KurtGieck
Reiner Gieck
Engineering
:
:
:
iFormiilfll
- 7th:EdililI
:
Digitized by the Internet Archive
in
2012
http://archive.org/details/engineeringformu7thgiec
'CIVIC CENTER
[
3 111101810 4297
AREAS B
SOLID BODIES C
ARITHMETIC D
CftfTfM]
FUNCTIONS OF A CIRCLE E
LIBRARY USE o:HANALYTICAL geometry
f
STATISTICS G
DIFFERENTIAL CALCULUS H
INTEGRAL CALCULUS
DIFFERENTIAL EQUATIONS
I
J
STATICS K
KINEMATICS
L
DYNAMICS M
HYDRAULICS N
HEAT O
STRENGTH P
MACHINE PARTS Q
PRODUCTION ENGINEERING R
ELECTRICAL ENGINEERING S
CONTROL ENGINEERING
T
CHEMISTRY U
RADIATION PHYSICS V
TABLES Z
1
23456789
10
ENGINEERING
FORMULAS
by
Kurt Gieck
Reiner Gieck
Seventh Edition
McGraw-Hill
New York
St.
Louis
San Francisco
Montreal
Toronto
Library of Congress Cataloging-in-Publication Data
Gieck, Kurt + Reiner
Engineering formulas.
Translation of: Technische Formelsammlung.
Includes index.
1
.
Engineering - Tables. I. Title.
620'.00212
TA151.G4713 1986
ISBN 0-07-024572-X
85-23153
English editions copyright ©
1997,1990, 1986, 1982, 1979, 1974, 1967
by
Gieck Publishing
D-82110 Germering, Germany
All rights reserved
ISBN 0-07-024572-X
First
published in the English Language under the title
A COLLECTION OF TECHNICAL FORMULAE
Seventh American edition
published by McGraw-Hill, Inc. in 1997
English translation by
J.
Walters B. Sc. (Eng.), M. I. Mech. E
R. Owen B. Sc,
M. Sc.
Printed in Germany
Preface
The purpose of this collection of technical formulae is to provide
a brief, clear and handy guide to the more important technical
and mathematical formulae.
Since the book has been printed on one side of the page only,
the facing pages are available for additional notes.
Each separate subject has been associated with a capital letter.
The various formulae have been grouped under corresponding
small letters and numbered serially. This method enables the
formulae used in any particular calculation to be indicated.
Preface
to the enlarged and revised 7
th
edition
A section on
CONTROL ENGINEERING
has been included in the new section T; RADIATION PHYSICS is
treated in section V.
Approximate solutions of equations of any degree to determine
zeros (roots) have been added to the ARITHMETIC section.
The section MACHINE PARTS with regard to the newest standards
has been revised and brought up to date.
Kurt Gieck
Reiner Gieck
Reference to BS, DIN and VDE
BS
•
British Standards Institution
(Address: 2 Park, St, LONDON
DIN
•
VDI
•
W
1
A 2 BS
Deutsches Institut fur Normung e.V.
(Address: D-10772 BERLIN)
Verein Deutscher Ingenieure
(Address: D-40001 DUESSELDORF, Postfach 10 10 54).
Method of Presentation and Use of Units
Most of the equations clearly reveal the physical relationships
which they describe and are valid regardless of the system of
units employed, provided that they are consistent.
Some of the equations are empirical in origin and the units quoted
must be used
in the formula to obtain the correct result, these
are mainly to be found in sections
and R.
It is
intended that the Stroud notation is used when evaluating
the formulae i.e. both the quantity and the unit is substituted for
a given symbol and the subsequent calculation involves manipulation of numbers and units together.
s
For example, taking equation
if
s (distance)
v (speed)
then
t
I
23:
t
=-
=
=
2-8 metres
8
metres/second
=
2-8
metres x second
8 metres
hence
t
=
0-35 seconds (time)
cancelling the unit 'metres'
It
is clear that t should have the units of time; if it does not,
then it is obvious that an error has been made and the working
should be checked. As a help, in many cases, the anticipated
units are quoted using the abbreviation "EU", Example-Unit.
When the
numerical values and the units are included in the
calculations, their equivalents or definitions are best written so
that they are dimensionless and have the value of 1-0. In this
form they are sometimes called "Unity Brackets" and their use
can be illustrated in three ways:
with consistent units,
equation a 6
~1
km =
1
10 3 m
becomes
1
ft
becomes
1
=
km"
10 3 m
equation a 62
"
"
12
in
=
1
ft
1
12 in
equation a 90
778-6 ft Ibf
=
becomes
Btu
1
1
778-6 ft Ibf
=
1
for example, to convert 14-7 lbf/in 2
w«
=
14 7
.
^22
w [i^T
in
n2
in the conversion
Ibf
1ft
i
in-
Btu
to lbf/ft 2
ft
-»«s
2
between different systems of units
equation a 36
N = 0-102 kgf
1
becomes
0-102 kgf
=
1
1
N
1
m
equation a 65
1
m = 3-281 ft
becomes
1
3-281
ft
equation a 110
1
Btu/lb
0-556 kcal lb
= 0-556 kcal/kg becomes
1
1
For example, to convert 1000kgf/cm 2 to
S.I.
kg Btu
units,
~
1000
kgf
=
1000
kgf.
cm
=
98-1
2
MN
m
2
'
9-81
1
N
"
"l0
kgf
1
_
4
mn"
cm 2
"1
m
10 6 N
2
in
the use of definitions:
1
Ibf is
Ibf
1
mass of
the force required to accelerate a
rate of 32-174 ft/s 2
=
1
lb at
1
the
.
lb x
32-174
%
becomes
=
1
32-174 lb ft
s2
1
s
2
Ibf
Similarly, the Newton is defined by the equation
N =
1
1
—-
kg x
which becomes
=
1
s2
and 1 kgf
=
1
—
kg x 9-81
becomes
1
N S'
1
kg m
=
1
m
9-81 kg
s*
1
2
kgf s
For example, to find the force in S.I. units required to accelerate a mass of 3 lb at the rate of 2-5 ft/s 2 proceed as follows:
,
= m a, equation m
3 lb
-5
3x2-5
0-4536
1
0-4536 kg
1m
11b
3-281
=
N
1
"l_N_s£l
ft
1
kg mj
-036 N
3-281
which is a unit of force.
Base Quantities and Base Units
of the International System of Measurement
base un it
base quantity
symbol
symbol
name
name
(italic
letters)
letters)
length
(vertical
m
mass
m
metre
kilogram
kg
time
t
second
s
electric current
I
absolute
temperature
ampere
A
T
kelvin
K
mol
cd
I
amount of
substance
n
mole
light intensity
/v
candela
Old units are put in ( ) brackets
List of symbols
Space and time
a.
Q
y angles
solid angle
/3,
b,
B
d,
D
h,
/,
£
extension, strain
E
H
length
G
/,
L
pitch
r,
R
direct stress
shear stress
normal pressure
breadth
diameter (diagonal)
height
p
o
p
q,
t
modulus of elasticity
(Young's modulus)
modulus of rigidity
(shear modulus)
M
bending moment
t
radius
distance covered,
perimeter
thickness
u,
U
Z
modulus of section
A
Am
A
V volume
Q
shear force, shear load
V
vertical reaction
W
weight or load, work
w
uniformly distributed
load
s
t
circumference
area, cross section
generated surface
surface area
S
center of gravity
T
torsional moment,
torque
time
v
velocity, linear
co
velocity, angular
/
moment of inertia,
second moment of area
a
acceleration, linear
/p
polar moment of inertia
a
acceleration, angular
J
torsion constant
g
acceleration, gravi-
Z
modulus of*section
tational
jU
coefficient of sliding
friction
Periodical and related
Ho
phenomens
T
period
/
n
frequency
co
A
angular frequency
wavelength
c
velocity of light
rotational speed
coefficient of static
friction
,Uq
coefficient of friction
of a radial bearing
;Ui
coefficient of friction
of a longitudinal bear-
ing
/
coefficient of rolling
friction
Mechanics
r\
dynamic viscosity
m
mass
v
kinematic viscosity
q
density
P
power
F
force, direct force
r\
efficiency
Heat
T absolute temperature
t
a
5
reluctance
temperature
A
linear coefficient
S
of expansion
a
magnetic conductance
length of air gap
temperature coefficient
of resistance
cubic coefficient
y
of expansion
y
conductivity
g
resistivity
q
heat current or flow
density of heat flow
quantity of heat per
unit mass
e
absolute permittivity
Q
quantity of heat
£r
relative permittivity
c
specific heat at
N
number of turns
\i
cv
constant pressure
specific heat at
constant volume
\x x
permeability
absolute permeability
relative permeability
y
ratio of c
p
number of pairs of
R
A
gas constant
thermal conductivity
z
number of conductors
a
heat transfer
Q
quality,
cp
p
p
to c v
e
permittivity,
dielectric constant
^o
poles
figure of merit
coefficient
k
coefficient of heat
S
loss angle
Z
impedance
C
transmission
radiation constant
X
v
specific volume
Ps
reactance
apparent power
fq
reactive power
Electricity and magnetism
/
current
current density
J
voltage
V
Vq source voltage
Cm moment constant
Light and related electromagnetic radiations
Ie
radiant intensity
R
resistance
/v
G
Q
conductance
<Pe
luminous intensity
radiant power, radiant
luminous flux
[flux
radiant energy
quantity of
<P V
electricity (charge)
Qe
Qy
C
capacitance
D
dielectric
quantity of light
irrediance
displacement
Ee
£v
electric field
He
radiant exposure
strength
// v
light exposure
Le
Lv
radiance
c
velocity of light
H
magnetic flux
magnetic induction
inductance
magn. field strength
n
refractive index
F
magnetomotiv force
/
focal length
(magnetic potential)
D
refractive power
E
<P
B
L
illuminance
luminance
UNITS
Ai
Decimal multiples and fractions of units
=
=
da
h
k
=
=
=
=
=
deca
hecto
kilo
M
=
=
=
=
=
G
T
P
E
mega
giga
1
10
10 2
10 3
10 6
10 9
10 12
10 15
10 18
=
=
=
tera
peta
exa
=
d
jeci
=
:enti
=
milli
=
-nicro
=
nano
=
Dico
=
f
emto
=
a
atto
=
c
m
=
=
=
u
n
P
10"
10' 2
10" 3
10" 6
10- 9
1
10 12
10 15
10' 18
Units of length
a
1
1
m
a
2
1
urn
a
3
1
mm
a
4
1
cm
a
5
1
dm
a
6
1
km
=
=
=
=
=
=
m
u.m
mm
cm
dm
km
1
10 6
1
10 3
10- 3
10 3
10 4
10 5
10 9
10 2
10" 4
10-
10
10 5
10- 2
10
10 2
10 6
1
10- 3
10- 9
10" 6
10" 5
10" 4
10 4
1
pm
(mX) 7)
10" 6
10" 3
icr 2
10"
1
10 3
1
1
1
10
1
10
10 5
Units of length (continue id)
mm
a
7
1
mm
a
8
1
urn
a
9
1
nm
a10
a 11
a 12
=
=
=
=
=
=
(1A)
1
Prn
(1mA)
\im
nm
(X)
3
6
7
10
1
1(T 3
10" 6
10" 7
10- 9
10
IO" 3
io- 4
IO" 6
-io
io-
10
10 4
10
10
10 3
1
7
,J
1
1010" 3
10~ 4
10 10
10 7
10 4
10 3
10
9
1
10
10 6
10 3
10 2
10" 2
10" 3
10"
cm 2
dm 2
km 2
2
10" 6
1
1
1
1
Units of area
m
a 13
1
m
a 14
1
fxm
1
mm
a 15
a 16
a 17
a 18
2
2
2
cm 2
dm 2
2
1 km
1
1
"
=
=
=
=
=
=
2
mm
12
6
10
1
10" 12
1
6
6
io~
10" 4
10" 2
10
10 8
10 10
10 18
10 6
X = Xngsl rom
|im 2
2
2
10
10" 6
4
1
10
10" 8
10" 2
10 2
10 4
10 12
10 2
10 10
1
>1mA = 1 XE = 1 X-un t
10
10 -io
10" 4
10" 2
1
10 8
10 -18
10" 12
10 -io
10" 8
1
A
UNITS
2
Units of volume
m3
m =
3
1
1
mm
1
cm 3
1
dm 3
1
km
3
cm 3
9
6
10" 9
10" 6
10" 3
10
9
dm 3
km 3
1 >
3
1
10
10- 3
10 3
10 6
10 18
1
10
10~ 6
10" 3
10 3
10 15
10 12
10
1
=
=
=
=
3
mm 3
1
10" 9
10" 18
10- 15
10" 12
1
Units of mass
1
kg
1
mg
1
9
=
=
=
=
dt
1
1
t
kg
mg
1
10 6
10" 6
10- 3
2
= 1 Mg =
10
10 3
g
dt
1
10 3
10" 3
10 3
10 8
10 9
10- 2
10" 8
10- 5
10 5
10 6
1
t
= Mg
1
10" 3
10" 9
10" 6
10-
10
1
1
Units of time
1
s
1
ns
1
us
1
ms
1
min
1
h
=
=
=
=
=
=
1
d
=
s
ns
^s
ms
min
1
10 9
10 6
10- 3
10 3
10" 6
10' 3
16.66*10" 3
16.66*10" 12
16.66»10~ 9
16.66'10" 6
10- 9
10" 6
10- 3
60
3600
1
10 3
10 6
60* 10 9
3.6*10 12
86.4*10 3 864*10 12
Units of fo rce
2
N
N =
kN =
MN =
1
1
1
dm = 1
1
10 3
10 6
3
11
1
»
l
= 1 liter
1
10 3
60' 10 6
3.6«10 9
86.4* 10 9
1
60'10 3
1
3.6*10 6
86.4'10 6
60
1440
(gravitational force also)
kN
MN
10~ 3
10" 6
0.102
1
io~ 3
0.102*10 3
0.102*10 6
10 3
(kgf)
1
2)
1
N =
1
(dyn)
10 5
10 8
10 11
kg m/s 2 = 1 Newton
UNITS
A3
Units of pressure
N/mm 2
Pa
a 39
1
Pa = N/m 2
a 40
1
N/mm 2
=
a 41
1
bar
=
a 42
(1
a 43
10
1
10 6
10 5
kgf/cm 2 = 1 at)= 98100
=
(1 torr)
133
6
bar
(kgf/cm 2 )
(torr)
10- 5
1.02*10- 5
0.0075
1
10
10.2
7.5 *10 3
0.1
1
1.02
750
736
9.81*10- 2
0.981
1
0.133*10" 3 1.33*10- 3 1.36*10-3
1 >
1
Units of work
kW h
J
a 44
a 45
=
1
= 3.60>1
1J2)
kWh
1
a 46
(1kgfm)
a 47
(1 kcal)
a 48
(1hph)
=
(kgf m)
0.278*10" 6
6
(hph)
(kcal)
0.239*10-3 0.378*10" 6
0.102
367* 10 3
860
1.36
2.345*10- 3 3.70*10" 6
1
1.58*10-3
4269
1
1
2.72*10- 6
= 4186.8 1.16*10- 3
= 2.65*1 6
0.736
9.81
0.27*10 6
632
1
Units of power
1
W3
a 50
1
kW
a 51
(1
kgf m/s)
a 52
(1
kcal h)
a 53
dhp)
a 49
=
=
=
=
=
>
a 54
1
W
kW
(kgf m/s)
(kcal/h)
(hp)
1
10- 3
0.102
0.860
1.36*10-3
1
102
860
1.36
1
8.43
13.3M0- 3
0.119
75
1
1.58*10-3
632
1
1000
|
1.16
9.81X10" 3
1.16*10-3
736
0.736
9.81
Unit of mass for jewels
carat = 200 mg = 0.2 x 10~ 3 kg = 1/5000 kg
Unit of fineness for precious metals
24 carat £ 1 000 00 %o
18 carat * 750 00 %o
1 4 carat ^
583 33 %o
8 carat ^ 333 33 %o
a 55
a 56
.
.
.
.
Units of temperature
a 57
r =(4 +27315 K
)
a 58
a 59
7R =
*
LB— k
boili ng point of
,
\
wa,e rat 760 torr/ J7J1b
=fi^ank
(^ + 459.67JRank = --Rank
-5ftt
anr
Rank
212
671.67
32-
491.67
273 , 5 .
+ 32) °F=(^B_- 459.67) °F
/
\Rank
/
Tr,
'F
100
-32)°C=^- 273.1 5j°C
a 60
T
°C
1
ft.
am thn Q mn ro.
t
abS0
'
Zer0 °
-273.15
-459.67
tures in the scales for Kelvin, Rankine, Celsius, Fahrenheit.
1)
2>
1
1
torr = 1/760 atm =
= 1
J = N
s
m
W
1
.333 22 mbar =1
I
3> 1
W=
mm Hg at
1
J/s
t =
°C
= 1 N m/s
UNITS
Conversion,
Anglo-American to metric units
Units of length
in
ft
yd
mm
1
0.08333
1
0.02778
0.3333
3
1
25.4
304.8
914.4
0.0254
0.3048
0.9144
1
in
1
ft
12
1
yd
36
1
mm
1
m
0.03937
39.37
1
km
39370
Units of area
sq in
sq in
1 sqft
1 sq yd
1
1
3281 *10~ 6 1094*10" 6
3.281
1.094
1094
3281
10" 6
1
0.001
10 6
1000
1
dm 2
nr
sqyd
cm'
0.772*10- 3
6.452
9
1
0.06452 64.5*10-
1
0.01
1
1.196
100
10000
00929
08361
00001
0001
100
1
cu yd
cm 3
dm 3
0.1111
1
cm 2
dm 2
0.155
1.076*10"
1.197x10~ 4
15.5
0.1076
0.01196
1
m
1550
10.76
cuft
2
0.001
sqft
1
1
1
1000
6.944x10-
144
1296
km
929
9.29
8361
83.61
Units of volume
cu in
=
5.786X10" 4 2144x10- 5 16.39
1 cu in
1
0.01639
= 1728
1 cu ft
1
0.037
28316
28.32
=
1 cu yd
46656
27
1
764555 764.55
3
= 0.06102 3532*10" 8 1.31*10- 6
1 cm
1
0.001
3
= 61.02
1 dm
0.03532
0.00131
1
1000
3
1
m - 61023
35.32
1.307
10 6
1000
1
64*10" 5
0.0283
0.7646
10" 6
0.001
1
Units of mass
dram
1
dram
1
oz
1
lb
1
g
kg
1
Mg
0.5643
564.3
564 3>10 3
1
kg
lb
Mg
0.0625
0003906
16
1
0.0625
2835
0.00177 1.77*10
0.02832 28.3-10
256
16
1
453.6
04531
0.03527
0002205
1
0.001
3527
2.205
1000
1
35270
2205
10 6
1000
1
1.772
4.53*10
10" 6
0.001
1
continued A 5
UNITS
As
continued from A 4
Units of work
lb
ft
a 85
J =
1356 376.8*10" 9 324*1 0' 6 1. 286*1 O 3
9807 2.725*10 6 2344*10~ 3 9.301*10" 3
ft
lb
=
a 86
1
kgf m
=
7.233
1
a 87
1J = 1Ws=
07376
0102
a 88
1
kW h
a 89
1
kcal
a 90
1
Btu
1
Ws
kgf m
0.1383
1
6
367.1*10 36*1
3
= 3.087*1
426.9
4187
778.6
1076
1055
= 2.655*1
277.8*1
1
3
kW h
6
kcal
0~ 9
Btu
239*1 0~ 6 948 4*10' 6
1.163*10" 3
0" 6
293*1
1
3413
3.968
0252
1
kW
kcal/s
Btu/s
860
1
Units of power
hp
kgf m/s
J/S=W
1
76.04
0.102
745.7
0.7457
9 807 9.807*10
10" 3
1
102
426.9
107.6
1000
4187
1055
1
hp
a 92
1
kgf m/s= 13.1 5*1
a 93
1J/s=1W= 1.341*10- 3
a 94
1
kW
a 95
1
kcal/s
=
a 96
1
Btu/s
=
a 91
0" 3
1.341
5.614
1.415
1
0.7073
0.1782
3
2 344*10 3 9.296*10" 3
0" 6
948.4*10' 6
239*1
1
0.9484
3 968
0.252
1
0.239
1
4.187
1.055
Other units
a 97
a 98
a 99
a100
a101
a102
a103
a104
a105
a106
a107
a108
a109
a110
a111
a112
a113
a114
a115
a116
1
mil
= 10- 3 in
sq mil = 10 _6 sq in
1 yard = 3 ft
1 English mile = 1760 yds
1 Nautical mile
1 Geographical mile
1 long ton = 2240 lb
1 short ton (US) = 2000 lb
1 long ton = 2240 Ibf
1 short ton (US) = 2000 Ibf
1 Imp. gallon (Imperial gallon)
1 US gallon
3
= 9.547 kcal/m 3
1 BTU/ft
1
BTU/lb = 0.556 kcal/kg
2
= 4.882 kgf/m 2
2
2
1 lbf/in
(p.s.i.) = 0.0703 kgf/cm
1 chain = 22 yds
1 Hundredweight (GB) (cwt) = 112 Ibf
1 Quarter (GB) = 28 Ibf
1 Stone
(GB) = 14 Ibf
,
1
1
Ibf /ft
0.0254 mm
645.2 urn 2
0.914 m
1 609 m
1852 m
7420 m
1.016 Mg
0.9072 Mg
9.96 MN
9.00 MN
4.546 dm 3
3.785 dm 3
39.964 kJ/m 3
2.327 kJ/kg
47.8924 N/m 2
0.6896 N/cm 2
20.11 m
498 kN
124.5 kN
=
62.3 kN
AREAS
B
square
A
=
a'
a
=
\a~
d
=
a\Y
A
=
a-b
rectangle
*
^-"
Var + V
d
•-
a
parallelogram
-.
=-'-/>
.
= a
:
s -
::
cot af +
=
\ ia - ft
-:
=
2
2
\ (a - h cot a) + /»
=
^-h = m-A
trapezium
^A
V
d-
-
-S*
£
I
\5Cs-a)(s-ft)i$-c)
i
•:
2;
2
•:
:
AREAS
B
equilateral triangle
5^
b 14
\F
b15
h
=
b16
A
=
|r 2 VlO + 2\/T
b 17
a
=
\r \/lO-2\AT
b 18
q
-
if v 6 + 2 v"^
1
construction
AB = 0,5 r, BC = BD, CD = CE
fa \^
b20
2a
^T>,
2
b21
b22
b23
hexagon
2
b 19
5
»
1.1555
d
«
0.866J
w
0.83
2
A
b24
=
2as
=
25
r
octagon
V^^
b25
a
=
5-tan22.5°
« 0.415 5
b26
^
=
d- cos 22.5°
« 0.924 d
b27
d
"
^TTo
^ 1083 5
cos 22.5
polygon
b28
b29
a-h-[ + b- h 2 + b
2
h3
AREAS
B
circle
A
=
d
J
4
2
nr 2
=
= 0.785 d 2
U
=
2nr
=
5 (o
nd
=
annulus
2
- <*
2
>
D-d
2
sector of a circle
360
fr=
c
2
Jt
180°
a in radian measure
a in degree
Jt
180°
segment of a circle
=
2r-sin|
=
^(3*2 + 4*2 )«^(a-S na)
=
2
8*
r(1-cos§)
a see formula b 39
ji
a b
=
|
tan^
SOLID BODIES
c
1
c
2
c
3
c
4
a be
c
5
2 {ab + ac + be)
c
6
cuboid
d
=
ya + b + c
2
2
2
parallelepiped
c
7
V -
A! h
(Cavalieri principle)
pyramid
A, h
3
frustum of pyramid
c
9
c 10
=
^(At +A 2 + VA -A 2
1
h
*l±A*
(for
Al
)
~A
2)
SOLID BODIES
cylinder
V
=
2
h
\d
4
A m = 2nrh
Aa =
2jir(r + /i)
hollow cylinder
%h(D 2 -d 2
)
^
V
-
f
m
=
V/^ + r2
nr(r + m)
'
AgiA, = x 2
frustum of cone
V
=
J
:
fc
-^h(D 2 + Dd + d 2
A m = -y m (D +
=
</)
VWsphere
V
=
I^T 3
=
i»
= 4.189 r
^'
)
2jrpm
SOLID BODIES
zone of a sphere
%h(3a 2 + 3b 2 + h 2
c26
V
c27
A m = 2nrh
=
n(2rh + a 2 +
c 28
)
2
ft
)
segment of a sphere
c29
c 30
2jirA
c31
2
2
+ 4*
f('
4
)
sector of a sphere
c32
V
=
^nr 2 h
c33
A
=
±r(4h+s)
c34
V
= £/i 3
c35
A
=
2nh(R + r)
c36
V
«
\nr 2 h
c37
A Q = 2nr(h + \lr 2 -£)
sphere with cylindrical boring
sphere with conical boring
SOLID BODIES
torus
Dd'
TM-
A
=
n 2 Dd
V
=
2
h
%d
4
sliced cylinder
A m = 71 d h
A
=
Ji r
/i 1
+ h2 + r +
^r + {h,-h 2 2 /A]
2
)
ungula
2
V
r
3
2
h
Am =
A
=
Am + Zr 2 +£ r yJr 2 + h 2
barrel
- ±h(2D 2 + d 2
)
prismoid
V
=
Z{Ai+A 2 + 4A)
This formula may be used for
calculations
shown
in
involving
fig.
thus spheres
spheres.
C1
and
.
.
.
solids
C3 and
parts
of
ARITHMETIC
Powers, Roots
Rules for powers and roots
general
numerical examples
1
pa ±q-a"
=
[p
d 2
a"<a"
=
a'
d
n
± q)a
3a 4 + 4a 4 =
n
a 8 /a
d 3
m n
(a
d 4
)
=
(a
d 5
b
(a
3 2
=
)
(a
= a8
~2
2 3
= a2
)
n
=
±q)y/T
(p
= ae
"3
=
Ma
11
^T
V^-W
tyTlT =
4^F+7\GT =
^16*81
= a6
1
^16'V^
-
iff
d 9
\b
&
VT
V2^
^~
d10
= a'
Va
d11
d12
4
n
P y/F ± q ya
d 8
_mn
)
1/a
d 6
d 7
n m _
n\m
=
2
7 a
^=T =
iVa^;
i
* N0UP
ScSSs
=
a
)
=
^^9^=
\^T
PeCial
3
•*
= a
V*
i
yj9
'
-
\
4
3
VF^=V^=+2; (V^) 2 = "2
Exponents of powers and roots have to be non-dimensional quantities!
Quadratic equation (equation of the second degree)
d13
Normal form
d14
Solutions
d15
Vieta's rule
x 2 + px + q »
x-\
e
2
2
p - -
±VF«
(jc-,
V 4
+ x 2 );
q
=
x v x2
Iterative calculation of an //-th root
d16
When
where a-
x = YyT",
then
H<»
D*c
r n-1
the initially estimated value of x. Repeatedly inserting the
obtained x as a new value of x gradually increases the accuracy of x.
is
ARITHMETIC
Powers, Roots -
D2
Binomial theorem
Expansion of general algebraic expressions
d 18
(a±b) 2 = a 2 ±2ab + b 2
(a±b) 3 = a 3 ± 3a 2 b + 3ab 2 ± b 3
d 19
(a + b)
d 17
n
-
n
a
+ ^a
n-1
^f^a
6 +
W (n-1)(/I-2)
a
1-2-3
2
d 20
(a+fe + c)
d 21
(a-b + cf
2
2
a - b
3
3
a + b
3
a 3 -b
n
n
a - b
d 22
d 23
d 24
d 25
=
=
=
=
=
=
n - 2
n-3
3
fe
6
2
b
2
2
I
+
+
n
2
a + 2ab + 2ac + b + 2bc + c
2
2
2
a -2afc + 2ac + b -2bc + c
(a + b) (a - 6)
- ab + b 2 )
+ ab + b 2 )
n_1
+ a n_2 fc + a n_3 b 2 + ...
(a-6) (a
n - 2
+ fc n
... + a6
)
(a + b)
{a
(a - b)
{a
2
2
1
Binomial theorem
d 26
(a + 6)
=
U/
•J aj
(2)an +(^ )a n
l
- 1
n(n-1) (n-2)
1-2-3
(n\
d 27
d 28
n
6+(j)a n
...
-2
&
fn-Jk+1)*
2
+(£)a n
"3
*a4
"3
3
+
fc
.*
.
.
J
... *
must be a whole number
(a + fc/
"1
b + *'\*~
a + 4a -6
2
- 1a 4 + ^a4
-
4
3
6a
+
2
b
2
2
+* \
4a-6
+
-fc
3
* + 6
4
+6 4
3
Diagrammatic solution
d 29
Coefficient - Pascal triangle
(a + b)°
(a + 6)
d 30
1
12
13
1
1
(a + b)
2
(a + fe)
3
(a + b)
4
(a + 6)
5
(a + 6)
6
4
1
|5
1
1
6
j
1
10
6
1
3
j
1
4|
10
1
20
15
1
1
5
15
1
6
1
Continue with each line starting and finishing with 1. The second and
penultimate numbers should be the exponents, the others the sum of
those to the right and left immediately above them.
Exponents: The sum of the exponents a and b in each separate term is
equal to the binomial exponent n. As the power of a decreases the power
of b increases.
(a + b) is always positive
d 31
Signs:
d 32
Examples:
(a - b) is initially positive
(a + bf
(a - b)
5
5
a + 5a*b +
=
= + a 5 - 5a 4 b +
and changes from term to term.
Wa 33 b + Wa 2 b 3 + Sab 4A + b
Wa b - 10aV + 5ab - b
2
5
2
5
ARITHMETIC
Partial fraction expansion of rational functions
Proper fraction rational function
a + a,jc + a 2 x<
P(x)
y(x)
2
+ by x + b 2 x +
Qix)
.
.
+ b n jc
.
n>m
r
n and m whole numbers
Coefficients a v 6^ can be real or complex. If n are the zeros
of the denominator Q(x) the factorized form from y(x) is
,
y
d 33
P(x)
y(x)
Q(x)
PM
m
a(x-n,)"-(x-n 2 )* 2 ...{x-n a k *
)
a
a constant factor.
Partial fraction expansion
For easy manipulation of y(x), e.g. for integration,
pansion y(x) into partial fractions is often appropriate
is
d 34
y(x)
m
PM
the
ex-
Au
(x-n,f
Q(x)
(x-n,Y
A 22
A 2k2
x-n 2
(x-n 2 Y
(x-n 2 y
Ag-\
Ag 2
(x-rc q kq
x-n q {x-n q 2
Complex zeros occunn pairs (conjugate complex numbers) when Q(x)
has real coefficients. For expansion these pairs are combined to
real partial fractions. If in d 33 the zeros az 2 = «i (fi2 is
conjugate complex to n-\) and if due to the pairwise occurrence
)
)
k\ = kz = k, the partial fractions of d 34 with the constants
^11 ••• ^2k2 can be combined to the following partial fractions:
B u x + Cu
d 35
2
x + ax + b
B, 2 x + C 12
(x
2
fl 1k
+ ax + bf
(x
2
x + C1k
+ ax + b)
k
To obtain the constants A-\-\ to A^ q resp. Bn, C\\ to #ik, Cn<
coefficients of equal power in x at the left side of the
equation are compared with those at the right side after having
converted to the common denominator Q(x).
Example:
2x^\
2x-1 _ Bn*+Cn
m
y[()'
(x+1 -2i)(x+1 +2i)(x+1)
2
Q(x)
A q2
>tgi
t
{
2
x+^
x + 2jc + 5
(x+1)
2
2x-1 = B u x(x+1)**C„(x+V 2 + A ,{x+V(x 2 + 2x + 5) + A (i2 (x2 + 2x + 5)
<i
Q(x)
Q(x)
2x--\ = (A^ + B u )x
2
2
+ (3/l q1 +A q2 +2B U + C u )x
+
+ (7,4 q1 + 2/t q2 + B u + 2C n )x+5i4 q1 + 5>t q2 +
Cn
Comparison of coefficients between left and right side:
S„ = -1/2; C„ = 1/4; A q = 1/2; A q2 = -3/4
,
If
there
are
single
zeros
n\,
the
constants
equation d 34 can be obtained by:
P(n 2 )/Q'(n 2 );
d 36
A^ = P{n,)IQ(n,)
A-\\,
A2\
PK)/Q'(n Q
^4qi
)
of
ARITHMETIC
D4
Logarithms
General
log to
system
d 37
iog a
d 38
logio
d 39
log e
d 40
log 2
terminology
the base of
a
log to base a
'g
10
common log
=
In
e
natural log
-
lb
2
log to base 2
The symbols in
x = b are called:
log a
a
base
x
antilogarithm
b
logarithm (log)
Rules for logarithmic calculations
d 41
d 42
d 43
d 44
!og a
(xy)
=
log a * + log a y
'OQa
f
=
log a x-log a ><
log a
xn
=
n
iog a
yr
•
log a x
log a x
Exponential equation
d 45
d 46
d
=
b*
=
e
xlnb
hieraus:
v
log a 6
Conversion of logarithms
d
47
igoc
lg
e
•
0.434 294
In x
ig_£
d 48
In
x
d 49
lb
x
2.302 585
Igjc
e
lg
1.442 695- In*
Base of the natural logs
In x
•
-
3.321 928
-
lg jc
e= 2.718281828459...
Key to common logarithm of a number
d 50
lg
0.01
d 51
lg
0.1
d 52
53
d 54
d
lg
1
lg
10
100
lg
=
=
-2
or
8.
...
-1
or
9.
...
=0
=
1
=
2
-10
-10
etc.
Note: The antilogarithm always has to be a non-dimensional quantity.
ARITHMETIC
Permutations, combinations
Permutations
An ordered
arrangement of r out
"permutation" of the n things taken
The number of these permutations is denoted by:
is
called
selection
or
a
P n = n(n-1)(/i-2). .(n-r+1),
d 55
.
T
If
of
r
n
at
things
a time.
n > r
r = n, this becomes
Pn = P
/i(n-1)(n-2) ... 1 =/i!*
Example: The n = 3 things a, 6, c can be permutated with
each other (i. e. at a time) in the following 6 ways:
a be
bac
cab
acb
bca
cba.
P3
d 56
=
Here r = n = 3.
1-2-3
=
3!
=
6
Special case: The number of permutations of n things taken all
together incorporating n-\ of one sort, n2 of another sort
and n r of a rth sort is:
«!
=
P
d 57
^1
•
!
ttg!
•
.
.
.
nr
\
Example: The n = 3 things a, a, 6 can be permutated the
following 3 ways:
aab
aba
baa.
3!
d 58
2!-1!
1-2-3
1
-2-1
Here
n = 3, %n-i = 2, n 2 = 1,
= 3
Combinations
A selection of r out « things without regard to order is called
"combination" of n things taken rata time. The number of these
combinations is denoted by
d 58
r
r\(n-r)\
'\{n-r)\
\r)
Example: The n = 3 things a, b, c taken together give
only the one combination ab c. Here n = 3, r = 3.
Hence
<* -
£) = ^fl - 1.
3
1-2-3
\3/
The table on page D 6 compares combinations and permutations
(with and without the things repeating).
*'
***
n\ is pronounced
„n factorial"
Symbol usual for binomial coefficient (see d27)
-
ARITHMETIC
6
Combinations, permutations
Combinations and Permutations
(Explanations see D 5)
d 59
c
c r o (o
CO
CO
-o
11
C 1-
CO C\J
ft,
^ — o *-
co
ft,
i
<o
-Q
cj
co
»B
CL~
E™
*
£o
D
t
<»
*
° Q-
CO
a)
„ II «
CO
*"•
•
5.
tr *- co
d 60
C\J
CSJICNJ
E ~ ~ 0)
«J -C
> Q)
D >
—
^
5 *
ii
co
S=
ii
c ft,
CD
II
cop
d 61
co
2 *£
to
~ o
co
*-*
eg
cnj
+
C
5 ^_
o
0-03
o cm
c -
co
c
—£
E.E
I
I
^o
CO
"O »_
« °
O) co
CD C
- o
li
d 62
.E
«
CO
Q.
sis
CO
CO
CO
5 05 05
co
CO
CO
I
co
C C
7
CO
o o o
(D O
iji-
— co
C\J
.0.0.0
E E E
3 3 3
s:
M
>,
Q. W
CO
CNJ
I
CO
CO CM
V
i:
II
coU
ii
§ o I 35
o =
a. \o
co
3 c c ^ » .2
« .2 a> £ o ~
O *- £
S8|dLUBX3
Q.
JO
ARITHMETIC
Determinants and linear equations
Second order determinants
d 63
an
•
x + a, 2
a?i
•
* + a22
•
a,,
a 12
a ?1
a 22
D -
y
y
c
Q\y
'
&Q2
0^\
'
Oy
insert r column in place of
column x
a 12
r,
d 64
D,
=
a 22
r
2
column y
=
r,
•
-r2
•
ajj
y
+
=
X
an
ri
a 21
r
"
a, 2
h
•
a,,
-r,
•
a21
2
+
5l
y
D
Third order determinants (Sarrus rule)
d 65
a„ -x + a 12 -y + a 13
a 21 x + a22 -y + a23
a 31 x + a32 7 + a33
•
•
d 66
O =
•
z
=
r,
= r2
z = r3
z
a ;i
a12
3 i>?
2?3
a"
a„
a 32
a 33
a 3l
^3?
3 12
ar
a'
•
3 13
a,,
,
•
a
a2 2 a 33 + a, 2 a23 s3
•
•
•
+ a 13 a21 a32 - a 13
•
a^ a 31
i 1
•
•
- a,
,
•
3j 3
•
a32 - a 12 a2
*
•
i
^33
insert r column for x column:
a 12
ri
d 67
01
=
a 13
r,
X.XX,
a 22
a23
"32
rf
a 12
°33
'22
.
+ an
'
Tj
'23
•
T
°12
"23
'3
a32
^3' a22' r3
a 32
a 12
'2
a 33
33
determine £>2 and D3 similarly by replacing the y- and z-column
by the r-columm
d 68
D
D
^
•
z=&
'
continued on D 8
ARITHMETIC
8
Determinants and linear equations
Determinants of more than the 2nd order:
may be used for determinants of
(The Sarrus Rule, see D 7,
higher order than the 3rd).
By adding or subtracting suitable multiples of two rows or
columns, endeavour to obtain zero values. Expand the determinant starting from the row or column containing most zeros.
Alternate the signs of terms, starting with an as +.
Example:
^23
d 69
a 34
Expand on 4th column:
'12
d 70
»32
»42
dl3
a 33
a 43
'12
*13
'22
*23
'42.
a 43
Further expand as:
d 71
D =
|
|
a32 a33
a 42 a43
a31
-a
|
°41
a* z
d 43
+a 13
N
|
|
a* 2
a 41 a42l
\
To form the determinants D-\, Z>2.
(see D 7l) substitute the
r column for the first, second,
column of D, and evaluate
in the same way as for D.
.
.
d 72
For determinant of the nth
mulae:
Di
.
.
.
.
order,
find
w-i ... n
f rorr
the for-
D2
D
Note: For
determinants of the nth order continue
determinants of the 3rd order have been obtained.
until
ARITHMETIC
Algebraic equation of any degree
Definition of an algebraic equation
An algebraic equation has the form:
/nW = a n x
d 71
n
+ a n _T jc
n_1
+
.
.
+ a 2 x 2 + a^x + a
.
any terms whose coefficients a^ are equal
be left out.
for
.
n
< n may
The solution of an algebraic equation involves the finding of
zeros (roots) of the equation for which fn (x) = 0.
Characteristics
"I.The algebraic equation fn (x) =
of
degree n has exactly
n zeros (roots).
2.
If
all
coefficients a v are real, only real or complex conjugate
zeros exist as solutions.
3.
If
a v are
all coefficients
>
there are no solutions whose real
part is > 0.
4.
If
n
is
of
odd at least one zero is real, assuming all coeffi-
cients a v are real.
5.
The relationships between the zeros x^ and the coefficients
are:
£*!
= - a n _-|/a n
for
d73
Z*j-Xj
=
a n _ 2 /a n _-|
for
d 74
Zjtj
= - a n _ 3 /a n _ 2
for
d72
i
i,j
where
•
Xj
•
xn
i
i,
x,
6.
d76
x2 -x 3
.
.
.
n
-x n = (-1)
•
i
.
.
.
n
=*1,2, ... n
=
j
=
j, k
where
d75
= 1, 2,
=
1 ,
2,
j
= k
.
.
.
n
a /a,.
The number of positive real roots of the equation in question
is equal to the number of sign changes of the coefficient series
an
,
a n _-|, a n _ 2
a2.aT.a0
or
this value less an even number (Descartes' theorem).
d77
Example: /3 (jc) = 2x 3 - 15x 2 + ^6x + 12 =
has the signs
+
+
+
and due to the 2 sign changes has either
2 or
positive real roots.
continued on D 10
.
ARITHMETIC
10
Algebraic equation of any degree
Continued from D 9
7. The
number of negative real roots of
question is found by substitution x = -z:
d78
equation
the
in
Here the number of sign changes of the coefficient series
a n *, a n _-,*, a n _ 2 *
a 2 *, a/, a * is equal to the number
of negative real roots or this value less an even number.
Applied to the example on D 9, point 6:
d79
3
2
/3 (z) = -2z - 15z - 16z + 12 =
has the signs
+
and therefore
equation d 77 due to only one sign change has only one
negative real root.
General solution
If
is
a root of an algebraic equation of wth degree
x-i
/n (x) = 0, the degree of fn (x) can be reduced by one degree
to /n _i (x) =
when fn {x) is divided by {x - x ). If another
root x 2 is also known, the equation can be reduced by one
degree further when divided by {x-x 2 ), and so on.
y
~
d80
/„(x)
= a n x n + a n _i * n_1 + a n _ 2 x n 2 +
d 81
/„/(*-*! )
=
d82
n 2
+ a n _2"x n 3 +
fn -i/(x-x 2 ) = /n _ 2 (x) = a n " x
etc.
/n _ 2 /(j:-X3) =
/
= a n * n_1 + a n _i'x n 2 +
= f
^/(jf-JCn)
.
.
.
.
+ a2 x + ay
.
.
.
+ a2 "x + a,"
'
~
.
d83
+ a2 x2 + a^x + a
..
.
-
n _i (x)
"
.
(x)
= a n ^.
a special case where the roots are complex conjugates; after division the degree of the equation is reduced
by 2 degrees. Division of the algebraic equation /„(*) by
(x-XjJ is easily carried out by using the Horner method in D 1 1
There
is
Homer method
The Horner method is a scheme for calculation which can be
applied to the nth degree polynom P
P n (x) = a n x n + a n _i -x n
d84
"1
+
.
.
.
+ a
1
x + a
to solve the following problems:
* Calculation of the value of P n {x) for x = x
.
* Calculation of the values of the derivatives P n '(x), P n "(x), etc.
up toP n
(0)
(x)forjt =x
.
* Reduction of the degree of P n (x) if there are known roots.
* Finding of zeroes (roots).
continued on D 11
ARITHMETIC
Dn
Algebraic equation of any degree
Horner method (see scheme below):
Set the coefficients a v to a v (0) and write the coefficients
of the polynomial Pn \x) - starting with the coefficient
which is related to the highest exponent - in the first line.
Positions with no covered exponents have entries of 0.
Scheme
x an
x
(0)
.(0)
x a _, m x a _
(1)
(0)
(0)
(1i
n
n
V2
(2)
a2 <°»
Q
2
3
a2
1
afl
2
<
1
<
>
(1)
oai
1»
ao
ai
WW
*0
*0 a n
*0 a n-2
x
xQ a n _r *o a n-2 w x a n _ 2
*0 a n-1
afl
m x am x
... x a
m= b
= P n (x
aT = b = MM-P n (x
:
an
0)
4
W
"
J3)
a•n-1
n-
a2
<
3
>
)
= fc 2 = 1/2!-P n "(jt )
(n)
(n)
1
fln^W-Vl-l/tl-l)!-^" '^)
a n = fc n = 1/n!-Pn
Example
'
n)
U
)
of the Horner method:
1
Calculation of the values P n (x), P n
P n "'(x)forx = x
x = 4:
'
{x),
P n " (x), and
;
P n (x)
3
=
a:
a3
<°>
-
+ 11*
- 6
a/°)
a <°>
-6
11
-6
4
-8
12
-2
3
8
6
a2
1
x = A
2
<°>
6jc
4
11
6 =
1
I
1
-
= P„(4)
P„'(4)
P n "(4)-1/2! P n "(4)
= P n '"(4)-1/3!;
;
=
1-2-6 = 12
P n '"(4) = 1-2-3-1
= 6
)
'
.
.
ARITHMETIC
12
Algebraic equation of any degree
Explanation to the Horner method
The value of a polynomial and
point x = x is to be calculated.
derivatives
its
a
at
fixed
The
with the multiplicants
results of multiplications of x
a n (1) a n-i (1) etc. given by the dotted lines are written in
(1)
(1)
= x a n ).
the 2nd line (e.g. x
an
.
•
Line 3 shows the results of the addition of lines 1
d106
.^ = a _,™ + x
an-r
e.g.
an
n
<
1
and 2.
where a r 1 = * <°>
<
>
;
>
a n - 2 (D =
d107
(
This means especially:
d108
a
<
1
=
>
aJM+xo-a^-bo-Pnixo)
the value of the polynomial at the point x = jc
Using the
same scheme - starting from
.
line
3 - with multi-
plications and additions leads to line 5 with
d109
ai
(2)
= b, = P n (x Q )
'
the value of the first derivative of P n (x) at the point x = x
This
scheme can be repeated
.
as a polynomial
Ai-times,
of
degree n has exactly n derivatives.
These calculations lead to:
d110
(x-x Q ) + a 2 (3) (x-x Q f
P n {x) = a (1) + a/
(*-*o)
(*-*o)
= P n (x Q ) + 1/1
d 112
+
.
.
.
!
•
1/(b-1)I
P n '(x
Pn
•
+
r
(n)
)
(x-x
(n - 1)
(jc
)
•
)
+ 1/2!
(x-x o
•
r +
P n " (jc
1//i!
•
)
]
•
Pn
ix-xof +
{n)
(x Q )
.
.
(x-x )
n
Example 2 of the Horner method:
Reduction of the degree
i.
e.
if
there
the finding of P n _-| {x) using:
d114
P n (x)/(x-x
)
is
Given:
d 116
Scheme:
'
a3
1
a2
known zero
(root)
x
,
P n ^{x).
=
P n (x) = x3 - 6X2 + 1 1 x - 6
d 115
a
with root x
= 1
"
d 117
d 118
^o=l
d 119
^n(1)-
d120
F n (1) =
shows, that x
Then P n ^(x) = Ax2 - 5x + 6.
Result:
= 1
is
a root of F n (x).
The roots of this last equation (x-, = 2 and* 2 = 3) can be determined
very easily using d 14.
ARITHMETIC
Approximate solution for any equations
13
General
the analytical determination of zeroes (roots) from
algebraic or even transcendental equations is only possible with
restrictions, in D14 to D16 the following methods for approximate solutions are given:
Since
Newton's method
Secant-method
Linear interpolation (Regula falsi).
Starting with an approximate initial value any degree of accuracy
can be reached by iteration.
Example of an algebraic equation (polynomial):
4
d122
jc
- 3X2 + 7x - 5 = 0.
Example of a transcendental equation:
jc-lgW- 1 =0.
d123
Procedure
• Graphical determination of the initial approximation by drawing
the curve from a table of known values.
• Choice of one of the 3 afore mentioned methods. Please
note that Linear interpolation is always convergent. For the
other methods convergence is only guaranteed under conditions
given in D14 und D15. The disadvantage of this additional
examination usually will be compensated by considerably
faster convergence.
• Improved convergence can often be reached by starting
with one method and continuing with another one; espe-
when no improvement in results
with one method after several iterations.
cially
has been reached
.
ARITHMETIC
14
Approximate solution for any equations
Newton's approximation method
The value x
the first approximation
is
for the root n
of the equation f{x) = 0.
The tangent is made at f{x Q ); the intersection of the tangent with the x-axis
is
a better value than the starting
point x
Calculation of jc-, is made by:
.
d125
x^
=x -f{x )/f'{x
The improved value x 2
d126
is
).
calculated using
x-,
a similar way:
in
etc.
*2-*i-/(*i)//'(*i)
Multiple repetition of this method leads to results of any desired
accuracy.
d127
General rule
* k+ =*k-/(*k)//'(*k)
* = o. 1.2. ...
Requirement for convergence of this method:
• n is a single zero
;
i
• between x
maxima or minima of the function f(x) are
und n
not allowed.
Convergence: Locally convergent.
Comment: The values /(jc k and f'{x k which are necessary for
)
)
Newton's method can be calculated very easily using the
Horner method given in D 1 1
d128
Example: f(x) = x
f(x) =
d129
1st step:
•
\gx - *\. The starting value for a zero to fulfil
= 3.
may be x
d125 requires the value of the derivative f'(x
)
to be
calculated:
d130
f'(x) = lg(x) + lg(e) = \g(x) + 0.434 294.
d 131
2nd step: Determination of an improved value jc
From d 125 the values x = 3, f(x ) = 0.431 364 and
d132
3rd step:
1
/' (x
)
:
= 0.91 1 41 5 lead to x, = 2.526 710.
Determination of an improved x 2
= 2.526 710; f(x,) = 0.017141
Using the values
and /' (jc-,) = 0.836 849 equation d 1 26 to
x 2 = 2.506 227; error +0.000 036.
Using x2 the zero has an error of 0.000 036.
:
^
d133
4th step:
If
the
accuracy with x 2
is
iterations must be carried out.
not
sufficient
further
ARITHMETIC
D 15
Approximate solution for any equations
Secant approximation method
The derivative /' {x) of Newton's approximation
is
replaced by the differential
quotient: Two adjacent points f(x Q ) and
/(jc-,)
are connected by a straight
'(*<>)
y
line.
The value x 2 at the intersection of this
line with the jc-axis has to be determined; x 2 is the first approximation for
the required zero n Q
.
X-\-Xp
X2 =*!
d140
/(*i)
/(*i)-/(*o)
In
the next step f(x^)
is
connected with f(x 2 ). The intersection
of this line with the jc-axis is the next approximation.
General iteration rule:
d 141
*k+1
~x
*k ~ *k-1
k ~/(*k)
fc« 1,2, ..
/M-/dk-i)
f(Xk )
* /(*„_!)
Comment: An especially fast convergence often can be reached
when the Secant and Newton approximation methods are used
alternately.
Convergence: Locally convergent.
d142
Example: f(x) = x Ig x 1; x = 4; Xi - 3.
•
f(x ) = 1.408240;
d143
1st approximation:^
=
%
f(xj = 0.431364.
3 - 0.431 364 (3 -4)/ (0.431 364-1.408240)
2.558425.
d144
Error
f(x 2 )
0.043 768
2nd approximation calculated with *, x 2 f{x^) and f{x2 ):
,
,
d145
x 3 = 2.558 425 - 0.043 768 (2.558 425 - 3) / (0.043 768 - 0.431 364)
d146
Error
= 2.508 562
f(x 3 ) = 0.001 982.
Instead of continuing with the Secant method, Newton's method
can now be used:
d147
d148
For this reason /' (x 2 ) has to be calculated: /' (x) = Ig x + Ig (e)
f'(x 2 )= lg(2.558425) + 0.434294 = 0.842267
d149
*3* = *2 ~f(x 2 ) //' (x 2 ) = 2.558 425 - 0.043 768/0.842 267 = 2.506 460.
d150
Error: f(x 3 *) = 0.000 230. * 3 * leads to a smaller error
than x 3 which was found by only using the Secant method.
,
ARITHMETIC
16
Approximate solution for any equations
Linear Interpolation (Regula falsi) approximation method
are chosen so that
'(*il
and /to) have different signs.
f(x
Between these two points at least one
zero n must exist. The intersection of
the line through f{x Q and /to) with the
Two values x and x
:
)
)
x-axis is the first approximation x 2
To determine the improved value x 3 a line
through f{x 2 ) and one of the earlier used points /(x ) or /to)
is made and the intersection of this line with the jc-axis has
to be calculated. From the earlier points the last one with a
different sign compared with/to) always has to be used!
d152
/to) 7to) <
Generale rule:
or
f(x 2 ) -f{x ) <
must be satisfied.
xw-Xt-fM- f,*)'**,^
d153
fc-1,2...
Q^j^k-l
/(*k)-/Uj)
Here
A*k) */(*,)
is the largest value smaller than k, for which
;'
/(*2)7Ui) < is valid.
Convergence: Always convergent.
d154
Exam pie :f(x) = x
d155
\gx -1; choice of x Q = 1 with/(x ) = -1
x, = 3 with /to) = +0.431 364
f{x ) -/to) <
here
d156
x^ -x
x 2 = x^ -/to)
is
f(x 2
)
satisfied*
3-1
ntnMMA
= 3-0.431364
n
2.397269;
0.431364 + 1
/(*i)-/(*o)
d157
and
= 2.397 269 lg 2.397 269 - 1 = - 0.089 71 7. This value represents the accuracy with which x 2 approaches the zero.
•
d158 As /to)*/to)<0 the line is made through f(x 2 and /to).
The intersection of this line with the x-axis is:
X 2 -*1
d159
x 3 =x 2 -/to):
2.501 044; f(x 3 = -0.004 281.
)
)
d160 As /to) -/to) >
f(x 3
)
/(* 2 )-/(*i)
but /(x 3 )-/to) <
and /to). The
the
line
is
intersection
made through
of
this
line
with the x-axis is:
*4 = x 3 - /to)
d161
d162
/to)-/W
= 2.505947
/to) = -0.000197 5.
a further increase in accuracy the intersection of the
through /to) and /to) and the x-axis has to be calculated. Since /to) -/to) >
the values
and /to) -/to) >
f(x 3 ) and/(x 2 ) cannot be used.
For
line
ARITHMETIC
17
Series
Arithmetic series
The sequence
d17l
10 etc. is called an arithmetic series.
(The difference d between two consecutive terms is constant).
w
Formulae: s n = | (a, + a n ) = a n + n ~ ' where d = a n -a n _i
1,
4,
an =
a-i
7,
'
y
CJ172
+ {n-A)d
Arithmetic mean: Each term of an arithmetic series is the arithmetic mean a m of its adjacent terms a m _-, and a m + -|.
d173
3m-1
Thus, the mth term is
(e.g. in the above series
3m + i
"*"
for
1
< m <n
4 + 10
a3
7)
Geometric series
The sequence
1, 2, 4, 8 etc is called a geometric series. (The
quotient q of two consecutive terms is constant).
d174
Formulae:
sn
= a,
^^-
ga "~ ai
=
q = ~^-
for
« "1
^^
<7" 1
d175
d176
Geometric mean: Each term of a geometric series is the geometric mean a m of its adjacent terms a m _., and a m+ v
Thus, the mth term is
am = Va m -1 a m + V
for
1 < m < n
'
(e.g. in the above series)
-
a3
For infinite geometric series
statements apply
= lim a n
d177
V2 8 = 4
'
(n-* 30
=
)
|^|<
;
;
=
sn
1)
following
trie
Mm s n
1
=
a!
1-9
Decimal-geometric series
Application for calculation of standardized number-series
Quotient of two consecutive terms is called ..progressive ratio q>".
d178
qp
b > 1,
= y/\0.
integer.
b determines the number of terms or number of standardized numbers
of a series within one decade. The values of the terms which should
be rounded up, are calculated according to d 77:
MVW
d179
i(l0
Starting with
10
Examples:
designation
6, 12, 24,
5, 10, 20,
•
1
an
d
:
:
E6, E12, E24,
R5, R10, R20,
term
final term
sn
:
difference between two
q
:
initial
consecutive terms
n
:
'/b)n-
or
a<\
= 100
or
note
intern. E-series, see Z22
DIN-series, see R1
number of terms
sum to n terms
quotient of two consecutive
terms
ARITHMETIC
Dl8
Series
Binominal series
d 180
/(,)
a
(i±x)«- ,±(°)x + ( )x*±(
2
-
a
)x> f
3
a may be either positive or negative, a whole
.
.
.
number or a fraction.
Expansion of the binomial coefficient:
a(a - 1) [a -2) (cr-3) ... (a-n+1)
1-2-3 ...
ai
/a\
\nl
•
for
Examples:
d 181
^—- =
(1
± jc)~
d182
yr±7 =
(1
± x)
1
'
v
2
=
1
+ x+ x 2 + x 3
=
1
± ^x-^-x 2 ± ^-x 3 - ...
2
1
d183
Ml r\~
(\JZXj
-
1/2
1
—
+
T ^1 Xr -r 3n Xr 2 +
T
+
2
VT17
Ixl
<1
l:rl
<1
IjcI
<1
16
8
5
I
\
...
Xr
3
+...
i
16
8
Taylor series
d184
+
= /(a)
/(*)
putting a =
d185
m
^(*-a)
+
^
(x-a) 2 +
t
gives the MacLaurin series:
./(o)
+
m), +
n^
+
...
for
Examples:
d186
e
x
=
1
+ * + *! +
2!
1!
d187
d188
a
x
Injc
to <i
+
all
X
...
3!
x-lna
(x-lna)
1!
2!
2
(x-lna)
3
all
X
1
=2 [^zl + 3i(£z|)
[x+1
d189
^+
3
3!
5
+
l(^) + ..l
x >
5 \x+^ 1
\jc+i /
„., _£ + £-£ + £ -
-1
...
< X
x-S+
1
d190
2
J
4
b
continuec on
D19
'
ARITHMETIC
Dl9
Series
Taylor series
(continued)
for
Examples
d191
d192
sin
5
3!
5!
7!
X
X2 + x 4
X6
— + ...
all
6!
X
X + x
—
x = x-
cos x =
3
4!
d194
cot x =
1
x3
2
3
3
X
3
_
.
1-3-5 x7
2-4-6 7
X
X
7
9
5
+ X
7
9
1x121
1x121
—
+ — + — + — +
3!
5!
7!
9!
x = x +
coth
X
all
"
x
+
d203 arsinh x =
d206 arcoth x =
x
3
+
3
1
x
2
3
6!
—
IxKf
45 *
0<lxl
lxl<jr
1
3 x
2-4 5
•
1
1
2X2
3x3
X
'
—
x x +
15"
"
315*
2
3
8!
5
3 *
d204 arcosh x - In9r
d205 artanh x = x +
4!
3^x
x = x
*
x =
all
"
cosh x =
tanh
1x121
1x121
5
? - Arctan x
2!
d202
1-3 x5
2-4 5
? - Arcsin x
d197 Arctan x = x
d198 Arccot x =
0<lxl
lxl<;r
945
45
3
d195 Arcsin x =
d196 Arccosx =
IxKf
l_l x _ JL X 3 _ ^.^5 _
x
d201
—
—
tan
d200
all
...
7
- x3 +
x = x
x5 +
x +
*
*+ 3 X
15 *
315
d193
sinh
x
—
+
1-—
2!
d199
7
7
945 *
5
1
•
3 5 x
7
1-3-5 1
2-4-6 6x6
579"
5x 5
7x 7
7
•
2-4-6
1*3 1
2-4 4x 4
2
lxl<1
lxl>1
lxl<1
lxl>1
.
ARITHMETIC
20
Fourier series
Fourier series
General: Each periodic function
f(x) whose period - x < x < ji
can be subdivided into a finite
number of intervals in such a
way that f(x) can be described
by a continuous curve in each
of these intervals, may be ex-
panded
in
interval
this
y\
into
convergent series of the following form {x = cot):
00
d207
y +2
-
fix)
[a n
cos {nx) + b n sin (nx)}
The various coefficients can be calculated by:
d208
ax =
n j
/W cos <**) d*
6k =
^ I /W sin (to) dx
-K
-It
with the index fc = 0, 1, 2
.
.
Simplified calculation of coefficient for symmetr. waveforms:
Even function:
f(x) = f(-x)
d209
d210
d211
d212
Odd-harmonic functions
Even-harmonic functions
d213
d214
fix)
/(f+*)
= f(-x)
and
= -fi^-x)
give:
= -fi-x)
and
/(§ + *)- /(f-*)
9^:
fix)
7l
d215
d216
d217
ak =
ak =
&k -
^
/(x) cos (/ex) dx
^k =
J
for
k =
for
k =
=
A:
for
1, 3, 5,
.
.
.
0, 2, 4,
.
.
.
1, 2, 3,
.
.
.
/2
¥ { /w sin
for
ak =
6k =
A:
=
k for * for
(/:J:)
d*
1, 3, 5,
.
.
.
0, 1, 2,
.
.
.
2, 4, 6, .
.
.
-
ARITHMETIC
21
Fourier series
Table of Fourier expansions
d218
d219
y
y
-
for
d220
d221
y =
y
<x < n
n < x < 2n
for
=
d222
M
-|s,n*
sinJ3x}
sinjc +
1
SiniSx]
+
3
a
for
-a
for
5
a < x <
-a
a.
n + a < x < 2:t-a
31
l
rr.
n 1
y
r
2
.i
a
3,1.1.1
:
!
d226
y =
4a
—
+
d224
d225
cosor
g- cos (5 a)
•
>
-/(2n + x)
for
•
= ax/b
= a
sin (5x)+
-,
S!niS=flQ cosx +
for
Olx^i
for
b ^ x ^ rc-b
> - a(jl~X)/b
X
"171
~i
2/1
-1
sl rl2
Jt
(
-a
]
x
"
A
N wr/z*
Id
3/1
)cos(2x)
—
/
\/i
Tl-fc^AT^Jl^' 6
for
X
L.
I
sin (3jc)
- sin 3 (;t -a) CQS {3x
d227
d228
d229
37t,
i
i
;
2^[«-o
r =
in
2^
O
d226
\
1
a < x < 2it-a
>
=
a
— cos (3a)
sin x +
•
!
'
\
i
—
!
h
Id
!
\3/t
/
n -b
d230
>•
=
4
—
a r 1
~2 sin 6
-r
1
•
sin x + -% sin {3b)
+
d231
y
=
^r
2 Jl
d232
>
=
f{2n + jc)
d233
-j>
sin (5ft)
•
sin (5x)
< x < 2x
for
a _ a fsinx
sin (2*)
sin (3jc)
^l
2
3
2
sin (3*)
•
1
""J
continued on D22
ARITHMETIC
22
Fourier series
Continuation of D21
d234
d235
d236
=
2ax/n
for
O^x^n/2
2a{n-x)/7i
for
tt/2 ^ x ^ n
yi\
-f
-f(n + x)
£.[*,-M + iS^-...]
d237
d238
d239
d240
axln
for
a(2n-x)/ji
for
71
^ x ^ Ti
^ x ^ 2ji
\ /\F^
/(2k+x)
k
d241
d242
d243
d244
-
„
4a cosx
a
cos (3jc)
["
a sin*
for
O^^ji
-a sin*
for
ii
jc
tt.
WYY
^ 2:
-
•
d246
for
d247
for
Am
^ ^ ti/2
3n
^ <- "2"
2 = ^ =
;c
7t
*
/(2ji+x)
d249
M
Ji
7T
[2
4
2
.
Jt
d251
/(-*) = /(2n+x)
for
^ x ^ 71
V
d255
a*/7t
^ x ^ ti
for
2
K
v^ re
/(2ji+jc)
a _ 2a
[
cosjc
2
2
tt
+
5rt
_ 4 [cos* _ cos(2;r) + cos (3jc)
T^L
4
2/1
AAA/
-7t
d252
2£F
cos(6x)
6^-1
cos (2x) _ cos (4jt)
2'-1
v-^
-ti
K
2
2
2a
d250
d253
d254
3*
in.
.
2a _ 4a F cos (2jc)
cos (4jc)
cos (6x)
+
+
+
n
1
3
3 5
5 7
L
d245
d248
^
cos (5x)
,
L
1
cos (3jc)
3
a^
fsinx _ sin (2 jc)
Ti
L
1
2
,
cos (5jc)
2
5
2
sin (3x)
3
.
ARITHMETIC
D 23
Fourier-Transformation
General
The Fourier-Transformation F {s(t)} based on the Fourier integral converts the time function sit) in a continuous spectrum
(spectral density) S(co) in a way that frequency a» corresponds
to the spectral density.
must have the following charac-
s(t)
teristics:
a)
be
smooth
piecewise
in
number
defined
a
of
inter-
finite
vals
d256
b)
d257
have defined values at the jumps s(t + 0) and s(t-O), so that
the value is equal to the average
s(t) = 1/2 [s(t-O) + s(t + 0)]
+ 0C
d258
c)
)\s(t)\
6t must be absolutely convergent.
-»
The inverse Fourier transformation F~
1
{S(co)} gives the time
function s(t).
Definitions
+ 30
|
d259
F{s(t)}
d260
F-'{S((o)}= s(t)
=
S(co)
= js(t)
^
1
-6t
J"S(co)-e
itl,t
i
;
-dco;
i
=
V1?
=
\CT
=
]pf
-00
+ 00
d261
Spectral energy
J
1
I
s{t)
2
I
•
dr =
oo
+0°
j- f\S{a>)
^ n -oo
2
I
»
•
dw
Calculation rules
d262
Time translation
F{s(t-t)} = 5(a»)-e
+
d263
Convolution
*i
_il0X
i
;
x
(0*%(0- Jji(T)-5 2 (r-r)-dr
=
d264
/^(^-^(r-Tj-dT
-oc
d265
F{*i(0**2 (0} = 5 (w)-5 2 (o;)
d266
F{5(0}
= S(co)
d267
F{s(at)}
=^ 5 (f)
d268
F{ 5l (0 + *2 (0}
- 5 (w) + 5 2 (w)
1
areal>0
1
continued on D24
ARITHMETIC
24
Fourier-Transformation
continued from
D23
equation d 259 calculated spectral densities are given
for some important time functions. Correspondence between time
Using
1
;
function and SDectral density:
1
d269
°°
iu,t
-dw
5(0 = =L /s(<u)-e
2n^
S(oj)
;
=
|5(r)-e-
Time function s(t)
Spectral density S((o)
S(u)
sit)
Dirac-Pulse
d272
A 6 (t)
,s(t)
polarity
•
A
(spectral density
constant over to)
is
d273 Rectangle funct.l A
=
S((o)
A6(t)
with change of
R J/2 (t-T/2)-
L^ R xu
.
( t
v
+TI2)
sin
d274
S(w) =
d275
S{oj)
3T
d276
-dr
2 AT- sin (o)T)/((oT)
A -R T (t)
d270 Rectangle function
d271
iwt
- 00
im .& mo sgxii
=
-\ 2
AT
^
2
oiT
2
4AT cos(2a)T)
sin (coT)
(oT
t
S((o)
=
A R w -(a))
aSM
[Rectangle
I
function
X
d277
2n
T
wo'
2n
U)
t
continued on D25
ARITHMETIC
25
Fourier-Transformation
continued from D24
Time function s(t)
Spectral density S(<o)
d278
d279
Modulated rectangle
d280 A R T (t)- cos (a)
d281
t)
=
TTA
/\
^ ^:
a
=
l
%(t)
;A
„
„
w
with
A
o
'
A; ART (t)
S{(jo)
=
sinT(a; + a;
A
sin T{o)-a>
d282
Gaussian-pulse
)
a) +co
)
a2 t 2
-«,2
S((o)
d283
=
£ v^
1
2k
d284
T
•s(t)
9
d285
S(oj)
\r°i
i-'£-)'
d286 cos 2 -pulse 4 2 -cos 2 (<y
d287
f)
with
w
= -=?
sin
5(w)
/I
•
;•*)
4
(- f)
* -
1
r 2 •O) 2
1
1-6 :t
d288 Exponential-pulse
d289
/I
S(io)
)(0 +
2
ARITHMETIC
26
Laplace-Transformation
General:TheLaplace-TransformationL{/M} based on thelntegral-
x
function
d290
F(s)
=
j fit)
e"
st
d/
o
d291
converts the time function f(t), which has to be zero for t<0
and which must be given completely for t>0, into a picture
function. The part e~ st in d290 is used as an attention factor
to get convergency of the integral for as many time functions
as possible; here is s = o+ito with o>0 a complex operation
variable.
In
this
picture-domain differential equations can
be solved and unique, non periodical processes (e.g. oscillating)
can be handled; the desired time behaviour is reached
finally by inverse transformation in the f-domain (see D 28).
Definitions
292/293
w
L{f(t)} = F(s) = j f[t) e- di
L-
1
{F(s)}=f(t)=£
\F(s)e
at
<is
ri
o
abbreviated description:
f{t)o—
—
abbreviated description:
F(s)
—
•—
F(s)
o /(/)
Calculation rules (operation rules)
d294
Linearity
d295
d296
L{Mt)+f2 (t)}
-
L{cf,(t)}
Translation
F,(s) + F2 (s)
c-Fi(s)
=
L{f(t-T)}
e-
Js
F(s)
t
d297
Convolution
J/i(r-r) /2 (r)dr
Mt)*f2 (t)
t
Sf:(r)f2 (t-T)dz
d298
d299
d300
d301
/1
(0*/2 (0
Variable
transform.
*•'(*))
Differen-
L{f'(t)}
—
-•
F,(s)
=
F(a-s)
=
s
F2 (s)
s-F(s)-f(0+)
tiation
d302
Linn)
d303
n
L{f (t)}
5
2
n
+
F(s)-s-f(0 )-f(0
-F(5)-2 /
k =
d304
Integration
L{jf(t).6t\
=
y^*)
(k)
+
(0 )5
+
)
n-k"1
ARITHMETIC
27
Laplace-Transformation
Application of the /.-Transformation to differential equations
Scheme
r-domain
d 205
Operation
!
normal equations
Differential equations
»J
I
for y(t) + start conditions
look at rules
for derivations
i
i
d 206
s-domain
f
for Y(s)
i
I
i
i
result of the solution of
the differential equations
solution of normal
equations for Y(s)
i
i
Inverse transform.
referring to D28
|
equations is transferred
to the inverse transformation. This can be simplified by expansion
from Y(s) into partial fractions (see D3) or into such partial functions,
for which in D28 conversions are given back in the time domain.
Difficulty of the solution of the differential
d 207
d307
Example:
Ty' + y
= f[t)
is
+
d301| Ts Y(s)-Ty(0
d305/
d306
y(t)
According
for
to
+
+ Y(s) = F(s)
)
M
—
• F{s)
[Here f(t)
y(t).
r
•
f(t)o
startfunction
= 2^ start condition
y(0 )
-
m
are
there
assumed
is
7>(0
+
_
Ms + Ty{V)
+ Ts
1
be
to
1
+ Ts
solutions
function. In
different
i
step
this case referring to d 313 F{s) = 1/^).]
Application \
of
D3
Y(s)
I
_
Ty(0 +
1
5(1
1
)
D28
Ty(0<
-VT
-v.
after
T
^+Ts
Ts)
y(t)
1
+ e
Application of the convolution rule to the
/.-Transformation on linear networks
The
originate function f-\(t) is changed to a response y(t) after
having passed a network. The network s defined by its transfer
function F2 (s). F2 (s) has the inverse transformate/^Wi
r-domain
d308
d309
/i(0
S
y(t)
Network
y(t)
= /i(0*/2 (0
domain
Y(s)
F^(s}
F2 (s)
-•
Y(s) =
F,(s)- F2 (s)
For a given network the response y(t) depends on f\(t). y(t) can be
found by d 305. After having found Y(s) calculation is continued
on line d 306. Comolete inverse transformation to the /-domain is
possible when F2 {s) is given as proper fraction rational function
in s and when L-Transformate F, {s) is given in D28.
C3:
ARITHMETIC
D28
Laplace-Transformation
Table o f correlation
O
DC
d310
F(s
= }/(/)-e- s, -d
)
+ ioo
(V(s)-e s1 -ds
271
with
=
£
O + \CD = O + 2ti/;
i
=
i
-ioo
V^
^-domain
/-domain
Laplace
original-
Laplace
original
transf. F(s)
function fit)
transf. F(s)
function f(t)
1
6{t) £ Dirac
d 311
d312
d313
d314
d315
d316
d317
d318
d319
d320
d321
d322
d323
5
^-domain
s2
2
(s
1/5
1
for
t
> o|?a|
for
t
<
1/s
r
2a
+
|
(s
2
3
exp (at)
-s.n(ar.+
— -cos(a/)
cos (at) -
Jf
+
-ft-s\n(at)
n-1
(1-1)!
M(s-a)
2
Dt5
/
n
1
^)
s
Ms 2
/-domain
b
for
1
(s-a)(s-b)
/•exp (at)
1
(5 + a)
a
s{s-a)
1
d326
d327
d328
exp (at) - 1
2
+ b2
— e~
1
at
+ r-5
1
a
S
2
±exp(-t/T)
1
sinh (at)
d330
d333
d334
cosh (at)
,
3
3/(4^T-/ /2)
5 + 6
In
a2
s
s
2
+ a2
1
d335
(s
2
2 2
+a )
d336
sin (at)
f(.--.-)
cos (at)
5
/
-cos (at)
2a*
5
d 337/338
2
(*
+ a2 ) 2
for
2
a > 0:
e
^5 sin(ar)-
1//sin(af)
arctan (a Is)
for
a
-avr
c
-a
4t
2/VttT
a > 0:
3
^
2vr
erfc-
5 e
(see G 8)
1
sin(a/)
J2a
VI"
5
s + a
s' +
r-1
-1/(2V5T-f /2)
sy/sT
a
d331
d332
'2
vr
-a*
s'-a2
-sin(b/)
1
/
s
d 329
a:
b-a
M(s-a)
2
d 324
d 325
=t=
e bt -e a1
,
2
V^ + a
,
.
J o\ al )
/
\
Bessel
function
2
ARITHMETIC
29
Complex numbers
Complex numbers
General
z
= re'* = a + \b
a = real part of z
b = imaginary part of z
r
=
<p
= or modulus of z
= argument of z
-*;
7
absolute value of z
a and b are real
^-T
d339
d340
=
-i
d341
j-2
=
-1
d342
j-3
=
+
d343
:-4
=
+1
j-5
=
-i
'
i
d344
_
,5
i
etc.
Note:
In
engineering
electrical
the
letter
j
is
used
for
to avoid confusion.
In the
Cartesian coordinate system:
d345
z
=
d346
*1
+ 22
=
(a^
+ a2 )
+
d347
Zy
~
Z2
=
(a^
- a2 )
+
Zl
•
z2
=
(a-, a 2
d348
a-\
£l
d349
a2
^2
d350
a
2
+ b2
=
(
- by b 2
a 2 + by b 2
)
a + \b
i
i
+
i
,
+
+ b
'
(by
+ b2 )
(by
- b2 )
(a-i
b2
+ a 2 by)
-ay b 2 + a 2 b
2
a2
+
T
b{
\b)(a-\b)
\fa^+b 2
d351
Where ay = a 2 and by = b 2 then Zy = z 2
,
continued on D30
)
ARITHMETIC
30
Complex numbers
Complex numbers
(continued)
In the polar coordinate
d353
z
d354
r
d355
=
Zi-z 2
d357
=
r
Y = r
2
d358
z
n
?nr
yz =
d359
"
[
•sin(g? 1 + (p2 )]
i
1
sin (/?<p)]
i
—
2nk
—^
(p +
:tL
1r<
\r(cos
(n
> 0, integer)
—
-
-sin
i
* 0)
w+2nk,
±:—-
.
+
(z 2
n - 1.
for k
=
0, 1, 2
k
=
this
If
+
C0S (<Pi -V2) + i-sin(g? - g?2 )]
r*[cos(ncp) +
if
a + \b
tang?
r 2 [cos(gp 1 + g? 2 )
2
=
=
-sing?)
cosg?
r
r
i
b
b
sin g?
d356
r(cosg? +
arctan
V
system:
the principal root,
is
k = 1,2
n-1 these are the adventigious roots (/i is an integer).
n
VTare the rc-th unit-roots.
The n-roots which fulfil z n = 1 are
d360
-
z*
d361
d362
n
for A; = 0,
1, 2,
cosg? +
sing?
cosg? -
e-'"
..2nk
cos^
i
i
•
n-'\
.
(n
1
-sing?
cosg? +
d363
d364
Ie
±lcp
|
=
e
cos g?
d365
Ty
iq>
+ e
g?
+ sin
=
r2
2
g?
=
i
-sing?
1
-icp
In r + \(cp +
Inz
If
y cos
2
an integer)
is
sing?
2nk)
(k
=
and cp^ = <p2 + 2nk,
Note: <p should be measured along the arc,
k is any arbitrary integer.
0,
then
±1, ±2,
z 1 = z2
.
..)
ARITHMETIC
31
Application of geom. series
Compound interest calculation
d366
kn
=
k
d367
20
9
'9
w
Annuity interest calculation
d368
d369
_
r
-rq
n
-1
n
<y
k
q
(k
-q n -k n )(q-:)
n
*-1
-1)<7
(9
r-g-k n (q-:)
l9
r-q-k (q-'i)
,
d370
m
n
Where k n = 0, we get the "redemption formulae'
Deposit calculation
(savings bank formula)
;
d371
n
-1
k Q -q" + r.q
{k n
d372
-k
n
-
n
(q
.
d373
<7
)(?-1)
-Dq
k n (q-V + r-q
Vkoiq-V + r-q
'9 9
Letters
initial
capital
number of years
+p
capital after n years
1
annual pensions
rate of interest
(withdrawals)
(e.g.
0.06 at 6%)
8
ARITHMETIC
32
Geom. construction of algebr. expressions
b'C
d374
d375
a:b
x
4th proportional
:
d376
d377
=
b:x
=
ya'b
a :b
x
:
3rd proportional
d378
x
/^90°Z'^
I
d379
a: x
=
x
/
b
.
/
/
/
x
:
x
2
a
1
+ b
x
/
Y
mean proportional
d380
/
—
\
I
6 .-
d
V-a- + b'
d381
x
:
hypothenuse of a rightangled triangle
d382
V3^
t
x
:
height of an equilateral
// \-\ \
</
•
1
triangle
d383
(V^-D
d384
d385
a:x
x
:
=
=
a -0.61
x:(a-x)
larger section of a repeatedly
subdivided line
(golden section)
K— "^
FUNCTIONS OF A CIRCLE
Basic terms
Circular and angular measure of a plane angle
Circular measure
Circular measure is the ratio of
the distance d measured along the
arc to the radius r.
It
is
given the unit
'radian"
which
has no dimensions.
e
(rad)
1
Unit: rad
Angular measure
Angular measure
the centre
as "degrees".
at
Unit:
e 2
e 3
is
a
of
obtained by dividing the angle subtended
circle into 360 equal divisions known
°
A degree is divided into 60 minutes (unit: '),
a minute is divided into 60 seconds (unit: ").
*
Relation between circular and angular measure
By considering a circle, it may be seen that
e 4
e 5
or
degrees
e 6
0°
360°
=
2 7i
rad
=
57. 2958°
1
15°
30°
12
6
0.26
0.52
CO
c
radians
45°
60°
75°
90°
ji
K
5n
jt
4
3
12
2
0.79
1.05
1.31
1.57
180°
K
270°
360°
3jt
2
2jt
co
3.14
4.71
6.28
FUNCTIONS OF A CIRCLE
General terms
Right angled triangle
opposite
e
7
e
8
e
9
_
c
adjacent
= b_
c
hypotenuse
opposite
=
tan a
a
hypotenuse
- _^_
b
adjacent
adjacent
cot a
opposite
Functions of the more important angles
e 10
angle a
0°
a
cos a
tan a
1
cot a
00
sin
15°
30°
45°
60°
75°
90°
0.259
0.500
0.707
0.366
0.966
1
-1
-1
0.966 0.866 0.707 0.500 0.259
1.732 3.732
0.268
0.577
1.000
3.732
1.732
1.000 0.577
180° 270° 360°
oo
1
oo
oo
0.268
oo
Relations between sine and cosine functions
Basic equations
e 13
Sine function
e 14
Cosine function
A sin (ka - y)
A cos (ka - cf)
sine curve
1
sine curve
1.5
cosine curve
1
or sine curve with a phase shift of
and
and
and
k =
1
k = 2
k =
1
•jx/2
FUNCTIONS OF A CIRCLE
p
£ 3
o,
Quadrants
e15
sin
e 16
e 17
cos
tan
e18
cot
e 19
sin (180°
e20
cos
e21
tan
e22
cot
e23
e24
e25
e26
sin
e27
e28
e29
e30
sin
(
-
90°
[
"
)
[
"
)
-
a)
»
)
"
)
[
I
-
(270°
a)
cos
tan
cot
i
"
i
-
(360°
a)
cos
tan
cot
i
e31
sin
e32
e33
e34
tan
)
_ a
(
cos
cot
"
)
i
"
)
!
"
)
!
"
)
+
+ cos a
+ sin a
+ cot a
+ tan a
sin
=
=
=
=
=
=
+ sin a
- cos a
- tan a
sin (180°
=
- cot a
=
=
=
=
- cos a
sin (270°
- sin a
+ cot a
+ tan a
cos (
"
)
tan
(
"
)
cot (
"
-
sin (360°
=
=
a
+ cos a
- tan a
- cot a
=
=
=
=
- sin a
+ cos a
- tan a
- cot a
sin (a ± n*360°)
a)
=
=
sin
90°
(
a)
cos (
"
)
tan
(
•
)
cot
(
»
)
+
a)
cos (
-
)
tan
(
"
j
cot
(
»
)
+
tan
(
cot
(
+ y<
=
=
=
=
- sin a
- cos a
+ tan a
+ cot a
=
=
=
=
- cos a
+ sin a
- cot a
- tan a
=
+
)
=
+ cos a
=
=
+
+
=
=
=
+ sin a
+ cos a
+ tan a
+ cot a
)
-
)
"
)
(a± n*180°)
cot (
+ cos a
- sin a
- cot a
- tan a
a)
"
cos (
tan
)
+
cos (
a)
=
=
=
=
=
)
••
sin
a
tan a
cot a
;
\
•
/
\
\
/
\
/
y
•
s
s
X
XX
X
N
•
s
•
/
X X
a
a
'
/\
/ \o
m;
-y
0°
/
90°
|
/
\*
180°
jt
270°
s-
\
360°
2ji
+
FUNCTIONS OF A CIRCLE
Trigonometric conversions
Basic identities
e 35
sin" a + cos* a
=
+ tan 2 a
=
e 36
1
1
tana -cot a
=
1
+ cot 2 a
-
-;
1
1
cos 2 a
Sum and difference of angles
e 37
sin (a ± fi)
=
e 38
cos(a±P)
-
e 39
tor,
tan
«\
/«
(a -i± /3)
a cos p ± cos a sin p
cos a cos p + sin g-sin/3
tan
tan P
m = cot a ncot^ *P + „1
= , -r gQ » ±„ to „£
cott (a ± p)
± cot a + cot
1 + tan a
tan
sin
.
„
,
,
;
,
.
•
Sum and difference of functions of angles
sin
a + sin
p
2 -sin
cos
sin
a - sin
£
2 -cos
sin
cos
cos
2 -sin
sin
cos a + cos
/S
=
cos a - cos £
=
tan a ± tan
p
=
cot a ± cot
/S
=
/S
=
sin
a
cos a
2
-
•
cos p
=
2
g^
•
cos p
sin 03 ± a)
sin
cos
2
sin (a ± P)
cos g
•
2
g
•
sin /S
1 sin (g + 0) + ^ sin (g - p)
2
1
2
cos (g + p) + 2 cos (g - P)
sin
a
sin
p
=
^cos(a-p) - 2-cos(g + £)
tan
a
tan
p
=
tan g + tan ff
cot a + cot p
cot
/3
=
cot g + cot jg
tan g + tan p
tan
£
=
cot g + tan fi
tan g + cot /S
cot a
cot a
-
tan g - tan
cot g - cot /S
m
.
_ cot g tan g _ cot a tan g -
cot ff
tan p
tan ff
cot p
Sum of 2 harmonic oscillations of the same frequency
a sin (of + <p<fi + b cos (<ot + q> 2 ) =
with c = a sin g?-|
<p = arctan J
T
J
+ & cos <p 2
and
;
rf
T = arcsin
cp
^ ^
/ _ f^ Sf
V c2 + ^ sin
"
+
= a cos q>-\ - b sin <j? 2
,
-\/
c 2 + ^2
I
c° n
H
must be satisfied
FUNCTIONS OF A CIRCLE
Trigonometric conversions
Ratios
between simple, double and half angles
sin
e 53
=
a
cos a
cos (90° - a)
e 54
V^
e 55
2 sin |
|
1
cot a
tan a
a
cos a
sin
cos 2 | - sin 2 t?
tan (90° - a)
1
cos a
a
sin
cot a
sin a
cos a
V1 + tan a
Vi + cot 2 a
VT - sin 2 a'
?i - cos 2 a
ycos 2 a - cos 2 a
1
\£_
V cos a
\/-^2
2
5/1 - cos a
VT"- sin 2 a
2
e 58
- sin a
cot a
=
cot (90° - a)
2
tan a
e 56
e 57
sin (90° - a)
V'l
cos
tan a
=
- c os 2 a
1
V
2
- 2sin 2
1
f
+ cos 2 a
\h
V
1
2
V sin
a
2
1
e 59
1
+ cot 2 a
V
2
tan
1
+ tan 2
1
V
+ tan 2 a
1
-
-tan 2
e 60
1
sin 2 a
e 61
=
2 sin a cos a
cot 2
f
+ tan 2 |
cos 2 a
- tan 2 |
2
tan 2 a
cot 2 a
1
=
2
cos 2 a -sin 2 a
1
e 62
2cos 2 a-1
e 63
-2 sin a
1
cot 2 a- 1
2 cot a
-~ cot
sin a
1
- cos a
V
a --p tana
cot
tan
+ cos a
1
V
|
2
e 64
1
+ cos a
1
- cos a
sina
1
e 66
tana
-tan 2 a
2
cot a -tan a
cos 2 =
e 65
=
cot
V
1
- cos a
+ cos a
sina
1
- cos a
1
+cosa
sina
cos a
V - cos a
1
FUNCTIONS OF A CIRCLE
6
Acute angle triangle
Oblique angle triangle
Sine Rule
e 67
a
sin
e 68
a
e 69
b
:
sin /?
b
=
:
sin y
_ _
:
sin/3
a
=
sin a
sin/?
a
e 70
Cosine Rule
b2 + c2 -
e 71
b2 =
e 72
c2
e 73
=
2b c cos a
'
+ a*
2ac cos
a2 + 6 2 -
2ab cos y
•
•
(for obtuse angles the cosine is negative)
Tangent Rule
e 74
a +
tan
a + b
a- b
tan
tan
a + c
a - c
2
a -
tan
a +
b + c
6 - c
a -
2
Half-angle Rule
e 75
Q
tan^ =
s -
tan
tan
a
|
idius of incircle and circumcircle
e 76
e 77
e 78
e 79
e 80
A
=
1
-j be -sin
a
2 acsin /8
\A (^ - a) (s - b){s- c)
(5 - a) (5 - 6) (5 - c)
=
-x aZrsin
=
ps
V
1
2
sin a
a + b + c
2
sin
2
,
'
c
sin y
y
tan ff+y
tan
£^
2
5
~
FUNCTIONS OF A CIRCLE
Inverse trigonometric functions
Inverse circular functions
k
2
&°
<H
o
o
<
>
\
\
Q.
O
=^1 CL
1
H
CO
o
a.
<
II
^1
1CL /7*
V
K
-a
-
1
X
Definitions
Funktion y
identical
e 81
with
defined
within
e 82
e
principal 1
83
arcs in x
arccos x
arctan x
arccot x
x = sin y
x - cos y
x = tan y
x = cot y
-1^x^ +
-1^x^ +
1
1
-oo <x < + oo -oo <x< + oo
)
values
y
2
-!<,<!
jt^y^O
2
,
n>y>0
Basic properties 1)
e 85
e 86
Arccos x
= ? ~ Arcsin x
Arcsin (-x)
Arctan (-*)
=
=
Arctan x
Arccot x
- Arcsin x
- Arctan x
= n- Arccos x
= n - Arccot x
Arccos (-x)
Arccot (-x)
Ratios between inverse circular functions 2)
e 87
Arccos x =
Arccos \/1 ~* 2
Arcsin \/l - * 2
*
x
Arctan
r
V1 -x
^
2
—
—
vi-*
*
Arctan
v/i
Arccot
/5
*
Arccot
1)
Arcsin
Al CCOS
2
*
X
Arcsm
^f^2
!
X
X
89
Arccot x =
Arctan x =
Arcsin x =
*
1
r.
V1 +x
*
1
Arccot X
5-
2
1
Vi~T^
—
*
Arrrnr
Arccos
rri
*
Arctan
-1
X
Principal values are marked by capital letters
2 ) The formulas marked with * apply for x
>
Continued on E 8
1
FUNCTIONS OF A CIRCLE
c
l_
L. b
inverse trigonometric functions
Addition theorems
T-
T-
T-
;
CO
T-
% %
r-
% %\
r~
vi
+ +!
vi
+ +
5v
CM
0J
«M
5v
*J
CM
CO
C
C
CO
CO
**
c +
CtJ
;
;
!
.2
;
a a
a a;
<N
CO
j
1
|
j
i
i
|
!
•
+
t5
c
o
;
«J
;
c C
:
CO
:
(0
O O
O Jr A OV
V A
o * "°J o * *
1
!
Jr
;.
wi
O O; yu O O
i
T-
:
!
o o
^ v
•
-Q
1
:
-Q
CQ
CQ
CO
CCJ
co
CCI
?T~
oP
i
1
co
to
1
>
i
.o
'
•
>
1
-Q
\
-Q
-Q
|
w
Q
i
CD
09
CD
>
5b
i
C r
1
o
SI?!
:
1
<D
5
I
II
II
II
|
8
<
II
|
t
cc
1
-a
V
}
#
#
«0
{
<0
CQ
>l
i
|
iUi
<o:
«0
|
«{ §
K
1
H
H
i
•Q:
i
2
?
"2
;
[
II
1
-Q
II
j
!
II
|
+i
Si
^
2£o
<
q'
i-Q
:
CO
»-
+
c
*~
t
1
\
|
co
;
'
1
•Q
<
+ is 5
+
B
}
1 5
+
C
K
'
:
<C
K
|
ii
ii
II
II
i
CO
1
a
\
1
^5
C
<>
1
H
Q
1
«
73 « i- c
*;< 8;
<j<
< t»j
+
to;
1
CD
it!
A
a
-Q
;
i
r-
i
^
|
S
n^'
O
K
;
i
m
8
co
co
o"
i
1
<
;
co
cc
c
i
1
U
c
2
V
Q
x
i
\
|S
s ci > 3; + >i
« c;l^
+
>
A
j
I
\
\
>i s
i
3b
i
O
1
v
•
!
ip
1
CO-
j
".
o
1
V
t-
1
1
5b
CQ
x»
*
y-
!
si
'.
1
j
>+
c
+
(Or
CO
O
o
r:
CQ
CO
A
A
.
^AVi^AV|++iAIVi^
CO:
c
i
j
•
CO
<
+
K
!
!
:
I
|
i
<
5
>
<
<
\
II
II
o
x»
o
o
o
o
<
<
"O
c
1
_
1
-a
:
,
CO
E
00
C
8
;
C
•
<
is
\<
+
|
I
i
co
j
«<J
J
£W
i£W
!
8
<
:
j
O
1
<
O
CO
CO
CO
CO
t-
O
is
<
5
i
\
+
*
8
O
g
<
:
O
18
|<
1
i
'
*
8
1
O
g
\
<
|
2
;
>
!
£i
'
j
i
;
i
j
•
l|
\< c
M
i
!
•
i
<c
i
!
+
1
to
co
O
O
u
{
«s
j
J3
j
!
c9
I
!
!
;
j
>
!
1 J
!
<c
j
<c
rr
CO
CO
CO
>
u)
a>
©
CD
C>
c9
CM
j
•
j
i
2.
O
O
2
<
<
CO
O)
N
CO
t
D
(D
ANALYTICAL GEOMETRY
Straight
line,
Triangle
Straight line
=
mx + b
t
1
Equation
y
f
2
Gradient
m = y2 — y
~
X2
Interc. form
^
a * 0;
for
=
b =t=
3
f
tana*)
JC-j
— c
Gradient m\ of perpendicular /Ifi
1
m, =
A?-Z
Line joining two points P
f
5
f
6
f
7
:
[x\,
vi)
and P 2 (x 2
,
>'2)
y-y\ = y2-yi
Line through one point P\ (jci, yi) and gradient m
y -y! = m(x - *i)
Distance between two points
J
=
y(.*2~ Jc i)
2
+ (.>'2 - .)'i) 2
Mid point of a line joining two points
yi +y2
y™ =
Point of intersection of
two straight lines
(see diagram triangle)
t>2~ £>1
f
9
*3
tri]
f
10
V3
— m2
Angle of intersection y
of two straight lines
tan cp -
Triangle
f 11
JCj
+x2 +x3
y<\
+> 2 + >'3
s
3
Centroid 5
f
12
f
13
s
=
,
3
Area
.
A
_
"
(*i y2~x 2
y^ + (X2y3-x 3 y2) + (x 3 y -*i ys)
A
2
Where x and y have same dimension and are represented
scales (see also h 1).
in
equal
ANALYTICAL GEOMETRY
Circle,
Parabola
Circle
Circle equation
centre
at the origin
f
14
f
15
f
16
elsewhere
|
(x-x f + (y-y o y
Basic equation
x
2
+ y 2 + ax + by + c
Radius of circle
r
=
2
V*o + y
2
- c
Coordinates of the centre M
f
17
f
18
Tangent Tat point P-\ (xi, y-\)
{x-x Q ){xi-x )
y^
>o
- y
Parabola
Parabola equation (by converting to
and parameter p may be ascertained)
f
19
20
f
21
2py
-2py
2
L: directrix
elsewhere
(x-x
(x-x
2
=
2p(y-y
)
2
=-2p(y-y
)
)
)
Basic equation
y = ax
equation the vertex
F: focus
vertex
at the origin
f
this
+ bx + c
f
22
Vertex radius
r
P
f
23
Basic property
PF
PQ
f
24
Tangent T at point Fi (x-i, y-\)
2{yi-y<>)(x-X'\)
>'i
ANALYTICAL GEOMETRY
Hyperbola
Hyperbola
Hyperbolic equation
point of intersection of asymptotes
at the origin
f
25
x
2
elsewhere
2
v
*—
-
(x-x
=
1
a
(y-y ) 2
2
)
2
2
ft
Basic equation
f
26
f
27
f
28
Ax 2 + By 2 + Cx + Dy + E =
Basic property
^7 - F\P
= 2a
^
Eccentricity
e =
7
V*
"°
Gradient of asymptotes
f
29
tan a =
m=±-
*J
—
p =
y
a
Vertex radius
f
30
Tangent 7
at Ft
b
72
to, yi)
to-x~ )(x-x,) +
,
"
y\
- y
Rectangular hyperbola
Explanation
bola
in
a rectangular hyperthus
a = b
Gradient of asymptotes
f
31
m= ±
tana
1
(a = 45°)
Equation (for asymptotes
to x and y axes):
parallel
point of intersection of asymptotes
at the origin
2
f
32
x-y =
f
33
Vertex radius
p = a
c
|
elsewhere
(x-x )(y-y ) = c i
(parameter)
*' Conditions according to note on page F 1
>1
„
n
_ °
"
ANALYTICAL GEOMETRY
Exponential curve
Ellipse,
Ellipse
Ellipse equation
point of intersection of axes
at the origin
f
x2
34
2
—+
v
•*-
a
b
2
2
-
elsewhere
1=0°
(jc-xo)
1
2
.
(y-y ) 2
2
a
b
2
Vertex radii
f
35
f
36
ru
=
—
Eccentricity
e
=
^a 2 -b 2
Basic property
37
FJ~P
F^P
+
2a
Tangent T at P^ (xm y<\)
_bg
f
38
.
Ui -Xp)^-^)
a2
yi
- >
Note: F-\ and F2 are focal points
Exponential curve
Basic equation
f
39
y
=
a
x
Here a is a positive
constant 4= 1, and x
is
i
a number.
Note:
exponential
curves
pass through the point
All
x = 0;y = 1.
The derivative of the curve
passing through this point
with a gradient of 45° (tan a + ) = 1) is equal to the curve itself. The constant a now becomes e (Euler number) and is the
base of the natural log.
e = 2 .718281828459
+ Conditions according to note on page F 1
)
1
ANALYTICAL GEOMETRY
Hyperbolic functions
Hyperbolic functions
Definition
x =
f
40
sinh
f
41
cosh x =
2
x
x
e + e"
xk.
f
f
tanh x =
42
coth x =
43
e
x
X
-e"
e
2x -1
e + e
x
x
e + e~
2x + 1
e
2x
+1
e
e~-e -
e
-X
x
//
V*
v
2
2
>o
1
-3
-2
1
1
/'
-1
1
3
2
1
,
,
,—
Basic properties
f
7
2
cosh * - sinh 2 * =
44
coth x =
f
45
tanh x
f
46
tanh x =
cosh*
v
1
sinh x
I
ft
1
1
^-
-tanh 2 * =
costr*
I
\
7
V
I
1
~1
-coth 2 * =
9
sintr*
I
Ratios between hyperbolic functions
cosh* =
sinh * =
2
±Vcosh *-l'
\/sinh
2
tanh*
48
*+1
sinh*
2
Vsinh *+i
Vcosh 2 *-1
1
*
Vl-tanh *
49
Vcoth
Vl-tanh *
*
1
2
I
*-!
coth*
Vcoth
2
I
*-
50
f
51
f
52
sinh (a ± b) =
f
53
cosh (a ± b) = cosh a
I
|
1
1
tanh *
+ cosh *
- coth *
Addition theorems
f
tanh
54
sinh a -cosh b
cosh b
±
±
(a ± b)
tanh a ± tanh b
tanh b
1 ± tanh a
(a ± b)
coth a coth b ± 1
coth a ± coth b
cosha«sinh6
sinha'sinhfe
•
•
f
coth
55
+)
cosh*
Vcosh 2 *-1
coth *
For the defined * values of f 58
sinh (-*) = -sinh*
cosh (-*)
tanh (-*) = -tanh*
coth (-*)
f
~
cosh*
2
*
_,
,
2
f
Vsinh 2 * + 1
sinh*
*
f
coth * =
tanh * =
Exponent x always has to be non-dimensional quantity
* Sign + for x > 0; - for x <
)
ANALYTICAL GEOMETRY
6
Inverse hyperbolic functions
Inverse hyperbolic functions
Definition
function y =
identical
f
56
f
57
f
58
with
logarithmic
equivalents
arsinh x
ar cosh x
ar tanh x
ar coth x
x = sinh y
x = cosh y
x = tanh y
x = coth y
ln(jt +
f
59
1-x
2
defined
-co
within
primary
value
VJr+1) = ±ln(x + yV-1)
<x < +
oo
_ (X><y< + 00
1
^X< +
oo
2
jc-1
<1
IjcI
>1
<y< + a>
lyl
>0
IjcI
_oo< y < + oo
Ratios between inverse hyperbolic functions
f
60
61
f
62
arsinh x =
arcosh x =
artanh x
±arcosh\/l +X2
arsinhu?-1
arsinh
artanh
arcoth
artanh
\T
MZ
arcoth x =
arsinh
V*2-1
iff
± arcosh-p=
arcoshc
VlT?~
arcoth
arcoth
2 -1
artanh-
For the defined x values of f 58
63
f 64
f
arsinh(-;t)
artanh(-jt)
= -arsinh*
= - artanh x
arcoth(-;t) = -arcoth*
Addition theorems
± arsinh b
arsinh (a\/^+T
±
2
f
65
arsinh a
f
66
arcosh a ± arcosh b
arcosh [aft ± V^a2 --!)^ 2 -"!)']
f
67
artanh a ± artanh b
artanh
f
68
=
a ±fc
1
arcoth a ± arcoth b
* Sign + for x > 0; - for x <
=
arcoth
± ab
ab ± 1
a
± b
b Va
+ 1
ANALYTICAL GEOMETRY
Vectors
Components, magnitude, direction cosines of vectors
Vector: Quantity with magnitude and direction
A: Coordinates of the origin
Unit vectors along
x v y^,z^
x2 y2 z2
of the vector a:
B: Coordinates of the end-point of the vector a
:
,
,
OX, OY, OZ:7,J, k
Components with magnitude
and direction
i\\oz
i
f
69
f
70
f
71
f
72
f
73
f
74
a x ay az
,
-;
,
a = a % +ay+az
"OY
=
\i\
Magnitude or norm of the vector: la
f
75
I
a*
I
=
2
Vax + ay
2
+ az
I
l/l
= \k\
=1
or a in engineering notation.
'
(I
a always > 0)
I
Direction cosines of vectors: cos a, cos /?, cos y
a, /3, y, angles between the vector a
(a.
f
76
ft.
cos a =
-z£-\
la
I
f
77
f
78
and the axes OX, OVand OZ.
y = 0° ... 180°).
cos^3 =
-^
la
;
cosy =
Calculation of the components when la
=
I
2
cos 2 a + cos 2 ft + cos y = 1
where
a„
—
la
I
I
a
•
I
cos a
I
la l'Cos/3
a, ft,
;
y are known:
a z = la
I
-cos y
Note: The components along OX, OY, OZ are used to determine the
magnitude, direction cosines, sum of vectors and product of
vectors.
ANALYTICAL GEOMETRY
8
Vectors
Vector sum (difference)
Vector sum s of two vectors a and b
f
79
s
= a +b
f
80
sx
= ax + bx
f
81
\T\ =
+s z -k
= s x -i +s -j
y
s
;
y
=a
+ by
y
sz =
;
b
az + b z
+ 5y2 + S 2
2
yfsx
Vector difference s of two vectors a and b
f
82
T
=
f
83
sx
= a x -b x
f
84
f
85
f
86
[-t)Ut*
]
a+(-~b)
\T\=
yjs x
2
\
2
+ Sy + s
87
f
88
f
89
90°
180°
2
270°
2
\a\*\V\ la*l + IM ]/\a\ + \~b\ \a\-\~b\ Vlal 2 + I6l 2
I
li*l = l?l
7
;
360°
0°;
2\a\
Vector sum s of vectors a, b
f
s z = a z -b
2
?£\£
Special
cases for
Is
s = a - b
y
y
y
= a* + b -c* +
.
.
.
= 5 x *i
a x + b x -cx +
i-V5
V
,
\a\yJY
la 1^2
- c, etc.:
+ Sy-j + sz -k
(Vector equations)
sz =
ay + b^
a z + b z -cz +
+ 5 W + 5,
Product of a scalar and a vector
Scalar: Quantity with magnitude only
Product of a scalar k and a vector a
f
90
c
=k-a
f
91
cx
=k-a x
If
/c
>
is
the vector c
(k = 0)
;
c
y
then
= fc-a
y
c
;
ft a
(Vector equation)
(c|0)
c = JMi*l
c z = k-a z
o
ie
a
»-
c
*H
k <
then
c || a ie
Example: Force Fa = mass m times acceleration a
If
f
92
m > 0;
fatta";
^ = m-a
';
Fa = m-a
*) The symbol ti denotes that the vectors (-D^l and
but opposite in direction.
(£>*)
are parallel
—
-
ANALYTICAL GEOMETRY
Vectors
Vector Products of 2 vectors
The scalar product of 2 vectors a and F is the scalar k.
Symbol for Scalar Product: Dot "•"
93
k
= a
f
94
k
= ax -b x + a -b + az -b z
y
y
f
95
<P
= b
b
ccos
•
1*
a = a b-cos q> =
f
a
—
x 'b x
1
I
-\b
I
-cos q>
{k = 0)
—
+ a v 'b v + a?'bz
*-zf
\a\-\b\
i^^\^>
Special
f
Ii*l-I6
96
0°; 360°
90°
180°
-cos qo + li*l-l£l
I
270°
-\at\-\b\
Example: Work done W, by a force F over distance s
97
f 98
f
W = force ^distance = F s
W = F-s -cos
(W = 0;
F,s^0)
qp
Vector product of 2 vectors a and b
is
the vector c
Symbol for Vector Product: Cross "x"
f
99
f
100
—*
—*
7""*
= axfc
c
i~T*
c
_L a and c*_L £*
a
,
b
c*
,
101
cx
= a b z - a z by
y
f
102
cy
a z' 6 x-a x -& z
f
103
f
104
yfc
Special
f
cases
105
I
-
sin qo
(c
= 0)
>180°... 060° -*
form a basis
f
\c\
+ cS + s
ic
2
Ir&^^JP
0°; 360°
90°
around the point
f
107
M
M =
- -» 5
= Radialvector x force = r xF = -(F * r )^
r
F sin<p
(M=0;
r,
F ^ 0)
270°
-\st\-\t\
Hi
O
- ->
->
106
180°
+ \a*\-\V\
la*!-! b l-singp
Example: Moment M of a force F
f
...180°-*
=-(ixa)
= a-b -sin qo = \~a\-\b
lc*l
U
~*\
fj
r^
STATISTICS
Basic theory of probabilities
g
1
Theoretical probability P(A)
If £ is the set of outcomes of an experiment all
of which are
assumed to be equally likely and an events is satisfied by a subset
A of them, then P(A) = n(A)/n(E).
Experimental probability P(A)
If an event A is satisfied by a certain outcome of an experiment
and, when the experiment is repeated n times under exactly the
same conditions, A occurs r times out of n, then
g
2
g
3
P(A) = limit (r/n)
Axioms to the probability
g
4
g
5
event A has the probability P(A)
P(A)
0,
h(A)
number of events in which A occurs
number of possible events
=
relative frequency
? P(A0 = 10.
The sum
of
the
probabilities
of
all
possible
events A, taking place must be 1-0.
g
6
P(AnB)*)
g
7
P(A/B)
g
8
g
9
P(A) + P(B) - P(AnB)*K
If A and B cannot take place at once, then
P(A) + P(B) and the events are said to be disjoint.
P(AnB)/P(B)* is called the probability of A conditional on B (the probability of the event A, given
that the event B has happened).
the events are independent (if the knowledge that
one event has occurred has no effect on*the probability
of the other occurring) assuming P(A) resp. P(B) 4= 0.
If
g 10
g 11
P(AnB)
P(AnA)
)
and
P(BIA) = P(B)
P(A/B) = P(A),
P(A) x P(B) if events are independent.
P(A) x P(A) = 0, as A and A are mutually exclusive.
Venn Diagrams
The rectangle
represents the sum of all events A
The large circle
represents the event A
The small circle
represents the event B
Hatched area shows the conjunction of the different cases.
A
AvB
A n B
A n B
("not" A)
(A "or" B)
(A "and'B)
(B "but not" A)
STATISTICS
General terms
The random variable A
The random variable A is a measurable quantity which can take
any number x, or a range of values with a given probability
distribution.
The cumulative distribution function F(x)
The cumulative distribution function F(x) shows the probability
of the random variable being less than a specified value x.
g 12
g 13
F(x) varies between
and 1.0.
F(-x) = o and F(x) increases with x.
for an experimental
F(x)
F(x)
distribution
Fix)
for continuous functions
or theoretical distribution
\Fix)
The probability density function f(x)
The probability density function f(x) shows the number of
times one particular value p\ or range of values fix) of the
random variable A occurs.
F(x) =
p\
2a
F(x) =
fix) for continuous functions
or theoretical distribution
for an experimental
distribution
i
f(x) dx
fix)
1
i
0,3'
0,2I
1
0.1-
1
t
Iff.
1
1
(
5
6
7
8
X
The hatched area under the probability density function curve
shows the probability that the random variable A lies between x-\
and x 2
.
g 16
a 17
P(x,
<A<x
2)
\f(x) dx
F(x 2 )-F(x,)
P(A <x 2 )-P(A <x,)
1
STATISTICS
Gs
General terms
Mean value x or expected mean E(x) or
Random variable A
Random variable A
p.
continuous
discrete
g
13
g
19
x
= x r p,+x 2 p 2 +
•
•
+x nPn
•
+ 00
n
g 20
fi
= \x-f(x)
dx
- 00
i
-
where p, and f(x) are probability densities.
Variance o 2
g
21
o2 =
Random variable A
discrete
continuous
(xi-x)
2
2
pi+(x 2 -x) -p 2 +
+ 00
+ ... + (x n -x) 2 p n
g 22
g
Random variable A
o 2 = \(x-n) 2 -f(x)-dx
= hx-,-X) 2 Pl
23
+ oo
= \x 2 -f{x)-6x-p. 2
g 24
- 00
g 25
where p\ and f(x) are probability densities and
the "Standard Deviation".
o
is
called
Central limit theorem (addition law)
When two or more random distributions A with expected values \i
and variances o 2 are combined
g 26
A
the random variable
2 A-.
=
i
= 1
n
n
g 27
g
28
g
29
the mean value
the variance
If
n
=
2
o
=
z, 0\
(x
p.\
=
2 *i)
',
the random variables have normal distributions, then
P(A < x) =
where
(^)
the cumulative distribution
is
function for the standard
normal distribution.
g
30
Example:
If 10 batches of components, each batch having a standard
deviation of 0.03 ^m, are mixed together, the standard
deviation of the whole at, is given by:
tf
t
2
= 10 a 2
;
a,
==
± aVTfJ« ± 0.095 urn
STATISTICS
Special distributions
STATISTICS
Special distributions
STATISTICS
Standard deviation o
Determination of a when discrete values are available
By calculation
Equation g 23 says:
= Z(*,-Jc)
g 41
i
-
2
2 Xj
x =
with
-Pi
p\
1
n
= ZXi
g 42
2
Pi-*
2
where x are measured values
random
of the
K
variable
A and
are the frequencies of their occurrance.
p\
By graphics
Standardise the distribution and choose four values of x\ spread
across the range, say x 4 x 6 x 7 and x 9 shown in the drawing.
,
,
For each of these plot the cumulative frequency against the value
of x\ e. g. 10% to value x 4 38% to value xq and so on.
,
If
a straight line can be drawn through these points, the distribu-
is proved to be normal. The values of the mean x
and the
standard deviation o are obtained as shown in the diagram.
tion
The mean value x is at 50%. The difference between the
value A at 84% and the value A at 16% gives 2o.
*^
^
\
r\
Jj^"^
J
1
*4
If
'
—T—
3
5
1
1
10
16
1
20
1
1
30
1
1
k0
(cumulative distribution)
1
1
50
1
1
60
—
1
1
70
% less than x\
1
1
80 8^
—
1
l-T-l
1
£
STATISTICS
Normal distribution (Gaussian)
Normal curve for probability density cp(k)
o 2 = 1 and fx = in g 39 leads to the
standardized
density
probability
with mean value A = 0.
g 43
cp(X) = _J
\[2tz
<p(A) is given in tables Z 26 and Z 27 for
be calculated from g 43.
< A < 1.99, but can also
The connection between standardized probability density cp(X)
and the real probability density f(x) for \x * and a2 4= 1 is
(x-n) 2
y(A)
g 44
2 a2
<7
where
0\[2jt
To use the table, first find the value of the standardized probability
density cf(k) corresponding to A. Divide by tf to get the real value of
the probability density fix) for the value of x (see g 44).
Normal probability curve(Probability distribution function)
o2 = 1 and
/.i
=
in
g 39 leads to the
standardized normal distribution
g45
0(A) = \tp(t)
df =
As limit 0(k) = 1
dt
-^=J<
for A -»
«>
and <pM is a symmetrical function
means that:
g46
&(-X) = 1-4>(A)
The relation between the standardized <P(X) distribution function
and the real distribution for u 4= and o 2 4= 1 is
47/48
F[x)
<P(A)
1
/2^rJ
o\[2k
"^
e
6t
where
A
=
—o
STATISTICS
Probability distribution
Gaussian or Error curve
The curve is based on the standardized
normal
distribution
using g 45 for o 2 = 1 and u = 0.
The area under the curve gives
the value of the distribution
function between -x and + x
of the symmetrical density function (f(t).
x
_,2
6t
g 49
Values of
and Z
@ (x) between 0<x<1.99 are given in tables Z 26
For greater values of x look at the approximation
the next paragraph. The connection between <P (x) and the
error function is <P (x) = erf (x/\/2~).
27.
in
g 50
Error function
g 51
•D c.v2)--y
erf(x)
dt
2
2
g 52
Si"
...
n
-(2n + 1)
Values of erf (x) between 0<x<1.99 are given in tables Z26 and
Z27.For x>2 values of erff.vj can be found approximately using
g 53
where
erf(x)
a = 0.515
a = 0.535
for
2 < x < 3
<x
<x
< 4
3
< 7
4
7 < x <a>
a = 0.56
for
Area beneath the error curve when erUx) is subtracted:
a= 0.545
g 54
erfc(x) =
1
y
- erf(x) _ _2
V5T.
o (x)
and [1-<P a;] in %
relation to the whole area
for special values
g 55
<Pn(X)/%
± a
68.26
±2o
9544
2.58 a
99
99.73
+ 3(7
3.29 a
99.9
[1-4> M]/°/c
6t
for
for
STATISTICS
Random sampling
General:
sive
When testing each individual component is too expen-
or not
possible,
test
samples must be chosen
all
by random sampling is used. The
equal chances for
arbitrarily to give
parts (i.e. good intermixing).
The aim
of the test by random sampling is to predict the
probability of the real failure rate of the whole lot on
the basis of measured failure or error numbers in a sample.
Hypergeometric distribution: A hypergeometric distribution
occurs when the sampling takes place without replacement.
The probability P(k) that in a lot of N using samples of n,
without
replacement, exactly k defective parts are found,
is
the assumed probability for a defective part
(i.e. pN is the number of real defective parts in N and is
a whole number).
(pN\(N^-p)\
when p
pN is a whole number
g 56
(If
The
probability
found is:
not
that
ZP(*)
g 58
more than
k
defective
parts
are
k
P(0) + P(1) +
x -
P(k)
P N\/N(l-p)
(
*
\
x )\
n-x
pN is a whole number
Example:
In
a batch of 100 screws, a maximum of 3 can be defective^ = 100,
pN = 3). Random samples of n = 20 are taken. How many defective
parts are allowed in the sample? - The probabilities I P(x) are
X
P(x)
2P(jc)
x =
:
1
2
3
The table shows that for
508
508
0391
0.899
0094
0007
0.993
!
1.000
90% probability one part may be de-
fective.
Further special distributions: Besides the hypergeometric distribution which takes much time for calculation there are
other special distributions for defined assumptions and conditions. In tables G 4 and G 5 these are shown together with
the hypergeometric distribution; special characteristics of
these are explained.
:
STATISTICS
G 10
Confidence statement; Operating characteristic
The confidence statement P(x>k)
From a lot N, a random sample of n
is taken and k defective
are found in it. If the probability of finding a defective part in the lot is p, the probability of finding more
than k defective parts in the sample n can be derived from g 57.
parts
g 59
P(x> k) = P(k + 1 + P(k + 2) +
)
.
.
.
2 P(x)
+ P(n) =
x = k+1
If N is
large which is true for most manufacturing processes,
and ;;<0-1 the Poisson distribution may be used:
R
g 60
P( ,
<^.,
>M .Y<^.e---1-i
X
x
x =
x = k+1
and if the size of the sample k is small, then
k
61
(np) x _. np
A
= 1
(npf
L£ + (np)2 +
2!
1!
'"'
k\
.
the confidence statement P(x>k) for the proportion defective in a lot N can be determined when there
are k defective parts found in the sample n, or g 61 may be
used to find the size of the sample required if with an error
probability of p = kin, k defective parts are allowed for a
given confidence statement P(x>k).
Using
g 61
,
The Operating Characteristic (OC)
A user needs to know whether a lot delivered by a producer
meets his quality requirements. Assuming a proportion p defective in the whole batch (p<p
he wants to know whether
to accept or reject the whole lot if in a random sample of
n parts, up to c are found to be defective. The probability
)
that the lot will
the sample
g 62
be accepted on the basis of the evidence of
is
L(p, c)
> 1 - a,
where a is the producer risk or
from g 57
g 63
L(p, c) = P(0) + P(1 )+.... +P(k - c)
or using the Poisson distribution:
g 64
^(£Zl! e
L(p, c)=2
it!
-np
=e -np[1 + n ^(^)!
2!
(npY
c!
.
continued on G11
STATISTICS
11
Operating characteristic; AQL-value
continued from G 10
Using equation g 64
the operating characteristics
may be plotted in two ways:
L(p,c)
,
Type A
constant;
c:
Type B
parameter
Example
T=t
1
2
3
1
<.
proportion defective
Note:
of
c
the
5
n:
parameter
Example
Cil00
6
1
* p%
2
3
l,
proportion defective
5
*-
p%
Note: The bigger the value of n,
the
steeper
is
the
operating
characteristic; when n - N the
curve is parallel to the ordinate
and every article is tested. The
steeper the curve the more stringent is the control, n must be > c.
The smaller the value
becomes, the nearer
operating
constant;
=
characteristic
approaches p = 0.
c must be < n
Acceptable Quality Level (A. Q. L): Agreement between the producer
and the user leads to the most important point on the operating
characteristic, the AQL-value. The manufacturer needs to be
assured that the method of sampling will accurately predict
the quality of the lot. If this has a probability of 90%,
then the producer risk from g 62
L(p,c) > 1 - a = 1 -0-9 = 10%,
but the method of sampling may increase the producer risk. To
overcome it, the producer may decide to hold his failure rate
well below the agreed value
Upc)>>
of the AQL to
say p ^
which gives a permitted
failure of c-i, in the sample
as shown in the graph of
L(p,c) against p, which is
:
less
than c2
value
the
originally
As a
required.
of
result
probability
the
success in the lot rises to
,
99%. In practice, the AQL
has a value of about 0-65.
n
c
:
:
p* p o proportion defective -»p%
number in random samples
number of the maximum admissible defective parts
STATISTICS
12
Reliability
General definitions
g 65
Reliability
R(t)
Probability to failure
F(t)
JX(X) dT
n(t)
1
-R(t)
Failure density
df
t
-fx(t)di
e°
A(f)
g 68
A(0 =
Failure rate
fit)
_
R(t)
1
d/?
R(t)
6t
MTTF (mean time to failure)
00
00
MTTF = \f{t)-t-6t = \R{t) -6t
g 69
systems which can be repaired, MTTF is replaced by the
mean time between two errors, the mean failure distance
m = MTBF (mean time between failures). Values of MTTF and
In
MTBF are equal.
MTTF = MTBF = m = \R{t)-di
g 70
Product rule for the reliability R s
When Ri ... R n are the reliabilities
:
ments 1
.
..
n,
of the single elethe reliability of the whole system becomes:
R s = R^ R 2
g 71
.
.
•
.
R n = n/?j
i
= 1
t
(T)+X 2 (T)...X n (T)]dt
_
e -J[X
~ p
o
g 72
1
Note
Expressions
for
the
reliability
functions R(t)
G 4 and G 5
distribution functions F(x) in tables
culation
use
g
66).
The
exponential
are
distribution,
to calculate, usually fulfills the requirements (A = const).
n(t)
:
number of elements at the time t
number of elements at the beginning
the
(for cal-
simple
.
STATISTICS
Reliability;
13
Exponential distribution
Exponential distribution used as reliability function
g 73
Reliability
R(t)
g 74
Probability to failure
F(t)
e"
xt
-xt
g 75
Failure density
fit)
g 76
Failure rate
»»-«-*
k
const.
(Dimension: 1/time)
g 77
Failure distance (MTBF)
Product rule for the reliability R s
/?s=
g 78
1
6t =
-J
:
"^ e"^
-
<
•
.
.
-(X1+ X 2 + ...+X n )1
g 79
g 80
A s = A-| + A 2 +
Cumulative failure rate
.
.
.
* An =
MTBF
For small values the failure rate can be calculated approximately
number of defectives
number of elements at the beginning x working time
g 81
/.-values are mostly related to working
g 82
Unit:
1
fit
hours
= 1 failure/10 9 hours
Typical examples for failure rate A in fit:
IC-digital bipolar (SSI)
IC-analog bipolar (OpAmp)
Transistor-Si-Universal
Transistor-Si-Power
Resistor-metal
0.2
10 Resistor-wire wound
5 Small transformer
100 HF-cool
10
10
5
1
10
3 Quartz
Diode-Si
Tantalum gl liquid L
|ectro|yte
capacitor ?! solid
Alu-electrolytic capacitor
10 Light emitting diode (Aluminous
intensity is reduced to 50%)
500
0.5
|
20 Soldered connection
0.5
Ceramic (multilayer) capacitor 10 Wrapped connection
Paper capacitor
2 Crimped connection
0.0025
0.26
Vulcanite capacitor
1
plug-in contact
Resistor-carbon > 100 kQ
5
plug-in socket per used contact 0.4
Resistor-carbon < 100 kQ
0.5
0.3
5 ... 30
plug-in switch
Note: Specifications for reliability see SN 29 500,
Standard), DIN 40 040 and DIN 41 611.
part
1
(SIEMENS-
DIFFERENTIAL CALCULUS
Differential coefficient
H
Differential coefficients (or derivatives)
Gradient of a curve
The gradient of a curve y = fix)
varies from point to point. By the
gradient of a curve at point P we
mean the gradient of the tangent
the point. If x and v have
equal dimensions - which is not
the case in most technical diagrams - and are presented at equal
scales, the gradient may be expressed by the tangent of angle a
between the tangent at point P and the horizontal axis:
at
m
=
tan a
Ay =
f(x + Ax) -f(x)
lim
oAx
Ax
6y
Always applicate is gradient:
rr
Difference coefficient
The difference coefficient or mean
gradient of the function y = fix)
between PP<\ is:
h 2
Ay m fjx + Ax)-fjx)
Ax
Ax
Differential coefficient
Where Ax is infinitely small, i.e.
where Ax approaches zero, the slope
at P becomes the limiting value
of the slope of one of the secants.
This slope is the "derivative" or
"differential
of
the
coefficient"
function at P.
h 3
-
y'
£
= lim
-
/•«
6x
fix)
I
DIFFERENTIAL CALCULUS
H
Meaning of derivative
Geometric meaning of derivative
Gradient of a curve
If,
for each point x of a curve, we plot its corresponding
gradient as an ordinate y\ we obtain the first gradient
curve y' = f'(x) or the first derivative of the original
curve y = f(x). If we now take the derivative of the first
gradient y' = f'(x) we obtain y" = f"(x) or the second derivative of the original curve y = f(x) etc.
minimum
Radius of curvature g at any point x
Q =
Vd-^y'
3
2
)
M
M
is
is
below the curve where g is
above the curve where g is
Centre coordinates for radius g
1
h 5
a
=
+y' 2
^— y
x
y
h 6
u
_
..
.
^y' 2
continued on H 3
DIFFERENTIAL CALCULUS
H
Meaning of derivative
Determination of minima, maxima and inflexions
Minima and maxima
The value x = a obtained for y' =
h
h
h
is
inserted in y".
7
For
y"(a) > 0,
there is a minimum at x =
a.
8
For
y"[a)<0,
there is a maximum at x =
a.
9
For
y"(a)
=0
see h 19.
Inflexion
The value x = a obtained for y" =
h 10
For \ "(a )*
inserted in y'".
is
=
there is an inflexion at x
a.
Shape of the curve y = f(x)
Rise and fall
h 12
y'{x)
y'(x)
> o
<
13
y'(x)
=
h
h
11
y(x)
y(x)
y{x)
o
increases as x increases
decreases as x increases
is
tangentially parallel
the x-axis at x
»
Curve
h 14
y" (x)
15
/'(*)
<
>
h 16
y"(x)
= o
h
convex (viewed from above)
concave (viewed from above)
with
a change of sign
flexion
withoutj /(jt)atxhasa jbottom point
y{x)
y(x)
is
is
I
]
Exceptional case
Where at a point* = a
h
17
h 18
,(n-1)
= y"(a) = /" (a)
(a) = 0,
ane of the 4 conditions is present:
y'(a)
y
n)
n = even number
h 19
{n)
y
(a)
y\\™
but
* 0,
(a)
in
>
(n)
y
/
|
(a)
/ max. \
^
'
= uneven number
//
<
{n)
y
(a)
i
>
y
».
I
(n)
(a)
i
<
te-
DIFFERENTIAL CALCULUS
H
Basic differentials
Derivatives
Basic rules
function
c-x
h 21
h 23
y
y
h 25
y
h 26
y
+
± v(x)
u(x)
v(x)
=
al*l
v(x)
=
V^
=
u(x)
n-1
en
u(x)
•
derivative
C
=
h 22
h 24
n
jc
±v'(x)
u'(x)
y'
=
+ u -v'
u' -v
v - u
V2
u'
•
V
2y[7
v{x)
M
vf^ +U
,.
|n
^
Derivative of a function of a function
(chain rule)
h 27
y
=
=
/["(*)]
f'(u)u'{x)
dy _ dy du
6x
6n dx
Parametric form of derivative
h 28
y
=
fix)
fit)
dy.dj =
fit)
dt
{
=
Derivative of inverse functions
The equation v = f(x) solved for x
tion x = (fly).
30
dx
X
=
(p(y)
gives the
<p'(y)
h
32
=
arccos x
y = fi x )
gives
x = cpiy) = cos y
x3
inverse func-
/'(*)-
Example:
h 31
y_
x
d 2 y _ x'y-yx
dx 2 ~
h 29
h
'
y-f(x)
1
\f'(x)
s\ny
^|^-x2
DIFFERENTIAL CALCULUS
Hs
Basic differentials
Derivatives
Exponential functions
function
h 33
y
h 34
y
h
35
y
h
36
y
h
37
h 38
h
h
39
40
y
y
y
y
-
e
y'
-
_x
y'
=
-e
y'
=
ae ax
y'
=
e
y'
=
=
e
=
e
ax
=
x
e
x
=
Ve^
=
x
=
=
derivative
x
x
e
x
-
(1
+ *)
•
In
a
2
x
=
a
y
=
n-a nx
y
=
a
y
y
=
cos x
=
-sin x
tan x
y
=
cot X
y
=
{kx)
y
cos (kx)
a
a
...
_x
y
a
=
= y"
nx
X?
x
x2
In
2x
•
a
In
a
Trigonometrical functions
h 41
h
42
y
y
43
y
h 44
y
h
h
45
h
46
h
47
h 48
h
49
h 50
h 51
h 52
y
y
y
y
y
y
y
y
=
=
=
=
sin *
cos X
=
a
=
a
=
sin
=
cos"*
=
tan
=
cot
=
•
sin
n
*
=
=
a
k
y
=
-a k
y
y
=
n
=
i
2
-(1 + cot *)
cos{kx)
sin (kx)
sin
n_1
x- cos*
-n cos n_1 x
n~1
sin
x
=
-n- cot n_1 x
(1
+ cot 2 *)
=
-cosx
y
y
n
sin
y
2
+ t*
Ic n *
+tan 2 *)
*
cos X
I
(1
n
y
1
x
=
1
-=\- =
x
*
sin*
—
5
sin
n
1
'
cos X
=
tan
2
*
sin *
2
cos *
DIFFERENTIAL CALCULUS
He
Basic differentials
Derivatives
Logarithmic functions
:
function
h 53
y
54
y
=
derivative
In jc
y
log a x
y
=
1
X
1
h
h 55
y
h 56
y
57
y
h
x
a
In
•
y
=
±1
1 ± X
y
=
n
X
y
=
y
=
coshx
cosh x
y
=
sinh x
tanh x
y
coth x
y
=
In (1
=
lnx
=
InV?
± x)
n
1
2x
Hyperbolic functions
y
=
59
y
=
h 60
y
h 61
y
h 58
h
=
sinh
x
1
cosh 2 *
=
-1
sinh
2
*
»
Inverse trigonometrical funct ons
1
h
62
y
arcsm x
y
h
63
y
arccos x
y
h 64
y
arctan x
y
^|^-x 2
1
=
1
=
1
h 65
y
=
arccot x
y
=
+x 2
1
1+JC
2
1
h
66
y
arsinh x
y
arcosh x
y
yjx
2
+ i
1
h 67
h 68
h 69
y
y
=
=
artanh x
arcoth x
y
y
y
2
V* -V
=
1
1-jc
=
2
1
1-x 2
INTEGRAL CALCULUS
Integration
Integration
Integration, inverse of differentiation
By
integral
tion
we mean the problem of finding a func= f(x), such that the derivate of F(x) is
calculus
Fix) given y
equal to fix).
Thus
fw - «£*-/w
hence, we define
the indefinite integral
i
2
F(x)
Here
C is an unknown constant which disappears on differen-
tiation,
since the derivative of a constant equals zero.
Geometric interpretation of the indefinite integral
As this figure shows, there are
an
number
infinite
of
curves
y = F(x) with gradient y' = fix).
All y = Fix) curves are the same
shape, but intersect the x-axis at
different points. The constant C,
however, establishes a fixed curve.
If the curve is to pass through the
point x /y then
,
>'o
F(x )
The definite integral
The definite integral is represented by
fix)
dx
F(x)
=
F(b) - F(a)
takes place between the limits a and b.
obtained by substituting b and a are subtracted
causing the constant C to disappear.
Here
The
integration
results
INTEGRAL CALCULUS
h
Integration rules
I
ntegration
Basic formulae
i
i
5
n
Jx dx =
6
/*-
i
7
i
8
J[u(x)± V (x)] dx
=
f^dx
u{x)
——r + C,
In
here
n
4=
-1
+ C
|jc|
Ju(x) dx ± Jv(x) dx
In |u(jc)|
+
C
+
C
J
2
i
9
Ju{x)
u (x) dx
= \[u(x)}
Integration by parts
i10
Ju (x)
v' (x)
dx = «(*)
•
v(x) - Ju'(x)
•
v(jc)
dx
Integration by substitution
i11
Jfix)
dx = //[?(*)]
here
x =
<p(z)
and
<P» dz
dx = g?' (z) dz
Example:
f
i12
F{x) =
]\j3x-5dx.
Where
3x-5 = z,
Thus
dx =
—
,
the derivative is
z'
=
3— =
3.
expressed in terms z, the integral becomes
o
F(A:).l/VFdz = fz Yz"+ C
expression:
F(x) =
Insert value of z in above
|(3jc-5) V3jc-5'+ C
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant
13
(1n
dx
J x
n-1
a°*dx
15
i
Injcdjc
16
(«=M)
r
1
1
a
b
ln|a|
= x In |jc| - x
=
(In xf-dx
J
x
jc(ln
2
- 2x-ln |jc| + 2a:
|jc|)
2
(In
i
17
i
i
i
18
!x
19
jx m
i
i
•
\r\x-6x =
In
ax
dx
/<e
22
xe^dx
Jxe^dx
23
x
=
3-3!
1
In |x|
n
m+
{m + '\)'
^
(m * -1)
^-(ax-1)
[x2 e ax dx = e ax
a
t
=
J
n
(In 1*1)3
|jc|)
22!
ln|(ln|jr|)|
x\nx
21
In |*|
2
xdx = x
20
J
i
+ In |jc|
(ln|jc|)|
In*
y
Q
]x e *dx=
(- - ^f2 + 43
\a
a
a2
t
a
*-dx
-'e^ax
±x e -jjx n-1
ax
n
n
25
(n* 1
26
27
i
dx
A*
a
11
fe
ab
28
./fc
+ c-e'
+
\r\\b
+ c-e ax
|
continued on
I
4
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant C)
dx
In
ac
Jbb + c
e
|e
ax
sin
|
b + c e
ax
|
ax|
dx
bx dx
(a sin bx
a2 + b2
- b cos bx)
*
e3
je** cos bxdx
-T
ax + b
dx
=
J(ax + b) n
ln
l«*»l
r
./(ax-Z>)
i
a(/i-1)(ax + &)
=i- |n l^" 6
./ax
ax-b
(a cos bx - b sin bx)
a2 + b2
n_1
l
(»> 1)
a(n-1)(ax-&) n_1
n
*
./
f (ax + 6) (ex + d)
bc-ad
/;{ax - b) (ex - d)
ad-bc
cx + d
\ax + b
cx-rf
xdx
/*
1
J (ax + 6) (ex
fjcdx
=
6c-aJ
+ d)
In
(foc-aJ * 0)
(flrf-6c
\ax-b
T6
a
(bc-ad * 0)
L_b
J
ax + b
/
J
x dx
- 2b(ax+ b) + b 2 In |ax + b\
=
J X(ax + bf
ax + b
a 12
J
ax + b
,
a
a2
'
X
'
a4 L
2
3
dx
[
m _±
Jx(ax + b)
b
.
In
a + -
x
a
dx
_ _1 + -7?
In a + Jx2 (ax+b)
6
b)
6x
f
* 0)
.
I
£>
J
—
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant C)
i
i
dx
x J (ax + ft)
45
(ax + b)
2
x
i
i
47
I
ax + b
2
dx
(ax + b)'
I
2a (ax + ft)
(ax + ft)
2
2x<
dx
x
46
J_ [ a 2 (n
2
2
a (ax+b)
X
\(ax+
a
J
In
|ax+ ft|
a
b)
-
- 2b In \ax+ b\
ax + b
L
48
(ax + ft)
i
dx
x
49
I(ax+bf
i
i
50
x
2
dx
/i(ax +
x 3 dx
(
51
bf
(ax + bf
i
i
i
dx
x(ax + bf
54
55
dx
f
3
i
57
f
58
f
]
i
i
59
60
3
L
2(ax + bf
ft|
\
'
+
-^- - 0/
a
2
\
I
x
I
ax + ft
b
2
In
b'(ax + b)
ab'x
b'
,
H a+
2 (ax + ft)'
£|
ax
ax+ft
*^|}
3
.a *. + (ax+ft]! _ 3a(ax+ft) 1
|«±ft|
+
3a m
4[
*
2x^
J
ax+ft
x
ft
L
2
1
I
4- arctan J
a
a
^dx
x- a arctan a-
2
3b<
1
b
\«
ax +
l
^-ln|a 2 +x2
a
b
2 (ax+bf
ax + b
—7 \(ax+ b) - 3b In \ax + b\
xdx
+x 2
56
'
i
(ax+bf
'a^x2
/*
i
[ln|ax +
X
a
dx
'x^ax + ft) 2
'x
i
ax + b
I
(
52
53
2
a
|
x 3 dx
^+~7
dx
dx
_
J_
J_ ln
a+x
^_a2 " a '2 |a-x|
= _l, n|a2_^
to -/.^
continued on
I
6
INTEGRAL CALCULUS
16
Basic integrals
Integrals
Q
(omitting integral constant
2
dx
fx
Ja2 -x2
fx
3
2
fx
_
ox
x + a— In
xt-a2
2
J:
"
dx
JaZ-x2
r(aW)
—^-^l
x dx
2
X
+ x- arctan —
9
a
2(a2 + x2 ) 2a
dx
tf+x2 2
)
x3 dx
f^ + J_ln|
2(a2 +x2 )
**2 2
2
(a
arctan a
-~
2a 3
2(a2 +x )
)
x
—
)
1
tf+x2 2
2
+
p-
2a2 (a2 + x2
2
)
2
2
a +JC
|
2
2/1-3
Jtf+x2 )"
2a 2 (n-1)(a 2 +^) n
-1
1
2a2 (a2 -x2 )
2
2 n -1
(
dx
2 2
/,(sf-x )
dx
f
2
2a (n-1) J(a + x
2a 3
)
\a±x\
ln
\a-x
2
dx
Ha2 -*2 2
)
f
2
x
~ 2{a2 -x2 )
dx
x
)
fyJTtx -
2a
)
2
|n
'"
la + *|
\a-x
fV?
ax + b dx
H
1
2(a2 -x2
Jtf-x2 2
3a
yjiax + bj*
aYTFdx = 2{3ax-2b)j/{ax + bf
2
'ax + b dx
2
2 (15a x -12afr;c + 86
105a°
fe = 2VF
2
)
V(as+6)
s
(n*1)
|
;
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant
dx
J\Jax + b
78
fxdx^ m 2{ax-2b) yT^T^
v
3a 2
J^/ax+b
2
79
[ x
J
2
m 2(3a x
dx
=
fxV>
82
J* Va
2
J*
2
| \/a
2
+ x2 +
2
IVc -*
+ x" dx
81
83
- 4abx + 8b 2 )\f(ax + b)
15a 3
'
i
2
'
yfax+F
2
2
fyja + x dx
i
Q
m 2y/(ax + bj
a
[
± arsinh
2 3
)
+x 2 dx - | Vfa 2 + x 2 3 - j- Uy/a'+x1 '* a 2 arsinh
)
y^S dx
-
^£±2?- ai^S
"
+x - a
84
85
i
i
86
87
dx
I*
13.
dx
/vfc
-
Va2
In
4
arsinh
Va 2 + x 2
2x 2
2a
2
2
J_ m a + Va +x
|
I
x
|
—J
89
90
Vjir+jr
dx
/
i
91
J. n
a
i
•'xVaV
dx
92
2
>^z
a + Va
|
x
2
+x2
|
VrW
continued on
I
8
INTEGRAL CALCULUS
8
Basic integrals
Integrals
(omitting integral constant
dx
f
*
=
_v2+
—
2X2 ia c
x 3 Va 2 +* 2
/V?7? d*
2'
i
=
2a
Q
V 2^ v
|a
In
2|
3
yja^+ a
x
[
a
2
-x 2 f
jx 2 ^a 2 -x 2 dx = -fy(a 2 -x 2 3 ~^(x\f
)
2
V(a -x
a'-jfdx
dx
f
x dx
f
x 2 dx
f
^
:
dx
a
x
\
\
Va -*
2
a?*
a + ya -x
/ * 3 \4
2a J
2a' x'
2
2
Jy^-a dx =
-a 2 dx
2
2
-|-
^V^ -a -a arcosh
=
1 \J(x -a
jx 2 yJx 2~^aT dx =
\*\Ptt*x -
^
)
1 n '+W-*
d*
2
-x 2 3
3
^%^-a^V^?
2
^T^
\x\lx
2
^*?
,
vS^7
I
yJ(a
o
5
x3 dx
VaT
a
:
tV*"
•/
)
5
arcsin
/ Va 2 -x 2
2 5
2
2
f \/(* " a
2 3
)
^Z^
5
y (* V* " ^ - ^ arcosh
a V^a!g
+
2 3 '+
2
)
2
3
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant
——
-a
<x
i
i
109
i
i
i
i
dx =
yx
Y*
~a 2
113
114
115
a2
+ arcosh -
- -g- cos ax
sin*
ax dx
/sin 3 ax dx
/
a arccos r
\
I
I
2
Q
3
2
- a2
1
Vx
a
v
o
+ <r- arccos -^
x
2x l
2a
111
112
Y
a
+r~2
4"
%^d*
110
2
i
,
sin
"
x
1
2
4a
sin 2ax
1
ax dx =
cos ax
•
1
sin"
ax +
rca
¥/*""
(a?
i
116
J
i
i
117
118
x
120
2x
Ix2
dx = -^ sin ax -
sin ax
dx
f sin a
J
x2
f sin ax
dx
ax dx
an integer > 0)
x cos ax
sin ax
fsinj
119
sin ax
x sin ax dx
is
(Zx2
6
j
2
a ~
i
2
g
c
1
sin ax - i-
£ cos
i
= ax .iixl + iaxl_iaxl +
3-3!
5-5!
7-7!
¥*•/'
dx
1
.
sin ax
dx
a
.
/"
cos ax
.„
121
122
I
cos ax dx
2
123
cos ax dx = 2 + 4^"
124
cos 3 ax dx = -=a
sin 2a *
sin ax
sin
3a
3
ax
continued on
I
10
INTEGRAL CALCULUS
10
Basic integrals
Integrals
Q
(omitting integral constant
—
n
n
Jcos ax dx = na sinax-cos
——
I
2x
x c cos ax dx =
/3X2
6
2" ~
—
=
:
ax dx
i
2 \
h] sin ax
ar
\a
sr
I* cos axdx
n -
sax +
^
i
sin ax
^H
3
(
\a
a
i
f cos ax
x2
f cos ax
x
J
/
/
'
r/i-1)x
-—
ax dx =
2
In
|
sinax-dx
X
a_
n"1
n-1
/"
sin axd.
tan
tn
cot ax dx = -j
n~1
ax
(
In
|
tan"
'
ax
1
(*=M)
dx
=
_
i
i
6x
(n*1)
tan
—
128
i
129
130
131
132
133
134
135
136
137
138
2
cot ax
Jsm n ax
i
i
.
127
L„„ ax
,„
In
a
/;
/*
dx
sin ax\
9°fl32L_L 0X »-2 ax
—
•
J
[cot°ax dx =
=
(n*U
i
cot ax
sm
n-1
i
cor ax dx
; sin a
ax
x
i
cos ax|
i(n-1)
/
i
— tan ax-
axdx =
tan' axdx =
/
/"
J
cos ax
=
.
n
tan
tan
cos ax _
X
dx
126
J
i
J
125
J
i
2
/x
cos ax +
-5-
n
sin ax
cos5 ax + x
ov- Hv =
xv ™e
cos axdx
ax + ^^fcos
1
1
139
cos ax
r
a(n-1)' sin "'ax
+
n-2 f dx
rt-Wsin n_2
(n>1)
140
INTEGRAL CALCULUS
11
Basic integrals
Integrals
(omitting integral constant
xdx
1
sin
f
i
2
x
1
cot ax + -*- In
ax
a
_*
=
142
t
dx
cos^ ax
143
a
|
sin ax\
a^
tan
ln
ri
?)l
t
tan ax
dx
dx
J cos
i
i
i
i
i
i
i
i
i
i
i
i
147
152
153
154
155
ax
dx
-
=
At
1 tan^
dx
->(f-f
T
^
sin ax
dx
\2
2
a
+„„
1
(n _l ax
cot
sin ax
sin bx
I*
dx
=
cosax cosbx6x ,
n
sin (ax + bx)
sin (ax - bx)
2{a + b)
2(a-b)
_ cos(ax + fex) _ cos (ax - bx)
2(a + b)
2{a-b)
cos bx dx
/sin ax
(|a|*|6|)
(|a|*IH
^^*5^(i«i*i»i)
axdx
dx
Jx sin ax
=
- |- cos ax +
x n cos axdx
=
|- sin ax +
J
(«>D
<
Itanf^-f
4
a
dx
cos ax
j
n-1 J
- tan ax + -* In |cos ax|
sin ax
149
151
1
fc
148
150
a(n-1) cos n
ax
x dx
cos ax
145
146
n
Q
j Jx
j-
J
n_1
xn
cos ax
dx
ax
dx
~
sin
J —^ =l|n|tanax|
_dx
X
f s]n ax
cos ax
_dx
156
/;
n^ax cos ax
|tanf^ +
4
^
2
contin.
on 12
I
INTEGRAL CALCULUS
12
Basic integrals
Integrals
(omitting integral constant
dx
3
sin ax cos ax
dx
cos' ax
,"
In
a
'in
—
2 cos
J
/
m ax-cos n axdx
m
ax
160
1
161
a(m + n)
dx =
cos ax
x dx = x
arcsin
arccos x dx = x
arctan x
dx = x
arccot x
dx = x
S[ "
mTn J
sin"
m
ax
cosn
"2
axdx
,T
arcsin x +
•
•
•
'
ax
dx =
sinh {ax)
sinh
2
n
y 1 -x
arccos x - y 1 -x 2
arctan x -
—
In
arccot x +
—
In
•
dx =
1
+ x2
1
+ x2
— coshx sinh n_1 -x
cosh (ax) dx = ^- sinh (ax)
163
i
164
i
165
166
I
i
i
*/ sinhnx >
•
[n
/
162
i
i
— cosh (ax)
•
i
2
dx = ^-sinh (2x) -|
x
x
(m*-1)
a (m + 1
2
Jsinh
J
159
ax
cot lax
cos 2 ax
J
J
2
n is an odd number, solution for the Remainder-integral:
sin
J
158
1
+
tan ax
In
I
•
+
/
i
I
dx
If
157
2s\n 2 ax
\
tan^U
cos a*
2
1
sinajc
•
dx
=
ax sin ax
sin' ax
sin
tan ax
|
Q
dx
i
167
168
169
0)
i
170
INTEGRAL CALCULUS
13
Basic integrals
Integrals
Q
(omitting integral constant
2
Jcosh
J
x
cosh n x
in (ax)
7
2
tanh
n
dx =
6x
=
/,sinh ax
d.r
/-
.'sinhr ax
f
dx
cosh ax
(
dx
/;coshr x
/
J
J
—
In
|
cosh (ax)
\
|
sinh (ax)
Jtanh
=
4a
--|- coth n_1 x +
n "2
Jcoth
ax
In |t<
2
coth x
2
c
arctan e
a
tanhx
arsinh
x - yx
2
•
+ 1
arcosh x dx = x
arcosh x - \/x
2
•
~ 1
artanh x
n_2
* 1)
x dx
(n
x dx
(n 4= 1)
|
x dx = x
arsinh
dx
(n>0)
= x - coth x
jcoth n x dx
J
•
n
dx - ^- In
coth 2 x dx
/
1 smh x cosh n_1 x + ^j— Jc Dsh n_2 x
-i-
1
n_1
x +
dx = - -|- tanh
x
coth (ax)
J
dx =
dx = x - tanh x
/« nh x
J
dx = ysinh(2x)
dx = x artanh x +
Jarcoth x dx = x
•
•
—
In
1
- x2
1
arcoth x + -r-ln|x
2
-1
|
INTEGRAL CALCULUS
14
Application of integration
Arc differential
Vd* 2 + dy 2
6s
H
189
1
+ y' 2
1
+
dx
surface area where the curve
rotates around the .v-axis
arc length
i
1
xjyi + y
dx
1
,d
dx
static moment of a curve
y-axis
.v-axis
b
b
t
i
190
Mx =Jy^+y' dx A/ = J*Vl +y' 2 \
2
y
a
a
coordinates of center of gravity
A/v
191
ys
s
volume of a
body where
body, the cross
section .4f of which
area A rotates
around the .v-axis
is a function of x
rotating
y dx
192
static
2
V = jv
J y
dx
moment of a curve
in relation to
the
u
dx
193
By-
J
xy dx
coordinates of center of gravity
194
tfv
V
= j A, {x) dx
—
INTEGRAL CALCULUS
15
Application of integration
Static moment of a body
(in relation to
195
=
Af yz
the y-z plane)
x yz
31
J
dx
Coordinates of center of gravity
i
196
Myz
—
V
=
JC,.
5
Pappus theorems
Surface area of a revolving body
Am =
arc
length
times
5
the
distance
covered
by
the
center
of gravity
=
197
2-JC-S -y a
(see also formulae
Volume of a revolving body
= area A times the
V
distance
covered
by
189 and
i
i
191)
center
the
of
gravity
=
2
Division
of
A
31
(see also formulae
y
192 and
i
i
194)
Numerical integration
area
into
an
even
number n of strips of equal
199
.
£1^2
Then, according to the
yn
W^f
!
t
^wIw^1v^^\J\\vnJs\
Trapezium rule
*>1
A
200
-
| (y + 2y, +2y 2 + ... + y n
)
Simpson's rule for three ordinates:
201
^1=|
(yo
+ 4y! + y 2 )
Simpson's rule for more than three ordinates:
202
A
=
+
+
|[>- +>'n 2(y2+>'4+---+}'n-2) 4(yi+>'3+---+>'n-l)
INTEGRAL CALCULUS
16
Application of integration
Moment of inertia
(Second moment of area)
General
By moment of inertia about an axis x or a point O, we mean
sum of the products of line-, area-, volume- or masselements and the squares of their distances from the jc-axis
the
or point O.
i
Second moment
of area
/** dm
203
dm, 6A
Moment
of inertia
-1 x2 dA
kg m'
Steiner's theorem (Parallel axis theorem)
For every mass moment of inertia, both axial and polar, the
following equation will apply:
/„ =
204
/
yy
Similar equations
line, area
kgm
m/.
will
2
apply for
and volume moments of
inertia:
2
/^ + Al s
Moment of inertia of plane curves
in
relation to the
x-axis
y-axis
>n
,b
7 oy
205
= /**
V 1+ J ,2d *
a
7 XX
I
m, A
/
s
:
:
:
:
moment of inertia about a general axis xx
moment of inertia about the centre of gravity
total length, area, volume, or mass
distance of centre of gravity from axis or point
J
.
INTEGRAL CALCULUS
17
Application of integration
Moments of inertia, centrif. moments of plane surfaces
By axial second moment of area of a plane surface in relation
to an axis x or v within the plane we mean the sum of the
products of the area-elements d.4 and the
Kii
squares of their distances from axis x or v,
respectively:
-/:
206
dA
A given function y
fix) yields:
v-axis
.v-axis
i
207
r,
-
cM
h = j
\f *
r
y
-j
dx
By polar second moment of area
of a plane surface in relation
to a point
within the plane we mean the sum of the products of the area-elements dA and the
y
squares of
i
their distances r
from point 0.
r
I
dA
208
ot
^
XU
C*
— —
,
X
Where the relative axes of / x and / y are perpendicular to each
other, the polar second moment of area in relation to the
pole
209
(intersection O of axis x and y) is:
ff
dA -
f (/
jT)
dA
=
/x +
L
By centrifugal moment (product of inertia) of a plane surface
in relation to 2 axes within the plane we mean the sum of
the products of the area-elements dA and
the products of their distances x and v
from the two axes:
x-y
210
-m
dA
One
of the relative axes being
plane surface results in / xy = 0.
an axis of symmetry of the
Conversion to an inclined axis x'\ Where moments / x
in
relation to two perpendicular axes x
yk
and y are known, the second moment of
area l u in relation to an axis inclined x'
by an angle a with respect to the .v-axis
can be calculated by:
211
Ix cos
2
a +
/
y
-sin^a
-
/ xy sin
2a
,
/
y
,
and / xy
INTEGRAL CALCULUS
18
Application ot integration
Examples in conjunction to second moments of area
on page 17
I
Rectangle
yM
,31
U'
212
i
-
-h
J
3 Jo
o
i
i
213
Z,
214
T
i
i
bh 3
«w(f)
b3 h
y
///XX////
Vy.'Vf/^rV
b^h
r
.
y
3
"
"
'
bh 3
k
216
2
_,_
b
3
f(b< + h ); /ps = ^(fcW)
2
As x' and/or / are axes of
h.
'
bh ,, 2
h _
2
symmetry, / x y is zero. Hence:
.
i
217
-xy
"
h
~2
±J^
12
'
215
b
2
y '\\
bh;
=
6y
y'-b
b h
{bh) =
(~2~
dA
\r 2 2K r 6r
.
Circle
i
i
i
218
frr
2
219
220
£L£! = £L»!
= Ze
" 2 "
64
4
as x and _y are axes of symmetry.
j
y
i
=
221
Semicircle
i
i
i
222
A
y -2JC
/ R
=
223
224
/
7p
2
2J> V^
2^-
dy
XR
A
= -r~y 2^ dy
jiR'
= 2
0,
as y is axis of symmetry.
8
Regular polygon
i
225
y
'x
y
2
V*
2-48
V
2
a
2
(6fl^ -a')
48
/vw =
r
K
:
:
radius of inscribed circle
a:
radius of circumscribed circle
n:
length of side
number of sides
INTEGRAL CALCULUS
19
Application of integration
Second moment of volume of a solid
'Moment of inertia' of a cuboid
Where ($£- +
^
)
is
rM7\
the polar
V12
12/
moment of inertia of a rectangle
(see
I
18),
the equation for the
Z-axis:
226
,z
J
\12
a
-^(b 2 + h 2
dz
12
)
Moment of inertia' of a circular cylinder
for the axis Z:
227
/¥
A
Jtr h
dz
h-~- \~~
for the axis X:
228
Dynamic moment of inertia (mass moment of inertia)
The mass moment of inertia / about a particular axis is the
product of the second moment of volume / v about the axis and
the density g.
229
i
i
230
where
J
=
Q
=
kg m
Jy g
2
,
N m s2 V A s3
,
kg m~ 3 kg dm" 3
y
,
e.g. for a cylinder about the axis Z:
jtr
231
/vz
4
2
h
r
m
mr 2
Kh
2
2
For other mass moments of inertia see M 3
DIFFERENTIAL EQUATIONS
1
General terms
Definition of the Differential Equation (DE)
A DE is an equation of unknown functions which contains derivatives (partial
derivatives)
of the
unknown functions and
independent variables. The different kinds are:
Ordinary
M
Differential
Equation
unknown functions
(ODE): the
depend only on one independent variable, e.g.:
y" + 2x 2y = sin*
y = f(x)
Equation (PDE): the unknown functions depend on a number of independent variables, e.g.:
Partial
Differential
2v W *x„dx
]2
Partial
sidered
x=f(u,v,w)
9v
9w
du 9v
Equations will not be specially conas methods for Ordinary Differential Equations
Differential
here,
can be applied.
Ordinary Differential Equations
J'3
F (x, y(x), y'(x),
y<n)(x)) = 0.
Where y (x) is the unknown function, y' ...yl n
Form:
...
)
are the
1
st
to
n th derivatives; x is the independent variable.
J4
Example: y'" (x) + m(x) y'(x) + n(x)y 2 (x) + p(x)y = q(x).
ODE»3 th order in
j5
Order: the highest derivative occuring
the above example.
J6
Degree: the highest exponent of the unknown function and
derivatives;
j7
in
the
its
2 nd degree in the above example.
ODE means, that the nighest exponent of the required
function is one; i.e. an ODE of degree 1.
Linear:
]8
Homogeneous ODE implies the forcing function,
)9
Inhomogeneous ODE implies the forcing function, q(x) = 0.
J10
Solution: y = y(x)
q(x)
= 0.
of an ODE means, that this function
derivatives satisfy the ODE.
and
its
)11
Integration of the ODE yields the solution.
J12
The general solution of an « th order ODE contains n constants
C,, C 2 ..., C n These constant are uniquely determined from n
boundary conditions.
.
J
13
y '(x Q ) =>>;
...
y (»-V(x ) =y
m-u
The particular integral of the ODE is a special solution.
DIFFERENTIAL EQUATIONS
Linear Differential Equations
Methods to solve an ODE
1.
Transform the ODE into one of the standard forms listed in J 6,
J 8 ... J 12.
2.
Application of a special method (cf. J 8).
Using this method ODE can often be reduced to a standard ODE
of lower order or degree (cf. J 9 ... J 12).
3.
Use
of
transformations,
particularly of the
Laplace-Transform
D18 ... D20.
cf.
Linear Differential Equations
Form:
+ p (x) y(«-V + ... + Pn _ (x) y' + Pn (x) y = q(x).
y<">
J
15
J
16
Here y = y(x)
i
17
to
/?
th
x
x
the required function, y' ... >>(") the 1 st
derivative of y(x) and P^(x) ... P n (x) are functions
is
of*.
General solution of a linear inhomogeneous ODE.
y = ^hom + JVpart
Solution of the homogeneous ODE v hom
_y
19
hom
is
determined by setting the forcing function q(x) =
C,...C n
.
Jhom = C y (x) + C2 y 2 (x) + ... +
20
x
x
J 9 ... J 12 give solutions for 1
tial
j
21
0.
Each linear homogeneous n-th order ODE has n linear independent solutions y-\,y~2---yn witn n independent constants
st
and
Cn yn (x)
2 nd order Linear Differen-
Equations.
Particular solution of the inhomogeneous ODE v part
y pan is determined for q(x) * 0. J 3, J 6 and J 7 suggest how
st
and 2 nd
to find solutions. J 9 and J 12 give solutions for 1
order Linear Differential Equations.
DIFFERENTIAL EQUATIONS
Linear Differential Equations
Particular Solution
Determination using "Variation of constants" when _y hom of a linear
« th order ODE is well known (cf. J 2, 20), the following formulation
always leads to a particular solution:
j
j
y P an = c iM y\ + ciM yi +•••+ cnW y n
23
Method to determine
J24
Cj(x),
-
C2 (x)...Cn (x):
Form the simultaneous equations
+ Ci(x) y 2 + ... + C'n {x) y n =
+ C'n (x) y'n =
C{(x) y 1
C[(x) y[ + Ci(x) y 2 + ...
Q(x) y 2 <n - 2> + ... + C'n (x) y n ("-V =
C[(x) yx (*-V +
C{(x) yi (n-D + Ci(x) y2 ("-V + ...
j
25
Determine C
'
:
for
(x)
i
+ C'n (x) yn (n-D = q(x)
= 1, 2... n
using the above equa-
tion system.
j
26
Integration of Cy (x) for
i
= 1, 2... rc yields the values of Cj fjcj
for the solution.
Example: Solution fory part of the ODE:
„
1
.
..
2x.
J27
28
1
Ace. to
j
J* -J»* dx + C,
121: ^hom
29
Inbcl
j
30
let
j
31
using
>>part
j
= C
\
hence
In
and y 2 (x) = 1
IjcI
2 (X) y 2
3^1
C[(x) Inlxl
Nqw
24
=
^ +C
fcj
QY*; = 2x2;
+ Ci(x)*l =
+ Ci(x)^0 = 2x
= -2*2 !nW
Integration of C A (x) and C2 (xj gives:
C (x) = |*3
t
SO
Jpart
;
C2
W = - Sx=
= 1*3* InW - |*3 (Inbcl
- j\
flnlxl
3
;
9
General solution:
j
+ C9
x
using y 1 fjcj
J32
C,
= Q y (x) + C2 y 2 (x)
35
3Wt= C l* lnW + C2 + 9* 3
>'hc
=
Check:
a:
-
£ + 1*2
jc2
3
Ci
x2
x = 2x
X
DIFFERENTIAL EQUATIONS
Linear Differential Equations
1
j
36
Form:
st
Order Linear ODE
+ p(x)y = q(x).
y'
15 for n = 1; the highest derivative
here is >•'. Solutions for y, _y hom and y pan are given in J 2 and J 9.
The form corresponds to J 2,
J
37
j
38
j
Example: y' + I = sin x
y = y hom + y pan
= I
= sin x.
from
j
110
from
j
109 the homogeneous solution is:
p(x)
d
C,e^
=
39
from
j
j
'
,n
= C1 e-
"
,
=
£
Cx |o.
with
110 the particular solution is
j
m = J sm xe
40
q(x)
x
e
d*
.Vpar
J(sin x e
,nlxl
dx e
)
Mxl
= J(sin
xx^i
= -sin jc - cos x
y = y hom + y?an =
J41
~.
,
Check:
y
v'
j
C
42
0;
X
x cos x- sin x +
Ci
sin x
= - -j+
^
r
.
,
+ 2- =
— Q
x
\{C + sin x) - cos x.
is
sin x
determined using the boundary condi-
tion e.g.
y(x ) = 1 for x
j
43
j
44
Then:
j
45
Gives
1
:
Cj
= ji/2
= ^(q + sin
f)
- cosf
= y- 1.
2 nd Order Linear ODE
j
46
y"
Form:
+ p (x) y'+p 2 (x) v = <?W
:
The form corresponds to J 2,
tive
is
/'.
and J 12.
Solutions for
y,
j
15, for n
= 2; the highest deriva-
y hom and y part are given
in
J 11
DIFFERENTIAL EQUATIONS
Linear Differential Equations
Linear 2 nd order ODE with constant coefficients
47
j
J48
Due
to the great importance of this ODE-type for oscillationproblems, special cases are considered.
y"
Form:
+ lay' + b 2 y =
q(x).
a and b are constants =t= 0,
is a forcing function
q(x)
49
j
General solution, according to J 2,
50
j
i
si
j
52
j
53
J
54
y -
j
j
56
j
57
j
58
Overdamped solution:
k2 = a 2 - b 2
6x
j
61
k2 = a2 - b 2 =
Critically damped solution:
Jhom =
y pan
CiS-^+Cixe—
= -e~ ax jx e™ q(x) dx + x e-™ Je *
^hom =
ypan
e-^lQ sm(o)x) + C2 cos (ax)]
oj
= \[b 2 -a 2
= e-"sin(ft«; | e « cos(w;cj
62
^
6x _
J> sm(o)x) q(x) dx
q(x)
= A
'
where:
/4
and:
y
= arccot
b
+ 4a 2 0) o 2
2 -0) 2
o
*)
s'\n(oj x)
S n (°>oX-r),
-Vpart = ^
\/(fc 2 -a> 2 ) 2
j
q(x) dx
k2 = a 2 - b 2 <
Underdamped solution:
Note: For the special case
60
>
.
- e-»cos(o>r;
j
:
r
with
59
15
.Vhom + ^part
a(-a+k)x
><-a+k)x
55
j
is:
DIFFERENTIAL EQUATIONS
Linear Differential Equations
Linear n xh order ODE with constant-coefficients
j
63
a n -y< n > + a n _ x -y(n ~^ + ...
Form:
+ a y' + a^ = q(x).
x
Solution of the homogeneous n xh order ODE with constant coeffij
64
j
65
cients <q(x)
Let
= 0).
y' = re rx
y = e";
.
;
.
.
y(n >
= r"-e rx
Substitution in the homogeneous ODE of 63 leads to the algebraic equation:
+ a n _ 1 rn ~ 1 + ... + a x r + a = 0.
j
j
a^
66
The roots
r1( r2 ... r n can be determined. Depending
the type of the roots, different solutions fory hom are found.
Case a)
j
67
j
68
:
on
r1t r2 ... rn are all real and different:
2 '*
+ - + Cn e r" x *>
?hom = c e "* + Q-e'
Case b): There are real single and multiple roots:
i
rx
= r2 = ... = rm
C
^hom -
i
QrvX +
,
rm+]
C2 -x-en* +
,
rm+2
rn
,
.
qp+r* +
+ Cm -x m -^e r r x + Cm+1 -e'™ + i* +
+
+ Cn -e r* x
...
...
*>
= e'r* (C + C2 x + ... + Cm -x m ~ ) +
+ Cm+l -erm*vx + ... + c„er» x
l
x
.
Case c):
There are conjugate complex roots:
r-j
J
70
y h0 m
= a + i/8;
r2
= a ~ P = 'TX
= C v e^+C2 -en-* •;
= e^-fA-cos /3x + fl-sin fix)
A = C + C2
x
;
5 = ifQ-C^
Particular solution of the inhomogeneous n th order ODE with constant-coefficients
y P an = SiM + SiM + ». +s*to-
The form of the particular solution depends on gfo). Some examples are given in J 7.
Using an appropriate form fory part the derivatives y'^, y"part etc.
are found and substituted in the ODE. By comparison of the
coefficients, the unknowns av and p can be determined (cf.
,
example on J 7).
*)
Cv C2
.
.
.
C n are arbitrary constants
DIFFERENTIAL EQUATIONS
J7
Linear Differential Equations
Linear n xh order ODE with constant coefficients
for q(x)
Form of y par =
t
A
J72
xm
73
J
J74
a
a + a x + a^x 2 + ... + a^ "
aQ + a x + c^x 2 + ... + a^c m
1
A Q + A X + A2X 2 +
X
x
+ A^"
1
...
J
75
A-e**
a-e^
J
76
a cos mx + /? sin mx
J
77
A cos mx
B sin mx
A cos mx + D sin mx
A cosh mx
B sinh mx
A cosh mx + B sinh mx
a cosh mx + ft sinh mx
A e^ cos mx
B e^ sin mx
A e^ cos mx + B e** s\n mx
a e^ cos mx + /te^ sin mx
+
+
]78
J79
J80
J81
J82
J83
]
84
j
85
j
86
j
87
j
88
+
+
+
+
Example: y"-y = cos 2x;
y = e
according to form of J 6,
=re n
y'
rjc
;
65 let:
j
y"=r2 e rx
;
Substitution in the ODE of example
85 gives
j
=0; r2 = 1; r x = 1; r2 = -1
= C, ei* + C2 &>* = C e* + C2 er2_i
y hom
x
%
Form of:
j
89
j
90
i
91
/part
sin 2x
= a cos 2x +
= ~ 2a s\n2x + 2B cos 2x
/part
= -4a cos 2x - AB s\n2x
ft
y parX
j
j
j
92
j
93
j
94
.
Equations 89 and 91 used in ODE
-5a cos 2x - 5/3 sin 2x = cos 2x.
Comparison of terms yields:
ft
=
;
a =
(line
j
85) yield
—| and therefore y pan = ~j cos 2x
General solution:
x
= Vl +
^Jv part.= C,1 e1 + C,2 e~
Jv
/horn
j
2x
-tcos
5
+y sin 2**2
x
e x - C2
e~ x
y" = Cj
e x + C2
e~ x + j
4
cos 2x
y" -y = Cj
e x + C2
e~ x + -j
4
cos 2x - C
Check: y' =
C
-C e- + -i-cos 2x = cos 2x
2
x
ex -
DIFFERENTIAL EQUATIONS
J8
Reduction of order
Reduction of order by variable-substitution
to solve an n th order ODE
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1
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v
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W ill 52 n|
,_
C Q 0)
oQ E w
V £
UJ
« o)
"Ew f.E
o3
<D .52
f
Z2t
f
821
f
STATICS
K
General terms
General
theory of equilibrium and with the
of external
forces acting on stationary solid
bodies (e.g. support reactions). The contents of page K1
K14
are applicable only to forces acting in one plane.
deals
determination
Statics
with
the
.
.
.
The most important quantities of statics and their units
Length /
Is a base quantity, see preface.
Units: m; cm; km.
Force F (see explanation on M 1)
Being a vector a force is defined by
its magnitude, direction, and point of
application.
Gravitational force G
Definition: force of earth's attraction
point of application; centre of grav. S
line
of action: vertical
line
intersec-
ting centre of gravity,
direction:
downwards (towards
perpendicular
earth's centre)
magnitude:
determined by spring
balance.
k_*
Support reaction FA
Force applied to body by support A.
Resultant force FR
Force representing the total action of
T*
T
several external forces.
Moment Mot a force F about a point
The perpendicular distance from point
to the
line
of
action of force
F is
called lever arm /.
k
2
and F2 form a pair of forces. The
moment may be represented by a vector.
F2 = -F,
F, = F
k
3
Moment:
F-i
;
M = ± Fl
Moment theorem: The moment of the resultant force is equal to
the sum of the moments of the individual forces.
STATICS
K
Forces composition
Graphical composition of forces
diagram of forces
force polygon
^1/
A number of forces and a common point of application
F3
k 5
k 6
'
link polygon
r n beam
JF
w -3
pole beam
link
A number of forces and any random point of application
link polygon^
^
link beam
^sf
pole beam
Construction of the link polygon
Draw force polygon and determine pole O so as to avoid any
link
rays
running
parallel.
Draw pole
polygon
rays.
Construct
link
such that link rays run parallel to corresponding
pole rays. Thereby each point of intersection in the link
polygon corresponds to a triangle in the force polygon (e.g.
triangle F-i-1-2 of force polygon corresponds to point of intersection F-i-1-2 of link polygon).
STATICS
K
Forces composition
Mathematical composition of forces
Resolution of a force
k
8
k
9
F cos a
Fv = F
+ F
tan a
sin a
Fx
signs of trigonometrical functions
of a see table k 16 to k 19)
F
(for
Moment M
of a force about a point
±F-l
M,
k 10
X
Fu
=
r*y
(for signs of trigonometrical functions
of a see table k 16 to k 19)
Resultant force FR of any random given forces
k
11
components
FRx = XFX
k
12
magnitude
k
13
FR = + y FRx 2 + FRy 2
FRv
direc-
angle of
k
14
T^
sign of
FR / R =
distance
k 15
—
tana R =
I
,
/
*
;
sin a R
r~\P^
[
=
F
Fr
—
|FR
(moment theorem)
|
sign of
k 16
.
k 17
k
aR
a,
.
90.
180 ..
270 ..
.
18
k 19
FRx FRy
,
x
,
a
,
/
,
y
aR
/
R
:
:
:
:
.
;
R
ZM
Signs of trigonometrical functions of x, y; Fx
Quadrant
Frx
cos»a R = -£=
FrR
Rvv1
^oi—
=
R
*)!*&
FRy = IF
cos a
,
Fy
;
FRx>
F,Ry
Rx r
tan a x. Fv
,
,
F
c
90
c
.
180
270°
.
360°
.
components of FR parallel to *-axis and y-axis
coordinates of F
angles Fand F R
distances of Fand FR from reference point
>. *>
FRy
STATICS
K
Equilibrium
A body
Conditions of equilibrium
when both
said to be in equilibrium
force and the sum of the moments of
any random point are equal to zero.
k
20
k
21
k
22
is
forces
graphical
with common point
of application
closed force
parallel to
force polygon
polygon
vertical axis
and link
arbitrary
polygon closed
the resultant
external forces about
all
mathematical
lF =0;lFy =0
x
IFx =0;IFy =0; ZM =
Simply supported beam with point loads W and W2
Find reactions R A and R B
Graph, solution
lution
:
:
:
:
Ra
1
1
s
R*
IV2
/
-•—
M b (z) = v* m H -m F
L
k 23
k 24
k
25
mL
mF
H
:
:
:
kN m, N cm, N mm
-
scale of length = true length/diagram length
scale of forces = force/diagram length
y* vertical distance between closing
pole spacing
line 5 and link polygon.
:
|
Mathem. solution: R A = w^/l + W2 2 ll\
l
R B = (W, + W2
)
- RA
Distributed loads are divided into small sections and considered as corresponding point forces acting through the centres
of mass of the sections.
Wall mounted crane (3 forces): Find reactions
problem
F,
STATICS
K
Lattice girder
Mathematical determination of member loads
(Ritter method - Method of Sections)
F,
O
upper boom
U
lower boom
diagonal
D
member
Determine the support reactions from K 4 (girder on two supDraw a line X
X through the framework to the bar in
question, but intersecting no more than 3 bars. Take tensile
forces as positive, so that compressive forces are negative.
ports).
.
.
.
Establish the equation of
moments ZM =
and internal forces taken
section of two unknown forces.
ternal
about
with
moments of ex-
the
point
Rule for moment signs
of
inter-
»
Where turning moment is counter-clockwise the sign is positive.
Where turning moment is clockwise the sign is negative.
Example (from the above girder)
problem
:
to find force F^ 2 in bar U2
solution
Draw a line X
and D 2 meet
.
:
.
.
.
X through
at
C,
2
this
is
section selected so that the
- D 2 - U 2 Since the lines () 2
.
the
point
relative
moment
2
of
inter-
and
D 2 may equal
F2
c(FA -Ftf
zero.
Proceed as follows:
k
26
+ a-F U2 + b
ZMq
F2 -c(FA - F,)
b
F[j2
+
STATICS
K6
Lattice girder
Graphical determination of forces in members
(Cremona method
- Bow diagramm)
k
27
Basic principles
*
Each
bar is confined by two adjacent joints.
forces only act through the joints.
The external
Procedure
Establish a scale of forces and determine the support reactions. Since each force polygon must not contain more than
two unknown forces, start at joint A. Establish identical
order of forces (cw or ccw) for all joints(e. g. FA - F^ - FS - FS2 )i
Joint A: Force polygon
a-b-c-d-a. Keep a record
of forces being tensile or compressional.
Joint C: Force polygon d-c-e-f-d.
etc.
Check
joint
in
the framework form
Forces acting through a single point
ces form a triangle in the framework.
in
the diagram of for-
Forces acting through a single
a polygon in the Bow diagram.
STATICS
K
Centre of gravity
Arc of circle
k
28
30
y
y
k 31
>>
=
29
k
r^_s
r
at
b
2a
=
r
at
2 a
=
r
at
2a
=
n a
=
=
k
180°
sin a
r
y
0.6366
0.9003
0.9549
c
180
c
90
c
60
Triangle
k 31
5 is the point of intersection
of the medians
Sector of a circle
2r sing 180°
k
32
y
k
33
y
y
y
..
3-n-a
=
=
=
0.4244
0.6002
0.6366
2r_s
3b
at
2a
2a
=
=
at
2 a
=
r
at
r
r
c
180
c
90
c
60
Trapezium
k
36
a + 2b
a + b
A
y
3
Sector of an annulus
k
37
R 3_ r 3 sin a
R 2 -r 2 area
R 3 -r 3 s
R 2 -r 2 b
2
y
3
2
k 38
3
Segment of a circle
k
39
s
y
12
3
A
fo r area A
see B 3
For determination of centre of gravity 5. see also
I
14
STATICS
K8
Centre of gravity
Determination of center of gravity
of any random surface area
Graphical solution
Subdivide the total area A into partial areas A-\, A 2
A n the
centers of gravity of which are known. The size of each partial
area is represented as a force applied to the centre of
area of each partial area. Use the force polygon (see K 2) to
determine the mean forces /1r x and A^ y operating in any two di.
rections
(preferably
The
angles).
right
at
the lines of application
sition of the centre of area A.
section
of
point
.
.
of
inter-
the
po-
areas
A-\,
indicate
will
y
,
11
W
f
*
# V.-J.—J-_.- h
fi_
h vr
!
t
A3
*2
|
i
'S 3
st
A R*
X
.-,
.
P
1
3^-s
1
a
2
1
*fi
Mathematical solution
Subdivide the above
A2
.
.
area
A
'
V-
into
partial
A n we now get
;
di-
stance
in
general
in
the above example
A,
x
A^
>'1
:
+
+
A 2 x 2 + A 3 x3
A
A2
>'2
+
^3
•
^3
A
>'s
Note:
total
i
a
In
the above example the distances *i, y 2 and y 3 each
equal zero.
STATICS
K9
Friction
Force acting parallel to a sliding plane
static friction
k
43
limiting friction
=
u
sliding friction
u =
u
>
k 44
k
45
k
46
k
47
< p-i (variable)<p
A
FZ
force
c
increasing
gradually
i
from zero is compensated by an increasing FW1 without causing the
body to move, until Fz-\ reaches the
rest
motion
c
o
value
rzo =
k 48
O u
o
.
As soon as this happens, the body
whereby Fz drops
starts
sliding,
to Gu. Any excessive force will
'-=
Fw
sliding
friction
now accelerate the body.
for oe acting
Force applied obliquely
The force F needed to set in motion a body,
k 49
weight
G {a > g
F = G
Vo
):
~
a- [1q cos a
sin
sin Q
sin(a-po)
The force needed to maintain the motion is
ascertained by replacing // by //. No motion
possible when result of F is negative (a < p
).
^Wl ^v
Fz,
<
^zo
friction force
•
Fz
'
traction force
-, fi
p-i
,
p
,
pi:
,
p:
friction coeff.
(
see
angle of friction I Z 7
I
1
STATICS
K 10
Friction
Inclined plane
General
The angle a
at
which a body
will
move easily down an
clined plane is the angle of friction p.
tan a
k.50
=
tan g =
n
Application in the experimental determination of the angle of friction
or the friction coefficient:
H
k 51
=
^T*
base
tan g
(horizontal)
Condition of automatic locking:
a < g
Friction properties
constant velocity maintained by
tractive force F parallel to
base
inclined plane
i
^
motion
upwards
k
^Sf^
/s\
< a < a*
sfv\\
g\
sin(a+p)
52
G T
F = G tan(a + p)
cosp
V
^5>^
downwards
^5r
< a< g
k
*//?<?
*/< /
g < a < a*
A.
F
54
=
Titling angle } of body
/ /^/.
/
v
^L
G
i
sin(a- p)
-
r-
'
F = G tan(a-p)
G- cosp
Note: For static friction replace pi by ,«
a*:
F = G tan(p-a)
cosp
downwards
G
k
sin(p-cr)
f " G
53
in-
A
*
md p by
/<
-
STATICS
K 11
Friction
/
1
v\'v\\\vvw
tan (g-i + g-\ ) + tan (g 2 + gj)
driving
k 55
in
loos-
k 56
k 57
ening
F2 m /7
F
- tan g 3 x tan (a 2 + g 2 )
1
tan(a 1 - g : ) + tan(a 2 - g 2 )
1 + tan
g 3 * tan (a 2 - g 2 )
automatic
a,
locking
= F tan(a + 2p)
1
F2 = Ftan(a-2p)
^ p 0l + ^q 2
+ a2
a ^ 2p
Screws
U M M n^ ¥ ¥
h
k 58
turning
raising
59
moment
when
lowering
k
k 60
conditions of
automatic locking
when lowering
k 61
efficiency
raised
M-,
=
when
1
=
2
a < p'
a < p
"
"
tana
ta r\(a + g)
»?
=
f
=
tan (a - g)
lowered
'
k 63
(tan a =
r-^)
64
angle of friction
(tan g =
fi)
k 65
angle of friction
(tan p' =
cos/3/2
:
mean radius of thread
tana
N m, [kgf m]
N m, [kgf m]
lowering moment
lead of thread
r
tana
tan(a + p']
tan(a-p')
tana
raising moment
k
Fr tan (a +p')
M = F r tan(a-p) M2 - Fr tan(a-p')
of screw
k 62
Fr tan (a +p) M
STATICS
K 12
Friction
Bearing friction
longitudinal bearing
radial bearing
P^§K
k
66
A/ R
t*q
:
V.
IfeJl'
r2
M R = H q -r-F
Mr - ft
^
U—
F
moment of friction
h
friction of a
|
bearing (not
constant values)
radial
coefficient of
I
^l
EB V^
V/7/7
/
longitudinal
|
Note: ,u q and fi\ are determined experimentally as a function of
67
k 68
k
bearing condition, bearing clearance, and lubrication. For run
~ ti L = ^q- Always use r-\ >
to allow for luin condition: (a
brication.
Rolling resistance
Rolling of a cylinder
k 69
k
70
F-£fn~£g
Fw < // FN
Rolling condit.:
F\fj\
/
:
force of rolling resistance
arm
lever
of
rolling
resi-
stance - value on Z 7 (caused by deformation of cylinder and support)
fi
:
coefficient of static friction between cylinder, and support
Displacement of a plate supported by cylinders
k 71
F
)G 1
= (/i^/2
sh
+nf2 G 2
27
fl
J
k
72
k
73
where /1 = /2 = / and n G 2 < G!
1;
\\v
M
IB
Gi
F
/t
r
,
:
G2
weight of plate and one cylinder
:
:
lever arms of force of rolling resistance
radius of cylinder
I
n
:
F
*
Q£srf
<
tractive force
and/2
:
~"1
number of cylinders
STATICS
K 13
Friction
Rope friction
traction force an d friction force for
lowering load
raising load
k 75
F,
= e»
FR
k 76
s
-G
F2 =
a
-1)G
FR = (1-e»a )G
= (e"
e^ 5 G
•
Formulae apply where cylinder is stationary and rope is moving at
constant velocity, or where rope is stationary and cylinder is
rotating at constant angular velocity.
k 77
Condition of equilibrium:
(F
:
F2 < F < F,
G e _t s < F < G e 5
*
*
•
•
|
1
force without friction)
Belt drive
k 78
k
79
k
80
wound-up^
k 81
k
82
Fy
FR
:
:
tangential force of driving wheel
frictional force of rope
Ma driving torque
:
a
(x
k
:
:
83
v
e
:
angle of contact (radians). Always introduce lowest value
into formula
friction
(value of experience for
ciefficient
sliding
of
leather belt on steel drum: \i = 0.22 + 0.012 v s/m
belt velocity
= 2.718 281 83
.
.
.
(base of natural logs)
STATICS
K 14
Rope operated machines
Rope operated machines
The following figures deal solely with rope rigidity, disregarding
bearing friction.
unknown
fixed
free
quantity
sheave
sheave
'/////.
1
pulley block
ordinary
differential
'////s
'/////,
ff£)
QQ
m>
m
GO
go
k
84
£
F^
•
e
G
1
+ e
1
+ e
n
Li
(£-1)
£
n
-1
e + 1
-U--1
k
F
85
k 86
iH°
i-«
k
87
2-h
k
88
mechanical advantage
n-h
force
effort
F-,
F
F
k 89
1+e \e
-1
:
:
:
£ =
f_
"
G
force required to raise load, disregarding bearing friction
force required to lower load, disregarding bearing friction
force, disregarding both rope rigidity and bearing friction
—
'
:
loss factor for rope rigidity (for wire ropes
and chains « 1.05)
efficiency
s
number of sheaves
h
:
:
path of load
path of force
KINEMATICS
General terms
General
Kinematics deals with motions of bodies as a function of time.
The most important quantities of kinematics
and their units
Length
see K 1
Units: m; km
/,
Rotational angle
Unit: rad
,
Time t
Is a base quantity, see preface.
Units: s; min; h
Frequency/
The frequency of a harmonic or sinusoidal oscillation is the
ratio of the number of periods (full cycles) and the corresponding
time.
number of oscillations
_
J
corresponding time
Units: Hz (Hertz) = 1/s = cycle/s; 1/min
Period T
The period T is the time required for one
reciprocal of the frequency/.
I
r
2
%
full
cycle.
It
is
the
"7
Units: s; min; h
Rotational speed n
Where an
coupled with the rotation of a
corresponds exactly to
one full cycle of the oscillation, the rotational speed n of
the shaft is equal to the frequency /of the oscillation.
shaft,
oscillation
is
tightly
and one revolution
of the shaft
n min
"
Units:
-
/:
/
-
lor
revolutions/second (1/s)
rev. /min. r.p.m. (1/min)
continued on L 2
KINEMATICS
L 2
General terms
continued from L 1
Velocity v
The
velocity v is the
respect to the time t:
=
v
4
1
derivative of the distance s with
first
ds
—
=
dt
s
Where the velocity is constant, the following relation applies:
5
1
-
•
i
Units: m/s; km/h
Angular velocity to, angular frequency to
The angular
velocity
to
the
is
first
derivative
of
the
angle
turned through q, with respect to the time t:
1
6
co
=
dw
—
-
w
*
=
dt
Hence, for constant angular velocity:
1
7
ao
=
SL
t
Where/= n (see
1
3), the angular velocity to is equal to the angular
frequency to.
1
8
=
to
2jzf
=
2nn
=
<p
Units: 1/s; rad/s; 1°/s
Acceleration a
The acceleration
a is the
with respect to the time t:
1
dv
—
a
9
derivative
first
=
v
=
dt
Units: m/s
2
;
km/h
lar velocity to with
;
dt
5
o
2
=
velocity
v
s
;
is
the
first
derivative of the angu-
respect to the time t:
dco
a
10
Units: 1/s 2 rad/s 2
d
—
the
2
Angular acceleration a
The angular acceleration a
1
of
2
•
1°/s
=
2
Tt
•
a
d2 w
"
d?
=
"
KINEMATICS
General
Distance, velocity, and acceleration
of mass point in motion
Distance-time curve
An s-t curve is recorded for the motion.
The first derivative of this curve is the
s(?)
instantaneous velocity v.
•
i
%
-
"
11
It
is
-i
ft
slope
the
tangent
the
6t
to
the
s-t curve.
Velocity-time curve
The velocity-time history is expressed as
a v-t curve. The first derivative of this
u(u)
curve is the instantaneous acceleration a.
Hence the acceleration is the second derivative of the distance-time curve.
12
a
8
m
Av
Al
a
=
%>
=
V
df
the
dt
the slope
v-t curve.
It
is
tangent
to
the
The shaded area represents the distance
travelled s (t).
Acceleration-time curve
The acceleration-time history is shown as
(a)
an a-t curve, which enables peak accelerations to be determined.
a
> 0:
acceleration
a
< 0:
retardation
a
=
constant velocity
(increasing velocity)
EF
(decreasing velocity)
0:
Note to diagrams
The
letters
in
see L 4 and L 6).
brackets
apply
to
rotations
(for
explanation
KINEMATICS
The most important kinds of motion
Linear motion
Paths
are
straight
lines.
All
points
a
of
^~Tv*
Vr
/ J
body cover congruent paths.
i
Special linear motions
uniform
uniform accelerated
f
|
u
motion
I
15
v =
v Q = const, a =
constant
a.
Rotational motion
Paths are circles about the axis. Angle
turned through <p, angular velocity to, and
angular acceleration a are identical for
all points of the body.
Special rotational motions
uniform accelerated
uniform
|
motion
I
16
(o
=
= const a = a
co
=
constant
|
Distance
velocity
s,
celeration
a f are
and tangential ac-
v,
proportional
the
to
radius:
I
I
=
v = rco
17
s
18
centripetal acceleration
rep
;
a = r a = at
;
an =
co
2
r
=
pendulum—;-} \
Harmonic oscillation
/•2 i«'
Paths
are
straight
lines
or
circles.
The
body moves back and forth about a position
The maximum deflection from this
of rest.
position is called "amplitude".
Instantaneous
position,
velocity,
and
ac-
celeration are harmonic functions of time.
h
KINEMATICS
Linear motion
Uniform and uniform accelerated
linear motion
uniform
uniform
a =
a =
u= const.
> 0)
< 0)
(a
(a
EU
un >
°
=
H,
accelerated
retarded
constant
n
t^r
m
I
vt
19
2a
2
I
I
I
25
V2^7=^ = ar
20
21
Vq.
+ V) = V Q t +
2
23
t
2^
i>
a
a
j
m/s
cm/s
km/h
- at
2s
t
_ 2s
v
v -Vq
a
-2
=yjv 2
V - Vg _
22
cm
km
2
+ 2 as
u + a t = \Au
v
const.
r
I
+ (v
25
m/s'
cm/h 2
km/h 2
2s
vQ + v
mm
V 2 -V0
2
h
Note
The shaded areas represent the distance s covered during the
time period t.
The tangent ft represents the linear acceleration a.
KINEMATICS
6
Rotational motion
Uniform and uniform accelerated rotation
about a fixed axis
uniform
II
a=0
a =
COq =
co = const.
(a > 0)
accelerated
retarded
constant
uniform
COq
i
(a<0)
w
^^
^T
EU
>
t
3
rnssfZt
t*J
I
f
t
2
I
m
25
f
2
m af_ m co
2
2a
co t
24
-
(2 a.tp
=
2(P
-f
at
-^(co Q +co) = co Q t +
-at 2
rad
COq + at
= y <y 2 + 2 a <p
1/s
m/ms
rad/s
- at = \J(i) 2 -2aq>
const.
26
27
ft)
_
f
"
2^ _ Q) 2
2
t
2cp
1/S
CO
-COq _ CO* -0)(
~
Z
mm/s 2
2<p
rad/s
!£.
min
t
s
28
f
=
2jp
CO -COq
CO
a
O)
+ OJ
h
Note
The shaded areas represent the angle
of rotation cp covered
during a time period t.
(Angle of rotation tp = 2 ji* number of rotations respective
360°* number of rotations).
The tangent /3 represents the angular acceleration a.
2
KINEMATICS
Oscillations
Linear simple harmonic motion
A body supported by a spring performs a linear harmonic oscillaFor this kind of motion, quantities s,
tion.
of time are equal to the projections
s,
v,
and a as functions
and a n of a uniform
v,
rotation of a point.
uniform
simple harmonic motion
rotation
position
29
I
cp = cut+cp
;
b=r{cot+cp
s
)
A -s\n (cot + cp
= Acq -cos(aU + cp
v = r- a)
df
31
)
acceleration-time curve
acceleration
I
)
velocity-time curve
velocity
30
=
2
^7 - -Aco
a= 0; a n =
at
•
s\n(cot
<Po)
Differential equation of simple harmonic motion
2
d 5 =
2
a =
~ (OS
32
6?
s
displacement
angular position at time r =
A amplitude (max. displacement)
angular position at timer
<p
r
a n centripetal acceleration
radius of circle
r
radius vector (origin: centre of circle; head: position of body)
B, C extreme positions of oscillating point
<P
:
:
:
:
:
:
:
:
1
)
KINEMATICS
Free fall, projection (without air resistance)
Free fall and vertical projection
free fall
vertical] upwards
u„ =
project.] downwards
v
v
>
<
+h
height
initial
h'
33/2
h=-g-t 2 /2=-vt/2=-v 2 /{2g) h=v -t-g-t 2 /2=(v + v)-t/2
= -2hlt = y/-2gh v=v -g-t
= \fv 2 -2gh
v=+g-t
33/3
t
133/1
I
I
height
=-2hlv=y/-2hlg
= + v/g
t
=(v -v)/g
= 2h/(v
v 2 /2g;
t hmax
+ v)
Vjg
Angled projection (upwards cp > 0; downwards cp < 0)
= v
1- cos cp
34
s
135/1
h = v -t-s\r\cp -gt 2 /2
I
I
36/2
I
I
I
I
I
=
35/2
136/1
37
38
-
tan cp - g
s
h±
s
•
2
v 2 cos 2 cp)
1 (2
•
v = \[v 2 -2gh'
= \/v 2 + g2 2 -2 g -v
-t
-
H= u
W= v
2
-sin 2 g?/(2g)
2
-s\n2cp/g
v -smcp/g;
39
^I
a
40
eS
45
°
|
?
s\n cp'
w = 2v -smcp/g
^ax = v
2
/g;
t
Hmn -v y/?/[2g);
141/1
H,
= W,- tan cp- g-W, 2 / (2 v 2 -cos 2
141/2
W,
= -v
141/3
tang? =
cos cp (v
•
^-v.yfi/g
t
sin cp ± Vsin 2 cp-v 2 - 2gH\
Wi
(2H, + £J\--*2
V
Vq 2
v 2
g-W,
v
Horizontal projection {cp = 0)
142/1
!
I
v
42/2
-g t 2
42/3
v = y/v 2 + g 2 -t 2
tH :
time for height
8
H
w time for distance W
f
:
v
v
velocity
:
initial
:
trajectory velocity
KINEMATICS
9
Motion on an inclined plane
Sliding motion on an inclined plane
excluding
un-
including
|
known
friction
param.
43
I
a
=
H >
=
li
(sin a -
sin a
g
44
I
at
45
I
a
y
r
cos a)
u
—
sin (a- g)
*
cosg
\j2as
t
at'
46
I
Ml
yr
2
2a
Qo
Rolling motion on an inclined plane
excluding
un-
known
/>0
/=0
sin a - - cos a
flf-r-
r2
I
I
I
r2
48
see above
49
see above
50
0.
.
.
a
51
\ar\a = Li
52
* .
I
I
+ fc 2
45
46
tana
r
r
^a
^
r
r2
max
:
tan a.
solid cylinder
ball
I
r
gr 2
+ k2
2
I
including
|
friction
param.
p
*-£
+ k 2 -f
k2
pipe with low
wall thickness
k2 =
r 1 2 + r2 2
2
^ r2
lever arm of rolling resistance see k 70 and Z 7
-2
acceleration due to gravity (g = 9.81 ms )
radius of gyration (see M2 and M3)
tilting
angle, where center of gravity S vertically above tilting
coefficient of sliding friction (see Z 7)
(see Z 7)
angle of sliding friction (/x = tan g)
angle of static friction (^ = tan g )
coefficient of static friction
I
I
53
54
[edge
KINEMATICS
10
Mechanism
Simple Conn-Rod mechanism
I
I
I
I
56
2
= r(1 - coscp) +
jf rs\n cp
= co r sin cp (1 + A cosg?)
57
=
58
= L
55
a)
2
r (cos cp +
4
/
I
6
"
2nnt
= cot =
59
A cos 2<p)
11
(A is called the crank ratio)
Scotch-Yoke mechanism
I
I
I
I
s
= r sin (cot)
61
co
= corcos(cot)
62
a
= - co 2 r s\n (cot)
co
=
60
63
— v -
2xn
(motion is simple harmonic)
Cardan joint
For uniform drive the oft-drive will be
uniform due to auxiliary shaft H
non-uniform
uniform drive
/-
M
being
located
in
cos 6
tan
T,
fi
-3—
**k
For
shafts
all
one
plane
the
following
tan cp3
= tan <p-|
relations apply:
I
I
64
tan cp 2 = tan cp,
65
co 2
= CO)
—
1
I
66
cp3
cos 8
-sin
2
C0 3
B- slrrcpi
<X2
=
2
= tan cp^
= CO)
sin 2 /?- cos 6- sin
<Ui
(1
co 3
= co-.
2^
-sin^-sin 2 ^) 2
Both axes A of the auxiliary shaft joints must
be parallel to each other.
The more the angle of inclination p increases, the more the max. acceleration a and the accelerating moment M a become; therefore, in
practice p < 45°.
DYNAMICS
M
General terms
1
General
Dynamics deals with the forces acting on bodies in motion and with
the terms "work, energy, and power".
The most important quantities of dynamics
and their units
Mass m (is a base quantity, see preface)
Units: kg; Mg =
g (see also A 2, A 3, A 4, A 5)
t;
mass of the international standard. A mass
measured by means of a steelyard.
kg
1
the
is
is
Force (gravitational force) F
The force F is the product of mass m and acceleration a.
F = ma
The gravitational force G
the force acting on a
is
mass m due
to the earth's acceleration g:
m 2
G
=
mg
Being a gravitational force the weight W is measured by means
of a spring balance.
Units: N;(kgf; Ibf; see also A 2)
N
is
m =
1
1
m s"
of 1
is
the force that, when acting on a body of a mass
kg for 1 s, accelerates this body to a final velocity
1
(i.
e. accelerates this body at 1
the gravitational force acting on a
m s~ 2
).
9.81
mass of
t
N (= 1 kgf)
kg due to
the earth's attraction.
Work W
The mechanical work is the product of force F and distance s,
where the constant force F acts on a body in linear motion in
a direction parallel to the distance 5 covered (W = F s).
Units: N m = Joule = J = W s; (kgf m; ft Ibf; see also A 3, A 5)
Where a force of 1 N acts over a distance of 1 m, it produces the
work (energy) of 1 N m (J).
Power P
The power P is the derivative of work with respect to time.
Where work (energy) increases or decreases linearily with time,
power is the quotient of work and time (P = Wit).
W (Watt); (kgf m s"
see also A 3. A 5)
an energy of 1 J is converted
a constant rate, the corresponding power is 1 W.
Un ts
i
:
1
;
H. P. ;
Where for a period of 1
s
1
W=
1
J/s
at
DYNAMICS
M
Mass, Mass moment of inertia
Definition of the mass moment of inertia
J
The mass moment of inertia of a body about
an axis has been defined as the sum of the
products of mass-elements and the squares
of their distance from the axis.
J =
Ir 2 Am
=
Jr
2
kg m
dm
2
N m s2
,
Steiner's Theorem (Parallel axis theorem) (see also J 9)
Where the mass moment of inertia of a body
of mass m about an axis through its centre
of gravity S-S is 7
the mass moment of
SS
,
about
inertia
a
O-O
axis
parallel
a
at
distance / s will be:
mL
kg m'
Radius of gyration k
The radius of gyration k of a body of mass m and mass moment
of inertia J is the radius of an imaginary cylinder of infinitely
small wall thickness having the same mass and the same mass
moment of inertia as the body in question.
hence
m, cm, mm
Vf
k
Flywheel effect
flywheel effect
G k2
mgk* = g J
(k
2
kg cm s *, N m^
formulae see M 3)
Equivalent mass (for rolling bodies)
kg
Basic formulae
linear moti on
formulae
m
9
Fa'
=
ma
rotational
N
,(kgf)
mot ion
formulae
units
units
N m
A* a = J a
,
(kgf m)
m 13
W Fs(F= const.) N m,(kgf m) W = M (A/ = const.) N m (kgf m)
(kgf m)
J
WK = \mv
J
.(kgfm) WK = \ja>
= Inn
mirr
s"
J
W P = Gh
.(kgfm)
W s (kgf m)
J
w = jFAl
.(kgfm) W = ±MA0
m 14
p
m 10
m 11
m 12
--
=
(p
2
,
2
,
1
(0
f
,
F
'3F-"
W ,kW
P
1
,
=
a t
W
,
kW
DYNAMICS
M
Mass, Mass moment of inertia
Mass moment of inertia about
axis b-b
passing through
centre of gravity 5
axis a-a
(turning axi
type of body
circular hoop
m 15
=
J
\mr 2
b\
2
m 16
.2
_ 1-2
2
cylinder
m 17
m 18
m 19
k
J
2
m 20
:
m 21
J
m 22
k
=
\m{R 2 + r 2
-
^(R 2 + r 2
2
)
hollow cylinder
k
)
w mr
= ^r
10
2
2
2
m 23
V
=
— {3R + 3r +
-
^™(Ar 2 + h
2
S
k
2
=
2
m(R 2 + |r 2
=
R 2 + ^-r 2
J
m 26
k
m 27
short
2
bar
thin
m 28
bar
y«*V
12
y =
+ 4/-
=
J
)
k
2
m 4/? + 5r<
= -l(4/? 2 + 5r 2 )
^/ 2
3
d,c<zl
*.--;!•
k
2
-
2
)
b
\r 2
-§«* + «*)
;
)
sphere
2
]
2
2
2
=
m 25
h
(4ff + *' )
=\mr
J
m 24
^(3r 2 + h 2
)
2
=
fcW + c
2
)
torus
b
UJ'-J
short bar, thin bar
DYNAMICS
M4
Rotation
Total kinetic energy of a body
W K = \mv* + \j co
m 29
2
s
J,
[kgfm]
J,
[kgfm]
Kinetic energy of a rolling body - - no sliding
m 30
m 31
W« = j (m + m red
vs
=
)
vs
2
m/s, km/h
(o r
Rotational torque
m 32
M m, [kgf m]
2 ;rn
co
Transmission ratios
Transmission ratio
33
driving
^2
z2
"1
Q)-\
d\
Z-i
rt
Q> 2
2
Q
Torque ratio
M
moment of force
moment of load
m 34
F
Ml
—
it)
/
h
6%
J
Efficiency
v
work produced
work applied
m 35
output
<>yy
driven
input
Overall efficiency for a series of transmissions
36
V
™red
=
V-\
•
V2
•
V3
•
•
•
see m 8
velocity of linear motion of centre of gravity
accelerating force
accelerating moment
kinetic energy
potential energy
energy of helical spring under tension
extension of helical spring
angular defection of spiral spring (in radians)
N, [kgf m]
N m, [kgf m]
J, [kgfm]
J, [kgf m]
J, [kgf m]
—
DYNAMICS
Ms
Centrifugal force
Centrifugal force
Fz
2
Fz = ma) 2 r
m 37
m 38
m 39
v
=
=
4n 2 mn 2 r
=
2jzrn
^f-
Ni
[
kg f]
N, [kgf]
tfr^L
\^
-J
m/s, km/h
r
i
m 40
=
(0
2n n
1/s,
\
1/min
v
J
Stresses in rotation bodies (appr formulae)
Disc
(JO
°z
'
2 r2
Q
3
V2 Q
F*~
3
N/m
2
,
[kgf/cm
2
]
v
„
Ring
m 42
a2 = —
^ (r^ + f!^ +
N/m
/s
:
e
:
Fz
:
Jq
:
J$
'
M^
:
2
,
2
F
z_
)
[kgf/cm
2
]
m, cm, mm
m, cm, mm
m, cm, mm
distance from centre of gravity
maximum pendulum swing
:
/
r2
fTV<
^ l^
Wj
instantaneous pendulum swing
centrifugal force
N, [kgf, gf]
kg m 2 [kgf m s 2
2
2
kg m
[kgf m s
mass moment of inertia about
mass moment of inertia about S
moment required to deflect spiral spring
by 1 rad = 57.3°
oz
T
^e
:
'
:
i>f-
Wke
]
N m, [kgf cm]
,
period of oscillation
{B to 5' and back)
velocity at E
,
s,
]
min
m/s, cm/s, km/h
velocity at F
kinetic energy at
]
,
N/m 2 kgf/cm 2 [kgf/mm 2
tensile stress
:
,
m/s, cm/s, km/h
£
M m, [kgf m]
I
DYNAMICS
M
Harmonic oscillations
6
Mechanical oscillation
General
m 43
period
m 44
stiffness
T= 2jz\jf
N/m, [kgf/m
Al
m 45
frequency
/ =
m 46
angular velocity
a>
=
—
(see L 1) s
2nf = V—
\ min
s"\ min
Critical speed n c of shaft
m 47
2n V
m 48
= 300
m
10cg mr
V 9.81 N
2-bearing shaft, load
symmetrical
asymmetrical
^££S
iSfl
overhung
(cantilever) shaft
,X- i m
^cz>X.
c/
/
48EI
m 49
3EI
2
a -b
A
/
m
I
:
:
:
2
:
3EI
3
/
deflection or elongation
second moment of area of shaft cross section
mass. When calculating the critical speed the mass m
(e. g. of a belt disc) is assumed to be concentrated at a
single point. The mass of the shaft should be allowed for by
a slight increase
c„
c Cq
~
stiffness for transverse oscillations
DYNAMICS
M
Harmonic oscillations
Pendulum
(Explanations see L 4)
Conical pendulum
m 50
m 51
T-2„\Jf -
2^'^ia
tan a
9
m52
-if
Simple pendulum
The arm of a pendulum has no mass, the
total mass is represented as a point.
m 53
= 2
s,
'>IF
m 54
Ve
m 55
WKE = mg 2/
-e'tff
vF =
2
Vf(e -/
J,
mm
m/s
2
)
km/h
N m, [kgf cm]
Compound pe ndulu m
m 56
s,
m 58
min
[kgf cm s
2
Nms ,[kgfcms
7s=G/s(-^-^)
91
4 Jt
2
N m s
m 57
2
,
2
]
]
\
body with centre of gravity S is
suspended from O, distant /s from 5, and
the period of swing determined the mass
If
a
moment of inertia about O can be calculated using
m 58.
J
Torsional pendulum
m 59
s,
For explanation of symbols see M 5
min
ft-?
DYNAMICS
M
Impact
8
Impact
m 60
When two bodies of mass m-\ and ni2 and velocities t'n and U21 collide,
the total momentum p = m v will remain constant over the whole
impact period (velocities become 1^2 and L'22):
p =
m 61
m-|
•
v-|i
+
m2
•
m-|
t>2i
•
m2
u 12 +
•
v22
Impact-direction
direct and
concentric
impact
velocities parallel
to normal to surfaces
at point of impact
oblique and
concentric
impact
any random
normal to surfaces
at point of impact
through centre of
gravity of both
bodies
any random normal
velocities
oblique and
excentric
impact
to surfaces at
point of impact
Types of impact
plastic impact
elastic impact^
relative
equal, before and
equals zero
velocity
after impact
after impact
m 62
velocity after
direct and
m 63
concentric
impact
m 64
Vn(mi -m2) + 2m 2
V21
_w
rri\
=
£
coeff. of restitution
1
+m 2 y2
w + m2
v 11
02
U2i(m2 -mi) + 2mi
»
i
1
+ WI2
£
1
Coefficient of restitution e
This indicates by what factor the relative velocities
=
will
vary
before (171) and after (172) impact:
m 65
E
=
tV2
here
«r1
normal to surface
/at point of impact
For an oblique, concentric, elastic
impact the velocity vector v is
split into a normal and a tangential
component. The normal component 17,
produces a direct impact (see above),
the tangential
component
has no effect on the impact.
0^ e^
'
v
t
1
HYDRAULICS
N
General
General
Hydraulics deals with the behaviour of liquids. With good approximation, liquids may be considered incompressible, which means
that the influence of pressure changes on their densities is
negligibly small.
Quantities
Pressure p
see O 1
Density p
see O 1
Dynamic viscosity
(Values see Z 5)
>/
The dynamic viscosity is a material constant, which
tion of pressure and temperature:
n
=
is
a func-
f(p,t)
The dependence on pressure can often be neglected. Hence
n 2
7]
=
(for figures see Z 14)
f(t)
EU: m 2/s = ( 1
Kinematic viscosity v
4
St) = ( 1
6 cSt)
The kinematic viscosity is the quotient of dynamic viscosity rj
and density g:
n 3
Hydrostatics
Pressure distribution in a fluid
n 4
P\
=
Po + gQh A
n 5
Pi
=
Pi +
=
p^ +
ggjh
— PiPz— /
'Pz-Py
t
-p<r\
gQ{h 2 -h^)
gQAh
—
-c"
_
-c
-
N
P
pressure
^
zL
ta
•c
gradient
1
h
'
continued on N 2
~
HYDRAULICS
N
Hydrostatics
Hydrostatic forces on plane surfaces
Hydrostatic
component
force is the
acting on the
surface which is caused by
the weight of the fluid
alone, i.e. without taking
into account the atmospheric pressure p
.
n 6
F
r v
„
Q-C _
~a^jr?
fc*
~~/
2~ J : _— /
= S G >'s A cos a = S Q h i A
•D
_
\h
yQ^
r>
y**
Hydrostatic forces on curved surfaces
The hydrostatic force acting on the curved surface
1,
2,
resolved into the
is
horizontal component FH and
the vertical component Fy.
equal to the weight
in (a) or the
equivalent
weight
of
fluid (b), above the surface 1,2. The line of action
runs through the centre of
Fy
is
of the fluid
gravity of the volume V.
gQ
n 8
Fh
is
N,
kN
equal to the hydrostatic pressure force acting on the
projection of the considered surface 1.2 on the plane perpendicular to FH
-
Calculation is accomplished by n 6 and n 7.
centre of gravity of area A
centre of pressure = point of action of force F
second moment of area A in relation to axis x
second moment
parallel
to axis
of area A in relation to an axis running
17
x through the centre of gravity (see
I
and P 10)
centrifugal
(see
I
17).
moment of area A
in
relation to
axes x and y
HYDRAULICS
N
Hydrostatics
Buoyancy
The buoyancy FA is equal to the weight
of the displaced fluids of densities
JTp
—
g
and o'.
FA =
gQV + gQ'V
kN-=~-—
N,
——
the fluid of density p' is a gas, the
following formula is valid:
If
n
FA ~ggV
10
1? >~p
kN
N,
With g k being the density of the body,
Q > Qk
Q = Qk
Q < QK
n 11
n 12
n 13
the body will float
'
"
"
"
remain suspended } in the heavier
fluid
sink
Determination of density g of solid and liquid bodies
Solid body of
smaller
greater
density than the
fluid
For
fluids
first
a fluid of known density g b
This yields:
used
1-
n 14
n 15
determine
F-i
and m of a deliberate body in
g= g
Q= Qf
n 16
.
mg
mg
77#»7,
..V
Q
Qb
m
F
resp.
"^^
M
,1
^_—
:
mass of the body remaining suspended in the fluid
:
equilibrium force necessary
^h: equilibrium force necessary
auxiliary body alone
Qf- density of the fluid used
in
the preliminary
trial
for the
:
HYDRAULICS
N
Hydrodynamics
Hydrodynamics
(of a steady flow)
Continuity equation
Rule of conservation of mass:
Ai V-\ Q<\ = A v g =
n 17
n
A 2 v2 £>2
kg
m = Vg
18
g
Rule of conservation of volume:
n
m° cnr
Hr,^
S
S
v = Av
1!
o
'
14
<j
Bernoulli's equation (Rule of conservation of energy)
No friction (ideal fluid):
)
£ + gzi + -£- = £ + gz +
n 20
-7-
^ + gz + - 2
=
2
pressure energy
per unit mass
potential energy
per unit mass
kinetic energy
per unit mass
datum line
Including losses (real fluid):
?+™
n 21
v
:
vy
"g" "
Pi +
p"
V
3*2 +
'
2
+ WR1.2
-y
velocity
w r-\,2
'•
resistance losses along path from
(Calculation see N 6)
1
to 2 (per unit
mass
HYDRAULICS
N
Hydrodynamics
Power P of an hydraulic machine
P
n 22
=
m w t1,2
kW,
W
technical work per unit mass:
J_
n 23
2
+ 9(^2-^1) + ^-(^2 -^i
7r(P2-pO
2
9 *" '" «"-«-"
2
)
n 24
for hydraulic machines:
u> t12
<
n 25
for pumps:
u? t12
>
+ wri,2
J/kg
Momentum equation
For a fluid flowing through a stationary reference volume the
following vector equation is valid:
IF
n 26
IF
=
m(yl - u|)
N, kN
the vector sum of the forces acting on the fluid contained in the reference volume. These can be:
is
volume forces (e. g. weight)
pressure forces
friction forces.
v2
is the exit velocity of the fluid leaving the reference
volume
is the entrance velocity of the fluid entering the reference volume.
t>i
Angular-momentum equation
M
In a steady state rotational flow a torque
is exerted on the fluid
flowing through the reference volume, given by:
M
n 27
f2iU and u 1u
m(v 2
v^,u^r^)
N m
are the circumferential components of exit velo-
city out of and entrance velocity into the reference volume.
r2
and
r-,
are the radii associated with v 2 and v v
HYDRAULICS
N6
Hydrodynamics
Friction losses in pipe flow
n 28
n 29
Friction loss per
hence
unit mass
Ap y
Pressure loss
=
p-tu R1i2
Determination of coefficient of resistance £ and coefficient of
shape a (Re: Reynolds' number):
n 32
*±*
=
Re
n 30
n 31
^
non circular pipes
circular pipes
=
Re
Where Re < 2000, the flow is laminar.
Where Re > 3000, the flow is turbulent.
Where Re = 2000 ... 3000, the flow can be either laminar or turbulent.
Flow
laminar
64
n 33
a =
for fittings,
1
f(Re, j~)
Re
a =
for straight pipes
n 35
turbulent*'
64
f(Re,*)
£ -
Re
n 34
Flow
laminar
turbulent*'
—
for straight pipes
unions and valves
Determination of coefficient q
n 36
n 37
For annular cross sections:
Dld\
1
<P
1.50
I
I
1.40
1.42
30
50
70
100
1.32
1.29
1.27
1.25
0.5
0.6
097
94
1.0
0.7
0.8
0.92 0.90 0.89
1.00
For rectangular cross sections:
a/b\
(p
|
0.1
0.2 0.3
1.50|l.34|1.20|1 1
I
I
0.4
I
1
1.02
I
d
internal diameter of pipes
length of pipe
/
d h = 4 All]
hydraulic diameter
A cross section perpendicular to fluid flow
U wetted circumference
kid and k/d u
relative roughness
k
mean roughness (see Z 9)
rj
dynamic viscosity (see N 1, values see Z 14)
*)£ is taken from diagram Z 8
:
n 38
10
3
1.47
:
:
:
:
:
:
:
HYDRAULICS
N
Hydrodynamics
Flow of liquids from containers
Base apertures
n 39
v
=
Cv yj2gH
40
V
=
C d A yj2gH
n
-2\m
Small lateral apertures
n 41
v
=
C^2gH
42
s
=
2 \[~H~h
n
/'
(ohne jegliche Reibwerte)
n 43
V
=
C d A\2gH
n 44
F
=
g Vv
&
Large lateral apertures
v
n 45
=
fc d 6v^(^-//^)
T
1
5^^
Excess pressure on surface of liquid
n 46
v
=
Cy \J2{gH+^L)
V
=
C 6 A\J2{gH+^)
Excess pressure applied to an outlet point
v
n 48
=
C
! Pex
T'
-<nnr
~-~r,
?%
^-'-T'
n 49
u
:
outlet velocity
m/s, km/h
2
/? ex :
Cd
Cc
Cv
b
F
V
:
:
:
:
:
:
2
N/m kgf/cm
pressure in excess of atmosphere pressure
(Cd = Cc x C v )
discharge coeff.
contraction coeff.
(Cc = 0.62 for sharp edge aperture)
(Cc = 0.97 for well rounded aperture)
,
velocity coeff. (for water C v = 0.97)
width of aperture
reaction force
volume of outlet flow
m, cm
N, kgf
m 3 /s, m 3 /h
HEAT
Thermal variables of state
Thermal variables of state are pressure p, temperature t, and
density g or volume per unit mass (specific volume), respectively.
EU: N/m 2 = Pa;
Pressure p
1
bar
Pa= 1 N/m2 =10- 5 bar =(7.5 x 10-3 torr)
Pressure is the quotient of force F and area A:
o
>-f
1
The absolute pressure can be interpreted as the total result of
the impacts of the molecules on the wall. The pressure measured
with a pressure gauge is the pressure differential Ap
in relation to the ambient pressure p u A state of pressure means
Ap > 0, vacuum means zip < 0. Thus, the absolute pressure/? can
.
be expressed by:
o 2
P
=
Pu +
Ap
Temperature T, t (Base quantity) see explanations at front of book.
The unit of temperature T is the Kelvin K, defined by equation
-^
1K
o 3
273.15
where T Tr is the temperature of pure water at the triple point.
addition to the Kelvin scale the centigrade scale is also
used. The temperature of this scale has been internationally
*
defined by:
In
O 4
t
Density g
=
(-? - 273.15
K
Vc
;
)
(£+
273.15) K
VC
EU: kg/m 3
Density is the quotient of mass
o 5
T =
'
(Values see Z 6)
°
-
m and volume V:
v
Volume per unit mass (specific volume) v
EU: m 3/kg
Specific volume is the quotient of the volume V and the mass m:
„
o 6
,
z,±Q
m
EU: m 3/mol
Molecular volume V m
Molecular volume is the quotient of volume Vand number of moles
contained in the volume:
o 7
Amount of substance n (Base quantity) see explanations at front of
book.
HEAT
Heating of solid and liquid bodies
Heating of solid and liquid bodies
Heat (thermal energy) Q
EU: J
Heat is energy exchanged between systems of different temperatures, where these systems interact with each other through
diathermal walls.
Heat per unit mass q
EU: J/kg
The heat per unit mass is the quotient of heat Q and mass m:
m
^
Specific heat r p
EU: J/(kg K)
The specif, heat r p denotes the amount of heat Q to be supplied
to or extracted from a substance of mass m to change its temperature by a difference At:
cp
The
specific
see Z 1
.
.
.
heat
is
=
-VmAt
a function
=
±
At
of
temperature.
For figures
Z 5.
Latent heats per unit mass /
EU: J/kg - (values see Z 10)
The supply or extraction of latent heat causes a bo%dy to change
its state without changing its temperature. The following latent
heats exist:
o 10
solid body of the fusing
fusion
temperature into a fluid
fluid of the boiling
vapourisation
the
heat
is
temper-
ature (dependent on pressure) into dry saturated
vapour
necessary
to
convert a
o 12
solid body of a temperature
below its triple temper-
sublima-
ature at the sublimation
tion
temperature (dependent on
pressure) directly into
dry saturated vapour
of the
same
temperature
.
HEAT
Heating of solid and liquid bodies
Expansion of solid bodies
A solid body changes its dimensions due to temperature changes.
With a being the coefficient of linear expansion (for figures
see Z 1 1) the following formulae are valid for:
o 13
o 15
I,
/
/
2
A2 ~
Area:
AA
o 16
o 17
»
Al =
Length:
o 14
Volume
AV
-/i «/i a(t 2 - *i)
2
A, [l+2a(^-^)]
A 2 -A, = A^2a{t2 -ty)
=
V
Vo~
o 18
[l+a(f2 -M]
1+3afe-t|)]
m K,
V, «
V,3a(t
3a (fj,-^)
y2 - Vi
2
=
'
><r^^
4
^''
1
^^
^-^
^
Expansion of liquid bodies
With y being the coefficient of volume expansion (for figures
see Z 1 1) the following formulae apply:
V2 -
o 19
AV
[1 +Y(h-t,)]
V2 -Vi - JVy(*2 -fi)
V,
=•
Mixture of liquids and liquids or/and solids
When several substances with masses m u m 2 m 3
the
corresponding temperatures ty t2 f3
and the specific
.,
heats cp1 cp2 c p3
are mixed - under the condition that
no heat is transferred outwards, no heat is fed from outside
and without change in the state of aggregation - the resulting
temperature t m will be (if necessary the quantity of heat of
the mixing tank has to be taken into consideration):
,
,
,
.
,
^ + m 2 c p2 t 2 + m 3 c p3 f3 +
m, c p1 + m 2 c p2 + m 3 Cp3 +
Z 5).
_ mi c p i
•
o 20
.
,
'
m =
•
•
(For c p values see Z 1
Bending due to heat A
.
.
•
•
•
•
•
•
.
.
.
.
.
Bimetallic strips are subject to bending due to heat. Bending
occurs towards the side of the metal with the lower coefficient
of expansion. With a b being the "specific thermal bending" the
bending due to heat can be calculating by (a b approx. 14 x 10" 6 /K,
for exact values see manufacturers
catalogues):
a b L 2 At
i
o 21
length at
length at
t
volume at
volume at
t
thickness
t
t
=
=
=
=
r-i
t
2
t^
t
2
area at
area at
t
t
= ^
= t2
temperature prior to heating
temperature after heating
temperature difference
HEAT
Thermodynamic states of gases and vapours
General equation of state of ideal gases
The state of ideal gas is determined by two thermal variables
of state. Thus, the third variable can be calculated using the
general equation of state. With R being the characteristic gas
constant (different values for different gases,
see Z12) the
equation reads as follows:
p v =
22
RT
mRT
or
p V =
is
related to the
the gas constant
reads
If
or
p = g RT
mole volume, the equation
o 23
where R m = 8314.3 J/(kmol K) is the universal gas constant
(valid for all ideal gases). R and /? m are related by
o 24
Rm
=
MR
where M is the molecular mass (see Z 12).
Thermal state of real (non ideal) gases and vapours
The thermal state of real gases and vapours is calculated using
special equations or diagrams.
Changes of state
Changes of state are caused by interactions of the system with
the surroundings. These interactions are calculated using the
1st and the 2nd law:
1st law for
open systems
closed systems
o 25
o 26
91.2 + «>1.2 " U 2
#1,2 +
2nd law for
all systems
w \\.2 = ^2~ ^1 + Ae
O 27
In
these formulae, energy input
is
positive (i.e.
91,2,
w\, 2. ^t 1. 2)
and energy output negative.
\s
entropy per unit mass
enthalpy per unit mass
internal energy per unit mass
w 2 external work done per unit mass (see = 7)
7)
to,'
continous external work done per unit mass (see
Ae changes in kinetic or potential energies
h
:
u
:
:
'
:
:
':
]
HEAT
Changes of state of gases
Changes of state of ideal gases
The table on page 6 shows the relations for different changes
of state,
which have been developed from formulae o 25 to o 27.
The following explanations apply:
Each change of state, may be represented by an equation
o 28
P v
n
=
const.
The various exponents // are given in column 1.
c pm and cvm are the mean specific heats for constant pressure and constant volume, respectively, in the temperature
range between
and t 2 There, the following relations apply
(values for c pm see Z 13):
Mi
|<2
_
Cp m
*2
t
pm
«2
t-\
o 29
-
-pm
o 30
o 31
=
The change
given by:
= c
is
o 32
7r
of entropy occurring
'1
,
pm
ln
[f-
during the change of state
,.
,
T2\
.
^{P2\
KM?
n
_ „
.
In
.
T2 \
,
+
,
,„fV2
RD In
T\
Changes of state of real gases and vapours
The table below shows the relations for different changes of
state,
which have been developed from formulae o 25 to o 27.
The thermal
variables of state, p,
v,
erties, u, h, s are generally taken from
change
external work
of state
const.
quantity
tf
-i
2
continuous
heat per
external work
unit mass
w i,2 = \v dp
dv
= - \p
T as well as the propappropriate diagrams.
«2 " "1
isochoric,
o 33
v = const
o 34
p = const.
?1.2
t
v(p z ~P^
[h 2
=
" *i) -
V{p 2 "Pi)
isobaric,
p{v, - v 2
h 2 - h^
)
=
isothermal, (u 2 -uj- T(s 2 ^s,)
o 35
o 36
-h,)-T(s 2 -s,)-
T = const.
(h 2
isentropic,
-(p 2 v 2 -p, Ui)
«2""1
5
= const.
{h 2
-h^-{p 2 v 2 -p^ v
(h 2
-h,)-T(s 2 -s,)
h 2 -A 1
: )
T{s 2 -Si)
-
HEAT
06
Ideal gases in open and closed systems
T-s-
^
diagram
CO
c
> o~
1
° °
\.
E
1
o
o °
~ Q.
I?
—
co
,r| /I
CO
Ci,
~ CD
^=
than
D)
1
to
to
;i:
T3
Q.
-«
isothermal
steeper
CD
1
c-T
per
mass
trans-
3
sS
fer
E-?
o
<?1,2
7
unit
heat
E
E
J*
*l c
el
E
1
cj
^ Yl '—
E ^JX i^3
f <£[£"
dp
v
/
system,
O
=
reversible
2
3
s as
-
^r
'
*
*
open
t1
II
w
-SJ3
'cm
7
^
•Ii
Hi
1
as
f~>
|T-
«a
l>~
ii
ii
Tic
f~r
f
2
system,
C- *'
1*
|
1
o
reversible
«. as
closed
'^m7~^'
&;
ft;
as
i
ii
3
and
1
2
II
II
£|^
S"|s E^|c4" f^l^
between
ii
ii
ii
relation
state
31 S
SI ci
state
£1 £
ii
37)
38)
(0
(o
const.
H
ii
II
39)
a s|s
o
(o
O
<Sl|o~
<S1|
ii
Sl| £X
41)
(0
const.
const.
const.
exponent
1
process
oo
details,
isothermal
isochoric
process
isobaric
=
=
f n
=
=
pn
=
=
S
T n
CO
"
=
ii
^ ~
ii
S| ^ sl^"
n
HEAT
Changes of state of gases and vapour
p-v diagram
For reversible processes the area be-
tween the curve of the variation of state
and the t'-axis represents the external
work per unit mass, the area between the
curve and the p-axis represents the
continuous external work per unit mass.
'////.
^ J^
n
iy
12
«>M,2
.2
111
T-s diagram
For reversible processes the area be-
tween the curve and the s-axis represents
the heat transfer per unit mass.
Total transfer of heat
The heat added to or removed from a closed system during a single
variation of state is:
o 42
01.2
-
™?1,2
J
The heat flow continuously added to or removed from an open
system is:
o 43
#1,2
=
01,2
=
mq^ 2
w
where m is the mass flow (EU: kg/s).
Total transfer of work
The external work added to or done by a closed system during a
single variation of state is:
Wi
o 44
mu) 1(2
The external power continuously added to or done by an open
system is given by:
o 45
?1,
'«1.2
W
1 1
HEAT
Os
Mixtures of gases
Mass m of a mixture of components m-\, ni2,
.
.
i
m
46
=
m
+
y
m2 +
.
.
- n
+ ra n = 21 m,
.
i
-
Mass fractions §j of a mixture
47
6
=
and
m
j?"*i -
1
Number of moles n of a mixture of components n-\, n 2
i
48
/i
=
n<t
+ n2
+
.
.
.
.
•
•
•
= n
+ n n = ^rtj
i
-
Mole fractions ip\ of a mixture
'" n
n
Vi =
o 49
X^i
and
7^
=
1
Equivalent molecular mass A/ of a mixture
For the molecular mass the following formulae apply:
"i-7?
50
M = g
and
where the equivalent molecular mass ;V/ of the mixture can be
calculated as follows:
M
o 51
=
i^Mi-Vi)
i=1
and
-j-
M
=
'i?(^)
i- 1 \A/|/
Conversion between mass- and mole-fractions
52
=
*'
M *'
Pressure p of the mixture and partial pressures p\ of the
components
53
p
=
2
p,
where
p
y
=
*p,
•
p
continued on page
9
HEAT
9
Mixtures of gases
Continuation of page O 8
Volume fractions n of a mixture
o 54
and
Vi
JE r,
Here, by partial volume V\ we mean the volume the component
would occupy at the temperature T and the total pressure p of
the mixture. For ideal gases the following formulae apply:
m.R.T
n-,R m T
P
P
o 55
^
and
Vi
V
=
Internal energy u and enthalpy h of a mixture
n
n
o 56
u
=
i: (£rui)
1
=
h
;
= 1.
i: (£r/zi)
1
=
1
Using these formulae, the temperature of the mixture can be
determined, for real gases and vapours by using diagrams, and
for ideal gases as follows:
internal
,
o 57
c Vm1 ./ 1
m 1+ cVm2 -/2 m 2 +
.
+ c
energy
v
o 58
enthalpy
CP m 71
i
»mn
-/
n
™n
m
m +C Pm2^2m 2 +
cD
m
1
.
t
Wn'«n
•
where the
specific
heats
of
the
mixture are determined
follows:
o 59
o 60
^.(*r«bJ
as
1
HEAT
10
Heat transmission
Due to the temperature difference between two points heat flows
from the point of higher temperature towards the point of lower
temperature. The following kinds of heat transmission must be
distinguished:
Conduction
in
in
plane wall:
the wall a pipe:
<P =
Q = XA
<P =
Q = XA
^w2
^w1
S
*w1 ~ *w2
The mean logarithmic area is:
A m = ji d m L
o 63
where
;
plane wall
pipe
d a -d,
d.
length of
the pipe
lnl
f.
Convection
By heat convection we mean the heat transfer in a fluid. Due to
their flow the molecules as carriers of the mass are also the
carriers of the heat.
Where the flow originates by itself, the convection is called natural convection. The
convection taking place in a flow is called
forced convection.
= Q = <xA(t-t w )
o 64
Radiation
This kind of heat transmission does not require mass as a carrier
(e.g. the radiation of the sun through space). For calculations
formula o 64 is used.
Heat transfer
By heat transfer we mean the combined
result of the different
processes con-
tributing to the heat transmission:
o 65
=
Q = kA(t,-t2
)
The
heat transfer coefficient k is
given by (for approx. values see Z 11):
1
plane wall:
o 66
k
i
-
1
o 67
pipe:
A
«
:
:
k A
«1 ^1
thermal conductivity (for values see Z 1
.
.
.
Z 5)
heat transfer coefficient (for calculation see O 12)
HEAT
11
Heat transmission
Heat exchanger
A heat exchanger transmits heat from one fluid to another. The
heat flow may be calculated by:
=
Q = k-A-At m
Here, At m is the logarithmic mean temperature difference. The
following formula applies for both parallel-flow and counter-flow
heat exchangers:
o 68
^ 'smalU
o 69
counterflow
parallel-flow
In counterflow operation Atg rea \ and At S ma\\< can occur on the
opposite ends of the exchanger, to that shown in the figure.
Symbols used on page O 12:
enveloped surface
C-,
(A,
enveloping surface J
inside diameter of pipe
outer diameter of pipe
,C 2
radiation constants
< A2
Nu
H
of
:
the
surfaces
C s = 5.67*1(T W/(m K
:
At =
V
V
A
y
exchanging rasee Z 1 2)
(for values
8
Pr
height of plate
length of plate
L
diation
o 70
Grashof number
Nusselt' number
Gr
)
2
4
)
:
radiation
constant of the black body
Prandtl-Number Pr =(r/c p )/A (for values see Z 14)
absolute temperature difference between wal
\t
and fluid in the thermally not affected region
ambient temperature (7^ = t^ + 273.15°)
'
:
|
kinematic viscosity (v =
rj
/g)
(for values see Z 14)
dynamic viscosity
dynamic viscosity at mean temperature of fluid
dynamic viscosity at wall temperature
(for values see Z 5, Z 6)
thermal conductivity of fluid
(see Z 11 and o 77)
volume expansion coefficient
temperature factor
|
v:
velocity
HEAT
12
Heat transmission
Calculation of heat transfer coefficient a "
For free convection (according to Grigull)
o 71
Nu =
on a
o 72
NuX
Nu
vertical
0.55
\JGrPr
,
for
1700 < GrPr < 10 8
= 0.13
$GrPr
,
for
GrPr> 10 8
gyAtm _ g Y At q? //3
plate
Gr
o 73
o 74
on a
A/u
horizon-
o 75
o 76
o 77
o 78
= 0.41
YCr-Pr'
GrPr< 10 5
for
,
WuA
tal
D
3
plate
Fluid properties must be related to reference temp.:
The expansion coefficient of gases is:
tB
y
=
fc
SL+
x
A/V
-
= MTc.
For forced convection inside pipes (according to Hausen)
a = NuX/d
0.0668 RePr
o 79
laminar
Nu =
3.65
Re Pr
+ 0.045
1
dY'2
/?<?<2320
10
o 80
o 81
> RePr^>
10
U
4
,
where
Re =
^4^
7/
turbulent
Re>2Z2Q
if
2320 < /?<? < 10 6
1
0.6 < Pr < 500;
;
< L/d< oo
With the exception of ry w all material values are related to the
mean temperature of the fluid.
For gases factor (r/ F i/r;w)
14 must
De omitted.
For radiation (heat transfer coefficient: orstr)
o 82
«Str
o 84
o 85
=
0*^1,2
c
parallel
o 83
between
enveloping
sur-
4
7i -72
faces
h-h
i,
q
4
Ci,
J_ + Al /J_ _ J_\
Q
')a in J/(m 2 s K) or W/(m 2 K)
For explanation of symbols see O 11
>4 2
JC 2 Cj
STRENGTH
General terms
Stress
Stress is the ratio of applied force F and cross section A.
Tensile and compressive stresses occur at right angles to the cross
section.
N/mm 2
o or f
p1
In
stresses are
positive
usually
negative
tensile
calculations
compressive
|
Shear stresses act along to the cross section
tot,
P2
N/mm'
-jf
Stress-strain diagrams (tensile test)
Materials with
yield point
plastic yield
(e.g. mild steel)
(e.g. aluminium alloy)
a
#m -
TW^
1/
*eH
"
RP
-
Jf
i
'
/
/
r
/
/
'
'
'
/
A
'
'
eP
€
'
'
/
€p
'
*
'
'
'
A
e
Notation The standard symbols are from BS 18 and DIN 50145.
:
P3
R m = tt-; \o B = —jo
F
A
L
:
tensile stress, where
J
tensile force
:
s o> [A
]
:
original cross section (of unloaded specimen)
Al
p4
1
Lo [lo]
AL; [Al]
;
:
:
00 %
strain,
where
original length (of unloaded specimen)
change in length of loaded specimen
continued on P 2
STRENGTH
General terms
continued from P 1 (stress-strain diagram)
ReH> [^Sol
Proof stress or yield strength (offset)
:
The limit of proportionality is sometimes
known as the elastic limit.
ep = 0.01
% =>
/?
p0-01
[rj
p]
;
Yield point (ferrous metals)
R eH> [o"so]
R eL
[a Su
;
]
:
:
upper yield stress or
upper yield point
lower yield stress or
lower yield point.
Proof stress (non-ferrous metals)
ep = 0.2
R ™ " 1^
P5
4L
p6
I
% => R pQ2
°b " ^T*
[o 02
;
]
tensile strength
I
100 %; fa -
percentage elonga-
100%
&!
tion after fracture.
For specimens with circular cross sections, percentage
elongations may be quoted, based on gauge lengths,
e. g.
A 5; [65]
is
based
,
on a gauge length of
5
y
^ mm
*
Permissible stress (allowable stress)
Must be below the elastic limit R n thus
p
the permissible stress is:
Rm
v
:
:
'
p
x
=
Km
—
yield strength of material
safety factor, always greater than 1.
Ultimate safety factor
against fracture)
Proof safety factor
(against yield or 0.2 proof)
Loads
type
load diagram
nature of stress
a
I
dead
-
s
—^
t
a
II
undulating
—^——
*"
III
alternating
\
l
*-f
STRENGTH
Tension, Compression
Modulus of elasticity E: The relationship between a and e (Hooke's
applicable to the elastic range, i.e. below the elastic
16/17 for values of £). £ is known as "Young's
law)
is
limit
(see Z
modulus ".
P 7
o = E
•
E Al
e =
Olr
E ~ e ~ ^l
Tensile and compressive stresses at and o7Cc
iF,
f/
Fx
P 8
^U
^n
Oc
Pi
Strain e under tension
P 9
I- h
Al
=
e
°1 =
E
Ft
"i±
E A
$
Compressive strain f c under compression
Fc
p10
£
/
E
fi
E A = tensile or com•
A
pressive stiffness
Transverse contraction under tension (Poisson s ratio)
For circular cross section
—"«
where e along
P 11
along
"
/- /o
_
and e
dp- d
UQ
dn
/<,
For most metals Poisson's ratio can be assumed to be \i = 0.3.
Thermal stresses: Tensile or compressive stress is caused by restricting thermal expansion (see also o 13/14):
P12
ath = E
At
is
% - E-a-At
(£, h
= a At)
the temperature difference between the unstressed
original state and the state considered.
At >0 tensile stress, positive
At <
compressive stress
For prestressed members subjected to thermal stress the total
strain comprises:
p13
«tot
-
+
£ei
fith
- FI(E-A) + a-4f
;
ee
=
,
FI(E-A)
Tensile and compressive stresses in thin-wall cylinders
(boiler formula):
o = p d I (2 s)
o = + p, 4 / (2 s) | vg|jd
o = -p a d a /(2s) )
Hoop stress
p14
p15
Tensile stress
Compressive stress
internal and external pressures
Pi and p a
inside and outside diameters
d and d a
:
:
x
wall thickness
s = 0.5{d a - d )
Tensile stresses in rotating bodies: see M 5.
:
t
da
d
x
^ 1 2
STRENGTH
Tension, Compression
Tensile stress in a shrunk-on ring (approximate formulae)
Shrunk-on ring on a rotating shaft:
The shrinkage force Fh of the ring
must be at least twice the centrip-
Fu
j_
petal force Fq.
p 16
Fh
Fc
P 17
m-y s
=
-
4
a)
2
^ 2 Fc
=
4n 2 m-y s -n'
•
R 3 - r3
R2 -
p 18
>'s
p 19
Cross section
p 20
Shrinkage allowance A =
3ji
A =
_Fh_
2-Pt
(A
1
n
= outside diameter of shaft - inside diameter of ring)
Shrunk-on ring for clamping
Split, rotating
Clamped parts
clamped parts.
Fc comprises:
Centripetal force FC r for ring
Centripetal force FCM for clamped parts,
Ring
or
P 21
FH ^ 2(FCR + FCM
)
then as p 19 and p 20
;
Energy of deformation U (Strain energy)
The energy stored in a deformed component is:
U = w- V where
p 22
;
-i
p 23
o- e
i-
2£
V:
volume of component
Limit cross section for similar types of stress
Where a tension (or compression) force is applied at a point
within the dotted core area, only tension (or compression)
forces will occur over the whole cross section. If applied at
any other point, bending stress, i.e. simultaneous tension
and compression stress will occur.
-SW]
p 24
5
Dm
:
:
centre of mass of half ring (see K 7)
mean diameter (D m = R + r)
STRENGTH
Loads in beams
Explanation
external loads on a beam (including support reactions and
own weight) produce internal forces and moments which stress
the material. By taking a section through the beam at a point x
is possible to show the internal
loads: Vertical shear forces V and bending moments M.
All
its
it
End loads P and torsion Tare considered separately.
[I
y
B
r ^f=i«^ttt:
(f
P
Referring to the x-y plane (z axis is at right angles):
end loads
P
r-axis
shear forces
bending moments
M
x-axis
torsion
T
Forces in
.v-axis
direction of
v-axis
Moments
about the
produce
V
Always consider the left-hand side of the section.
In
each part of the beam there must be equilibrium between all exmoments:
Considered separaterly:
ternal and internal forces and
V +
p 25/26
M
p 27/28
^L V;
hi.
=0
P +
T +
Method of calculation
1.
2.
Calculate the reactions.
Section the beam at the following places:
2.1 Points of action of point loads
and beginning and end
of distributed loads w.
2.2 Points where the beam axis changes direction or the cross
section changes.
2.3 Any other convenient places.
W
continued on P 6
STRENGTH
Loads in beams
continued from P 5
3.
Find the forces and moments on the
section as in p 25 ... p 28
4.
Plot shear force and
Relations between h
m
p 29/30
,
_
hand side
of the
bending moment diagrams.
V and M at any point x
dV
v
dx
left
dx
Rules:
M
is
In
sections with no loads
a maximum when
V =
V =
constant
Example: Simply supported beam with end load. (Fixed at A)
The reactions are:
/?
A = 2.5kN;
Pa = 3 kN;
R B = 1.5 kN
*V=2kN
r
——j—3kN
/>,»
m
w=
1
kN / m = const.
XL
3m
•|iv|
Shear force V in kN
3.0
End load P in kN
Calculations see P 7
!
1
STRENGTH
D
IT 7
Loads in beams
continued from P 6
c)
<x <
1
from
equation p
27
26
29
m
1
from
equation p
.
25
27
4 m < x < 6 m
m <x<4m
26
!
29
from
equation p ...
27
26; 29
.
25
*
H
25
1
11 E|
z.
*
"?
"*
=
=
kN
kN
0.5
0.5
Z
z
in
to
H
Z
c
o
o
to
c
o
o
z
z
cvj
in
in
=
-*
co
o
1
il
"
ii
a?
Z
J*
|
CO
1
ft?
ft?
II
II
*
^
i
II
:
i
*
i
a?
II
*H
+
Z
E
ft,
i
z
ii
a.
i
-o|-3
o
to
d
'
ii
5
^
ii
+
H
1
CSJ
H
.
,
in
+
E
^
II e
»-
d
11
+
in
Z
E"
*~
in
,*
+
^
+
ft?
ft?
^
^
aight line
+
ft,
i
z
.*
CO
* !S
1
1
if)
ii
!
I
i
z
+
^i E
°
z
ii
E
z
"
i"
I
s
i
8?
1
qT
ft,
_*
7
o
II
+
ftT
'
i
i
k
ii
o
o
CO
ftT
+
a?
i
7'°
H
H
"
1
H
k
'
CM
5
i
o
II
^
I
,—
o
«q
i
in
O
E
H
CO
e"
ii
*
|
jl ;II e
CM
Z
|
cvj
E
CNJ
E
1
in
i-
!
ii
-^
io
ii
-*
ii
cvj
CNJ
z
i
CVJ
i
cvj
i
o
i
11
I
i
Til
o
ii
ii
Z
II
in
d
i
he: 7
=
1
«£
i
const.
const.
E
i
'
+
+
i
o
ft,
5
ai
^
o
a.
**' Parabola
con inued on P8
—
STRENGTH
Analysis of forces
continued from P 7
Example:
Curved cantilever beam
(r = const.)
The limits are:
o ^ y ^ 90°
o^j^rj
or:
p 31
p 32
Bending moment:
+F| -r (1 -cos qp) + F2 r-sin cp =
M = — F-| r + F-| r cos cp - F2 r sin cp
M
•
•
•
At the section q>, F
•
•
and F2
:
are resolved into tangential
and radial components.
Shear force (radial):
p 33
Fq + F-|
p 34
Fq = - F-\
p 35
p 36
Fq
sin cp + F2
•
•
sin cp -
dM =
= -
—
6s
cos <p =
F2 cos cp
•
— -6M
1
r
;
or from p 30:
,.
(because s = r
•
6<p
1
d(-Fi-r+Fi •/••cosy-^-r-sincp)
r
6<p
•
cp
^
F-,
;
j =
6s
r
j
x
dcp)
^'
sin^-Z^cosg?
Normal force (tangential):
p 37
Fn - F-i
•
cos g? + F2
•
sin
<p
p 38
Fn - jF,
•
cos <p - F2
•
sin
<p
For a graphical method of determining bending moment see K 4
STRENGTH
9
Bending
Maximum bending stress
p 39
A*->W
_
u btmax _
,
^
'xx
M
p 40
= Pb
Values for p b see Z 16/17
>W (tension)
distance from surface fibre to the x-axis
passing through the centroid S or neutral
(compression)
7 XX
neutral axis
Second moment of area about S z or about plane of neutral
:
axis.
Bending stress at distance y from the neutral axis
p 41
=
°\>
M
Yy
Section modulus Z mm
p 42
/_
= -i^max
Z
Second moments of area
Axial second moment of area see
I
Polar second moment of area see
Product moment
I
see
I
17 and table P 10
17
17
Principal second moments of area and principal axes
The
principal
second moments
of
area
h = Anax and h = Anm are applicable
to asymmetric sections, when the
principal
axes are rotated through
the angle cp
p 43
*1
2
p 44
Anax
-.mm
FH
A
^
^^
-
.
-/x )2 + 4 /xy 2
x )±|y(/
o U'y
*
y
2/xv
tan 2<p
\
7/
For calculations of 7xy see
117/18.
The principal axes are always perpendicular to each other.
The axis of symmetry of a symmetrical section is one principal
= 7X
axis e.g.
I-\
.
STRENGTH
10
Bending
Values of / and Z for some common sections
(see p 41 and p 42)
For position of centroid 5 (or neutral axis) see K 7
Z x and Z
and / y
/x
b-h 3
P 45
Zx
12
h
p 46
b
-ft
ft
Cross section
v
h~ *.—
2
%
6
h-b 2
3
»>
12
p 47
n-d'
64
/v«/«
/x = /'y = £(£> 4
y
64
p 48
-d 4
Jt
,
p 51
p 52
p 53
/x
= K-a-b
-a
3
•
•
3
Z, =
4
jt
p 54
b
7
^•y
p 56
24
6 2 -ft
-h
24
48
p 57
a+6
36
h
3
p 58
ft
p 59
2
6-ft 2
36
b3
fc
4
2
= jt-a Z>
4
b-h 3
p 55
10
0.1203- j 3
0.6250 R 3
0.1042 -s 3
0.5413 R 3
= 0.06014 -s A
= 0.5412
RA
y
d3
32
P 4 -rf 4 _, fl 4 -<i 4
10£>
D
p 49
/
K-d 3
z x = zy
)
32
p 50
M
2a+b
a + b
h2
12
(a+b) 2 +2ab
2a+&
a+2b
centroid axis
Steiner's theorem
(Parallel axis
theorem for second
moments of area).
p 60
I
J B -b
=
/x +
^-a 2
1
I
STRENGTH
11
Deflection of beams in bending
Beams of uniform cross section
Equation of the elastic curve
The following
^S.
apply to each section of
the beam (see P 5, Method
9TV
Y
i\
of calculation, Item 2):
p 61
d2 y
El
^M
_ ±
R
EI-y" = -M
p 62
£./^=
EI-V = -[Mdx + C,
dX
p 63
Ely = -JftJ M 6x dx + C, x + C
p 64
•
R
p 65
_M_
„
y
dx 2
y'
:
•
2
radius of curvature of the elastic curve at point x.
= tan q>
inclination of the tangent to the elastic curve
:
at point x.
v
= deflection of beam at point x.
C, and C2 are constants of integration and are deter-
mined from known factors,
e. g.
v
Vj
y'
v'j
=
at the support.
= y j+ i at junction between sections / and (/ + 1).
=
at the support of a cantilever beam and at
the centre of a beam with symmetrical loading,
= y' + 1 at the junction between sections i and (/
•+
i
Strain energy due to bending U:
For a beam of length I:
p 66
U = ~2
JE-J 6x
A beam with discontinuous loads may be
divided into n length:
p 67
"m-JMlIt*
STRENGTH
12
Deflection of beams in bending
Tote.
+ co
IK
*_
p 68
cnjIco CCCNJIcM
ICO
ii
6
-
^ -O ^ -C
+
— !•© —I
+
cc
+
*'*
+
,
p 69
*'IcN
>lcO >lcO
co
CD
or
^.
3
|
^N
CM
col"""
co
II
II
II
y-
C\J
Ico
Ico
II
II
r-
C\J
O<
O
CD
££
p 70
c o £ £
§2
<
^
-C|^ CCS
i
P 71
CD
o *T
o
.
„
s:
ii
«S
-o
l
<
ll
*?
II
CM
I
%jTih
Q-O
II
I
£
£ ^ ^
OS
££
1
JW
II
<
00
a:
az
i«
i"
£
-* ^
5 5
* *
STRENGTH
P13
Deflection of beams in bending
«5
P 72
Uj
~
£ I«
"
OcM
*
|S
II
^r
3
EH
E
2^
>lco
%\^,
CVJ
tfc
CO
+
p 73
CO
2
co
II
II
°xl^
CD
ICO
E »
p 74
E.E
5 2
-ICVJ
SjCNJ
+
p 75
II
Q-o
3 CD
CO
<D
t0|— v^H,
*
coioo
n
*
*
mioo t-ico ^ ^ £ ^
ii
<
cd
as os
*4m -
<
tn
^ ^
.
STRENGTH
14
Deflection of beams in bending
Mohr's analogy
Graphical method
1. Determine the bending
polygon (see also K 4).
moment curve by constructing a link
Original beam
Position diagram
1
2.
Fig.
Force diagram
1
Construct a link polygon as the equivalent distributed load w*
on the "equivalent beam". Another link polygon will give the
tangents to the elastic curve.
Equivalent beam
Equivalent force diagram
Fig. 2
Deflection of the original beam at point x:
p 76
y
=
/»'
HH* m
EI
...
m A m L3
w .
F
•
•
Slope at support A and B:
tan yA = Rff-pr—;
P 77
m F m A m L2
•
resp. tan <p B
.
H
EI
rrip •
mA
•
rri\_
.
Mathematical method
1
p 78
Calculate the equivalent support reaction /? A * of the
"equivalent beam" carrying the equivalent distributed load
w* = A-i +... + A n (see fig. 2).
2. Calculate
p 79
the equivalent bending
;
(see fig.
1
+2)
Distance between center of gravity of equivalent distributed load A and the section x.
Slope y = V*/E-I
3. Deflection
y = M*IEI\
xA
p 80
moment M* and the
equivalent shear force V* at the point x:
V* = R,
M* = /? A*-2-A-z A
:
continued on P 15
STRENGTH
Pl5
Deflection of beams in bending
continued from P 14 (Mohr's analogy)
Choice of equivalent beam
The supports of the equivalent beam must be such that its maximum equivalent bending moment A/* max coincides with the point
of maximum deflection in the original beam.
Simple beam
A
Original beam
Equivalent beam
A
AA
A A
A8
V
A %
%
D
.
Cantilever beam
Beam of varying cross section
Fig. 1
Original beam
e. g. shaft
z
I
Plot the bending moment curve as the equivalent distributed load
w*(z) on a uniform equivalent beam of cross section equal to the
maximum second moment of area /x max of the original beam. (See
P 14, item 1).
Plot w*(z) according to the ratio
/xmax
.
p 81
Fig. 2
Equivalent beam
of the shaft
in Fig.
1
J1
I
Then calculate according to P 14 (items 2 and 3) or p 78 ... p 80.
STRENGTH
P16
Beams of uniform strength
maximum
typical
maximum
section
dimension
x = resp. v =
deflection
dimension
p 83
t bp
bt
SWl
6W-z
6 W( I V
h2 Pbt
h 2 PbX
b-E\h I
'
p 84
8W( l\
bE\hl
6W-z
6W-I
V b-Pbx
aI
p 82
-
yl
3w/
V
ft-Pbt
type of beam
/=
2
\
bl- PbX
lwl( I V
b
2_wl 2
p 85
•
~hTp^
V^p
v Ab Pm
t
3
V / 32
p 87
V
W
•
J_3jWi/
W
p 86
E\h )
3vv / z 2
h 2 -l- Pbi
•
/
^-Pbt
3
,
/
32
*</'
4z^\
1
bt
l
2
/
W z
•
V n 'Pb\
W
point load
w
uniformly distributed load
Pbt
permissible bending stress
/
64-
EI
192_
W-/ 3
5
'End*
kN
kNm
N mm 2 (see Z 17)
*
|
STRENGTH
17
Statically indeterminate beams
Fig. 1
Convert
a
— — — — c /[
I
indeterminate
statically
I
I
I
I
i
|
beam
(fig.
into a statically deter-
1)
minate one (fig. 2) by replacing one
support by its support reaction {R c in
^ A
Divide into two separate beams or sub-
systems. Determine the deflections at
the point of the statically indeter-
/*
F,9- *
J
fig- 2).
V
I
I
I
IcKI
I
^
I.
I
A
Wl
/W
1s
stt subsystem
su bsystem
minate support (see P 11 to P 15) from
each subsystem, in terms of Rq.
Since no deflection
support C.
p 88
>ci
can
occur
at
2nd subsystem
A^
- ycz
Hence, calculate the support force at C, R c and then the remaining
support reactions.
Method of solution for simple statically indeterminate beams
Statically in-
1st
2nd
subsystem
subsystem
Statically
determinate beam determinate beam
w,
—
Rc
]%] w
yC 2
\m
c
i
B
IV,
4
JV2
^r
5&
*»-~±^" e
J/?C
s
2
W;
^2
c
^Xt 4
I
V\\C\
r ^~
2
I
I
I l'T™
fr^- rf
^
l/? A
'
^3
*C ^S
A
l"I)
4
--*^
2
statically indeterminate support reactions and
ffA
g"
rc\
V4
Ma
moments
J±Sn
C|
Mb
STRENGTH
18
Shear
Hooke's law for shear stress
p 89
T =
or
q
G
y
G: shear modulus
y shear strain
:
Relation between shear modulus and modulus of elasticity or
Young's modulus
G = —— £
p 90
- 0.385
•
E
;
2 (1 + A*)
Mean shear stress
p 91
q
f
or
</ f
or r
correspond to P 3 with u = 0.3
f
T,
Permissible shear strength p q (values for R p02 see Z 16)
dead
^P0.2 /1.5
type of load
(see P 2)
undulating
^p0.2 /2.2
alternating
^p0.2 /3.0
Ultimate shear stress q £
_ vmax
92
for ductile metals
A
Shearing force
Q
applied by
guillotine shears
parallel cutting
cross cutting
cutting tools
(punch
1°
1-2 9a
Q - 2 tan
p 93
^\-\\\\\v \\^\
<o
-!°-j
i
G- 1-2 ^/-.v
a
tan a = 5//;
/
< L
Theorem of related shear stresses
The shear stresses on two perpendicular
faces of an element are equal in magnitude, perpendicular to their common edge
and act either towards it or away from it.
p 94
q =
q
q
:
:
q'
transverse shear stress (transverse to
beam axis) resulting from shear forces Q
axial shear stress (parallel to beam
axis)
0.8 /?„
for cast iron:
"complementary shear"
.
.
.)
STRENGTH
19
Shear
Axial shear stress due to shear forces
p 95
1 =
QM "
TT
p 96
A/=
AA-y s
q =
q max
1
for
ab = abmax
occurs when ab = 0,
e. on neutral axis.
.
i.
Max. shear stress for different cross sections
<?max
= q
Ta
4
3
'
d a 2 + d a -dj + dj 2
d 2 + d2
p 97
for thin-wall tubes: 2
(<*a«4)
Strain energy u due to shear
1
p 98
q-y =
2G
Shear deflection of a beam
x -pr—i +
p 99
G -A
C
K^+C
£v4
Determine the constant C from known factors, e.g. v =
at
the
supports.
p100
The factor
»-*/[£
d/l
allows for the form of cross
(A)
section. For examp ie:
H =
1.2
®
I 80
I 240
I 500
1.1
2.4
2.1
2.0
shear force of point x of the beam
bending moment at the section A
second moment of area of the total cross section A about
the z axis
width of section at point y
5 AA
:
centre of area of section A
STRENGTH
20
Torsion
General
T a <
Tfmax
—
•
p 101
p 102
Shear stress due torsion r
t
=
=
T - t
Torque
=
2-Ji-n
CO
power
p«
Fa
Tfmax
distance from surface fibre to centre of mass
constant; formulae see P21 (attention: torsion
constant is not the polar moment of inertia; only for
circular cross section J = /
a = D/2).
p
torsion
;
Bars of circular cross section
Angle of twist (f (see e 5)
p 104
<P
III m 180/
-
!
p
-G
7-/
k
Stepped shafts:
T
p 105
u
"
i
-
l\
1
'pi
"
T
180°
p 106
polar moment of
inertia / n
p 107
32
P 108
32
(D 4 -^ 4 )
Vi"
"
D*-d 4
"
'
Bars of non-circular, solid or thin-wall hollow section
p 109
Angle of twist
<p
Tl
180°
Ix -G
Jt
G
position and
constant J
magnitude of r
1
Tt
:
K
- Ttmax c2-
Tt2 = c3 .Ttmax
rt3 =
in 3:
1
1.5
2
3
0.141
0.196
0.852
0.858
0.229
0.928
0.796
0.263
0.977
0.753
0.675
1.000
cross
section
t
C^h-b 2
in 2:
=
-|
=
c^h-b 3
p 110
C1
•
torsion
in
/i '6
T-l
/t
'
4
0.281
0990
0745
6
0.298
0.997
0.743
CO
0.307
0.999
10
0.312
1.000
0.333
1.000
0743
0743
743
8
continued on P 21
1
1
STRENGTH
21
Torsion
position and
magnitude of
torsion
constant
cross
section
J =
at 1
46.19
m
26
p 113
t
3
h?
at 2:
T t2 =
at
T t1
0.1154-j4
0.0649 d A
p 114
t i
a
p 112
p 115
- r max
20 T ^ 13 T
T
:
a4
p 111
1
:
T tmax
~~
= 5.297
-8.157
p 116
at 1: Tt1
_k_
16
p 117
D 3 -d3
'
=
* 5.1
D^d
T
at 2:
p 120
D/d = DJd = n^ 1
,
r-i
tmax
-
= 5.1
n2 +
16
2
r"t max --^-
-
t
at 1: Tt1
p 118
p 119
'tmax
Dd
2
7-d
n (rf 4 - 4
4
)
at 2: r, 2
with varying wall thickness:
at 1: Tt1 =
p 121
Ttmax
7
p 122
2/1 m 'mm
at
p 123
Tt2 -
with thin, uniform wall:
AA
p 124
r
p /4 m (
*
i
r
2-f
2 A,
"imax
J|n=3
M
[""
1
/4
5
t
s
m
:
:
<
_[_ rp
rectangular
1.0
1.12
tion of
+
ness b max (e. g. po-
n=3
n=2
< 1.3 1.17
sition
sec-
max. thick1
sketch).
in
the
-§*
y?7?7Z
I-JT
area enclosed by the median line
length of median line
wall thickness, (min. wall thickness)
part length of median line when wall thickness t, = const.
(f min ):
:
7;
the
long side h of the
n=2
|
«1.3
up of
rectangular
cross
sections
r
midway along
Foppl's factor:
p 125
profiles built
T-b,
2>i 3 '*i
STRENGTH
22
Buckling
Euler's formula
Applies for elastic instability of struts.
which buckling occurs:
EI
p 126
Minimum
Pe
load
at
r
Pa
lt+i
p 127
L= 2-
p 128
Permissible working load
F = P e /v
p 129
Slenderness ratio
W-'V^
/
/ic
=
L = 0.707-/
/
Limit 1 based on R pO.01
p 130
,m
L = 0.5-/
Limit 2 based on R p02
^lim
-"V^
* V* P
«
Tetmajer formula
Valid in the range
p0.01
— A ~ ^P°- 2
Material of strut fails due to bending and compression
p 131
Px
=
=a - b-k + C A 2 = P'V k
A
Material
a
US:ASTM
GB-Standard
\
b
»
\
c
N/mm 2
valid
for /.
=
BS 970, 050 A20 A283Gr.C 289 0.818
80... 100
BS970,080M30 A 440
589 3.918
60... 100
BS 1452 -220
A 48 A 258 776 12000 0.053 5... 80
timber
30 020
2. .100
beech or oak
38 025
0...100
mild steel
mild steel
cast iron
Calculation method
determine
the
second
moment
of
Euler
for-
area using
mula:
the
caused by buckling
mini-
ffpO.01
-
2
p 132
material failure
First
mum
_ *Vfk
i/
min ~
2
r
.
'
- v p
Pe k
.
Then select a suitable cross
section, e.g. circular tube,
solid rectangle, etc., and
find / and A.
0.01
continued on P 23
STRENGTH
P 23
Buckling
Continued from P 22
vk = 3
5
vk = 4
6
H
s ... 8
vk = 6
.
.
.
.
.
.
|
in
the Tetmajer range
.
,
in
_
—
larae
kfor
for small
,
|
the Euler range
M
1
structures
,,
|
|
|
Present
1
slenderness ratio
A
1
Limiting
A| im0 01
.
|
|
and A| im0 -2
j calculate
from
p 129
p 130
Determine buckling and compressive stress as follows:
If
Pe < F
If
•
A
>
*«„ tun
^MmO.01
>
^
^
<
^Iim0.2
vk
use p 127
^ ^limO.J use p 131
>
use p
8.
redesign with larger cross section.
,
Method of buckling coefficient co (DIN 4114)
Specified for building and bridge construction, steelwork and
cranes.
Buckl in
9 1 CO =
coeffic ientj
&
permissible compressive stress
Pk
buckling stress
Oj^Pc
mild steel
BS
A
T3
050 A 20
CO
mild steel
c
USA U) A 283 Grade C
20
40
60
80
100
120
140
160
180
200
1.04
1.14
1.30
1.55
1.90
2.43
3.31
4.32
5.47
6.75
where co = /(A)
3uckling coefficient io for
BS970 alum, alloy
cast iron
080 M 30 BS L 102
BS 1452-220
ASTMA440
alum, alloy
1.06
1.19
1.03
1.39
1.99
3.36
5.25
1.41
1.79
2.53
3.65
4.96
6.48
8.21
10.31
cast iron
AA2017 ASTM-A48A25B
7.57
10.30
13.45
17.03
21.02
1.05
1.22
1.67
3.50
5.45
////////////
not valid
in this
range
Calculation method:
Estimate co and choose cross section calculate A, 7 mjn and A
from p 134. Then from table read off co. Repeat the calculation
initial
with
the
appropriate
new value
and final values are identical.
of
co,
until
the
1
STRENGTH
24
Combination of stresses
Combination of direct stresses
Bending in two planes with end loads
The stresses a arising
from bending and end loads must be added together.
p 135
FK = F- cos a
p 136
Fy = F cos £
p 137
Fz = F- cos y
p 138
where cos 2 a + cos 2 /3 + cos 2 y
For any point P(x, y) on the cross section Bi B 2 B3 B 4 the resultant
normal stress is in the z-direction:
Fyl
p 139
Fx-l
A,
Note the sign of x and
and y w '" D ©
in
functions see
E 2.
y.
If
Fz is a compressive force,
a,
(3
quadrants. For the sign of cosine
different
Long beams in compression should be examined for buckling.
Neutral axis of o =
p 140
is
the straight line:
y
Fy
x
/
y
Fy A-
which intercepts the axes at:
p 141
Xq
Fz
h
7y
Fx 'Al
With asymmetrical cross section
the principal axes (see P 9).
>'o
F resolve in the directions of
Bending in one axis with end load
Either Fx or F
y
in
formulae p 139 ... p 141
tension
Bending with
is zero
displaces the
compression
neutral axis
compression
towards the
tension
-
,
STRENGTH
25
Combination of stresses
Stress in curved beams (R < 5 h)
The direct force Fn and bending
moment A/ x (see P 8) act at the
most highly stressed cross sec-
a
tion A.
For the stress distribution over
the cross section:
p 142
o
t
=
AR
A
The stresses
the
at
C
R+y
inner
and
—
outer surfaces are:
p 143
p 144
Fn
A/x
.
M
x
Mx-R
Icil
C
R+\e<i\
M R
\e 2
X
AR
*Pt
\
R-\e 2 ^Px
C
\
Formulae for coefficient C
+
1
p 145
C = b-R 3
In
d
2R
d
A.
R
2R
p 146
j
VMi
C = e 2 nR'
:
+
p 147
a-b /
f-(i)
a e + b e2 \
/'
K(a-fc)
•
1
V
fc
1
Position of the center of mass, see K 7
3 +
p 148
3/j
(»*«i)
b_ b-h
R 2R 2
2h
To
STRENGTH
26
Combination of stresses
Combination of shear stresses
The stresses arising from shear and torsion at any cross section
must be added vectorially.
The maximum shear stress rres occurs at point 1 and acts in the
cross sectional plane. A complementary shear acts perpendicular
to it.
Maximum torsional stress r res
r
5.1
1.7/
~
p 149
~d 2
where T =
F|
d*
5.17D
D 4 -d*
,
,
n
_
D + Dd + d
2
T7
' t
f|
PqX
:
2
-\. 7 -d(D + d)
F 4.244 D +
p 152
Pqt
2
D*~^d*
where T =
Pqt
:
4.244
p 150
p 151
cross section
point 1
in
D*-d*
Pqt
For thin wall tubes:
5.1
p 153
TD
2.55-F
D 2 -d 2
2
+ d2
2D
2.55 F-
Pqt
T + 1.5F
<
~ p«
2h^>)
PqX
D A -d A
p 154
d4
££
p 155
c,'b 2~h
~b~h
Pqt
where T
p 156
p q{
t
rq
F
T
permissible shear stress (see Z 16)
:
shear stress
:
:
:
calculated maximum torsional sheat stress
force producing torsion
:
for c-,
torque produced by F
and c 2 see P 20
STRENGTH
27
Combination of stresses
Combination of direct and shear stresses
Material strength values can only be determined for single-axis
stress
to
Therefore,
conditions.
multi-axis
stresses
av
P
single-axis equivalent stresses
(see
are
converted
The following
29).
then applies, according to the type of load:
<7V
^
P\
Pb\
Pc
<
Stresses in two dimensions
An element is subject to
shear stress
direct stress
az in z direction
in y-z
ov in v direction
plane
yz
By rotating the element through the angle <p a the mixed stresses
can be converted to tensile and compressure stresses only, which
are called
Principal stresses
au o 2
p 157
Direction
0.5 ( az +
of
tan 2 <po
highest
± 0.5 l/(az - oy ) 2 + ki 1
principal
stress
a-i
at
angle of
Uorw the original position is:
rotation (p
p 158
the
ov )
2r
=
*>
(where the shear stress is zero).
Oz
Rotating the element through the angle tpT gives the
Maximum shear stresses
p 159
T max ,
T™
=
± 0.5 ]/( oz
2
av )
+4r
2
0.5 (a, - o 7 )
The direct stresses act simultaneously:
p 160
OM =
0.5 (C7Z +
Oy ) =
0.5(O, + C7 2 )
Direction of the maximum shear stress r max is
oz
p 161
tan 2 <p r
2r
The Principal stresses and Maximum shear stresses lie at 45
to
each other.
*
}
The solution gives 2 angles. It means that both the Principal
stresses and the Maximum shear stresses occur in 2 directions at right angles.
STRENGTH
28
Combination of stresses
Stress in three dimensions
The stress pattern can be replaced
by the
Principal stresses o,, o2
o3
They are the 3 solutions of the
,
equation:
p 162
o3 -Ro 2 + So-T=
p 163
where
R = ax + <7y + oz
ax ay + ay
+ oz -ox - rxy
p 164
S =
p 165
T = ox oy az + 2 rxy Ty.T^ - crx Tyz 2 - o^zj - crz Txy 2
•
•
<jz
•yz
Solve the cubic equation p 162 for a1t o2 and o3 as follows:
Put equation p 162 = y (or substitute y for
on the right
side), then plot y = f(o). The points of intersection
with the zero axis give the solution. Substitute these values
in p 162 and obtain more accurate values by trial and interpola-
hand
tion.
The case where o-\ > o2 > o3 gives the Maximum shear
Tmax = 0.5 (ai - a3
stress
).
Bending and torsion in shafts of circular cross section
According to the theory of maximum strain energy:
= \jfbx 2 + 3(a
-/qt ) 2
p 166
Equivalent stress
oE
p 167
Equivalent moment
M E = yM b + 0.75(a
2
-
< p bt
T) 2
To find the diameter of the shaft, determine the necessary section
modulus Z from:
P
I'
z
=Mi
Pbt
tensile stress due to bending
shear stress due to torsion
bending moment
torque
according to P 29
n
r 29
STRENGTH
Combination of stresses
;
N
1
in
co
b
1
w
b
u
1
=
= =
tT
E
CD
—
?
E
CM
+
CO
>
3
CD
.^
+
CNJ
w.E
c •-._
c c
CO O
E ~
£
co
0.^
II
2
«
CO
c
t:
co
co
CD
b*
b*
<S"
b~
CM
ipt
3
b
11
11
1?
g
b
"co
TO
b
CD
9
w
0)
II
1
cB
CL
i2 d)
CD
b~
+
E
3 CO
c CO
.E v
x .3
Q.
°-2w.E
CX
^>
o~
•
00
CO
CM
3§
-
' ?« £
11
+
CM
c
0.0
a
*-
CM
CM
c
£
CO
b~
s +
00
j?
.CM
+
CO
^-~
c
fj
co
2.
+
5 E
='
- = =
,-
CO CO
E
11
b
b
T3
«
E
0)
CO
CO
6
CO
11
.J
b
1
1
CO
b
b
11
11
z
b
z
tr
CD
CM
CM
(-
C -P
S
*
-<r
CM
b
b
H
b"
V
b
£?
c
J
.2
Q.
E
55
~§
O
+
= = = =
— =£ — =
tT
II
E
lr
b
11
£ «
Is
X!
c
tf
:
01
in
Q.
i
c
°
s
II
<
+
00
3
O
CO
| 5CO
1
CM
b
CD
3
c
O
c
ro
0)
O D) 0)
E
a
+
»
is
-%%%%
SiSS
.
N
fc
in
OS
O
-*
b
s
«=
E-o c c 3
C <D .2
O CO rC
T3
„_
CM
2
A
b
£ CD
C
f£
CD
A
W "D
*=
-
IS)
Cfl
-2.
II
+
0,-°
woD
l
11
1
cm
c°
cm
..
C/3
CD
X
11
-
to
c
to
13
-
8
C
O
5 cm
B
CM
^
CO
a;
*"
O
3
O
w £
* t.
E
-
to
"ro
c
J>
-E
O
-—
Cfl
c
O
C
Q)
2 S3 to
~ £ to
E
CO
>
E "co
1 E
m
°
a)
2
a
01 >^
II
,
O W
TO
C 3
3 cr
3
»»!
CO
at
•
ra
00
w to
S
CP
g g
<D
co
(D
0)
a3
9 «
c
E
CO
C7
<u
Q- i:
>^ 00
0)
co
C
c
-
— O c
TO
(1)
co
oj
C
"D
-0
03
^
cyj
3
.
—
X •«—
a>
co
CD
*"
Jr
~
r>
CD
oj
co
£
E
>
(5
1
«
to"
)
MACHINE PARTS
Screws and bolts
Lead screws see k 11
Fixing bolts
Bolted joints (approximate calculation)
Prestressed
axial working load Fa
q
Fc req
A3 « As =
(1.3... 1.6) FA
Fmax =
q 2
shear load Fs
calculation for friction effect:
Ao ~ A<
1
(0.25... 0.5) /? p
(using load-extension diagram,
.
2
see P 1
q 3
= (0.25
Pi
(allowing
for
.
.
.
v-Fs
0.5) /? p02
fere
(U-
and
torsion
(for values of [/. see Z 7)
factor)
safety
m -w
High-stress bolted joints see VDI 2230
Bracket attachment (precise calculation not possiPle)
Practical
F
assumption:
\
F
m p.
/
E
T
i
that the centre
pressure
of
^
is
& ^>
the point of ro-
-0-
tation,
a
e. g.
q 4
\-
« h/A.
^
& O-
i
For a rigid attachment:
F
q 5
q 6
=
/
A1
•
FA1
•
b,
/ A2
•
+ FA2
•
FAn
b2 +
.
.
.
FAn
br
•
b^\ b 2
Allow for the extra shear load Fs = F There must be compressive
stress over the whole attachment plane, when under load.
A3
As
core cross section
r
L4 S =
stress cross section
^creq required clamping force
:
:
Id
"7'y^k
—d
3
:
m
:
no. of bolts
w = 3;n =
i
e.g
n
v
Rp
p
:
:
no. of joint faces
02
:
:
x
I
proof stress
permissible stress
|
j
I
d2
d3
:
:
p-
'
i
safety factor against slipping [v =
1
.5
.
.
.
1
m = 3;n = 2
(2)]
outside diameter of bolt
core diameter of bolt
MACHINE PARTS
Axies and Shafts
Axles and shafts (approximate calculation)
Stability
q
7
q
8
q
9
fixed 2
solid axle
required
section
of circular
modulus
cross section
for bending
(Z^^/10)
Axis
permissible
bending
stress 3
M
Pbt
'
m a/ 10-M
V Pbt
Pbt
rotating
(3... 5)
Q'bt A
Pbt
(3 ... 5)
permissible
diameter for
Shafts
torsional
solid shaft
stress 3
pure torsion
q 11
q 12
'
Qbtu
'
Z =
q 10
1 '
P^
V
torsion and bending
Pqt
n
Pqx
'
(3 ... 5)
-
TtU
(10... 15)
Bearing stress
on shaft \
q 13
extension]
r
F
/bm = d-b
'
(Pb see Z 18)
Shear due to lateral load: Calculation unnecessary when
> d/A
>
325 h
/
/
.
for all shafts
|
... 8iU
with
|
for fixed axles
circular
cross
rectangular
sections
Deflection due to bending see P 12
due to torsion see P 20
Vibrations
see M 6.
1)
For precise calculation see DIN 15017
2)
Formulae are restricted to load classes
3)
pbt and pqt allow for stress concentration-, roughness-, size-, (see DIN
15017), safety-factor and combined stresses
/
:
M, T
I
+
II
(P 2)
arm of force F
bending moment, torque
:
/bm (Pb) mean (permissible) bearing stress [see Z18)
/bmax see d. 47 f° r hydrodynamically lubricated plain bearings,
values see Z16.
[other cases see Z18.
Obt
MACHINE PARTS
Shaft-hub joints
Friction-locked joints
Proprietary devices (e.g. annular spring, Doko clamping device,
Spieth sleeve, etc.): see manufacturer's literature.
For interference fits see DIN 7190 (graphical method).
Clamped joint
T- v
q 14
fid
imaginary joint,
not to stiff
Taper joint
q 15
Taper from
D-d
tan a =
For
shaft
I
tapered
see
DIN 1448, 1449.
Approximate formula for axial force FA
extensions
on the nut:
q 16
Fa
=
tan
-'&
+ *)
D+d
q 17
Specially machined joints
Proprietary splined fittings, hub to shaft e.g. polygon:
see manufacturer's literature.
Plain key (approximate calculation)
is based on the bearing pressure on the side of the
the weaker material. Allowing for the curvature of
the shaft and the chamfer r-\, the bearing height of the key can be
Calculation
keyway
in
taken approximately as t2
.
tj
The bearing length / is to transmit
a torque T:
q
t
Dimensions
precise
nigung
DIN 6885,
form A.
to
fillets with
For
I
-
\^ x^
SV*±-
2 T
d-h'Pb
U
r77777%777777A
((^L^^ j
K^
preferably
calculations refer to
Antriebstechnik
e.V.
heft 26, 1975.
For symbols see Q 4
Sheet
1.
Mielitzer,
Allowance
for
ForschungsvereiForschungs-
Frankfurt/M.,
continued on Q 4
MACHINE PARTS
Shaft-hub joints
continued from Q 3
Splined shaft
Shaft
-
q 19
/
q 20
dm =
q 21
h
=
2
•
~
2
The load is not shared equally between the splines so allowance
for unequal bearing
made with the factor q?:
is
Type of location
<P
shaft located
0.75
hub located
0.9
For cross section dimensions refer to DIN 5462
.
.
.
5464.
Hub dimensions
Use diagram on page Q 5 to determine dimensions of hub.
Example:
Find the length L and radial thickness s of a hub
to transmit a torque T of 3000 N m, made in cast
steel fitted with a plain key.
*
needed
1.
Determine the appropriate range "hub length L, CS/St,
group e", follow the boundary lines to T = 3000 N m.
Result: L = (110 ... 140) mm.
2.
Determine the appropriate range "hub thickness s, CS/St,
group I", follow the boundary lines to T = 3000 N m.
Result: s = (43
.
.
.
56) mm.
normal force of transmitting surface
bearing length of joint
no. of splines
coefficient of sliding friction (see Z7)
safety factor (see Q1)
angle of friction (g = arctan /i)
permissible bearing pressure. For approximate calculation
material
(gray cast iron)
CS (cast steel), St (steel)
CI
[p^ in N/mm
50
100
2
/higher values possibl*
\
in
special cases
—
MACHINE PARTS
Shaft-hub joints
Diagram to obtain hub geometry for Q 4
These empirical values are for steel shafts made of steel to
ASTM A 572 Grade 42 - resp. to BS 4360 43 B - but not for special
cases
(such
as
high
centrifugal
force,
etc.).
when there are other forces or moments being carried.
u
CD
a
o
<D
CL
£
CO
X
w ® o
JD -*
—
c
en
CO
QC
LU
o
^
£en O
.
C/D
c
-Era
*C -D "O©
'
.C
-
(j)
C
CD
Q.
\
tarQ
\\
\
\
\\ \ \
Increase
L
MACHINE PARTS
Springs
Spring rate R and spring work
W (Strain energy)
Characteristic
general
const
6F
q 22
6s
q 23
/' 6
W*
s
JSfor
J*
R\
s^f/wwflww;V
Rr.
fh,
/
Ri
q 24
5 tot
q 25
^tot
= S;
J,
=
52
F,
+
F2 + F3 ... + F,
S3 ...
R tot = /?! + F 2 + R 3
q 26
.
.
.
+
53
+
F2 = F3
=
S2
+ R
s
i
F\
1
R
t
:
R2
'
R,
Springs in tension and compression
e.g. ring spring (Belleville spring)
Springs in bending
Rectangular, trapezoidal, triangular springs
q 27
q 28
permissible load
q 29
deflection
-h 3
b
bi/bA
V
|
1
1
>
I
0.8
l
0.6
1.000 1.054 1.121
1
I
1
1
0.4
1.202
E
0.2
I
1
315
'
|
1.5
2)
|
Rectangular spring
Triangular spring
continued on Q 7
MACHINE PARTS
Springs
continued from Q 6
Laminated leaf springs
Laminated leaf springs can be imagined as trapezoidal springs
cut into strips and rearranged (spring in sketch can be replaced by
two trapezoidal springs in parallel) of total spring width:
b Q = z-b
q 30
no. of leaves.
Then (as q 28):
F =
q 31
b-h 2
-
Pbx
6 /
leaves 1 and 2 are the same
length (as in the sketch):
If
q 32
b,
= 2b
The calculation does not allow for friction. In practice, friction
increases the carrying capacity by between 2 ... 12%.
Precise calculation according to Sheet 394, 1st edition 1974
*
Beratungsstelle fur Stahlverwendung, Dusseldorf.
/=
1
Disc springs (Ring springs)
characteristics
Different
/,=,
can be obtained by combining
n springs the same way and
springs the opposite way:
/
^r
/'
**'[/
=
q 33
q 34
n-FB
'single
Deflection s
DIN 2092: Precise calculation of single disc springs.
DIN 2093: Dimensions and charact. of standard disc springs.
Material properties: Hot-worked steels for springs to ASTM A 322
e.g. for leaf springs 9255; 6150 - resp. to BS 970/5 e.g.
250 A 53; 735 A 50 - (Modulus of elasticity: E = 200 000 N/mm 2 ).
p bt
:
static 910 N/mm
oscillating (500
2
± 225) N/mm 2 scale removed and tempered
continued on Q 8
MACHINE PARTS
8
Springs
continued from Q 7
Coiled torsion spring: The type shown in the sketch has both
ends free and must be mounted on a guide post. Positively
located arms are better.
q 35
Perm, spring force
FD « —-^
q 36
Angle of deflection
a
q 37
Spring coil length
/
i
IE
= D m -i
no. of coils.
:
f
F-r-l
(Additional correction
is
needed for deflection of long arms).
For precise calculation, see DIN 2088.
Springs in torsion
Torsion bar spring
Shear stress
Torque
T
q 38
d3
Angle of twist
#
d3
spring length as shown in sketch.
=
Tls
lor-/s
GI D
G-d 4
Stress p qt and fatigue strength Tf in N/mm 2
oscillating 2
static
q 39
not preloaded
Pqt
preloaded
rm
:
rA
:
|
700
1020
Tf = Tm ± TA
d = 20mm
d = 30mm
*
500 ±350
500 ±240
mean stress
alternating stress amplitude of fatigue strength
Precise calculation see DIN 2091, especially for the spring
length.
1
'Not allowing for the stress factor arising from the curvature
of the wire.
2)
Surface ground and shot-blasted, preloaded.
continued on Q 9
MACHINE PARTS
CN
Springs
continued from Q 8
Cylindrical helical spring (compression and tension)
*
% Fb
,*ith
<-.
Compr. spring s f
Normal coil ratio:
D = (D
Did = 5 ... 15;
compression spring
D known
8 fctheoiH
d > /
D unknown estimate Did
d ^l/ 8 ^ctheor
tension spring
-|
d
D
q 42
max. perm, deflection
3A
q 43
sum of min. distances
sA =
with x ={0.21.
10.7
D/d=l 4
120
•;
+ £>j)/2
> j/ QFn -D
r
q 41
without
preload
3
q 40
—
Tension spring s f
Static stress:
Fn
Fy
^-Pqt
8 Fn
d > l/
P'
x-dn
between coils
at
q 44
no. of effective coils
q 45
permissible shear stress
I.
n
solid length
G-d A
s
F
Pqt=Pqtc = 0.56xfl
see diagram
'
r
8~D~5
p q = 0.45 x /? n
t
diagram x 0.8
For higher relaxation requirement see DIN 2089.
Cold-wound compr. springs
Theoretical spring i
deflec-l
iperm.
1200
f
,orce
tion
I
Istress
when just solid
(compression spring)
*c
Circ. spring steel wire ASTM-
|^ctheor| Pqtc
spring wire
~1000
spring wire
§
^
8 °°
1600
Oscillating stress:
^ 500o
u
Include the coefficient k for curvature
of the wire and
use the fatigue
(see DIN 2089) in the calculations.
strength
of
spring
steel
1
MACHINE PARTS
10
Bearings
Rolling bearings
Use the formulae from the manufacturer's literature which gives
load capacities and dimensions, e.g. S.K.F., Timken.
Journal bearings
Hydrodynamically-lubricated plain journal bearing
Bearing must be
running at proper
temperatures and
2£J£
without excessive
wear, i.e. separation of the journal
and bearing by a
film of lubricant.
Pressure distribution in transverse
and longitudinal sections
e* = b/d
Length/diameter ratio
0.5
fete^
1.5
1.0
*-B*
2.0
I
vJ///////,
Auto-
Pumps
mobile
engines
Machine
'/>////////,
Marine
Grease
bearings
lubrication
tools
Aero
engines
Steam
Gearing
turbines
General p roperties
Short bearings
Long bearings
Large pressure drop at each end, Small pressure drop at each end,
therefore
good
cooling
with
therefore high load capacity
adequate oil flow.
Excellent
Good at low rotational speeds.
for
high
rotational
speeds.
Low
load
Poor cooling facilities.
capacity
tational speeds.
Symbols see Q 14
at
low
ro-
Danger of edge loading.
continued on Q 1
MACHINE PARTS
Q11
Bearings
continued from Q 10 (journal bearings)
Bearing pressure p, p
q 46
F
p
bearing
pressure
q 47
r
D B
Pmax
sf*dF
<7dF unknown, use R
p02
Pmax depends mainly on the
If
minimum
lative
lubrication
re-
film
thickness h* min
(See Sommerfeld number q 56).
.
The adjacent diagram shows the
ratio
of
maximum
pressure
to
mean bearing pressure in relation
to the
thickness of the
(According to
relative
lubrication
film.
2
0.3
0.4
0.5
Bauer, VDI 2204).
Bearing clearance
C,
relative bearing clearance
D R - D,
q 48
y>
is
ing
basically the
operation
y
CID
relative
(including
bearing clearance established durthermal expansion and* elastic de-
formation).
q 49
Typical values y = (0.3 ... 1
... 3)
10~ 3
1)
Criteria for the choice of \p:
Lower value
q 50
phosphor)
soft (e.g. white metal) hard (e.g.
viscosity
relatively low
relatively high
peripheral speed
relatively low
relatively high
bearing pressure
relatively high
relatively low
length/diameter ratio
B* < 0.8
B* > 0.8
support
self-aligning
rigid
Minimum values for plastics
q 51
q 52
Upper value
bearing material
sintered metals
1)
Grease-lubricated plain bearings
Symbols see Q 14
4)10" 3
2)10" 3
3)
10" 3
continued on Q 12
MACHINE PARTS
Q 12
Bearings
continued from Q 11 (journal bearings)
Minimum permissible lubricant film thickness
during operation h Um in urn.
Theoretical
q 54
> h
+
Actual
shaft deflexion
v
,
bearing
distortion
+ sum
of peak-to-valley heights.
(*z.B
+ *zj)
l*
_
mm
"min
C/2
_
i
I
J
U*
special
cases,
some
automobile
J
crankshaft
^
Relative minimum
lubricant film thickness //* min
q 55
10... (15)]
3.5
[(D
e.g.
bearings
small-
1
large
shaft
diameters
Relative eccentricity e
2 AZmin
V D
•
For statically loaded bearings:
otherwise instability
q 56
Sommerfeld
number So
Nr.
BID
1
1/1
(Dimensionless)
2
3
4
5
6
1/2
So =
PV2
Qetf
Inserting
V
1/3
1/4
1/6
1/8
So
the adjacent
diagram gives
/i* min and therein
fore also h min
.
To a first estimate ?/ eff viscosity is based on
mean temperature of bearing.
A
better
esti-
mate is:
7"
ef
f=o.5(ren +re:
0.002
0.1
Symbols see Q 14
0.2
continued on Q 1 3
MACHINE PARTS
13
Bearings
continued from Q 12 (journal bearings)
Lubricant flow rate
to hydrodynamic pressure development Q
3
The
q 57
flow rate required to maintain
theoretical
hydrodynamic
(exactly values see DIN 31652):
lubrication is
e3 = 0.5 5u(C-2A mln
)
Rules: Oil inlet to the expanding part of the bearing.
Oil velocity u:
q 58
In
supply lines: v
q 59
In
return lines:
Oilbags,
oil
slots
o
= 2 m/s
= 05 m/s;
;
(depth
=
p
p
= 0.05 ... 0.2 MPa
= 0.
2 C) never
in
loaded zones,
no connection with the surface area of the bearings, of
higher pressure only short oil slots or -bags; bigger
bags only in special cases for greater heat removal.
oil
Heat removal
Requirement:
q 60
P = f-Fu = P th (rate of heat removal)
/by the use of the following diagrams and So by q 56)
Friction power
(Calculate
500
f
-I
0.01
0.02
0.05.
10
0,
So
50
20
100
So
200
500
1000
*
Heat removal by convection at the housing surface A.
q 61
Ah, amb
k
W/(m^K)
_ k'A(Ts
= 7+12
Tamb)
Wamb
tm
s-1
with empirical formula for k:
(equation scaled\
{
for units
/
continued on Q 14
Symbols see Q 14
MACHINE PARTS
14
Bearings
continued from Q 13 (journal bearings)
If
the area of the heat removal surface
approximate value for step-bearings:
is
unknown,
Bu + Sh
n-H,
q 68
A
running oil temr\
is temperature-sensitive and the
perature is initially unknown, use iteration method for
preliminary and successively improved estimates of rB by q 61
Since
until,
Pth amb = P
according to q 60,
Heat removed by lubricant
P th L
if
necessary with
conduction are neglected):
Oil circulation,
q 69
f
.
:
oil
cooler (convection, heat
L= QCpQiTex - Ten)
^th,
Guide values for simple calculations for mineral oils:
q 70
c
c
.
p
/
:
/i
min
k
p en
:
.p« (1.6... 1.8) *10 6 Jm
_3
K"
coefficient of friction (values see Z 7)
minimum lubricant film thickness during operation
heat transmission coefficient
lubricant feed pressure
mean, max. bearing pressure
P> Pmax
u
peripheral speed of bearing journal
:
warnb
1
specific heat (values see Z 5)
:
:
D
:
»
velocity of air surrounding the bearing housing
external width of bearing housing in axial direction
C
nominal clearance, concentration, chamfer
nominal bearing diameter
bearing force (nominal load)
F
height of bearing housing
H
rotational frequency (revolutions per time unit
/V
Pth amb heat flow rate to the ambient
D
H
I
:
:
:
Q
:
lubricant flow rate, volume flow rate
mean value of surface finish C.L.A. of shaft mating surface
average peak-to-valley height of bearing sliding surface
ambient temperature
am
7en> Tex lubricant temperature at bearing entrance, exit
Tq
bearing temperature
r]
dynamic viscosity of the lubricant (values see Z 14)
^z.B
J-
:
:
:
rj
e ff
p
Od f
:
effective dynamic viscosity of the lubricant
:
density (val.s. Z 5)
:
crushing yield point
I
\p
:
co
:
relative bearing clearance
(val.s. Q11)
angular velocity (co = 2n \)
MACHINE PARTS
15
Cross head guide, Clutches
Crosshead guide
Crosshead guide will operate smoothly only when
L
q 71
tan a
<
'////////////////A
(2-h + l)-fi
F cos a
the length ratio is
l_
q 72
2 n
,
h
If
tana
- n tana
1
the above
Y////////////////A
conditions
for
tan a are
not satisfied
there
is
a danger of tilting and jamming.
Friction clutches
Energy loss and slip time per operation
Clutch
Drive side
II
TMi w,
/if
Driven side
Ji,
TL ,
o/ 2
II
A
model with the following conditions is sufficent
for approximate calculation:
to 102 = &>i,
Acceleration of driven side from C02 =
= const.; T\_ = const.; 7c = const. > T\_. Then, per
a>i
simplified
operation:
q 73
energy loss
q 74
slip time
W,
h- 2
(**Y&d
Calculating the area of the friction surface
flat
single-
|
twin-
cylin-
multi-plate
plate
drical
clutches
clutches
III!
i
TTT
ii
/^r
^
Number and size of friction surfaces depend on the permissible
contact pressure p^ and the permissible thermal
per unit area q p
capacity
.
continued on Q 16
Symbols see Q 17
MACHINE PARTS
16
Clutches
continued from Q 15 (friction clutches)
Calculation of contact pressure /» b
(for values see Z 19)
For all types of friction surfaces:
i-A
q 75
^m
R a 3 - /?j 3
R a 2 - R?
Pb ^dyn
'
q 76
2
where
3
'
'
conical
flat
cylindrical
friction surfaces
q 77
operating
Fa = A-p
Fa = A p sin a 2
force
(axial)
q 78
to prevent
for multiplate
clutches usually:
q 79
IT = 0,6.
Calculation for a shaft:
.
«a = R\ = Km
locking:
.0,8
tan a > /i stat
^stat
T
x
i"dyn
Calculation permissible temperature rise
In HEAVY-LOAD STARTING the maximum temperature is reached
in one operation. It depends on the energy loss, slip time, heat
conduction, specific heat and cooling. These relationships
cannot be incorporated in a general formula.
CONTINUOUS OPERATION constant temperature is only
established after several operations. There are empirical
values for permissible thermal capacity per unit area q p with
continuous operation (see Z 19).
With
q 80
q 81
Friction power
Condition
Symbols see Q 17
PF =
Wr z
Wr z
i-A >
MACHINE PARTS
17
Friction clutches and brakes
Friction brakes
clutches can also be used as brakes. But there are also:
All friction
Disc brakes
with caliper and pads.
6
Braking torque 7"B
:
TB = 2pF6 jrn
q 82
Expanding-shoe drum brakes
(Drawing of simplex brake showing, simplified, the forces acting on
the shoes).
Leading
|
Trailing
shoes
Fs-l
Fm =
a- fx-r
(Servoaction)
Braking torque T&:
(Fn1 + Fn2 ) ii r
TB =
q 85
Leading
Trailing
shoe
shoe
Band brakes see K 13
Notation for friction clutches and brakes (Q15
A
:
.
.
.
Q17)
area of friction surface
radius of friction surfaces
R
R m R a /?, mean, outside, inside radius of friction surface
operating torque
7C
:
,
,
:
:
TL
TM
T
t
W,
:
:
:
:
i
:
/
:
z
:
fx,
nd
co
:
load torque
motor torque
transfer moment of clutch
energy loss per operation
no. of friction surface
no. of calipers on a disc brake
operating frequency
fx sXaX
:
(EU: s"\ h
_1
)
friction-, sliding friction-, static coefficient of friction
angular velocity
(for properties of friction
materials see Z19)
MACHINE PARTS
Q 18
Involute-tooth gears
Involute-tooth gears
Spur gears, geometry
q 86
Gear ratio
q 87
Transmission ratio
na
0) b
Transmission ratio of multi-stage gearing:
q 88
=
q 89
invor =
i,
•
•
/„
•
;„,
tana — a
Showing the transverse path of contact
(see ISO R 1 122)
Active
flank area
of wheell
^&\°jS Drive
If A and E do not fall between Tt
and 7"2 interference will occur
and "modified" gears as in Q 20
,
should be used.
^Negative for external gears
because rotation is opposite.
Positive for internal gears. The
sign can normally be disregarded.
Standard gear > with g« iaring according to DIN 867
spur
q 90
q 91
normal pitch
q 92
circular pitch
q 93
q 94
q 95
normal module
q 96
P =
circular module
q 97
addendum
dedendum
q 98
bottom clearance
helical
— m
=
•
=
Pi
~ cos |8
ji
m « =2i = J-cos/8
mn _ d
m
= d
*- s
m a -K
m n -n
Po
z
<
" cos j8
z
m
h =
p = m + c
c = (0.1
0.4) m « 0.2 -m
h a = h aP =
t
/j
f
.
.
continued on Q 19
For symbols see Q 29, suffixes see Q 23
:•
MACHINE PARTS
19
Involute-tooth gears
continu 3d from Q 18 (spur gears)
Standard gears
spur
q 99/100
reference diameter
q 101
tip diameter
q 102
root diameter
q 103
pressure angle
helical
m-z
=
d
d =
da
= d + 2-h a
d
=
d-2-hf
a = a n = a = ap
On = «p
f
t
tan a 1t =
q 104
q 105/106
q 107
base diameter
dcos a
db =
m -z
=
cos pc
x
tan a n
?
cos p
d b = d- cos a
t
1
equivalent no.
of teeth
"cos 2 /3b -cos/3
table see DIN 3960
z
q 108
cos 3 /?
min. no. of teeth
q 109
to avoid
2
theory
sm 2 a
9
(undercut)
for
when produced
z gs
^
~
'
17- cos 3 p
a P = 20°
by toothed
q 110/111
q 112
rack tool
practice
zg
'
~
1
*-
4 -*
m M-cos 3 /3
Zgs'
gp = 6 tan
spread
-
|
/8
|
Stands ird gearing
spur
q
113/114
+ do
d\
length of path
of contact
36
116/117
transverse
contact ratio
'" n
2
r
9a= ^V^1
2-cosi3
.
2
-^1
2
+ V<*a2
2
-4>2 2
'-
- Wbi + d b 2) -tan a x )\
(total length)
I
+ Zl
centre distance
,
q 115
helical
l\
ea -
9«
p cosa
9a
^
q 118
overlap ratio
to
q 119
contact ratic )
e
p
Y
p
—
•
t
cos a
t
fc-sinl/Sl
m n -Jt
'
= e a + ep
continued on Q 20
For symbols see Q 29, suffixes see Q 23
MACHINE PARTS
20
Involute-tooth gears
continued from Q 19 (gearing)
modified gears
spur
p,
m,
q 120/121
Pn
Px
,
mn
,
rrii,
,
z,
zr
d,
dy
helical
I
see standard gears
profile offset
q 122/123
-"-mm
to avoid
interference
,
frao-paoO -sin a)
_ z-sin2g
t
+
COS fi
/iao-paoC -sin gn
*min -
p
+
2
•
)
can be up to 0.17 mm
ditto 1
14-z
^ 14-(z/cos 3 ff)
'
17
17
to give a specific
centre distance
q 126
* 1 + X2 ~
(z :
+Z2)-('nvaw t-invat )
(z-i
+ zo)
2-tanan
(total)
a wt calculated from
q 127
or
q 128
cos a,
-
invawt
2-a
invot +
centre distance
q 129
addendum modifica-
q 130
k*
addendum
h a = h aP +
q 132
dedendum
hf
= h iP
2-f
q 133
outside diameter
d a = d + 2-h a
root diameter
d<
q 135
length of path of
=
COSOft
"
i
^
n
{x
-
2 '+
2 rf
b1
9a = -2-[V^1
x
+ x2 )
2)
V d a2 -^2
2
2 '-
- (d b1 + rf b2 )-tana wt J
e a = 9a'(P
•
cosa)
£ a = 9jiP\ -cos at)
= ft-sin|/j|/(w n -jc)
overlap ratio
£p
q 139
contact ratio
£y =
2)
-tana n
d-2-hf
q 138
1)
2
x-m n + k*-m n
x-m n
-
q 134
transverse
contact ratio
x
mn= a- a6 -m
q 131
contact
•
C0S at
a,
3d COS
Owt
a
9
tion coefficient
—m
ea + £p
If tool data unknown take ap = 20°.
Note the sign. With external gears k x m n < 0! When k < 0.1 addendum
modification can often be avoided
For symbols see Q 29, suffixes see Q 23
MACHINE PARTS
Q
Involute-tooth gears
21
Spur gears, design
The dimensions are derived from
load-carrying capacity of the tooth root
load-carrying capacity of the tooth flank,
which must be maintained independently.
Gearing design is checked in accordance with DIN 3990. By conversion and rough grouping of various factors it is possible to derive
some approximate formulae from DIN 3990.
Load capacity of tooth (approximate calculation)
Safety factor .S> against fatigue failure of tooth root:
q 140
^Flim^ST ^NT' ^SrelT' ^RrelT ^X
_
c
•
•
m
b
^ s
t
K.K,
r
Using the simplification:
{KFa 'YB 'Y$~V,
(^fcrelT' ^RrelT" ^x)
*t
q 141
(^A ^v)
•
.
'
^Fg
•
Y ST = 2;
~ 1
Ym ~ 1;
Y Fs Sf min
•
2-o> lim
b
FFS tooth form factor for external gearing (see diagram)
:
q 142
KA -Ky = 1 ... 3, rarerly
5.2-
more, (allowing for
external
shock and
torque,
exthe rated
additional
internal
dynamic
irregularitis
ceeding
forces arising from
and
tooth
errors
circumferential velocity).
q 143
5 Fmin =
1.7 (guide value)
q 144
<7Flim
guide values see table on Q 22
continued on Q 22
For symbols see Q 29, suffixes see Q 23
|
MACHINE PARTS
Q 22
Involute-tooth gears
continued from Q 21 (spur gears, design)
Load capacity of tooth flank (approximate calculation)
Safety factor 5 H against pitting:
q 145
°H lim Znt
_
c
'
•
V^
{Z\.- Zy
Zp) -Zw Zx
'
*Sl
Zh 2 E Z z Zp y^A ^v Ky
•
'
•
•
"
For metals the factor of elasticity Z E is simplified to:
where
VO-175
q 146
2
£
•
E
E2
y
Ei
Therefore, the approximate formula becomes:
* 1
q
147
>
d
fe,!^ 0175 £
,
g,'
,
z
cos
H^E'V^'V^'V% ,SHp
s
Appr oxima te values for strength
(D/a grams in DIN 3990 part 5)
Sp scification to
Mat.
Norm
6
A 48-50 B
80
360
4
ASTM A 536-20-90-02 230 560
A 572 Gr. 65
200
400
3
-
1064
220
620
2
-
AS
SAE 4140
290
670
ASCH
3240
cs
20°
-
2.5CI
=
7
^liml^lim
N/mm 2
Grade
=1
q 151
only valid for a n
28
—^\0
h**,v\\\
.^^^S^nN
Crts"^****^.
\ \
^
\ \ ^
v \ \
jSw
1
500 1630
2.0-
0.05
9
0.08
cast iron
CI:
CS:
carbon steel
AS:
alloy steel
K A Ky
-
:
^^S^,
O.l^"" "*-*-"^
1
8
ASCH: case hardened alloy steel
_ __
continued on Q 23
^^**s^S
I
I
1.7
-
J
J skew angle for helical gears
ZH for a n = 20°
(pitch cylinder)
\
)
a
P ~*
see load capacity of
tooth root (q 142)
q 148
q 149
q 150
q 151
5 Hmi n - 1.2 (guide value)
guide values see table
a H iim
Z H zone factor (see diagram)
(Z L Z v Z R ) « 0.85 for hobbed or planed gearing
» 0.92 for ground or shaved teeth with average
:
:
•
•
peak to valley height R z -[qo ^ 4 pirn
For symbols see Q 29, for suffixes see Q 23
MACHINE PARTS
f
VjC 23
Involute-tooth gears
continued from Q 22, (spur gears, design)
In q 141, q 145 and q 147 b or b and d must be known. The
following ratios are for estimating purposes and should be
used for the initial calculation:
Pinion dimensions
Either:
Or:
from gear ratio i
and a specified
centre distance a
(seeq 113-114-129)
^shaft 1
q 152
q 153
pinion integral with shaft
1.2 ... 1.5
pinion free to turn on shaft
2
Tooth width ratios
Tooth- and bearing-quality
q 154
teeth smoothly cast or flame cut
b_
b
m
dt
6 ... 10
teeth machined; bearings supported each
q 155
q 156
on steel
overhung
side
construction
or
pinion
(6) ...
10... 15
teeth well machined; bearings supported
each side in gear casing
15 ... 25
teeth precision machined; good bearings
q 157
each side and lubrication in gear casing:
< 50 s-
«1
20 ... 40
1
.
q 158
overhung gearwheel
£
q 159
fully supported
$ 1.5
Suffixes for Q 18 ... 25
a
:
b
:
m
n
:
:
:
,
:
v
:
1
:
2
:
driving wheel
driven wheel
tooth middle for bevel gears
normal
tool
tangential
on back cone (or virtual cylindrical gear)
small wheel or pinion
large wheel or wheel
For symbols see Q 29
0.7
MACHINE PARTS
24
Bevel gears
Bevel gears
Bevel gears, geometry
Equations q 86... q 88
are applicable and also:
cone angle d:
q 160
q 161
q 162
q 163
tan (5-,
sin I
cos I + u
I = 90° => tan 61
tan 6
sin I
cos I + 1/w
(I = 90° => tan 6 2
angle between
q 165
y
z
J
Only the axial and radial
forces acting on mesh
wheel 1 are shown
cone distance } *e = 2 -sin 6
Y
q 164
shafts
I
external pitch
Development of the back cone to examine the meshing conditions
and for determination of the load capacity gives, the virtual
cylindrical gear (suffix „v" = virtual) with the values:
q 167
straight bevel gear
cos 6
^v1
dm
cos (5
Formulae q 92, q 95 ... q 100 are also applicable to the surface
of the back cone (suffix "e").
Bevel gears, design
The design
(suffix
is
referred
to
the
MID-POINT OF THE WIDTH
b
"m") with the values:
169/170
Re -
171/172
2-fl,
dm
z
sin
2-T
<5
dm
continued on Q 25
For symbols see Q 29, for suffixes see Q 23
MACHINE PARTS
25
Bevel gears
continued from Q 24
Axial and radial forces in mesh
q 173
axial force
Fa
=
Fmt -tan a
q 174
radial force
F,
=
Fmt -tan a cos <5
•
sin 6
•
Load capacity of tooth root (approx. calculation)
Safety factor 5 F against fatigue failure of tooth root:
q 175
° F lim ^ST ^6 rel T ^R rel T ^X
_
c
'
'
b ef -m
'
^FS
'
_
'
*
•
r
With the exception of YSJ the factors
substitute spur gears (index „v").
Y are determined for
,
Giving the approximate formula:
FmX
...
^•y s-*WWY
q 176
S Fl
Y E' Y K' (^A K v K Fa A" F p)
Sfi
K -(KA .tfv)tf F|r
F
^ST (^6 rel T ^R rel T ^x)
"
*
0.85-6
Y FS
* ^
'
"
'
aF
I
T1
2
Substitute the number of teeth of the complementary spur
gear z v The graph for spur gears on page Q 21 is then also
applicable to bevel gears.
For all other data see q 142, q 143 and q 144.
:
.
Load capacity of tooth flank (approximate calculation)
Safety factor 5 H against pitting of tooth surface.
q 177
OH lim
Sh
•
(
Z L Zy Zr) -^X
•
'
^
c
— ^H
min
'
.
*
V:^
~b
2 H ^E Z e Z K
\
'
"
"
V-^A
'
-^v
"
*
^H a -^H
*
f
For metals the factor Z E is simplified to:
q 178
Z E = Yo.175 E
with
Giving the approx. formula:
2- Ty cosd]
q 179
<mi^V
Kh
0.85
q 180
Mv + 1
•
E2
E,+E 2
=1=1
s.q181
s q 180
^H Zk ZeV ^Ha V^A ^v V^Hg ^H min
'
0,175-E
'
'
•
(Z L
"v
•
'
Z v Z R Zx
•
•
)
sTqTSI
fe
'
:
H im
|
= 1
K H $ = A" Fp ~ 1.65 for fully supported pinion and wheel
ZH
« 1 .88 for fully supported and overhung
V
* 2.25 for overhung pinion and wheel
see diagram for Z H (page Q 22), but only valid for
= m
to + x 2 ) (zi + z 2 with
Z H = 2.495 for a = 20° and standard or modified gearing.
)
q 181
2 Ey
E =
.
For all other data see q 148 ... q 151.
For symbols see Q 29, for suffixes see Q 23
MACHINE PARTS
Epicyclic gearing
26
Velocity diagram and angular velocities
(referred to fixed space)
q 182
q 183
q 184
MACHINE PARTS
27
Worm gearing
Worm gearing, geometry
(Cylindricalworm gear-
normal module in
ing,
BS 2519,
axial section,
angle between shafts
I = 90°).
Drive worm
All
the
ing
on the teeth
forces
actin
mesh are shown by
the three arrows Fa
,
F and F
r
x
In
.
the example:
z-i
=
2,
right-hand
helix.
Gear tooth ratio and transmission ratio as q 86 ... 88
Worm, suffix 1
q 185
module
q 186
pitch
q 187
mean diameter
Worm wheel, suffix 2
|
m-n = p 2 = d2 nl z 2
rf
=
m1
2-r m1
(free to chose, for normal values see DIN 3976)
q = d my lm
q 188
form factor
q 189
centre helix angle
q 190
pitch diameter
q 191
q 192
addendum
dedendum
hu
q 193
tip clearance factor
cf = (0.167
q 194
outside diameter
4n
q 195
tip groove radius
tany
=
=
m
a1
tooth width
b^
q 197
root diameter
rffi
q 198
centre distance
Profile offset factor x for
otherwise x = 0.
mm _ ii
+0
= m(1
=
^
<*mi
.
.
.
m 22
=
d2
/i
q 196
'>
»
j
•
h a2 = m(1
/if 2
0.2
+x)
1 >
= m(1 -jc+c2*)
*
.
.
3) = c 2
.
+ 2/i a1 d a2 = d 2 + 2- h a2
W
" d22
rk
= a - d a2 /2
b2
« 0,9-d m1 -2m
= d m1 - 2 /z f1 rff2 =
a = (</ m1 + d 2 )/2 +
f/2
— ^
xm
^f2
1>
check of a pre-set centre distance,
continued on Q 28
For symbols see Q 29, for suffixes see Q 23
MACHINE PARTS
28
Worm gearing
continued from Q 27
Worm gearing, design (worm driving)
Worm
q 199
peripheral force
Worm wheel
*"»
q 201
q 202
axial force
tan (y+p)
cosptana p
,1
sin (y+p)
rubbing speed
Fm
i
Fa2 = ^t1
*
Fr1
radial force
1
Fn
*.i
=
Fx2
"ml
q 200
- Ft
= *"r2
0>!
V9
cos y m
2
Efficiency
Worm driving
q 203
Worm wheel driving
tan ym / tan (ym + p)
= tan (ym - p) / tan yn
77'
(7m <?)==> self-locking!
Coefficient of friction (typical values) \i = tan
vg
«
worm teeth hardened
and ground
1
p
m/s
~ 10 m/s
vg
0.04
0.02
0.08
»0.05
worm teeth tempered
and machine cut
For calculation of worm shaft deflection see P 12
Calculation of module
q 204
•
;
q 205
q 206
m
Load capacity of teeth root and flanks and temperature rise
are combined in the approximate formula:
where b 2 ~0.8 d m1
Fx2 = Cb 2 P2
p 2 = m-n.
;
Fx2 = 2 T2 /d2 = 2 T2 /(m z 2
•
-i£
.8-7-2
CpernVtf
'
22
•
•
)
q
* 10 for
i
= 10,20, 40
9
» 17 for
1
= 80, self-locking
Assumed values for normal, naturally-cooled worm gears (worm
hardened and ground steel, worm wheel of bronze):
v9
I
Cperm
|
m s~
N mm
1
I
1
|
8
2
I
2
I
5
I
10
3.5
I
15
2.4
I
20
2.2
When cooling is adequate this value can be used for all speeds:
q 207
Cnerm - 8 N mm
~2
For all symbols see Q 29, for suffixes see Q 23
MACHINE PARTS
29
Gears, Gearings
Notation for Q 18 ... Q 28 (suffixes see Q 23)
centre distance (a d standard centre distance)
facewidth
effective facewidth (root /flank) for bevel gears
:
b ef I fe eH
:
addendum of cutting tool
addendum of reference profile (e. g. DIN 867)
dedendum of reference profile
change of addendum factor
no. of teeth
z
peripheral force on pitch cylinder (plane section)
application factor
dynamic factor (with regard to additional dynamic forces
caused by deviat. of the gearing and tooth bending vibrat.)
K Fa / Kf$ transverse load- /face load factor (root stress)
K Ha / K H a transverse load- /face load factor (contact stress)
total pitch cone length (bevel gears)
*e
mean pitch cone length (bevel gears)
Fx
*A
:
:
T
torque
tooth form factor (force applied at the tip of a tooth)
tooth form factor for external gearing
»s
stress correction factor (applied at the tip of the tooth)
stress correction factor
life factor
index "T": for standard^R rel T relative surface factor
conditions
Ybre\J relative sensitivity factor
YK /Y /Y E size- /helix-angle- /contact ratio factor ior tooth root
z E /z H /z L elasticity- / zone- / lubricant factor
Znj/ Za I Z. life under standard conditions- / roughness- / speed
factor
Zw
work hardening factor
Z K /Z X bevel gear- /size factor
Z$/Z z helix-angle/contact ratio factor for flank
>Sa
>'ST
]
:
:
fi
:
:
:
an
aP
aw
:
yS
:
normal pressure angle
:
:
/3 b
:
g
p a0
:
:
m
:
CTHMm
:
<7FM
reference profile angle (DIN 867: a p = 20°)
operating angle
skew angle for helical pitch cylinder
skew angle for helical gears base cylinder
sliding friction angle (tan g = ji).
tip edge radius of tool
fatigue strength
Hertz pressure (contact pressure)
Precise calculations for spur and bevel gears: DIN 3990.
DIN 3960
spur gear and gearing
Terms and
or BS 2519
DIN 3971
bevel gears and gearing
definitions for
DIN 3975
straight worm gearing
]
I
PRODUCTION ENGINEERING
Machining
Machine tool design: general considerations
Components of machine tools which are subjected to working
stresses (frames with mating and guide surfaces, slides and
work spindles with bearings) are designed to give high
tables,
accuracy over long periods of time. They are made with generous
bearing areas and the means to readjust or replace worn surfaces
should become necessary.
The maximum permissible deflection at the cutting edge (point
of chip formation)
is
approximately 0.03
flection refer to formula
Cutting
(main
drives
mm. For spindle de-
P13 and for cutting forces see r 4.
with
drives)
v
=
const,
over the entire
working range (max. and min. workpiece or tool-diameter) are
obtainable with output speeds in geometric progression:
=
nk
n 1 <pk-i
The progressive ratio cp for the speeds n-\
n k with k number
output speeds
r
2
are calculated by:
<P
and the preferred series is selected.
Standardized progressive ratio cp: 1.12-1.25-1.4-1.6-2.0
20 /
—
Speed basic series R20 where cp = yiO =1.12:
.100-112-125-140-160-180-200-224-250-280-315-355-400.
.
-450-500-560-630-710-800-900-1000-..
.
rpm.
Cutting gears are designated by the number of shafts and steps.
Example: A III/6 gear drive incorporates 3 shafts and provides 6 output speeds. Representation of gear unit as
= 180; « k = 1000):
shown (for k = 6; cp =1.4;
m
Network of scales (symmetrical)
Gear arrangement
Speed diagram
/
1
A
>
to
<-
jf-JH
in
CM
/^
m
m
CO
<
<
L
\y $^$
Jj AJ
&\
%\'
3
3
-
1
3
8
<*
For explanation of symbols refer to R 5
1
n
PRODUCTION ENGINEERING
Machining
Cutting power
r
3
Pc
Cutting power
General
Fc
Pc =
*?mech
r
4
Cutting force
Fc =
-
'
Drilling
v
Fc {D + d) n n
7?electr
2
•
'
^mech
1 - mC
.
i
(/c
-mc
1
c
see Z 1 7)
°«-
5
*?electr
mm 2e
^•*c,,*( 4)
Tables of values for K, b, h, z e
'
I
0Q
o = 7 w
O
O iI
<N
NO
ffl'c
O "5 *C
6
E
CO
<o
B-
fc>|CM
o x
^ ^
X
x
t-
-CO
t-
x
\i
^
I
J^&
en
»-
—
E S o
= "° £
o
r
}no-dn pue jno-u/v\op
BujHjuj
in
8UB|d
For explanation or symbols refer to R 5
PU3
1
PRODUCTION ENGINEERING
Machining
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O C
£:
Q)
"•*-
For explanation or symbols refer to R 5
o
t- in
't c\i
in -<fr <* co co co
I
aiqei
PRODUCTION ENGINEERING
FU
Machining
Feed drives
Feeds in geometrical progression with progressive ratio
q> = 1.12- 1.25-1.4-1.6-2.0.
Feed rate
Method
r
15
r
16
Notes
Feed rate
Turning, longitudinal
u = n-s
(external and internal)
u = n-sz -z s
Drilling
for twist drills
Ze = 2 S = 2
sz = 0.5 5
17
r
r
18
r
19
Planing, shaping
u = V
Milling, plane milling
u
= n-s z -z s
and end milling
Cutting times
tc
where
f,
= / + /'
When calculating the cycle and machining times for each workpiece, the feed and infeed travels and also the lengths covered during non-cutting motions, divided by the corresponding
speeds, must be taken into account.
Feed power P v
r20
Feed power
r21
Feed force
r22
Friction force
u(FR + Fv )
rV
^mech
Fv «
^R
=
*
V alectr
0.2 Fc
;
(Fc from
r
4)
rn b -g-n
where m b is the mass moved, e.g. in the case of
machines the sum of the table and workpiece masses.
milling
must be determined whether the feed power as calculated
under r 20 is sufficient to accelerate the moving components
to rapid motion speed ue within a given time t b (in production machines «e ~ 0-2 m/s).
Otherwise the following applies:
It
r23
Pv = u E -m b (fi.g + !^.\
\
For explanation or symbols refer to R 5
f
b / '/mech
LH electr
PRODUCTION ENGINEERING
Machining
Explanation of symbols
used on pages R 1
infeed
/
K
.
.
.
R 4
effective grain distance
:
width of chip
see table 2
effective width
speed
rough plane grind. b w = 5 S /1.
smooth grinding
minimum output speed
maximum output speed
b w = B s /3
milling width
B2
:
cutting power
milling width measured
feed power
from tool centre
feed
width of the buffing disk
feed per cutting edge
diameter of pre-drilled hole
working-part - outside,
acceleration time
inside diameter resp.
cutting time
u
tool diameter
uE
friction force
v
cutting force
ze
:
:
:
:
feed force
feed rate
rapid traverse speed
cutting speed
number of cutting edges
in action
slenderness ratio (es = a Is)
gravitational acceleration
£s
chip thickness
*?eiectr
number of output speeds
^mech
j:
basic cutting force related
:
x
to area
H
method factor
o
cutting travel
(p
work travel
<p s
:
:
:
:
'
overrun travel at both ends
:
electrical efficiency
1
mechanical efficiency
setting angle
friction coefficient, see Z 7
drill tip
angle
progressive ratio
incident angle for milling,
cutting or grinding resp.
with feed rate u
number of cutting edges
HM
carbide tip
per tool
Ihss
high-speed tip
PRODUCTION ENGINEERING
Metalworking
Cold working of sheet
Deep drawing
Initial
r24
blank diameter D
w
°Am
4
D
LA,
are the surface areas of the finished item which can be
calculated from the following formulae b 30, c 12, c 16, c 21,
c 25, c 27 or c 30. The surface areas at the transition radii
for both drawing and stamping dies are calculated as follows:
r25
\
* +
A m = ±[2nd,r z + 4(K-2)r z
Example: (assume
r26
D
rs
/L
m = f (27td 4 + 8rs )r s + ^d 4
= rz = r)
= }Jd 4 2 + d 6 2 -d 5 2 +
fl
4d^
+ 2nr(di + d 4 )+4nr
2
zd
1st and 2nd stages
r27
ft
r28
I
r29
-7
«1
02
'<*1
0.001
FD1 =nd^skim ^q>^
*7ei
r30
r31
r32
r33
<p<i
=
In
1
-<*,!•
cf,
max = 0100 + 0-1
*"
5
2nd stage
1st stage
—
2
2
\/0-6^i -0.4
ftmax
= 37"
=^00 + 0.1-^-0.001
FD2 = ~2~ + red 2 s ktm2<P2-rf(p2
without
=
In
I
2
V0-6^2 -0.4'
_ frl + fc f2
inter-
mediate'
with
annealing
"fm2 - g)
2
continued on
R7
PRODUCTION ENGINEERING
R7
Metalworking
continued from
R6
The work w, related to the volume and the yield strength k
is
f
obtained from the deformation curves for the appropriate value
of logarithmic deformation ratio cp (see Z 20).
Blank holding forces FB and FB2
i
2nd stage
1st stage
r34
*--tf-tf!&[<*-
i )2
+
f]
Bottom tearing occurs if
r35
r36
_ F D1 +0.1 FB1
„
_ F D2 + 0.1 FB2
Maximum drawing conditions (3 and R m
Material - Sheet metal
Specification to
U.S.A.
BS Grade
366-79
Carbon steel
^ 619-75
H
< 620-75
CO
]~
o>
-*
*
283 Gr. C
Stainless steel
(18% Cr; 9% Ni)
SAE 3310
AA 6004
AIMgSi soft
soft
/?100
1490
annealing
1.7
-
-
390
360
350
340
410
S20
2.0
1.2
1.8
600
LM 5
2.05
1.4
1.9
150
1.7
1.2
1.5
4
1.8
1.2
1.6
3
1.9
1.25
1.65
2
2.0
1.3
1.7
321
Notation for R6 and R 7
A mx surface area
^bi. FD2 drawing force in 1st and 2nd stage
A: fm i or k
mean yield strength, 1st and 2nd
im2
yield strength for g?-, and q>2
*fii ^f2
r
radius
rs
radius of stamping die
rd
radius of drawing die
WQrk f deformation
;
w
work per unit volume = —
*
:
'•
:
:
:
:
:
!
:
;
forminged volume
drawing ratio, 1st and 2nd stage
fa, fa
=
ratio
for
1 mm and d = 100 mm
max.
drawing
s
fa 00
fa maxmax max drawing ratio, 1st and 2nd stage
process efficiency, 1st and 2nd stage
*7ei. *7e2
logarithmnic deforma tion -a No, 1 st a nd 2nd st age
<Pi, qp 2
'
:
&
:
:
:
-
^m
& max N/mm
15
1
970
without] with
intermediate
2
PRODUCTION ENGINEERING
8
Metalworking
Extrusion
r37
Remodel force, pressure force
r38
Remodel work
r39
Mean strength for deformation
F
=
A'k 1m -<pA
±
w
i
<Pa
full
Extrusion forward
hollow body
Extrusion
body
backward
'"^S<^
~S^
<P\do
.^i*.'
*
I
7TA
'**
.
r40
A
A
d
=
f(d
2
-d, 2
)
^0
VI
%
r
41
<Pa=
r42
In
^
d?
4 -'o "o
r43
r/
F =
0.7 ..
.
0.8
V =
T] F
2
J h {d Q -d
= 0.6
.
.
.
0.7
=ln
,
do 2
d^l?
fdo 2 h
2
)
r]
F
= 0.5
0.6
Maximum logarithmic ratio of deformation <pAmax
without intermediate annealing
Steel
"^material
AI99.5
methoa\
3.9
3.0
backward
4.5
40
1.2
V
:
:
:
:
:
low
1.2
0.9
0.8
0.7
1.1
1.1
0.95
0.8
used area
logarithmic deformation ratio
deformation efficiency
remodelled volume
volume related remodel-work according to curves Z 20
Ah: depth of stroke
alloyed
alloyed
forward
<pA
w
C<0.1% C<0.15% 00.15%
1.4
A
7/ F
AIMgSi
soft
ELECTRICAL ENGINEERING
General terms
The most important electrical quantities
and their units. - Basic rules
Note regarding capital and small letters used as symbols
Electrical engineering quantities that are independent of time
are mainly denoted by capital letters. Quantities that vary
with time are denoted by small letters or by capital letters
provided with the subscript t
Examples: formulae s 8, s 9, s 13
Exceptions: /, a), i, v, PFeio
Electrical work
W
W
is equivalent to mechanical work W as explained
on M 1. Energy conversion, however, is subject to losses.
Units: J; Ws (wattsecond); kW h; MW h; see also A 3, A 5)
Electrical work
1
Further the following
plained on S1 and S2:
relation
Ws =
1
applies,
1
N m
quantities
ex-
=
Joule
using
J
1
=
2
W = I-V-t = ¥-t = I Rt
2
s 2
Electrical power P
Electrical
power P
explained on
is
power
equivalent to mechanical
P,
as
M 1. Energy conversion, however, is subject to
losses.
Units:
W (Watt); kW; MW
;
see also A 3, A 5)
1W
Further the following
plained on S1 and S2:
relation
see L 1
Period T
see L 1
= 1
using
P
s 3
Frequency/
applies,
_
±s = 1*LS1
s
quantities
ex-
Yl
2d
(T=1//)
Angular frequency co, angular velocity to see L 1
Current /
Is a base quantity (see preface and instructions)
Units: A (ampere); mA; kA
The current of 1 A has been defined by means of the attracting
force which two parallel current-carrying conductors exert on
each other.
continued on S 2
ELECTRICAL ENGINEERING
s2
General terms
continued from S 1
Current density J
S 4
-i
>
Applicable only where
over cross section A.
Units: A/m 2
;
distribution
of
current
/
uniform
is
A/mm 2
Potential difference V
v-i
s 5
Units: V (volt); mV; kV
Where a direct current of 1 A through a conductor converts energy at
a rate of 1 W, the voltage across this conductor is 1 V.
IV
-
1$
1^
-
1AO
-
-
,${.
Resistance R
R - j
s 6
(Ohm's law)
Units: Q (ohm); kQ; MQ
Where a voltage of 1 V across a conductor causes a current of 1 A to
the flow through it, resistance is 1 Q.
= J_V =
1 A
W
1
=
A2
1
_W_» =
s A2
_Nm_
2
s A
Conductance G
Conductance G is the reciprocal of resistance R.
G
=
MR
1/Q
=
[1
s 7
Unit: 1/Q
Mho]
Quantity of electricity, charge Q
s 8
q
=
fi'dt
Q
=
It
C
=
1
(see s1)
For direct current:
s 9
Unit: C (coulomb):
1
As
continued on S 3
ELECTRICAL ENGINEERING
General terms
continued from S 2
Capacitance C
The capacitance C of a capacitor
electricity
Q stored in
it
the ratio of quantity of
is
and voltage V across it:
s 10
C
V
Units: F (farad); jxF; nF; pF
Where a capacitor requires a charge of 1 C to be charged to a voltage
of 1 V,
its
capacitance is 1 F.
1
F =
C.
1
=
As
<
1
A2 s _
'"
V
V
W
tf_£ =
1
J
A2 s 2
1
Nm
Magnetic flux
s
11
<P
X
=
jj
f
u df
(seesl)
Here N is the number of turns of a coil and u the voltage induced, when the magnetic flux <Pt linked with the coil varies
with time.
Units: Wb (weber) = Vs = 10 8 M (maxwell)
1
Wb is the magnetic flux which, linking a circuit of
it
a voltage of 1 V as it is reduced to zero at a uniform rate in 1 s.
1
turn, induces in
Magnetic induction (flux density) B
The magnetic induction in a cross section A is:
B
s 12
~
&
A
Here A is the cross-sectional area traversed
by the homogeneous magnetic flux <P.
Units: T (tesla); nT; nT; V s/m 2
1T =
;
perpendicularly
G (gauss)
= 10- 4 -Vs = 10 4 G
1 V-f[
=
1Q
4_M,
Where a homogeneous magnetic flux of 1 Wb perpendicularly traverses an area of 1 m 2
,
its
magnetic induction is 1 T.
continued on S 4
1
ELECTRICAL ENGINEERING
General terms
continued from S 3
Inductance L
Ny
L =
s 13
N -j1
=
(seesl)
Here / is the current flowing through a coil of N turns and <P
the magnetic flux linked with this coil.
Units: H (henry); mH
1 H is the inductance of a closed loop of one turn which, positioned in
vacuum and passed through by a current of 1 A, enclosed a magnetic
flux of 1
Wb.
=
iM
A
1H =
1
Vs.
A
Magnetic field strength H
S
jr
14
B
_
Units: A/m; A/cm; A/mm; (Ampere Turn/m)
Magnetomotive force F
s
F = NI
15
Units: A; kA; mA; (Ampere Turn)
Magnetomotive force F\ in the f-th section of a magnetic circuit:
s
16
Fj
here
l\
is
-
Hv
l\
the length of this section.
2>i - F
s 17
i
-
Reluctance S of a homogeneous section of a magnetic circuit:
„
s
S =
18
<P
Units: 1/H; = A/V s; (Ampere Turn/Wb)
for symbols see S 16
/
equivalent to
\
magnetic circuits/
\
Ohm's law for
-£-
ELECTRICAL ENGINEERING
Electric circuits
Basic properties of electric circuits
Directions of currents, voltages, arrows representing them
Direction of the current and of arrows
representing positive currents in
s 20
s 21
I
I
generator
--» +
load
+-*-
Direction of the potential difference and of
arrows representing positive voltages always
s 22
I
(
Directions of arrows representing currents or voltages
where calculation results in a
where function
of element
determine
(generator or
load) as well
as polarity is
directions
value, direction with respect
of arrows
to arrow of current
as stated
—
—
equal
opposite
s
23
known
s
24
unknown
positive
negative
or voltage is
above
at
random
Special rule
Arrows representing voltage drop across a resistor and current
causing it, should always be determined in same direction (as
R > 0).
Ohm's Law
RI
Current through a resistor:
s
25
s
26
I
=
V
El
(see also s 6)
R
Resistance R of conductor
/
A
Y-A
Resistance R of conductor at temperature
(in degrees centigrade)
S
R
27
=
/?2o[l+a(0-2O°C)]
Electric heating of a mass
Vl-t-7]
s 28
a
:
Y
g
:
=
&
m
c-m-A&
temperature coefficient
(s.
A&: temperature change
conductivity
(s.
t
resitivity
c
:
specific heat
rj
:
efficiency
(s.
Z 21)
Z 21)
(s. Z 21)
Z1 ... Z4 and 02)
/?2o
:
:
time
resistance at
# = 20°C
continued on S 6
ELECTRICAL ENGINEERING
Resistor combinations
continued from S 5
1st Kirchhoff Law
The algebraic sum of all currents entering a branch point (node) is zero.
s
II
29
=
/_/, _/ 2 _/ 3
into
Here currents
the node
are considered
out of
positive
Ratio of currents
Where several resistors are connected
in parallel, total current and partial
currents are inversely proportional to
their respective resistances.
,. J
j
1
/1 * /z
'
.
30
.
*
_
,
1
1
.
" R
/3
1
.
R,
'
1
.
R2
R3
Current division
Partial currents of 2 resistors connect-
ed in parallel:
s 31
h
G
+ G2
:
Ri
2nd Kirchhoff Law
The algebraic sum of all voltages around
a closed mesh (loop) is zero.
s
IV
32
=
0.
Here voltages traversed
in
accordance
with (opposite to) direction of arrow are
considered positive (negative).
Ratio of voltages
Where several resistors are connected
in
series,
the
ratio
of
partial
volt-
ages is equal to the ratio of the res
33
spective resistances.
V,
V2
V3 =
:
:
^
:
R2
:
r,i
/?,;
R3
Voltage divider
Partial
voltages
across
2
resistors
connected in series:
s
34
v,
=
<^2
v
Gi
,
= TV-
#1
R2
rS5n
ELECTRICAL ENGINEERING
Combination of resistances
Series connection
Total resistance R s
(according to s 26)
generally:
s
35
s
36
Rs
R) + R 2 + R 3 +
=
.
for n equal resistances R:
nR
=
R,
Parallel connection
Total resistance R p (according to s 30)
generally:
s
1.1.1
1
37
s
Gp
38
R,
R^
Rr
=
j*1
tyl
Gt + G;
resistances
several resistances
s
s
39
R,R 2
/?! + R 2
Rp
R-\
P
/?1 /?2
"*"
R] R3
1
1
40
R 2 /?3
"2 "3
"*"
G, + G 2
G-j
-^^p
for « equal
for 3
for 2
ft*
+ G2 + G3
R
-
£
1
nG
Multiple connection
A multiple connection of several known resistances is subdivided
into parallel and series connections, proceeding outwards. These
are separately converted as to be conveniently combined again,
e.g.:
s 41
s
42
s
43
/
'3
=
R2 + R3
Ri R 2 + R-\ R 3 + R 2 R 3
=
R-\
R2 +
I\-\
R 2 + R-\ /?3 + R 2 R% V
Gl (G2 + G3)
+ G 2 + G3
G1 G3
G-j + G2 + G3
G^
V
G-| + G2 + G3
G-\
A3 + R 2 R 3
R2R3
R-\
y =
=
ELECTRICAL ENGINEERING
8
Networks
Solutions of Linear Networks
General: There are special methods which allow the calculation
of unknown voltages and currents in a network more easily than
mesh or node analysis, e.g.:
Use of the Superposition Theorem: In a general network, let all
voltage
and current 2 sources be successively applied to the
network, compute the voltages and currents caused by each
source acting alone.
1
'
'
• The remaining voltage sources are short-circuited.
• The remaining current sources are open-circuited.
The complete solution is the sum of these partial solutions.
General procedure to compute Vx in a general network with voltage sources V ... Vx and current sources I ... /
S 44
S45
r
Vx = a
r
-V + a -V +... + a v -V v +
1
,
1
^xaO+^xal
^xbO+^xbl
\X
|^
;
,
|P^
xbn
Computation of the partial solut.
S46
S47
where /
.
.
.
Vy = 0, with Vq * 0, and /
=0, with /q 4= 0, and V
1^
S48
a -V + a : V : +b
-I
VX a0+^xa1+^xbO
S49
Equivalent networks for computation of each partial solution:
V 4=0; V-i = 0; I = V = 0; VA * 0; 7 = V = 0; V, = 0; 7 4=
s50
/n*
R< Ro
S51
Required voltage
(Cf.
V, -
s 48)
Explanations cf. S 9
(£ +
£
7 >
R
.
^^
i\1
/\2
R
/R
continued on S 9
ELECTRICAL ENGINEERING
Networks
Use
of Thevenin's
theorem: Consider a general network convoltage
and current 2 sources. It is required to
compute the voltage Vx across resistance R in branch AA'.
This may be achieved by replacing the rest of the network
by an equivalent voltage-source V, and resistance R..
taining
1
'
'
,
«j5
'jSjlcV
» Q|*
Produce to compute R and V
:
x
t
Remove the branch AA' from the network.
is the resistance between A and A'.
/?i
V\ is the voltage at AA'.
Note:
If
/?,
is
known, V may be computed using V[ = /sc R r
which flows when A and A' are connected.
Isc is the current
Hence:
s52
/<=,/?;
sc
R + Ri
i//?i + 1//?
Computation of R
Computation of /S(
Ri
ft
4
rf
Equivalent circuit
"
*.
A
-$ + £ + <
Hence: V
s53
V
Using s52:Kx = (-^
V
+
,
-
/?
1
4c «i
1
/o)
1//?!
1
1/^ + 1/^2
+ UR 2 + MR
cf. result
of S8, s51.
Explanations:
ap-ay
'
2
'
coefficients
voltages
of
currents
voltage source
with internal
resistance
current source
which are determined by the
resistors in the network
R\
ft
=
{
ELECTRICAL ENGINEERING
10
Resistor combinations
Transformation of a
delta to a star-circuit and vice versa
^10 ^20 + ^10 ^30 + ^20 ^30
^30
10
*10 *20 + /?irj- R30 + R 20 #30
^20
120
•
S
54
s
55
s
56
i? 12
-
'
R,o-Ri
'
/?23 + R)2 + R-\3
R23' R-\2
R23 + R^2 + R^3
^10 ^20 + ^10 ^30 + R20 R30
R23' R-\3
'
l
R-\o
30
#23 + #12 + ^13
Potential divider
Potential dividers are used to provide reduced voltages.
—
I
#V
#1 Ry #2#\
/?2
s 57
,/?l/?2
s
58
"*"
"*"
For applications, where Vy has to be
approximately proportional to s. the
condition R y > 10 (R + R 2 ) has to be
A
satisfied.
s:
distance of sliding contact from zero position
voltage
divider
¥x
ELECTRICAL ENGINEERING
11
Resistor combinations
Applications in electrical measurements
Extending the range of a voltmeter
- 1
RM
=
s 59
£&
Extending the range of an ammeter
A.
s
60
max
'Mmax
CD-
Wheatstone bridge for measuring an unknown resistance R x
A slide-wire Wheatstone bridge may be used for measuring resistances of between 0.1 and 10 6 ohms. The calibrated slide
wire is provided with a scale
reading a/ (I- a). The sliding
contact is adjusted, until the
detector current /b is zero.
Then
s
_a_
R*
61
R
s
62
l-a
Rx
and hence
=
R ~—
l-a
Wheatstone bridge used as a primary element
In many types of measuring equipment Wheatstone bridges
serve as comparators for evaluating voltage differences.
R A sensor resistor,
:
riation
of
portional
which
to
the vais
pro-
the quanti-
ty x to be measured (e. g
temperature,
distance,
angle etc.)
R 2 zero value of /?i
Approx. the relation applies
:
s 63
vM ~ AR~ x.
internal resistance of the measurement
«
ELECTRICAL ENGINEERING
12
Electric field
Capacitance C of a capacitor
S
f
a
£
C
64
•
r
A
—
Quantity of electricity Q (see s 8)
Electrical work
cv
Wc
65
i<
Wc stored in an electric field
Capacitors connected in parallel
Where capacitors are added in parallel,
the total capacitance C increases.
C
66
=
Ci + C2 + C 3
C3
Capacitors connected in series
Where capacitors are added in series,
the total capacitance C decreases.
s
67
C
c3
C3
C2
1
c2
Capacitance of two coaxial cylinders
V o
s 68
s
69
2
£r
£
'•
:
A
:
a
:
r-[
:
r2
:
/
:
•
Jt •
£o
-r
r2
(see Z 22)
relative permittivity
absolute permittivity
plate area (one side)
thickness of dielectric
radius of inner cylinder
radius of outer cylinder
length of cylinders
£ Q = 8.85 x 10" 12
As/(Vm)
ELECTRICAL ENGINEERING
13
Electro-magnetic rules
s
Deflection of a magnetic needle
70
The N-pole of a magnetic needle is attracted by a magnetic S-pole
and repelled by a magnetic N-pole.
Fixed conductors and coils
Magnetic flux about a current-carrying conductor
Assuming a corkscrew were screwed in the direction
of the current, its direction of rotation would indicate the direction of the lines of magnetic flux.
s 72
Magnetic flux within a current-carrying coil
Assuming a corkscrew were rotated in the direction of
the current through the
the direction of its
axial motion would indicate the direction of the lines
of magnetic flux through the coil.
coil,
^
—
"Z-
~
N
Movable conductors and coils
73
Parallel conductors
Two parallel conductors carrying currents of the same
direction attract. Two carrying currents of opposite
-)(-
direction repel each other.
S
74
Two coils facing each other
Where two coils positioned face to face carry currents
of the same direction, they attract, where they carry
currents of opposite directions, they repel each
J^l
t^. s
**c^-i
J^~f
Machines
75
S 76
N
t=J-S
other.
S
N
Right Hand Rule (generator)
Where the thumb points in the direction of the magnetic flux and the middle finger in the direction of
motion, the index finger indicates the direction of
current flow.
Left Hand Rule (motor)
Where the thumb points in the direction of the magnetic flux and the index finger in the direction of
current flow, the middle finger indicates the direction of motion.
r?i
ELECTRICAL ENGINEERING
Magnetic field
S 14
Quantities of magnetic circuits
Magnetic flux
s
NI
77
(see also s 11)
S
Magnetic induction (flux density) B
R
s 78
H li
T
H
(see also s 12)
Inductance L
s
L
79
Nf = *A - f
=
(see also s 13)
For calculation of L see also s150 through s 156
Magnetic field strength // (Magnetising force)
s
80
//
-»-
=
=
(see also s 14)
t*rt*o
Magnetomotive force /
n
F
s 81
;
NI
-
2 F;
=
(see also s 15)
Magnetomotive force F\
s
82
(see also s 16)
Reluctance S
s
S
83
-^
=
I
=
(see also s 18)
H H A
r
Energy W m stored in a magnetic field
s
W m = 1 ^/0 = ±LI 2
85
Leakage flux
L
Part of the total
is
thus
ful flux
s
86
lost
flux <P leaks through the air and
desired effect. &\_ is related to the use-
magnetic
for the
<V Hence the leakage
coefficient is:
j
total flux
_
<p u
For symbols see S 18
useful flux
(1.15... 1.25)
ELECTRICAL ENGINEERING
15
Magnetic field
The magnetic field and its forces
Force F m acting between magnetic poles
In the direction of the magnetic flux
a tensile force Fm occurs:
B2 A
1
fm
'
2"
Mo
Forces F\ acting on a current-carrying conductor
A conductor carrying a current / encounters
force
over
F|
its
length
/
perpendicular
a transverse
the lines of
to
magnetic flux:
s
89
=
F,
Bl I
When applied to the armature of a DCmachine, the moment is:
s
90
Af
j
1
=
£z
^-0
2
a
I
Jt
/
(p
:
:
flux per pole
Induced voltage V\ (induction law)
Where a coil of N turns and resistance
magnetic flux <P that varies with time,
R\
is
threaded by a
an open-circuit voltage
s
92
V,
=
N -ry1 (see also s 11)
induced across its terminals. This
voltage causes a current through an
is
external load resistor Ru.
voltage induced by
motion
duction
dicular to flux
s 93
s
conductor loop
erator armature
in magnetic field
(p
V
{
= (D<Pn
sin cot
IdB
94
s 95
rotation of gen-
rotation of
conperpen-
of
Voltage V\ due to self-induction:
Continued on page S 18
V;
= L-6i/6t
For symbols see S 11
ELECTRICAL ENGINEERING
16
Alternating current
General terms relating to alternating-current circuit
Sense of phase angles
In vector diagrams arrows are sometimes used to
represent
phase angles. Here counter-clockwise arrows are taken positive,
clockwise arrows negative.
Example:
sense of phase angles
A--
96
s 97
s
<P\- <Pz =
=
<Pi
\
360° =
<?2
Peak values (see also s 1)
Current i and voltage v of
an alternating current vary
periodically
usually
maximum
are
with
time
sinusoidally.
values
i
t,
The
and v
peak values. At
an
angular
frequency
co = 2
/ the angle covered in time t is:
called
,-r
s
98
a = to t = 2nft
Hence at this time
s
3
99
the current is
=
100
the voltage is
= 6-sin (cot)
i- sin (cot)
Root-mean-square (rms) values
These are used for practical
i
•
sin a
v-sin a
calculations
indicated by meters
generally
s 101
/
=
/eff
-
102
V
=
veff
-Vi/w
s 103
Vt/'-«
and
are
usually
for sine waves
7
=
v
=
With these values the relation P = V I
ternating current, if cos cp = 1 (see s 115).
=
/eff
vM
V?
=
^
also applies for
al-
continued on S 17
/
ELECTRICAL ENGINEERING
17
Alternating current
continued from S 16
Phase shift, phase angle q>
Where different kinds of
load (resistance, inductance and/or
capacitance) are present in an alternating-current circuit,
a phase shift between current and voltage occurs.
vector diagram
waveforms
S 104
i
s\n(o)t - (p)
Q factor, damping factor tan 6, loss angle S
The Q factor of a circuit has been defined by:
s
Q
106
^**
=
Here w is the peak value of the energy stored
and VVyp the loss energy dissipated in one period.
in
the circuit
The reciprocal of Q factor is called damping factor
s
106
tan <5= 1/(2
For a choke
combination
125 and s 128)
(s 126 and s 129)
(s
((5 is
and
for a
this
definition
the loss angle)
capacitor-resistor
results
in
the
simple relations:
s 107
Q= ltan<pl
s 108
6 = 90° - \q>\
s 109
I
tan 6
=
=
=
MQ = 1/ltan
IVw /V b
I
7W /
b
l
l
<p\
(for series connection)
(for parallel
connection)
For formulae regarding tan cp see S 19 and S20.
Formulae s 138 and s 139 applicable for resonant circuits are
not so simple.
B
ELECTRICAL ENGINEERING
18
Alternating current
Basic equations for single phase alternating current
Impedance
s 110
Admittance
Z
Y
see S19 and S20
= 1/Z
s 111
Voltage across impedance
V
=
s 112
Current through impedance
/
s 113
Apparent power
Reactance
s 115 Active power
s 114
s 116
S
117
s 118
Reactive power
Power factor
flux in a coil
=
VI
=
Zsin<p
P
=
Q
=
VI -cos q>
VI -sin cp
=
P_ =
)2
^/p^+
S
^
J
V
z
X
cos cp
Alternating magnetic 1
IZ
=
=
I'Z
I'R
2
I
X
P_
VI
S
AAA Nf
s 95 continued (V\ due to self-induction)
Where the current i flowing through a coil changes with time,
the magnetic field caused by this current also changes. Thereby
a voltage V\ is induced in the coil. Its direction is such
that
it
counteracts
the
instantaneous
change"
of
current
(Lenz's law).
Symbols used on page S 15
119
absolute permeability
permeability
relative
(fi
=
Anx 10" 7 Vs/Am)
for vacuum, gases, fluids and most solids: n,
for magnetic materials take \i t from Z23
number of parallel paths through winding
length of magnetic circuit
number of turns of coil
number of pole pairs
number of conductors
resistance in
series
parallel
inductance in
equivalent circuit
of choke
parallel
c 19
W
ELECTRICAL ENGINEERING
.
Alternating current
Components, series- and parallel connections
carrying alternating current
e-
o
c
8
Ha
8
1
—
"
o
Ha
3
Q£
D
-~
T3
3
9
><
N
E
-t-
x°
OS
N
N
NJ
o
8II
V
Q. C
&
^
8-
8-
^o
<B
°-
co
a
—
CO
to
^
i-.
CO
^
^ (0°O
°*
o
JO
a>
V** I**-/
1
H
i.
%
\V
^
1
lit-
v° =
r
*r
S =5
— c
°
>
O i
<D
E
o
o
V
OJ
I
I
D
S
_
o
c
CO
CO
§!
"I
120
V-
S?
~> *•a> £Q)
> >
cd
S
8
Q.
Jo
o
S 121
3
S
o
22
4I
i.
:».
^|,
-l°
'3
Oc-E.*
A
r-
3
CD
o
> o
> *- °
CO 3 CD
'^
<0 "O
CD C CD
—
—c w
OJ
o
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si? I s s€|i
5
il||l l»
sii>3
1-,
1
^
co
CD
co
C
a)
No.
-i°
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-C O)
—
c
-
-|
*ft*
•U>
co
n
1
^nt^
a.
8
>*
-°
^r
-A
-1
1
O
c
a>
c
o
8,
J5
i^
rsT
-A
*
^
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^/
2±
Ix,
i^
CO
CO
o
--.
2 2
E
8-
o
CO
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V
ov
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o
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o
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o
II
CO
o
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CD
ov
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o
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o
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3
C
C
O
o
a:
ii
N
CD
CO
CO
CD
+
ii
Q.
o
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i
ii
<D
CNJ
r«
'"la
I^J
-J
o
Ik
3
124
5r
co
!
S125
I
^
ELECTRICAL ENGINEERING
20
Alternating current
continued from S 19
1
<K
^ 8 £*
^ "o
re
oc
g>
to
a>
^o re|
sz —
CO
c
°
^ „ „>
9 o
o
cc
<»
o
T3
CO
0)
—<D C
J5
— i3>. ^
re
i_
*£
re
a ro
© .E
E
o re
5 $ re
Jrexo
X3
.:=
g
|S.2
*~
^
re
° 852
-r,
V
*'5
© _ o_
> a
ore
£ ore
o.
re
re
re
OT
0)
S126
^
re
1 > $
n
o -o
re
*-
ra
o 3
.1 *1 ^ o
.re
£ a. + a
w
—
."K
c 5 &
re 2 3
° c © c
<d
.E
c
i
eg
'5
ro
Q.
>
O .Q
S127
S128
s129
s 130 131
ELECTRICAL ENGINEERING
21
Alternating current
Resonant circuits
series-
parallel-
|
resonant circuit
symbol and
see s123
general vector
see s127
diagram
Ac
vector
V'VR
diagram
at resonance
Wc
I = Io
k
resonance
condition
-
(o r
co r Lc
a) r
resonant
frequency
2
LH C =
2ji
C
wfLpC =
1
yL R C
1
2k V l p c
where line frequency/ = /r resonance occurs
,
current at
It
=
resonance
V b = K L - Vc =
at
(p
Q factor
loss angle 6
from
s 140
Qr =
tan <5 R
at
Rr
(o r
J_ =
Gr
r
R*
LR
tan St
300-10 6
c_
resonant
period
2it
(D r Lf
J_
0p
co r
wavelength
RP
Q P = (O CR P
CR R
(O r
CR
r
m
frS
/r
s 141
7C =
/b
=
V^r C
2ji
V^p C
Tank Circuit
A parallel resonant circuit has its maximum impedance Z max at
resonant frequency. Therefore
its
currents of this frequency.
142
RR C
for symbols see
S 18
it
acts
as
and current
a
rejector
for
ELECTRICAL ENGINEERING
22
Alternating current
Alternating-current bridge
AC bridges are used to determine capacitances and inductances.
For balancing the bridge variable capacitor C2 and resistor R 2
are adjusted until the sound in the low resistance headphone K
reaches its minimum or vanishes. The following circuits are independent of frequency.
measurement of
capacitance
inductance
EH
R*
^3
Cx
C2
Rz
R* R
tan 6*
Rx<oCx
Determination of an unknown impedance by measuring the voltages
across this impedance and an auxiliary resistor:
R
S
146
Pwz
s
147
cos<pz
s
148
«
s
149
"
=
-CZh
V 2 -Vn 2 -Vz 2
2R
—
A/vz
Vr
e
z
h© 6 &4
-4
select auxiliary resistor R such that V R «
I
Vz
I
loss angle, see S17
<5 X
unknown capacitance
R2
unknown inductance
a- known resistances
unknown resistance of coil or capacitor
I
|
:
.
calibrated adjustable capacitance
unknown impedance (inductive or capacitive)
ELECTRICAL ENGINEERING
23
Alternating current
Inductance L from impedance and resistance
Calculating /. from impedance and resistance
s
150
Pass an alternating current (J = I/A « 3 A/mm 2 through a coil and
)
measure the terminal voltage V, current /, active power P:
s 151
Z =
impedance
resistance
L- ±^*-W
s 152
Calculating /. for a toroidal coil
s
153
2:r
r.
Calculating L for a square coil
armatures must be circular
D
s 154
S
s
155
156
<
inductance
1.05
1
> 1
£ 3
1.05
m
V
HH
\u
2
^/V
VV (-)
m
\u
^H
'
values become unreliable
1
|iH
=
10" 6
^
thickness of winding
cross section of wire
width of coil
external diameter of wire and insulation
mean diameter of coil
internal length of armature winding
157
mean length of armature winding (/ m =
number of turns
+ jra)
/
circumference of coil cross section
ratio a
s
158
:
b
degree of loosing of turns
(/3
=
ab
.
ELECTRICAL ENGINEERING
24
Alternating current
Non-magnetic coils with specified inductance L
High frequency coils
D
formula
u
here:
s
159
s 160
,j
<1
ep--igf) w-**
>1
©*«"
-
2VA^
4d + «) Vf[
Low frequency coils
Assuming that
- 1
s 161
and
s 162
V -
975^
s 163
a
i.±V?
D = u,
then
Calculation of number of turns
\ of a coil
From cross section
s 164
TV
« ab
Ar
« RA
From resistance
s 165
Using reference coil
Position unknown coil of N x turns and reference coil of N
turns at short distance on closed
separation gap
lrcn core
iron core. Magnetize core by alapplied
to
V
ternating voltage
e
magnetizing
voltages
coil
Ne
Vx and V
.
Measure
using high
impedance voltmeter. Then
s 166
«.£
For explanation of symbols see S 23
rt
SB.
^
**>
ELECTRICAL ENGINEERING
25
Alternating current
Hysteresis
Remanent-flux density B
A residual magnetism of flux density B
remains in the iron core, after the external magnetic field strength H has been
removed.
r
T
Coercive force H c
The coercive force H c has to be applied to
reduce the flux density B to zero.
W
Hysteresis work
H
The energy
H dissipated during a single cycle of the hysteresis loop is equal to the product of area of the hysteresis
loop w H and core volume V Fe
w
:
Hysteresis power PV h
WH f = W H V F ef
Eddy currents
According to the induction law alternating voltages are also induced inside an iron. core. Depending on the resistivity of the
core iron these voltages cause induction currents called eddy
currents. They are kept small by lamination (making up the core
of thin metal sheets, which are insulated from each other).
Core losses (iron losses)
Core losses per unit mass pp e
They are the combined hysteresis and eddy-current losses per
unit mass. They are measured at a peak induction B = 1 T = 10 kG
or 1.5 T = 15 kG and at a frequency / = 50 Hz and are then
denoted? 1.0 or P 1.5 respectively. For values see Z 24.
Total core losses P Fe
=
m Fe
:
mass of core
I
x
:
F1.0
B
L.
T
50 Hz
m Fe (1 + x)
addition for punching ridges etc. (0.1 ... 1.0)
ELECTRICAL ENGINEERING
26
Alternating current
Choke coil
Choke coil used as a dropping impedance
It
is
used
in
an ac
circuit to
reduce the
line
voltage
V down
to a value Vv for a restitive load with minimum losses.
S 170
choke
Z D -y]R* 2 + (coL R
total circut
z
-
impedance of
S 171
-
-
V(Rr + /? v
2
)
f
+ (a >L R
f
*-±V(l-VPl -(^v + ^r)
a rough calculation of Lr neglect the unknown resistance Rr
of the choke. After dimensioning the choke 7?r is known, and Z
In
may be determined exactly. Check Vv by
s 173
V„ =
VRr
Z
and repeat procedure, if necessary.
Choke of constant inductance without core
t
Dimension according to S 23. Make preliminary assumptions
regarding values r2 /n (toroid coil) or D/u (straight coil).
In
case of unfavourable results repeat procedure. Determine
resistance of choke according to s 26.
Choke coil of constant inductance with iron core
The
iron
core essentially serves
for guiding the magnetic flux and
air gaps
should incorporate as many single
air
gaps
6^
should be
as possible. These
filled
with
insulating
layers and should not exceed 1 cm
in length.
The m.m.f. required to
magnetize the core is neglected.
Peak values of H and B are used
for calculations.
inductance
The variation of
winding to be
distributed on both legs
L R my be expressed
continued on S 27
ELECTRICAL ENGINEERING
27
Alternating current
continued from S 26
terms of the
of inductance
in
S
174
_
_
gL
maximum relative current-depending variation
I^Rtot
--^r
.
A
—= —
fe fi fe d
1
1
.
(
^R
H Fe fe
l
gL
ti
AL
If
repeat dimensioning with greater
gL > gL requ
smaller B Fe at unchanged product Af e x B Fe
,
+ 1
and
Af e
.
Dimensioning. Given: Lr, /, g Lre qu. ^Leff or /effl then the
preliminary
final
dimensions are
S
s
effective
175
cross section
of core
176
A F e= VK /
with
-
I eU
VLe „'
VLe „
ef(
2ji/Lr
number
s 177
V
of turns
cross section
s 178
of air gap
take AfQ from standards
or determine a and b by
\A Fe = 0.9 a b ~ A F e
I
^Lefl
.
4A4f8 Fe A Fe
Ai= ab + 5 cm (a + b) [A L= ab + 5{a + b)6A
abnN n
2
s 179
length
total
of air
s 180
gap
single
s 181
diameter
of wire
s 182
6
=
by
=
u
nL B -5N 2 n {a + b)
< 1 cm
d'/n
'•^
Aw
l
= din <
cm
1
Use next standard values for
d, d a including insulation
cross section
of winding
length of limb
(5,
1.12 da
=
2
N
%
to determined from dimensions of core
s
section and A w
Choke coil of current-depending inductance
This type of choke employs an iron core without an air gap. It
is used only for special purposes, e. g. as a magnetic amplifier.
power coefficient of choke
» 0.24 cm 4 /VA for air-cooled chokes
sb 0. 15
c
m 4 /VA for oil chokes
core section see S 26
core section increase values by 75%
preliminary current density for air-cooled choke J' = 2 A/mm^
for
|a o|
2
k
for oil choke /'— 3 ... 4 A/mm
core induction (take approx. 1 ... 1.2 T)
strength in core corr. to
according to material employed
field
flFe
t0
be
from
taken
number of single air gaps, increase reduces stray flux
resistance of winding according to s 26
resistance of choke includ. core losses (Kr ~ 1.3 Rc u
mean length of magnetic path through iron
)
Z
23
N
ELECTRICAL ENGINEERING
28
Alternating current
Transformer
Designation of windings
distinction by
function in circuit
nominal voltages
(direction of power transfer)
winding with
higher
|
input
lower
high-end
output
winding
nominal voltage
low-end
winding
secondary (index 2)
winding
primary (index 1)
|
|
Nominal values (index n)
s
183
s 184
rated power
nominal transformation ratio
5N
1
=
=
V 1N -/ 1N
=
^1 N ' ^20
=
^ N -/2N
1 2 N / ^1
By the rated secondary voltage V2 h we mean the open-circuit
secondary voltage (V2 n = ^20). not the one at nominal load.
Core losses Pf e and open-circuit measurements
In.
The core losses /Ye only depend on primary voltage
V-\
and
frequency/, not on the kind of load.
s
185
^10
=
^Fe
Core losses Pf e and nominal transformation ratio ii are determined
by open-circuit measurements (see circuit diagram: secondary
open, values provided with index o). The primary current's
resistive component /RFe covers the core losses, its reactive
component is the magnetizing current 7 m The copper losses are
negligibly small. The core losses Pf e a r © required for calculating operational power dissipation and efficiency.
.
continued on S 29
ELECTRICAL ENGINEERING
29
Alternating current
continued from S 28
Copper losses Pqu and short-circuit measurements
1.2
7.2
Pcu depends only on the primary current /i
and is determined by short-circuit measurements (see circuit diagram, values provided with index k). With
the secondary shorted, the primary voltage is adjusted to a
value V-ik, which causes the rated currents to flow. V-|« is so
small that /RFe and I m are negligible. Then the short-circuit
primary power Pi« is equal to the rated copper losses PcuN of the
total
transformer at rated currents. P-|« is required for calculat-
ing operational power dissipation and efficiency.
s
186
"
^1 K
^Cu N
The values measured are used for calculating the relative shortvoltage v^, which, for bigger transformers, is always
indicated on the name plate:
= 100(V 1K /V 1N )%.
circuit
s
187
The following
s 188
may be determined
quantities
diagram:
= ^r/^1 N
^Cu
L = V L /co/ 1N
Operating conditions
For calculating the operational secondary voltage
V2 for a given load all
secondary quantities are
first computed into those
of
an equivalent transformer having a transfor-
;
cos <p 1K =
Vo = i-V
\
U
=
hn
simplified
simplified
vector diagram
_/ 1^.
Rv
L
Vrv
Vl
Jl
Vi
I
1
>
Load-dependent variation AV of V£
(approximation for i'« = 4%)
s 190
AV = K 1K -(cos<p 1K cos<p2 + sin<p 1K -sin<p2) /2 //2n
s 191
« V 1K -COS(g?iK-<P2 )-/2/^2N
Secondary voltage V2
V2 = V, - AV
'
VR /Vi
equivalent circuit
mer ratio of i = 1 (index '):
s 189
using the vector
V9 '/i
J
ELECTRICAL ENGINEERING
30
Three-phase current
Basic connections
Star
VD
s 192
^M?
v,
*—
1=1
s 193
\v
^ .
uW
Ph
^Y
1
f
Delta
S 194
V - V,P h
s 195
/
1
=
V3
/ Bhh
P
Measuring three-phase power
Load balanced
with neutral point
(star connected)
—•- I
from
mains
rr\
v
.
^s5i_l
conn< sction
without neutral point
(delta connected)
-*- I
L\
to
from
L2
load
mains
(w)
L3
196
total
'° ad
3
{c
power
to
h
JLJl
"1
N
s
ȣi
.
x
\r
""
simulated neutral
V3V-/-cos(p
3/ly
Load unbalanced (Two wattmeter method)
For delta connected without
neutral
point.
(Also
for
from
/.,
~^
balanced load without neutral
V\
point).
s 197
total
/ ph
/
L
N
line current
:
:
,
L2 L3
,
:
outer conductors
neutral conductor
Fwph
:
l3
P, + P 2
power
phase current
:
^=^r
active power of one phase
yph
V
:
phase voltage
:
line voltage
ELECTRICAL ENGINEERING
31
Three-phase current
Reactive and active power, power factor
(for symmetrical load)
s 198
reactive power
q
= y/2'v-I-s\x\
s 199
active power
p
=
200
power factor
s
<p
y/3V-l-cosq)
P
V3V-/
Power factor correction
(for inductive consumers)
General
to power factor according to current rate, usually
cos q> = 0.8 ... 0.9. Adjust large consumers separately and directly and small consumers centrally to main or subdistributors.
Adjust
Calculating the required capacitor power
Calculate power factor cos cp as above, use wattmeter (see
connection in S30) or a current meter to determine P.
capacitor power
Q
=
(tan q>,
inherent consumption of condenser
Pc
«
0.003 Q
- tan q>2 ) P
Table (numerical)
cos q
tan q
cos q
0.42
2.161
062
1
0.44
2.041
0.46
1.930
0.48
tan q)
cos q
tan q
cos q
tan q
265
0.81
0724
91
0.64
1.201
0.82
0.698
92
456
426
1.828
066
068
1
0.50
1.732
0.52
1.643
1.138
83
0.672
0.93
0395
078
0.84
0.646
0.94
0.363
70
1.020
0.85
0620
0.95
0.329
0.72
0.964
0.86
0.593
0.96
0.292
0.54
1.559
0.74
0.909
0.87
0567
0.97
0.251
0.56
1.479
0.76
0.855
0.88
0.540
98
0.203
0.58
1.405
0.78
0.802
0.89
0.512
0.99
0.142
0.60
1.333
0.80
0.750
0.90
0.484
1
0.000
tan <pi or tan q) 2 can be calculated from the above table, cos q-\
representing the required power factor and cos q- 2 the consumer
power factor.
ELECTRICAL ENGINEERING
32
Motors
Direct-current machine
(motor and generator)
General
c
203
moment constant
s
204
rotational source voltage
C M <Pa> = 2nC M 0n
s
205
torque
c M <Pia
s
206
armature current
s
207
terminal voltage
s
s
208
speed
s
209
internal power
--fh
± (V ~ Vq)
K,±'aKa
V + /a Ra
*'
**'
2jrC M <*>
= M; m =
Pi
to generator
s 210
*»
Vq /a
Pa =
v/,,
mechanical power supplied
by motor
s 211
r)Vl
x
Shunt motor (for circuit diagram see S33)
%
Easy starting, speed is fairly independent of load and, within
certain limits, easy to regulate.
Series motor (for circuit diagram see S33)
Easy starting with powerful starting torque. Speed depends
greatly on load. When running free may become unstable.
Compound wound motor (for circuit diagram see S33)
Operates almost like a shunt motor.
ensures a powerful starting torque.
number of armature pairs
number of pole pairs
2
:
Ra
:
Main
circuit
number of conductors
armature resistance
magnetic flue
*'
+ motor
- generator
")
winding
- motor
+ generator
\
ELECTRICAL ENGINEERING
D. C.
33
machines with commutating poles
motors
generators
counter-
counter-
clockwise
clockwise
S202
—
1
L
id!
S203
^
II
(I
[]
J
t@
S204
Id
II
I
ELECTRICAL ENGINEERING
34
Motors
Three-phase motor
Speed
At a given frequency / the speed
of pole pairs p.
s
215
is
determined by the number
speed
ns
= J-
=
™l-s-
p
p
.
-U
mm
Switching
Where both terminals of each winding are accessible on the
switchboard, the three-phase motor can be connected either
in star or in delta.
phase voltage
in star
s
in
delta
216
A 400/230 volt motor operates with its nominal values of current,
torque and power, when connected to
s
217
V = 230 V in delta, meaning
s
218
V_ m 400 V
=
V = 400 V in star, meaning Vph - 77=;
.)i,
= 230 V
VS
V3"
Vph = V
= 230 V
Star-delta connection
Higher powered motors usually operate in delta. Yo avoid excessive inrush currents, particularly in relatively low current
networks, the motor is started in star and then switched over
into delta. If, for instance, a 400/230 volt motor is connected
in star to a 230/135 volt network, it is supplied with only 1/|/3
times its nominal voltage.
Induction motor
The
rotating field of the stator causes voltage and current to
be induced in the armature winding. Due to slip the rotational
speed of the armature is about 3 to 5% lower than that of the
rotating field; it remains almost constant under load.
Synchronous motor
Requires direct current for excitation and is synchronized
with the speed of the rotating field by means of an auxiliary
squirrel-cage armature. Can be used directly as a generator.
ELECTRICAL ENGINEERING
35
Transformer switch groups
Switch groups generally used for transformers
type
sign
key-
switch
numb
group
switch diagram
PV
SV
PV
ratio
SV
Threephase-output transformers
s
219
M3
D d
W^iw
X
IU
l^2V
X
W^-^IW
2V
IU
IV
s
220
s
221
s
222
s
223
D
z
n/-^iw
2U /^-2W
2W-{
d
5|
224
s
225
IV
IW
LU
N2
IW
2U2V2W
iu^iw
m
?>N2
urn
2U2V2W
2W^2U
6
iu*-^iw
IV
s
226
Y
y
6
2V
2W*^2U
iu^iw
2V
IV
s
227
D
z
6
s
228
D
y
11
s
229
D d
11
s
230
Y z
11
2W~^j2U
lU^IW
*2V
\—2W
iu^Aiw
r
V?N2
Yin,
N2
2N,
YJN2
s2
iu-^ IW
D d
1l
111
£A?3
IV
AN
2
2U 2V 2H
1/1/1
IU
IV
s
IW
2
IV
IV
IV
2U 2V 2V»
IU
IV
IW
y/j
IU
IV
IW
IU
IV
IW
LU
2U 2V
2V
2W
JL
N2
3 2W
ri/i/i
2U 2V jW
M/3
L
Wn
2
IV
11
\>2W
w^^iw
2U 2V
IV
2W
*V2W
lUy^lW
N2
2N<
Single phase-output transformers
1
i
PV: pri nary voltage
D
SV: sec ondary voltage
d
numbers
\
i
*
2.2
1.2
2.2
1
1.2
Key
2.1
2.1
;.;
1
s 231
|Y|
star
dSlta
are
N2
|y|
|
used
to
calculate
zig-zag
1
the
z
phase
angle
(= key number x 30°) between the primary and secondary voltage,
e. g. for Dy5 the phase angle is 5 x 30 = 150°.
Note: Use the framed switch groups for preference.
ELECTRICAL ENGINEERING
S 36
Measuring instruments
The most important measuring instruments
*
~?1
o.§
CO
CO
il
l
1
co
CD
w
-1
1
c
o
~<"
1
i
1
2 «
3 CO
E
i
« CO
1
2 3
£
n
i
CO
11
^8
O
o
co
kJ-
to
to
CO
CD
CO
>
co
co
*l
1
~l
1
Si™**
^
CO
(D
i_
O — x_S ~c o l|
2- 5
>.-
CO
co
CO
!
o
c
CO
CD
c
-3
CO O"
.2 -a
CO
CO
"
CD
*co
CO
a)
E "°
2 Too
> &
o
o E
£
"co~
CD
a3
0>
.
©
co
Q>
2 5
4*
E CD
o «
2 o
w
|o
o
o
X
b §
2
X
o-
Soco
co
CO
"CO
"O
o
CO
3
C
c
c X
O N
of
b«
mov-
CO
fixed
contact
springs
tension
h- O)
permanent
current
c
o
O CO
of
other
2
b
a
of
instruments
close
leads
moment
T-
in
bands
coil
in
current
and
fixed
coil
non-uniform
2
a or
bo
b
en
o>£ E
'>
1 5 o
o
E
E
c
O)
magnetic
with
c
'>
O
E
in
counter
spiral
-o
c
coil
leads,
E c
o
magnet
CO
CD
co
2
moving
as
-o
c iS
°b>
bands
coil,
CO
in
o
o
co
co
o
o
V
^
O
uniform
o
2
moving
CO
Q.
'>
coils
co
CO
CO
c£
heater.
Thermocouple
moment
O c
T3 O 3
c
o
CO-D O
°8
moment
springs
— ©
ing
permanent
tension
and
Thermocouple
magnet.
to
o
o
screen
.— fc
Q. O
field
CO
or
field
*l|
o
x <o E
application
each
—
thermal
without
serve
uniform
of
Q.
feeds
spiral for
Q.
°.E
moving
radial
to
current
counter
counter
field
c
o
>
CO
c 3
>
E
o §
CO
C 3
2 5
E
CO
is
E
O
CD
&
CO
CD
E
E
F Tl
o i. £
O
CD
Q-
ffl
in
c
o
b
CO
i
o
||
2 -g"
|
2 .2
b
CO
c
"co
« 3
T3 O
o t
co
O
-Q
E
CO
o Qj cs cH c #
OT
o
-Ih °7o
«-
CO
ELECTRICAL ENGINEERING
37
Installation
Current rating /z
PVC insulation, unburied copper conductors including overload
protection devices at an ambient temperature of 30 °C 1)
Nominal
cross section
1.5
mm 2
A
A
7 Z in A
7 n in A
7Z in A
7 n in A
10
2.5
50
/ n in
Group 2
Group 3
Nominal Cu-wire
diam. mm, approx.
:
2:
3:
18 26
10 2 20
108 135 168
80 100 125
>
80
1.1
1.4
1.8
50
35
103 132
80 100
7 Z in
Group 1
Group 1
25
16
2.3
2.8
3.6
129 158 198
100 125 160
multiwire
One or more single core cables laid in a conduit
Multi-core cables (including ribbon conductor)
Single-core cables in free space (spaced at least one wire-
diameter apart).
1
>The 7Z value decreases (increases) by about 7% per 5 °C temperature
increase (-decrease).
50 °C should not be exceeded!
2 ) For
cables with only two current carrying conductors an over-load
protection device. (L) or (gl_) with 7n = 16 A should be used.
Switches
single pole switch
two pole and cross-over switch
2 loads
1 switch
1 load
3 switches*)
PE
N
1-
._
XX
Nominal current
rating
of type
gL fuses and type L automatic
circuit-breakers.
Current rating of the cable. Also the maximum allowed value of
overload protection devices of type B, C or K (for / n iz ).
PE: Earth
Neutral
N:
Live
*)
each additional switch requires an extra cross-over
CONTROL ENGINEERING
Control engineering terms
Control
Control
a process by which a quantity called the controlled
variable (the quantity which has to be controlled) is recorded.
This controlled variable is then compared with another quantity, the reference variable. The controlled variable is then
influenced in such a way that it equals the reference variable.
The main characteristic of control is the closed action path
in
which the controlled variable is continously influencing
itself (for action flow see below).
is
Preliminary note:
Names and definitions
of the following terms very strongly
follow those given in the norm DIN 19226, version 2/1994.
Functions, Quantities and Symbols to describe the
behaviour of transfer elements and systems
Input variable u
The input variable u is a quantity acting upon the considered
system without being influenced by it.
Output variable v
The output variable v is a quantity of a system which can
be influenced only by itself or its input variables.
Delay time T, time constant T
The P-Tj element, the first order delay element,
is
a func-
tional unit with the transfer behaviour:
v(t)
where T
+ Tv(t) =
Kp u(t)
also called time constant.
the solution of this differential equation see J 4, J 9).
is
the delay time,
(For
Characteristic angular frequency co damping ratio #
The P-T2 element, the second order delay element is a functional unit with the transfer behaviour:
,
t2
v(t)
Here
(o
damping
+ (20/a>o) Ht) + (1/a)
2
)
v (t) =
K P u(t)
the characteristic angular frequency and
ration. (For the solution of this differential
is
# the
equa-
tion see J 5, J 11).
Eigen angular frequency cod
The eigen angular frequency co d is given by the formula below
in
which the characteristic angular frequency to and the
damping ration & are used.
t3
a> d
-
aJo-Vl-fl 2
Explanation of the symbols see T35
CONTROL ENGINEERING
Control engineering terms
Step response
The step response
progress in
time of the output
variable of a transfer element when a
step function is used
input
variable
as
the
is
Steady stc te/^TX Step response v
Step function u
(see fig. 1).
/
/
1
The equivalent dead
Tu
\l
value
Equivalent dead time Tu
J>7
defined
as the time between t
and the point of intersection
the
of
first inflexion of the
step response with
the x-axis
time
Transient value
\
tolerance
0\ ^ershoot
u,v
is
Tu
Tg
f
\
Trise
Settling time
Fig. 1
k
Step response of a
transfer element
(see fig. 1).
Build up time Tg
The build up time
Tc ,s defined as the time between the point
intersection of the first inflexion of the step response
with the x-axis and the point when this first inflexion
reaches the steady state value.
of
Overshoot vm
The overshoot vm is the widest deviation of the step response
from the steady state value.
Ramp response
The ramp response is the progress in time of the output variable when a ramp function with given rate of change is used
as input variable (ramp generation).
'<i
MO
ik
Fig. 2
r(t)
T
t
4
t
5
r
H
e(t)dt
Ramp generation
C/q
te(t)
T is the ramp time, e (t) the unit step generation:
r
£{t)
=
/
Ofor t <
I
1
for r
^
Unit step response
A step response related to the step amplitude of the input
variable leads to the related step response, called unit step
response h (f).
continued on T3
t6
CONTROL ENGINEERING
Control engineering terms
characterises the dynamic behaviour of the transfer
element. For unit step responses for the most important transh (t)
fer elements see T 14 to T 17.
Transfer Function F(s)
The transfer function F(s)
is the ratio of the Laplace Transform v(s) of the output variable and the Laplace Transform u(s)
of the input variable of a transfer element. For transfer
functions for the most important transfer elements see T 14
toT17.
Frequency Response F(jco)
The frequency response F(\co)
is the ratio of the pointer of
the sinusoidal output variable and the pointer of the applied
sinusoidal input variable of the transfer element in its
periodic steady behaviour in dependence or co or /.
Amplitude response F(co)
The amplitude response F{(o) is the magnitude of the frequency
response F{]co) in dependence of the angular frequency co.
Phase response arc F(jcu)
The phase response is the argument arc F(\(o) of the frequency
response F(\co) in dependence of the angular frequency co.
Frequency response characteristics, Bode diagram
The frequency response characteristics (Bode diagram) are obtained when the absolute value (logarithmic or in dB) and
phase response (proportional) are plotted together in dependence of co or the standardized angular frequenoy oo/oo^.
Corner angular frequency co n
The corner angular frequency (o n {n = 1, 2, 3
is that angular frequency co, where the asymptote line of the absolute
up or downwards - by
value in the Bode diagram breaks off
an integer multiple of 20 dB per decade.
.
.
.)
The action diagram
The action diagram
symbolic
is the
tions in a considered system.
of
illustration
all
opera-
Elements of the action diagram
Elements of the action diagram are the action
line,
the func-
tional block, the addition and the branching point.
Fig. 3
Functional
block
Action
line
u, w-i,
u 2 input stimuli
:
Addition
= ± u 1 ± u2
v
v
:
output result
T
Branching
point
CONTROL ENGINEERING
Control engineering terms
Basic structures of the action diagram
The basic structures of the action diagram are the series-,
parallel- and circle structure.
Rule for the addition in an action diagram
An addition has only one single leaving action
line
(output
variable).
Rules for the representation of a system by an action diagram
Each equation of the system is shown only once in the action
diagram.
A
negation (change of sign, reversion of polarity) must be
at an existing or at an extra summation point; it is
not allowed to hide in the coefficient of a block.
shown
In
the action diagram of a passive system no positive feed-
back occurs.
To have a
clear form of the final drawing of the action
diagram the shortest path (forward path) between input variable (upper left side) and output variable (upper right side)
should be a horizontal line.
Derivative elements should be avoided.
tions of the loop should be reordered.
To achieve this, equa-
Components of the control loop and its quantities
Fig.
trol
4 shows a typical action diagram of a closed loop consystem including its functional units.
Typical action diagram of a closed
loop control system
Controlled system
The controlled system
is to be influenced.
is
that part of the control loop
p-
*
which
CONTROL ENGINEERING
Control engineering terms
Point of measurement of the controlled variable, controlled variable x
The point of measurement of the controlled variable is the location in the controlled system where the value of the controlled variable x is obtained (see fig. 4). The controlled variable x is the variable of the controlled system which is recorded for controlling and is fed to the controlling system via
the measuring equipment, x is the output variable of the controlled system and input variable of the measuring equipment.
Formation of the final controlled variable, final controlled variable xA
The final controlled variable xA is a quantity which it is
the task of the closed loop control to influence. If easy to
obtain by measurement, then jca is identical to the controlled
variable x and will be fed back via the measurement equipment
to the comparing element. Only if it is not possible or
only possible with great difficulty to obtain xA will it occur
as an independent quantity beside the controlled variable*.
In fig. 4 - typical action diagram of a closed loop control
system - the formation of the final controlled variable a:a is
made by the controlled variable x, usually by attachment
to the controlled system. Here the final controlled variable
appears outside the control loop, the control of influencing
disturbance variables during the formation is not possible.
Example: Final controll. variable: Temp, of the contents of a pot.
Temp, of the hotplate.
Controlled variable:
The final controlled variable *A can also occur within the con-
system, that means within the control loop, m this case
controlled variable is formed via the final controlled
variable; influencing disturbance variables can be controlled.
trolled
the
Example:
Final controlled variable: Mixing ratio of two liquids.
Specific resistance.
Controlled variable:
Measuring equipment, feedback variable r
The measuring equipment is the sum total
of
the func-
all
recording,
transferring,
for
determined
elements
adapting and distributing the variables (see fig. 4). The
feedback variable r is the variable which results from the
tional
measurement of the controlled variable x.
The reference variable adjuster, reference variable w
The reference variable adjuster
outputs a reference variable
is
a
functional
unit
which
w derived from a user defined
target variable w* (see fig. 4).
The reference variable w is not influenced by the related closed
loop control; the output variable of the closed loop control
shall follow the reference variable in the given dependence.
Note: Goal and reference variable are very often identical.
CONTROL ENGINEERING
Control engineering terms
Forming device for the reference variable, target variable w*
The forming device for the reference variable builds from
a target variable *• - applied at the input - an output
reference variable w. This forming process ensures that
the reference variable w, or time derivatives of it will not
exceed critical values (see fig. 4). The target variable w*
is
externally defined and is not influenced by the considered closed loop system; the final controlled variable w of
the closed loop system should follow the target variable
in the given dependence.
Comparator, error variable e
The comparator produces the error variable e depending on
reference variable w and feedback variable r (see fig. 4).
e
= w - r.
Controlling element, controller, controller output variable \ R
The controlling element produces the output variable y R of
the controller using the error variable e from the comparator. The process ensures that the controlled variable x
in the control loop - even at the occurrence of disturbance
variables - follows the reference variable w as quickly and
precisely as possible. The controller consists of the comparator and the controlling element (see fig. 4).
Actuator
The actuator
is
a functional unit which uses the controller
output variable y R to form y. The variable y is necessary
for modulation of the final controlling element (see fig. 4).
Final controlling element, manipulated variable j
The
final controlling element is located at the input of the
controlled system and influences the energy flow. Its input
variable is the manipulated variable y (see fig. 4). y transmits
the controlling result of the system to the controlled system.
Final controlling equipment
The
final
controlling
final controlling
equipment consists of the actuator and
element.
Controlling system
The
has
controlling system is that part of the action path which
to influence the controlled system via the final con-
trolling
element.
Manipulating point
The manipulating point is the point of application of the
manipulated variable y.
Point of disturbance, disturbance variable z
The point of disturbance is where the externally applied
disturbance variable z affects the intended influence
closed loop control (see fig. 4).
in
the
CONTROL ENGINEERING
Quantities and functions
Quantities and functions to describe the dynamic
behaviour of control loops
Open loop transfer function F (s)
transfer function F {s) is the product of
transfer functions serial of a control loop or a loop.
The open loop
al
Example:
u(s)
:
9
^
D
y
t
M
F2 (s)
1
"«
.
F{
' 1 (W
F
(5)
= F,(5)
-F2 (j)
*J
Open loop gain V
The open loop gain V
is the value of the open loop transfer
function F {s) in the case when the Laplace variable 5 = 0.
This term is only applicable for control loops and loops
without behaviour. The higher the open loop again the more
precise the closed loop control.
I
Control factor H F (0)
The control factor F F (0) is given by
1
R F (0)
10
= 1/(1 + V )
Gain crossover angular frequency co D
The gain crossover angular
frequency co D is the open „, a
loop frequency where the | £
absolute value (amplitude) H 8.
of the open control loop <»o
is equal to 1.
Phase crossover angular frequency co
The phase crossover angular
frequency coK is the open
loop frequency when the
phase response of the open
controlled loop is -180°.
Phase margin
<5
The phase margin 5 is the
angular difference between
the phase response of the
open control loop at the
angular
crossover
gain
frequency (Dq and -180°. The
necessary sign change in
the control loop is not
taken in consideration.
Gain crossover angular frequency
c^
:
Phase crossover angular frequency
Fig. 5
Diagram of absolute value and
phase response (not logarithmic)
of an open controlled loop
CONTROL ENGINEERING
8
Quantities and functions
Gain margin e
The gain margin e
is the reciprocal value of the absolute
value (amplitude) of the open control loop at the phase
crossover angular frequency (o n
.
Time to reach lower tolerance r8tart
^start is tne time duration which begins when the value of the
controlled variable x - after applying a step function of
the reference variable w or a step function of the disturbance
variable z - leaves a given tolerance field of the controlled
variable and ends when it enters this field for the first time
Steady state
Overshoot
(see fig. 6 and 7).
„
|
esired value deviation
Desired value
Steady state value
t
Agreed tolerance
Tstart-
field
Variation in time of the
controlled variable after
applying a step function of
the reference variable w
7j
dead time
A
step function of the reference variable also produces a
step in the tolerance field of the controlled variable.
Overshoot xm of the controlled variable
The overshoot x m of the controlled variable x is the maximum (interim) deviation from the desired value during the transition from one steady state to another when a step function
of the reference variable w or of a disturbance variable z (see
Steady state
fig. 7) is applied
Overshoot
desired value deviation
/
Step response
Fig. 7
i
Tstart
Variation in time of the
controlled variable after
a step function of the
disturbance variable z
t
3
t
Time to reach steady state 7 end
Tend is the time duration which begins when the value of the
controlled variable x - after applying a step function of the
reference variable w or a step function of the disturbance variable z - leaves a given tolerance field of the controlled variable
and ends when it enters this field permanently (s. fig. 6 + 7).
CONTROL ENGINEERING
9
Rules
Rules to determine the transfer function
of the whole control loop
The complete transfer function
transfer element.
built
by using each individual
-^
F(s) = F,(s)-F2 (s)
is
Series combination
U(S)
^(S)
t11
—
v(s)
F2 (S)
-
Parallel combination
u(s)
1
12
—
—
*(s) -Zl
I
F2 (S)
..
v(s)
F{s) = Fi(s)+F2 (s)
—I
Feedback rule
u(s)
,(s)
K(s)
1
13
F^(s)
F(s)
1+F,(s)-F2 (s)
F2 (s)
-
Note: The sign of the denominator of F{s) is the opposite of the
sign at the addition point in the action diagram.
A "+" sign at the addition point means positive
feedback.
A "-" sign at the addition point means negative
feedback.
and/or F2 (s) contain sign changes, then negative (positive)
feedback will result when the number of sign changes in the whole
If F-i
(s)
loop is odd (even).
Special case: F2 {s) = 1 (direct feedback).
u(s)
»
p_ *
~-\
V(SJ
/"-,
' 1
/c)
(S)
F^{s)
I
1
14
F(s)
1
+ F< (s)
CONTROL ENGINEERING
10
Rules
Extended feedback rule
If there
are no addition points between the branches of an
action diagram, then the transfer function Fres (s) can be
determined very easily using the following formula:
1
^w-44--^^
+XF
15
u(s)
F0i (s) means
with FVre8 (j) = IlFv
oi (s)
1
loop transfer function of the single conloops in the applied action diagram; please
bear in mind that with positive feedback loops F0i is written
with a negative sign in the sum of the denominator of Fres
trol
loops
the
or
.
1
16
m
^Vres( 5 ) = n FVk is the product of all transfer functions of the
k - 1
transfer elements which lie in the forward path.
The overlapping of action lines does not
tion of the extended feedback rule.
affect the applica-
Example:
u(s)
kH f
vi (s)
r^9
HF s
v2 ( )
F R ,(s)
F R2 (S)
t
17
F
(s)= Jd^l ^
tres(S)
u(s)
Fw(s)-Fv 2 (s)-Fy3 (s)
1+F,7V2(^) •Fm (s) + F}n (s)-Fyn (s)-1FV3(S)-FR2(S)
_
Determination of the transfer function using the back
annotation method
Using this method start at the output v(s) and follow the
action diagram in the direction of the input variable or of
the reference addition point B. Determine and note the
Laplace Transform of each respective time function before
and after every transfer element. Finally at the reference
addition point B the transfer function Fres (s) can be determined using the known Laplace Transform at this location.
CONTROL ENGINEERING
T11
Rules
Example:
To describe the back annotation method the Laplace Transform
of the
1
1
18
respective time functions are determined at the points
to 6 in the following example:
X;(S) =X${S)+X2(S) =
xm (s)
v{s);
FV2 FV3
F\j9
^V3
•
Fv
v(s)
v(s)
Fw(S) L^QJ^Fy 2 (s)--Y^Fyz(s)
Fr,(s)
<f
1
19
FR2 (s)
x z (s) = ^-v(s)
x s (s) = FH2 -v(s)
xi(s)_
t20
1
*V1 (s)
1
FV2 (s)-Fsj 3 (s)
^
F
+
li£)l
v(s)
V; lis)}
At the reference addition point B there
is
the following
rela-
tionship:
t
u (s) - x® (s) = x% (s)
21
;
for x s (s) and x t (s) use the
values obtained above:
u(s)-FR2 -v(s)
t22
1
FV i(*) I^V2(^)-^V3(^)
^R1 (5)
+
Fv3 (s)
V(s)
The solution Fres (s) of this equation is the same as that already
found in t 17:
t23
Fresis)
=
Fyi (s)-Fv 2 {s)-Fy 3 {s)
v(s)
u(s)
1
+ Fv2 (s)
FR1 (s) + FV1 (5) FV2 (5) FV3 (s) FR2 (5)
•
•
•
CONTROL ENGINEERING
12
Rules
Rules for the normalized form of the transfer function F(s)
If
the normalized form of the transfer function is used,
the type and characteristics of the transfer element can
readily be seen. To transform a transfer function into its
normalized form, well judged expansion, bracket removal and
combining are necessary to ensure that nominator and deminator of the transfer function are polynomials of the
Laplace variable sor products of polynomials of s in which:
a)
no negative powers of s occur (meaning that neither denominator nor nominator of the transfer function may
contain a fraction where s occurs in the denominator).
b) the lowest power of s has a coefficient of 1
c)
and
there is no polynomial common factor.
Example:
t24
(1
+ ajS + a 2 s 2 +
° (1
+ fc s + & 2 *2 +
v{s) m
F:(s)
u{s)
.
.
.) (1
+c,s+c 2 s 2 +
-
•
•)
+<M + d2
2
.y
+
.
.
..)(.. .)
..)( )
or PID factor occurs, another form in which
this factor [\/(T„s) + 1 in the example] still remains,
Exception:
If
a
is
an approved normalized form.
F2 (s)
t25
1
PI
vis± =
u{s)
1/(Tn s) + 1 _ K P
„
p
1
+ Ts
Tn s
^+Tn s
1
+ Ts
From the table showing the most important transfer elements
the following type can be found for this equation:
(PI )
_T ^|-(PD)-T
1
1
.
The following table shows different types of normalized form:
Type of the
normalized form
Product
normalized form
Sum
normalized form
Representation
denominator and
nominator are
factorized form
denominator and
nominator are
in
Application
Bode diagram and
series stabilization
of control loops
Hurwitz criterium
written in sum
denominator as far as
Mixed
possible divided into
normalized form
factors. Nominator in
in the form of a sum
Determination of
the step
and ramp response
The product normalized form has precedence over the other
representations as the others can be determined when the
single factors of the product normalized form are multiplied.
To make reductions possible later on, terms in brackets
should be multiplied as late as possible - if at all.
CONTROL ENGINEERING
13
Rules
Example of the determination of the normalized form of a
transfer function
For the following action diagram the relationship:
F(s)-b(s)/Fk (s)
t26
is
to be determined
B
B
Fk
b
Fk
>B
bR
tys)
P
-bqR
mm
o
-%
(s)
fys)
1/(nB)
Fjs)
-6
F^(s)
1/(rm)
step by step
t27
t28
t29
F,(s) = F„ B (s)/b QR (s)
F2 (s) = b qR (s)/F k (s)
and F3 (s) = b R (s)/F k (s)
are determined; then
F(s) = B + F2 (s) +
t30
t
31
^1 (s) =
Fns(s)
b qR (s)
t32
F2 (s) _
F3 (s)
1/(1
A+M{nB)-M{rms 2
M(qRs)
b q R(s)
fkW
qRs
+ nrmBs 2
1
_
F, (s)
1
)
q Rs + ^1
W ?#*
1
+ nrmBs 2
2
nrmBs
qRs [1 + nrmBs2 + rmsl[qR
t33
t34
t35
F3 (s)
F(s)
bn(s)
Fk ($;
1
M(ms)
+R/{ms)
1
s
/?5 [1+mj/R]
1
= 2? +
+
nrmBs 2
qRs-[l + rms/(qR) + nrmBs2
Rs["\+ms/R]
This description of F(s) shows that the system which leads
diagram is a parallel combination built
action
this
to
from a P element (B), a -T-, element (first fraction) and a
l-(PD)-T 2 element (second fraction).
I
CONTROL ENGINEERING
14
Primitive transfer elements
First order delay element
I
dent if[er
Symbol in the
Structure
Equation
in the time domain
Examples
action diagram
Kp
v = KP
•
u
Proportional element
v= K, fudt
= K Ju dt+ v(0)
{
v =
Kr u
Intergral element
V =
M&
fvdt =
Kn
•
— L^—
U
u
K D -u
Derivative element
-M^
i
v[t)-u(t-T
t )
^cn^
Dead time element
KP
P-Ti
v +
KP
5
T v = KP
u
1/r
r^p^rr
T
P-T, element
Explanation of the symbols see T35
CONTROL ENGINEERING
Tl4
Primitive transfer elements
First order delay element
Unit step response
Equation h(t) =
Transfer
function F(s)
diagram
KP
"(t)|
KP
K
([h]
•
t
•
6 (t)
t
= Unit of h)
KD
"W'l
Kn-S
for
t
< T
x
;
1
for
t
> T
x
/»(t)4
-T,s
1
K 9 ^-e' n
)
a rt(t)
0.95Kp'
1
+ Ts
0.63KP
-
OT
37
Explanation of the symbols see T 35
CONTROL ENGINEERING
15
Second order delay element
Parallel combinational element PI
Identifier
Structure
Equation
Symbol in the
in
the time domai
action diagram
Examples
%
Kp
^o
P-Tp
79
'^v
+ tgL) 2 *-
= Kp-u
P-T,
KP
1/72
1/7i
t±ft±r
K\judt + Kp-u
MJ^
pi
1/Tn
df + u
Kp
Kf
Tn
with
rn = a:p /a:.
Kp/rn
Explanation of the symbols see T 35
o*
i
^_
CONTROL ENGINEERING
Second order delay element
15
Parallel combinational element PI
Unit step response
Equation h{t) =
Transfer
function F{s) =
diagram
*p[1 "^e^'-cos^ /-©)]; <od =a>Q \j\ - &
2
= Arcsin #
<6>< 90°
KP
KP (l+e"'rtanC:')
+
'con
s +
bi
\a)r
< #<1
a: p
Kp
n+20
3ti+20
5n+20
2<y<j
2£Dd
2wd
- [T,e^-T2 e^
1-
+ 7, *) (1 + T2 s)
(1
i
1
IW)
9>
l
Kp
=
^1,2
V^-D
JC^
.1 -
K
,/ xf
i
J"
>
*
1
7 lnK
Ks^/r,
1
tf,f +
Kp
^7^
Kp
Tn -s
(1
Kp =
K-1
T
1
InK
K-1
-
|T +7 2
'
K P (1 +=r)
W)
+1
+ V*)
Explanation of the symbols see T35
CONTROL ENGINEERING
Tie
|
ce~:'e-
Eca: :-
S.-rc
« :-e
= ::
aran
z-"
z
v =
"
KP u + Kq
= k p (ii + r„
-1—
-
=
-
•
•
u
u)
*
^i
—-
—
rlt—
-—
.
-&
f r
LjLa
= :
lit
-
hP^
—3^—1
—
r
l-T,
-ZZ-
-^—^—
D-T.
-
-~ZrExr ="= z~
z~
e
:
•
-
-
s
see ~ 2 r
/
CONTROL ENGINEERING
Parallel combinational elements PD,
T16
PID
Series combinational elements I-T-, and D-T-,
Unit step response
Equation h(t) =
Transfer
function F{s) =
diagram
* P + a: d<5(0
i
Kp + Kq
1
•
s
K P (1 + T^-s)
-
~^0
K^+ K + KD -s
P
K,f + K P + K D <5(0 =
for
<
—<
oo
r
K P \jr + l + Tv 6(t)}
;
1
*Pk I
jnk '*
hjnk -5)(i + rvk -5)
^nk
=
^vk
«
\
= ^rn (i-Vi-4r
v /rn )
V
*Pk ~ *,
•
*
'o
-Tn
f
7nk
A-,(f-r+ r-e~' /T
i I
)
W)
K\
s(1 + 7-5)
37K|TN
*~
f
7
^ D (1/7-)e-'
KD .
Kd-s
/T
im
T
1+7-5
0.37^
~^0
7-
"
f
Explanation of the symbols see T 35
CONTROL ENGINEERING
17
Series combinational element
Group combinational elements (PD)-"^ and (PID)-T.,
Identifier
Structure
Equation
Symbol in the
in
the time domain
Examples
action diagram
-/w
D-T5
v+ 2
Kp
= KP
(PD)-l,
rv
1
u + r¥ u
•
j
Kp T,,T
Kp TnTv
v +
I"
1
7 v = AT| u dJ J
+ Kp
K P u + Kp
•
li
(PIDM,
Kp
7-,rn ,rv
X
r.
+ u + rv u
•
r
=
—
Explanation of the symbols see T 35
^
i
r
CONTROL ENGINEERING
Series combinational element
17
Group combinational elements (PD)-Ti and (PID)-T-)
Unit step response
Equation h(t) =
Transfer
function F(s)
diagram
Mo
-0O)
"" 0,
of
,
^Dag--e
-sinfi\ J r
ATd-5
Q_
rt(t)
u
2<yd
+
Kp + *p
„
1
+ 7V -J
1
+ r-5
,|o_^j e-; =
1
- 7) -5
+
rv - r = rv
1
r-j
^^^-iJeT
Kp- KJ+ K^+ \KJ- K P + K D
Ts
+
\
KP (0.63 + 0.377v /r)
,''(t)
KpTv /7
j
TTT^
"(t)A
'
rn
2<y d
*
n rv -7Tn«
rn (i + r-5)
7-
,
a
2<u d
Kp7,/r
T
v
KP + K
Kp
2
5
(7*
=
5<
*
+ T'S
1
=
= a> \/l - # 2
6) = Arcsin
1+2^5+,,
•
o) d
;
= Xp /K,
;
Kp [0.37(^-+-k)+0.63]
Kp[l+— ln(1--^ + 7"nV
7"n
'n
,
'
7V = Xq //Cp
r* =Tn -T
Explanation of the symbols see T35
CONTROL ENGINEERING
18
Methods for determining stability
Stability of the control loop and calculation
for a controller
(for linear control
loops)
Definition of stability
Stability is attained after the alteration of the reference variable or the occurrence of a disturbance variable, when the controlled variable reaches a steady value.
Comment: When
a controlled variable reaches its limiting
values directly after switch on, the cause is very often a
wrong polarity connection of the feedback variable at the comparing element (comparator).
Methods for checking the control loop for stability
Assumptions:
* The reference or disturbance transfer function of the closed
control loop or unit step function are known.
* The loop transfer function of the open control loop is known.
1.
Hurwitz criterium
Only when the reference or disturbance transfer function of
the closed control loop is known in the form of a polynomial
can stability be determined using the Hurwitz criterium.
Stability is attained when the coefficients of the characteristic equation (when the polynomial of the denominator of the
transfer function = 0)
t
a + a-i s + a 2 s* +
118
.
.
.
+ a n sn =
satisfy the following conditions:
• All coefficients a v must be
* The coefficients
>
(see also D 9)
themselves have to fulfil special depend-
encies.
The conditions for equations up to degree 5:
coefficient conditions
Equation
1st degree
2nddegree
a
a
and a^
,
a-|,
a2
>
>
>
a2 - a3 a
2
a 2 a 3 - a3 a
3rd degree
4th degree
a-\
5th degree
A = a a2 a 3 a 4 + a
a-|
- a* a4 >
a 4 a 5 - a-, a 2 a 5
-
B — 3q d-\ a 4 85 + 3q a 2 a 3 a§ — ag a 3 a 4
--
1
D n _! = A-B>0
a-i
a?a A 2 >0
2
2
a5 >
a
For polynomials of higher degree see „Ebel, Tjark, Regelungstechnik, 6. Aufl. Stuttgart Teubner 1991, S. 38 ff."
continued on T 19
CONTROL ENGINEERING
Methods for determining stability
19
Advantage: The method
leads to a swift and accurate
statement concerning the stability of a given control loop.
Disadvantage: The Hurwitz criterium gives neither a statement as to the resilience of the control loop to instability nor
to the result of changes of its characteristics nor to its
dynamic behaviour;
for
these reasons other methods are
usually preferred.
2.
Reduction into single polynomials
Transform
the reference or disturbance transfer function
into a sum of single polynomials of 2nd order maximum (see
partial fraction expansion D 3):
In a stable control loop there are only stable transfer elements. These are usually pure P or delayed P elements and
delayed PD elements.
If
a
I
,
I-T-,
or l-(PD) element occurs the control loop
will
become unstable.
Advantages: In
the stable and the unstable cases the
evaluation of the transformed reference or disturbance
transfer function leads to a conclusion as to the degree
of stability or instability of the control loop. To gain
this information the transfer functions of all single ele-
ments have to be superimposed.
Disadvantages:
The effect of the introduction of a
defined control element and the knowledge as to which
of the characteristics must be changed to meet the required
behaviour of a control loop cannot be seen. After each
change to the controlling element a new calculation for the
arithmetical transition from the open control loop to the
closed control loop must be made.
3.
Nyquist criterium
Nyquist criterium states that the (closed) control
loop is stable when the frequence response locus F [\(d)
of the open control loop - in the sense of higher values
of the angular frequency a> - always has on its left the
plane. The greater the
in the complex
critical point -1
distance between the frequence response locus and the
the control loop is
more
robust
-1,
the
point
critical
under the effects of unexpected variations in the charac-
The
teristic data.
A measure of how close the system is to becoming unstable
is
given by two characteristic values:
continued on T 20
CONTROL ENGINEERING
20
Stability
Choice of the controlling element type
The phase margin <5 (see (T 7) and the gain margin e (see T 8).
The determination of the actual
J*lmF
(jco)
n
values for both of these characteristics and the realization of
their required values by insertion of a suitable controlling
element is usually made via the
Bode diagram.
Re F Ga>)
Fig. 8
30°
Recommended values for the phase margin
Recommended values for the gain margin
e: 8 dB
60°
16 dB
<5:
(corresponds to the factors 2.5 ..
Advantages: Examination
.
6.3).
the transfer function F (s)
of the open control loop - especially the related frequency response F (\(o) (replacement of s by jw) leads very easily to a statement of stability and shows
the resilience to instability - particularly when there are
unexpected changes to the characteristics of the control
loop. Also the effects of changes to the type and
characteristics of the controlling element - using a simple
series insertion in the control loop - together with
the resulting dynamic behaviour of the control loop
can be seen very easily when this method is used.
of
Choice of the type of the controlling element
General
In most control loops the controlled system and the measuring equipment are together of type (PD)-T n meaning a series connection of a number of PD elements and delayed elements. The rate times Tv = K D / Kp of the PD elements are
always essentially < than the delay times of the delay
elements: In real systems by a factor > 10.
The most important controlling elements
PI
loops only the P
(PD)-^ and
control
In
linear
(PID)-^ elements are of real importance.
Characteristics of a control loop with a P or (PD)-^
controlling element
;
,
,
When there is an influence of disturbance variables applied
between
controlling
Point of
element and point of
u
Controlling _
measurement only a """ ps.
"T^T element
is
accuracy
finite
This
accuracy is given by
the
of
the
value
control factor R F (0).
possible.
measurement
Controlled
system
Comparator
Measunng
element
continued on T 21
CONTROL ENGINEERING
Graphical determination of a controller
Characteristics of a control loop with a PI
controlling element
21
or (PID)-T 1
Complete
compensation for the influence of disturbance
between controlling element and point
of measurement is possible. If the controlled system contains an
element with no negative feedback disturbance
variables applied between the
element output in the controlled system and the point of measurement, there will be
complete compensation even when there is no factor.
Note: Disturbance variables applied between the point of
measurement and the output of the controlling element can never be compensated for.
applied
variables
I
I
I
Graphical determination of a linear controller
based on the Nyquist criterium
General
The procedure is carried out via the Bode diagram. For this
one Bode diagram both the construction of the series connection of the controlled system and measuring equipment and
also construction of the controlling element are necessary.
The Bode diagram of the whole circuit is found by the addition (multiplication) of the amplitude and phase responses
of the single series transfer elements (see T 22 and T 23).
This is possible due to the logarithmic nature of the amplitude response after the conversion to dB.
For the graphical illustration semi-logarithmic paper with
4 decades on the x-axis should be used.
Procedure:
* Determination of the area of the angular frequency w, for
which the Bode diagram should be made; plotting of all break
points of interest.
factors of the frequency response of type I, P, D,
P-T 1( PD and P-T 2 with attenuation # < 1 are permitted
(see T 22 and T 23).
* After extraction of the
factor, PI elements are converted
into an l-PD structure and PID elements with T n /T v > 4 are
converted into an l-PD-PD structure.
* Only
T
t,
I
the serially occuring Integral - (K,),
Derivative (K D ) action coefficients are
one single action coefficient.
•All
or
Proportional-(Kp)
summarized
into
« Representation of amplitude response.
* Representation of phase response.
* Completion by the controlling element.
The tables on pages T 22 and T 23 show the Bode diagrams for
D, T P-T 1t P-T 2 and PD elements. These diagrams are used
P,
for the' determination of the amplitude and phase responses.
I,
t
,
CONTROL ENGINEERING
Bode diagrams for basic elements
22
and P-Ti elements
Amplitude response F(co) =
Phase response Arc F{\o)) = q> =
Diagram (Ampl. logarithm.)
Diagram (Phase linear)
Symbol
KP
|
-oo
F |dB
< n < +oo, integer
1Q
K 40
n
1Q
n+1
10
n^2
10
n^3
1Q
,
n+4
>{oA
20
10
n
10
n+1
10
n+2
10
n+3
10
n+4
M
-oo
Kr M(D
F ldB
<i
-oo
< n < +oo, integer
n
10
n+1
10
n+2
10
n+3
10
n 4
l ,„
-oo
< n < +oo, integer
-oo
< n < +oo, integer
Kn-O)
-oo
c
n <p/1°
10
dB/decade
n F ldB
< n < +oo, integer
-90
-90"
< n < +oo, integer
10
n
10
n+ 1
10
n+2
1Q
n +3
1Q
n+4
\co\
io7io n+1 i(f
1/K
D
=(o
+2
io
n+3
io
n+4
D
" "
10- 2 10" 1
10" 2 10- 1
10°
10
1
10
P-T.
40
Xp|dB
20
~^f
"V/ ~¥+ *
/2
lO^IO'^ 10°
H
i- 20
"
\o°/l0
1
lO
2
^
- Arctan {Toy)
,.,.
F |dB~"
"iv.
x
n ;4
2
Kp/\/l +_{T0)f
l
-0)
-T
" "
"
"p/~i»
F
l V
dB/dec.
dB/d«
Tto
/2
1Q
1
W
2
^
CONTROL ENGINEERING
Bode diagrams for P-T2
23
and PD element
Amplitude response F((o) =
Phase response Arc F(\(o) = <p
Diagram (Ampl. logarithm.)
Diagram (Phase linear)
Symbol
< 0<1
(*)'
Arctan
KP
i
V[ -(*)T
90
20%+
" (*)'
c
,
< #< 1
co
>
P-T;
Kp
1
I
(T <*>/ wn
V
P dB
Vl-2 9
2
yKP 2 + (K D -cof
= K P V1 + (rv w) 2
with
Arctan (Tv w)
Tv = /Cd/A'p
PD
i^ldB
1
40
20 dB/dec.
L<P/1°
90
45
7>
10" 2 10" 1
|
10°
Iff
io
IO"
2
-90
2
1
IO"
10°
10
1
10
5
CONTROL ENGINEERING
24
Bode diagrams for (PD)-"^ and
(PID)-"^ element
Amplitude response F(co) =
Phase response Arc F[]co) = cp =
Diagram (Ampl. logarithm.)
Diagram (Phase linear)
Symbol
K P + (K D o))
2
V
2
+ (Tea) 2
1
Arctan {Tw co) - Arctan (Tco)
(T,coy
(Tea)'
Kpyl
1Q- 2/p- 1
ip°
-20dB/dec
(PD)-L
l
10
1
lO
2
^^
T>TV
>-j
\<plf
1
45
T.
10
-2
io 0/ io
10' 1
—
—71
1
1
10
2
'
I
T<T,
^^^
10" 2 10' 1
7-v*
10°
10
TJT
1
\10 2
TJT
Arctan 7> —— ]- Arctan (To)
Tn
w>o
o<r /r <oo
-
K P 2 + (KpCD- KJco) 2
1
+ (Tco) 2
g Wi*P>-i /(7»r
_
P
V
1
+ (To;
co\
L
n
= Arcsin
v
7>-i/(7»
Vl+[7>-1/(7»]
- Arctan (7a)
o < rn / r„ <
F
r>4r
idBi
r k >7
(PIDJ-T,
^) K ldB/7
/
V
K|_
i
1
_J
_
,
2
1_
°°
w>o
CONTROL ENGINEERING
25
Amplitude response
Method for the depiction of the amplitude response for the
whole circuit
Preliminary note: the break points of the amplitude response are marked by arrows at the top of the paper.
An arrow with one head pointing down characterizes an incline of -20 dB/decade which is made by a P-Ti factor
(see T14) at point 1/7.
An arrow with two heads pointing down characterizes an incline of -40 dB/decade which is made by a P-T2 factor
(see T15) with # < 1, P-T 2 factors with $ > 1 are divided into
two P-T-, factors.
PD-factors (+ 20 dB/decade at point 1/TV ) with positive incline are characterized by upper-pointing arrowheads.
After estimation of the variation of the amplitude response
within the 4 decades of the angular frequency ca, the scaling
should be chosen so that the area of interest is shown with
maximum possible resolution; the expected maximum value of
the amplitude response will define the origin.
Next is the determination of the value of the amplitude
response at the left margin of the drawing. If the series
connection to be illustrated contains an
or D factor,
the starting value and starting incline is determined by
or D, otherwise by the P factor. The starting value is
found when using Fl dB = 20 Ig (K^co) (I factor), F| dB = 20 Ig (K D (v)
(D factor) or Fl dB = 20 Ig K P (P factor) see T22 and T23.
I
I
Comment: The physical
omitted
here
and
units
of
K
it
K D or K P should be
added afterwards when evaluation
is
carried out.
of the amplitude response up to the first
- 20 dB/dec. (I factor), + 20 dB/dec. (D factor) or
(P factor). The asymptotic curve of the amplitude response
drawn from one break point to the next; for this the
is
above changes corresponding to the entered arrows have to
be considered. The Tt factor has no influence on the amplitude response. Finally, corrections are made at the break
points: when the P-T-, factor is used the asymptotic curve
is corrected at the points co E /2 and 2 co E by -1 dB and at
the corner angular frequency co E by -3 dB.
The
break
gradient
is
When the PD factor (inverse to the P-T
corresponding corrections have to be
point o) E = 1/TV upwards (i.e. + 3 dB)!
factor)
1
made
at
is
used the
break-
the
Around the break point of a P-T 2 factor with & < 1 the asymp2
+ (2 dw/(o 2 ].
totic curve is corrected by -10 Ig [1 - ((o/co
)
)
CONTROL ENGINEERING
26
Phase response
Representation of phase response
The phase response of the whole circuit is determined by
the addition of the single phase responses of the serially
connected transfer elements or frequency response factors.
The single factors contribute the following values (see also T22
and T 23).
Factor
Phase response
0°
P
- 90°
+ 90°
I
D
Tt
-T (o
P-Ti
-Arctan (Tco)
Table A
t
PD
Arctan (Tv co)
P-T2 with # < 1
PI =
I
- PD
Comment:
- 90° + Arctan {Tn to)
whilst
Tt factor has a
creases with co.
the P factor results in no effect, the
very significant contribution which in-
Remark: after drawing up the constituent factors into a sum,
the values of the phase response are then determined e.g.
by using the memory of a pocket calculator. Usually the
phase response of the Tt factor must be replaced with the
factor 180 /n in order to obtain a common result io degrees.
Finally the scale and the origin for the illustration of the
phase response have to be chosen so that the area of greatest
interest is shown with the maximum possible resolution. The
phase response is drawn on the same diagram as the amplitude
response. The scaling of the y-axis is made on the right hand
margin of the drawing.
Determination of the controlling element
Problem: The result must be such that the requirements
phase margin <5 and gain margin e are satisfied.
of
Referring to T 20, figure 8, this means that the frequency
response pointer of the final open control loop around the
point -1 has to satisfy two requirements: at the
critical
angular frequency with an amplitude of 1 (gain crossover
angular frequency) the phase distance to the negative part
of the real axis must be at least <5 and at the angular frequency where the phase is -180° (the phase crossover angular frequency) the amplitude has to be at most Me.
continued on T 27
CONTROL ENGINEERING
26
Phase response
Representation of phase response
The phase response of the whole circuit is determined by
the addition of the single phase responses of the serially
connected transfer elements or frequency response factors.
The single factors contribute the following values (see also T22
andT23).
Factor
Phase response
0°
P
D
- 90°
+ 90°
Tt
-T co
P-Ti
-Arctan {Toy)
+ Arctan (Tv co)
I
Table A
x
PD
P-T 2 with #<1
PI =
I
- PD
Comment:
"-"(t^S 2 -"-)
- 90° + Arctan (Tn (o)
whilst
T factor has
t
a
the P factor results in no effect, the
very significant contribution which in-
creases with a).
Remark: after drawing up the constituent factors into a sum,
the values of the phase response are then determined e.g.
by using the memory of a pocket calculator. Usually the
phase response of the Tt factor must be replaced with the
factor 180 /ji in order to obtain a common result in degrees.
Finally the scale and the origin for the illustration of the
phase response have to be chosen so that the area of greatest
interest is shown with the maximum possible resolution. The
phase response is drawn on the same diagram as the amplitude
response. The scaling of the y-ax\s is made on the right hand
margin of the drawing.
Determination of the controlling element
Problem: The result must be such that the requirements of
phase margin 6 and gain margin e are satisfied.
Referring to T 20, figure 8, this means that the frequency
response pointer of the final open control loop around the
has to satisfy two requirements: at the
critical point -1
angular frequency with an amplitude of 1 (gain crossover
angular frequency) the phase distance to the negative part
of the real axis must be at least 6 and at the angular frequency where the phase is -180° (the phase crossover angular frequency) the amplitude has to be at most Me.
continued on T 27
CONTROL ENGINEERING
29
Gain margin condition
Realization of the Gain margin condition
Determ. is very similar to that of the phase margin condition:
- Determination of the phase (pt = -<jp r * - 180
j
.
- Determ.
phase crossover angular frequ. a)ne where
in the phase response q> ((o)*) of the series connection of
v
the controlled syst. and measuring equipm. the phase <pe is
reached; at this frequ. the phase response (p(a>) of the
final determ. open control loop goes through the -180° line.
- Determ. of the position of the break points when the above
mentioned ratios are applied to the value of the phase
crossover angular frequency determined from the diagram.
- Determ. of the incline m F of the inverse amplit. response
Fr^) -1 *) of the controlling element around the phase crossof the
,
over angular frequ. (onE An asymptotic section with this
incline is laid through the point of the amplit. response
.
Fy {(o)*) of the controlled system and measuring
value of
equipment at the phase crossover angular frequency a>nt
In the distance f| dB - above this asymptotic section - the
of the controlling
inverse amplitude response FR (<y)~
element is plotted and evaluated in the same way as done
before for the realization of the phase margin.
.
-
1
-
In
element,
the case of a (PID)-T| controlling
follow
the
same method as (D.
Remark: It is recommended
in addition to determine the
gain crossover angular frequency <w De This frequency lies
at the intersection point of Fyico)^ of the series connection of the controlled system and the measuring equipment
-1
and the inverse amplitude response Fr^o)) ** of the controlling element.
.
Choice for one of the two determined controlling elements
Not only the realization of the phase - but also of the gain
margin condition leads to the determination of a controlling
element. The element with the smaller proportional - or integration action coefficients will be chosen. Disregarding
some rare exceptions, by this choice the controlling element
also meets the other determinaton requirement.
Comparison between controlling elements determined by the
choice of different ratios
Using the choice of different ratios the best behaviour of
the determined controlling element is given for the element
with the maximal value of the gain crossover angular frequency (o D This angular frequency - determined for the open
control loop - is a quantity indicating the speed by which
loop
control
.^. .^^
is .w«w..«n.
reached.
closed ^^..
mc umaou
vaiue in the
K .w
trie final
nnai value
the
.
t
*>
Explanation for index R and index y see footnote on T 28
CONTROL ENGINEERING
30
Examples
w
3
CONTROL ENGINEERING
T31
Examples
Example 1: Determination of a PI or P controlling element
Problem: A PI controlling element with given Tn = 10/ a> D
Tn =
;
has to be determined for a control loop with
the series connection of controlled system and measuring
equipment of P-T 3 behaviour (P-T r T2 ). The phase margin 6
shall be at least 40°, the gain margin e at least 3.16
(corresponds to 10 dB).
"lO/toji
4
F 5 = rlA =
i
)
2
2
(1+10sec-s)[l+0.8/(5sec)s + 1/(25sec )5 l
y(s)
Solution: According
T13
T17
the character= 5 sec
istic data K Py = 4; T = 10 sec; (o
are read out of
the transfer function F(s) of the series connection; next
according to T 22 the amplitude response Fy and the phase
response <p of the series connection have to be constructed.
to
to
first
-1
y
The next steps for determination are given as numbers (in
T27...T29. These step
circles) at the left margin on
numbers are also plotted in the Bode diagram on T30 at
the corresponding positions. The following results are found
at the step numbers:
6°
<p R (o) D ) = (p R (io n ) = - 90° + Arctan [(10/eo D ) w D ] « 3 q> b = -180° + 40° + 6° = -134°
-1
4 a) D6 = 3.4 sec
5 Due to the given Tn = 10/ D -> 1/rn6 = 0.34 sec" -» Tnb = 2.94 sec
1
8 K PR& in dB = 16.5 -> /C PR& = 6.68 -> K IR6 = 2.23 sec"
11 <p e = -180° + 6° = -174°
1
12 o) ne = 4.8 sec"
-1
1
Due to the given rn = 1 0/o) n -» 1 / Tne = co ne /1 - 0.48 sec
-* 7nE = 2.08 sec
=
14
F
1
16 K PRe in dB = 9 -> A: PRe = 2.82 -> ^| Re = 1.32 sec"
1
18 (D Dz = 1.2 sec"
21 AT PRe < K PRi) -> K PR = 2.82
2
•
1
m
the required controller has the characteristic
Tn = 2.08 sec.
from
this
data:
K PR = 2.82;
list
The speed by which the controlled loop using this controller
-1
reaches its final values is characterized by cj d = 1.2 sec
.
The determination of a P controlling element is done in a
similar way compared with that for the Pl-controlling element:
The result at the step numbers:
= -180 o + 40° = -140°
3
<p 6
4
8
= 3.4 sec
K PR6 indB = -16.5-»K PR8 = = 6.7
1
o) Db
11
<pt
12
a»j,
= -180°
E
= 5 sec"
1
16 /Cp RE indB = -10^K PRE = 3.16
continued on T 33
CONTROL ENGINEERING
32
Examples
o
1
o
CM
O
CO
o
1
I
"a-
o
in
7
CONTROL ENGINEERING
33
Examples
18
ci>
1
De = 1.3 sec
^PRe < ^PR6 ~~* ^PR " 3.16;
V = K PR -K Py = 3.16-4 = 12.6->K F (0) = 1/(1 + V = 7.3%.
The steps 1, 2, 5, 6, 9, 13, 14 are not applicable
for the determination. Steps 7 and 15 are only made
within the Bode diagram!
Comment: The control factor /? F (0) is the factor which
reduces a disturbance applied between the output of the
controlling element and the measuring point. With a
P-controlling element disturbances are not fully compen-
21
)
sated as in the cases of a Pl-controlling element.
Example 2: Determination of a (PID)-T., controlling element
Problem: For a control loop with the same series connecexample 1 a (PID)-T-i controlling element shall be
determined when the following values are given.
i/r6 =
6w D6
i/rvk6 = (» D6 /4
i/r„ M =o) D6 /i2
i/re 6a>„ e
i/rvkE = <o ne /4
^/Tnke =Q) ne /^2
Phase and gain margin shall have the same values as in
tion as in
example 1.
Solution: Amplitude and phase response can be taken from
example 1. The single determination steps following the
numbers (in circles) at the left margin on T 27 ... T 29 must
be carried out; these step numbers are also plotted on T32.
The results at the step numbers are:
2 <p R6 = -90° + Arctan (12 co D /co D ) + Arctan (4 coq/wq) - Arctan [(1/6) (a) D /a?D )] = 62°
3 <p b = -180° + <5- <p R = -180° + 40°- 62° = -202°
•
-1
4 co D6 = 6.0 sec
-1
Tnkb = 2 sec;
5 1/Tnk6 = a> D6 /12 = 6/(12 sec) = 0.5 sec
1/7"
= coq/4 = 6/(4 sec) = 1,5 sec 1 7vk6 = 0.67 sec;
vk6
-1
=
-» Tb
28 msec
1/r6 = 6 w D = 36 sec
8 AT PkR6 indB = 12^ AT PkR6 = 4
=
5.34
9 AT PR6 = (4/2) -2.67
;
;
7n6 = (2 + 0.67) sec = 2.67 sec; 7"v6 = 2
11
<p e
•
0.67 sec/2.67 = 0.5 sec
=-180 o -<p R1 =-242°
12 £Djt e = 10 sec
=0.83 sec" 1 7nk£ = 1.2 sec.
13 1/rnke = ft>nE /12 = 10/12 sec
1
1
1/7vke = (onz /A = 10/4 sec" = 2.5 sec" 7"vkE = 0.4 sec;
1
Tt = 17 msec
1/7e = 6 (onz = 60 sec"
= -20 dB/Dek.
14
F
16 AT PkRe indB= 17^ K PkRE = 7.1
1
A- PRE = 7.1/1 .2 (1 .2 + 0.4) =9.47; 7nE = (1.2 + 0.4) sec - 1 .6 sec;
7VE = 1.2-0.4/(1.2 + 0.4) sec = 0.3 sec.
n
18 ct) DE = 6. 2 sec
continued on T 34
21 K PRb < Kp Re -> K> R = 5.34
;
;
;
m
•
^^HBBCU*
CONTROL ENGINEERING
34
Setting rules
Continued from T 33
The desired
teristic data:
(PID)-Ti controlling element has the
PR = 5.34; Th = 2.67 sec; T = 0.5 sec.
charac-
K
The speed
for the desired control loop to reach the
-1
value is characterized by co D = 6.0 sec
final
.
Proven setting rules for P PI and PID controlling elements
For controlled systems with one first order delay element
and a dead time element - that means controlled systems
without part or factors ZIEGLER and NICHOLS recommend
the following characteristic data for the above mentioned
,
I
I
controlling element types.
AT Py
,
7y and 7*ty of the controlled system are known:
Table C
conTy
^PR
troller
TV
P
K Py Tty
•
n n
°J
PI
Ty
3.3 rty
AV Py T
lty
.
-
1"
PID
^
2 7\
A'py • Tty
q.5 rty
y
Characteristic data of the controlled system are unknown:
Table D
con-
K PR
troller
Tn
rv
P
0.5 A'pRc-jt*
PI
0.45 tfPRcrK*>
0.83 7crit **)
PID
0.6 KpRcrit*
0.5 7crit
•'A'pRcrit :
K PR value when
the control loop.
permanent
Oscillation period
in
,)7
crit
:
tion.
">
0.125 rcrit ">
oscillation
occurs
in
the case of permanent oscilla-
1
CONTROL ENGINEERING
35
Abbreviations and formulas
Types of transfer elements
D element:
Derivative element
D-T 2
Derivative element with 2nd order delay
:
Integral element with 1st order delay
l-T :
t
PID element: Proport. integral derivat. elem.
P-T, element: 1st order delay element
P-T 2 element: 2nd order delay element
element: Integral element
I
element: Proportional integral element
PI
D-T,: Derivative element with 1st order delay
(PD)-T, element:
PD elem. with 1st order delay
P element: Proportional element
(PIDJ-T, elem.:
PD element:
T element: Dead time element
Proportional derivative element
PID elem. with 1st order delay
t
List of symbols for Control engineering terms
:
Error variable
:
Incline of the
r
:
Bode diagram
Feedback variable
u
:
v
:
Output variable
vm
:
Overshoot of the
e
mF
Tu
Equivalent dead time
r¥
Rate time
rnk ,(ryk ): Reset time (rate
:
amplitude response
in
the
:
ries
v
time)
of
the
step function of
in
the
se-
PID element
.
Tnkb>(TrtbY- Rese ' time ( rate time
unit
a transfer element
w
representation
with Tn > 4 r
Input variable
)
in
^e se-
PID element
ries
representation
with
Tn > 4 Tv determined according
of
the
to the phase requirement
:
Reference variable
iv*
:
Target variable
x
:
Controlled variable
ries
representation
xA
:
Final controlled variable
with
7n >
xm
rnk£1 (Tvke): Reset time (rate time) in the se4
rv
of the
PID element
determined according
to the gain margin requirement
:
Overshoot of the controlled variable
>
:
Manipulated variable
e
z
:
Disturbance variable
6
:
Gain margin
:
Phase margin
F(\(o)
:
Frequency response
F(s)
:
Transfer function
Phase of the series connection - controlled system, measuring equipment - at
F((o)
:
Amplitude response
the gain crossover angular frequency a>
ip
F (jw): Frequ. response of the open contr. loop
FQ (s) Transfer fund, of the open contr. loop
F ((o) Ampl. response of the open contr. loop
Fn (w) Ampl. response of the controllg. elem.
Fy (o) Ampl. response of the series connec:
:
8
<p e
and meeting the phase njargin 6
Phase of the series connection - control-
:
led system, measuring equipment - at the
:
:
:
tion of the controlled
system and the
:
Derivative action coefficient
K,
:
Integral action coefficient
%
uring equipment
:
i?
:
K Pk (a>): Proport. act. coeff. in the series re-
(o
present, of the PID elem. with
action
coefficient
> 4 Tv
a)
of
the
co
of
the
7"
n
controlled element
KPR
Proportional
:
action
controlled element
T
Tg
Th
Tn
Tend
b
<y D
coefficient
c>
E
:
Angular frequency
:
Characteristic angular frequency
:
:
:
:
:
Eigen angular frequency
Gain crossover angular frequency
Corner angular frequency
Gain crossover angular frequency when
realizing the phase margin condition
:
Build up time
:
Half-life period
:
Reset time
u) x
Time to reach steady state
o)
Time to reach lower tolerance
Damping frequency
(i)
Delay time
:
:
w D6
:
7,,,,,:
:
(a)):
K P Proportional action coefficient
Control factor
R F (Q)
Km (o)): Integral
phase crossover angular frequency wn
Phase response
Phase response of the open control loop
((o) :
Phase response of the controlling elem.
qk
%[(o): Phase response of the series connection
of the controlled system and the meas-
measuring equipment
KD
<p(a))
D(
Gain crossover angular frequency when
realizing the gain margin condition
nt
:
:
Phase crossover angular frequency
Phase crossover angular frequency when
the gain margin e is met.
CHEMISTRY
Ui
Elements
atomic
element
symbol
atomic
mass
in
element
symbol
mass
144.240
neon
Nd
Ne
nickel
Ni
58.71
u
in u
aluminum
Al
antimony
argon
Sb
121.75
Ar
39.948
As
Ba
Be
74.9216 niobium
Nb
92.906
137.34
nitrogen
9.0122
osmium
N
Os
190.2
bismuth
boron
bromine
Bi
208.980
oxygen
O
B
10.811
palladium
106.4
Br
79.909
phosphorus
Pd
P
cadmium
Cd
112.40
platinum
Pt
195.09
caesium
Cs
Ca
132.905
potassium
K
39.102
40.08
praseodymium
Pr
140.907
226.04
arsenic
barium
beryllium
calcium
carbon
cerium
26.9815 neodymium
20.183
14.0067
15.9994
30.9738
chlorine
CI
35.453
rubidium
chromium
Cr
51.996
ruthenium
Ra
Rh
Rb
Ru
cobalt
Co
Cu
58.9332 samarium
Sm
150.35
63.54
scandium
44.956
Er
167.26
selenium
Sc
Se
copper
erbium
C
12.0112 radium
Ce
140.12
rhodium
18.9984 silicon
fluorine
F
gadolinium
Gd
Ga
Ge
157.25
silver
69.72
72.59
Au
He
H
196.967
1.008
In
114.82
gallium
germanium
gold
helium
hydrogen
indium
iodine
1
Si
102.905
85.47
101.07
78.96
28.086
107.870
sodium
Ag
*
Na
strontium
Sr
87.62
sulfur
180-948
tellurium
S
Ta
Te
thallium
Tl
204.37
126.9044 thorium
thulium
192.2
Th
232.038
Tm
168.934
118 69
4.0026 tantalum
22.9898
32.064
1276
iridium
Ir
iron
Fe
55.847
tin
Sn
krypton
Kr
83.80
titanium
Ti
47.90
lanthanum
La
138.91
tungsten
W
183.85
lead
Pb
207.19
Li
6.939
magnesium
manganese
Mg
Mn
24.312
uranium
vanadium
xenon
mercury
Hg
200.59
zinc
U
V
Xe
Y
Zn
238.03
lithium
molybdenum
Mo
95.94
zirconium
Zr
91.22
u
:
54.9381 yttrium
atomic mass unit (1 U = 1. 66 x 10~
27
kg)
50.942
131.30
88905
65.37
CHEMISTRY
U2
Chemicals
Chemical terms
chemical
trade
chemical
formula
name
acetone
acetone
acetylene
acetylene
ammonia
ammonia
(CH 3 2
C2H2
CO
•
)
NH 3
NH4OH
ammonium (hydroxide of) ammonium hydroxide
C6 H 5
NH 2
aniline
aniline
bauxite
bleaching powder
hydrated aluminium oxides Al 2 3 2 H 2
calcium hypochlorite
CaCI (OCI)
blue vitriol
copper sulfate
CuS0 4
borax
sodium tetraborate
Na 2 B 4
butter of zinc
zinc chloride
ZnCI 2
cadmium sulfate
cadmium sulfate
CdS0 4
calcium chloride
calcium chloride
CaCI 2
carbide
calcium carbide
phenol
carbon dioxide
silicon carbide
CaC 2
C 6 H 5 OH
KOH
NaOH
CaC0 3
cinnabar
potassium hydroxide
sodium hydroxide
calcium carbonate
mercuric sulfide
ether
di-ethyl ether
(C 2 H 5 ) 2
fixing salt or hypo
glauber's salt
sodium thiosulfate
sodium sulfate
glycerine or glycerol
glycerine
Na 2 S 2 3 -5 H 2
Na 2 S0 4 10 H 2
C 3 H 5 (OH) 3
graphite
crystaline carbon
green vitriol
ferrous sulfate
C
FeS0 4
gypsum
calcium sulfate
CaS0 4
heating gas
propane
C3 H8
hydrochloric acid
hydrochlorid acid
HCI
hydrofluoric acid
hydrofluoric acid
HF
hydrogen sulfide
hydrogen sulfide
H 2S
iron chloride
ferrous chloride
FeCI 2
iron sulfide
ferrous sulfide
laughing gas
nitrous oxide
FeS
N2
lead sulfide
lead sulfide
carbolic acid
carbon dioxide
carborundum
caustic potash
caustic soda
chalk
•
•
5 H2
•
•
7
10 H 2
3 H2
•
C0 2
SiC
HgS
•
7 H2
•
•
2 H2
4 H2
PbS
<
continued on U 3
CHEMISTRY
Us
Chemicals
continued from U 2
chemical
trade
chemical
formula
name
limestone
magnesia
marsh gas
calcium carbonate
magnesium oxide
CaC0 3
methane
CH 4
2 PbO Pb0 2
MgO
minimum or red lead
plumbate
nitric acid
nitric acid
phosphoric acid
potash
potassium bromide
ortho phosphoric acid
HNO3
H 3 P0 4
potassium carbonate
potassium bromide
KBr
potassium chlorate
potassium chloride
potassium chromate
potassium cyanide
potassium chlorate
potassium chloride
potassium chromate
potassium cyanide
KCIO3
potassium dichromate
potassium iodide
K 2 Cr 2
prussic acid
potassium dichromate
potassium iodide
hydrogen cyanide
pyrolusite
manganese dioxide
Mn0 2
quicklime
calcium monoxide
potassium ferrocyan.
CaO
•
K 2 C0 3
KCI
K 2 Cr0 4
KCN
7
Kl
HCN
salammoniac
silver bromide
ammonium chloride
K 3 Fe(CN) 6
NH4CI %
silver bromide
AgBr
silver nitrate
silver nitrate
AgN0 3
slaked lime
Ca(OH) 2
soda ash
sodium monoxide
calcium hydroxide
hydrated sodium carb.
sodium oxide
soot
amorphous carbon
C
stannous chloride
stannous chloride
SnCI 2
sulphuric acid
sulphuric acid
H 2 S0 4
table salt
sodium chloride
NaCI
tinstone, tin putty
stannic oxide
trilene
trichlorethylene
urea
urea
white lead
basic lead carbonate
Sn0 2
C 2 HCI 3
CO(NH 2 2
2 PbC0 3 Pb (OH) 2
white vitriol
zinc sulphate
ZnS0 4
yellow prussiate of potass.
potass, ferrocyanide
K 4 Fe(CN) 6
zinc blende
zinc sulphide
ZnS
zinc or Chinese white
zink oxide
ZnO
red prussiate of potassium
Na 2 C0 3
Na 2
10 H 2
2 H2
•
)
•
7 H2
•
3 H2
CHEMISTRY
U
II 4
A
Acids, Bases
pH values
The negative log of the hydrogen-ion-concentration c H - indicates its
pH value:
pH
cH +
1
10"'
=
10" 2
-log c H -
10- 7
u
|io-
12
|io- 13 1fJ
-14
{
pH-y a\ue
2
1
7
|
1
-
_
"
..
nnirt
-
12
13
|
14
neutral
1
Establishing /?// values by using suitable indicators.
Acid-base-indicators
Indicator
Coiour change
pHRange
from
to
thymol blue
[benz.
p-dimethylamino-azo-
1.2... 2.8
red
yellow
2.9. .4.0
red
orange-yellow
bromophenolblue
3.0... 4.6
yellow
red-violet
congo red
3.0... 4.2
blue-violet
methyl orange
brom cresol green
3.1. ..4.4
red
3.8.5.4
yellow
red-orange
yellow-(orange)
blue
methyl red
4.4.. .6.2
red
litmus
5.0... 8.0
red
bromocresol purple
5.2. ..68
yellow
brom phenol red
bromothymol blue
5.2. .6.8
orange yell. purple
6.0...7.6
yellow
phenol red
6.4. .8.2
yellow
red
neutral red
6.4.. .80
(blue)-red
cresol red
7 0. .88
meta cresol purple
7.4.. .9.0
yellow
yellow
orange-yellow
purple
purple
thymol b lue
phenolpl ital Bin
8.0. .96
alizarin > ellc
w 66
82.9.8
yellow
colourless
(orange)-yellow
blue
purple
blue
blue
red-violet
100. .12.1 light-yellow light brown-yellow
1
CHEMISTRY
u5
Reagents, Equations, Freezing mixtures
Reagents
reagent
indicator
colouration
blue litmus paper
red phenolphthalein
yellow methylorange
acids
red
colourless
red
blue
red
yellow
red litmus paper
bases
colourless phenolphthalein
red methylorange
ozone
potassium-iodide starch paper
blue-black
lead iodide paper
brown-black
hydrochloric acid
white fumes
calcium hydroxide
sediment
H 2S
ammonia
solution
carbonic
acid
Preparation of chemicals
use reaction
to prepare
co 2
-2NH 3 +
CO(NH 2 2 + H 2
H2
NH 4 OH + HCI -* NH 4 CI +
NH4OH
+ H2
NH 3
+
H 2 S0 4
CdS0 4 +H 2 S --Cds
+ CaCI 2 + H 2
CaC0 3 + 2 HC -*C0 2
+ CaCI 2 + H 2
CaOCI 2 + 2 HC -*CI 2
+ %
+ Zn
ZnS0 4
H 2 S0 4
2
+
H2 S
FeCI 2
+ 2 HC
FeS
ammonia
)
ammoniur n chloride
ammoniur n hydroxide
cadmium sulfide
—
carbon di Dxide
chlorine
^H
—
hydrogen
hydrogen sulfide
lead sulfic ie
Pb(N0 3 2 + H 2 S
-*PbS
oxygen
sodium h) 'droxide
2 KCIO3
— NaOH
zinc sulfic e
ZnS0 4
—>3
)
Na 2
+ H2
+ H S
2
+
+
2 HN0 3
2 KCI
+
H 2 S0 4
2
-*ZnS
2
Freezing mixtures
Mixture
Drope in t< jmperature
from
°C
+
+
10
10
+
8
+
15
(The figures stand for
proportions by mass)
to
°C
- 12
- 15
- 24
4 H2
- 21
3-0 ice (crushed)
- 39
- 55
- 78
1-2 ice (crushed)
H2
1 H2
1
+
+
+
1
1
1
KCI
NH 4 NO
NaN0 3
1-4 ice (crushed)
1
methyl alcohol
c
+
+
+
+
+
-
1
1
NH 4 CI
NACI
2CaCI 2
•
2 CaCI 2
•
1
6 H?
6 H2
C0 2 solid
CHEMISTRY
u
Moisture, Drying agents, Water-hardness
Atmospheric relative humidity in closed containers
Relative humidity above
the solution (%) 20°C = 65°F
Supersaturated
aqueous solution
92
86
80
76
Na 2 C0 3
63
55
45
35
NH4NO3
Ca(N0 3 2
•
10 H 2
KCI
(NH 4 2 S0 4
NaCI
)
)
K2CO3
•
CaCI 2
•
4 H2
2 H2
6H 2
Drying agents (desiccants) for desiccators
water remaining
desiccant
after drying at
name
25°C (77°F), g/m 3 air
copper sulfate, dehydrated
CuS0 4
0.8
zinc chloride
calcium chloride
ZnCI 2
CaCI 2
0.14 ... 0.25
0.16
0.008
1
formula
1.4
NaOH
sodium hydroxide
magnesium oxide
MgO
0.005
0.003
0.002
calcium sulfate, dehydrated
hydrated aluminum
potassium hydroxide
CaS0 4
Al 2
0.001
silica gel
(Si0 2 x
0.000025
phosphorus pentoxide
P2O5
)
Hardness of a water
10 mg CaO
Vd
water
German hardness
1
1°d
=
I
7.19 mg
1
I
MgO
water
1.25° English hardness = 1.78° French hardness
17.8 American hardness (1.00 ppm CaC0 3
)
Classification of hardness
4°d
8°d
12°d
very soft
soft
slightly hard
rather hard
12 ... 18 d
18 ... 30°d
hard
above 30°d
very hard
Mixture rule for fluids (mixture cross)
capa-
starting
city
mixed
of the
admixture
for water is b
in
fluid
3
KOH
weight-%
= 0.
Example: a = 54%; b = 92%; c shall become 62%.
One should mix thus 30 weight-sharings of a with 8 of b.
I"
RADIATION PHYSICS
V
Photometry and Optics
General
For every photometric quantity there is a corresponding radiationsame relationships apply to both. They
are differentiated by different suffixes, v for visual and e for
energy.
physical quantity and the
Photometry
Quantity
Symbol
Units
luminous
luminous
flux
4> v = fi-/ v
radiant
cd
intensity
lumen
Im = cd sr
radiant
power
lumensecond
quantity
of light,
luminous
energy
candela
'v
intensity
Radia tion physic s
Quantity
Symbol
Units
Gv=<V'
W
U
sr
*,-fl-/.
W = J/s
Q e =4>*t
J
radiant
energy,
Im s,
quantity
also Im h
of radiation
Lv =
W
cd
luminance
/v
radiance
m2
A, -cos e.
A^ COS £f
lux
illuminance
light
exposure
lx=
—
m^
Im
irradiance
radiant
ff v = £ v • t
Ixs
= Ws
exposure
Ee
W
A
2
H e = E e -t
Definition of the base unit "candela" (cd)
The luminous intensity of a surface of 1/600000
sr m'
m2
Ws
m2
m 2 (= 1% mm 2 )of
a black body at a temperature of 2042 K.
Photometric radiation equivalent
A = 555 nm.
1 watt = 680 Im for wavelength
Luminous flux consumption for lighting(values see Z 25)
A surface A
lit
to an
illumination
flux of
Z<Py
For symbols see T 2
A-Ey
E v will
require a luminous
RADIATION PHYSICS
Distance law, Refraction of light
Optical distance law
The illumination of a surface
is
inversely
proportional to
the square of its distance from
the light source:
£y1
A,
£v2
Where two light sources produce equal illumination of a surface,
the ratio of the squares of their distances from the surface
surface
equal to the ratio of their
/ ..
is
_
(
luminous intensities:
t=;
hi
/V 2
Light refraction
\>-~i
\*
np
sin a
n a ~ sin /?
= const, for all angles.
Where sin fi ^
—?-
total reflection
|\
dense
oc ' urs
thin
medium
medium ft\
Refractive index for yellow sodium ligh tning A = 589.3 nm
solid matter
|
in relation to
fluid
plexiglas
1.49 water
quartz
1.54 alcohol
crown glass
diamond
2.41
A-\
:
A2
'•
1.56 glycerine
benzol
in
relation to vacuum
hydrogen
oxygen
1.47 atmosphere
1.000292
1.000271
1.50 nitrogen
1.000297
1.33
1.36
1.000292
area of radiating surface
area of illuminated or irradiated surface
projection of the radiating surface A<\ perpendicular to the
cos £^
A-i
gasous matter
matter
atmosphere
:
direction of radiation
n a< (n b)
refractive index of thin (dense) medium
angle between emergent beam and normal to radiating surface A-\
£1
Q solid angle Q is the ratio of the area M k intercepted on a sphere
2
of radius r k to the square of the radius: Q = -4 k /r k
:
:
:
;
unit sr =
m 2/m 2
.
The solid angle of a point is Q = 4 x sr = 12.56 sr
n
:
luminous efficacy (see table Z 25)
.
RADIATION PHYSICS
Wavelengths, Mirror
Wavelengths (in atmosphere)
Wavelength A = elf
Type of radiation
v16
X-rays
hard
0.0057 nm
soft
ultra-soft
0.08
nm
2.0
nm..
ultra-violet radiation
violet
blue
visible radiation,
green
light
yellow
red
IR-A
IR-B
IR-C
infra-red radiation
2.0
.
37.5
nm
nm
nm
mm
100 nm
UV-C ... IR-C
UV-C
UV-B
UV-A
optical radiation
0.08
.
100 nm
280 nm
315 nm
280
315
380
380 nm
420 nm
490 nm
530 nm
650 nm
780 nm
420
490
530
650
780
nm
nm
nm
nm
nm
nm
nm
nm
1.4 nm
3.0 nm
Mirrors
Plane mirrors
The image is at the same distance behind the mirror as the
object is in front of it:
u = -v
v17
Concave mirrors
i.i + i
v18
v
u
f
Depending upon the position of
object, the image will be real
or virtual:
u
v
00
f
at focal point
>2/
f<v<2f
real, inverted,
2/
2/
2/ >u > f
>2/
00
f
negative
<f
smaller
real, inverted, of equal size
real, inverted, larger
no image
virtual, larger
Convex mirrors
Produce only virtual and smaller
images. Similar to concave
mirror where:
v19
/= -r/2
•)c = 299 792 458 m/s « 0.3 x 10
9 m/s (velocity of light)
For explanation of symbols for mirrors refer to V4
|
/: frequency 1/s
RADIATION PHYSICS
V
Lenses
Lenses
Refraction D of a lens
Unit: 1 dpt = 1 dioptrics
v20
Lens equation (thin lenses only)
v21
'>£ + t8
v22
5
v23
Where two lenses with focal depths /1 and /2 are placed immediately one behind the other, the equivalent focal length/, is given
11^1
by
v24
/i
Magnifying lens
general
h
._ _ object
where object
is in
-j
focus
m = j+-\
v25
Microscope
total
magnification
& *
§
j
m
v26
t-s
fvh
v27
v28
Macro photograpy
camera extension
v29
a = f(m + 1)
distance of object
B
F
/
G
:
n^-k)
size of image
n
:
focus
r
:
:
focal length
t
:
:
size of object
m
:
:
refractive index (see V2)
radius of curvature
optical length of tube
magnification factor
range of vision (= 25 cm for normal vision)
RADIATION PHYSICS
Ionizing radiation
Ionizing radiation
Ionizing
radiation
any
is
which causes direct or
permanent gas.
Accumulated
radiation
indirect
v30
As
absorbed
energy
(measured
kg
value)
1
rontgen
1
R
258^
J
t
kg
a
Units
kg
m
absorbed
dose,
*-?-£m
D-f.J
=
of
rate of
absorbed
dose
v33
particles
excitation
absorbing
energy
kg
v31
v32
or
time rate
values
Units
values
amount of
charged
of
ionization
1
gy_
vv
s
kg
"
31.56X10 6
i
kg a
W
m
kg
6.242 x10
dose
equivalent
(theoretical
value)
v34
v35
H=D = q-.
q
1
^y
0.01
sv
dose
VAs
Ws
equivalent,
kg
kg
-rate
1
sievert
=
1
1
w
gy
kg
[100 rem
1
sv]
=
= q-f-J
q-D
Ionization current /; When air molecules are ionized by radiation
and a voltage is applied, an ionization current / flows.
(Instrument: the ionization chamber).
Charge Q: When an ionization current / flows for a time t it produces a charge
Q = It
v36
Units in (
)
are earlier units
continued on T 6
RADIATION PHYSICS
V
Ionizing radiation
Dose J: Dose / is a value related to mass m, e.g. J = Q/m.
Radiation energy W: W is the radiation energy necessary for ionization.
Each pair of ions in the air molecule requires the energy
= 33.7 eV
L
(Charge of one electron: 1 e = 1 .602 x 10~ 19 As)
19
As x 1 V = 1 .602 x 10~ 19 J)
(1 electron volt: 1 eV = 1 .602 x 10"
W
The activity A
Activity A:
is
number of atoms of a radioactive
the
substance that disintegrates per unit time.
A = -dN/dt = \N.
Units: bq (becquerel) [1 curie = 1 ci = 37 x 1 9 bq]
1 bq is 1 disintegration of a radioactive atom per second.
Decay X:
X = In 2/Ty2
The half life is Ty2 the time taken for one half of the radioactive mass
to decay.
Units: s~ 1 min -1 h~ 1 d _1 a -1
,
,
,
,
.
Half lives of some natural and artificial isotopes
relative
atomic 1
'
number Z
Element
atomic 2
atomic 1 >
half life
>
number Z
mass A r
3
40
42
60
90
tritium
potassium
cobalt
strontium
iodine
12a
caesium
caesium
12.4 h
radium
5.3a
thorium
29 a
uranium
134
137
226
232
238
8.0d
Plutonium
239
131
:
q
:
:
atomic 2
|
Ty2 half
:
:
2.1a
30 a
1600 a
14xi0 9 a
4.5 x I0 9 a
24000 a
= 1
q = 1
<7
ionization constant for tissue
...
20
/-/l
/=(1 ...4)/L
for bone
fL
FA
life
number of radioactive atoms
quality factor for (3 -, y - und X-rays
for other radiation
/
half life
>
mass A r
1.3xl0 9 a
Symbols used
m mass (base unit)
N
Element
fL = WL/e = 33.7 V)
ionization constant for air
Notation of units used
A: ampere
|
C: coulomb
J: joule
|
a:
annum (1 annum = 1
J
31.56x 10 6 s)
= 1
86 400 s)
Exposure to radiation (dose equivalent): In 1991 the average person in
the Federal Republic of Germany would have been exposed to the
I
follwing radiation:
d: dies
(1
dies
h in
Type
r
msv
from natural sources
for medical reasons
<2.4
< 0.5
t
other artificial radiation
<0.1
*permitted by law
< 0.3
')
number of protons
2)
I
number of protons and neutrons
[m rem]
240
50
< 10
< 30
TABLES
Zi
Properties of solids
Reference conditions
Density pat t = 20°
C
point t: The values in brackets refer to sublimation,
direct transition from the solid to the gaseous state.
Boiling
i.e.
Thermal conductivity A at t = 20° C
Specific heat c p for the temperature range
andp = 1.0132 bar.
density
Substance
o<r< 100
meltinglboiling thermal
point
conduc-
specific
heat
cp
9
t
t
tivity A
kg/dm 3
°C
°C
W/(m k)
agate
2.6
2.6
aluminum, rolled
2.7
2600
2300
2200
2200
10.89
aluminum bronze
aluminum cast
1600
1040
658
658
300
127.9
209.4
209.4
0.436
0.904
0.904
630
815
1440
22.53
0.209
0.348
1.357
0.816
0.29
7.7
amber
1.0
antimony
6.67
arsenic
5.72
artificial
wool
1.5
asbestos
barium
2.5
3.59
1300
704
barytes
4.5
beryllium
1.85
1580
1280
bismuth
9.8
271
boiler scale
2.5
borax
1.72
1200
740
brass, cast
8.4
brass, rolled
8.5
0.17
1700
0.13
1.2. ..3.5
0.80
0.996
900
900
1100
1100
113
113
0.385
0.385
0.92
63
1.8
8.83
-7.3
910
brown iron ore
5.1
1570
cadmium
864
321
calcium
carbon
1.55
3.51
cast iron
7.25
850
3600
1200
cerium
6.77
630
chalk
1.8
165
1.0
W/(m K) = 0.858 9 kcal/(h rr K)
2300
64
0.58
765
1439
92.1
2500
58
8.9
0.92
2>
|
»
0.80
8.1
3.14
1
kJ/(kg K) 2
0.46
bromine
bronce (Cu Sn 6)
»
»
2970
1560
2800
brick
1
1
1
kJ/(kg K)
1.02
0.37
0.67
0.234
0.63
0.854
0.532
0.84
= 0.2: 188 kcal (kg K)
TABLES
z2
Properl ties of solids
meltingj boiling
thermal
specific
po int
conduc-
heat
density
Substance
charcoal
P
t
t
tivity A
kg/dm 3
°C
°C
W/(m K)
0.084
0.4
chromium
7.1
clay
1.8. ..2.1
cobalt
8.8
coke
1.4
concrete reinforce
constantan
copper, cast
copper, rolled
cork
2.4
diamond
8 89
8.8
8.9
1800
1600
1490
1600
1083
1083
0.9. ..1.0
duralium
ebonite
2700
2980
3100
2400
2500
2500
2.8
0.84
0.452
1
0.88
69.4
0435
0.184
0.84
0.8. ..1.7
0.88
23.3
384
384
0.41
0.394
0394
2.0
0.52
[3540]
650
350
2000
129.1
40. ..50
kJ/(kg K) 2
69
0.05
0.2... 0.3
3.5
dripping, beef
1 >
0.88
0.92
0.17
1.2. ..1.8
electron
1.8
650
1500
162.8
1.00
emery
4.0
2200
2000
700
3000
2900
11.6
0.96
0.47
0.88
brick
fire
1.8. ..2.2
glass, window
2.5
glass-wool
gold
0.15
19.29
1063
2700
310
0.130
graphite
2.24
3800
168
0.71
ice
0.92
ingot iron
7.9
1460
47. ..58
iodine
4.95
113.5
4200
100
2500
184
4800
2500
58
0.532
46. ..58
0.461
0.58
1740
0.67
0.130
iridium
22.5
2450
5.1
1200
1200
1570
lead
11.3
327.4
leather
09. ..1.0
iron, cast
iron,
7.25
forged
7.8
iron-oxide
limestone
2.6
lithium
0.53
magnesia
1.74
magnesium, alloy
1.8
2)
1
W/(m K)
1
kJ/(kg K)
0.84
0.04
0.84
0.44
2.09
0.49
0.218
59.3
0.134
2.3
34.7
0.15
1.5
2.2
0.909
179
1372
301.2
0.36
657
650
1110
1500
157
1.05
70. ..145
1.01
3.2. .3.6
magnesium
')
0.81
:
=
0.3598 kcal/ (h m K)
2388 kcal (kgK)
>
TABLES
zs
Properties of solids
melting boiling
thermal
specific
po int
conduc-
heat
t
tivity A
cp
°C
°C
W/(m K)
1221
2150
density
Substance
Q
kg/dm
manganese
t
3
7.43
marble
mica
2.0... 2.8
molybdenum
10.2
nickel
8.9
osmium
22.48
2.8
oxide of chrom
palladium
12.0
paper
0.7. ..1.1
5.21
paraffin
peat
0.2
8.8
1.82
7.0. ..7.8
pinchbeck
8.65
pitch
1.25
pit coal
21.5
porcelain
2.2... 2.5
potassium
0.86
quartz
2.5
radium
2500
2300
1552
5300
52
5
2930
300
280
2500
1300
0.381
71
0.14
88
0.24
rubber, raw
0.95
125
39
1550
2>
= 0.8598
1 W/(m K)
1
kJ/(kg K)=
0.2: $88
kcal/(h
1.02
0.13
70
0.092
•
1)
0.38
0.80
0.7
100... 300
105
0.54
127.9
1.07
silver
52.3
159
2300
5500
rosin
3.12
0.36
0.80
2500
silicon, carbide
1.9
110
9.9
1960
2.33
3.26
2230
1140
12.3
silicon
1.336
0.26
1470
960
rhodium
4.4
0.24
0.14
1500
220
1420
960
m K)
kcal/( <gK)
0.92
0.8... 1.0
762-2
21.4
2.1. ..2.5
70.9
0.24
rhenium
sandstone
selenium
0.130
075
4400
950
3175
1.52
0.272
0.461
1770
1650
63
8.8
1.4. ..1.6
0.87
145
59
0.13
red metal
rubidium
0.84
0.08
900
44
1560
1000
8.6. ..9.1
sand, dry
0.46
2.8
0.35
0.42
1.35
platinum
red lead
5500
2730
0.9
phosphorbronce
phosphorus
pig iron, white
2600
1452
kJ/(kg K) 2
1 »
1
0.32
*
1
1.30
0.2.0.35
700
2230
688
2600
2170
58
0.33
0.58
0.80
2.3
0.71
0.20
0.33
83
0.75
15.2
0.67
407
0.234
>
TABLES
z4
Properties of solids
meltinglboiling
thermal
specific
point
conduc-
heat
cp
density
Substance
Q
t
t
tivity A
kg/dm 3
°C
°c
W/(m K)
slate
2.6. ..2.7
2000
snow
0.1
sodium
97.5
0.98
soot
1.6... 1.7
steatite
2.6. .2.7
1600
7.85
1460
115
2990
steel
sulfur, cryst.
2.0
tantalum
16.6
tar
1.2
tellurium
6.25
thorium
11 7
timber, alder
"
,
ash
,
birch
,
larch
"
"
0.49
0.20
0.70
0.138
- 15
455
1800
4000
38
0.201
0.14
1.3
2.5
1.4
1.6
1.9
1.4
,
red beech
0.8
0.19
0.14
,
red pine
0.65
0.15
1.5
white pine
walnut
0.75
0.65
0.15
0.15
1.5
,
tin,
cast
7.2
tin,
rolled
7.28
232
232
2500
2500
64
64
3200
5900
3800
3300
130
28
2500
2100
906
1000
906
wax
0.96
1670
3410
1133
1890
60
welding iron
white metal
7.8
1600
titanium
4.5
tungsten
19.2
uranium
vanadium
19.1
6.1
7.5. ..10
300 .400
zinc, cast
6.86
zinc, die-cast
6.8
419
393
419
zinc, rolled
1
0.19
4.9
2.4
1.28
,
1
54
0.17
pockwood
"
»
0.83
47. .58
0.14
,
"
1
2
2500
445
4100
300
1300
1.6
0.85
0.75
"
2>
1.26
0.84
0.16
pitchpine
"
0.07
0.75
,
"
0.76
4.187
0.17
0.16
0.14
0.12
,
,
126
kJ/(kg K) 2
0.55
0.75
0.65
0.75
maple
oak
"
0.5
100
880
1 >
W/(m K)
kJ/(kg K)
7.15
=
-
0.85 98 kcal/(h
0.23 88
m K)
kcal/(kc JK)
15.5
31.4
0.084
1.3
1.4
0.24
0.24
0.47
0.13
0.117
0.50
3.43
35. .70
0.515
0.147
110
140
110
0.38
0.38
54.7
0.38
>
T_
TABLES
£m 5
Properties of liquids
Reference conditions
Density g at t = 20°C and p = 1.0132 bar.
Melting point and boiling point I at p = 1.0132 bar.
Thermal conductivity A at t = 20°C. For other temperatures see Z 15.
Specific heat c p
for the temperature range
p = 1.0132 bar.
Q
thermal
specific
point
conduc-
heat
t
/
tivity A
W/(m K) 3
kg/dm 3
°C
°C
acetic acid
1.08
16.8
118
acetone
0.79
alcohol
benzene
0.79
0.89
benzine
0.7
-150
50.. .200
chloroform
1.53
61
diesel oil
0.88
-70
-5
ether
0.73
117
gas oil
0.86
glycerine
1.27
heating oil
f 10%
hydrochlor
acid
1 40%
hydrofluoric acid
0.92
-95
-130
5.4
3>
-30
-20
-5
-14
1.05
linseed oil
machine oil
mercury
13.6
0.8
-98
nitric
1
W/(m K)
1
kJ/(kg K)
0.12
102
0.50
2.26
0.15
0.29
2.43
19.5
3.14
2.33
0.15
2.09
-20
150. ..300
-10
-20
-70
-160
-10
160
119
0.10
1.80
150.300
0.159
0.905
2.09
40. ..70
0.14
1.76
338
0.5
1.38
1.49 '
-73
-10
0.88
1.47
4°
1 at
-94.5
110
87
100
0.14
1.59
1.40
>
175.. .350
0.14
-41
sulfuric acid 50%
2»
0.13
2.1
357
66
86
sulfuric acid cone.
1
175
35
200... 300
290
2.43
1.80
380... 400
0.80
0.67
1.84
>
1.62
4
trichlor ethylene
0.137
0.16
-5
petroleum
petroleum ether
toluene
0.17.. .0.2
-38.9
0.87
sulfurus acid
78.4
80
316
of turpentine
perchlor ethylene
oil
56.1
-20
3
1.56
0.94
acid
of resin
water
-92.5
0.91
methyl alcohol
cp
kJ/(kg K) 2
>
»
1.20
0.99
0.96
oil
< 100°C and
t
melting| boiling
density
Substance
<
-^86
= 0.8598 kcal/(h m K)
- 0.23JJ8 kcal/(k JK)
0.126
1.68
8.4
0.138
2.51
0.26
0.15
1.72
1.34
0.16
1.30
0.58
4.183
at
t
=
'at
t
=
3)
4
0°C
-20°C
>
TABLES
Ze
Properties of gases
Reference conditions
Density g at t = 0°C and p = 1.0132 bar For perfect gases g
can be calculated for other pressures and/or temperatures
from: g = p/(R x T).
Melting point and boiling point at p = 1.0132 bar.
Thermal conductivity A at t =
C and p = 1.0132 bar.
For other temperatures see Z 15.
Specific heat c p and c v at t = 0°Cand/7 = 1.0132 bar.
c p at other temperatures see Z 13.
/
Substances
acetylene
melting|boiling
thermal
point
conduc-
density
t
I
°C
°c
W/(m K)
1.17
- 83
81
0.018
1.293
-213
ammonia
0.77
argon
1.78
-192.3
- 77.9 - 33.4
-189.3 -185.9
blast furnace gas
1.28
-210
-170
butane, isobutane, ncarbon di-oxide
2.67
-145
-135
-
carbon disulfide
carbon monoxide
3.40
chlorine
coal gas
ethylene
air,
atmosphere
heat
tivity A
9
kg/m 3
-
specific
cp
cv
|
kJ/(kg K) 2
1 '
>
1.616 1.300
1.005 0.718
0.02454
0.022
0.016
0.02
2.056 1.568
0.52 0.312
1.05 0.75
1.25
- 78.2 - 56.6
-111.5
46.3
-205.0 -191.6
0.015
0.0069
0.023
0.816 0.627
0.582 0.473
1.038 0.741
3.17
-100.5 - 34.0
0.0081
0473 0.36
0.58
-230
1.26
-169.3 -103.7
- 270.7 - 268.9
-111.2 - 84.8
0.017
0.143
0.013
1.47
1.173
5.20
3.121
0.171
14.05 9.934
0.013
0.0088
0.030
0.046
0.992 0.748
151
0.25
2.19 1.672
0.024
0.024
1.038 0.741
0.909 0.649
2.70
1.97
helium
hydrochlor acid
0.18
hydrogen
hydrogen sulfide
0.09
1.63
10
1
-210
krypton
3.74
methane
neon
0.72
-2592 -252.8
- 85.6 - 60.4
-157.2 -153.2
-182.5 -161.5
0.90
-2486
nitrogen
1.25
1.54
-246.1
2.14
1.59
0.795 0.567
1.03
0.618
oxigen
1.43
-210.5 -195.7
-218.8 -182.9
ozone
propane
2.14
-251
2.01
-187.7 -
42.1
0.015
1.549 1.360
sulfur dioxide
2.92
- 75.5 -
10.0
water vapor3
0.77
0.0086
0.016
0.0051
0.586 0.456
1.842 1.381
0.16 0.097
'
xenon
>
2>
5.86
0.00
100 00
-111.9 -108.0
1
W/(m K)
=
0.8598 kcal/(h m K)
1
kJ/(kg K)
=
02 388 kcal/(k gK)
1
-112
3)
at t
--
=
100°C
TABLES
Friction numbers
Coefficients of sliding and static friction
sliding friction \x
static friction \i
on
material
material
-Q
bronze
bronze
0.20
0.18
0.18
cast iron
steel
oak
oak
0.20. ..0.40
0.15. ..0.35
||
oakf
cast iron
rubber
cast iron
steel
0.17. ..0.24
asphalt
0.50
0.60
concrete
hemp rope
timber
leather
oak
belt
cast iron
PE-W
0.014
0.10. ..0-30
PTFE
W PE-W
PTFE
POM
POM
0.18
031
0.16
0.10
0.10
0.02. ..0.05 0.18. ..0.24
0.30
0.50
0.20. ..0.50
1 »
PTFE 2
PA 66 3
0.10
0.19
0.10 0.05. ..0.15 0.40. ..0.60
0.50
0.08 0.04. ..0.12
0.40
ice
POM
0.11
0.20
0.30
0.50
steel
PE-W
0.06
0.08
0.07
0.50
0.40
0.40
wood
steel
0.10
0.50 0.12
0.26 0.02. ..0.10 0.50. ..0.60
0.027
0.11
0.02. ..0.08 0.15. ..0.30
0.10
0.40. ..0.50
)
0.03. ..0.05
0.10
>
0.30. ..0.50
4
>
0.35... 0.45
2
)
0.035. ..0.055
1)
0.2. ..0.7
0.50. ..0.70
4>
0.40. ..(
Rolling friction
K 12 and L 9)
(for section
lever arm / of f rictional
material on material
force in
mm
0.10
0.15
0.50
rubber on asphalt
rubber on concrete
lignum vitae on lignum vitae
0005. ..0.01
steel on steel (hard: ball bearing)
steel on steel (soft)
elm on lignum vitae
0.05
0.8
movement with grain of both materials
perpendicular to grain of sliding body
-H-: movement
:
II
Lupolen from BASF)
126 from Dupont)
1)
polyethylene with plasticizer
(e.g.
2)
polytetrafluorethylene
(e.g. Teflon C
polyamide
polyoxymethylene
(e.g. Ultramit CA from BASF)
3)
4)
(e.g. Hostaflon C 2520 from Hoechst)
TABLES
8
Friction factors for flow in pipes £
fi-e'r-wtb
H==
gffioo
K 10
f
(Re, $)
TABLES
9
Water pipes, Hydrodynamics values
Galvanized Steel Tubes, suitable
for Screwing to B.S. 21 Pipe Threads
(Approximate values for Medium pipes,
colour code - blue, to B. S. 1389)
Nominal bore (inches)
Vb
Threads per inch
28
19
%
'/a
19
14
14
%
V/a
1
11
11
11
11
Outside diam. of pipe (mm) 10.2 13.5 17.2 21.3 26.9 33.7 42.4 48.3 60.3
Inside diam. of pipe
(mm) 6.2
12.3 16 21.6 27.2 35.9 41.8 53
(mm 2
Flow area
)
Ration of flow area (mm 2 )
to nominal bore (inches)
30
242
119
243
581 1012 1371 2206
201
317 402
488
581
810 914 1103
Roughness k
(according to Richter, Hydraulics of Pipes)
Material and
kind of pipe
Condition of pipe
new seamless typical rolled finish
rolled or drawn pickled
(commercial)
steel pipes
used steel
pipes
pipes
pipes folded
mm
0.02. .0.06
»
cleanly galvanized (dipping process)
0.03. ..0.04
0.07.. .0.10
commercial galvanized
0.10. ..0.16
uniform corrosion pits
about 0.15
medium corrosion, light incrustation
medium incrustation
0.15. .0.4
heavy incrustation
cleaned after long use
cast iron
k in
about 1.5
2. ..4
0.15. ..0.20
new, typical cast finish
new, bituminized
used, corroded
0.6
0.2
..
0.1
...0.13
1
...1.5
incrusted
1.5
cleaned after several years of use
mean value in urban sewerage installations
heavily corroded
0.3
new, folded
about 0.15
...4
...1.5
1.2
4.5
and riveted of new, depending on kind and quality
sheet steel
of riveting
light riveting
about 1
heavy riveting
to 9
25 years old, heavily incrusted, riveted pipe
12.5
TABLES
Zio
Heat values
Latent heat of fusion per unit mass
material
kJ
material
kg
aluminum
377
164
168
46
126
134
243
172
113
antimony
brass
cadmium
cast iron
chromium
cobalt
copper
ethyl ether
kJ
/.<
kJ
material
kg
kg
glycerine
176 paraffin
147
gold
67 phenol
109
113
manganese
335 platinium
205 potassium
23 silver
155 sulfur
mercury
11.7 tin
naphthaline
151
nickel
234 zinc
ice
iron
lead
59
109
38
59
Wood's alloy
33.5
117
Latent heat of evaporation per unit mass / d
at 101.32 kN/m 2 (= 760 torr)
material
kJ
kJ
material
kg
kJ
kg
material
kg
alcohol
88C hydrogen
503 oxygen
ammonia
141C mercury
281
carbon dioxide
59J methyl chloride
chlorine
29C nitrogen
406 toluene
201 water
)
214
402
sulfur dioxide
365
2250
Calorific value Hu
(average values)
Solids
Hu
MJ
Liquids
Hu
MJ
Gases
Hu
MJ
kg
kg
kg
Hu
MJ
anthracite
33.4
alcohol
26.9
acethylen
48.2
bituminous coal
31.0
benzene
40.2
butane
45.3 122.3
56.4
brown coal
9.6
Diesel fuel oil
42.1
coal gas
4.1
5.2
furnace coke
30.1
gasoline
42.5
hydrogen
119.9
10.8
gas coke
29.2
heating oil
41.8
methane
50.0
36.0
non coking coal
31.0
methyl alcohol
19.5
municipial gas
18.3
11.3
peat, dry
14.6
methylspirit (95%)
25.0
natural gas, dry*
43.9
39.0
wood, dry
13.3
petroleum
40.8
propane
46.3
93.1
•Provenance: USA (Panhandle)
3
For Great Britain (Leman Bank): 48.7 MJ/kg resp. 37.0 MJ/m
1
kWh = 3.6 M J (Cf. K3)
i
TABLES
7
£m 11
Heat values
-inear coefficient of expansion i in 1/K
at/ =
a/10 -6
material
...
100°C
a/10 -6
material
23.8
German silver
18.0
porcelain
13.5
gold
14.2
quartz glass
brass
18.5
ead
29.0
silver
bronze
17.5
molybdenum
5.2
steatite
cadmium
30.0
nickel
13.0
steel,
0.5
8.5
mild
cast iron
10.5
nickel steel
15.2
= Invar 36% Ni
1.5
ungsten
copper
16.5
platinum
9.0
line
12.0
23.0
in
Cubic coefficient of expansion )
at/ = 15°C
y/lQ-z
4.0
19.7
constantan
material
a/10- 6
material
aluminium
bismuth
y/10~ 3
material
in
4.5
30.0
1/K
y/10-3
material
alcohol
1.1
glycerine
0.5
Detroleum
benzene
1.0
mercury
0.18
oluene
1.08
ether
1.6
oil
1.0
water
0.18
of turpentine
1.0
Coefficient of heat transfer k in W/(m 2 K)
(Approx. values, natural convection on both sides)
material
thiol* ness
3
10
50
lc iyer
i mrr
100 120 250 380 510
4.3
3.7
of in sula ing
20
reinforced concrete
i
3.5
2.4
1.2
0.7
0.5
1.6
0.9
0.7
1.7
1.0
0.7
3.1
2.2
1.7
1.4
3.4
2.3
foam mortar
%
(e.g. thermalite)
ac = 2.45 N/mm 2
ac = 4.90 N/mm 2
ac = 7.35 N/mm 2
glass
5.8
5.3
4.1
2.4
glass-, mineral-
wool, hard foam
timber wall
1.5
0.7
0.4
3.8
2.4
1.8
chalky sandstone
gravel concrete
4.1
3.6
1.7
slag concrete
2.7
1.7
1.4
1.0
brick
2.9
2.0
1.5
1.3
double or treble glazing
single window, puttied
double window, 20 mm spacing, puttied*'
double window, 120 mm spacing, puttied*
tiled roof without/with joint packing
*'
also for wind ows wit i seal< ?d air gaps
2.6 or 1.9
5.8
2.9
2.3
11.6 5.8
TABLES
"7 <,„
£- 12
Heat values
Gas constant A' and molecular mass M
material
acetylene
air
ammonia
carbonic acid
carbon monoxide
R
M
J
kg
kg K
kmol
319
287
488
189
297
26
29
hydrogen
material
17
oxygen
sulfuric acid
28
water vapor
M
J
kg
kgK
kmol
4124
297
260
130
462
nitrogen
44
R
2
28
32
64
18
Radiation constant C at 20 C
material
C
W/(m 2 K 4
C
material
W/(m 2 K 4
)
)
polished
0.17 x 10~ 8
copper, oxidated
3.60 x 10~ 8
aluminum, polished
0.23 x 10" 8
water
3.70 x 10
-8
copper, polished
0.28 x 10~ 8
timber, planed
4.40 x 10
-8
brass, polished
0.28 x 10" 8
porcelain, glaz
5.22 x 10
-8
zinc, polished
0.28 x 10~ 8
glass
8
5.30 x 10"
polished
0.34 x 10" 8
0.34 x 10" 8
brickwork
soot, smooth
5.30 x 10"
8
5.30 x 10~
aluminum, unpolished
0.40 x 10~ 8
zinc,
unpolished
8
5.30 x 10~
nickel, polished
0.40 x 10
iron,
unpolished
5.40 x 10
brass, unpolished
8
1.25 x 10"
absolutely
ice
3.60 x 10" 8
black surface
silver,
iron,
tin,
polished
Dynamic viscosity
/
SAE
1
in
r\
°C
-8
8
5.67 x 10
of motor oils in N s/m 2 x
0-5*>
1
20
50
100
10
0.31
0.079
0.020
0.005
0.007
20
0.72
0.170
0.033
30
40
1.53
0.310
0.061
0.010
2.61
0.430
0.072
0.012
50
382
0.630
0.097
0.015
N s/m 2 = 1 kg/(m s) = 1 Pa s - 1000 cP
-8
-8
TABLES
z 13
Heat values
Mean specific heat c pm
in
t
°c
of various gases
kJ/(kg K) as a function of temperature
CO
C0 2
H2
H2
N2
1 >
pure
N2 2
>
o2
so 2
air
1.039
0.8205
14.38
1.858
1.039
1.026
0.9084
0.607
1.004
100
1.041
0.8689
14.40
1.874
1.041
1.031
0.9218
0.637
1.007
200
300
400
1.046
0.9122
14.42
1.894
1.044
1.035
0.9355
0.663
1.013
1.054
0.9510
14.45
1.918
1.049
1.041
0.9500
0687
1.020
1.064
0.9852
14.48
1.946
1.057
1.048
0.9646
0.707
1.029
500
600
700
800
900
1.075
1.016
14.51
1.976
1066
1.057
0.9791
0.721
1.039
1.087
1.043
14.55
2.008
1.076
1.067
0.9926
0.740
1.050
1.099
1.067
14.59
2.041
1.087
1.078
1.005
0.754
1.061
1.110
1.089
14.64
2.074
1.098
1.088
1.016
0.765
1.072
1.121
1.109
14.71
2.108
1.108
1.099
1.026
0.776
1.082
1000
1100
1200
1300
1400
1.131
1.126
14.78
2.142
1.118
1.108
1.035
0.784
1.092
1.141
1.143
14.85
2.175
1.128
1.117
1.043
0.791
1.100
1.150
1.157
14.94
2.208
1.137
1.126
1.051
0.798
1.109
1.158
1.170
15.03
2.240
1.145
1.134
1.058
0.804
1.117
1.166
1.183
15.12
2.271
1.153
1.142
1.065
0.810
1.124
1500
1600
1700
1800
1900
1.173
1.195
15.21
2.302
1.160
1.150
1.071
0.815
1.132
1.180
1.206
15.30
2.331
1.168
1.157
1.077
0.820
1.138
1.186
1.216
15.39
2.359
1.174
1.163
1.083
0.824
1.145
1.193
1.225
15.48
2.386
1.181
1.169
1.089
0.829
1.151
1.198
1.233
15.56
2.412
1.186
1.175
1.094
0.834
1.156
2000
2100
2200
2300
2400
1.204
1.241
15.65
2.437
1.192
1.180
1.099
0.837
1.162
1.209
1.249
15.74
2.461
1.197
1.186
1.10.4
1.167
1.214
1.256
15.82
2.485
1.202
1.191
1.109
1.172
1.218
1.263
15.91
2.508
1.207
1.195
1.114
1.176
1.222
1.269
15.99
2.530
1.211
1.200
1.118
1.181
2500
2600
2700
2800
2900
3000
1.226
1.275
16.07
2.552
1.215
1.204
1.123
1.185
1.230
1.281
16.14
2.573
1.219
1.207
1.127
1.189
1.234
1.286
16.22
2.594
1.223
1.211
1.131
1.193
1.237
1.292
16.28
2.614
1.227
1.215
1.135
1.196
1.240
1.296
16.35
2.633
1.230
1.218
1.139
1.200
1.243
1.301
16.42
2.652
1.233
1.221
1.143
1.203
1)
at low presj >ures
2
>d jrived fr om air
Calculated f 'om figi jres gi /en in E. Schn -lidt:
Einfuhrun g in die Techni scheTflermod ynamik, 11. Ai iflage. Berlin/
Got tingen Heidelt erg: S pringer 1975.
I
TABLES
Zl4
Heat values
Liquids*)
Q
C
kg
kJ
m3
kgK
t
Substance
°C
water
octane C 8 H 18
ethane C 2 H 5 OH
20
50
100
999.8
998.3
4.217
4.182
988.1
958.1
4.181
200
864-7
4.215
4.494
-25
738
719
ammonia NH 3
spindle-oil
insulating oil
3
13.44
6.99
3.57
1.75
0.90
547.1
2093
2232
0.183
0.177
3241
1786
1201
701
3707
200
661
20
50
100
200
885
867
839
793
672
1.612
1.717
1.800
1.968
2.617
1435
1383
1296
1.33
20
50
695
636
609
4.45
2.395
0173
2.801
0.165
0.152
11.11
22.52
16.63
11.90
326
7.41
0.144
0.134
0.127
0.108
649
436
7.79
5.93
4.04
0.144
0.136
0.128
0.108
773
586
419
269
133
8.65
7.14
5.55
4.14
3.22
0.212
0.199
0.177
368
304
234
2.09
4.61
0547
0540
4.74
5.08
0477
317
169
138
103
2.58
1.44
1.26
1.10
0.144
0.143
0.139
13060
5490
2000
168
79
32
31609
7325
3108
482
125
60
1.729
1821
1.968
137
1.48
0.141
0521
261
113
2.31
196
20
50
100
871
185
852
820
2.06
2.19
20
60
100
866
842
818
2.09
229
0.124
0.122
0.119
13546
0.139
9.304
1558
0.02
1260
2.366
0.286
1.5*10 6
1.24x10"
mercury Hg
glycerine C 3 H 8
1791.8
1002.6
0.5620
0.5996
0.6405
0.6803
14.62
3.454
561
-
1020
714
879
847
793
20
50
Pa S
0.144
0.137
20
50
100
-50
Pr
2.131
806
789
763
716
sulfur dioxide S0 2
10 6 *7
281.7
134.6
20
50
100
toluene C 7 H 8
A
W
mK
06685
2.064
-25
benzene C 6 H 6
P
20
*) Explanation of the s ymbol. 5 Cf.
11
TABLES
Z 15
Heat values
Gases (at 1000 mbar)*)
A
10 6 r?
kg K
W
mK
Pa s
1.006
1.006
1.007
1.012
1.026
1.069
0.023
0.025
0.026
0.032
0.039
0.053
16.15
17.10
17.98
21,60
25.70
32.55
0.71
20
100
200
400
1.377
1.275
1.188
0.933
0.736
0.517
-30
2.199
0.013
0.015
0.016
0.022
0.030
12.28
13.75
14.98
18.59
26.02
078
t
Substance
°C
air,
-20
dry
carbon dioxide C0 2
Q
C
kg
kJ
m3
1.784
1.422
1.120
3.13
2.87
2.29
0.473
0.477
0.494
0.0081
25
100
0.0093
0.012
12.3
13.4
16.8
0.72
0.69
0.69
25
100
200
0.76
0.70
0.56
0.44
2.056
2.093
2.219
2.366
0.022
0.024
0.033
0.047
9.30
10.0
12.8
16.5
0.87
0.87
0.86
0.83
-50
1.73
0.903
0.909
-
1.41
16.3
19.2
0.73
29
0913
1.03
0.81
0.934
0.963
0.024
0,026
0.032
0.039
0.586
0.607
0.662
ammonia NH 3
25
100
200
1
sulfur dioxide S0 2
2.88
2.64
25
100
2.11
nitrogen N 2
1.23
1.13
0.90
25
100
200
hydrogen H 2
0.71
-50
0.11
25
100
200
0.09
0.08
0.07
0.05
50
100
200
300
0.0049
0.0830
0.5974
7.865
46.255
water vapour
(at
saturation)
0.70
0.70
0.69
0.68
0.66
25
100
200
chlorine CI
2
Pr
0.800
0.827
0.850
0.919
0.997
1,951
oxygen
P
*) Explanation of tr ie symt )OlS cf
.
011
0.78
0.78
0.77
0.76
-
24.3
28.8
0.71
0.71
0.71
0.0086
0.0099
0.014
11.7
12.8
16.3
0.80
0.78
0.77
1.038
1.038
1.038
1.047
0.024
0.026
16.6
17.8
20.9
24.7
13.50
14.05
14.34
0.141
0.171
0.181
0.211
0.249
14.41
14.41
1.864
1.907
2.034
2.883
6.144
0.031
0.037
0.0165
0.0203
0.0248
0.0391
0.0718
2J3.3
7.34
8.41
8.92
10.4
12.2
9.22
10.62
12.28
15.78
19.74
0,72
0.71
0.70
0.70
0.70
0.69
0.71
0.71
0.71
1.041
0.999
1.007
1.163
1.688
TABLES
Z16
Strength values in N/mm 2
CO
CD
28 o o oo
"3-
VI
VI
-I
o>
co
3
VI
V|
Vj
d
CD
cb
T-
d
*
co
T- ft
A A
A
A A
o o
o o o o o
3
iCD - O)
-> a.
o
CM
CM CM
CO
—
o CM
o o CD
o O O O
O
O
O
O
O o
O
o
O) W t- s
CO CM
o
CO
o
o
«
CM CO
o
o O O
O E.
CD CO
S O
CM CM
^~
^ "K
i=
O
O
O O
CO
o _0)
1^
co m co ^j
o
o i—
O O
CO
in m
in
o O
s
•^-
I
CM
^
"tf
ci.
3 o,
c
©
d)
vj
C
o
£
-a
£
Vj
<
1
<
CM CO
"*
in in
CO
o
o
o
in
•<*
in
r*»
2 < S
<J
a
^
CO CO
in
o o
o o o
o
o
m
co t- in
o
en 00
t- i- CM CM
CO
CM CM
o
CO
CO
m m m s o
o
CO 1^
CO
<t CO
in
CO Is- CJ> CO
CM CM CM CO
> Q-cvj t?
CM CO CO
E
co
cd
JS
._
o
o
o o Ev.
o
CO CM 00
CM CO CO
CD
p
o o
o
o
he
t
o
o
in in
130
180
210
250
380
440
«>
Jfi
reat
for
heat-1
f- CO
18)
<«•
elasticity
o
CO
CO
-i-
actor
o
O iO O
O o o
co
h- r^
m
co -t -t m
CD
CO CO
strength
of
CO
<D
C\J
i
Tensile
Modulus
o
en
O
O O O CO
o o
o
1^ O "* CO
im
i- CM CM CM
CO
CO CO
m
m
in in
o
o
co
CD
o) co
en m
CM CM CM CO
-o
CM
210000
210000
210000
210000
o
o
o
o o
o
o o
o
in
o in
o
o
CM CM
CM
o
o
CD
o
o
o
o
CM
o
o
o
en
o o
o o
o
o
o o
in m
o
o
CM CM
/vith
P
400
600
700
safety
and
a
2
F
Allow
(see
specially
165000
170000
es
185000
ameter,
*)
d
C D 55 65
the
Grade
Grade
Grade
Grade
CO
Material
A-Standard
283
3
IUOJJ SUOI}
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aterial
dard
^
572
572
A A A A o
5
LU
H
<
CO
CO
<
o
20 22 30 40
in
A A M M
o
co
o
050
§
284
060
080
080
*
o
in
CM
CM
60-40-18
80-55-06
O
5
100-70-03
pends
de
LU
<
CO
1^
<
o
00
o
LU
CO
<
CO
o
CO
<
§
1-
<
o
in
CM
536
ASTM
A
seeP2
igth
2 CO CM ^-^2
oo O 0)0) to
* CD S .E .= CD
DOS * £2 H
o
WOJJ SUOjJ
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;s
CO
Jj
co
on
O
Alter
CO CM
CO
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CO
o
en
U:Undi
BS
2789
Note:
A:
TABLES
17
Strength and Machining values
Allowable bending and torsional stresses; E and G moduli
for elastic materials in
N/mm 2
Modulus of Type of
Material
elasticity
loading
Modulus of
1 '
80000
110000
200 100
150
100 50
42000
142000
300
250
200
55000
110000
200 100
150
100 50
42000
120
100
80
117000
300
220
150
45000
200
180
150
Nickel Silver
HV160
ASTM-B 122(752)
Tin Bronze
CDA-419 HV190
Phosphor Bronze
CDA-529
Pqt
1000
750
500
HV150
65-18
G
210000
Yellow Brass
ASTM-B 134(274)
rigidity
B
A
Spring Steel
SAE1078;
hard. + temp.
Pbt
E
HV190
for simple springs
for bent and shaped springs
for springs with no hysteresis effect
650
500
350
120
100
80
200
180
150
(safety factor
(
(
— 1-5)
-
"
- 3)
"
"
^10)
v For explanation refer to P 1.
2)
For cylindrical helical springs use diagram on page Q 9.
Characteristic quantities for machining
(for turning outside longitudinally)
Material
ASTM - A572iGrade
UNS - K 04600
SAE
SAE
SAE
SAE
SAE
SAE
SAE
- 1045
- 1060
- 5120
- 3140
- 4135
- 4140
- 6150
SAE - L6 annealed
SAE - L6 tempered
Mehanite A
Chilled cast iron
ASTM - A 48-40 B
Strength in N/mm 2
mc
or hardness
520
720
670
770
770
630
600
730
600
940
ASTME18-74-HRD54
360
ASTME18-74-HRD60
ASTME18-74-HRD33
1
- mc
0-26
0-30
0-14
0-74
0-70
0-86
0-18
0-26
0-30
0-82
0-74
0-70
0-21
0-26
0-26
0-79
0-74
0-74
0-24
0-24
0-26
0-76
0-76
0-74
0-19
0-26
0-81
fcci.1
N/mm 2
0-74
Specified values apply directly for turning with carbide tip
Cutting speed v = 90 ... 125 m/min
< h < 2-5
Ratio of slenderness
Chip thickness h = 005
Normal side-rake angle y = 6° for steel, 2° for cast iron
mm
mm
|
1990
2260
2220
2130
2100
2260
2240
2500
2220
1740
1920
1270
2060
1160
TABLES
18
Strength values in N/mm 2
Permissible contact pressure p b in N/mm 2
Bearing pressure of joint bolts (Building construction DIN 1050)
Load characteristic
material
material
pb
pb
|
|
main load
main and additional load
|
ASTM-
206
A 283 Grade C
235
ASTMA440
304
343
Journals and bearings, bearing plates (see q 13)
Hydrodynamic lubrication see q 47.
Mixed lubrication, shaft hardened and ground: 1)2)
ms
Material
Pb
8. .12
(836) Leaded
m j. Red Brass
(938) Leaded
Tin Bronze
sintered iron
Pz
m/s
Cast Tin Bronze CDA90I
gray cast iron
20 3)
3
15 3
grease lubrication
quality bearings
4. ..12
6G
^0
PA66 (polyamide)
>
15
009
dry
...1
grease lubrication
<1
5
0-35
'
HDPE
2. ..4
(high-density
sintered iron
with copper
sintered
002
polyethylene)
PTFE (polytetra-
12
bronze
-*0
30
fluorethylene
006
enclosed)
tin-bronze
PTFE - lead
+ bronze
20
graphite
(DEVA metal)
90 4
(GLACIER-DU)
>
<0005
1404)
0-5. .5
General, non-sliding surfaces: Max. values are possible up to
the compressive yield point at the material (o6f
R e ). But
normal values for good p b are lower.
:
Material
—
Normal values of p b under
shock load
undulating load
dead load
bronze
15
30
12
.. 30
.. 50
10
20
..
cast iron
gunmetal
8
..
20
30
malleable iron
steel
..
11
(p x L')p e rm are closely related to heat dissipation, load, bearing pressure,
type of lubrication.
21
Sometimes a much higher load capacity with hydrodynamic lubrication
is
possible
3)
Limited life (wearing parts).
41
Specially developed metals
5)
I
For shell thickness 1
mm
TABLES
2:i9
Data for clutches and brakes
Properties of friction materials
(Q 15 ... Q 17)
"CO
CO
^~
CO
CM
cvj
CO
CO
C\J
in
CVJ
iri
CVJ
in
CO
in
O)
iri
CM
CVj
C\j
o
2
Contact
pressure
o
tn
«-
CVJ
c
'c
c
B T3
O_
-
Pb
o
c
-
CO
-
c
N/mm
in
d
d
o
°
T
(O X
o
m
Q)
CO
oo
CO
CD
o
O
°C
transient
o
o
o
o
perature
o
°
-r
co -
o
o
CO
o
m
CO
o
o
*
o
m
o
m
O
w
*O
CO
1
ten
-1
Max.
CO
°C
continuous
in
o o
o
o
CD
CD
CD
O o
O
o
in
in
o
m
O
O
o
o
CO
CVJ
'
4
coefficient
Sliding
m
CD
d
d
CVJ
m
CO
d
/"slide
friction
d
"to
II
m
o
m
o
CO
o
o
CVJ
o
o
o
CVJ
cvj
CO
CD
d
d
CVJ
CO
d
d
co
o d d* d
UO
i>
c
H
CO
o o
o m
d d
in
-Q
6 d
_3
C
o
0>
CD
iron
_c
CO
cast
Q.
£
2
*
5
friction
E
3
Q.
0)
CO
5?
CO
o
en
Q)
O
a)
o
E
o
C
-Q
a
CD
CD
organic
ll |
CO
general
lining/steel
c
CO
Q)
or
0)
CD
O
C
o
o
en
0)
CD
a3
a>
J5
o
E
o
D
up
cvi
~CD
N
C
o
-D
n
o
TJ
T3
CD
CD
to
C
C
CO
CO
C
;
co cq
o
1
75
w
II
II
=
0)
CO
03
o
/
N
C
o
m o o
c
£
to
"CD
a)
CO
00
(9^ JM
CO
«
^
^ Q=L 3.
co
~ ST m
TABLES
20
Work w and yield strength kf
0,5
1,0
2,0
1,5
SAE3310
0,25
<p
0,50
0,75
•
1,00
0,5
:
logarithmic deformation ratio
:
strain energy per unit volume
|
w
1,0
1,5
<P
For other materials see VDI 3200
Af
:
yield strength
2,0,
TABLES
Z21
Electrical properties
Electrical specific resistance g
and specific conductance y of conductors at
Q
material
aluminium
antimony
brass - 58% Cu
brass - 63% Cu
Q mm 2
m
m
Q mm
/
= 20 C
9
Y
material
2
y
Q mm 2
m
m
Q mm 2
0.0278
36
iron (pure)
0.10
10
0.417
2.4
lead
0.208
4.8
0.059
17
23
14
magnesium
manganese
0.0435
0.071
0.423
2.37
1.063
cadmium
0.076
13.1
mercury
0.941
carbon
40
0.025
mild steel
013
7.7
1
1
nickel
0.087
11.5
2.0
cast iron
chromium-Ni-Fe
constantan
copper
0.10
10
nickeline
0.5
0.48
2.08
platinum
0.111
9
0.0172
58
silver
0.016
62.5
German silver
0.369
2.71
tin
0.12
8.3
gold
0.0222
45
tungsten
0.059
17
graphite
8.00
0.125
zinc
0.061
16.5
Electrical resistance g of insulators
material
9
materia
Q cm
14
baskelite
10
glass
10 15
marble
mica
10
10
paraffin oil
paraffin wax (pure)
plexiglass
brass
carbon
constantan
copper
German silver
graphite
manganese
15
10 18
10
porcelain
10
14
17
pressed amber
10
18
10
18
vulcanite
10
10
18
water, distilled
10
material
1/Kor 1/°C
+ 0.00390
+ 0.00150
mercury
- 0.00030
- 0.00003
nickel
+ 0.00380
+ 0.00070
-0.00020
±0. 00001
16
7
= 20°C
O20
a20
aluminium
10
polystyrene
Electric temperature coefficient « 2 o at /
material
Q cm
»
mild steel
nickeline
platinum
silver
tin
zinc
1/Kor 1/°C
+ 0.00090
+ 0.00660
+ 000400
+ 0.00023
+ 0.00390
+ 0.00377
+ 0.00420
+ 00370
TABLES
7,
£- 2i
Electrical properties
Dielectric constant e r
insulant
insulant
Er
36
araldite
atmosphere
1
bakelite
casting
3.6
insulant
quartz
micanite
nylon
5
5
shellac
4.5
3.5
slate
4
paper
soft rubber
2.5
steatite
6
3.5
2
25
olive oil
4
3
2.3
sulfur
teflon
5
8
transformer oil
mineral
transformer oil
vegetable
turpentine
caster oil
4.7
paper
ebonite
25
paper,
glass
5
guttapercha
hard paper
4
paraffin oil
paraffin wax
2.2
(laminated)
insulation of high
voltage cables|
insulation of telephone cables
4.5
petroleum
phenolic resin
2.2
plexiglass
polystyrene
porcelain
3.2
vulcanised fibres
3
vulcanite
1.5
4.4
water
marble
8
pressed board
4
1
|
impregnated
4.2
£<
6
oil
compound
£r
mica
2.2
2.2
2.5
2.2
2.5
80
El ectro-motive series
(potential difference 3 with respect to hydro gen electrode)
material
V
V
material
volt
potassium
calcium
-2.93
-2.87
sodium
-2.71
iron
magnesium
-2.37
-1.85
-1.66
-1.19
-0.76
cadmium
beryllium
aluminum
manganese
zinc
chromium
-0.74
-0.58
-0.41
-0.40
-0.28
-0.23
-0.14
-0.13
tungsten
cobalt
nickel
tin
lead
V
material
volt
volt
hydrogen
antimony
cooper
0.00
+ 0.10
+ 0.34
+ 0.80
+ 0.85
+ 1.20
+ 1.50
+ 2.87
silver
mercury
platinum
gold
fluorine
Standardize! J numbers using progi ession ratio
according to E-series
Shown for E 6 ... E 24
(
24 /
1
E 6 series (« |/l0)
1.0
2.2
4.7
E
1
2 series (= ^10)
1.0
2.2
4.7
E 24 series (~ l/i0)
1.0
1.1
1.2
2.7
5.6
1.2
2.0
3.0
3.3
3.6
3.9
4.3
10
22
1.3
1.5
3.3
6.8
1.5
3.3
6.8
1.5
1.6
1.8
10
22
etc
47
10
3.9
22
etc.
82
47
2.2
2.4
2.7
1.8
etc.
4.7
5.1
5.6
6.2
6.8
7.5
8.2
9.1
47
TABLES
23
Magnetic properties
H
Magnetic field strength
and relative
permeability // as a function of induction B
r
steel casting
cast ron
allow sheet and strips steel
P1.0 = 1.3*
kg
H
T.34
m
2
G
[
Mr
A/m
-
215
243
H
kg
A/m
-
[gauss
1000
2000
3000
440
740
980
4000
5000
6000
1250
1650
2100
7000
8000
9000
3600
5300
7400
97
230
10000
11000
12000
10300
14000
19500
77
63
295
370
49
520
2690
2360
1830
13000
14000
15000
29000
42000
65000
36
750
1250
2000
1380
890
600
16000
17000
18000
3500
7900
12000
363
19000
20000
21000
19100
30500
50700
79
22000
23000
130000
218000
0.4
0-5
0.6
0.7
0.8
0.9
1.0
1.3
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
-
Mr
A/m
-
I
0.1
0.3
H
Ur
tesla
0.2
dynamo
|
P1.0 = 3.6^
induction B
alloyed
i
and dynamo
practica limit
181
2650
2650
2980
8,5
4180
3310
3410
65
90
125
4900
4420
3810
3280
3350
3110
170
220
280
3280
2900
2550
355
460
660
2240
1900
1445
820
1260
495
265
254
100
241
120
227
140
154
170
120
190
29
18
9390
6350
5970
30
60
80
171
119
52
25
40
*
2250
4500
8500
13100
21500
150
103
39000
115000
39
67
14
33
13
4
Note
P1.0 s eeZ24
TABLES
Z24
Magnetic properties
D ynamo sheet properties
mild
sheet
type
and
allow sheet and strips steel
strips
steel
low
medium
3.0
2.3
high
stray power
at 1:0 T in W/kg
thickness
3.6
mm
density kg/dm
3
1.3
1.5
0.35
0.5
7.8
7.75
7.65
P 1.0
3.6
3.0
2.3
1.5
1.3
P 1.5
8.6
7.2
5.6
3.7
3.3
7.6
core
losses
per unit
at/= 50Hz
W/kg
(max.)
#25
induc-
#50
tion
(min.)
#100
#300
Vs/m 2
1.53
1.50
1.47
1.43
[gauss]
[15300]
[15000]
[14700]
[14300]
Vs/m 2
1.63
1.60
1.57
1.55
[gauss]
[16300]
[16000]
[15700]
[15500]
Vs/m 2
1.73
1.71
1.69
1.65
[gauss]
[17300]
[17100]
[16900]
[16500]
Vs/m 2
1.98
1.95
1.93
1.85
[gauss]
[19800]
[19500]
[19300]
[18500]
Explanations
2
2
fl 2 5 = 1-53 Vs/m indicates that a minimum induction of 1.53 Vs/m
[or 15300 gauss] is reached with a field strength of 25 A/cm.
Thus a flux length of e.g. 5 cm requires a circulation of
5 x 25 A = 125 A.
P 1.0
P 1.5
describes the core losses per
unit mass at / = 50 Hz and an
induction of
1.0 Vs/m
2
= [10000 G]
1.5 Vs/rr
2
= [15000 G]
\
TABLES
Z 25
Lighting values
Guide values for
il
lumination Es/ in Ix = lm/m 2
lighting
or location
rough
workshops
according
to work done
medium
precise
very precise
normal
offices
medium
lighting
streets and
light
medium
traffic
heavy
factory yards
light
with traffic
heavy
bench
general
100
200
300
500
50
100
200
300
200
500
1000
1500
200
500
bright
squares with
lighting
only
500
750
open
rooms,
living
General and spec.
General
Type of establishment
20
50
100
20
50
Lumin ous efficac
Colour o f illuminated surface
medium
dark
Type of ighting
light
direct
0.60
0.45
indirect
0.35
0.25
0.30
0.15
deep bowl
street and
widespread
reflector
0.40
0.4 5
Lum inoui flu: (
i
V
of la mps
Standard lamps
Pel
w
15
25
40
60
75
100
with single coiled
filament
(at operating voltage)
<P V
klm 0.12
0.23
0.43
0.73
0.96
1.39
150
200
300
500
1000 2000
klm 2.22
3.15
5.0
8.4
18.8
W
Pel
<Z> V
40.0
tubular
Fluorescent lamps
values for
diameter
26 mm
Pe\
V
'Warmwhite'
'Daylight'
38 mm
High-pressure
W
18
36
58
klm
1.45
3.47
5.4
W
15
20
40
65
klm 0.59
1.20
3.1
5.0
Pel
W
125
250
400
700
1000 2000
</>
klm
6.5
1 4
24
42
60
Pel
V
lamps filled with
mercury vapour
|
v
125
TABLES
26
—
Statistics
1
<p(x)
,
I
0-398 942
0398 922
0398 862
*oW
df;
e
erf (x)
A
cp(x)
o (x)
erf (x)
0000 000
0007 979
0.000 000
0.50
0352 065
0011 283
051
350 292
0.52
0.348 493
0.346 668
0.344 818
0.520 500
0.529 244
0.537 899
0.546 464
0.554 939
0.55
0.56
382 925
0.389 949
0.396 936
0.403 888
0.410 803
417 681
0.424 521
0.571 616
0.431 322
0.579 816
0.438 085
0.444 809
0.595 937
015 957
0.023 933
0398 623
0398 444
0.031 907
0.022 565
0.033 841
0.045 111
0.039 878
0.056 372
398 225
0.397 966
0.397 668
0.397 330
0.396 953
0.396 536
396 080
0.395 585
0.395 052
0.394 479
0.393 868
0.393 219
0047 845
0067 622
0.055 806
0.063 763
0.078 858
0.090 078
0.101 281
057
058
0.112 463
0.123 623
134 758
0.145 867
0.156 947
0.60
0.167 996
0.179 012
0.189 992
0.200 936
0.65
0.66
0.67
0.68
0.69
0.391 806
£K-
$oW
0.398 763
0.392 531
erf(x) =
0.071 713
0079 656
0.087 591
0.095 517
103 434
0.111 340
0.119 235
0.127 119
0.134 990
0.142 847
0.150 691
0.158 519
0.166 332
0.174 129
0.211 840
053
0.54
0.59
0.61
0.62
0.63
064
0.342 944
341 046
339 124
0.337 180
335 213
333 225
0.331 215
0.329 184
0327 133
0.325 062
0.322 972
320 864
0.318 737
0.316 593
0.314 432
0.312 254
0.310 060
0.563 323
587 923
0.451 494
603 856
0.458 138
0.464 742
0.611 681
0.471 306
0.477 828
0.484 308
0.490 746
497 142
0.503 496
0.509 806
0.516 073
0.522 296
0.528 475
0.534 610
0.540 700
0.546 745
0.552 746
0.558 700
0.564 609
0.570 472
0.619 412
0.627 047
0.634 586
0642 029
0.649 377
656 628
0.663 782
670 840
0.222 702
0.233 522
0.244 296
0.70
0.181 908
255 022
073
0.189 670
0.265 700
0.74
0.197 413
0.205 136
0.212 840
0276 326
0.75
0.76
0.77
0220 522
0307 880 078
228 184
0.235 823
0.243 439
0.251 032
0.258 600
0.266 143
0.318 283
0.79
0.292 004
0.328 627
0.80
0.81
0.289 692
0.287 369
0.576 289
0338 908
582 060
748 003
0.349 126
0.82
0.83
0.84
0587 784
0.753 811
0.759 524
0.765 143
366 782
0.273 661
0.281 153
0.288 617
0.296 054
0.303 463
0.310 843
0.318 194
0.365 263
0.363 714
0.362 135
0.360 527
0.358 890
0.357 225
0.355 533
0.353 812
325 514
0.332 804
0.340 063
0347 290
0.354 484
0.361 645
0.368 773
0.375 866
379 382
389 330
0.399 206
0.409 009
0.418 739
0-428 392
0.437 969
0.447 468
0.456 887
0.466 225
0.475 482
0.484 656
0.493 745
0.502 750
0.511 668
285 036
0.282 694
0.280 344
0.277 985
0.275 618
0.273 244
0.270 864
0.268 477
266 085
263 688
0.261 286
0.391 043
0.390 242
0.389 404
0.388 529
0.387 617
0.386 668
0.385 683
0384 663
0.383 606
0.382 515
0.381 388
0.380 226
379 031
0.377 801
0.376 537
0.375 240
373 911
0.372 548
0.371 154
0.369 728
0.368 270
0.286 900
0.297 418
359 279
0369 365
0.71
0.72
0.85
0.86
0.87
0.88
0.89
0.90
091
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
0.307 851
0.305 627
0.303 389
0.301 137
0298 872
0.296 595
294 305
0.258 881
0.256 471
0.254 059
0.251 644
0.249 228
0.246 809
0.244 390
0.593 461
0.677 801
0.684 666
0.691 433
698 104
704 678
0.711 156
0.717 537
723 822
0.730 010
736 103
0.742 101
599 092
604 675
610 211
0.615 700
0.770 668
0.776 100
0.621 141
0.786 687
0.626 534
0.791 843
0.631 880
0.796 908
0.637 178
0.642 427
0.647 629
0.652 782
0.801 883
0.657 888
0.662 945
0.667 954
0.672 914
0.677 826
0.781 440
0806 768
0.811 563
0.816 271
0.820 891
825 424
0.829 870
0.834 231
0.838 508
TABLES
f
2I 27
Statistics
2
X
<p(x)
1.00
0.241 971
0.239 551
0.237 132
0.234 714
0.232 297
0.229 882
0.227 470
0.225 060
0.222 653
0.220 251
0.217 852
0.215 458
0.213 069
0.210 686
0.208 308
0.205 936
0.203 571
1.01
1
02
1.03
1.04
1.05
1.06
1.07
1
08
109
1.10
1.11
112
1.13
1.14
1.15
1.16
1.17
1.18
1.19
0.201 214
0.198 863
0.196 520
1.20
0.194 186
1.21
0.191 860
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
0.189 543
0.187 235
0.184 937
0.182 649
0.180 371
0.178 104
0.175 847
0.173 602
1.30
0.171 369
1.31
0.169 147
0.166 937
0.164 740
0.162 555
0.160 383
0.158 225
0.156 080
0.153 948
0.151 831
0.149 727
0.147 639
0.145 564
0.143 505
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
1.41
1.42
1.43
1.44
0.141 460
1.45
1.46
1.47
1.48
0.139 431
0.137 417
0.135 418
0.133 435
.49
0.131 468
1
2
*
fW-^yi*
f
--JU fe" 2
V2FJ
-
df;
erf(jc)
=
-im V'.d,
V>T.
<Po(x)
erf (x)
X
<f(x)
<P (x)
erf (x)
682 689
0842 701
150
0.846 810
0.850 838
0.854 784
0.858 650
1.51
0.129 518
0.127 583
0.125 665
0.123 763
0.866 336
0.868 957
0966 105
0.687 505
0.692 272
0.696 990
0.862 436
0866 144
1
0.869 773
0.873 326
0.876 803
1.57
1.58
1.59
1.60
0.745 714
0.880 205
0.883 533
0.886 788
0889 971
0.893 082
0.749 856
0.753 951
0.757 999
0.762 000
0.765 953
0.896 124
0.899 096
0.902 000
0.904 837
0.907 608
165
166
0.769 861
0.773 721
0.777 535
0.910 314
0.912 956
1.70
0915 534
0.781 303
0.918 050
0.920 505
1.72
1.73
1.74
0-701 660
0.706 282
0.710 855
0.715 381
0.719 858
0.724 287
0.728 668
0.733 001
0.737 286
0.741 524
0.785 024
152
0.873 983
0.876 440
0969 516
0.121 878
1.55
0120 009
0.878 858
0.971 623
56
118 157
0.116 323
0.114 505
0.112 704
0.110 921
0.109 155
107 406
0.105 675
103 961
0.102 265
0.100 586
0.098 925
0.097 282
0.095 657
0.094 049
0.092 459
0.090 887
0.089 333
0.087 796
0086 277
0.084 776
0.083 293
0.081 828
0080 380
0.078 950
0.077 538
0.076 143
0.074 766
0.073 407
0.072 065
0.070 740
0.069 433
0.068 144
066 871
0.065 616
0.064 378
0.063 157
0.061 952
0.060 765
0.059 595
0.058 441
0.057 304
0.056 183
0.055 079
0.881 240
0972 628
0.883 585
0.885 893
0.888 165
0.973 603
0.974 547
0.975 462
0.976 348
0.977 207
1.53
1.54
1.61
1.62
1.63
1.64
1.67
1.68
1.69
1.71
0.788 700
0.792 331
0.795 915
0.799 455
0.802 949
0.922 900
0.925 236
0.927 514
0.929 734
0.806 399
0.809 804
0.813 165
0.816 482
0.819 755
0.934 008
0.936 063
0.938 065
0.940 015
0822 984
0826 170
0.943 762
0.945 562
1.85
0.829 313
0.832 413
0835 471
0947 313
1.87
1.88
0.838 487
0.841 460
0844 392
0.847 283
0.850 133
0.852 941
0.855 710
0.858 438
0.861 127
0.863 776
0.931 899
0941 914
949 016
950 673
0.952 285
0.953 853
0.955 376
0.956 857
0.958 297
0.959 695
0.961 054
0.962 373
0.963 654
0.964 898
0.871 489
0.967 277
0.968 414
1.75
1.76
1.77
1.78
1.79
1.80
1.81
1.82
1.83
1.84
186
189
1
90
1.91
1.92
1.93
1.94
195
1.96
1.97
1.98
1
99
0.890 401
0.892 602
0894 768
0896 899
0.970 586
0.898 995
0978 038
0978 843
0979 622
901 057
0.980 376
903 086
0.981 105
0.981 810
0.905 081
907 043
0.908 972
0.910 869
0.912 734
0.914 568
0.916 370
0918 141
0.919 882
0.921 592
0.923 273
924 924
0.926 546
928 139
0.929 704
931 241
0932 750
934 232
0.935 687
937 115
0.938 516
0.939 892
0941 242
0.942 567
0.943 867
0.945 142
946 393
0.947 620
948 824
0.950 004
951 162
952 297
0.953 409
0.982 493
0.983 153
0.983 790
0.984 407
0.985 003
0.985 578
0.986 135
0.986 672
0.987 190
0.987 691
0.988 174
0.988 641
0.989 090
0989 524
0989 943
0.990 347
0.990 736
991 111
0991 472
991 821
0992 156
0992 479
0.992 790
993 090
993 378
993 656
0.993 922
994 179
0.994 426
0.994 664
0.994 892
0.995 111
.
INDEX
acceleration
- angular
,
-
due to gravity
time curve
Acceptable Quality Level
L 3
,
,
L
L
L
L
2
2
9
7
G 11
U
T
T
T
T
acid-base-indicators
action diagram
basic structures
- - rule for addition
,
,
- line
active power
4
3
4
4
3
S29
T
actuator
6
addendum
Q 18 Q20
admittance
S 18
D 9
D 10
algebraic equation, definition
general solution
,
,
relationship between
D
D
D
zeros and coefficients
,
,
solution
zeros, roots
9
9
9
9. .D 12
bars of circular cross section
P20
non-circular cross section P 20
basic differentials
H 4 H 5
- integrals
3 ...I 13
beam of uniform strength
P 16
- of varying cross section
P 15
beams
P 11. ..P 16
- of uniform cross section
P 11
bearings
Q 10. .Q 14
bearing clearance
Q11
- friction
K 12
bearing journal
Q10
- pressure
Q11
bearing rolling
Q 10
- stress
Q 2
Becquerel
V 6
Belleville spring
Q 6
I
bending
- in one axis
of any degree
- expressions
D
D 2 D21
two planes
alternating current
S 16. ..S29
- moment
- of uniform beams
- stress
S22
bridge
American hardness
amplitude response
for the whole circuit
P 5, P 6
analysis of forces
Angstrom
angle of twist
angular frequency
annuity interest calculation
annulus
apparent power
A.Q.L.
arc differential
- length
U 6
T 3
T24
,
P
7
A 1
P20
S16
D25
,
I
I
14
14
F
7
3
3
3
,
bolts
Q
Q
boring
R
bottom clear
- tearing
Bow diagram
bracket attachment
brakes
Q18
brake torque
branching point
U 6
P 19
buckling
axiomes to the probability
G
1
buoyancy
axles
Q
2
band brakes
K 13
bankholder forces
R
7
calculation for a controller
calculation of module
calorific value
barrel
C
4
candela
axial shear stress
2
bolted joints
A 1
D 17
D 17
asymptote
atmospheric relative humidity
K
P 1 1 .P 13
P 9 Z 17
Bernoulli Differential Equation
J 10
Bernoulli's theorem
N 4
bevel gears
Q 24 Q25
bimetallic strips
O 3
binomial series
D 18
B 3
S 18
-
units of
P24
P28
P24
shafts
G 11
K
N
,
P 9 P 10
- theorem
D 4
- Bode diagram
T3, T22.T23, T 29 T31
boiling point
Z 1. ..Z 6
- of circle
Archimedes principle
area expansion
arithmetic mean
- series
K13
belt drive
build up time
R
K
1
1
2
7
6
Q
Q 17 Z 19
Q17
1
T
P 22
3
P23
T
N
2
3
T18
Q28
Z10
V 1
S3 S 12
capacitance
- of coaxial cylinders
capacitor power
S 12
carat
S31
A 3
Cardan joint
L 10
Cartesian coordinate system
D23
Cavalieri principle
center of gravity
14,
central limit theorem
centrifugal force
changes of state of ideal
1
C
I
K
15
G
M
gases
1
8
3
5
5
characteristic angular
T
S
U
U
frequency
charge
chemical elements
- terms
U 2
choke coil
S26 S27
,
.
.
,
arc of
sector of
segment of
circumcircle
Clairaut' Differential equation
clamped joint
clutches
coeff. of elongation
friction
Q28
- - heat transmission
2
7
7
7
E 6
J 11
Q 3
S 12
P 2
Z 7
Z 11
restitution
M
rolling friction
Z
Z
Z
sliding
static friction
non-magnetic
cold working of sheet
combination of stresses
combinations
comparator
8
7
7
7
S25
S24
S24
Q 8
S24
coercive force
coil, high frequency
- low frequency
coiled spring
coils,
1
3
Q 15 Q 16 Z 19
coaxial cylinders, capac. of
- -
2
B 3 F
K
B3 K
B3 K
circle
-
1
R 6
P24. .P29
D 5 D 6
T
complex numbers
D29
components of the control
M
P3
P
7
4
2
conditional probability
G
1
conductance
S
2
conduction of heat
conductivity, thermal
M
10
Z 1. ..Z 6
7
list of symbols
T34
control factor
T 7
control loop, rules to determine
the transfer function
T 9
control loops, components
T 4
quantities and
functions to describe the
dynamic behaviour
T 7
controlled system
T 4
- variable
T 5
T 8
overshoot
- point of measurement
T 5
T 6
controller
- output variable
T 6
controlling element
T 6
choice of the type
T20
,
,
.
,
,
,
determinations
T 25 ,T26
proved
setting rules for P-, PI- and
T33
T20
(PID)-types
the most important
,
- system
T
6
10
4
3
convex mirrors
14
Q
13
Q
cooling
D 18
coordinate system. Cartes.
D 19
polar
copper losses
S29
core losses
S 25 S28
T 3
corner angular frequency
E 6
cosine rule
convection of heat
conversion of logarithms
D
V
,
,
coulomb
S
counterflow exchanger
Cremona method
critical
speed
by vibrations
cumulative distribution
- pendulum
- wound motor
compression
compressive stress
2
G 10
conical pendulum
control engineering terms,
P
D 12
C
confidence statement
S32
4
compound interest
2
Q 15
,
crosshead guide
cross section, nominal
- switch
cube
cubic coeff. of expansion
cuboid
T
loop
6
D30
C
cone
- clutch
- frustum
function
current
- density
current rating
- source
2
11
K
6
6
Q 10
Q 15
M
S37
S37
C
1
Z 11
C
G
G
S
S
1
9
2
1
2
S37
S
9
curved beams
P25
cutting drives
force
R
R
R
R
R
R
-
gears
power
-
round
times
C
C
C
cylinder
-
,
hollow
,
sliced
1
2
1
2
3
3
2
2
4
direct feedback
direction cosines of vectors
T
9
F
7
disc brakes
- springs
distance law, optical
- -time curve
disturbance variable
Q17
Q 7
drilling
V
1
L
3
T
6
R2 R 3
Q27
,
drive worm
drying agents for diseccators
U 6
dynamic moment of inertia
18
N 1 ,Z 12
dynamic viscosity
I
damping ratio
D.C. machines with
commutating poles
S33
decimal-geometric series
D 17
T
1
Q18 Q20
dedendum
deep drawing
R
P 12
deflection
deformation efficiency
- ratio, logarithmic
delay time
density
N 1, Z 1
- determination of
deposit calculation
..
,
derivative
derivatives
H 4
R
R
T
.Z
N
8
8
1
6
3
D25
H
H
..
D
Descartes theorem
desiccants
determinants
determination of o
6
P20
,
U
D7 D
G
3
6
9
6
8
6
a (PID) -T, controlling
element (example)
T31 ..T33
a PI- or P-controlling
element (example)
diagram-hub geometry
diameter, base
-
,
reference
,
root
•
tip
Q19
Z
H
H
,
order
- - ,2 nd order
th
n order
1
,
,
J 5
J 2
,
.
.
.
.
reduction of order
with constant
J 5
coefficients
D 17
DIN-Series
dioptrics
direct-current machine
1
1
J 12
J
linear
partial solution
2
1
J 10
J 4, J 9,
J 4, J 5, J 11
,
- -
.
.
definition
st
,
J 1
Z23
eddy currents
Q 12
S25
Eigenangular frequency
T
elastic curve
electric circuits
P 11
J 12
J
J
J
J
,
- networks
calculation
,
.
.
,
Euler formula
-
4
1
S32
S13
Z22
U
1
B3 F
4
U
6
4
4
,
Q26
.
D 16
D26
K 4
P 15
T
2
G
G
8
8
8
6
- molecular mass
-
differential equation
expansion, area
-
,
linear
of gases
of liquid bodies
of solid bodies
,
volume
1
S 12
epicyclic gearing
equations of any degree, approximate solutions D 13 ..
equilateral triangle
B2
equilibrium
7
R
V
S
enthalpy
entropy
3
8
7
1
Z21
Z21
English hardness
8
J
9
9
S
,
electro-magnetic rules
-motive series
elements, chemical
error curve
5
S8 ,S
S8 S
electric field
equivalent beam
- dead time
1
S12
S11
- power
- specific resistance
- temperature coefficient
- work
electrical work in an
function
variable
,
S
- field
electrical measurements
ellipse
T30 ..T32
Q 5
Q 19
Q 19
Q 19 Q20
dielectric constant
difference coefficient
differential coefficient
- equations
dynamo sheet properties
eccentricity, relative
T
P22
J 11
3
3
4
3
3
3
G
G
expected mean
experimental probability
exponential curve
- equation
3
1
F
4
D 4
H 5
T 10
- functions
extended feedback rule
S 3
R 4
R 4
R 4
T 9
T 10
T 5
T 5
farad
feed drives
power
rate
feedback rule
- rule, extended
- variable
controlled variable
- controlled variable,
final
formation
- controlling element
- controlling equipment
first order delay element
T 1
,
T 5
T 6
T 6
T 14
fixing bolts
Q
flow, equation
friction loss
N 5
N 6
N 6
N 6
N 4
N 4
S 3
S 12
,
-
,
,
laminar
turbulent
fluid, ideal
-
,
flux,
-
,
real
density
magnetic
M
flywheel effect
focal point
F
F
focus
Foelmer formula
Q
force between magn. poles
-
2
4
2
2
S 15
M
M
moment
polygon
K
K
units of
A
2
2
K
3
T
5
1
T 6
D 22
D 25
D 20
- transformation
D 23
- - calculation rules
D23
- - convolution
D23
correspondence
D 16 D25
..
..
time translation
free fall
freezing mixtures
frequency
- angular
- resonant
locked joints
losses in pipe flow
materials, properties
D23
L
U
.
N 6 Z 8
Z 19
Z 15
,
K13
,
rope
,
sliding
K
frustrum of cone
C
C
-
of pyramid
functional block
functions, exponential
-
,
,
,
,
3
3
K13
Q 15 Q 17
K9 ,Z 7
Q 3
numbers for fluids
hyperbolic
inverse trigonometrical
logarithmic
trigonometrical
9
2
1
T
H
H
H
H
H
6
5
T
7
3
5
6
6
gain crossover angular
frequency
margin condition,
-
,
gain margin
,
S21
5
5
04 ,Z 12
08
9
Z14
gases, tables
laws
O
S
gauss
Gaussian curve
G
Q 18
general equation of state
generators
geometric mean
- series
application of
..
4
3
8
Q29
Q 5
4
S33
D17
D 17
D31
German hardness
U
Grashof number
group combinational elements
(PD)-T, and (PID)-!",
012
gravitational force
K 1
N 2
V 5
R 3
6
golden section
D32
gradient of a curve
H 1 H 2
graphical determination of a
T 21 .. .T32
linear controller
- pressure
grinding
1
2
A
A
,
gray
L
S
,
- United States
gas constant
- mixtures
8
5
,
T28
T8 T20
realization
-
gallon, Imperial
,
forming device for the referFourier series
T
T
.
K 12
bearing
clutches
coefficient
gearing, epicyclic
,
ence variable
,
gears
,
,
-
1
,
trolled variable
<9
5
centrifugal
gravitational
forces composition
K 2
formation of the final con-
response characteristics
response
1
,
,
-
friction
T17
half-angle rule
E
hardness of water
harmonic oscillation
U
L
M6 M
- oscillations
heat conduction
-
7
integral, definite
O 11
O 7
O 2
O 2
O 10
Q13
Q 14
O 10
flow
latent
of sublimation
radiation
,
removal
by lubricant
-
involute toothing
10
10
exchange
exchanger
,
6
6
4
transfer
coeff.
10,
-
transmission
values
helical spring
henry
Z 14
11
10,0 11 012
Z 10
..
.Z13
Q 9
S
S37
S
-
indefinite
,
1
D 10
Hurwitz criterium
D 12
..
N4
..
hydrostatics
N 1
..
N
Z
N
hyperbola
F
F 5, F6
hypergeometric distribution
H
hyperbolic functions
- power
- work
9
9
,
E
F
7
Kelvin
A
9
3
3
6
9
key joints
3
6
6
Q
S
Kirchhoff laws
N
laminar flow
laminated leaf spring
Laplace transformation
Q
D26
convolution
correlation-table
Laplace Transform
Laplace transformation
differential equations
- - translation
,
S 19
S20
,
E
inclined plane
K 10 L
6
9
G
1
,
independent events
induced voltage
inductance
induction motor
inertia, dynamic
S
,
moment of
moment of
inhomogeneous differential
equation
4
16,
variable transf.
02 ,Z 10
lattice girder
K5 K 6
S14
leakage flux
left
S14
S34
length, units of
lens equation
I
J
19
19
1
T 10
D27
D26
D26
latent heat
S 15
I
I
6
7
D28
D26
D26
D28
..
calculation rules
4
8
8
8
1
%
N
incircle
6
8
,
M
M
M
7
D23
Johnson formula
V1 Z21
,
Z22
5
5
5
5
5
5
illumination
- elastic
- plastic
impedance
2
,
ideal fluid
impact
15
P
S
G
S25
S25
S25
hysteresis
2
2
joule
Q 4
T12 ,T18
hydrodynamics
hydrodynamics values
19
V5 V 6
S25
ionizing radiation
isothermal state
isentropic
isobare
isochore
isothermal
hub dimensions
I
- hyperbolic functions
isentropic
2
I
- enthalpy of a mixture
International E-series
D 17
inverse circular functions
- Fourier transformation
iron losses
J
I
I
1
equation
Horner method
..
I
2
homogeneous differential
14
numerical
4
P 18
1
rules
L
S24
1
I
internal energy of a mixture
B
C
1
,
-
6
I
11,12
integration
- application of
- by parts
- by substitution
hexagon
Hooke's law
hollow cylinder
Q23
..
instruments, electrical
hertz
high frequency coil
Q 18
installation
,
light refraction
S 13
A 1
V 4
V 4
V 2
lighting values
Z25
linear coefficient of expansion
Z 11
hand rule
lenses
linear controller, graphical
determination
T 21 ... T32
- differential equations J 2 ... J 7
- equations
D 7, D 8
- expansion
O 3
- interpolation approximation
rule
D 16
- networks, calculation
S 8, S 9
methods
S 8, S 9
,
O
liquid bodies, heating of
- pressing
backward
,
2
R 8
R 8
R 8
Z 14
,
forward
liquids, tables
measuring instruments electr.
S36
melting point
Z 1 .. Z 6
metalworking
P
H
8
6
D
4
S22
Q 12
Q 13
Z25
V1, Z25
Z25
lubrication film
lubricant flow rate
luminous efficacy
flux
of lamps
- intensity
- quantity
V
V
Mac Laurin series
1
1
D 18
V 4
macro photography
S 14, S 15
- strength
S 4, S 14, Z23
- flux
S 3. S 14
- induction
S 3. S 14
magnetomotive force
S4. S 14
V 4
magnifying lens
magnetic field
M 1,
mass
,
mirror
mixture rule for flu ds
- cross
module, circular
- normal
rigidity
Mohr's analogy
- stress circle
molecular volume
- mole fraction
1
8
moment of a force
K
inertia
I
16
.
- theorem
I
L4
motion, linear
,
rolling
,
rotational
,
simple harmonic
,
sliding
L
L
L
L
L
L4
motor, compound wound
- direct current
- induction
- shunt
,
,
,
-
.
K
,
synchronous
,
three phase
S32
motors
multi plate clutch
1
19
1
5
9
6
7
9
S32
S32
S34
S32
S34
S34
S34
Q 15
S37
D 13, D 14
S 24
equivalent
moment of inertia
non-magnetic coils
normal curve for probability
,
units of
18,
maximum shear stresses
M2, M 3
A 2
P27
maxwell
mean shear stress
- specific heat of various
gases
- value
mechanical oscillation
mechanisms
meaning of derivative
measure of plane angle
measuring equipment
P13
Newton's approximation method
fraction
,
I
-
4
4
3
U 6
U 6
Q 18
Q 18
P 18 Z 17
Z 17
P 14
circuit switch
manipulated variable
manipulating point
-
V
R2 R
V2 V
milling
-
7
T18
control loop for stability
microscope
Q21, Q22, Q24, Q25
loads in beams
logarithmic functions
logarithms
low frequency coils
9
P23
method of bucklinc coefficient
methods for check ng the
modulus of elastic ty P 3.
load capacity of tooth
-
S8 S
R6 R
mesh-work
S 3
P 18
density
- distribution
- probability curve
G
G 8, G
G
normalized form of the transfer
T 12
function, type mixed
- product
T 12
- sum
T 12
15
numerical integration
Nusselt' number
O 12
T 19, T 21
Nyquist criterium
,
Z 13
G
M
3
6
L 10
H 2, H 3
E 1
T 5
7
7
7
,
I
oblique angle triangle
octagon
S2 S
Ohm's law
transfer function
operating characteristic
optical distance law
optics
PI- or (PID)-^ controlling
T
T
element, characteristics
of a control loop
pinion, dimensions
pitch, circular
V
plain key
plane, inclined
7
7
3 10 G 11
V 2
ordinary differential equat on
harmonic
mechanical
overshoot
- of the controlled variable
oscillation,
-
5
Q 13
Q 13
,
bags
- slots
open loop gain
oil
,
1
-
normal
,
-
J
1
L
4
mirrors
planing
M
6
2
8
plastic yield
point of disturbance
Poisson distribution
T
T
T20
Q23
Q18
Q18
Q
,
R2
G4 G 10
,
P
T
- second moment of area
T20
Pappus theorems
I
parabola
parallel combination
- combinational element PI
elements PD, PID
15
F
2
T
9
T15
T16
parallelepiped
parallel-flow-exchanger
- resistances
C
S
B
partial fraction expansion
D 3
- differential equation
J
1
,
S
S
6
2
8
M
1
power
- active
- factor
1
J
1
J
2
D
4
S31
S31
S31
S31
A3 A 5
,
correction
S21
particular
2
polytropic state
potential difference
- divider
- reactive
- units of
powers
Prandtl-number
- resonance
17
B
1
parallelogram
I
polygon
11
6
1
6
D30
polar coordinate system
P- or (PD)-"^ controlling
element, characteristics
of a control loop
3
K 10
V 3
R 4
,
,
,
D
1
12 ,Z 15
U 5
11,
preparation of chemicals
pressure
- hydrostatic
- in a fluid
N 1
1
N
N 1 N
,
1
Pascal-triangle
pendulum
- compound
- conical
- simple
- torsional
M
M
M
M
M
7
7
7
7
7
primitive transfer elements
principal stresses
T14
P28
prismoid
pentagon
B
2
period
permissible stresses
L
1
progressive ratio
proof stress
projection, angled
C 4
G 2
G 2
D 17
,
,
,
,
P2, P 18, Z 16
.
.
permittivity, absolute
-
,
relative
permutations
phase crossover angular
frequency
- margin condition,
realization
- margin
- response
-
Z22
properties of gases
vertical
T
7
proportional, 3rd
T
3
- shift
photometric radiation
S 17
equivalent
,
liquids
solids
T25
photometry
pH-values
horizontal
6
representation
,
,
D5 D
T27
T7 ,T20
V 1
V 1
U 4
*
probability density function
- distribution
Z 18
S 12
,
,
P
L
L
L
Z1
:
Z
Z
.Z
,
C
C
,
quadratic equation
quantity of electricity
2
8
8
8
6
5
4
D32
D32
D32
- ,4th
- mean
pyramid
- frustrum
Q factor
2
1
1
S 17 S21
D 1
S 2
,
radians
E
radiation
- constant
10
Z 12
radius of gyration
M
ramp response
random variable
range of an ammeter
- - a voltmeter
T
2
2
G
3
G
ratio of slenderness
reactance
2
1
rectangle
reference variable
- - adjuster
refractive index
5
4
1
5
5
2
T
relative frequency
G
Q 12
S4 S 14
rem
remanent flux density
V
,
moment of
,
rolling
5
S25
S2 S 5
resistance
- of a conductor
-
2
1
- eccentricity
S
5
7
P
K 12
S6 S
S6 S
resistances in series
- parallel
resisting moment
resistor combinations
,
7
7
P20 P21
S6, S 7 S 10 S 11
resonant circuits
- frequency
S21
S21
restitution, coefficient of
M
reversible processes
8
7
6
2
Reynolds' number
right angled triangle
N
- hand rule
S 13
K 5
V 5
E
Ritter method
Rontgen
- unit
A
1
L9 K 12 Z
7
rolling friction coefficient
- resistance
root-mean-square
root diameter
roots
rope friction
- operated machines
rotation
rotational motion
rough plane
roughness in pipes
2
7
9
8
- product
Scotch-Yoke mechanism
L10
screws
K 11
- and bolts
Q 1
secant approximation method D 15
second moment of area
19 P 9
- - - volume
19
- order delay element
T 1 T 15
section modulus
P 9
sector of an annulus
K 7
- - circle
B3 K 7
segment of a circle
B3 K 7
related step response
reluctance
P
D
Z 17
S 18
D 16
Regula falsi
safety factor
9
Sarrus rule
U
N
B
T
T
V
real fluid
T12
T
S 11
S 11
S31
reactive power
reagents, chemical
rules for the normalized form
of the transfer function
- to determine the
transfer function
K 12
S 16
Q 19, Q20
D
1
K 13
K 14
L7, M
L
R
Z
4
4
3
9
F8
scalar
,
F
F
I
I
,
C
sphere
3
self-induction
series
D 17 ..
- arithmetic
series combination
series combinational element
S 15
D 19
D 17
and D-T,
geometric
T16
D17
Mac Laurin
D 18
,
I-T,
-
,
,
S32
motor
resonance
Taylor
shaft-hub joints
shafts
,
shaping
shear
- modulus
- stress
due torsion
shearing force
- deflection of a beam
shrinkage allowance
shrunk-on ring
shunt motor
side-rake angle
Simpson's rule
T 9
T 17
S21
D 18 D 19
Q3
Q
Q
R2 R
..
5
2
4
P 18 P 19
P 18
P 1
P20
P 18
P 19
P 4
P 4
S32
Z 17
I
15
6
sine rule
single phase-output
E
transformers
slenderness
S35
P22
sliced cylinder
slider crank chain
sliding friction coefficient
C
- motion
smoothing
4
L 10
Z
7
L
9
3
R
solid bodies, heating of
2
Q 12
Sommerfeld number
G
special distributions
spectral energy
specific conductance
- heat
- heat, liquids
9
D23
2,0 7, Z 1
Z
.Z
1
6
-
,
P 1
compressive
in
,
,
three dimensions
Q
- heat of various gases
- lighting
- resistance
speed diagram
- rational
Z15
static
strain diagram
tensile
stress, torsional
Z 17
stresses in curved beams
Z21
rotating bodies
shafts
sphere
C
C
C
C
- with conical boring
cylindrical - zone of a
,
spread
3 6
..
spring, coiled
spring, disc
-
,
helical
,
laminated leaf
,
leaf
,
torsion-bar
1
L
1
2
3
3
3
Q 4
Q 19
Q 9
Q 8
Q 6
Q 9
splined shaft
springs
R
Q
Q
Q
rate
7
7
6
8
Q
Q 18 Q21
spur gears
square
B
- coil
1
S23
of the control loop
T 18
T 18
standard deviation
standerdized numbers
G 3
Z22
- number series
star-delta connection
state and variations of state of
D 17
stability, definition
-
S32
gases
4
statically indeterminate beams
P 17
static friction coefficient
Z20
- moment of a body
14
14
Q 9
I
a curve
- stress
statistic-tables
Steiner's theorem
stiffness
steel tubes
I
Z26 ,Z27
1
16
,
M 2 P 10
M 6
,
-
Q
target variable
Taylor series
D
6
3
6
8
A
3
E
T
temperature
1
- units of
tensile force
,
- stress
tension
P 1
,
P3
,
tesla
theorem of rel. shear stresses
P 1
P 3
P 4
S 3
P 18
P22
Tetmajer formula
- stresses
theoretical probability
P
3
G
1
thermal conduction
10, Z 1
.
.
Z6 ,Z15
- energy
*
- state of real gases
- stresses
- variables of state
thermodynamic diagrams
Thevenin's theorem
2
P
4
3
S
6
9
1
threephase current
S30 S31
- motors
S34
- output transformers
S35
- power, measurement
S30
T 8
time to reach lower tolerance
steady state
T 8
,
9
2
tonne
A
A
tooth width ratios
Q23
time, units of
toroidal coil
1
torque
torsion
P 4 P 11 ,Z20
Z 16 .Z 18
P 1
P3 Z 17
,
2
8
tangent rule
taper joint
1
bending
I
S34
3
,
5
Q 2
P27
S
F
-
P25
M
synchronous motor
P
.
9
P 1
P 3
Z 17
superposition
straight line
strain
- energy
,
- in two dimensions
substitution method
Stokes
strength values
stress
P 1
,
Z
T
N
step response
3
P 2
P 18
permissible
shearing
9
Z14
P
Q 9
P28
specific heat of a mixture
..
,
oscillating
- bar
- in shafts
- stress
torsional pendulum
2
2
S23
P20
P20 P21
Q 9
P28
P20 Z 17
,
,
M
7
torus
C
total reflexion
V
T
transfer function
-
4
2
3
determination using
the back annotation
,
,
- ramp response
T
5
2
variances
G
3
variations of state of deal
method
- -
T 10 ,T 11
T 12
normalized form
determination
T12
star-circuit
gases
real gases
T12
vectors
vector difference
- equation
S 10
- product
- sum
T 12
rules
type
transformation of a delta to a
- of a star to a delta-circuit
S 10
velocity
transformer
S28
S35
S35
S35
M 4
- angular
- of light
- time curve
Venn diagram
P
3
vibrations
viscosity, dynamic
-
,
single phase output
switch groups
threephase output
transmissions ratios
transverse contraction
trapezium
B 1
K
7
- rule
triangle
I
15
F 1
K
7
6
-
,
B 1
acute angle
,
centroid
E
F
,
equilateral
B2 D32
,
,
,
oblique angle
radius of incircle
1
circum circle
right angled
trigonometric conversions
,
turbulent flow
turning
R2
two coils facing each other
types of transfer elements
vertex radius
voltmeter
,
S21
Z 9
S 3
K11
S11
3
9
1
A
S
2
wavelengths
A1,V
3
watt
F
2
7
A 1 A
A
A 1 A
A 2 .. A
A3 A
A
A
A2 A
4
2
4
4
5
3
2
4
,
S 11
N 1
S 2
S 7
S 9
S 36
units of
,
T
volume
N1.Z16
water pipes
per unit mass
Wheatstone bridge
work
- units of
,
6
wave trap
fraction
-
wedges
pressure
time
4
M
T34
-
4
3
power
1
F2 ... F
volume expansion
P 18
.
6
5
9
8
8
9
8
2
2
3
7
G
- divider
- source
C
A
mass
V
L3, L
,
ungula
force
length
F
F
L
L
- kinematic
voltage
ultimate shear stress
- step response
- vectors
F
F7, F
,
weber
unit of fineness
F 7, F8, F
6
6
6
2
5
N 6
Z 17
S 13
E
E
E
E
E
,
units of area
A3, A
units of work
,
worm gearing
M
2
1
A 5
Q 27, Q 28
X-rays
V
3
yield point
P
1
Young's modulus
Z20
Z16
zone of a sphere
C
- strength
3
McGraw-Hill
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ISBN 0-07-024572-X