 c
a=
f180° ^ (180° ^
n
/ = D • s in = 8 0 m m - s i n ^ —— J = 4 0 m m
Corner angle
d = V D 2 - / 2 = A/6400 m m 2 - 1 6 0 0 m m 2 - 69.282 mm
P= 18 0° -a
„ n - l - d 6 • 40 mm • 69.282 m m 4156.92 m m 2 ,
>4 = — - — =
C a lc u l a t io n o f r e g u l ar p o l y g o n u s i n g t a b l e v a l u e s
No. of Area A « Diameter of Diameter of Length of side I «
Vertices n circumscribed circle D « inscribed circle d
0.325 • D2 1.299 d2 0.433 • I2 1.154 2 000 0.578 • / 0.500 0.867 1.732
0.500 • D2 d2 1.414 1.000 / 0.707 0.707
1 000 d2 1.000 I 2 1.414 1 000
0.595 D2 1.702 0.50808
0.908 1.721 • / 2 1.215356 0.383 0.75727
0.309 0.414
0.649 D2 0.866 d2 2.598 • I2 2.000 1 082 1.377362 • // 00.880696 0.259 0.325
0.707 • D 2 0.829 • d2 4.828 • I 2 2.614 1.052 2.414 • / 0.924 0 268
10 0.735 • D2 0.812 d2 7.694 • / 2 3.236 1.035 3 . 0 7 8 • I 0.951
12 0.750 • D2 0.804 • d2 11.196-1 2 3.864 3.732 • / 0.966
Example: O c t a g o n with / = 20 mm A = ?; D = ?
A * 4.828 • I 2 = 4.828 • (20 m m ) 2 = 1931.2 m m 2 ; D « 2.614 • / = 2.614 • 2 0 m m = 52.28 m m
Circle
A area C circumference Area
d diameter
Example:
d = 60 m m ; A = ?; C = ?
. ji-d2 JT • (6 0 mm )2 2 82 7 m m 2 Circumference
A= =
44 C=K- d
C = JI • d= • 60 m m = 188.5 m m
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28 Ma the m atics : 1. s
Circular sector, Circular segment. Circular ring. Ellipse
Circular sector
A area / chord length Area
d diameter r radius
l a arc length a angle at center
Example:
d= 48 mm; a = 110°; la = ?;A = ?
L, = n-r-a = J I- 24 m m -110°= 46.1 m m
180° 180°
„ = JLi — =• r 46.1 mm -24 m m- = 553 m m_2
A
22
Circular s egm ent A area w width of segment Area
d diameter r radius
C i rc u l ar s e g m e n t w i th a < 180c /a arc length a angle at center
/ chord length
Example:
r= 30 m m; a = 120°; / = ?; w=?;A = ?
. a 0 . 120°
/ = 2 - r - s i n - = 2-30mm-sin = 51.96 mm
/ a 51.96 mm 120° ^ ^
w = - - t a n - = tan 14.999 mm = 15 mm
=2i-d 2 4 a l-(2r-w) 4
"" 4 ~ 360° 2
n • (60 m m)2 1203 _ 51.96 mm • (30 mm - 15 mm)Height of segment
4 360° 2 Ia
= 552.8 m m 2 Arc length w = 2- •tan—4
Radius
w I2 J I • r-a w = r- I * - * -
L =
r = — + 8-w 4
180°
2
Circular ring
A area d m mean Area
D outside diameter diameter A = n-dm • w
d inside diameter w width A = -- (D2 - d2)
Example: 4
D= 160 mm; d= 125 mm ; A = ?
A = - • D2 - d2) = - • (1602 m m 2 - 1 2 5 2 m m 2 )
44
= 7834 m m 2
Ellipse
A area d diameter Area
EDx a ml epnlget:h
C Jt D-d
Circumference
D = 6 5 m m ; d = 2 0 m m ; A = ? Circumference
D . n -D d n • 6 5 m m • 2 0 m m C «JtD• +d
A = =
4 4
= 1021 m m 2
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M a t h e m a t i c s : 1 .7 V o l u m e a n d S u r f a c e a re a 29
Cube, Square prism, Cylinder, Hollow cylinder. Pyramid
Cube V volume / length of side Volume
>4S surface area
/ L_ V=l3
Square prism Example:
Cylinder 1 = 20 m m ; V=?;AS = ? Surface area
V = / 3 = (20 mm ) 3 = 8000 m m 3
A s = 6 • I 2 = 6 • (20 m m ) 2 = 24 00 m m 2 As = 6 •I2
V volume h height Volume
As surface area w width
/ length of side V= I •w h•
Example: Surface area
/ = 6 cm ; w=3 cm ; h = 2c m ; l / = ? 3 A r = 2 • [I •w +I •h + w •h)
V = I •w •h = 6c m • 3 c m • 2 c m = 36 cm
V volume d diameter Volume
As surface area h height
Ac cylindrical surface area
Example: Surface area
d= 14 m m ; h = 25 m m ; V= ?
= Ji • d • / l + 2 •C-d2
jt-(14mm)2 Cylindrical su rface area
•25 m m
Ac = J I •d •h
= 3848 m m 3
Hollow cylinder
V volume D, d diameter Volume
As surface area
h height
Example: Surface area
D =42 m m ; d = 20 m m ; h = 80 mm;
V=?
V = — - (D2-d 2)
re4• 80 m m ( J _ 0 0 0. A; = n-(D + d) -•(D-d) + h
= (422 m m 2 - 2 0 2 m m 2)
4
= 85703 m m 3
Pyramid
V volume / length of base Volume
h height li edge length
hs slant height w wid th of base l-w-h
v= 3
Example: Edge length
/= 1 6 m m ;w = 2 1 mm;/? = 45 mm; V= ? h s 2 + 4^
V = -w-h 16 m m • 21 m m • 45 mm
Slant height
= 5040 m m 3
hs=y h, 2 h —4I2
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30 M a t h e m a t i c s : 1 .7 V o l u m e a n d S u r f a c e a re a
Truncated pyramid. Cone, Truncated con e, Sphere, Spherical segment
Truncated pyramid
V volum e >4i area of base hs slant height Volume
ly,l2 lengths of surface h height
w 1( w2 widths
base A2 top surface
Example:
/w•i2= =4015mmmm; l;2 h= =2520mmmm; w; -V\ ==?28 mm; Slant height
h s= J h 2 + l 1'-'2
3
50 m m • (1120 + 330 + 71120-330) m m 2
= 34299 m m 3
Cone
V volume height Volume h
slant height
Ac conical surface area V = n-d 2
d diameter
Example: Conical surface area
d = 52 mm ; h= 110 mm ; V= ?
Ti-d2 h A =n-d-hs
V =
4 *3 Slant height
^ • ( 5 2 m m ) 2 110 m m
" s = J T + " 2
= 77870 m m 3
Truncated cone V volume diameter Volume
Sphere Ac conica l surface area of top
D odfiabmaseeter V = —2 .(D2+d2+D-d)
shleaingth th e i g h t
Example: Conical surface area
D =1 00 m m ; d= 6 2 m m ; h =80 m m ; V= ? A c= K 'h *. (D + d)
V = --(D2+d 2 + Dd) c 2
2
Slant height
tt • 80 m m (1002 + 6 22 +100 • 62) m m02 mhs = \jh2 + j 2
12
= 419800 m m 3
AVs vsoul urfmacee area d diameter of sphere Volume
Example:
Surface area
d =9 m m ; V=?
A, = Jt •d2
y_Tc-d 3 j t - ( 9 m m ) 3 = 38 2 m m 3
Spherical segm ent
V volume d diameter of sphere Volume
A\ lateral surface area
A s surface area h height V = ii-h2- d_h
2 3
Example: [Surface area
d = 8 m m ; h = 6 m m ; V= ?
I As = 7i •h • (2 • d- h)
o? ? ( 8 mm 6 mm Lateral surface area
= it • 6 Z m m z •
3 [ A\ = k •d •h
I2
= 226 m m 3
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Mathematics: 1. a 31
Volumes of composite solids, Calculation of mass V2+...-V3-Vt
Volumes of co mpos ite solids
V total volum e Total volume
V-\, V2 partial volum es V= V, +
Example:
Tapered sleeve; D = 42 m m ; d= 2 6 m m ;
d-i = 16 mm; h = 45 m m ; V= ?
tt = 7±J±.(D*+d2+Dd)
12
jt • 45 m m ( J _ 0 __ 0 t n 0
(422 + 2 6 2 + 42 • 26) m m 2
12 3
. . = 41610 m m J I - 1 6 2 m m 2 ,
JI - D F , 45 mm = 9048 m m J
Vj = — •h
2 4
4
V = Vy-V2 = 4 1 6 1 0 m m 3 - 9 0 4 8 m m 3 = 3 2 5 6 2 m m 3
Calculation of m ass
Mass, general
m mass q density Mass
V volume
m = V •q
Example:
Values for density of
Workpiece made of aluminum; solids, liquids and gases:
V= 6.4 d m 3 ; e = 2.7 kg/d m 3; m = ? pages 116 and 117
m =V • g =6 A d m 3 • 2.7kg 3
= 17.28 k g dm
Linear m ass density m mass / length Linear m ass density
m= m' • I
Area mass density m' linear mass density
Application: Calculating
/< Example: the mass of profile sec-
tions, pipes, wires, etc.
Steel bar with d = 15 mm ; using the table values for
m ' = 1.39 kg/m; / = 3.86 m; m = ? m'
m = m'.l = r 39k ma— - 3 . 8 6 m Area mass density
= 5 .37 k g m= m • A
m mass A area Application: Calculating
the mass of sheet metal,
m" area mass density foils, coatings, etc using
the table values for m"
Example:
Steel sheet
f =1.5 mm; m" = 11.8 kg/m 2;
A = 7.5 m 2 ; m = ?
m = m "->4 = 1 1.8 m^ -2- 7 . 5 m 2
- 88.5 kg
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32 M a th em a tic s: 1. nts
Centroids of Lines and Plane Areas
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30 A /J I Table of Contents 33
m
20 LA/ / J 2 Physics 34
10 35
2.1 Motion
0 Uniform and accelerated m otion 36
Speeds of machines 36
1 2 3 4s5 37
time t • 2.2 Forces 37
A d d i n g and resolving force vectors
7><7i We ight, Spring force 38
Lever principle, Bearing forces 34 09
F = F r Torques, Cen trifugal force
2.3 Wo rk, Pow er, Efficiency
Mechanical work
SP iomwpelre amn da cEhfifniceise n c y
Fu 2.4 Friction 41
Friction force 41
fR ' 1 E Coefficients of friction 41
Friction in bearings
^ 2.5 Pressure in liquids and gases 42
Pressure, defin ition and types 42
^P Buoyancy 42
Pressure changes in gases
L—
2 .6 S t r e n g t h o f m a t e r i al s
/1 A/
Load cases, Load types 43
Safety factors, Me chanical strength properties .. 44
Tension, Com pression , Surface pressure 45
Shear, Buck ling 46
SBheanpdeinfga,cTtoorrssioinn streng th 44 78
Static mom ent, Section modulus, Mo men t of inertia . 49
Com parison of various cross-sectional shapes .. 50
2.7 Thermodynamics 51
Temperatures, Linear expansion, Shrinkage 51
Quantity of heat 52
Heat flux, Heat of combustion
© 2.8 OElhemct'rsicLitayw, Con ductor resistance 53
Resistor circuits 54
Types of current 55
Electrical wo rk and pow er 56
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34 Physics: 2.1 Motion
Uniform motion and uniformly accelerated motion
U n if o rm m o t i o n
Linear motion
Displacement-time diagram v velocity Velocity
t time
30 V = —S
t
1m > s displacement
1 20 - m m o c km
jt Example: 1— =60 =3.6 —
c1z 10 /to vl* v =4 8 km/h; s = 12 m; t= ? s min h
1 k m i c cc-7 m
CaQEJJi 3 4s5 _ km 48000 m „ 0 0 0 m 1 — = 16.667
"a. —
u Conversion: 48 — = • = 13.33 — h min
= 0.2778 —
0 ( 1 2 t = * = 12m h 3600 s s s
= 0.9 s
v 13.33 m/s
XJ time t
Circular motion
v circumferential velocity, n rotational speed Circumferential
cutting speed r radius velocity
a) angular velocity d diameter j i • d • n
Example: V = co • r
Pulley, d = 250 mm ; n= 1400 m in - 1 ;
v = ?; a) = ? Angular
Conversion: n = 1400 m in - 1 =1400 = 23.33 S "1 velocity
60s
v =ji . d • n = n • 0.25 m • 23.33 S"1 = 18.m3 — i . . . 2 • ii • n
to =2 • x • n = 2 it -23.33s~1 - 146.6 s " 1
1 •i 1
sFeoer apacguetti3n5g. speed of a circumferential velocity = min -1 =
m in 60 s
Uniformly accelerated motion
Linear ac c elerated mo tion
Velocity-time diagram The increase in velocity per second is called accel- The following applies
eration; and a decrease is deceleration. Free fall is to acceleration from
tl uniformly accelerated mo tion on which gravitational rest or deceleration to
acceleration g is acting. rest:
*4
\ \^ / v term inal velocity (acceleration), Terminal or initial
t M = 3 s-n.2 Vt ^ r - or initial velocity (deceleration) velocity
i— a
/ s displacement t time
V a acceleration g agcrcaevlietartaiotinoanl V =•a-t
2 3 4s5 1st exam ple: V = ^ 2-a-s
time t
Object, free fall from s = 3 m; v = ? Dis plac ement due to
Displacement-time diagram acceleration/
m deceleration
\ 1 2 a = g = 9.81 —
1
m s2 2 •f
V = V2 • a • s = y j l • 9.81 m /s 2 • 3 m = 7.7 —
123
time t s
2nd example:
Vehicle, v= 80 km/h; a = 7 m /s 2 ; 1 2
Braking distance_s_ =km? 80000 m „ „ „ „ m S = 2 3• f
Conversion: v = 80 — = = 22.22 — V2
s=
h 3600 s s
2a
v = V 2 • a • s
2 —(22 2- .722mm/-s/sr2)- 2= 3o5c.3 ,m
v a
s = 2 =
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Physics: 2.1 Motion
Speeds of machines
Feed rate Vf feed rate Feed rate
n rotational speed for drilling, turning
Turning f feed
Vf = n • f
Milling fNx nfeuemdbpeer rocfuctutitntgingedegdeges, or
number of teeth on the pinion Feed rate
for milling
P thread pitch
vf = n • ft • N
p pitch of rack and pinion
fFoeresdcrreawte drive
1st example: = n -P
Cylindrical milling cutter, z= 8; ft = 0.2 m m; Feed r ate for
rack and pinion
n = 45/min; v f = ?
Vf =n • N - P
v. = rr t • N = 45 — • 0.2 m m • 8 = 72 — Vf = 71 'd • n
m in min
Screw Cutting speed
drive vc = n • d n
2nd example:
Feed drive with threaded spindle,
P = 5 m m ; n = 112/min; v f = ?
Threaded spindle __ . 1 n . mm
with pitch P
v< — n • P — 112 5 m m = 560
m in min
Rack and 3rd example:
pinion
Feed of rack and pinion ,
n = 80/min; d = 75 mm ; v f = ?
1
Vf = Ji • d • n = j i • 75 m m • 80
m min
= 18850 m in =18.85 m i n
C u t t i n g s p e e d , C i r c u m f e r en t i al v e l o c i t y
C u t t i n g s p e ed vc cutting speed
v circumferential velocity
d diameter
n rotational speed
Example: Circumferential
Turning, n= 1200/min; d= 3 5 m m ; vc = ? velocity
vc = n-d • n = it • 0.035 m • 12001 v = it • d • n
min
Circumferential
velocity = 132 m
min
A v e r ag e s p e e d o f c r an k m e c h a n i s m
va average speed Average speed
n number of double strokes V a = 2 • S • n
s stroke length
Example:
Power hacksaw,
s = 280 m m; n = 45/min; va = ?
1
va = 2 - s • n = 2 - 0.28 m • 45
min
= 25.2 m
min
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36 Physics: 2. o
Types of forces
Add ing and resolving forces
Chosen for the following Fy, F2 com ponent forces / vector magn itude Vector magnitude
exam ples /ty = 10
Fr resultant force (length) /=
F:
Fr Representing forces s ca le of fo rc es M<
Forces are represented by vectors. Sum
F r = F 1 + F 2
The length / of the vector corresponds to the
D i f f er e n c e
magnitude of the force F. Fr=F,-F2
Adding c ollinear forc es ac ting in the s am e direc tion
Example: F 1 = 8 0 N ; F 2 = 1 6 0 N ; Fr = ?
Fx = ^ + F 2 = 80 N + 160 N = 240 N
Subtracting collinear forces acting in opposite directions
Fr Example: F 1 = 240 N; F 2 = 90 N; F r = ?
Fr = ^ - F 2 = 240 N - 90 N = 150 N
Addition Ad dition and res olution of forc es Solv ing a forc e diagram by
X V v -1 1f to c wh os e lines of ac tion inters ec t adding or resolving
(force vectors)
Resolution Example of graphical addition:
Shape of Required
^ = 120 N; F 2 = 170 N; y = 118°; the force trigonometric
M f = 10 N/mm; F r = ?; measured: / = 25 m m diagram function
F r = / • /V7f = 25 mm • 10 N/m m = 250 N
Force diagram sine,
Example of graphical resolution: with right
cosine,
F r = 260 N; a = 90°; £ = 15 °; M f = 10 N/mm; angles
Ft = ?; F 2 = ?; measured: /t = 7 mm ; l 2 = 27 m m tangent
Ft = /t • /W f = 7 m m • 10 N/m m = 70 N
F 2 = / 2 • M f = 27 mm • 10 N/mm = 270 N Force diagram Law of sines,
with oblique Law of
angles cosines
Forces of acceleration and deceleration
A force is required to accelerate or decelerate a mass. Acceleration force
F= m • a
F acceleration force a acceleration
m mass
Example: 4 H • 4 1 m
m = 50 kg; a = 3sm—2 ; F = ? 1 N = 1 kg • —
F = m • a = 50 k g • 3 ^ = 150 k g • ^ = 150 N
s 2 s 2
Weight
Gravity generates a weight force on a mass. Weight
| F w = m ~
Fw weight g gravitational
~ rn „ „ m
m =1kg r m mass acceleration <7 = 9.81—s 2r - «10—sr2-
Calculation of mass:
' F w = 9,81 N Example: page 31
I-beam, m = 1200 kg; Fw = ? Spring force
F = R • S
F w = m • g = 1200 kg • 9.81 ^ = 11772 N
Change in spring force
Spring force (Hooke's law)
AF= /?• As
The force and corresponding linear expansion
of a spring are proportional within the elastic range.
F spring force s spring displacement
R spring constant
Example:
0 10 20 mm 40 Compression spring, R = 8 N/mm; s = 12 mm ; F = ?
spring F = R • s = 8 — • 12 m m = 96 N
displacement s
mm
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Physics: 2. orce 37
Torque, Levers, Centrifugal force
Torque and levers
Single-ended lever The effective lever arm is the right angle distance M o m e n t
between the fulcrum and the line of application of M= F • I
the force. For disk shaped rotating parts the lever
arm corresponds to the radius r.
M mom ent F force Lever principle
DM, = l M r
/ effective lever arm
2M \ sum of all counter-clockwise m omen ts
lM r sum of all clockwise mom ents
Example: Lever principle wi th
only 2 applied forces
Angle lever, Fy = 30 N; l<\ = 0.15 m; l 2 = 0.45 m;
F2 = ? F , - h = F 2 - l :
F , = F,-/-, 30 N -0.15 m
= 10 N
U 0.45 m
Bearing forces
Ex ample of bearing forc es A bearing point is treated as a fulcrum in calculating Lever principle
TF bearing forces. /, / l f l2 effective | 2My = Mr
F A , Fb bearing forces
Torque in gear drives
F 1f F2 forces lever arms
Centrifugal force
Example:
Overhead travelling crane, F| = 40 kN; F 2 = 15 Bearing forc e at A
kN ; /•, = 6 m; l 2 = 8 m; / = 12 m; F A = ?
Solution: B is selected as fulcrum poin t; the
bearing force F A is assumed on a single-
ended lever.
Ft • A, + F2 • I2 40 kN • 6 m + 15 kN • 8 m
FA = I = 30 k N
12 m
The lever arm of a gear is half of its reference diame- Torques di
te r d. Different torques result if two engaging gears M-| = F t 1
do not have the same number of teeth. 2
Driving gear Driven gear
F t1 tangen tial force F t2 tangen tial force Ft2-d 2
M1 torque M 2 torque m2 = 2
di reference diame ter d2 reference diame ter
Zy num ber of teeth z2 number of teeth
n- 1 rotational speed n2 rotational speed m 2 = • i- M<\
m2 = £2
/ gear ratio
Example:
Gears, /'= 12; M^ = 60 N • m; M 2 = ? m2 = m
M 2 = /• M-i = 12 • 60 N • m = 720 N • m MI n2
For gear ratios for gear drives see page 259.
Centrifugal force Fc when a mass is made to move
along a curvilinear path, e.g. a circle. Centrifugal force
F c centrifugal force w angular velocity F r= m • r • o)
m mass v circumferential velocity
r radius
Example: m • v'
Fc =
Turbine blade, m = 160 g; v= 80 m/s;
d = 400 mm ; F c = ?
0.16 kg • (80 m/sP = ^ kg_m =
c r 0.2 m s2
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38 Physics: 2.3 Work, Power, Efficiency
Work and Energy
M e c h an i c al w o r k , l if t in g w o r k a n d fr ic t io n a l w o r k
Work is performed whe n a force acts along a distance. Work
W = F • s
F force in direction of travel W work
Fw weight s force distance
FF nr fnroicrtmioanl ffoorrccee /sj, h choeeigffhict ieonf tlifot f L iftin g w o r k
friction
1st example: [ W= Fw • h
F= 300 N; s = 4 m ; W = ?
Fric tional work
Krl W = F - s = 3 0 0 N - 4 m = 1200 N • m = 1200 J
FR F = F R 2nd example: 1 J = 1 N • 1 m
Frictional work, F N = 0.8 kN; s = 1.2 m; /x = 0.4; W = ? kg • m 2
• W = n • FN • s = 0. 4 • 800 N • 1.2 m = 384 N • m = 384 J
= 1 W • s = 1
1 k W • h = 3.6 MJ
Energie of po sition
Energy of Wn Energie of position is stored work (energy of position,
position spring energy).
Energy of pos ition
F G E, Wp energy of position R spring constant | W p - f w - s '
Fw weight s, h trave l, lift or fall
F force
height, spring
displacement
Spring energy Example: Energy of th e spring
Drop hammer, m = 30 kg; s = 2.6 m ; W p = ? P 2
W 0 = F W • s = 30 kg • 9.8m1 — • 2.6 m = 765 J
Kinetic energy Kinetic energy is energy of motion. Kinetic energy
o f l i n ea r m o t i o n
Linear motion E, l/l/k kinetic energy or work v velocity
m Kinetic energy
o f r o t at i o n a l m o t i o n
co angular velocity m mass
2
J mass mo me nt of inertia
l/K = J-co
Rotational motio n (rotation) Example:
Drop ha mm er, m = 30 kg; s = 2.6 m; W k = ?
v = j2-g-s = yj2-9.81 m /s 2 • 2.6 m = 7.14 m/s
m, m ' v 2 30kg- ( 7.14m/s)2 ^ ,
W^ = ——— = = /bo J
Golden Rule of Mechanics
"W hat is gained in force is lost in distan ce". Golden Rule
of Mechanics
W, input work W 2 output work
F2 outp ut force W<\ = W 2
Fy input force s 2 displacem ent of
s^ displacement of ^ • ST = F 2 . s2
force F2
force Ft rj efficiency FT • St = F\N • h
Fw weight Allow ing for
h height of lift friction
Example: W2
ri
Lifting device, Fw = 5 kN; h = 2 m ; F = 300 N; s = ?
F w-/ 7 5000 N • 2 m
s= = 33.3 m
300 N
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Fixed pulley1) Physics: 2.3 Work, Power, Efficiency
Block and tack le11 Simp le machines
Movable pulley11
F-\ = Fw W Fi = —
s-i = h c St = 2 • h
W 2 = FSN -h a [ W2 = Fw • h
n no. of load-bearing i o) a angle of inclination
ropes, pulleys Ft • ST = F w • h
1
Fi~-F wi
Inclined plane11
F2=Fvj
F 2 = F w L . U [ s<\ = n • h
W e d g e11 | w 2 =
F2=FW B o l t 11
angle of inclination P thread pitch
tan/3 incline / lever arm
For 1 full turn
jy* 'Vi iy xi
[ i • 2 • j i • / = Fo • P
. , i q|K
| St = 2 • jt • /
s 2 = s-i • tan/? W i = F i • 2 • j i • /
H o is t in g w i n c h 11 Wo = Fo • h G ear w i n c h 11 I w 2 = F 2 • P
/ crank length l crank length
d drum d drum
diameter diameter
n D number of turns / gear
of the drum ratio
F^ -d
fw-d
Fi •/ = Fi • / •/ =
II F 2 = F w h = j i • d • Hp /=
W 2 = FSN -h W 2 = F\m • h
1 ) The formulae apply to a hypothe tical frictionless cond ition, wherein the ou tput work 14/- is equal to the input wo rk 41/431
W2.
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40 Physics: 2.3 W ork , Pow er, Efficiency
Power and Efficiency
P o w e r i n li n ea r m o t i o n
Power is work per unit time. Power
P power s displacement in
the force direction
W work
v velocity t time
1st example:
Forklift, F = 15 kN; v= 25 m/m in; P= ?
P = F v - 15000 N- = 6250 = 6250 W = 6.25 kW
60s s
2nd example: 1 W =1 J-
s
Crane lifts a machine, m = 1.2 t; s = 2.5 m; N • m
f = 4.5 s; P = ?
Fw = m • g = 1200 kg • 9.81 m / s 2 = 11772 N 1 k W = 1.36 PS
P = filLf.11772N.Z5m
t 4.5 s
For power in pumps and cylinders see page 371.
P o w e r i n ci rc u l ar m o t i o n
P power s displacement in the force direction Pow er
M torque t time F> =F - V
F tangential force n rotational speed
v veloc ity cu angular veloc ity P = F - 7i • d - n
Example:
Belt drive, F= 1.2 kN; d = 200 mm ; n = 2800/min; P = ? P = M •2-:rc • n
P = F-n-d-n
= 1.2 k N . * . 0 . 2 m - ^ = 3 5 . 2 ^ = 35.2 kW P = M • w
60s s
or:
Numerical equation: Power
Enter —> M in N • m, n in 1/min
Result —> P in kW M-n
P=
For cutting power in machine tools see pages 299 and 300.
9550
Efficiency
input output Efficiency refers to the ratio of power or work output to the Efficiency
power power power or work input.
PQ2 = P2 Pi input pow er P 2 output power '1
Wi input work W2 output work n= —
?7i, rj2 partial efficiencies
rj to ta l efficie ncy
-f fl- =
motor gear- Example:
box
Belt drive, Py = 4 kW; P 2= 3 kW; = 85%; rj = ?;rj2 = ? Fotal efficiency
1
12 «1 = P22 = 3 kW = 0.7_5, T/o = n = 0.75 „_ M 7 = ^71 -V2-V3---
rj = rjvi72 Pi 4kW 2 ^ 0.85 =0.88
Efficiencies 7 (approximate values)
Brown coal powe r station 0.32 Gas oline engine 0.27 Screw thread 0 . 30
Coal pow er station 0.41 Automobile diesel engine (partial load) 0.24 Pinion gear 0.97
Natural gas pow er station 0.50 Au tom obile diesel engine (full load) 0.40 Worm gear,/'= 40 0 .6 5
Friction drive 0.80
Gas turbine 0.38 Large diesel engine (partial load) 0.33 Chain drive 0.90
Wide V-belt drive 0 . 85
Steam turbine (high pressure) 0.45 Large diesel engine (full load) 0.55 Hydrostatic transm ission 0 . 75
Water turbine 0.85 Three phase AC mo tor 0 . 85
Cogeneration 0.75 Machine tools 0 . 75
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P h ys ic s: 2. r i c i n 41
Types of friction. Coefficients of friction
Friction force The resulting friction force is dependent on the normal Fric tion forc e for s tatic
force FN and the and sliding friction
Sta tic fr ic tio n , s lid in g fr ic tio n
• type of friction, i.e. static, sliding or rolling friction
IF* • frictional condition (lubrication condition):
N • sduryrf,amceixeroduogrhnviesscsous friction. Ff = ^ • FN,
• • material pairing (material combination)
Friction force
Sta tic fr ic tio n , s lid in g fr ic tio n These effects are all incorporated into the experim entally f or r o l li n g f r i c t i o n 1'
Fn determined coefficient of friction /z.
F c = • FNt
Rolling friction Fn normal force f coefficient of rolling friction
Fp friction force fi coefficient of friction r radius
1st example:
Plain bearing, FN = 100 N; \x = 0.03; FF = ?
/=p = M . F n = 0.03 • 100 N = 3 N
2nd example:
Crane whe el on steel rail, Fn = 45 kN; d = 320 m m ; 1 ) caused by elastic
f = 0 .5 m m ; F F = ? deformation be-
tween roller body
F f= n 0.5 mm -45000 N= 140.6 N and rolling surface
•
Fc r 160 m m
Coefficients of friction (guideline values)
Material pairing Ex ample of applic ation C o e f f i c i en t o f s t a t i c f r i c t i o n /jl C o e f f i c i en t o f s l i d i n g f r i c t i o n n
dry lubricated dry lubricated
steel/steel vise guide 0 20 0.10 0.15 0.10-0.05
steel/cast iron
machine guide 0.20 0.15 0.18 0.10-0.08 2
sstteeeell//PCbu--SSnn aallllooyy
0.20 0.10
steel/polyamide
steel/PTFE sshhaafftt iinn msoullitdilapylearinpblaeinarinbgearing 0.15 0.10 00.1100 00 .. 00 56 -- 00 .. 00 33 2 ''
steel/friction lining
steel/wood shaft in PA plain bearing 0.30 0.15 0.30 0.12-0.032'
low temperature bearing 0.042)
shoe brake 0.04 0.04 0.04
part on an assembly stand
0 60 0.30 0.55 0.3-0.2
0.55 0 10 0.35 0.05
wood/wood underlay blocks 0.50 0 20 0.30 0 10
cast iron/Cu-Sn alloy
rubber/cast iron adjustment gib 0.28 0 16 0 20 0.20-0.10
rolling element/steel
belts on a pulley 0.50
anti-friction bearing3', guideway3'
0.003-0.001
2 > The significance of the material pairing decreases with increasing sliding speed and presence of mixed and viscous
friction.
3 ) Calculation performed in spite of rolling m ovem ent, because it is typically sim ilar to calculations of static or sliding
friction.
C o ef f ic i en t s o f r o l l in g f r i c t i o n (g u i d e l i n e v a l u e s )41
Material pairing Example of application Coeffic ient of rolling fric tion f in m m 4 ) Data on coefficients
of rolling friction can
steel/steel steel whe el on a guide rail 0.5 vary considerably in
plastic/concrete caster wheel on concrete floor 5 technical literature.
rubber/asphalt car tires on the street 8
F r ic t io n m o m e n t a n d f r ic t i o n p o w e r i n b ea r in g s
M friction moment coefficient F ri c ti o n m o m e n t
of friction •d
F N normal force diameter
P friction power
Example: rotational speed 2
Steel shaft in a Cu-Sn plain bearing, /x = 0.05; Friction power
F N = 6 k N ; d= 160 mm ; M = ?
F F / Y F N d-n
M = fx-FN-d 0.05 • 6000 N 0.16 m= 24 N m
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42
Physics: 2.5 Pressure in liquids and gases
Types of pressure
Pressure
p pressure A area Pressure
F force P=
Example:
^^ P F= 2 MN; piston d = 400 m m ; p = ? Units of pressure
521 — F 2000000N ^ N 1 Pa = 1m-^2=0.00001 bar
n P = ^ r = — ^ — ^ = 1 5 9 1 — ? = 1 5 9 1 b a r
ir A JI • (40 c m )2 cm2 1 bar =10 ——r- = 0.1 ———
For calculations on hyd raulics and pn eumatics see page 370. cm' mm"
1 mba r = 100 Pa = 1 hPa
Gage pressure, air p ressure, absolute pressure
21 Q. p e gage pressure (excedens, excess) Gage pressure
P a m b a ' r pressure (ambient, surroundings) Pe ~ P a b s P a m b
bar bar <D p a b s absolute pressure
P a m b = 1 - 0 1 3 b a r « 1 b a r
1 0_ i- Tpho esi tgi vaeg,ei fppr easbssu r>e pi s m b and
(standard air pressure)
0-L-1 gj a
oO(0)) tam- air
negative, if p a b s < p a m b (vacuum)
~>qj WaQ>- Cip.Praems sur e Example:
b Car tires, p e = 2.2 bar; p a m b = 1 bar; p a b s = ?
Pabs = P e + P a m b = 2.2 bar + 1 bar = 3.2 bar
Ocu) Oto) 3
CO)t/3 - vacuum
Hydrostatic pressure, bu oyancy
p e hydrostatic pressure, FB buoyant force Hydrostatic pressure
inherent pressure V displaced volume
h depth of liquid I
q density of the liquid
Pe = 9- Q- h
g gravitationa l acceleration
IBuoyant force
^- V Example:
I F * = 9 • e • V
density q
g = _9. 8_1„ -mr » 1 0 -mr -
pressure m p.= g e- / I = 9 . 8 1 22 • 1000 m— 9 10 m
s 3 For density v alues, see page 117.
= 98100 kg = 98100 Pa « 1 bar
m • s"
Pressure ch anges in gases
Compression Condition 1 Condition 2 Ideal gas law
condition 1 condition 2 P a b s i absolute pressure Pabs2 absolute pressure Pabsi Pabs2 V2
volume V2 volume T-, T2
Pabsi P a b s T2 absolute
l/ 2 T2 7"i absolute
temperature temperature
h Example: Special cases:
constant temperature
A com pressor aspirates V^ = 30 m 3 of air at | Pabsi = P a b s 2 •V 2
P a b s i = 1 bar and f-| = 15°C and com press es
Boyle's Law it to V2 = 3.5 m 3 an d t2 = 150°C. constant volume
5 Wh at is the pressure p a b S 2 ? Pabsi Pabs2
T, T2
bar Calculation of absolute temperatures (page 51):
r n = fn + 273 = (15 + 273) K = 288 K constant pressure
i 4 T2 = t2 + 273 = (150 + 273) K = 423 K
> P a bs i - V r T2
y
Pabs2 " TVV2
1 2 3 dm 3 5 1 bar • 30 m 3 • 423 K T, T 2
volume V 288 K • 3.5 m 3 = 12.6 bar
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Physics: 2. r oeri 43
Load cases Types of loading Material properties, Stress limits
Load cases dynamic loading
static loading pulsating alternating
stationary
time — • Load case II Load case III
Load case I The load increases to a maximum The load alternates between a posi-
Magn itude and direction of the load value and then falls back to zero, tive and a negative maximum value
remain the same, e.g. for a weight e.g. for crane cables and springs. of equal magnitude, e.g. for rotating
load on columns. axles.
T y p es o f l o a d i n g , m a t e r ia l p r o p e r t i es , s t r es s l i m i t s
Material prop erties Stand ard stress lim its <7|;m
for load c ase
Type of load Stres s Strength Limit values Deformation
for plastic II III
Tension tensile
stress deformation
'///////<
tensile material pulsating alternating
M strength tensile tensile
yield strength elongation ductile brittle fatigue fatigue
Rm Re strength strength
£ (steel) (cast
0.2%-yield ^tpuls
point elongation iron)
R,02 at fracture
Ra Rn
Rp0.2
Compression compres- compres- natural material pulsating alternating
sion sion compres- compres-
Y77777Z stress compression compres- ductile brittle
Bending Or strength sion sion
yield point sion set (steel) (cast fatigue fatigue
m. bending °cB iron) strength strength
stress °cF
bending °cB ^cpuls
tfb strength 0.2%-offset compressive
yield strength failure tfcO.2
ObB
OcO.2 £cB
bending deflection bending pulsating alternating
limit limit bending bending
fatigue fatigue
°t>F strength strength
abpuls tfbA
m
Shear shear shear shear
Mm stress strength strength
TsB rSB
Torsion torsional torsional torsional angular torsional pulsating alternating
limit deflection limit torsional torsional
stress strength fatigue fatigue
*tF *tF strength strength
1
Ttpuls rtA
Buckling
Mt
buckling buckling buckling
stress strength strength
^bu °buB
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44 Physics: 2.6 Stre ngth of M aterials
Mech anic al strength properties, A llo w able stresses, Safety factors
M e c h a n i c a l s t r e n g t h p r o p e r t i es i n s t a t ic a n d d y n a m i c l o a d i n g 11
Type of load Tension, Comp ression Shear Bending Torsion
Load case I II III I I II III I II III
Stress Rp0.2 ^ tpul s t fb F ° b puis °bA *tF Ttpuls TtA
limit orHm °cF> °c0.2 ° c p u l s
Material Stress limit a|jm in N/mm 2
S235 235 235 150 290 330 290 170 140 140 120
S275 275 275 180 340 380 350 200 160 160 140
E295 295 295 210 390 410 410 240 170 170 150
E335 335 335 250 470 470 470 280 190 190 160
E360 365 365 300 550 510 510 330 210 210 190
C15 440 440 330 600 610 610 370 250 250 210
17Cr3 510 510 390
16MnCr5 635 635 430 800 710 670 390 290 290 220
20MnCr5 735 735 480
18CrNiMo7-6 835 835 550 880 890 740 440 360 360 270
C22E 940 1030 920 540 420 420 310
C45E
C60E 960 1170 1040 610 470 470 350
46Cr2
41Cr4 340 340 220 400 490 410 240 245 245 165
50CrMo4
30CrNiMo8 490 490 280 560 700 520 310 350 350 210
580 580 325 680 800 600 350 400 480 240
650 630 370 720 910 670 390 455 455 270
800 710 410 800 1120 750 440 560 510 330
900 760 450 880 1260 820 480 630 560 330
1050 870 510 1000 1470 930 550 735 640 375
GS-38 200 200 160 300 260 260 150 115 115 90
GS-45
GS-52 230 230 185 360 300 300 180 135 135 105
GS-60
260 260 210 420 340 340 210 150 150 120
300 300 240 480 390 390 240 175 175 140
EN-GJS-400 250 240 140 400 350 345 220 200 195 115
EN-GJS-500
300 270 155 500 420 380 240 240 225 130
EE NN -- GG JJ SS -- 67 00 00 430600 335350 210950 760000 556000 457200 320700 322900 320755 116705
1 ) Values were determined us ing cylindrical samples having d < 16 m m wit h polished surface. They apply to struc-
tural steels in normalized condition; case hardened steels for achieving core strength after case hardening and
grain refinement; heat treatable steels in tempered condition.
The compression strength of cast iron wit h flake graphite is o c b « 4 • R m .
Values according to DIN 18800 are to be used for structural steelwork.
A l l o w a b l e s t r e s s fo r (p r e -)s i zi n g o f m a c h i n e p a r t s
For safety reasons parts m ay only be loaded with a portion of the stress lim it <7|jm which will
lead to permanent deformation, fracture or fatigue fracture.
fallow allo wa ble stress C7|jm stress limit depe nding on
v safety factor (table below ) type of loading and load case Allowable stress
Example: (preliminary design)
What is the allow able tensile stress fallow for a hexagonal bolt ISO 4017 - M12 x 50 - °allow —
10.9, if a safety factor of 1.5 is required with static loading?
NN a |im 900 N/m m 2 N
a „ m - / ? e - 1 0 . 9 . 1 0 - —m-' = 900 m m '- ; = —vv
af,,a ll ow 11..55 = 600 — 3
mm
For mechanical strength properties for bolts see page 211.
Safety factors v for (pre-)sizing mach ine parts
Load case 1 (static) II and III (dynam ic)
Type of material ductile materials, brittle materials, ductile materials, brittle materials,
e.g. steel e.g. cast iron e.g. steel e.g. cast iron
Safety factor v 1.2-1.8 2.0-4.0 3-41> 3-61)
1 ) The high m argins of safety in part sizing relative to the stress limits are intended to comp ensate for yet unk now n
strength-redu cing effects due to part shape (for shape-related streng th factors see page 48).
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Physics: 2. r oeri 45
Tensile stress, Compressive stress, Surface pressure
Tensile stress
The calculation of allowable stress only applies to static Tensile stress
loading (Load case I).
F
0[ tensile stress Re yield strength
F tensile force Rm tensile strength
S cross-sectional area v safety factor
fallo w allowable tensile stress F a n o w allowable tens ile force Allow able tensile force
Example:
Round bar steel, <7 t,allow = 1 3 0 N / m m 2 Allowable tensile stress
°t, allow = Re
( S 2 3 5 J R , v = 1.8); F aiiow = 13.7 kN; d= ?
V
s = 13700N mm2 for
steel ° t , allow = V
a t , allow 130 N /m m 2
for
c =12 m m (according to table, page 10) cast
iron
For mechanical strength properties Re and R m see pages 130
to 138. For calculation of elastic elongation see page 190.
Compressive stress The calculation of allowable stress only applies to static Compressive stress
loading (Load case I).
Surface pressure Allowable
A'l-b tfcF compression yield point F compressive force compressive force
compressive stress ^aiiowallowable comp. force
S cross-sectional area
fallow allowable comp. stress R m tensile strength
v safety factor
Example:
Rack made of EN-GJL-300; S= 2800 m m 2 ; Allowable
compressive stress
v = 2.5; F.llow = ?
°c , allow °"cF
_ a 4 - a
4-a
Fallow c, allow ^ ~ •S for
steel °c, allow
4 - 3 0 0 N / m m 2 • 2800 m m 2 =1 344000 N
2.5 for
cast
For mechanical s trength prop erties see page 44 and pages 160-161 iron
F force contact surface,
p surface pressure projected area
Example: Surface pressure
Two metal sheets, each 8 mm thick, are joined with a
bboeltaDppINlie 1d4g4i5v-e1nOah 1m1a xx i1m6uxm30a. lHloowwabglreeastuarfafocreceprmesa-y
sure of 280 N/mm2?
N
F = p. A = 280 m m ' • 8 m m • 10 m m
22400 N
A l l o w a b l e s u r fa ce p r e s s u re f or j o i n t s w i t h p i n s an d b o l t s
m ade of steel (standard v alues)
Assembly type Pres s fit s mo oth pin Fit wit h n otc hed piec e Sliding fit s m ooth b olt
Load case I II III I II III I II III
Component material allowable surface pressure in N/mm 2
S235 100 70 35 70 50 25 30 25 10
E295 105 75 40 75 55 30 30 25 10
cast steel 85 60 30 60 45 20 30 25 10
cast iron 70 50 25 50 35 20 40 30 15
CuSn, CuZn alloy 40 30 15 30 20 10 40 30 15
AlCuMg alloy 65 45 25 45 35 15 20 15 10
For reference values for allow able specific be aring load of various plain bearing materials see page 261.
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46 Physics: 2.6 Stren gth of M aterials
Shear and buckling stress
Shear stress
The loaded cross-section must not shear. Shear stress
r s shear stress Fallow allow able shear force
alow allowa ble shear stress S cross-se ctional area
r s B shear strength v safety factor
Example: Allowable
shear stress
Dowel pin 0 6 mm, single-shear loaded,
•s, allow
E 295,v = 3; Fallow =? TsB
Ts,allow __ r s B _ 3 90 N/ mm 2 1 3 0 N V
mm m
v ~ 3- - m'
Ti-d2 jt • (6 )2
= 28.3 m m 2
single- double- allow = S T s, alow 28.3 mm 2-130 N = 3679 N Allow able shear force
shear shear f a l l o w ~~ S ' 7 s, al l ow
mm"
For mechanical strength properties r s B and safety factors see page 44.
Cutting of materials
The loaded cross-section must be sheared. Maximum
s hear s trength
TsB max max. shear strength S shear area
TsBmax ~ 0-8- /?m, max
Rm m a x max. tensile strength F cuttin g force
k \ \ \ \ y V\KV S3 Example:
, S=C-s
Punching a 3 mm thick steel sheet S235JR; Cutting force
f = n'd d= 16 m m ; F= ?
flmmax = 4 70 N /m m 2 (Table page 130) F= S • r s B m a x
^sBmax * 0.8 • H m m a x = 0.8 • 4 7 0 N / m m 2 = 37 6 N / m m 2
2
FS == jSi •• rds B•m sa =x j=i •1 1560.8mmm m• 23 •m 3m7 6=N1/5m0m.82 m= m56701 N
= 56.7 kN
For mechanical strength properties Rm max1o r steel, see pages 130 to 138
Buc k l i ng s t r es s (Euler c o l um n s )
Load case and free buckling Calculation for buckling of Euler columns applies only to
lengths (Euler colum ns)
thin (profile) parts and within the elastic range of the
workpiece. A l l o w a b l e b u c k l in g
Load case F b u aiiow allowable buckling force E Modulus of elasticity force
II III IV / length / Mo me nt of inertia
/b U free buckling length
v safety factor (in mac hine cons truction « 3-10)
IK — Example:
Beam IPB200, / = 3.5 m; clamp ed at both ends;
v = 10; F b u a i i o w = ?; E = 210000 N /m m 2 = 21 • 1 06 N / c m 2
(table below); 71* = 2000 c m 4
n 2 o i rc2 • 21 • 1 06 - i l L • 2000 c m 4
c " _ cm '
r b u , a l lo w /g u • v (0.5 • 350 cm )2 • 10
- MJ = 1.35 - 1 0 6 N = 1.35 MN
free buckling lengths 1 ) f or m om ent s of i ner t ia of an area ( 2nd m om ent ) , see pages
/bu=2-/ /b u= / l bu=0.1-l /bu=0.5-/49 and 146-151. Special calculation methods are stipulated
for structural steel according to DIN 18800 and DIN 4114.
M odul us of elas ti c it y Ei n k N/ m m 2
steel EN-GJL- EN-GJL- EN-GJS- GS-38 EN-GJMW- CuZn40 Al alloy T1 alloy
150 300 400 350-4
196-216 80-90 110-140 170-185 210 170 80-10 0 60-80 112-130
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7/18/2019 Mechanical and Metal Trades Handbook 47
Physics: 2. r oeri
Bending and to rsional stress
Bending stress
Tensile and compressive stresses occur in a member
during bending. The maximum stress is calculated in
boundary areas of the mem ber; they may not exceed
the allowable bending stress.
<M7bb bbeennddini ngg smtreosms e n t Ff bdeenfldeicntgionforce Bending stress
W axial section modulus
Example: Allowable bending
Beam IPE-240, W= 324 cm 3 (page 149); clamped at
stress a b anow
one end; concentrated load F = 25 kN; / = 2.6 m ; a b = ? from page 44
( ^ Wu 324 c m 3 * f l r tu n 2 0*N0UI N
cm mnr
Bending load cases in beams Beam with a uniformly distributed load
Beam loaded wi th a concentrated load
fixed at one end /W k= F-/ fixed at one end F •/
/ F -F • I Mk =
F • /3 F • /3
f= f=
8 EI
3 • E • I
supported at both ends supported at both ends
5 • F • I3
f=
384 • E • I
fixed at both ends fixed at both ends F • /
r = r • i M k = 12
F •/
F -/3 y W j i y J J f - F •/3
f= I 384 •E •/
192 • E • /
E Mo dulus of elasticity; values: page 46 / 2nd mom ent of inertia; formulae : page 49; values: pages 146 to 151.
F' Distributed load (load per unit length, e.g. N/cm) / Length of distributed load
Torsional stress
Mt torsional mom ent r t torsional stress
Wp polar section mo dulus Torsional stress
Example: Tt = W n
Shaft, d =32 m m ; Mt = 420 N • m; r t = ? Astlrleoswsabr tl ea n toowr sfiroonmal-page
44 or page 48
rc-d3 n-(32 m m ) 3 3
16 16 = 6434 mm
_ M t_ 420000 N • m m _ N
= 65.3
6434 mm 3 mm"
For polar section m odu li see pages 49 and 151 49/431
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48 Physics: 2.6 Stre ngth of M aterials
Shape factors in strength
S h a p e -r e l at e d s t r e n g t h a n d a l l o w a b l e s t r es s fo r d y n a m i c l o a d i n g
Shape-related strength is the fatigue strength of the cross-section of a dynamically loa- Shape-related strength
ded mem ber w ith an additional allowance for the strength reducing effects of the com- (dynamic loading)
ponent's shape. Important factors include
•b 2
• the shape of the com pon ent (presence of stress con centration)
^s
•• mstaocchkindiinmgenqsuiaolnitsy ((msuermfabceer rtohuicgkhnneessss)). r s r l i m • •b 2
When compensating for the required safety factor this yields the allowable stress nee- Allowable stress
ded to verify the strength of a member w hich is dynamically loaded. (dynamic loading)
cr s shape-related strength /?1 surface con dition factor ^allow ~
VF
<7|jm stress limit of the unnotc hed b2 size factor
cross-section, e.g. o b a or r t p u | S (page 44) p k stress conc entration factor rallow —I s
vf safety factor for fatigue fracture a (r ) a M o w allowab le stress vp for steel « 1.7
Example:
Rotating axle, E335, transverse hole, surface roughne ss Rz= 25 pm ,
rough part diameter d = 50 mm , safety factor v F = 1.7; cts = ^aiiow = ?
< r b w = 280 N/m m 2 (page 44); Q8 (flm = 570 N/m m 2, diagram below);
b2 = 0.8 (diagram below); y3k = 1.7 (table below)
„ s = ^ w b l b 2 = 2 8 0 N W . 0 . 8 . 0 . 8 = 1 0 5 N / m m 2
Pk 1-7
"allow = °s ' v F = 10 5 N / m m 2 /1.7 = 62 N /m m 2
Stress con centration and stress co ncentration factors /3k for steel
Example: Stress distribution Unno tched cross-sections have an uninterrupted distribu tion of forces and there-
for tensile loading fore a uniform stress distribution . Changes in cross-sections lead to c oncen trations
of lines of force where stresses are concentrated. The ens uing reduction o f streng th
engineering is primarily influenced by the notch shape, but also by the notch sensitivity of the
stress in material.
unnotched part
Notch shape Material Stress concent ration factor
bending torsion
Shaft with shoulder S185-E335 1.5-2.0 1.3-1.8
Shaft with semicircular notch S185-E335 1.5-2.2 1.3-1.8
Shaft with retaining ring groove S185-E335 2.5-3.0 2.5-3.0
Key way in shaft S185-E335 1.9-1.9 1.5-1.6
C45E+QT 1.9-2.1 1.6-1.7
50CrMo4+QT 2.1-2.3 1.7-1.8
Woodruff key way in shaft S185-E335 2.0-3.0 2.0-3.0
Spline shaft S185-E335 1.6-1.8
* stress Shaft interface to snug fit hub S185-E335 2.0 1.5
F T concentration in Shaft or axle wit h transverse S185-E335 1.4-1.7
through hole S185 -E335 1.3-1.5 1.4-1.8
notched part
Flat bar with hole tensile loading
1.6-1.8
Surface condition factor b^ and size factor bz for steel
t \ ensio n, cor npres sion
I 0.9 . t )endir g/to rsion
(SI
-Q
o 08
t_) 0.7
ro
400 600 800 1000 1200 1400 06 50 75 100 125 150 mm 200
tensile stength R m in N/mm2 *
0 25
stock diameter d •
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