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
A COLLECTION OF TECHNICAL FORMULAE
Seventh American edition
published by McGraw-Hill,
Inc. in
1997
English translation by
J.
Mech. E
Sc, M. Sc.
Walters B. Sc. (Eng.), M.
R.
Owen
Printed
B.
in
I.
Germany
title
Preface
The purpose
of this collection of technical
a brief, clear and handy guide to the
and mathematical formulae.
formulae
is
to provide
more important technical
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
A
and revised 7 th edition
section on
CONTROL ENGINEERING
has been included
treated
in
in
the
new
section T;
RADIATION PHYSICS
is
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
has been revised and brought up to date.
to the
newest standards
Kurt Gieck
Reiner Gieck
Reference to BS, DIN and
BS
•
British
Standards
Institution
(Address: 2 Park, St,
DIN
VDI
•
•
VDE
LONDON
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
Method
of Presentation
54).
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
in the formula to obtain the correct result, these
are mainly to be found in sections
and R.
must be used
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.
For example, taking equation
if
=
=
s (distance)
v (speed)
then
t
=
I
23:
t
s
=-
2-8 metres
8
2-8
metres/second
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
three ways:
in
with consistent units,
equation a 6
~1
km =
1
10 3
m
becomes
1
becomes
1
=
km"
m
10 3
equation a 62
"
"
12
in
=
ft
1
ft
1
12
in
equation a 90
778-6
ft
Ibf
=
becomes
Btu
1
1
778-6
=
1
for
example, to convert 14-7
w
in
«
=
to lbf/ft 2
w [i^T
14 7
.
^22
lbf/in 2
in
n2
Ibf
1ft
i
in-
Ibf
ft
Btu
ft
-»«s
2
the conversion 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
becomes
ft
1
3-281
ft
equation a 110
1
Btu/lb
=
0-556 kcal/kg
0-556 kcal
becomes
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
lb
1
N
"
"l0
kgf
1
_
4
cm 2
m
2
"1
mn"
10 6 N
in
the use of definitions:
1
Ibf is
Ibf
1
mass
the force required to accelerate a
rate of 32-174 ft/s 2
=
1
of
32-174
lb x
%
becomes
=
1
32-174
s2
Similarly, the
N =
1
Newton
1
kg
1
kgf
=
1
kg
which becomes
=
1
1
1
—
9-81
x
2
lb
ft
Ibf
defined by the equation
is
—-
x
s
1
s2
and
the
lb at
1
.
becomes
=
1
N
S'
kg
m
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,
3
equation
m
-5
lb
3x2-5
0-4536
1
1m
0-4536 kg
=
N
"l_N_s£l
3-281
11b
1
-036
ft
1
kg mj
N
3-281
which
is
a unit of force.
Base Quantities and Base Units
System of Measurement
of the International
base un it
base quantity
symbol
symbol
name
name
(italic
letters)
letters)
length
I
m
mass
time
t
electric current
I
absolute
temperature
T
amount
light intensity
Old units are put
metre
kilogram
in
(
)
m
kg
second
s
ampere
A
kelvin
of
substance
(vertical
n
mole
/v
candela
brackets
K
mol
cd
List of
Space and time
a.
Q
y angles
solid angle
/3,
b,
B
d,
D
h,
extension, strain
G
r,
R
t
modulus
of elasticity
(Young's modulus)
modulus
of rigidity
(shear modulus)
radius
distance covered,
perimeter
thickness
u,
U circumference
A area, cross section
A m generated surface
A surface area
V volume
t
£
length
pitch
direct stress
shear stress
normal pressure
H
L
o
p
q,
E
/,
t
/,
breadth
diameter (diagonal)
height
p
s
symbols
M
bending moment
S
center of gravity
T
torsional
moment,
torque
Z
modulus
Q
shear force, shear load
V
vertical reaction
W
weight or load, work
w
uniformly distributed
load
time
v
velocity, linear
co
velocity,
a
acceleration, linear
/p
a
acceleration, angular
J
g
acceleration, gravi-
Z
tational
jU
/
moment
of section
of inertia,
second moment
angular
of area
polar moment of inertia
torsion constant
modulus of*section
coefficient of sliding
friction
Periodical and related
Ho
phenomens
T
period
/
n
frequency
A
speed
angular frequency
wavelength
c
velocity of light
co
rotational
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
mass
v
kinematic viscosity
q
density
P
power
F
force, direct force
r\
efficiency
m
viscosity
Heat
T
t
a
absolute temperature
temperature
5
reluctance
A
linear coefficient
S
magnetic conductance
length of air gap
temperature coefficient
of
expansion
a
of resistance
cubic coefficient
y
of
expansion
y
conductivity
g
resistivity
q
heat current or flow
density of heat flow
quantity of heat per
e
absolute permittivity
Q
quantity of heat
£r
relative permittivity
c
specific heat at
N
number
cp
unit
mass
e
permittivity,
dielectric constant
\i
cv
constant pressure
specific heat at
constant volume
\x x
of turns
permeability
absolute permeability
relative permeability
y
ratio of c
p
number
R
A
gas constant
thermal conductivity
z
number
a
heat transfer
Q
quality,
p
p
to c v
^o
of pairs of
poles
conductors
of
figure of merit
coefficient
k
coefficient of heat
S
loss angle
Z
impedance
C
transmission
radiation constant
X
v
specific
Ps
reactance
apparent power
fq
reactive
volume
Electricity and magnetism
/
current
current density
J
voltage
V
Vq source voltage
Cm moment
power
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
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
E
<P
B
L
illuminance
exposure
luminance
power
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
c
m
=
=
=
u
n
P
f
a
jeci
=
:enti
=
milli
=
-nicro
=
nano
=
Dico
=
emto
=
atto
=
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
10 3
10- 3
10 2
10" 4
10-
10
10 5
10- 2
10- 3
10- 9
10" 6
10" 5
10" 4
10" 6
10" 3
1
10 3
10 4
10 5
10 9
icr 2
10"
1
10 3
1
1
10
10 2
10 6
1
10
1
10
10 5
1
10 4
1
Units of length (continue id)
mm
a
7
1
mm
a
8
1
urn
1
nm
a
9
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
1
io- 4
IO" 6
-io
io-
7
pm
1010" 3
10~ 4
(mX)
10
10 7
10 4
10 3
10
10
10 6
10 3
10 2
1
1
7)
10
9
10
10 4
10
10
10 3
1
,J
10" 2
10" 3
10"
cm 2
dm 2
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
1
1
1
"
2
2
2
cm 2
dm 2
km 2
=
=
=
=
=
=
|im 2
2
10
1
10" 12
1
6
mm
10
10 8
10 10
10 18
10 6
2
>1mA =
2
4
6
10
10" 6
6
io~
10" 4
10" 2
X = Xngsl rom
12
10
10" 8
10" 2
1
10 2
10 4
10 12
1
XE =
2
10
10 -io
10" 4
10" 2
1
10 2
10 10
1
X-un
1
10 8
t
km 2
10" 6
10 -18
10" 12
10 -io
10" 8
1
A
UNITS
2
Units of
m =
3
1
1
mm
1
cm 3
1
dm 3
1
km
3
10
1
=
=
=
=
3
10" 9
10" 6
10" 3
9
10
volume
mm 3
m3
9
1
10 3
10 6
10 18
Units of
1
kg
1
mg
1
9
=
=
=
=
dt
1
1
t
=
1
10 6
1
10" 6
10- 3
1
10
10 3
Mg =
km 3
>
10" 9
10" 18
10- 15
10" 12
10
10~ 6
10" 3
10
10- 3
1
10 3
10 15
1
10 12
1
mass
g
dt
10 3
10" 3
10- 2
10" 8
10- 5
10 3
10 8
10 9
2
1
3
6
mg
kg
dm 3
cm 3
1
10 5
10 6
t
= Mg
10" 3
10" 9
10" 6
101
1
10
1
Units of time
s
1
s
1
ns
1
us
1
ms
1
min
1
h
=
=
=
=
=
=
1
d
=
1
10- 9
10" 6
10- 3
60
3600
ns
^s
ms
min
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
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 =
3
11
1
»
1
10 3
10 6
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
1
2)
1
(dyn)
(kgf)
N
=
1
kg m/s 2 =
10 5
10 8
10 11
1
Newton
UNITS
A3
Units of pressure
N/mm 2
Pa
a 39
1
Pa = N/m
2
10
1
a 40
1
N/mm 2
=
a 41
1
bar
=
a 42
(1
a 43
(1 torr)
10 6
10 5
(kgf/cm
10- 5
2
(torr)
)
1.02*10- 5
10
1
0.1
kgf/cm 2 = 1 at)= 98100
=
133
1
bar
6
0.0075
10.2
7.5
9.81*10- 2
0.981
1
0.133*10" 3 1.33*10- 3 1.36*10-3
>
*10 3
750
736
1.02
1
1
Units of work
kW
J
a 44
a 45
=
1
= 3.60>1
1J2)
kWh
1
a 46
(1kgfm)
a 47
(1 kcal)
a 48
(1hph)
=
h
(kgf
0.278*10" 6
6
1
2.72*10- 6
= 4186.8 1.16*10- 3
= 2.65*1 6
0.736
9.81
Units of
a 49
a 50
1
W3
1
kW
=
=
=
=
=
>
a 51
(1
kgf m/s)
a 52
(1
kcal h)
a 53
dhp)
a 54
1
W
kW
1
10- 3
1000
736
0.736
632
1
power
(kgf m/s)
(kcal/h)
(hp)
0.102
0.860
1.36*10-3
102
860
1.36
1
8.43
13.3M0- 3
0.119
75
1
1.58*10-3
632
1
1
1.16
(hph)
(kcal)
0.27*10 6
|
9.81X10" 3
1.16*10-3
9.81
m)
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
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
a 58
a 59
*
LB— k
=(4
)
=fi^ank
7R =
(^ + 459.67JRank = --Rank
r
+27315 K
-5ft-
°C
boili ng point of
,
\
wa,e rat 760 torr/ J7J1b
-32)°C=^-
273.1
100
273 , 5 .
'F
Rank
212
671.67
32-
5j°C
°F=(^B_- 459.67) °F
/
\Rank
/
am thn Q mn ro.
+ 32)
a 60
T
Tr,
t
anr
1
ft.
tures
in
1)
= 1/760 atm =
= N m = 1 W s
2>
1
1
torr
J
t
abS0
'
Zer0 °
-273.15
-459.67
the scales for Kelvin, Rankine, Celsius, Fahrenheit.
1
.333 22
mbar =1
3 1
W=
I
>
mm
1
Hg
=
J/s
at
1
t
=
N m/s
°C
491.67
UNITS
Conversion,
Anglo-American
to metric units
Units of length
in
ft
yd
mm
0.08333
0.02778
0.3333
25.4
304.8
914.4
1
in
1
ft
12
1
1
yd
36
3
1
mm
1
m
1
km
1
0.03937
39.37
39370
Units of area
sq
1
1
1
sq in
sqft
sq yd
in
1
144
1296
1
3281 *10~ 6 1094*10" 6
3.281
1.094
1094
3281
sqft
6.944x101
9
sqyd
cm'
6.452
0.1111
1
cm 2
dm 2
0.155
15.5
0.1076
0.01196
1
m
1550
10.76
cuft
2
Units of
1
1
1
1
1
1.197x10~ 4
dm 2
9.29
83.61
in
1.196
100
cu yd
cm 3
dm 3
1
kg
lb
1
oz
16
1
lb
256
16
0.5643
564.3
564 3>10 3
0.03527
0002205
3527
2.205
1000
35270
2205
10 6
1
Mg
00929
08361
00001
0001
1
1
64*10" 5
0.0283
0.7646
10" 6
0.001
1
mass
dram
1
nr
0.01
100
10000
dram
g
kg
1
0.06452 64.5*10-
929
1
1
1000
5.786X10" 4 2144x10- 5 16.39
cu in =
1
0.01639
cu ft = 1728
1
0.037
28316
28.32
=
cu yd
46656
27
1
764555 764.55
3
cm = 0.06102 3532*10" 8 1.31*10- 6
1
0.001
dm 3 = 61.02
0.03532
0.00131
1
1000
3
m - 61023
35.32
1.307
10 6
1000
Units of
0.001
1
8361
1
10" 6
0.001
volume
cu
1
1.076*10"
10 6
0.772*10- 3
1
1
1
1000
km
0.0254
0.3048
0.9144
1
0.0625
1
0003906
0.0625
1
1.772
2835
453.6
1
Mg
0.00177 1.77*10
0.02832 28.3-10
04531
0.001
1
1000
4.53*10
10" 6
0.001
1
continued A 5
UNITS
As
continued from A 4
Units of work
lb
1
ft
a 86
1
kgf
a 87
1J =
m
kgf
lb
ft
a 85
1
=
7.233
a 88
1
a 89
1
kcal
a 90
1
Btu
s
kW
kcal
h
Btu
0102
277.8*1
1
3
367.1*10 36*1
426.9
4187
6
0~ 9
239*1 0~ 6 948 4*10' 6
1
3413
3.968
0252
1
kcal/s
Btu/s
860
1
1076
1055
hp
kgf m/s
J/S=W
kW
1
76.04
power
1
hp
a 92
1
kgf m/s= 13.1 5*1
a 93
1J/s=1W=
a 94
1
kW
a 95
1
kcal/s
=
a 96
1
Btu/s
=
a 91
3
=W
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
1.163*10" 3
0" 6
293*1
3.087*1
778.6
Units of
J
1
1Ws= 07376
kW h = 2.655*1 6
=
m
0.1383
=
1.341*10- 3
0.102
745.7
0.7457
9 807 9.807*10
10" 3
1
1.341
102
426.9
107.6
1000
4187
1055
0" 3
5.614
1.415
1
3
0.7073
0.1782
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
1
a 98
1
a 99
1
a100
1
a101
1
a102
a103
a104
a105
a106
a107
a108
a109
a110
a111
a112
a113
a114
a115
a116
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
mil
= 10- 3 in
sq mil = 10 _6 sq in
yard = 3 ft
English mile = 1760 yds
Nautical mile
Geographical mile
long ton = 2240 lb
short ton (US) = 2000 lb
long ton = 2240 Ibf
short ton (US) = 2000 Ibf
Imp. gallon (Imperial gallon)
US gallon
BTU/ft 3 = 9.547 kcal/m 3
BTU/lb = 0.556 kcal/kg
2
= 4.882 kgf/m 2
lbf/in 2 (p.s.i.) = 0.0703 kgf/cm 2
chain = 22 yds
Hundredweight (GB) (cwt) = 112
Quarter (GB) = 28 Ibf
Stone (GB) = 14 Ibf
,
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
Ibf
=
62.3 kN
AREAS
B
square
A
=
a'
a
=
\a~
d
=
a\Y
A
=
a-b
rectangle
*
^-"
Var +
d
V
•-
a
parallelogram
-.
.
= a
:
s -
=-'-/>
d-
=
\ ia -
-
=
\
:
(a
ft
cot
::
af
+
- h cot a) 2 +
trapezium
=
^A
-
^-h
=
m-A
-S*
£
I
\5Cs-a)(s-ft)i$-c)
i
•:
2;
2
•:
V
2
/»
:
AREAS
B
equilateral triangle
5^
b 14
b15
h
=
\F
b16
A
=
|r 2 VlO
b 17
a
=
\r \/lO-2\AT
b 18
q
-
if v 6
1
+
2\/T
+ 2 v"^
construction
AB =
0,5
BC = BD, CD = CE
r,
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
b26
^
=
d- cos 22.5°
«
0.924 d
b27
d
"
^TTo
cos 22.5
^ 1083
5
5
polygon
b28
b29
a-h-[ + b- h 2
2
+ b h3
AREAS
B
circle
A
U
2
nr 2
=
=
J
4
=
0.785 d
=
2nr
=
=
2
5 (o
-
d
2
nd
annulus
2
<*
>
D-d
2
sector of a circle
360
fr=
c
2
Jt
180°
a
a
Jt
180°
segment
of
a circle
in
radian
in
degree
measure
=
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
c
5
2 {ab + ac + be)
c
6
cuboid
d
=
be
ya
2
+ b 2 + c2
parallelepiped
c
7
V -
A! h
(Cavalieri principle)
pyramid
A, h
3
frustum of pyramid
c
9
c 10
=
^(At
h
+A 2
*l±A*
+
VA
(for
1
-A 2
Al
)
~A
2)
SOLID BODIES
cylinder
V
=
2
h
\d
4
Am
=
2nrh
Aa
=
2jir(r +
/i)
hollow cylinder
%h(D 2 -d 2
)
^
V
-
f
m
=
V/^
nr(r + m)
'
+
AgiA, = x 2
frustum of cone
r
2
J
:
fc
V
=
-^h(D 2
Am
=
-y
m
+
(D +
Dd + d 2
=
</)
VWsphere
V
=
I^T 3
=
i»
= 4.189 r
^'
)
2jrpm
SOLID BODIES
zone
c26
V
=
%h(3a 2 + 3b 2 + h 2
c27
Am
=
2nrh
n(2rh + a 2 +
c 28
sphere
2
ft
)
segment
c29
c
of a
)
of a
sphere
2jirA
30
2
2
+ 4*
f('
4
c31
)
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
AQ
=
2nr(h + \lr 2 -£)
sphere with cylindrical boring
sphere with conical boring
SOLID BODIES
torus
Dd'
TM-
Dd
A
=
n2
V
=
2
%d
4
Am
=
71
A
=
Ji r
sliced cylinder
h
d h
+ h2 +
/i 1
+
r
2
2
^r + {h,-h 2
ungula
2
V
r
3
2
)
/A]
h
Am
=
A
=
Am + Zr 2 +£ r yJr 2
-
±h(2D 2 + d
=
Z{Ai+A 2 + 4A)
+ h2
barrel
2
)
prismoid
V
This formula
may be used
calculations
involving
shown
in
fig.
thus spheres
spheres.
C1
and
.
.
.
for
solids
C3 and
parts
of
ARITHMETIC
Powers, Roots
Rules for powers and roots
general
d
1
pa
n
numerical examples
±q-a"
=
[p
a"<a"
=
a'
d 2
± q)a
3a 4 + 4a 4 =
n
a 8 /a
d 3
m
(a
d 4
n
)
=
(a
d 5
b
P y/F
±
(a
3 2
=
)
(a
= a8
~2
2 3
= a2
)
n
q ya
=
±q)y/T
(p
= ae
"3
=
Ma
11
^T
= a6
1
V^-W
4^F+7\GT
=
^16*81
^16'V^
-
iff
d 9
\b
&
VT
V2^
^~
d10
= a'
Va
d11
d12
4
n
tyTlT =
d 8
_mn
)
1/a
d 6
d 7
n m _
n\m
=
2
7 a
^=T
=
iVa^;
i
* N0UP
ScSSs
Exponents
of
=
a
)
=
^^9^=
\^T
PeCial
3
•*
= a
V*
i
yj9
'
-
VF^=V^=+2; (V^) 2
powers and roots have
to
\
4
3
=
"2
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 );
Iterative calculation of
d16
When
where
x = YyT",
then
H<»
an
//-th
D*c
q
=
x v x2
root
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.
a-
is
ARITHMETIC
Powers, Roots -
D2
Binomial theorem
Expansion
of general algebraic expressions
(a±b) 2 = a 2 ±2ab + b 2
(a±b) 3 = a 3 ± 3a 2 b + 3ab 2 ± b 3
d 17
d 18
d 19
(a
+
b)
n
-
a
n
^a n
+
-1
6 +
^f^a
W (n-1)(/I-2)
a
1-2-3
2
d 20
(a+fe +
d 21
(a-b + cf
c)
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
I
+
+
b
n
2
2
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
(a
+
-
b)
{a
b)
{a
2
2
...
1
Binomial theorem
(a + 6)
d 26
=
(n\
d 27
-
-
"3
(2)an +(^ )a n
6+(j)a n 2 & 2 +(£)a n
n(n-1) (n-2) ... fn-Jk+1)*
1-2-3 ... *
l
1
3
+
fc
.*
.
.
J
U/
•J
d 28
n
aj
must be a whole number
(a
+ fc/ - 1a 4 +
-
4
a +
^a4
4a
3
"
1
b +
*'\*~ 2
6a
+
-6
2
b
2
2
4a-6
+
-fc
"3
*a4
+* \
3
* + 6
4
+6 4
3
Diagrammatic solution
d 29
Coefficient - Pascal triangle
(a
(a
(a
(a
(a
(a
(a
+ b)°
+ 6)
+ b) 2
+ fe) 3
1
12
13
1
1
+ b) 4
+ 6) 5
+ 6) 6
|5
1
1
Continue with each
d 30
4
1
j
1
10
line starting
1
3
j
1
4|
1
10
1
20
15
6
6
1
5
15
and finishing with
1
6
1.
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)
(a
-
d 31
Signs:
d 32
Examples:
(a
+ bf
(a
-
b)
5
is
always positive
b) is initially positive
5
a + 5a*b +
=
= + a 5 - 5a 4 b +
and changes from term
Wa 33 b
Wa b
2
2
+
-
Wa 2 b 3 +
10aV
+
to term.
Sab A
5ab 4
+ b5
- b5
ARITHMETIC
expansion of rational functions
Partial fraction
Proper fraction rational function
y(x)
P(x)
a +
Qix)
+
a,jc
by
+ a 2 x<
2
x + b2 x +
.
.
+ bn
.
n>m
r
jc
m
n and
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)
a
is
PM
m
a(x-n,)"-(x-n 2 )* 2 ...{x-n a
Q(x)
k
*
)
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
d 34
PM
m
y(x)
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\
^11
kz = k, the partial fractions of d 34 with the constants
^2k2 can be combined to the following partial fractions:
B, 2 x + C 12
fl 1k x + C 1k
•••
B u x + Cu
d 35
2
2
2
k
x + ax + b
(x + ax + bf
(x + ax + b)
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
Example:
y[()'
common denominator
2x^\
m
Q(x).
2x-1 _
Bn*+Cn
Q(x)
x +
2
2
2jc
2x-1 = B u x(x+1)**C„(x+V 2 + A ,{x+V(x 2 + 2x +
Q(x)
5)
+
(A^
+
B u )x
2
+ (3/l q1
+ (7,4 q1 + 2/t q2 +
Bu
+
+A q2 +2B U + C u )x2
2C n )x+5i4 q1
Comparison of coefficients between
S„ = -1/2; C„ = 1/4; A q =
,
there
(x+1)
2
A (i2 (x2 + 2x + 5)
Q(x)
2x--\ =
If
{
x+^
+5
<i
A q2
>tgi
t
(x+1 -2i)(x+1 +2i)(x+1)
are
single
zeros
n\,
the
left
+
+ 5>t q2 +
and
1/2;
A q2
constants
equation d 34 can be obtained by:
P(n 2 )/Q'(n 2 );
d 36
A^ = P{n,)IQ(n,)
Cn
right side:
=
-3/4
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
=
In
e
-
lb
The symbols
common
log to base 2
2
x = b
log a
in
log
natural log
are called:
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
x-log a
'OQa
f
=
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
d
47
igoc
lg
e
•
d 48
In
x
lb
x
Base
Key
lg
0.01
d 51
lg
0.1
lg
1
lg
10
100
lg
e=
common
to
d 50
53
d 54
2.302 585
1.442 695- In*
of the natural logs
In
•
x
Igjc
e
lg
d
of logarithms
0.434 294
ig_£
d 49
d 52
x
In
=
=
3.321 928
2.718281828459...
-
logarithm of a
or
8.
...
-1
or
9.
...
=
1
=
2
lg
jc
number
-2
=0
-
-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
Pn
T
r
= n, this
= n(n-1)(/i-2).
becomes
Pn = P
r
.(n-r+1),
.
/i(n-1)(n-2) ...
1
n
of
a
d 55
If
or
n
at
>
things
a time.
r
=/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:
be
bac
cab
acb
bca
cba.
a
P3
d 56
=
Here r = n =
1-2-3
=
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!
•
Example: The
n = 3 things a,
following 3 ways:
aab
aba
.
.
a,
nr
Here
1-2-3
2!-1!
1
\
6 can be permutated the
baa.
3!
d 58
.
-2-1
n =
%
3,
n-i
=
2,
n2 =
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,
Hence
The
table on
(with
*'
***
<* 3
£)
\3/
=
^fl
1-2-3
-
=
3.
1.
page D 6 compares combinations and permutations
and without the things repeating).
pronounced „n factorial"
Symbol usual for binomial coefficient (see d27)
n\ is
r
-
ARITHMETIC
6
Combinations, permutations
Combinations and Permutations
(Explanations see D
d 59
c
c r o
CO
11
(o
C
1-
CO
ft,
^—
5)
CO
-o
o
*-
£
o
C\J
co
ft,
i
<o
-Q
cj
CO
a)
„ II
CO
•
«
*"•
co
»B
CL~
E™
*
5.
tr *- co
d 60
D
t
*
°
<»
co
S=
Q-
C\J
E
~~
«J
^D >>
Q)
CSJICNJ
0)
-C
—
5
ii
*
ii
c ft,
CD
II
cop
d 61
2
o
co
*£
~
*-*
eg
cnj
+
C
5 ^_
t-
o
o
-
o
0-03
cm
co
c
—£
E.E
« °
O) co
CD C
- o
I
I
^o
.E
«
5
CO
Q.
co
c -
CO
"O »_
li
d 62
sis
CO
CO
co
CO
CO
CO
I
co
05 05
C C
7
iji-
—
E E E
3 3 3
s:
II
CO
Q.
>,
W
I
CO
CO CM
V
i:
M
CNJ
co
C\J
.0.0.0
coU
ii
CO
o o o
(D O
o
=
a. \o
§
o I
«
.2
a>
*-
£
35
3 c c ^ »
O
£
S8|dLUBX3
o
Q.
co
.2
~
JO
ARITHMETIC
Determinants and
linear
equations
Second order determinants
d 63
an
•
x +
a?i
•
* + a22
a, 2
•
D
y
y
insert r
a,,
a 12
a ?1
a 22
-
c
column
in
a 12
d 64
D,
a 22
r
2
r,
•
-r2
•
ajj
=
an
ri
a 21
r
0^\
"
a, 2
h
-r,
y
+
X
&Q2
'
Oy
column y
=
=
'
place of
column x
r,
Q\y
•
a,,
•
a21
2
+
5l
y
D
Third order determinants (Sarrus rule)
d 65
a„ -x + a 12
a 21 x + a22
a 31 x + a32
•
•
d 66
O
=
,
+ a 13
+ a23
+ a33
•
•
•
z
=
r,
= r2
z = r3
z
3 13
a ;i
a12
3 i>?
2?3
a"
a„
a 32
a 33
a 3l
^3?
ar
column
a 12
ri
01
7
3 12
insert r
=
-y
a,,
a'
d 67
-y
for
a
a2 2 a 33 +
•
i
1
a, 2
+ a 13 a21 a32 - a 13
•
•
- a,
,
•
3j 3
•
•
a23 s3
•
a^ a 31
•
a32 - a 12 a2
*
•
i
^33
x column:
a 13
r,
X.XX,
a 22
a23
"32
rf
a 12
°33
'22
.
+ an
'
Tj
'23
•
T
°12
a 32
"23
'3
^3' a22'
a32
a 12
'2
r3
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
Determinants and
8
equations
linear
Determinants of more than the 2nd order:
(The Sarrus Rule, see D
higher order than the 3rd).
may be used
7,
for determinants of
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.
an
Alternate the signs of terms, starting with
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:
= a n x n + a n _T
/nW
d 71
n_1
+
jc
.
.
+ 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
complex conjugate
coefficients a v are real, only real or
zeros exist as solutions.
3.
If
4.
If
>
coefficients a v are
all
part
n
is
>
is
of
whose
there are no solutions
real
0.
odd
one zero
at least
is
real,
assuming
coeffi-
all
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,
j,
x,
6.
x2 -x 3
The number
is
.
=
j
.
n
-x n = (-1)
•
i
equal to the
number
an
this value less
,
of sign
.
.
n
,
2,
.
.
.
n
= k
a /a,.
changes
a n _-|, a n _ 2
.
j
1
of positive real roots of the equation
d76
d77
.
1, 2,
=*1,2, ... n
=
k =
where
d75
=
in
question
of the coefficient series
a2.aT.a0
or
an even number (Descartes' theorem).
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
question
is
of
negative
real
roots
equation
the
of
in
found by substitution x = -z:
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.
d78
Applied to the example on D
9,
point
6:
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
d79
negative real root.
General solution
If
/n
to
is
x-i
=
0,
/n _i
(x)
(x)
root of an algebraic equation of wth degree
the degree of fn (x) can be reduced by one degree
=
when fn {x) is divided by {x - x y ). If another
a
is also known, the equation can be reduced by one
degree further when divided by {x-x 2 ), and so on.
root x 2
d80
/„(x)
= a n x n + a n _i * n_1 + a n _ 2 x n
d 81
/„/(*-*! )
=
d82
n
fn -i/(x-x 2 ) = /n _ 2 (x) = a n " x
etc.
/n _ 2 /(j:-X3) =
/
n _i (x)
.
d83
^/(jf-JCn)
= f
.
~2
+
= a n * n_1 + a n _i'x n
'
~2
+ a n _2"x n
+ a2 x2 + a^x + a
..
.
-2
"3
+
.
.
.
+ a2 x + ay
+
.
.
.
+ a2 "x + a,"
.
(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
d84
P n (x) =
an x
n
+ a n _i -x n
"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)
m
_,
(1)
(0)
(0)
(1i
x an
V2
(2)
x a n _2
m
a2
*0 a n-2
x
xQ a n _r *o a n-2 w x a n _ 2
*0 a n-1
<
x a2
1
<
>
afl
4
1»
ao
ai
aT
W
"
J3)
a•n-1
n-
a2
<
3
>
=
fc
2
(1)
xoai
WW
*0 a n
an
afl
m
xQ a3
...
1
*0
0)
a2 <°»
m
= b: =
= b
=
Pn
MM-P n (x
= 1/2!-P n "(jt
(x
)
)
(n)
(n)
1
fln^W-Vl-l/tl-l)!-^" '^)
an =
Example
fc
n
= 1/n!-Pn
'
n)
U
of the Horner
1
)
method:
Calculation of the values P n (x),
P n "'(x)forx = x
x = 4:
P n (x) = a: 3 - 6jc 2 + 11* -
Pn
'
{x),
Pn "
(x),
and
;
6
<°>
a 3 <°>
a 2 <°>
1
-6
11
-6
4
-8
12
-2
3
8
6
x = A
4
a/°)
11
6 =
1
I
1
a
-
= 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
.
•
d106
Line 3
shows the
e.g.
an-r
results of the addition of lines
.^
= a n _,™ + x
=
a n - 2 (D
d107
(
This
means
d108
an
;
<
1
jc
1
>
= *
<°>
.
same scheme -
- with
starting from line 3
additions leads to line 5 with
and
d109
ai
(2)
the value of the
Pn
= b, =
multi-
(x Q )
'
P n (x)
derivative of
first
scheme can be repeated
This
<
aJM+xo-a^-bo-Pnixo)
=
>
the value of the polynomial at the point x =
Using the
2.
where a r
>
especially:
a
plications
and
1
1
<
x = x
at the point
.
as a polynomial
Ai-times,
of
degree n has exactly n derivatives.
These calculations lead
d110
P n {x)
= a
(1)
to:
r
(n)
(*-*o)
(*-*o)
= Pn
d 112
+
(x Q )
.
.
.
+
1/1
+
(x-x Q ) + a 2 (3) (x-x Q f
+ a/
!
•
1/(b-1)I
P n '(x
Pn
•
)
(x-x
(n - 1)
(jc
)
•
+ 1/2!
)
(x-x o
r
]
•
Pn "
(jc
+ 1//i!
ix-xof +
•
)
•
Pn
{n)
(x Q )
.
.
(x-x
n
)
Example
2 of the Horner method:
Reduction of the degree if there
i.
e.
Pn
the finding of
d114
P n (x)/(x-x
= P n ^{x).
P n (x) = x3 - 6X2 + 1 1 x -
d 115
Given:
d 116
Scheme:
a3
a
6
with root x
(root)
x
,
)
'
1
known zero
is
using:
{x)
_-|
a2
=
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:
The roots
of this last equation
very easily using d 14.
(x-,
=
= 2
1
is
a root of
and* 2
=
3)
Fn
(x).
can be determined
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
Starting with an approximate
can be reached by iteration.
Example
jc
Example
- 3X2 + 7x - 5 =
0.
of a transcendental equation:
jc-lgW-
d123
value any degree of accuracy
of an algebraic equation (polynomial):
4
d122
initial
falsi).
1
=0.
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
one method after several
cially
with
in
results
iterations.
has been reached
.
ARITHMETIC
Approximate solution
for
14
any equations
Newton's approximation method
The value x
for the root
The tangent
the
is
n
is
approximation
first
of the equation f{x) = 0.
made at f{x Q ); the inter-
section of the tangent with the x-axis
is
a better value than the starting
point x
Calculation of jc-, is made by:
.
d125
=x -f{x
x^
)/f'{x
The improved value x 2
d126
).
calculated using
is
etc.
*2-*i-/(*i)//'(*i)
Multiple repetition of this method leads
x-,
a similar way:
in
to results of
any desired
accuracy.
d127
General rule
* k+ =*k-/(*k)//'(*k)
Requirement for convergence
• n
is
* =
;
i
of this
o.
1.2. ...
method:
a single zero
• between x
und n
maxima
minima
or
of the function f(x) are
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
d129
Example:
1st step:
d130
f(x) =
f(x) =
x
•
\gx -
*\.
may be x
The
=
starting value for a zero to
fulfil
3.
d125
requires the value of the derivative f'(x
calculated:
=
lg(x) + lg(e) = \g(x) + 0.434 294.
f'(x)
)
to be
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 not
must be carried out.
iterations
sufficient
further
ARITHMETIC
Approximate solution
for
D 15
any equations
Secant approximation method
The derivative
/' {x) of Newton's approreplaced by the differential
quotient: Two adjacent points f(x Q ) and y
/(jc-,) are connected by a straight 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
ximation
is
'(*<>)
.
X-\-Xp
X2 =*!
d140
/(*i)
/(*i)-/(*o)
In
the next step f(x^)
is
of this line with the jc-axis
General iteration
d 141
*k+1
~x
k ~/(*k)
connected with f(x 2 ). The intersection
is
the next approximation.
rule:
*k ~ *k-1
fc« 1,2,
/M-/dk-i)
f(Xk )
*
..
/(*„_!)
Comment: An
when
especially fast convergence often can be reached
the Secant and Newton approximation methods are used
alternately.
Convergence: Locally convergent.
d142
Example:
f(x) =
x
•
Ig
x 1;
d143
1st
approximation:^
=
=
x
f(x
)
4; Xi
-
3.
= 1.408240;
3 - 0.431 364
(3
%
f(xj = 0.431364.
-4)/ (0.431 364-1.408240)
2.558425.
d144
Error
f(x 2 )
0.043 768
2nd approximation calculated with
*,
,
x2
,
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:
(x 2 ) has to be calculated: /' (x) = Ig x + Ig (e)
f'(x 2 )= lg(2.558425) + 0.434294 = 0.842267
d147
d148
For this reason /'
d149
*3* = *2 ~f(x 2 ) //' (x 2 ) = 2.558 425 - 0.043 768/0.842 267 = 2.506 460.
Error: f(x 3 *) = 0.000 230. * 3 * leads to a smaller error
d150
than x 3 which
,
was found by only using the Secant method.
ARITHMETIC
Approximate solution
for
16
any equations
Linear Interpolation (Regula falsi) approximation method
values x and x 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
:
)
)
x-axis
the
is
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!
or
d152
< must be satisfied.
f(x 2 -f{x
/to) 7to) <
)
Generale
)
rule:
xw-Xt-fM- f,*)'**,^
d153
fc-1,2...
Q^j^k-l
/(*k)-/Uj)
Here
A*k) */(*,)
the largest value smaller than k, for which
is
;'
/(*2)7Ui) < is valid.
Convergence: Always convergent.
d154
Exam pie :f(x)
d155
\gx -1; choice of x Q = 1 with/(x ) = -1
x, = 3 with /to) = +0.431 364
= x
here
d156
x2 =
f{x
x^
x^
)
-x
=
-/to)
-/to)
<
is
d157
f(x 2
)
satisfied*
3-1
ntnMMA
3-0.431364
n
2.397269;
0.431364 +
/(*i)-/(*o)
and
1
= 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
with the x-axis
*4 = x 3 - /to)
d161
d162
is
intersection
made through
of
this
line
is:
/to)-/W
/to) = -0.000197
the
line
= 2.505947
5.
a further increase in accuracy the intersection of the
line 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
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,
7,
'
y
CJ172
an =
+ {n-A)d
a-i
Arithmetic mean: Each term of an arithmetic series is the arithmetic mean a m of its adjacent terms a m _-, and a m + -|.
d173
Thus, the mth term
(e.g. in the
3m-1
is
above series
3m + i
"*"
for
1
< m <n
4 + 10
a3
7)
Geometric series
The sequence
1,
4,
2,
8 etc
called a geometric series.
is
quotient q of two consecutive terms
d174
Formulae:
sn
= a,
^^« "
constant).
ai
is
ga "~
=
q = ~^-
for
^^
<7" 1
1
(The
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)
'
30
(n-*
|^|<
;
=
;
=
sn
following
trie
1)
Mm
sn
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
y/\0.
>
integer.
1,
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
10
Starting with
Examples:
5, 10, 20,
•
an
d
:
:
a<\
E6, E12, E24,
R5, R10, R20,
term
term
initial
n
final
sn
:
q
:
difference between two
consecutive terms
:
= 100
or
note
designation
6, 12, 24,
1
'/b)n-
or
intern. E-series,
number
see Z22
DIN-series, see
R1
terms
sum to n terms
quotient of two consecutive
terms
of
ARITHMETIC
Dl8
Series
Binominal series
d 180
/(,)
a
a
(i±x)«- ,±(°)x + ( )x*±( )x>
2
-
f
.
.
.
3
a may be either positive or negative, a whole
number
Expansion
a(a -
/a\
\nl
or a fraction.
of the binomial coefficient:
[a -2) (cr-3)
1)
1-2-3
...
...
(a-n+1)
ai
•
for
Examples:
d 181
^—-
=
(1
±
jc)~
d182
yr±7
=
(1
±
x)
v
1
'
2
=
1
+ x+ x 2 + x 3
=
1
±
1
Ml r\~
(\JZXj
-
1/2
1
—
\
r
^ X
2
VT17
3
-r
I
n
<1
...
l:rl
<1
+...
IjcI
<1
-
16
8
1
T
+
Ixl
...
^x-^-x 2 ± ^-x 3
2
d183
+
Xr
2
5
T
+
Xr
3
i
16
8
Taylor series
d184
putting a =
d185
m
^(*-a)
+
= /(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
i(£z|)
+
3 \jc+i
[x+1
d189
^+
3
+
3!
l(^)
5 \x+^
/
„., _£ + £-£ + £
5
+ ..l
x
-1
-
>
1
...
< X
x-S+
1
d190
2
J
4
b
continuec on
D19
'
ARITHMETIC
Dl9
Series
Taylor series
(continued)
for
Examples
3
d191
sin
x =
X
—
x-
+ x
cos x =
7
x
—
X
X2 + x 4
X6
—
all
4!
6!
1-—
d194
cot x =
l_l x
x
d196 Arccosx =
1
x3
2
3
?
^.^5
_
3
1-3-5 x7
2-4-6 7
X
X
5
7
Arctan x
? -
.
—
3!
x = x +
+
—
5!
7
9
+ X
1x121
9
1x121
+
—7!
+
—
all
"
+
9!
X
all
cosh x =
"
2!
x
+
x
d204 arcosh x - In9r
d206 arcoth x =
x
6!
—
x
15"
+
1
1
2X2
3x3
5
1
3 x
2-4 5
•
2
3
8!
-
45 *
3
3
2
d205 artanh x = x +
3
3 *
1
d203 arsinh x =
4!
3^x
x = x
*
x =
1x121
1x121
5
d198 Arccot x =
coth
0<lxl
lxl<;r
- Arcsin x
X
3
tanh
IxKf
945
1-3 x5
2-4 5
_
d197 Arctan x = x
d202
—
_ JL X 3 _
45
3
d195 Arcsin x =
d201
...
X
—
tan
d200
+
7
- x3 +
x = x
x5 +
x +
*
*+ 3 X
15 *
315
d193
sinh
...
7!
2!
d199
all
+
5!
3!
d192
5
X
'
—
x
7
315*
945 *
5
1
•
3 5 x
7x 7
2
7
•
2-4-6
1*3 1
2-4 4x 4
IxKf
0<lxl
lxl<jr
7
1-3-5 1
2-4-6 6x6
579"
5x 5
"
+
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
into
interval
this
y\
convergent series of the following form
{x
=
cot):
00
d207
y
-
fix)
The various
d208
ax =
2
+
[a n
cos {nx) + b n
sin (nx)}
coefficients can be calculated by:
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:
fix)
= -fi-x)
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,
.
.
.
and
/(§ + *)- /(f-*)
/2
¥ { /w sin
for
ak =
6k =
9^:
for
for
A:
=
k * -
(/:J:)
d*
1, 3, 5,
.
.
.
0, 1, 2,
.
.
.
2, 4, 6,
.
.
.
-
ARITHMETIC
21
Fourier series
Table of Fourier expansions
d218
d219
y
y
-
d220
d221
y =
y
=
d222
for
<
for
n < x
M
-|s,n*
<
<
x
n
1
2 n
sinJ3x}
sinjc +
SiniSx]
+
3
a
for
-a
for
5
a < x <
-a
a.
n + a < x < 2:t-a
l
rr.
n
y
31
r
2
.i
a
1
3,1.1.1
:
!
d226
y =
4a
—
+
d224
d225
cosor
cos
g-
(5 a)
•
—
1
x +
cos (3a)
>
-/(2n + x)
for
-,
S!niS=flQ cosx +
- sin 3
d227
d228
d229
= ax/b
= a
(;t
for
Olx^i
for
b
> - a(jl~X)/b
^
x
•
X
L.
I
sin (3jc)
X
^
-a)
"171
~i
sl
rl2
Jt
-a
]
CQS {3x
(
N
Id
3/1
x
)cos(2x)
"
A
—
/
rc-b
2/1
-1
wr
\/i
Tl-fc^AT^Jl^' 6
for
37t,
i
i
;
2^[«-o
r =
in
2^
O
d226
\
sin (5x)+
a < x < 2it-a
>
=
a
sin
•
'
\
i
—
!
!
h
Id
!
/z*
\3/t
/
n -b
d230
>•
=
4
—
a
r
-r
1
1
~2 sin 6
•
sin
x + -%
+
d231
y
=
^r
2 Jl
d232
>
=
f{2n +
d233
<
for
x
-j>
sin (5ft)
sin (3*)
•
•
sin (5x)
< 2x
jc)
a _ a fsinx
sin (2*)
^l
2
2
sin {3b)
1
sin
(3jc)
3
""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
^
yi\
^ n
x
-f
-f(n + x)
£.[*,-M + iS^-...]
d237
d238
d239
d240
axln
for
a(2n-x)/ji
for
71
^x ^
^x ^
Ti
\ /\F^
2ji
/(2k+x)
k
d241
d242
d243
d244
-
„
4a cosx
a
cos
["
a sin*
for
O^^ji
for
ii
2a _ 4a
n
tt.
F
cos
L
1
(2jc)
•
d246
for
d247
for
^
+
^
jc
-
(4jc)
3
(5x)
cos (6x)
+
2
Am
^ ^ ti/2
3n
^ <- "2"
= ^ =
*
/(2ji+x)
d249
M
Ji
7T
[2
4
2
.
d250
Jt
d251
/(-*) = /(2n+x)
for
^
x
^
71
V
d252
d253
d254
d255
_ 4 [cos* _ cos(2;r) + cos
a*/7t
^
for
x
^
a _ 2a
[
cosjc
2
2
tt
+
L
1
cos
3
(3jc)
a^
fsinx _ sin (2 jc)
L
1
2
cos
2
Ti
2
K
v^ re
ti
,
5rt
(3jc)
/(2ji+jc)
4
2/1
AAA/
-7t
T^L
2£F
cos(6x)
6^-1
cos (2x) _ cos (4jt)
2'-1
v-^
-ti
K
2
2
2a
+
5 7
5
;c
7t
3*
in.
.
WYY
2:
cos
3
cos
,
-a sin*
d245
d248
(3jc)
(5jc)
5
2
sin (3x)
3
.
ARITHMETIC
D 23
Fourier-Transformation
General
The Fourier-Transformation F
based on the Fourier
{s(t)}
inte-
gral 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~
function
1
{S(co)} gives the time
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
_il0X
F{s(t-t)} = 5(a»)-e
+
d263
Convolution
*i
i
;
x
(0*%(0-
Jji(T)-5 2 (r-r)-dr
=
/^(^-^(r-Tj-dT
d264
-oc
d265
F{*i(0**2 (0} = 5 (w)-5 2
d266
F{5(0}
= S(co)
d267
F{s(at)}
=^ 5 (f)
d268
F{ 5l (0 + *2 (0}
- 5 (w) + 5 2 (w)
1
(o;)
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^
|5(r)-e-
A
function
Dirac-Pulse
A 6
d272
2
(t)
AT- sin
(o)T)/((oT)
S(u)
(t)
,s(t)
funct.l
polarity
A R J/2
R xu
L^
•
A
(spectral density
constant over
is
with change of
=
S((o)
A6(t)
d273 Rectangle
(
v
t
+TI2)
sin
S(w) =
d275
S{oj)
3T
im
.& mo sgxii
to)
(t-T/2)-
.
d274
d276
-dr
Spectral density S((o)
s(t)
-R T
sit)
d271
iwt
- 00
Time function
d270 Rectangle
=
S(oj)
;
=
-\
2
4AT
AT
^
2
oiT
2
cos(2a)T)
sin (coT)
(oT
t
S((o)
=
A Rw
-(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
Spectral density S(<o)
s(t)
d278
d279
Modulated rectangle
d280 A R T (t)- cos (a)
d281
d282
t)
=
TTA
/\
^
l
%(t)
;A
„
„
w
with
A
o
=
^:
a
'
A; ART (t)
Gaussian-pulse
a2
t
S{(jo)
=
sinT(a; +
A
a;
)
sin T{o)-a>
)
a)
+co
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)
1
1-6 :t
d288 Exponential-pulse
d289
/I
S(io)
)(0
+
2
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
5
(t)}
2
n
F(s)-s-f(0
-F(5)-2 /
k
d304
Integration
L{jf(t).6t\
=
y^*)
=
(k)
+
)-f(0
+
(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
for y(t)
»J
I
+
conditions
start
look at rules
for derivations
i
i
d 206
I
i
solution of normal
equations for Y(s)
i
Inverse transform.
referring to D28
|
equations
Difficulty of the solution of the differential
d307
for Y(s)
i
i
result of the solution of
the differential equations
i
d 207
s-domain
f
is
transferred
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.
to the inverse transformation. This
Example:
Ty'
+ y =
f[t)
y(0
d301| Ts Y(s)-Ty(0
d305/
d306
y(t)
According
to
+
r
•
M
start
—
[Here f(t)
• F{s)
assumed
case referring to d 313 F{s) =
is
are
+
_
1
i
+ Ts
solutions
function. In
step
1/^).]
T
1
)
Ty(0<
^+Ts
Ts)
-VT
-v.
D28
Ms + Ty{V)
different
be
to
Ty(0 +
1
5(1
7>(0
there
this
after
condition
+ Ts
1
y(t).
I
2^
m
-
for
Application \
_
Y(s)
of D3
startfunction
=
)
+ Y(s) = F(s)
)
f(t)o
is
+
y(t)
1
+ e
Application of the convolution rule to the
/.-Transformation on linear networks
The
originate
having passed
function F2 (s).
function f-\(t) is changed to a response y(t) after
a network. The network s defined by its transfer
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
-•
Y(s) =
(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
+
O
=
£
+
Laplace
original-
transf. F(s)
function fit)
transf. F(s)
^-domain
d 311
\CD
/-domain
6{t)
1
d312
d313
d314
d315
d316
d317
d318
d319
d320
d321
d322
d323
=
2ti/;
= V^
^-domain
Laplace
5
£
i
1
for
for
2
Ms 2
^)
(s
2
2a
—
-cos(a/)
cos
b
for
1
(s-a)(s-b)
(at)
/•exp (at)
1
(5
a
s{s-a)
exp
(at)
-
+
2
a)
1
+ b2
— e~
1
at
d 329
+ r-5
1
a
S
2
±exp(-t/T)
1
sinh (at)
d333
d334
+
a2
s
s
2
+ a2
1
d335
(s
2
+a
2 2
cosh
(at)
,
sin (at)
cos
+ a2 ) 2
+a
a
>
e
^5
5
2
+6
s
sin(ar)for
/
-cos (at)
a
/2)
/2)
2
a
c
-a
4t
2/VttT
0:
(see
,
sin(a/)
2
V^ + a
3
^
2vr
erfc-
5 e
1
J2a
3
5
1//sin(af)
0:
-avr
>
VI"
f(.--.-)
(at)
2a*
(*
5
arctan (a Is)
for
5
d 337/338
r-1
3/(4^T-/
In
)
d336
-sin(b/)
-1/(2V5T-f
sy/sT
a
s'
'2
vr
-a*
s'-a2
d331
d332
a:
1
/
s
d330
=t=
b-a
M(s-a)
2
1
d326
d327
d328
-
e bt -e a1
d 324
d 325
(at)
-ft-s\n(at)
n-1
exp
f(t)
-s.n(ar.+
3
Jf
+
(1-1)!
M(s-a)
2
+
|
/
r
1
Dt5
<
t
n
original
> o|?a|
t
s
1/s
/-domain
function
s2
Dirac
(s
1/5
-ioo
i
,
.
J o\ al )
G
/
\
8)
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 = absolute value of z
= or modulus of z
<p
-*;
7
= argument of z
a and b are real
^-T
d339
d340
=
-i
d341
j-2
=
-1
d342
j-3
=
+
i
d343
:-4
=
+
1
j-5
=
-i
'
i
d344
_
,5
etc.
Note:
In
engineering
to avoid confusion.
In
electrical
z
=
d346
*1
+ 22
=
(a^
+ a2 )
+
d347
Zy
~
Z2
=
(a^
- a2 )
+
Zl
•
z2
=
(a-,
a 2 - by b 2
a-\
a 2 + by b 2
£l
d349
a2
^2
d350
letter
j
is
used
for
the Cartesian coordinate system:
d345
d348
the
a
2
+ b2
=
)
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
= z2
continued on D30
)
ARITHMETIC
30
Complex numbers
Complex numbers
(continued)
In
the polar coordinate system:
d353
z
d354
r
d355
r(cosg? +
arctan
V
g?
i
a + \b
cosg?
r
Zi-z 2
=
r
d357
Y
=
r2
d358
z
n
=
r*[cos(ncp) +
d359
?nr
yz =
2
=
-sing?)
b
b
sin
d356
=
r
r2
(p 2 )]
C0S (<Pi -V2) + i-sin(g? -
g? 2 )]
g? 2
1
[
+
+
[cos(gp +
"
)
•sin(g? 1
i
1
=
k
=
sin (/?<p)]
i
—
2nk
—^
1r<
\r(cos
for k
If
tang?
(p
:tL
+
+
this
1.
the principal root,
is
k = 1,2
n-1 these are the adventigious roots (/i is an integer).
if
n
VTare
the rc-th unit-roots.
The n-roots which
d360
-
z*
d361
A;
d362
i
cosg? -
e-'"
i
n
=
•
1, 2,
n-'\
.
(n
d363
d364
Ie
|
cos
d365
=
1
-sing?
e
Note:
iq>
g?
+ e
+ sin
2
g?
Ty
<p
k
In r
=
r2
+
=
i
-sing?
1
-icp
sing?
g?
Inz
If
y cos
2
an integer)
is
sing?
cosg? +
±lcp
are
1
n
= 0,
cosg? +
z
..2nk
cos^
for
fulfil
\(cp
and
+ 2nk)
cp^
=
<p 2
(k
=
+ 2nk,
should be measured along the
is any arbitrary integer.
arc,
0,
then
0)
0, integer)
—
-
-sin
i
n -
>
*
w+2nk,
±:—-
.
2
0, 1,
(n
(z 2
±1, ±2,
z 1 = z2
.
..)
ARITHMETIC
31
Application of geom. series
Compound
d366
interest calculation
kn
=
k
d367
20
9
'9
w
Annuity interest calculation
d368
d369
_
r
-rq
n
k
q
(k
-q n
l9
d370
m
n
Where
kn
=
0,
we
*-1
-k n )(q-:)
n
-1)<7
(9
,
-1
n
<y
r-g-k n (q-:)
r-q-k (q-'i)
get the "redemption formulae'
Deposit calculation
(savings bank formula)
;
d371
k Q -q"
{k n
d372
-k
n
-
n
(q
.
d373
n
-1
+ r.q
<7
)(?-1)
-Dq
k n (q-V
Vkoiq-V
'9
+ r-q
+ r-q
9
Letters
initial
capital
capital after n years
number
1
of years
+p
annual pensions
rate of interest
(withdrawals)
(e.g.
0.06 at
6%)
8
ARITHMETIC
Geom.
32
construction of algebr. expressions
b'C
d374
d375
a:b
x
4th proportional
:
d376
d377
a :b
x
:
b:x
=
3rd proportional
d378
x
=
ya'b
/^90°Z'^
I
d379
a: x
=
x
/
b
.
/
/
/
x
:
mean
/
Y
—
proportional
d380
x
2
a
1
+ b
x
/
\
I
6 .-
d
V-a- + b'
d381
x
:
hypothenuse
of a right-
angled 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
Circular
and angular measure
of a plane angle
measure
measure is the ratio of
the distance d measured along the
arc to the radius r.
Circular
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:
is
a
of
obtained by dividing the angle subtended
circle into 360 equal divisions known
°
e 2
A degree
is
e 3
a minute
is
divided into 60 minutes (unit:
divided into 60 seconds (unit:
Relation between circular and angular
By considering a
circle,
e 4
e 5
or
degrees
e 6
0°
it
that
360°
=
2
rad
=
57. 2958°
1
15°
30°
12
6
0.26
0.52
CO
c
").
measure
may be seen
7i
*
'),
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
_
hypotenuse
adjacent
e
8
e
9
hypotenuse
tan
opposite
=
a
a
c
= b_
c
- _^_
b
adjacent
adjacent
cot a
opposite
Functions of the more important angles
e 10
angle a
a
cos a
tan a
0°
sin
cot
a
1
00
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 A cos (ka -
y)
cf)
sine curve
1
sine curve
1.5
cosine curve
1
or sine curve with a phase shift of
and
and
and
k
= 1
= 2
k
=
k
1
•jx/2
FUNCTIONS OF A CIRCLE
p
£
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
(
=
=
=
cos
(
"
)
tan
(
•
)
cot
(
»
=
=
=
+ sin a
- cos a
- tan a
sin (180°
=
-
=
=
=
=
- cos a
sin (270°
+
+
sin
cos
(
"
)
tan
(
"
)
cot
(
"
-
sin
a)
=
=
cot a
a
cot a
tan a
90°
a)
)
+
a)
cos
(
-
)
tan
(
"
j
cot
(
»
)
sin (360°
=
=
a
+ cos a
- tan a
- cot a
=
=
=
=
- sin a
+ cos a
- tan a
- cot a
sin (a
cos
(
tan
(
cot
(
+
a)
)
+
=
=
=
=
+ cos a
- sin a
- cot a
- tan a
=
=
=
=
- sin a
- cos a
+ tan a
+ cot a
=
=
=
=
- cos a
+ sin a
- cot a
- tan a
a)
=
+
)
=
+ cos a
=
=
+
+
+ sin a
+ cos a
+ tan a
+ cot a
"
)
-
)
cos
(
tan
(a± n*180°)
=
=
=
cot
(
)
=
+ y<
o,
3
± n*360°)
"
)
••
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"
e 36
a + cos* a
=
+ tan 2 a
=
1
1
±
sin (a
e 38
cos(a±P)
e 39
tor,
tan
/«
(a
±
-i-
«\
/3)
tan
g
,
1
Q»
tan
-r
+
Sum and
-;
difference of angles
sin
-
=
1
-
a cos p ± cos a sin p
cos a cos p + sin g-sin/3
=
fi)
=
+ cot 2 a
1
cos 2 a
Sum and
e 37
tana -cot a
1
±
„
a
tan P
„£
to
tan
•
.
;
„
,
cott (a
±
,
a+
sin
p
2 -sin
cos
sin
a -
sin
£
2 -cos
sin
/S
=
cos a - cos £
=
tan
cot
sin
a±
a ± cot
a
cos a
•
•
=
/S
cos p
=
p
=
tan
a
tan
p
=
cot
a
cot
/3
cot
a
tan
£
=
=
±
±
a)
sin
•
<p<fi
= a
sin
/S
(g + 0) +
<p = arctan J
T
J
- p)
sin (g
+ 2 cos (g - P)
^cos(a-p)
- 2-cos(g + £)
cos (g +
g
a
cot g
tan g
cot g
tan g
tan
cot
(<ot
+ & cos
and
^
p)
+ b cos
g?-|
2
g^
P)
+
+
+
+
+
+
tan
cot
+
<p 2
_ cot
tan
.
_ cot
tan
/S
=
q> 2 )
;
m
fi
rf
T = arcsin
cp
tan
cot
tan ff
cot p
cot jg
tan p
harmonic oscillations
a sin (of +
with c
1
2
sin
of 2
1 sin
2
a
Sum
g
2
cos p
•
sin 03
sin
cos
sin
sin (a
=
sin
-
2 -sin
-
cos g
/S
.
,
2
cos
cos
2
=
p
tan
cot a cot P + 1
*
„
^ cot
± cot n
a +
=
difference of functions of angles
sin
cos a + cos
m
p)
of the
,
-\/
q>-\
"
sin
tan
cot
cot
tan
tan
cot
^
- b sin
/ + ^2_
c2
-
/S
ff
p
ff
p
same frequency
V c2 + ^
= a cos
g
g
g
g
a
g
<j?
+
^
2
f^ Sf
I
c° n
H
must be satisfied
FUNCTIONS OF A CIRCLE
Trigonometric conversions
Ratios
between simple, double and
a
sin
e 53
e 54
e 55
V^
2 sin
cos
|
|
a
sin
|
a
- c os 2 a
2
1
- a)
2
\h
V
1
a
sin a
cos a
tan
a
cos a
sin
a
sin
sin 2 a'
\£_
V cos a
1
+ cos 2 a
tan (90° - a)
cot
VT -
a
cot
1
a
f
angles
=
cot (90° - a)
sin 2 t?
- 2sin 2
half
a
tan
a
+ cot 2 a
Vi
2
=
-
cot
ycos 2 a - cos
V
-
cos 2
+ tan 2 a
1
sin (90°
V'l
tan
V1
e 58
cos a
cos (90° - a)
e 56
e 57
=
a
cos a
- cos 2 a
?i
\/-^2
1
2
V sin
2
5/1 - cos a
a
VT"- sin 2 a
2
1
e 59
V
1
+ cot 2 a
2
V
tan
1
+ tan 2
1
+ tan 2 a
1
-
-tan 2
e 60
1
sin 2
e 61
2 sin
a
=
a cos a
+ tan 2
cot 2
f
|
cos 2 a
- tan 2
1
=
tan 2
2
cos 2 a -sin 2 a
1
e 62
2cos 2 a-1
e 63
-2 sin a
1
sin
- cos a
V
-~
cot a --p
tana
1
+ cos a
cot
1
V
a
+ cos a
- cos a
sina
1
e 66
cot 2 a- 1
2 cot a
tana
-tan 2 a
a -tan a
|
cot 2 a
tan
1
1
=
cot
2
e 64
V
a
2
cot
cos 2 =
e 65
2
|
1
- cos a
+ cos a
sina
1
1
- cos a
+cosa
sina
V
1
cos a
- cos a
FUNCTIONS OF A CIRCLE
Acute angle
6
triangle
Oblique angle triangle
Sine Rule
e 67
a
sin
e 68
a
e 69
b
:
sin
/?
b
=
:
sin y
_ _
:
sin/3
a
=
a
sin
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
•
•
obtuse angles the cosine
(for
is
y
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
tan^ =
s
Q
- a
idius of incircle
e 76
A
=
1
-j
be -sin a
-
e 77
\A
(^
e 78
V
(5
2
sin
e 79
e 80
tan
tan
a) (s
a) (5
2
acsin
- b){s- 6) (5 -
c)
/8
=
-x
=
ps
a + b +
c
2
aZrsin y
c)
1
a
|
and circumcircle
sin
2
,
'
c
sin 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
83
arctan x
arccot x
x = sin y
x - cos y
x = tan y
x = cot y
-1^x^
1
+
1
-oo
)
y
2
Basic properties
e 86
arccos x
-1^x^ +
1
values
e 85
arcs in x
jt^y^O
2
= ? ~ Arcsin x
Arcsin (-x)
Arctan (-*)
=
=
Arccos x =
*
*
Arctan
r
x
*
Arctan
v/i
Arccot
/5
—
—
vi-*
*
Arccot
1)
2)
oo
n>y>0
Arccos x
= n - Arccot x
2)
Arcsin
Al CCOS
2
Arccot x
*
X
*
Arccot
Arcsm
^f^2
!
X
X
89
+
Arctan x
Arctan x =
Arcsin \/l - * 2
^
<x<
n-
=
Arccos (-x)
Arccot (-x)
~* 2
2
V1 -x
,
Arccot x
- Arcsin x
- Arctan x
Arcsin x =
\/1
-oo
oo
1)
Arccos x
Arccos
+
-!<,<!
Ratios between inverse circular functions
e 87
<x <
1
r.
5-
2
V1 +x
1
X
=
1
Vi~T^
—
*
Arrrnr
Arccos
rri
*
Arctan
-1
X
marked by capital letters
The formulas marked with * apply for x >
Principal values are
Continued on E 8
1
FUNCTIONS OF A CIRCLE
c
l_
L. b
inverse trigonometric functions
Addition theorems
T-
T-
T-
;
r-
% %\
r~
vi
+ +!
vi
5v
CO
CM
0J
**
c +
;
5v
*J
;
% %
+ +
CM
CO
<N
CO
j
1
|
j
i
i
|
!
•
+
t5
C
C
CO
CO
O
Jr 1
1
A V
;
wi
«J
;
!
Jr
O "°J
O;
c C
:
CO
:
(0
O O
V A
;.
o *
o * *
O O
yu
i
T1
:
:
!
•
o o
^ v
-Q
-Q
CQ
CQ
CO
CCJ
co
CCI
?T~
oP
i
1
co
to
O
o
c
r:
1
>
+
>
i
•
\
-Q
-Q
|
*
y-
w
>
-Q
CD
>
5b
i
>
Q
C r «
1
s
c;l^
1
m
<D
5
i
it!
I
II
II
II
|
8
<
II
|
t
V
;
co
co
;
SI?!
:
1
A
a
cc
;
1
-a
V
}
#
#
«0
{
<0
CQ
|
•Q:
>l
i
i
iUi
<o:
«0
>
ci
|
«{ §
1
H
H
i
i
i
K
-Q
S
n^'
3; + >i
2
?
"2
+
1
73
!
;
<
[
II
1
-Q
*;<
t»j
II
j
1
H
8;
i-
|
+i
q'
i-Q
Si
^
c
*~
t
\
<
+ is 5
+
1
:
CO
»-
+
1
<>
C
to;
Q
1
«
c
<
<j<
II
«
+
2£o
CD
K
A
i
r-
8
^
|
i
O
<
co
o"
i
\
|S
1
U
c
2
cc
c
\
>i s
i
3b
Q
x
\
i
O
o
j
I
•
!
ip
1
CO-
1
v
t-
1
1
+
CQ
j
".
o
si
'.
1
j
5b
(Or
V
!
x»
.o
'
CO
1
i
CD
09
CQ
CO
A
A
.
^AVi^AV|++iAIVi^
CO:
c
i
j
•
CO
c
o
CtJ
;
!
.2
O
«M
T-
a a
a a;
\
|
co
;
CO
1
'
1
a
<
•Q
+
B
}
1 5
+
K
'
:
<C
K
|
ii
ii
II
II
i
^5
C
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
co
j
«<J
J
i£W
!
£W
8
<
:
j
O
1
<
O
CO
CO
CO
CO
t-
O
is
<
i
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
{
j
J3
j
!
c9
I
!
+
1
to
co
O
O
!
!
>
!
1 J
<c
j
<c
j
i
u
«s
rr
CO
CO
CO
>
u)
a>
©
CD
C>
c9
CM
j
•
j
i
2.
O
O
2
<
<
CO
O)
N
CO
D
(D
t
ANALYTICAL GEOMETRY
Straight
line,
Triangle
Straight line
mx
t
1
Equation
y
=
f
2
Gradient
m
= y2
Interc.
form
y^
X2 —
a
for
+ b
~
*
=
b
0;
=t=
3
f
tana*)
JC-j
Gradient m\ of perpendicular
—
c
/Ifi
1
m, =
A?-Z
Line joining two points P
f
5
f
6
f
7
[x\,
:
vi)
and P 2
yi)
and gradient
(x 2
>'2)
,
y-y\ = y2-yi
Line through one point P\
y
(jci,
-y! = m(x -
Distance between two points
Mid point of a
line joining
m
*i)
J
=
y(.*2~ Jc i)
2
+
(.>'2
2
.)'i)
two points
y™ =
yi
+y2
Point of intersection of
two straight lines
(see diagram triangle)
t>2~
f
9
*3
f
10
£>1
V3
m2
—
tri]
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
s
=
,
3
Area
.
f
A
13
Where
_
"
x and y have
scales (see also h 1).
(*i
y2~x 2 y^ + (X2y3-x 3 y2) +
(x 3 y A
-*i ys)
2
same dimension and
are represented
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
Radius
r
+ y 2 + ax + by + c
of circle
=
2
V*o +
2
-
y
c
Coordinates of the centre
f
17
f
18
M
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)
2py
-2py
(x-x
(x-x
2
=
2p(y-y
)
2
=-2p(y-y
)
f
19
20
f
21
y =
f
22
Vertex radius
r
f
23
Basic property
PF
f
24
)
)
Basic equation
Tangent T
L: directrix
elsewhere
ax2 + bx +
equation the vertex
F: focus
vertex
at the origin
f
this
c
at point Fi
P
PQ
(x-i, y-\)
2{yi-y<>)(x-X'\)
>'i
ANALYTICAL GEOMETRY
Hyperbola
Hyperbola
Hyperbolic equation
point of intersection of asymptotes
at the origin
f
x
25
2
elsewhere
2
v
*—
-
(x-x
=
1
a
2
2
(y-y
)
)
„
n
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
to-x~
b
72
to, yi)
)(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
*' Conditions
|
elsewhere
(x-x )(y-y
)
= ci
(parameter)
according to note on page F
1
+
,
>1
_ °
"
ANALYTICAL GEOMETRY
Exponential curve
Ellipse,
Ellipse
Ellipse equation
point of intersection of axes
at the origin
f
x2
34
—+
a
b
2
Vertex
f
35
f
36
2
v
•*2
elsewhere
1=0°
-
(jc-xo)
2
.
1
(y-y
2
a
b
2
)
2
radii
=
ru
—
Eccentricity
^a 2 -b 2
=
e
Basic property
37
F^P
+
FJ~P
Tangent T
_bg
f
38
2a
at P^
(xm
Ui
.
a2
Note:
F-\
y<\)
-Xp)^-^)
yi
- >
and F2 are focal points
Exponential curve
Basic equation
f
39
y
=
a
x
Here a
constant
is
is
4=
positive
a
1,
and x
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"
x
e + e
x
e + e~
e
-1
2x +
e
2x
+
e
-X
x
e~-e
2x
-
//
V*
v
2
2
>o
1
1
-3
-2
1
1
/'
-1
1
3
2
1
,
,
,—
1
e
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
Vl-tanh *
49
Vcoth
50
f
51
f
52
f
53
f
2
I
*-!
coth*
Vcoth
2
I
*-
I
|
Addition theorems
sinh (a ± b) = sinh a -cosh b
cosh (a ± b) = cosh a cosh b
tanh
54
(a
±
b)
coth
+)
(a
±
b)
~
1
tanh *
+ cosh *
- coth *
cosha«sinh6
sinha'sinhfe
•
coth a coth b ±
coth a ± coth b
cosh*
Vcosh 2 *-1
1
tanh a ± tanh b
± tanh a tanh b
•
55
±
±
*
_,
coth *
For the defined * values of f 58
sinh (-*) = -sinh*
cosh (-*)
tanh (-*) = -tanh*
coth (-*)
1
f
*
cosh*
Vl-tanh *
*
1
f
Vcosh 2 *-1
,
2
1
sinh*
2
Vsinh *+i
1
2
f
Vsinh 2 * +
sinh*
*
f
coth * =
tanh * =
1
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
arsinh x
identical
f
f
56
57
logarithmic
equivalents
ln(jt
f
58
59
VJr+1)
= ±ln(x +
<x <
+
_ (X><y< +
00
within
1
^
X
_oo<
y
+
<
oo
+
2
<1
<y< +
oo
x = coth y
1-x
IjcI
x
ar coth
x = tanh y
yV-1)
<
x
ar tanh
2
oo
-co
primary
value
+
cosh x
x = cosh y
x = sinh y
with
defined
f
ar
a>
jc-1
IjcI
>1
lyl
>0
Ratios between inverse hyperbolic functions
arcosh x =
arsinh x =
f
60
61
f
62
±arcosh\/l +X2
artanh
arcoth
arsinhu?-1
artanh
\T
MZ
artanh x
arsinh
arsinh
V*2-1
iff
± arcosh-p=
arcoshc
VlT?~
arcoth
arcoth
For the defined x values of
arsinh(-;t)
f
f
f
65
arsinh a
f
66
arcosh a ± arcosh b
arcosh
f
67
artanh a ± artanh b
artanh
f
68
artanh(-jt)
=
b
arcoth(-;t) = -arcoth*
arsinh
(a\/^+T ±
[aft
arcoth a ± arcoth b
* Sign + for x
>
0;
-
for
x
arcoth
± ab
ab ±
a
<
b Va
2
+
1
± V^a2 --!)^ 2 -"!)']
a ±fc
1
=
-1
artanh-
Addition theorems
± arsinh
2
58
= -arsinh*
= - artanh x
63
64
f
arcoth x =
±
1
b
ANALYTICAL GEOMETRY
Vectors
Components, magnitude, direction cosines
Vector: Quantity with magnitude
and
A: Coordinates of the origin
x v y^,z^
x2 y2 z2
of the vector a:
B: Coordinates of the end-point of the vector a
Unit vectors along
of vectors
direction
:
,
,
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
f
75
I
a*
I
=
Vax
2
+ ay
2
of the vector: la
+ az
(a.
f
76
/3,
y,
ft.
77
f
78
=
y
cos^3 =
Calculation of the
a
•
I
cos a
-^
la
cos 2 a + cos 2
I
cos
/?,
=1
engineering notation.
a always
I
>
0)
cos y
OVand
OZ.
0° ... 180°).
-z£-\
la
where
=
a,
= \k\
angles between the vector a and the axes OX,
cos a =
a„
in
(I
I
f
or a
'
Direction cosines of vectors: cos
a,
I
l/l
ft
;
cosy =
+ cos 2 y =
components when
—
la
I
I
1
la
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
sum
Vector
Vector
sum
s of two vectors a
f
79
s
= a +b
f
80
sx
= ax + bx
f
81
= sx
2
\T\ =
s
;
+
yfsx
5
+s z -k
+s y -j
-i
=a + b
y
y
y
2
+
y
S
(difference)
and b
b
= az + b z
sz
;
2
Vector difference s of two vectors a and b
f
82
T
=
f
83
sx
= ax
f
84
\T\=
-b x
yjs x
2
85
f
86
cases for
Is
87
f
88
f
89
s
\
+
S
y
2
+
y
7
s
= az
sz
;
-b
2
\a\*\V\
360°
0°;
la*l
90°
IM
+
]/\a\
2
180°
+ \~b\
2
270°
\a\-\~b\ Vlal 2 + I6l 2
I
= l?l
li*l
Vector
f
]
= a -b
y
y
?£\£
Special
f
[-t)Ut*
a+(-~b)
sum
= a* + b -c* +
ax + bx
i-V5
V
2\a\
s of vectors a, b
+
.
.
.
= 5 x *i
-cx +
+
5W
-
,
+
ay
\a\yJY
Sy-j
+
la
1^2
c, etc.:
+ sz -k
(Vector equations)
b^
sz
= a z + b z -cz +
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
(k
;
>
= fc-a
y
c
y
then
c
;
ft a
=
cz =
is
the vector c
(Vector equation)
0)
c =
k-a z
k <
then
c || a ie
Example: Force Fa = mass m times
a
»-
c
*H
o
ie
(c|0)
JMi*l
If
f
92
m >
0;
*) The symbol
but opposite
fatta";
ti
in
^
=
acceleration a
m-a ';
Fa
=
denotes that the vectors (-D^l and
direction.
m-a
(£>*)
are parallel
—
-
ANALYTICAL GEOMETRY
Vectors
Vector Products of 2 vectors
The scalar product
Symbol
f
93
k
= a
b
f
94
k
= ax
-b x
f
95
<P
= b
•
+ ay -b
ccos
y
a
—
-b z
the scalar
is
k.
"•"
a = a b-cos
+ az
F
and
of 2 vectors a
for Scalar Product: Dot
1*
q>
=
{k
=
1
-\b
I
I
-cos
q>
0)
—
+ a v 'b v + a?'bz
-
x 'b x
*-zf
\a\-\b\
i^^\^>
Special
f
Ii*l-I6
96
-cos
I
Example: Work done W, by
f
f
97
98
W
W
360°
0°;
90°
= force ^distance = F
= F-s -cos qp
(W
F
a force
f
99
f
100
—*
—*
7""*
= axfc
c
= a-b
lc*l
=
F,s^0)
0;
-sin qo
=
a
,
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
+ cS +
cases
105
la*!-!
I
-
sin qo
f
107
r
F
sin<p
>180°...
90°
360°
180°
+ \a*\-\V\
M of a force F
= Radialvector x force =
=
0)
060°
-*
ic
0°;
around the point
M
M
=
s2
->
106
(c
b l-singp
Example: Moment
f
...180°-*
£*
Ir&^^JP
Special
f
\~a\-\b
form a basis
f
\c\
U
=-(ixa)
_L a and c*_L
,
the vector c
is
~*\
i~T*
c
b
over distance s
s
Cross "x"
for Vector Product:
270°
-\at\-\b\
Vector product of 2 vectors a and b
Symbol
180°
+ li*l-l£l
qo
-\st\-\t\
Hi
O
r
->
xF
(M=0;
=
r,
-(F *
F^
0)
270°
-» 5
r )^
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) =
Axioms
limit (r/n)
to the probability
A
P(A)
0,
number of events in which A occurs
number of possible events
g
4
h(A)
g
5
? P(A0 =
g
6
P(AnB)*)
g
7
P(A/B)
g
8
g
9
=
event
relative
has the probability P(A)
frequency
10. The sum of the probabilities
events A, taking place must be 1-0.
of
all
possible
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
assuming P(A) resp. P(B) 4= 0.
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.
If
g 10
g 11
of the other occurring)
P(AnB)
P(AnA)
)
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.
o and F(x) increases with
F(-x) =
F(x)
for
an experimental
F(x)
for
distribution
x.
continuous functions
or theoretical distribution
Fix)
\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\
for
2a
F(x) =
continuous functions
or theoretical distribution
an experimental
fix) for
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
Random
variable
discrete
g
13
g
19
= x r p,+x 2 p 2 +
x
mean
value x or expected
•
•
E(x) or
A
variable
A
continuous
+x nPn
•
+ 00
n
g 20
p.
Random
= \x-f(x)
fi
dx
- 00
-
i
where
p,
and
f(x) are probability densities.
Variance o 2
Random
variable
discrete
g
21
o2 =
+
g 22
g
(xi-x)
+
...
2
A
(x n
-x)
-p 2
A
+
+ 00
= \(x-n) 2 -f(x)-dx
o2
pn
= hx-,-X) 2 Pl
23
variable
continuous
2
pi+(x 2 -x)
2
Random
+ 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
and variances o 2 are combined
g
26
A
the random variable
2
=
i
=
A
with expected values
A-.
1
n
n
mean
g
27
the
g
28
the variance
g
29
If
the
value
random
n
=
2
o
=
z, 0\
x) =
=
2 *i)
',
(^)
the cumulative distribution
is
(x
p.\
variables have normal distributions, then
P(A <
where
\i
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
a2
;
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
x =
with
-Pi
Xj
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.
,
a straight line can be drawn through these points, the distribution is proved to be normal. The values of the mean x and the
standard deviation o are obtained as shown in the diagram.
If
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
1
(cumulative distribution)
1
50
k0
%
1
1
60
—
1
1
1
80
70
less than
x\
1
8^
—
1
l-T-l
1
£
STATISTICS
Normal distribution (Gaussian)
Normal curve for probability density
o 2 = 1 and fx = in g 39 leads to the
standardized
probability
mean value A = 0.
cp(k)
density
with
g 43
cp(X) =
_J
\[2tz
<p(A) is given in tables Z 26 and Z 27
be calculated from g 43.
<
for
A
<
1.99, but
can also
The connection between standardized probability density
and the real probability density f(x) for \x * and a2 4= 1 is
cp(X)
(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
o2 =
and
1
/.i
=
in
distribution function)
g 39 leads to the
standardized normal distribution
g45
0(A) = \tp(t)
As
limit
means
g46
0(k) =
df =
1
dt
-^=J<
for A -»
«>
and <pM
is
a symmetrical function
that:
&(-X)
=
1-4>(A)
The relation between the standardized <P(X)
and the real distribution for u 4= and o 2 4=
47/48
F[x)
<P(A)
1
/2^rJ
o\[2k
"^
e
6t
distribution function
1
is
where
A
=
—o
STATISTICS
Probability distribution
Gaussian or Error curve
The curve
based on the stan-
is
dardized
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
g 50
@
(x) between 0<x<1.99 are given in tables Z 26
For greater values of x look at the approximation
in the next paragraph. The connection between <P (x) and the
error function is <P (x) = erf (x/\/2~).
Values of
and Z
27.
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)
Area beneath the error curve when
g 54
erfc(x) =
1
y
- erf(x) _ _2
V5T.
o (x)
and [1-<P a;] in %
whole area
relation to the
for special values
g 55
<Pn(X)/%
± a
68.26
±2o
9544
a
99
99.73
2.58
+ 3(7
3.29 a
99.9
[1-4> M]/°/c
a = 0.515
a = 0.535
for
2
3
a= 0.545 for 4
7
a = 0.56
for
erUx) is subtracted:
6t
for
<
x
<
<x <
<x <
3
4
7
< x <a>
STATISTICS
Random sampling
General:
When
each individual component is too expentest by random sampling is used. The
samples must be chosen arbitrarily to give equal chances for
or not
sive
all
parts
(i.e.
The aim
testing
possible,
good
intermixing).
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.
of
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
g 56
is
a whole
number
(If
The
probability
found
not
that
more than
k
defective
parts
are
is:
k
ZP(*)
g 58
P(0) + P(1) +
x -
P(k)
P N\/N(l-p)
(
*
\
n-x
x )\
pN
is
a whole
number
Example:
In
maximum of 3 can be defective^ = 100,
of n = 20 are taken. How many defective
the sample? - The probabilities I P(x) are
a batch of 100 screws, a
pN =
3).
Random samples
parts are allowed
in
X
P(x)
2P(jc)
x =
:
1
2
3
The
table
shows
that for
508
508
0391
0.899
0094
0007
0.993
90%
!
1.000
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(n) =
P(x)
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
k
(np)
x
61
_. np
=
is
small, then
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
L(p,
^(£Zl! e -np =e -np[1 + n
c)=2
it!
^(^)!
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
of
c
the
3
<.
5
6
operating
2
1
* p%
parameter
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
characteristic
approaches p =
c
n:
Example
Cil00
1
2
proportion defective
Note:
constant;
=
0.
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
number in random samples
c
number of the maximum admissible
p*
po
proportion defective -»p%
:
:
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)
MTTF (mean
1
d/?
R(t)
6t
time to failure)
00
00
MTTF
g 69
= \f{t)-t-6t = \R{t) -6t
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
g 70
Product rule
When
ments
Ri
1
.
MTBF
for the reliability
...
..
=
n,
Rn
m
=
Rs
= \R{t)-di
:
of the single elethe reliability of the whole system becomes:
are
Rs
g 71
the
reliabilities
R2
= R^
.
.
•
.
Rn
= 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
distribution,
to calculate, usually fulfills the requirements (A
n(t)
:
number
number
of
of
elements
elements
at the
time
at the
beginning
t
are
=
the
(for cal-
const).
simple
.
STATISTICS
13
Exponential distribution
Reliability;
Exponential distribution used as
g 73
Reliability
R(t)
g 74
Probability to failure
F(t)
reliability
e"
function
xt
-xt
g 75
Failure density
fit)
g 76
Failure rate
»»-«-*
k
const.
(Dimension: 1/time)
g 77
(MTBF)
Failure distance
Product rule
for the reliability
Rs
/?s=
g 78
1
6t =
-J
:
"^ e"^
-
<
•
.
.
-(X1+ X 2 + ...+X n )1
g 79
g 80
Cumulative
As =
failure rate
+ A2 +
A-|
.
.
.
*
An =
MTBF
For small values the failure rate can be calculated approximately
number
g 81
number
of
elements
/.-values are mostly related to
g 82
Unit:
1
fit
=
IC-analog bipolar (OpAmp)
Transistor-Si-Universal
Transistor-Si-Power
failure/10 9 hours
1
10
in
fit:
0.2
Resistor-metal
10 Resistor-wire wound
5 Small transformer
100 HF-cool
10
5
1
10
3 Quartz
Diode-Si
Tantalum gl
beginning x working time
working hours
Typical examples for failure rate A
IC-digital bipolar (SSI)
of defectives
at the
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
Vulcanite capacitor
Resistor-carbon
Resistor-carbon
>
<
100 kQ
100 kQ
0.26
1
plug-in contact
5
plug-in socket per used contact 0.4
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
varies from point to point.
=
fix)
By the
gradient of a curve at point P we
mean the gradient of the tangent
at 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:
m
Always applicate
is
gradient:
=
tan
a
rr
Difference coefficient
The difference coefficient or mean
gradient of the function y = fix)
between
h 2
Ay
Ax
PP<\
is:
+ Ax)-fjx)
m fjx
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
-
/•«
Ay
oAx
= lim
f(x + Ax) -f(x)
6y
Ax
6x
fix)
I
DIFFERENTIAL CALCULUS
Meaning
H
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
Q =
of curvature g at
Vd-^y'
1
a
=
M
M
3
2
)
Centre coordinates
h 5
any point x
for radius
+y'
is
below the curve where g
above the curve where g
is
is
g
2
^—
x
is
y
y
h 6
u
_
..
.
^y' 2
continued on H 3
DIFFERENTIAL CALCULUS
Meaning
H
of derivative
Determination of minima, maxima and inflexions
Minima and maxima
The value x = a obtained
h
h
h
>
for y'
=
is
7
For
y"(a)
0,
there
is
a
minimum
8
For
y"[a)<0,
there
is
a
maximum
9
For
y"(a)
=0
see h
19.
The value x = a obtained
for
inserted
at
at
x =
x =
in
y".
a.
a.
Inflexion
h 10
For
)*
"(a
\
there
an inflexion
is
Shape
Rise and
y'{x)
y'(x)
>
<
o
13
y'(x)
=
o
h
11
of the
inserted
is
at
=
x
curve y
in y'".
a.
= f(x)
fall
h 12
h
y" =
increases as x increases
decreases as x increases
y(x)
y(x)
y{x)
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:
0,
y'(a)
y
n)
n = even number
h 19
{n)
y
(a)
y\\™
but
*
(a)
in
>
(n)
y
/
|
/
(a)
max.
'
= uneven number
//
<
{n)
y
(a)
>
y
(n)
(a)
<
\
^
i
».
I
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
u'(x)
y'
=
u'
±v'(x)
-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
=
e
=
e
=
x
=
=
=
=
derivative
x
x
y'
-
_x
y'
=
-e
ax
y'
=
ae ax
y'
=
e
y'
=
e
x
Ve^
e
x
-
(1
+ *)
In
a
2
x
x
=
a
y
=
n-a nx
y
=
a
y
y
=
cos x
=
-sin x
tan x
y
=
cot X
y
=
{kx)
y
=
cos (kx)
y
=
-a k
y
y
=
n
y
=
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
=
=
•
sin
cot
n
n
n
*
*
*
1
sin*
=
1
cos X
'
—
5
1
I
+
i
t*
Ic
n
2
*
cos X
=
-=\- =
sin
x
a
k
2
-(1 + cot *)
cos{kx)
sin (kx)
sin
n_1
-n cos n_1
x- cos*
x
x
sin
x
(1
+tan 2
y
=
-n- cot n_1 x
(1
+ cot 2 *)
y
=
-cosx
n
sin
y
=
sin
tan
2
*
*
2
cos *
n~1
*)
DIFFERENTIAL CALCULUS
He
Basic differentials
Derivatives
Logarithmic functions
:
function
h 53
y
54
y
=
derivative
jc
y
log a x
y
In
=
1
X
1
h
h 55
h 56
h
57
y
y
y
x
a
In
•
±1
± X
y
=
y
=
y
=
y
=
coshx
cosh x
y
=
sinh x
tanh x
y
coth x
y
=
In (1
=
lnx
=
InV?
±
x)
n
1
n
X
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
equal to
we mean
calculus
=
Fix) given y
f(x),
the problem of finding a func-
such that the derivate of F(x)
is
fix).
Thus
fw
we
hence,
-
«£*-/w
define
the indefinite integral
i
2
F(x)
Here
C
tiation,
is an unknown constant which disappears on differensince the derivative of a constant equals zero.
Geometric interpretation of the indefinite integral
As this figure shows, there are
an
infinite
number 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
fix)
is
)
represented by
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
dx
=
f^dx
u{x)
-
J[u(x)±
V (x)]
——r
+ C,
In
+
|jc|
here
n
4=
-1
C
Ju(x) dx ± Jv(x) dx
In |u(jc)|
+
C
+
C
J
2
i
9
Ju{x)
u
= \[u(x)}
dx
(x)
Integration by parts
i10
Ju
(x)
v' (x)
dx = «(*)
•
v(x) - Ju'(x)
•
dx
v(jc)
Integration by substitution
Jfix)
i11
dx = //[?(*)]
x =
here
and
<p(z)
dz
<P»
dx =
dz
g?' (z)
Example:
f
i12
F{x) =
Where
Thus
]\j3x-5dx.
3x-5
dx =
= z,
—
,
the derivative
expressed
in
terms
z,
is
z'
=
the integral
3— =
3.
becomes
o
F(A:).l/VFdz
expression:
=
fz Yz"+
F(x) =
C
Insert value of z
|(3jc-5) V3jc-5'+ C
in
above
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant
13
(1n
J
dx
x
n-1
a
1
a°*dx
b
15
i
Injcdjc
16
= x
xf-dx =
(In
J
x
(«=M)
r
1
ln|a|
In
- x
|jc|
jc(ln
2
- 2x-ln
|jc|)
|jc|
17
i
i
i
18
!x
19
jx m
i
i
\r\x-6x =
•
In
xdx
20
J
i
In
|jc|
ax
dx
/<e
22
xe^dx
Jxe^dx
x
22!
]x e
3-3!
1
m+
{m +
^
'\)'
(m * -1)
=
=
^-(ax-1)
[x2 e ax dx = e ax
a
(In 1*1)3
t
In |x|
= xn
J
n
In |*|
2
2a:
|jc|)
ln|(ln|jr|)|
x\nx
21
23
+
(ln|jc|)|
In*
y
+
2
(In
i
Q
*dx= ±x
(-
4
+
- ^f
a2
a 23
\a
n
e
ax
t
n
a
*-dx
-jjx n-1
-'e^ax
25
(n*
26
27
i
1
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
Jbb +
In
ac
c
e
|e
ax
sin
b + c e
|
ax
|
ax|
dx
bx dx
(a sin
a2 + b2
bx - b cos bx)
*
e3
bxdx
je** cos
-T
ax + b
dx
=
J(ax + b) n
./(ax-Z>)
cos bx - b
sin bx)
l«*»l
ln
r
i
a(/i-1)(ax +
l^"
|n
=i-
./ax
ax-b
(a
a2 + b2
6
n_1
&)
l
*
f (ax + 6) (ex +
d)
bc-ad
/;{ax - b) (ex - d)
ad-bc
./
cx + d
\ax + b
cx-rf
xdx
/*
J
1
(ax + 6) (ex +
fjcdx
J
=
ax + b
6c-aJ
d)
In
L_b
a2
a
*
0)
(flrf-6c
*
0)
(bc-ad *
0)
'
X
J
ax + b
a4
2
3
L
dx
[
m
Jx(ax + b)
_±
dx
_
Jx2 (ax+b)
_1
f
(foc-aJ
T6
a
x dx
- 2b(ax+
=
J X(ax + bf
ax + b
a 12
,
1)
\ax-b
J
/
(»>
a(n-1)(ax-&) n_1
n
.
In
a + -
x
b
b)
6x
+
a
.
-7? In
6
I
£>
a + -
b)'
+ b2
In
|ax + b\
J
—
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant C)
i
i
dx
x J (ax +
45
(ax +
2
x
i
i
47
i
i
i
49
i
i
\(ax+
J
ft)
In
|ax+
ft|
a
b)
- 2b
In
\ax+
-
b\
ax + b
L
dx
+
58
59
60
a
ft|
+
\
'
—7 \(ax+
bf
dx
x(ax + bf
dx
'x^ax +
- 3b
b)
-^-
-
ax + b
In
b
0/
2 (ax+bf
\«
3
'a^x2
f
f
a
xdx
+x 2
2
^dx
ax +
1
b
2
\
x
I
I
ax + ft
b
2
In
2
(ax+bf
3b<
\ax + b\
l
b'(ax +
ft)
dx
f
56
57
[ln|ax +
L
2(ax + bf
(
52
55
ax + b
I
3
bf
x 3 dx
(
51
54
2
X
a
x
50
53
ft)
a
]
i
X
a
2
'
i
2
a (ax+b)
I(ax+bf
/*
i
2
(ax +
ft)
dx
x
'x
i
2a (ax +
I
2x<
2
2
dx
(ax +
i
ax + b
48
/i(ax
i
b)
(ax + b)'
(ax +
i
I
(n
dx
x
46
J_ [ a 2
ft)
b)
ab'x
b'
,
H a+
2 (ax +
£|
ft)'
ax
ax+ft
*^|}
3
2
1
4[
ft
3a m
L
.a *. + (ax+ft]! _ 3a(ax+ft) 1
|«±ft|
+
*
2x^
J
ax+ft
x
I
4- arctan J
a
a
^-ln|a 2
x-
+x2
a arctan
|
a
x 3 dx
^+~7
dx
dx
^_a2
to
^
-/.
_
"
=
J_
J_
a '2
ln
a+x
|a-x|
_l, n|a2_^
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—
xt-a2
In
2
J:
"
dx
JaZ-x2
r(aW)
—^-^l
x dx
x
arctan
-~
2a 3
a
2
)
dx
2
**2
)
X
+ x- arctan —
9
a
2(a2 + x2 ) 2a
dx
x
(a
—
2
2(a2 +x
2
)
tf+x2
3
+
)
1
tf+x2
2
p-
2a2 (a2 + x2
2
f^ + J_ln|
2(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 )
dx
(
dx
2 2
/,(sf-x )
f
2
2
2
2a (n-1) J(a + x
2a 3
ln
n -1
)
\a±x\
\a-x
2
dx
Ha2 -*2 2
~ 2{a2
)
f
2
x
dx
Jtf-x2
H
)
x
-
b
1
2(a2 -x2
2
)
fyJTtx
ax +
-x2
)
2a
2
|n
'"
fV?
dx
aYTFdx
'ax + b
dx
3a
=
yjiax + bj*
2{3ax-2b)j/{ax + bf
2 2
2 (15a x -12afr;c +
105a°
fe
=
+ *|
\a-x
la
2VF
86 2
)
V(as+6)
s
(n*1)
|
;
INTEGRAL CALCULUS
Basic integrals
Integrals
(omitting integral constant
78
79
[
dx
J\Jax + b
m 2y/(ax + bj
a
fxdx^
m
3a 2
2
2
[ x
J
m 2(3a x
dx
i
15a 3
=
2
J*
J*
| \/a
2
+ x2 +
IVc
+ x" dx
fxV>
82
83
- 4abx + 8b 2 )\f(ax + b)
'
'
81
v
2
yfax+F
2
2
fyja + x dx
i
yT^T^
2{ax-2b)
J^/ax+b
Q
2
2
-* 2
±
arsinh
3
)
Va 2 +x 2
dx -
| Vfa
+ x 2 ) 3 - j-
y^S
dx -
^£±2?- ai^S
Uy/a'+x1 '*
2
a arsinh
"
+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 2 +x2
x
|
VrW
continued on
I
8
INTEGRAL CALCULUS
8
Basic integrals
Integrals
(omitting integral constant
dx
f
*
x3
2'
_v2+
=
—
2X2 ia c
Va 2 +* 2
/V?7? d*
i
=
2a
jx 2 ^a 2 -x 2 dx
x dx
f
x 2 dx
f
a
yJ(a
o
1
a
:
dx
^*?
,
'+W-*
n
x
\
\
Va -*
2
a?*
d*
a+
/ * 3 \4
2
2
dx =
-a 2 dx
-|-
=
=
\*\Ptt*x
-
^V^
2
1 \J(x
2
2
f \/(*
-a 2 -a 2
-a 2
arcosh
3
)
" a 2 ) 3 '+
^Z^
5
ya -x
2a J
2a' x'
jx 2 yJx 2~^aT dx
^
3
)
:
2
^T^
\x\lx
-x 2
3
^%^-a^V^?
^
Jy^-a
2
5
vS^7
I
)
5
arcsin
x3 dx
VaT
-x 2 f
2 5
V(a -x
tV*"
•/
2
)
a'-jfdx
f
a
2
2 3
-fy(a -x ~^(x\f
=
2
dx
2|
[
a
/ Va 2 -x 2
In
3
yja^+
x
Q
V 2^ v
|a
+
a
2
y (* V* " ^ - ^ arcosh
V^a!g
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
sin*
ax dx
/sin 3 ax dx
/
sin
"
a2
a arccos
r
+ arcosh
-
\
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
,
cos ax
-g-
x
1
2
4a
2ax
sin
1
ax dx =
cos ax
1
sin"
•
ax +
rca
¥/*""
(a?
i
116
J
i
i
x
sin
117
118
x
120
sin
2x
ax dx = -^
sin
ax dx
fsinj
119
f sin a
J
x2
f sin
sin
ax dx
ax
dx
ax
sin
Ix2
ax -
.iixl
i
0)
sin
g
ax -
sin
c
1
i-
£ cos
i
+ iaxl_iaxl +
3-3!
1
.
>
2
5-5!
7-7!
¥*•/'
dx
ax dx
an integer
x cos ax
(Zx2
6
j
2
a ~
= ax
is
ax
dx
a
.
/"
cos ax
.„
121
122
I
cos ax dx
123
2
cos ax dx = 2 + 4^"
124
cos 3 ax dx =
-=-
a
sin
sin
2a *
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
xv ™e
cos
I
ov- Hv
axdx
——
cos5 ax + x
=
2x
x c cos ax dx =
sin
cos
I*
axdx
6
~
:
ax dx
i
ax
2 \
h] sin
ar
\a
—
=
-
i
sr
/3X2
2"
n
^^fcos
n J
ax +
2
/x
cos ax +
-5-
1
sax +
^
ax
i
^H
3
(
\a
a
sin
ax
i
cos ax
dx
x2
f
cos ax
f
x
J
/
/
'
r/i-1)x
-—
ax dx =
2
tan'
axdx
—
=
axdx
tan
tn
=
In
|
tan
n~1
/
cot ax dx = -j
|
=
sm aax
; sin
n-1
/"
sin
x
i
i
axd.
n-1
(n*U
i
ax(
i
tan"
'
ax
•
dx
J
(*=M)
sin ax\
1
i
9°fl32L_L 0X »-2
ax
.
6x
(n*1)
tan
—
dx
=
_
128
i
129
130
131
132
133
134
135
136
137
138
2
cot ax
Jsm n ax
127
L„„ ax
,„
In
a
/;
i
i
cot ax
—
i
cos ax|
ax
cor ax dx
[cot°ax dx =
/*
In
sinax-dx
X
a_
n"1
i(n-1)
/
/"
J
cos ax
=
.
n
tan
tan
cos ax _
X
126
J
i
J
125
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
J
i
i
i
ax
_*
=
146
147
i
i
i
i
i
i
i
152
153
154
155
ri
a(n-1) cos n
tan ax
dx
dx
-
1
-
tan ax + -*
=
Itanf^-f
4
a
In
At
dx
->(f-f
T
^
j
ax
n
J
2
a
ax
sin
bx dx
=
sin (ax
+ bx)
2{a +
- |- cos ax +
x n cos axdx
=
|- sin ax +
J —^
_dx
X
cos ax
_dx
/;
ax
n^ax cos ax
- bx)
(|a|*|6|)
2(a-b)
(|a|*IH
^^*5^(i«i*i»i)
=
156
_l
sin (ax
b)
axdx
ax
dx
f s]n ax
(n
_ cos(ax + fex) _ cos (ax - bx)
2(a + b)
2{a-b)
cos bx dx
ax
sin
+„„
1
cot
cosax cosbx6x ,
Jx
(«>D
<
\2
dx
/sin
J
|cos ax|
1 tan^
ax
n-1
ax
dx
cos ax
sin
sin ax\
t
sin
I*
|
?)l
fc
149
151
tan
ln
dx
cos n ax
148
150
In
a^
a
sin
i
cot ax +
dx
cos^ ax
x dx
cos ax
145
t
Q
1
-*-
a
142
143
i
2
x
-
j Jx
j-
J
n_1
xn
cos ax
dx
ax
dx
~
sin
=l|n|tanax|
|tanf^ +
4
^
2
contin.
on 12
I
INTEGRAL CALCULUS
12
Basic integrals
Integrals
(omitting integral constant
sin
3
dx
ax cos ax
dx
cos' ax
'in
sin
ax
2 cos
n
sin
/
J
J
/
axdx
159
160
1
161
S[ "
mTn J
m
ax
cosn
"2
axdx
an odd number, solution for the Remainder-integral:
is
m
ax
dx =
cos ax
x dx = x
arcsin
arccos x dx = x
arctan x
dx = x
arccot x
dx = x
sin"
,T
arcsin x +
•
•
•
ax
'
a (m +
dx =
sinh {ax)
2
2
y 1 -x
arccos x - y
-x 2
1
arctan x -
—
In
arccot x +
—
In
sinh
J
n
x
•
dx =
1
+ x2
— coshx
•
cosh
(ax)
1
+ x2
i
163
i
164
i
i
-|
sinh
dx = ^- sinh (ax)
162
n_1
165
166
I
i
-x
*/ sinhnx >
•
[n
/
i
i
— cosh (ax)
dx = ^-sinh (2x)
x
(m*-1)
1
2
Jsinh
158
ax
a(m + n)
J
J
2
cot lax
cos 2 ax
m ax-cos n
i
1
+
tan ax
In
I
•
+
If
cos a*
I
—
157
2s\n 2 ax
\
tan^U
2
1
dx
sin'
tan ax
|
sinajc
•
dx
=
ax sin ax
,"
In
a
Q
dx
i
167
168
169
0)
i
170
INTEGRAL CALCULUS
13
Basic integrals
Integrals
Q
(omitting integral constant
Jcosh
J
cosh
2
x
dx = ysinh(2x)
n
x
dx =
in (ax)
7
2
tanh
n
6x
=
/,sinh ax
d.r
.'sinhr
f
J
ax
dx
cosh ax
(
dx
/;coshr x
/
J
J
—
In
=
|
cosh
4a
(ax)
x
dx
\
|
sinh (ax)
Jtanh
n_2
*
x dx
(n
x dx
(n 4= 1)
1)
|
coth
n_1
x +
n "2
Jcoth
ax
In |t<
2
coth x
2
c
arctan e
a
tanhx
•
arcosh x dx = x
•
artanh x
In
--|-
x dx = x
arsinh
n_2
= x - coth x
jcoth n x dx
/-
Dsh
(n>0)
dx - ^-
coth 2 x dx
/
•
1
n_1
x +
dx = - -|- tanh
x
coth (ax)
J
smh x cosh n_1 x + ^j— Jc
n
dx = x - tanh x
/« nh x
J
dx =
1
-i-
yx 2 +
1
2
arcosh x - \/x ~
1
arsinh
x -
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
6s
H
189
+
1
y'
2
moment
static
1
1
+
dx
surface area where the curve
rotates around the .v-axis
arc length
i
+ dy 2
xjyi
dx
1
+ y
,d
dx
of a curve
y-axis
.v-axis
b
b
t
i
190
Mx
2
=Jy^+y'
dx
A/
=
+y' 2
J*Vl
y
\
a
a
coordinates of center of gravity
A/v
191
ys
s
volume
body where
rotating
A
area
rotates
around the
y dx
192
static
V
moment
=
2
jv
J y
.v-axis
dx
of a curve
in relation to
the
u
dx
193
By-
J
xy dx
coordinates of center of gravity
194
tfv
of a
body, the cross
section .4f of which
is a function of x
V
= j A, {x) dx
—
INTEGRAL CALCULUS
15
Application of integration
Static
moment
195
=
Af yz
body
of a
the y-z plane)
(in relation to
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
covered
distance
by
the
center
of gravity
2-JC-S -y a
=
197
Volume
V
(see also formulae
A
189 and
i
191)
body
of a revolving
= area
i
times
distance
the
covered
by
center
the
of
gravity
=
2
A
31
(see also formulae
y
i
192 and
i
194)
Numerical integration
Division
of
number
area
into
an
even
n of strips of equal
199
.
yn
W^f
£1^2
Then, according to the
Trapezium
!
t
^wIw^1v^^\J\\vnJs\
rule
*>1
A
200
Simpson's
201
202
A
=
|
(y
+ 2y,
+2y 2
+
...
+ yn )
rule for three ordinates:
^1=|
Simpson's
-
(yo
+ 4y! + y 2 )
rule for
more than three ordinates:
+
+
|[>- +>'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
about an axis x or a point O, we mean
the products of line-, area-, volume- or masselements and the squares of their distances from the jc-axis
or point O.
sum
the
i
203
of inertia
of
dm, 6A
Moment
Second moment
of inertia
of area
/**
Steiner's
dm
-1 x2
kg m'
theorem
(Parallel axis
For every mass moment of
following equation will apply:
/„ =
204
/
yy
line,
will
area and volume
theorem)
inertia,
kgm
m/.
Similar equations
dA
both axial and polar, the
2
apply for
moments
of
inertia:
2
/^ + Al s
Moment
in
of inertia of plane curves
relation to the
x-axis
y-axis
>n
,b
7 oy
205
=
/**
V 1+ J ,2d *
a
7 XX
:
I
m,
/
s
:
A
:
:
moment
moment
about a general axis xx
about the centre of gravity
total length, area, volume, or mass
distance of centre of gravity from axis or point
of inertia
of inertia
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
h
207
-
r,
cM
j
fix) yields:
v-axis
.v-axis
i
=
\f *
-j
r
y
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
from point 0.
their distances r
r
I
dA
208
ot
Where
other,
pole
209
^
XU
C*
— —
,
X
the relative axes of / x and / y are perpendicular to each
the polar second moment of area in relation to the
O
(intersection
ff
dA
-
of axis
f (/
x and y)
jT)
dA
=
is:
/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
One
of
the
-m
dA
axes being an axis of symmetry of the
relative
plane surface results
in / xy
=
0.
to an inclined axis x'\ Where moments
in
relation to two perpendicular axes x
yk
and y are known, the second moment of
Conversion
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
/x
,
/
y
,
and
/ xy
INTEGRAL CALCULUS
18
Application ot integration
Examples
in
moments
conjunction to second
on page 17
of area
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
215
k
216
_,_
b
3
bh
h _
h.
'
,,
2
f(b< + h
2
);
217
-xy
"
~2
/ps
As
x' and/or
symmetry, / x y
2
.
h
b
2
±J^
12
'
2
i
y '\\
bh;
=
6y
y'-b
b h
{bh) =
(~2~
dA
\r 2
=
^(fcW)
/
are axes of
is
zero. Hence:
.
Circle
i
i
i
218
frr
2
2K
r
6r
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
/
=
223
224
/
7p
A
y -2JC
R
2
2J> V^
2^-
A
~y 2^ dy
=
dy
XR
-r-
jiR'
= 2
0,
as y
Regular polygon
i
225
is
axis of symmetry.
8
y
'x
V*
2
a
2
(6fl^
y
2
2-48
V
-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
'Moment
of inertia' of a
^
Where ($£- +
(see
I
)
volume
of a solid
cuboid
is
rM7\
the polar
12/
V12
moment
of
of inertia of a rectangle
18),
the equation for the
Z-axis:
226
,z
J
Moment
a
-^(b 2 + h 2
dz
\12
12
)
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
The mass moment of inertia / about a
product of the second moment of volume
the density
i
230
/v
is
the
about the axis and
g.
J
229
i
of inertia)
particular axis
where
Q
e.g. for a cylinder
=
Jy g
=
y
kg
/vz
For other mass
N
,
kg
m
moments
4
m
h
2
of inertia see
r
M
2
3
Kh
s
2
,
VA
s
3
m~ 3 kg dm" 3
,
about the axis Z:
jtr
231
m2
mr 2
2
DIFFERENTIAL EQUATIONS
1
General terms
Definition of the Differential Equation (DE)
A DE
an equation of unknown functions which contains deri-
is
vatives
derivatives)
(partial
the
of
unknown
independent variables. The different kinds
Ordinary
Equation
Differential
M
unknown functions
(ODE): the
depend only on one independent
and
functions
are:
variable, e.g.:
y = f(x)
y" + 2x 2y = sin*
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
Form:
F
Where y
y(x), y'(x),
(x,
(x)
n th derivatives; x
J4
Example: y'"
y<n)(x))
...
unknown
the
is
=
0.
function, y' ...yl n
)
are the
1
+ m(x) y'(x) + n(x)y 2 (x) + p(x)y =
(x)
st
to
the independent variable.
is
j5
Order: the highest derivative occuring
the above example.
J6
Degree:
in
the
ODE»3
th
q(x).
order
the highest exponent of the unknown function and
2 nd degree in the above example.
in
its
derivatives;
j7
Linear:
ODE
function
is
means, that the nighest exponent of the required
i.e. an ODE of degree 1.
one;
]8
Homogeneous ODE
)9
Inhomogeneous ODE
J10
Solution: y = y(x)
Integration
The general solution
C2
...,
Cn
.
0.
=
0.
ODE
means, that
this function
13
y '(x Q ) =>>;
...
its
yields the solution.
of
an « th order
ODE
contains n constants
These constant are uniquely determined from n
boundary conditions.
J
and
ODE.
ODE
)11
J12
C,,
=
of an
derivatives satisfy the
of the
q(x)
implies the forcing function, q(x)
implies the forcing function,
y (»-V(x
The particular integral
)
=y m-u
of the
ODE
is
a special solution.
DIFFERENTIAL EQUATIONS
Linear Differential Equations
Methods
1.
ODE
Transform the
one
into
an
to solve
ODE
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
3.
Use
transformations,
of
D18
cf.
degree
J 9
(cf.
...
J 12).
particularly of the
Laplace-Transform
D20.
...
Linear Differential Equations
Form:
y<">
J
15
J
16
Here y =
i
17
to
/?
+ p
x
(x)
y(«-V +
...
+
Pn _
(x) y'
x
+
Pn (x) y
=
q(x).
y(x) is the required function, y' ... >>(") the 1 st
derivative of y(x) and P^(x) ... P n (x) are functions
th
of*.
General solution of a linear inhomogeneous ODE.
y = ^hom +
Solution of the
homogeneous ODE
hom
is
C,...C n
.
Jhom =
20
Cy
x
x
(x)
+ C2 y 2 (x) +
12 give solutions for
Equations.
J 9 ... J
tial
j
21
v hom
determined by setting the forcing function q(x) = 0.
Each linear homogeneous n-th order ODE has n linear independent solutions y-\,y~2---yn witn n independent constants
_y
19
JVpart
is
determined
to find solutions. J
and
...
2 nd
+ Cn yn (x)
order Linear Differen-
inhomogeneous ODE v part
* 0. J 3, J 6 and J 7 suggest how
9 and J 12 give solutions for 1 st and 2 nd
Particular solution
y pan
1
st
of the
for q(x)
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
23
Method
J24
to
Form
y P an = c iM y\ + ciM yi +•••+
determine
Cj(x), C2 (x)...Cn (x):
cnW y n
-
the simultaneous equations
C{(x)
y 1 + Ci(x) y 2 +
+ Ci(x) y 2 +
C[(x) y[
+ C'n {x) y n =
+ C'n (x) y'n =
...
...
yx (*-V + Q(x) y 2 n 2 + ... + C'n (x) y n ("-V =
+ Ci(x) y2 ("-V + ... + C'n (x) yn (n-D = q(x)
= 1, 2... n using the above equaDetermine C (x)
for
C[(x)
<
>
C{(x) yi (n-D
j
25
'
i
:
tion system.
j
26
Integration
of
Cy
(x) for
=
i
2...
1,
rc
yields the values of Cj
fjcj
for the solution.
Example:
Solution fory part of the
„
1
.
..
2x.
J27
28
Ace. to
j
29
j
30
let
j
31
using
-J»* dx +
J*
121: ^hom
C,
C,
Inbcl
+ C2 y 2 (x)
Q
and y 2
using y fjcj = In
= C
3^1 + C2 (X) y 2
=
1
ODE:
IjcI
>>part
j
^
\
C[(x)
Nqw
24
1
Inlxl
hence
QY*; =
Integration of
C
SO
C
A
2x2;
(x)
and
C2
C2
= -2*2 !nW
W = - Sx=
= |*3
Jpart
= 1*3* InW - |*3
;
=
Ci(x)*l
(xj gives:
(x)
t
+
- j\
flnlxl
(Inbcl
3
;
9
General solution:
j
35
3Wt= C l* lnW + C2 + 9* 3
>'hc
=
Check:
a:
-
£ + 1*2
jc2
3
(x)
+ Ci(x)^0 = 2x
fcj
J32
+ C9
y x (x)
Ci
x2
x = 2x
X
=
1
DIFFERENTIAL EQUATIONS
Linear Differential Equations
1
j
36
Form:
st
+ p(x)y =
y'
The form corresponds
here
J
37
j
38
j
Example:
_y
y = y hom + y pan
sin x
= I
x
=
j
110
from
j
109 the homogeneous solution
'
= C1
.Vpar
J
sm xe
= -sin
~.
,
Check:
y
+
v'
C
42
"
,
=
e-
= -
r
0;
is:
£
Cx |o.
with
is
e
dx e
)
sin x.
Mxl
= J(sin
xx^i
- cos x
jc
=
2-
— Q
x
,n
d*
,nlxl
y = y hom + y?an =
J41
q(x)
110 the particular solution
j
m =
40
d
C,e^
=
39
p(x)
J(sin x e
j
j
from
from
j
y,
+ I =
y'
15 for n = 1; the highest derivative
hom and y pan are given in J 2 and J 9.
to J 2,
Solutions for
is >•'.
ODE
Order Linear
q(x).
is
Ci
-j-
sin
\{C +
sin x)
X
+
,
x cos
- cos
x- sin
x
+
x.
.
sin
x
^
x
determined using the boundary condi-
tion e.g.
j
43
y(x
j
44
Then:
j
45
Gives
)
=
1
=
^(q + sin f) - cosf
=
y-
Cj
:
1
for
x
=
ji/2
1.
2 nd Order Linear
j
46
y"
Form:
+ p
:
(x)
y'+p 2 (x)
The form corresponds
tive
and
is
/'.
J 12.
v
to J 2,
Solutions for
y,
ODE
= <?W
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
47
j
J48
Due
the
y"
Form:
+
+
lay'
=
b2 y
i
50
y -
si
Overdamped
j
52
j
53
J
54
.Vhom
+
j
j
56
j
57
j
58
j
61
62
>
.
k2
Critically damped solution:
Jhom =
y pan
Underdamped
^hom =
ypan
=
a2
q(x)
=
b2
dx + x e-™ Je
k2
solution:
e-^lQ
=
a2
+ C2 cos
sm(o)x)
\[b 2 -a 2
= e-"sin(ft«; | e «
cos(w;cj
= ^
b2
^
6x
q(x)
= A
'
= arccot
b
+ 4a 2 0) o 2
2 -0) 2
o
_
*)
s'\n(oj x)
S n (°>oX-r),
/4
y
<
J> sm(o)x) q(x) dx
special case
-Vpart
where:
oj
-
q(x) dx
*
(ax)]
=
and:
-
CiS-^+Cixe—
= -e~ ax jx e™
\/(fc 2 -a> 2 ) 2
j
:
6x
Note: For the
60
15
r
- e-»cos(o>r;
j
j
k2 = a 2 - b 2
solution:
with
59
to J 2,
^part
a(-a+k)x
><-a+k)x
55
q(x).
a and b are constants =t= 0,
is a forcing function
q(x)
General solution, according
j
with constant coefficients
great
49
j
ODE
importance of this ODE-type for oscillationproblems, special cases are considered.
to
is:
DIFFERENTIAL EQUATIONS
Linear Differential Equations
ODE
Linear n xh order
j
63
a n -y< n >
Form:
64
j
65
=
cients <q(x)
Let
x
+ a
...
y'
x
+
a^ =
ODE
order
q(x).
with constant coeffi-
0).
y = e";
= re rx
y(n > = r"-e rx
homogeneous ODE of 63
y'
Substitution in the
braic equation:
j
with constant-coefficients
~^ +
-y( n
homogeneous n xh
Solution of the
j
+ an _
.
.
leads to the alge-
j
a^
66
.
;
+ an _
rn
1
~1
+
...
+ a
r
x
+ a =
0.
The roots
r1( r2 ... r n can be determined. Depending
the type of the roots, different solutions fory hom are found.
j
Case
a)
Case
*>
+ - + Cn e
?hom = c e "* + Q-e'
b): There are real single and multiple roots:
67
r1t r2 ... rn are
:
and
real
different:
r" x
2 '*
i
=
rx
j
all
on
=
r2
...
= rm
QrvX
,
rm+]
,
rm+2
rn
,
.
qp+r*
C
+
+
+
+ Cm -x m -^e r r x + Cm+1 -e'™ + i* +
^hom -
68
C2 -x-en*
i
=
+ C2 x + ... + Cm -x m ~
+ Cm+l -erm*vx + ... + c„er» x
e'r* (C x
l
+
+ Cn -e r*
...
...
)
x
*>
+
.
Case
There are conjugate complex roots:
c):
= a +
r-j
J
70
r2
i/8;
= a ~ P=
X
= C v e^+C2 -en-* •;
= e^-fA-cos /3x + fl-sin
y h0 m
A = C + C2
x
Particular solution of the
;
'T-
fix)
5 = ifQ-C^
inhomogeneous n th order ODE
with con-
stant-coefficients
y P an =
The form
given
les are
SiM + SiM +
of the particular solution
».
+s*to-
depends on
gfo).
Some examp-
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
*)
Cv C2
.
.
.
Cn
J 7).
are arbitrary constants
DIFFERENTIAL EQUATIONS
J
Linear Differential Equations
ODE
Linear n xh order
for
q(x)
J74
AQ + A
X
X
+ A2X 2 +
J
75
J
76
A
cos
J
77
B
sin
]78
J79
J80
J81
J82
J83
]
84
j
85
j
86
j
87
j
88
t
a
xm
73
J
89
90
i
91
j
92
j
93
A
cos
mx
mx
mx + D
mx
sin
A e^ cos mx
B e^ sin mx
A e^ cos mx + B e**
94
=re n
y'
;
ODE
the
of
example
=0; r2 = 1;
ei* + C2 &>* = C
mx
/te^ sin
J 6,
65
j
mx
let:
e rx
85 gives
j
= 1; r2 = -1
e* + C2 e-
rx
x
%
sin 2x
= a cos 2x +
= ~ 2a s\n2x + 2B cos 2x
= -4a cos 2x - AB s\n2x
/part
Equations
-5a cos 2x
;
a =
used
sin 2x = cos
89 and
j
-
5/3
j
91
Check:
y'
=
=
ODE
(line
j
85) yield
2x.
of terms yields:
and therefore y pan =
—|
C
x
y" = Cj
-y
.
in
General solution:
x
= /horn
Vl
+
^Jv part.= C,1 e1 + C,2 e~
Jv
y"
sinh
ft
/part
=
ft
of:
y parX
ft
sin
y"=r2
;
r2_i
C,
mx
/?
a e^ cos mx +
+
+
according to form of
rjc
=
1
...
+
+
2x;
in
mx +
a cosh
mx
s\n
Example: y"-y = cos
y = e
+ a^ "
+ a^c m
...
+
+
A cosh mx
B sinh mx
A cosh mx + B sinh mx
y hom
mx +
a cos
Comparison
j
x
1
A-e**
Form
j
a + a x + a^x 2 +
aQ + a x + c^x 2 +
a-e^
A^"
+
...
Substitution
j
with constant coefficients
Form of y par =
A
J72
7
Cj
~j cos
2x
j
-tcos
5
+y sin
2x
2**2
ex
- C2 e~
ex
+ C2 e~ x + j
4
cos 2x
ex
+ C2
4
cos 2x - C
-C e2
+
x
e~ x +
-j
-i-cos 2x
= cos
2x
x
ex -
DIFFERENTIAL EQUATIONS
J
Reduction of order
8
Reduction of order by variable-substitution
to solve an n th order ODE
o
<J
E
o
E
o
6
E
E
o
o
E
o
*>
Sec
c o
O-
61x5
^ £
W ^
.Z
II
-O
3
ft)
(i,
*
|
Q,
ft.
II
II
V,
>s
ft.-ol"o
II
+
II
V
"ss
"ft.
II
II
+
XX
c
o
C/)
3
4-
i
i
H
<n
<?
<3
$
+
J
«
'
CQ
s
+
»
^
o
t=
&
S
£
c
i
-Q
+
>,
W
ci,
J?
y. u|cm
^
<d
1
*
r'r
«?
tf
II
H
ft.
>s
1
0)
^
1
^
U
*
+ +
o
^7
®
r~
'-.HKiU
H
II
^
QJ
II
1
?sQ
>%
o>
_>> OJ
E
3
>>
X
••1
CX
^
ii
|
II
II
+
tf
H
Ol-D
ol-D-ol-a
II
+
1
ft,
CO
c$$$
*
^
|
loo coo coo
c
o
^
*
..
-^
II
^
+
—
— !5
!5
—
8q.«
X >
°
<D
:»!
CO
<->
co
S-'l
>,
0>
CO
=
^
>
£
tv
%
co
-2
~ ra
O
> CO
« (DO
•^ -o c
?
®
o
„
11
u U
II
jm
„
•
><
u
•
°
"
0)
,r
^
LU
T
O
-
Q
.c
o
E
o
LL
^
«
•
c
c
X
5
<5 *
o
^
a>
^
+
E
;*
to
H
II
s
3s,
"?
X X
,„
S
SS
>>
II
Q.
£
U
U
U
U U
oV ©JE^OOtflOUl
-S?
-,-
II
jj
^
^
c
~
l
-ii
E
5
X^
II
"ex,
ftj
II
II
a,
ft.
=N pn
II
1
<P
II
II
II
II
cxX >,V "K^
>s
i;
i
V
ok
©
ft.
^w ^
Uj
O
000)000000
t-
CM
CO
•<*
lO
DIFFERENTIAL EQUATIONS
1
st
J
order Differential Equations
v
3
J
into
and
"re
—
.2
5=
equation.
c
c
c.2
«
x
OJ
E
E
o
of
c/>
right
separated
and
.1=
9
2,
~ ^>s
of
109
03
j
E
o
O^
J
using
§§
c
°
particulary
cf.
cf-
the
Determination
left
of
be
variation
- 2>
W—
C
variables
O
Oi3F
O
.Q
the
CO
side
The
solution
^hom
o
o*
™
x:
oj
constants
the
<n
Q.
can
h~*
1
E
o
y
.c
J
=^
*
H**
r-
,5.
i
Hh
+
c
g
D
O
H
+
D
T3
n
0J
H
i
1
ii
+
'<D
u
CO
g
T3
0J
Iq.
3
£
oj
H
73
+
ex
y
II
1
1
<
PS
T3
OJ
"*
^-r
§
0J
«
,
II
II
ii
ii
60
3
c
o
I
,->
s
31*
II
,
or)
.Q
+
SI
H
"Ol-D
.
oa,
CO
oItd
^
+
a
+
+
E
-icq.
II
II
^ 2 H
I
e "ol-o
II
"*
^
oj
V
ii
2.
St4<^
^ u
II
V^ NX
'oj
i
ii
®
II
ii
4-
E
o
3B
II
oa.
+
II
u_
^7>
a
T31T3
<=;
ii
+
ii
+
OJ
"O
-Q
03
—
LU
SlTO 18
IS
<D
OJ
CO
CO
901
>-LU
ElQ
wi50
!
zo
K
0J
,_
O)
03
-f
63.
F O
PW
o5
"P LU
03
11?
8
O—
O
"O LU
^
601.
f
OU
f
OJ
1*8 H
tn
X
80
OJ
II
Ul
f
—
»
DIFFERENTIAL EQUATIONS
st
1
c
ft
-£•i
E
E
o
i
des
1
111
w°
"or
o
_y.
§?||»
curv
integr
E
<5o2o
c ** £*~
and
-
w Q.
o =Etj,
w.E
©
arameter
x
T3
gular
0- **
•«-
~
rr
9.
r.
J
one
b
solu
known
must
in
z;
Inhomog.
cular
as
in
~
©*v
nvelope
ion
UJ
solu
O C
elimination
.„
10
J
order Differential Equations
least
(0
(0
r
X
^
1
D>^
.£,
2.
X
n
5
ca-
1?
s
c
o
+
a,
H
o
>
CO
^
s
1
^tf
^®
+
II
ft
II
H
•^
HI ft
dItd
ft,
1
1
II
II
H
c
o
3
Pn
a;
""P
II
|i-
K)
1-
<D
<U
If
^
+
+
x
$ *
jft,
'
PN
>
ii
>ii
*
>,
HI
ft,
ft,
^
ft,
CO
II
.o
3
CO
II
II
II
V
V,
""
II
II
N
+
E
X
«;
R
3^
o
II
1
Cf
Ps
H
II
Pn
?S
>
aoQ T
*^
3U
c
f
siii U
l ^
^
iO
Q.O Ej
J=
<°
ei
<
fl
1
O
HI
ft
£
+
§
+
X
ps
to
©
=
f
. 0>
05
£
3UJ-D 0)UJ
OT3 o
Eo
© to x:
CO
ii
N
^<~
+
II
.2 "2 uj
(0
+
*»
t>
H
X
-D
c
PS
GO
ii
C
V
+
o
T3
iu
*-
II
_^
£
ii
11
II
PS
o
r^
u_
,_
J
f
N
II
+
" 5
^
^p-
ii
^ft,
| H
I
Pn
S£:
J=^
—
^
T
1
^ t
I
^-n
->
ll
i
ll
N
2?
(0
OC
UJ "D CJ)LU
CO
T3
"-"s;
QU
f
9U
f
s
DIFFERENTIAL EQUATIONS
J
2 nd order Differential Equations
—
CD
c
o
E
E
o
cd
"frf^
c-Qc v —
o
ro
3
ro
CD
O
ill
"D
«
~
*
w
+ T *£-*
>
c
o
3
O
CO
ii
i
H
>
^i
+
^
«
II
^
+
^i
*
.E
&
a>
*
7 7
*
£
II
•:
*
-Q
>
+
f?
"O
5!
O
**
C
CD -H-
CD
H
||
^||
1
CD
o ^U C% ^
o
II
S
£
II
II
>%
?s
II
£
II
>,
?s
ro
,*
p^
>*
>s
Q.
?s
+
S
H
a
© ®
is.
x.
CD
v.
ii
ii
II
i
+
II
T3
"
1
.
CO
"
+ ^
c
&^
i.
II
or
e
I
**
^
II
+
Q.
CO
Ps
H
^ ^U 3*3*S*
^
II
O
ii
$
>
<N
c
o
3
*.
o + +
'
K,
-C
o
u
H
* ^0)
+1
.
E
o
«
^—
C"
+
II
+
«*
H
+
II
o
cs
r
<D
1"
O
O
J
T ^
OT
II
1
5
>
-a
* 3.1
+
-C
_
o
sz
1
V
p^
£"
II
PS
O
CO
sz
X3
-£
—v
^
ii
E
i-
<D
-p
£
O
A
n±
+
CO
go
5 w
t-O «}•»-:£
o " C
«
ii
a-
H
Ocx>
o — c
if
•i:
,
o
7
—®
c
O
o
3 3
-^
•
>,
o
+
c
+ ~ ci
ra
co
'
^
c
^U
CO
1*8
m * c
-
II
II
«
*; en
O J
1
11
E
o
a
i:
""sL
six
tj|t3
?s
3
ii
CO
^
>s~X ps
O
II
II
II
^
II
V
Ps.
^
o
o
Gr-
?s
o
II
^o
a
+
E
o
5
X
+
PS
+
2
+
+
^
II
**
V,
"ps.
clslsSi ?E
C
CO
CO
CO
igliil
1
Uf
v»
cB
£
8LL
f
wO
6U
^ E
c E
0j!0
F
£-w
?
x
1
+
Ps
a)
^co c
T3
II
<3
2
^
+
II
X X
^
>^
" o
o
f
021-
f
ra
5s»
t2Lf
V
DIFFERENTIAL EQUATIONS
J
2 nd order Differential Equations
W
O
c *"
«—
m—
o)
c
CD
>
O
o£.£
CO
E
E
o
OT
c
3^3
||,lfi
i
-^S-s"
^ O C °o ^
CcioroE^
^2 b - o g
>#
(0
$o"
o
+
=
a)
-S
CT)
—
•
111
E
(0
o
raio a5 ->
.H ^ .E -o
r
o *;
„
ja
o
c
o >
c
c5
CD
ea
12
,.,
T3
+
+
'"*"
3
c
o
3
O
+
^
T3
+
3
T3
«
ii
>
"O
—
CO
^
>
II
1$
II
+
5
,i
'cd
,
SI*
f ^
"J?
O,
a3
^
«
II
II
>; p^
?s
-31^
1
>
v
3
X
5
'qj
,
^
31
3|X
i3:
~r*
IS
ii
ii
fir
a>
ex,
+
's
§
-*
•
o
+
II
"°
ex.
5 3
s
3
3
ii
c
3
II
o
31 PS
3^
T3|"0
i
a.
3^ +oE
CO
3
•-
II
II
II
"^
II
II
3 3
II
II
K
M
VV
V, V,
s
V^
Ls^
3 ^
II
II
xv,v
<*-,
3
-ol-O
II
U
II
II
3
Ps
V, Vs
II
>.
II
—
C*
+
II
V
3
^
II
ll
H ?
>
o
-^
u.
II
II
\\
<*-»
>s
^H
E
o
53 33^;
>
Is
<
s
3"ofc>
CO
"3
o
II
II
V
V
<*-,
<*-,
ex
+
+
V
>,
=
II
X
X
+ +
V
V
05
fe
g
§
"D
~
k:
E o i,«
m E
C
X
11
ii
11
"2 JB
-55
.cE
221
f
§o> Eg E|
E-a
®|ri
V £
o
yE
E
>S
^.2
.2
cnQ»
300 jepjo pu g
O
o3
j5 fe
a)
!=
en
«
oQ
E
eft
ezt!
1
W ill 52 n|
921.
o)
"Ew f.E
w
<D .52
"V
=^.<2
UJ
,_
Q
mO"E
C
0)
O m °
s?
I
°
CM
E
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.
G
Gravitational force
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
T*
T
total action of
several external forces.
Moment Mot
a force F about a point
The perpendicular distance from point
to the
line
called lever
of
arm
action of force
F
is
/.
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
the sum of the moments of the individual forces.
is
equal to
STATICS
K
Forces composition
Graphical composition of forces
diagram of forces
force polygon
^1/
A number
and a common point
of forces
of application
F3
k 5
k 6
'
link polygon
A number
link
r
of forces
polygon^
n
JF
link
w
-3
beam
pole
and any random point
of application
beam
^sf
^
link
Construction of the link polygon
Draw force polygon and determine pole O so
link
rays
running
parallel.
Draw pole
polygon
rays.
pole
beam
beam
as to avoid any
Construct link
such that link rays run parallel to corresponding
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 inpole
rays.
tersection F-i-1-2 of link polygon).
STATICS
K
Forces composition
Mathematical composition of forces
Resolution of a force
k
8
F cos a
k
9
+
Fv
F
F
=
a
tan
sin
a
Fx
signs of trigonometrical functions
of a see table k 16 to k 19)
F
(for
M
Moment
(for
of
of a force
±F-l
M,
k 10
about a point
Fu
=
X
r*y
signs of trigonometrical functions
k 16 to k 19)
a see table
Resultant force FR of any random given forces
XFX
k
11
components
FRx
=
k
12
magnitude
FR
= +
k
13
direc-
k
T^
14
distance
k 15
Rv
tana R =
,
/
*
|FR
FR / R =
sign of
;
2
+ FRy
sin
k 16
.
k 17
k
.
90.
180 ..
270 ..
.
18
k 19
FRx FRy
,
x
a
/
,
,
,
y
aR
/
R
:
:
:
:
r~\P^
[
F
Fr
—
Rvv1
=
;
R
Frx
cos»a R = -£=
FrR
(moment theorem)
|
sign of
aR
a,
2
aR
ZM
Signs of trigonometrical functions of
Quadrant
*)!*&
IF
=
^oi—
=
R
y FRx
F
—
angle of
I
FRy
cos a
x,
y;
Fx Fy
tan
,
a
x.
;
FRx>
F,Ry
Rx r
,
Fv F
,
c
90
.
c
.
180
270°
.
360°
.
components
of
FR
coordinates of F
angles Fand F R
distances of Fand
parallel to *-axis
FR from
and y-axis
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
with
is
forces
graphical
common
closed force
point
of application
x
force polygon
vertical axis
and
link
polygon closed
arbitrary
Simply supported
Find reactions
mathematical
lF =0;lFy =0
polygon
parallel to
the resultant
external forces about
all
beam
R A and R B
Graph, solution
lution
IFx =0;IFy =0; ZM =
W and W2
with point loads
:
:
:
:
Ra
1
1
s
R*
IV2
/
-•—
M b (z)
k 23
k 24
k
25
mL
mF
H
:
:
:
mL H
= v*
-
-m F
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
+
W
2
l
2
ll\
RB
= (W, +
W
2
)
-
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
problem
(3 forces):
Find reactions
F,
STATICS
K
Lattice girder
member
Mathematical determination of
(Ritter
method - Method
loads
of Sections)
F,
O
upper
U
lower boom
diagonal
D
boom
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).
.
.
.
moments
Establish the equation of
and internal forces taken
section of two unknown forces.
ternal
Rule for
moment
Where
Where
turning
turning
Example (from
problem
:
ZM
=
about
with
moments
the
point
signs
of
of exinter-
»
moment
moment
is
counter-clockwise the sign
is
clockwise the sign
is
is
positive.
negative.
the above girder)
to find force F^ 2 in bar
solution
Draw a line X
and D 2 meet
U2
.
:
.
.
.
X
at
through
C,
2
this
is
section selected so that the
-
D2
the
- U 2 Since the
.
point
relative
moment
2
and
lines () 2
of
inter-
D 2 may
equal
zero.
Proceed as follows:
k
26
+
a-F U2
+ b
ZMq
F2 -c(FA -
F,)
b
F[j2
F2
+
c(FA -Ftf
STATICS
K
Lattice girder
Graphical determination of forces
6
members
in
(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
Joint A: Force polygon
all
joints(e. g.
FA -
a-b-c-d-a.
F^
- FS - FS2 )i
Keep a record
of forces being tensile or compressional.
Joint C: Force polygon
d-c-e-f-d.
etc.
Check
Forces acting through a single
a polygon in the Bow diagram.
joint
Forces acting through a single point
ces form a triangle in the framework.
in
in
the framework form
the diagram of for-
STATICS
K
Centre of gravity
Arc of circle
k
28
k
29
k
30
k 31
sin
r
y
a 180°
r^_s
n a
y
y
=
=
>>
=
0.6366
0.9003
0.9549
b
r
at
2
at
2
a
a
=
r
r
at
2a
=
=
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
k
39
of a circle
s
y
12
fo
3
A
r
area
A
see B 3
For determination of centre of gravity
5.
see also
I
14
STATICS
K
Centre of gravity
8
Determination of center of gravity
of any random surface area
Graphical solution
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 directions (preferably at right angles). The point of intersection of the lines of application will indicate the position of the centre of area A.
Subdivide the
.
.
.
y
,
11
fi_
h
A3
*2
|
W
f
*
i
# V.-J.—J-_.- h
!
t
.-,
.
'S 3
st
vr
P
A R*
X
1
3^-s
1
a
2
1
*fi
Mathematical solution
Subdivide the above
A2
.
.
A n we now
;
di-
stance
in
total
area
A
'
V-
into
partial
areas
A-\,
get
general
in
A,
A^
the above example
x
:
>'1
+
+
A 2 x2
A
A2
>'2
+
A 3 x3
+
^3
•
^3
A
>'s
Note:
i
a
In
the above example the distances *i, y 2 and y 3 each
equal zero.
STATICS
K
Friction
9
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
<
A
(variable)<p
p-i
FZ
force
from zero
move,
to
until Fz-\
value
rzo
As soon as
starts
rest
reaches the
motion
c
= O u
o
.
happens, the body
Fz drops
excessive force will
this
sliding,
Any
Gu.
now
in-
the
o
k 48
to
increasing
compensated by an
is
FW1 without causing
creasing
body
c
gradually
i
'-=
Fw
whereby
sliding
friction
accelerate the body.
for oe acting
Force applied obliquely
The force F needed
weight
k 49
F
G
G
=
> g
{a
to set in
~
Vo
a- [1q
sin
motion a body,
):
cos a
sin Q
sin(a-po)
to maintain the motion is
ascertained by replacing // by //. No motion
possible when result of F is negative (a < p ).
The force needed
^Wl
Fz,
<
^v
^zo
friction force
•
Fz
'
traction force
-,
p-i
,
fi
p
,
pi:
,
p:
I
friction coeff.
(
angle of friction
I
see
Z 7
1
STATICS
K
Friction
10
Inclined plane
General
The angle a
clined plane
is
tan a
k.50
which a body will
the angle of friction p.
at
=
tan g =
move
k 51
=
n
^T*
base
tan g
(horizontal)
a < g
Condition of automatic locking:
Friction properties
constant velocity maintained by
tractive force F parallel to
base
inclined plane
i
^
motion
upwards
k
52
G T
F
cosp
< a<
^5>^
f
g
<
"
F
a*:
=
/ /^/.
/
v
^L
G
i
sin(a- p)
body
F
G- cosp
static friction replace
Titling angle } of
F = G tan(p-a)
cosp
A.
G
Note: For
sin(p-cr)
G
*//?<?
*/< /
a < a*
54
tan(a + p)
^5r
g
53
G
=
V
downwards
k
sfv\\
g\
sin(a+p)
downwards
k
^Sf^
/s\
< a < a*
pi
by
,«
in-
A
Application in the experimental determination of the angle of friction
or the friction coefficient:
H
down an
easily
*
md
=
p by
-
'
G
/<
r-
tan(a-p)
-
STATICS
K
Friction
11
1
/
v\'v\\\vvw
tan (g-i +
driving
k 55
in
k 57
1
loos-
k 56
+ tan (g 2 + gj)
- tan g 3 x tan (a 2 + g 2 )
ening
F2 m
/7
)
g-\
a,
locking
^
+ a2
=
F
F2
=
Ftan(a-2p)
1
tan(a 1 - g : ) + tan(a 2 - g 2 )
1 + tan
g 3 * tan (a 2 - g 2 )
automatic
F
tan(a + 2p)
a ^ 2p
p 0l + ^q 2
Screws
U M M n^ ¥ ¥
h
k 58
k
59
turning
raising
M-,
=
Fr
tan (a
moment
when
lowering
M
=
F
tan(a-p)
2
conditions of
automatic locking
when lowering
k 60
k 61
raised
efficiency
of
+p)
M
M2
1
=
-
"
"
tana
+ g)
»?
=
f
=
ta r\(a
tan (a - g)
lowered
'
moment
lowering moment
(tan
a = r-^)
64
angle of friction
(tan
g =
k 65
angle of friction
(tan p' =
fi)
cos/3/2
r
:
mean
radius of thread
p'
tana
tan(a + p']
tana
N m,
N m,
lead of thread
k
tan (a +p')
tan(a-p')
tan(a-p')
tana
raising
k 63
Fr
Fr
a <
a < p
screw
when
k 62
r
[kgf m]
[kgf m]
STATICS
K
Friction
Bearing
friction
longitudinal bearing
radial bearing
P^§K
k
66
A/ R
t*q
moment
:
^l
Mr
F
bearing (not
constant values)
radial
h
friction of a
|
^
U—
- ft
of friction
coefficient of
I
V.
r2
-r-F
= H
q
EB V^
V/7/7
/
IfeJl'
MR
12
longitudinal
|
determined experimentally as a function of
,u
q and fi\ are
bearing condition, bearing clearance, and lubrication. For run
~ ti L = ^q- Always use r-\ >
to allow for luin condition: (a
Note:
k
67
k 68
brication.
Rolling resistance
Rolling of a cylinder
k 69
k
70
F-£fn~£g
Fw <
Rolling condit.:
F\fj\
/
:
//
FN
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
Displacement
k 71
F
between
27
fl
J
k
72
k
73
where
/1
= /2 = / and n
G2 <
G!
1;
\\v
M
IB
Gi
F
/t
r
,
:
G2
~"1
weight of plate and one cylinder
:
:
lever
arms
radius of cylinder
of force of rolling resistance
number of cylinders
n
I
:
F
*
Q£srf
<
tractive force
and/2
:
and support
sh
+nf2 G 2
)G 1
= (/i^/2
cylinder,
supported by cylinders
of a plate
STATICS
K
Friction
Rope
13
friction
traction force an d friction force for
lowering load
raising load
k 75
F,
= e»
FR
k 76
s
= (e"
a
-G
F2 =
e^ 5 G
-1)G
FR
(1-e»a )G
=
•
Formulae apply where cylinder is stationary and rope is moving
constant velocity, or where rope is stationary and cylinder
at
is
rotating at constant angular velocity.
k 77
Condition of equilibrium:
(F
:
force without
F2 < F <
F,
|
G
e _t
*
•
s
< F< G
•
e
*
1
5
friction)
Belt drive
k 78
k
79
k
80
wound-up^
k 81
k
82
Fy
FR
Ma
a
(x
k
:
frictional force of
:
driving torque
:
:
83
v
e
tangential force of driving wheel
:
:
rope
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
Rope operated machines
14
Rope operated machines
The following
bearing
figures deal solely with rope rigidity, disregarding
friction.
unknown
fixed
free
quantity
sheave
sheave
'/////.
pulley block
ordinary
differential
'////s
1
'/////,
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
k
87
k
88
2-h
n-h
force
mechanical advantage
F-,
F
F
:
:
:
£ =
\e
iH°
i-«
effort
k 89
1+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
«
efficiency
number
of
1.05)
s
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
and their units
of kinematics
Length
see K 1
Units: m; km
/,
Rotational angle
Unit: rad
Time
,
t
a base quantity, see preface.
Units: s; min; h
Is
Frequency/
The frequency of a harmonic or sinusoidal oscillation is the
ratio of the number of periods (full cycles) and the corresponding
time.
number
_
J
Units: Hz (Hertz) =
Period T
The period T
1/s
=
of oscillations
corresponding time
cycle/s; 1/min
%
is
the time required for one
full
cycle.
It
is
the
reciprocal of the frequency/.
I
"7
r
2
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
tightly
is
and one revolution
of the shaft
n min
"
Units:
revolutions/second
-
/:
/
-
lor
(1/s)
rev. /min. r.p.m. (1/min)
continued on L 2
KINEMATICS
L
General terms
continued from L
Velocity v
The velocity v is the
respect to the time t:
Where
the velocity
5
1
1
derivative of the distance s with
first
=
v
4
1
2
ds
—
dt
=
s
constant, the following relation applies:
is
-
•
i
Units: m/s; km/h
Angular velocity
The angular
to,
angular frequency
velocity
to
the
is
to
first
derivative
turned through q, with respect to the time
1
6
co
=
dw
—
-
of
the
angle
t:
w
*
=
dt
Hence, for constant angular velocity:
1
7
ao
=
SL
t
Where/=
n (see
frequency
1
1
3),
the angular velocity
8
=
to
Units:
equal to the angular
to is
to.
2jzf
=
2nn
=
<p
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
1
2
;
is
the
dco
a
1/s
dt
first
with respect to the time
10
Units:
rad/s 2
d
—
5
o
2
the
=
velocity
v
s
2
Angular acceleration a
The angular acceleration a
lar velocity to
of
2
•
;
1°/s
=
2
Tt
•
derivative of the angu-
t:
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
instantaneous velocity
i
"
11
It
-i
ft
slope
the
is
v.
%
-
•
s(?)
the
6t
tangent
to
the
s-t curve.
Velocity-time curve
The
velocity-time history
a v-t
curve
The
curve.
is
first
is
expressed as
u(u)
derivative of this
the instantaneous acceleration a.
is the second de-
Hence the acceleration
rivative of the distance-time curve.
12
a
8
m
Av
Al
a
=
%>
=
V
dt
the slope
v-t curve.
It
is
df
the
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
=
0:
(decreasing velocity)
constant velocity
(increasing velocity)
EF
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
|
I
15
v =
v Q = const, a =
h
u
motion
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
Harmonic
a = a
an =
t
co
2
r
=
oscillation
pendulum—;-}
\
/•2 i«'
Paths are straight lines or circles. The
body moves back and forth about a position
of rest. The maximum deflection from this
position
is
called "amplitude".
Instantaneous position, velocity, and acceleration are harmonic functions of time.
KINEMATICS
Linear motion
Uniform and uniform accelerated
linear motion
uniform
accelerated
retarded
constant
uniform
a =
a =
u= const.
EU
un >
°
=
H,
> 0)
< 0)
(a
(a
n
t^r
m
I
+
vt
19
2a
2
I
I
I
25
V2^7=^ =
20
21
Vq.
+ V) = V Q
u + a
2
-
2^
23
a
i>
a
_ 2s
v
j
cm
km
+ 2 as
-2
at =yjv 2
-V0 2
V2
25
t
2s
t
+
= \Au 2
t
V - Vg _
22
t
m/s
cm/s
km/h
v
const.
r
I
ar
(v
v
-Vq
2s
vQ + v
a
m/s'
cm/h 2
km/h 2
mm
h
Note
The shaded areas represent the distance
time period
The tangent
s
covered during the
t.
ft
represents the linear acceleration
a.
KINEMATICS
6
Rotational motion
Uniform and uniform accelerated rotation
about a fixed axis
uniform
II
a=0
co
a =
= const.
COq
=
COq
i
>
(a
accelerated
retarded
constant
uniform
EU
>
w
^^
^T
0)
(a<0)
t
3
rnssfZt
t*J
I
f
t
m af_ m
2
co t
24
2
I
m
25
(2 a.tp
=
2(P
f
2
co
2a
at
-f
-
-^(co Q +co) =
co Q t
+
-at 2
rad
COq
+ at =
2
y <y
+2
a <p
1/s
m/ms
rad/s
- at =
const.
26
27
ft)
_
f
"
2^
2
t
_
Q)
2
2
\J(i)
-2aq>
1/S
CO
2cp
-COq _ CO* -0)(
~
2<p
t
Z
mm/s 2
rad/s
s
28
f
=
2jp
CO
CO
-COq
a
!£.
O)
+
OJ
min
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, v, and a as functions
of time are equal to the projections s, v, and a n of a uniform
tion.
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
v=
= Acq -cos(aU +
r- a)
df
31
a=
0;
cp
)
acceleration-time curve
acceleration
I
)
velocity-time curve
velocity
30
=
2
^7 - -Aco
an =
at
Differential equation of simple
•
s\n(cot
<Po)
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
projection
fall,
Free
free
(without air resistance)
and
fall
vertical projection
fall
u„ =
vertical]
upwards
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
=(v -v)/g
t
v
Angled projection (upwards
34
I
135/1
I
36/2
I
I
I
I
37
38
1-
-
cos
=
tan
s
v = \[v
= \/v
H
W=
cp
-g
s
•
-2gh'
2
+ g2
2
-
-sin
v
2
-s\n2cp/g
^I
a
1
v
(2
2
•
cos 2
s\n
-t
?
= v
^ax
°
|
2
/g;
H,
= W,- tan cp- g-W, 2 / (2 v
W,
= -v
cos
Vq 2
cp (v
•
sin
cp
Hmn
2
v
-v
-cos 2
y/?/[2g);
2
^-v.yfi/g
t
± Vsin 2 cp-v
2
- 2gH\
Wi
+ £v
Horizontal projection
{cp
=
0)
142/1
I
42/2
-g
42/3
v = y/v
tH :
:
8
+ g2
-t
2
H
time for height
for distance
w time
f
v
t2
2
0)
cp)
(2H,
J\--*2
V
g-W,
!
<
cp'
t
141/2
tang? =
cp
w = 2v -smcp/g
141/1
141/3
downwards
g?/(2g)
eS
45
2
-2 g -v
2
v -smcp/g;
40
0;
+ v)
Vjg
t hmax
-gt 2 /2
= u
39
>
= 2h/(v
/2g;
h±
2
2
cp
cp
h = v -t-s\r\cp
35/2
136/1
I
= v
s
2
W
v
v
velocity
:
initial
:
trajectory velocity
KINEMATICS
9
Motion on an inclined plane
on an inclined plane
Sliding motion
excluding
un-
including
|
known
friction
param.
43
I
a
=
li
=
g
a
sin
(sin
H >
a - u
cos a)
—
(a- g)
cosg
sin
44
I
at
45
I
a
y
r
*
\j2as
t
at'
46
I
Ml
yr
2
2a
Qo
Rolling motion on an inclined plane
excluding
un-
known
/>0
/=0
I
I
gr 2
+ k2
see above
49
see above
50
0.
.
.
a
51
\ar\a =
^a
^
Li
52
*
.
I
I
+
fc
2
45
46
tana
r
r
r
r2
max
:
p
+ k2
tan a.
-f
k2
pipe with low
solid cylinder
ball
I
r2
48
2
I
cos a
a - r
sin
flf-r-
r2
I
including
|
friction
param.
wall thickness
*-£
k2 =
r 1 2 + r2 2
2
^
r
2
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
I
I
53
54
angle,
where center
of gravity S vertically
Z
(see Z
coefficient of sliding friction (see
7)
coefficient of static friction
7)
angle of sliding friction (/x = tan g)
angle of static friction (^ = tan g
)
above
tilting
[edge
KINEMATICS
10
Mechanism
Simple Conn-Rod mechanism
I
I
I
I
56
2
= r(1 - coscp) +
jf rs\n
= co r sin cp (1 + A cosg?)
57
=
58
= L
55
a)
2
(cos
r
cp
4
/
I
crank
(A is called the
2<p)
6
"
2nnt
= cot =
59
+ A cos
11
cp
ratio)
Scotch-Yoke mechanism
I
I
I
I
s
= r sin
61
co
= corcos(cot)
62
a
= -
co
=
60
63
co
(cot)
2
—
v
-
r s\n (cot)
2xn
(motion
is
simple harmonic)
Cardan
joint
For uniform drive the oft-drive will be
uniform due to auxiliary shaft
non-uniform
uniform drive
/-
T,
H
M
fi
-3—
**k
For
being
located
cos 6
tan
shafts
all
in
one
plane
the
following
relations apply:
I
I
64
tan
65
co 2
cp 2
=
CO)
= tan
—
1
I
66
cp,
= tan
cp3
cos 8
-sin
2
B<X2
C0 3
slrrcpi
=
2
sin 2 /?-
<Ui
(1
=
cp^
tan
CO)
cos 6-
sin
cp 3
= tan
co 3
=
<p-|
co-.
2^
-sin^-sin 2 ^) 2
Both axes A of the auxiliary shaft
be parallel to each other.
joints
must
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
the terms "work, energy, and power".
The most important quantities
and their units
m
Mass
(is
of
in
motion and with
dynamics
a base quantity, see preface)
Mg =
t; g
(see also A 2, A 3, A 4, A 5)
mass of the international standard. A mass
measured by means of a steelyard.
Units: kg;
kg
1
the
is
Force (gravitational force) F
The force F is the product of mass m and acceleration
F = ma
The
G
gravitational force
the force acting on a
is
is
a.
m
mass
due
to the earth's acceleration g:
m
G
2
mg
=
Being a gravitational force the weight
W
is
measured by means
of a spring balance.
Units: N;(kgf;
N
1
m
is
=
1
m
see also A 2)
Ibf;
when
the force that,
kg for
s"
1
1
s,
acting on a body of a
accelerates this body to a
accelerates this body at
of
1
is
the gravitational force acting on a
(i.
e.
1
m
s~ 2 ).
mass
final
9.81
of t
N
mass
velocity
(=
1
kgf)
kg due to
the earth's attraction.
W
Work
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
Where work (energy) increases or decreases linearily with
power is the quotient of work and time (P = Wit).
Un
i
t
s
:
W (Watt);
Where
(kgf
m
s"
1
;
a constant rate,
H. P.
;
see also A
3.
an energy of
the corresponding power
for a period of
s
1
1
W
=
1
J/s
A
1
is
time.
time,
5)
J
1
is
W.
converted
at
DYNAMICS
Mass, Mass moment of
Definition of the
mass moment
The mass moment
of inertia
of inertia of a
M
inertia
J
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.
Ir 2 Am
J =
=
Jr
2
dm
kg
m2 N m
,
s
2
Theorem (Parallel axis theorem) (see
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
Steiner's
also J 9)
,
about
inertia
distance
/s
a
axis
parallel
O-O
at
mL
Radius
a
will be:
kg m'
of gyration k
The radius
of gyration k of a
body
mass
of
m
and mass moment
the radius of an imaginary cylinder of infinitely
small wall thickness having the same mass and the same mass
of inertia J
moment
is
of inertia as the
hence
body
in
k
Vf
k2
mgk*
question.
m, cm,
mm
Flywheel effect
G
flywheel effect
= g J
(k
Equivalent
mass
(for rolling
2
kg cm s *, N m^
formulae see M 3)
bodies)
kg
Basic formulae
linear moti on
formulae
m
9
m
10
m
11
Fa'
m
12
m
13
W
WK
WP
w
m
14
p
f
--
ma
=
N
Fs(F= const.) N
=
rotational
,(kgf)
m,(kgf m)
A* a = J
a
W = M (A/ = const.)
WK = \ja> 2
(p
=
\mv
=
Gh
J
.(kgfm)
(0
=
jFAl
J
.(kgfm)
W F = ±MA0
2
'3F-"
J
W
.(kgfm)
,kW
mot ion
formulae
units
P
=
Inn
m
N
m
J
s"
t
,
(kgf
m)
,
(kgf
m)
,
(kgf
m)
,
mirr
,
(kgf
1
Ws
W
=
a
units
N
,
kW
1
m)
DYNAMICS
Mass, Mass moment of
Mass moment
axis
of inertia
M
inertia
about
axis b-b
a-a
passing through
centre of gravity 5
(turning axi
type of body
circular
m
=
J
15
\mr 2
hoop
b\
2
m
.2
16
_ 1-2
2
cylinder
m
17
m
18
k
m
19
J
m
20
:
m
21
J
m
22
k
m
23
m
24
m
25
J
m
26
k
m
27
2
=
\m{R 2 + r 2
-
^(R 2
2
=
r
k
)
w mr
^r
10
2
2
2
V
2
=
R 2 + ^-r 2
short
bar
bar
+
y«*V
12
y =
|r 2
+
2
— {3R
-
^™(Ar 2
S
=
=
J
)
k
4/-
+ h
=
2
2
+ 3r
(4ff
=\mr
k
m(R 2
^(3r 2
)
;
m
4/?
2
+ 5r<
)
b
fcW + c
torus
b
= -l(4/? 2 + 5r 2 )
-
2
\r 2
3
2
)
sphere
28
k
2
2
^/ 2
d,c<zl
+ h
+ *' )
-§«* + «*)
]
2
+ h
2
*.--;!•
2
hollow cylinder
2
=
=
)
J
thin
m
+
2
2
)
UJ'-J
short bar, thin bar
DYNAMICS
M4
Rotation
Total kinetic energy of a
m
WK
29
=
\mv*
+
body
\j s co
Kinetic energy of a rolling
m
m
30
31
W«
=
j (m + m
vs
=
(o r
red )
vs
2
body
--
J,
[kgfm]
J,
[kgfm]
no sliding
2
m/s, km/h
Rotational torque
m
32
M m, [kgf m]
2 ;rn
co
Transmission ratios
Transmission
33
Torque
m
ratio
driving
^2
z2
"1
Q)-\
d\
Z-i
rt
Q> 2
Q
ratio
M
moment of force
moment of load
34
2
F
Ml
—
/
h
it)
6%
J
Efficiency
m
v
work produced
work applied
35
output
<>yy
driven
input
Overall efficiency for a series of transmissions
36
V
™red
=
see
V-\
•
V2
•
V3
•
•
•
m
8
velocity of linear motion of centre of gravity
accelerating force
accelerating moment
kinetic
energy
J,
potential energy
J,
energy of helical spring under tension
extension of helical spring
J,
angular defection of spiral spring
m]
m]
[kgfm]
[kgf m]
[kgf m]
N, [kgf
N m,
(in
radians)
[kgf
—
DYNAMICS
Ms
Centrifugal force
Centrifugal force
Fz
2
m
37
m
38
m
39
Fz = ma) 2
v
=
r
=
4n 2 mn 2 r
=
2jzrn
^f-
Ni
[
kg f]
N, [kgf]
tfr^L
\^
-J
m/s, km/h
m
40
=
(0
2n n
1/s,
Stresses
in rotation
v
r
i
\
1/min
bodies
J
(appr formulae)
Disc
(JO
°z
2 r2
'
V2 Q
Q
3
F*~
3
N/m
2
,
[kgf/cm
2
]
v
„
Ring
m
a2 =
42
—^ (r^ + f!^ + r
2
N/m
/s
:
e
:
Fz
:
Jq
:
J$
'
M^
:
^e
:
'
:
i>f-
Wke
F
z_
)
[kgf/cm
2
]
instantaneous pendulum swing
centrifugal force
N, [kgf, gf]
mass moment of
mass moment of
moment required
1
inertia
inertia
kg
kg
rad = 57.3°
E
velocity at
F
kinetic
energy
m2
m2
,
[kgf
,
[kgf
m
m
s
s
2
]
2
N m,
[kgf
]
cm]
N/m 2 kgf/cm 2 [kgf/mm 2
,
period of oscillation
{B to 5' and back)
:
about
about S
to deflect spiral spring
tensile stress
velocity at
mm
mm
mm
m, cm,
m, cm,
m, cm,
maximum pendulum swing
by
oz
T
,
2
distance from centre of gravity
:
/
2
fTV<
^ l^
Wj
,
s,
]
min
m/s, cm/s, km/h
at
£
m/s, cm/s, km/h
M m, [kgf m]
I
DYNAMICS
Harmonic
M
oscillations
6
Mechanical oscillation
General
T=
m
43
period
m
44
stiffness
2jz\jf
N/m, [kgf/m
Al
m
45
frequency
/ =
m
46
angular velocity
a>
m
47
Critical
speed
=
—
\ min
1) s
V—
=
2nf
s"\ min
n c of shaft
2
m
(see L
48
n
= 300
m
V
10cg mr
V
9.81
iSfl
N
2-bearing shaft, load
symmetrical
asymmetrical
^££S
overhung
(cantilever) shaft
,X- i m
^cz>X.
c/
/
m
3EI
48EI
49
2
a -b
A
/
m
I
:
:
:
c Cq
~
2
3EI
3
/
deflection or elongation
second moment
mass.
When
of area of shaft cross section
calculating
the critical
assumed
speed the mass
m
be concentrated at a
single point. The mass of the shaft should be allowed for by
(e. g.
of a belt disc)
is
a slight increase
c„
:
stiffness for transverse oscillations
to
DYNAMICS
Harmonic
M
oscillations
Pendulum
(Explanations see L
4)
Conical pendulum
m
50
m
51
-
T-2„\Jf
tan
2^'^ia
a
9
m52
-if
Simple pendulum
The arm of a pendulum has no mass, the
total mass is represented as a point.
m
53
m
54
Ve
-e'tff
m
55
WKE
=
= 2
mg
mm
s,
'>IF
vF =
m/s
2
2
Vf(e -/
N m,
J,
)
km/h
[kgf
cm]
s,
min
2/
Compound pe ndulu m
m
m
m
56
N
57
58
m
s
Nms
7s=G/s(-^-^)
91
4 Jt
2
,
2
[kgf
cm
s
,[kgfcms
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
lated using
m
about
O
can be calcu-
58.
J
Torsional pendulum
m
59
s,
For explanation of symbols see
M
5
min
ft-?
DYNAMICS
M
Impact
8
Impact
m
60
m
61
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
p =
m-|
•
v-|i
+
m2
•
L'22):
m-|
t>2i
•
m2
u 12 +
•
v22
Impact-direction
direct
and
velocities parallel
normal to surfaces
at point of impact
concentric
impact
oblique and
concentric
impact
any random
of
any random normal
velocities
oblique and
excentric
impact
Types
normal to surfaces
at point of impact
through centre of
gravity of both
bodies
to
to surfaces at
point of impact
impact
impact^
elastic
velocity
after
m
62
m
63
concentric
impact
m
64
coeff. of restitution
equals zero
after impact
impact
Vn(mi -m2) + 2m 2
velocity after
direct and
V21
_w
rri\
=
1
v 11
02
U2i(m2 -mi) + 2mi
£
w
»
m
(171)
and
+m 2
+ m2
y2 i
£
1
=
will
vary
after (172) impact:
65
E
=
tV2
here
«r1
the tangential
component
has no effect on the impact.
0^ e^
'
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),
1
+ WI2
Coefficient of restitution e
This indicates by what factor the relative velocities
before
impact
plastic
equal, before and
relative
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
The dynamic
(Values see Z 5)
>/
viscosity
tion of pressure
a material constant, which
is
a func-
is
and temperature:
=
n
f(p,t)
The dependence on pressure can often be neglected. Hence
n 2
=
7]
(for figures
f(t)
Kinematic viscosity v
The kinematic
and density g:
EU:
viscosity
is
m 2/s
=
4
(
1
St)
=
see Z 14)
(
1
6 cSt)
the quotient of dynamic viscosity
rj
n 3
Hydrostatics
Pressure distribution
in
a fluid
n 4
P\
=
Po +
gQh
n 5
Pi
=
Pi +
gQ{h 2 -h^)
=
p^ +
A
gQAh
ggjh
— PiPz— /
'Pz-Py
t
-p<r\
—
-c"
_
-c
-
N
P
pressure
^
zL
ta
•c
gradient
1
h
'
continued on N 2
~
HYDRAULICS
N
Hydrostatics
Hydrostatic forces on plane surfaces
Hydrostatic
force is the
acting on the
component
surface which is caused by
the weight of the fluid
alone, i.e. without taking
into account the atmospheric pressure p
.
n 6
F
= S G
•D
>'s
A cos a
= S Q hi
r
„
Q-C
2~ J
_ _
v
~a^jr?
_—
:
A
\h
fc*
~~/
/
yQ^
r>
y**
Hydrostatic forces on curved surfaces
The hydrostatic force
act-
ing 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
N,
kN
Fh is 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 of area A in relation to an axis running
17
parallel to axis 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
and
FA =
gQV
+
gQ'V
the fluid of density p'
following formula is valid:
If
n
FA
10
JTp
—
g
o'.
kN-=~-—
N,
is
——
a gas, the
1? >~p
~ggV
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
and
a
m
first
1-
Q=
body
known density g b
This yields:
fluid of
used
n 14
n 15
determine
of a deliberate
g=
Qf
F-i
in
.
mg
g
n 16
mg
77#»7,
..V
Q
Qb
m
F
:
mass
:
equilibrium force necessary
of the
resp.
"^^
body remaining suspended
^h: equilibrium force necessary
auxiliary body alone
Qf- density of the fluid used
in
M
,1
^_—
in
the fluid
the preliminary
trial
for the
:
HYDRAULICS
N
Hydrodynamics
Hydrodynamics
(of a
steady flow)
Continuity equation
Rule of conservation of mass:
Ai
n 17
n
18
V-\
Q<\
=
m
=
A
A2
v g =
v2
£>2
kg
Vg
g
Rule of conservation of volume:
n
v
1!
=
m° cnr
Hr,^
S
S
Av
o
'
14
<j
Bernoulli's equation (Rule of conservation of energy)
No
friction (ideal fluid):
£ + gzi
n 20
+ -£- =
)
£
+
gz
+
-7-
^
=
+ gz 2 + -2-
pressure energy
per unit mass
potential energy
per unit mass
kinetic
energy
per unit mass
datum
Including losses (real
?+™
n 21
v
:
line
fluid):
vy
"g" "
Pi +
p"
V
3*2 +
'
2
-y
+ WR1.2
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_
2
2
+ 9(^2-^1) + ^-(^2 -^i +
7r(P2-pO
2
9 *" '" «"-«-"
n 23
)
n 24
for hydraulic
n 25
for
Momentum
machines:
pumps:
u> t12
<
u? t12
>
wri,2
J/kg
equation
flowing through a stationary reference volume the
following vector equation is valid:
For a
fluid
IF
the vector sum of the forces acting on the fluid contained in the reference volume. These can be:
IF
n 26
=
m(yl -
u|)
N,
kN
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
rence volume.
t>i
fluid
entering the refe-
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
city out of
r2
and
r-,
m(v 2
are the circumferential
and entrance velocity
N
v^,u^r^)
components
of exit velo-
into the reference
are the radii associated with v 2 and v v
m
volume.
HYDRAULICS
N
Hydrodynamics
6
Friction losses in pipe flow
n 28
Friction loss per
unit
n 29
hence
mass
Ap y
Pressure loss
=
p-tu R1i2
Determination of coefficient of resistance £ and coefficient of
shape a (Re: Reynolds' number):
non
circular pipes
n 32
*±*
=
Re
n 30
n 31
=
Re
Where Re < 2000, the flow is laminar.
Where Re > 3000, the flow is turbulent.
Where Re = 2000 ... 3000, the flow can be
^
circular pipes
either laminar or turbulent.
Flow
laminar
64
n 33
a =
n 35
1
—
for fittings,
unions and valves
1
I
for straight
pipes
q
10
3
1.47
1.50
<P
I
1.40
1.42
30
50
70
100
1.32
1.29
1.27
1.25
0.5
0.6
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
097
94
1.0
0.7
0.8
0.92 0.90 0.89
d
internal diameter of pipes
length
/
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
a =
pipes
For annular cross sections:
Dld\
n 37
f(Re, j~)
Re
for straight
Determination of coefficient
n 36
turbulent*'
64
f(Re,*)
£ -
Re
n 34
Flow
laminar
turbulent*'
:
:
:
:
:
:
:
of pipe
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
=
Cd
n 44
F
=
g
Large
v
n 45
&
A\2gH
Vv
lateral apertures
=
fc d 6v^(^-//^)
Excess pressure on surface
n 46
Cy
\J2{gH+^L)
V
=
C6
A\J2{gH+^)
v
n 48
=
C
5^^
-<nnr
=
to
1
of liquid
v
Excess pressure applied
T
~-~r,
an outlet point
! Pex
T'
?%
^-'-T'
n 49
u
/? ex
Cd
Cc
Cv
b
F
V
:
:
:
:
:
:
:
:
m/s, km/h
outlet velocity
pressure
in
excess
of
atmosphere pressure
discharge coeff.
contraction coeff.
velocity coeff. (for water
width of aperture
reaction force
volume of outlet flow
N/m 2 kgf/cm 2
,
(Cd = Cc x C v )
(Cc = 0.62 for sharp edge aperture)
(Cc = 0.97 for well rounded aperture)
Cv
= 0.97)
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.
Pressure p
Pressure
o
N/m 2 = Pa; bar
N/m2 =10- 5 bar =(7.5
force F and area A:
EU:
1
is
Pa=
1
the quotient of
x 10-3
torr)
>-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
Temperature
The
T,
t
Pu +
(Base quantity) see explanations
unit of temperature
T
is
at front of
book.
the Kelvin K, defined by equation
-^
1K
o 3
where T Tr
Ap
273.15
the temperature of pure water at the triple point.
In addition to the Kelvin scale the centigrade scale is also
used. The temperature of this scale has been internationally
*
defined by:
O 4
=
t
is
(-? - 273.15
K
Density g
Density
is
Vc
o 5
°
Volume per
Specific
unit
mass
volume
is
(specific
'
(£+
VC
273.15) K
EU: kg/m 3
m
-
and volume V:
v
volume) v
EU:
the quotient of the volume
„
o 6
T =
;
)
(Values see Z 6)
the quotient of mass
,
Molecular volume V m
Molecular volume is the quotient
contained in the volume:
m 3/kg
V and
the
mass m:
z,±Q
m
EU:
of
m 3/mol
volume Vand number of moles
o 7
Amount
book.
of
substance n (Base
quantity) see explanations at front of
HEAT
Heating of solid and
bodies
liquid
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
The heat per unit mass
EU: J/kg
is
the quotient of heat
Q
and mass m:
m
^
Specific heat r p
EU: J/(kg K)
denotes the amount of heat Q to be supplied
to or extracted from a substance of mass m to change its temper-
The
specif, heat r p
ature by a difference At:
cp
The
see Z
specific
1
.
.
.
Z
heat
is
=
-VmAt
a function
=
±
At
of
temperature.
For figures
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
fusion
body
of the fusing
temperature into a
fluid of the boiling
vapourisation
the
heat
is
fluid
temper-
ature (dependent on pressure) into dry saturated
vapour
necessary
to
convert a
sublimao 12
tion
of the
same
body of a temperature temperature
below its triple temper-
solid
ature at the sublimation
temperature (dependent on
pressure) directly into
dry saturated vapour
.
HEAT
Heating of solid and
bodies
liquid
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
Length:
/
o 15
»
AA
Volume
/
*i)
[l+2a(^-^)]
A 2 -A, = A^2a{t2 -ty)
=
AV
-M]
-/i «/i a(t 2 -
2
A,
Vo~
o 18
[l+a(f2
I,
A2 ~
Area:
o 16
o 17
2
Al =
o 14
=
V
1+3afe-t|)]
y2 -
V,
Vi
m
«
'
V,3a(t
3a (fj,-^)
K,
2
><r^^ ^
^''
4
1
^^
Expansion of liquid bodies
With y being the coefficient of volume expansion
see Z 1 1) the following formulae apply:
V2
o 19
AV
-
^-^
(for figures
[1 +Y(h-t,)]
V2 -Vi - JVy(*2
V,
-fi)
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):
=•
,
,
,
.
,
.
,
o 20
'
_
m =
cp
^ + m2
m, c p1 + m 2
Z 5).
mi
•
•
i
•
(For c p values see Z 1
Bending due to heat A
.
.
•
c p2
•
•
t
m 3 c p3
m 3 Cp3 +
2 +
c p2 +
•
•
•
f3 +
.
.
.
.
.
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
volume
at
t
at
t
thickness
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:
RT
p v =
22
the gas constant
reads
is
mRT
p V =
or
If
p = g
or
RT
mole volume, the equation
related to the
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
where
M
Thermal state
The thermal
=
mass
the molecular
is
of real (non ideal)
state of real gases
MR
(see Z 12).
gases and vapours
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
o 25
o 26
+
91.2
«>1.2
" U2
2nd law for
all systems
law for
open systems
closed systems
#1,2
+
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.
h
:
u
:
w
:
enthalpy per unit mass
internal energy per unit
2
to,'
Ae
:
entropy per unit mass
done per unit mass (see =
7)
continous external work done per unit mass (see
changes in kinetic or potential energies
'
:
':
\s
mass
external work
7)
]
HEAT
Changes
of state of
gases
Changes of state of ideal gases
The table on page 6 shows the
of state,
relations for different changes
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
'1
pm
is
o 32
Changes
The
7r
of entropy occurring
,
ln
during the change of state
,.
,
T2\
.
^{P2\
KM?
n
[f-
_ „
.
T2 \
.
In
state,
T as well as the propv,
taken from appropriate diagrams.
continuous
external work
const.
quantity
tf
-i
2
w
dv
= - \p
work
external
t
i,2
=
isobaric,
p{v, - v 2
isothermal, (u 2
5
= const.
[h 2
"
*i)
=
-
V{p 2 "Pi)
p = const.
isentropic,
mass
?1.2
«2 " "1
v(p z ~P^
v = const
const.
heat per
unit
\v dp
isochoric,
o 36
In
variables of state, p,
change
T =
,„fV2
gases and vapours
below shows the relations for different changes of
which have been developed from formulae o 25 to o 27.
of state
o 35
RD
of state of real
erties, u, h, s are generally
o 34
,
table
The thermal
o 33
,
+
T\
(h 2
T(s 2 ^s,) =
-h,)-T(s 2 -s,)-
-(p 2 v 2 -p,
«2""1
{h 2
h 2 - h^
)
-uj-
-h^-{p 2
v2
(h 2
-h,)-T(s 2
Ui)
-p^ v
h2
: )
-A
1
-s,)
T{s 2 -Si)
-
HEAT
Ideal
T-s-
gases
in
^
diagram
CO
>
c
o~
1
° °
\.
;i:
E
1
o
°
~o
Q.
I?
—
co
,r| /I
CO
Ci,
to
to
CD
~
^=
than
D)
1
06
open and closed systems
T3
Q.
-«
isothermal
steeper
CD
1
c-T
per
mass
trans-
3
sS
fer
E-?
o
<?1,2
7
unit
heat
E
E
J*
E
^ Yl '—
E ^JX i^3
f <£[£"
dp
v
/
system,
O
=
reversible
2
3
s
as
-
^r
'
*
*
open
t1
II
w
f~>
«a
-SJ3
'cm
7
^
•Ii
Hi
1
as
|T-
l>~
ii
ii
Tic
f~r
f
2
system,
C- *'
1*
«. as
closed
'^m7~^'
|
1
o
reversible
c
*l
el
1
cj
&;
ft;
as
i
ii
3
and
1
2
II
II
£|^
S"|s E^|c4"
between
f^l^
ii
ii
ii
relation
state
SI
state
31
ci
S
£1
£
ii
39)
a s|s
o
(o
O
<Sl|o~
37)
38)
(0
(o
const.
H
ii
II
<S1|
S|
1
oo
details,
isothermal
isochoric
process
isobaric
=
=
f n
=
=
=
=
p
n
T
n
S
CO
"
=
ii
^ ~
sl^"
41)
exponent
process
^
(0
const.
const.
const.
ii
ii
Sl| £X
n
HEAT
Changes
gases and vapour
of state of
p-v diagram
'////.
For reversible processes the area be-
^ J^
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.
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
The heat flow continuously added
system
to or
#1,2
m
is
the
=
mass flow (EU:
Total transfer of work
The external work added to or
single variation of state
Wi
o 44
01,2
=
o 45
is
mq^ 2
w
kg/s).
done by a closed system during a
is:
mu) 1(2
The external power continuously added
system
removed from an open
is:
o 43
where
J
to or
done by an open
given by:
?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
+
.
.
+ ra n =
.
i
Mass
fractions
§j
6
of
-
of a mixture
47
Number
- n
21 m,
moles n
=
and
m
of a mixture of
j?"*i -
components
n-\,
1
n2
48
/i
=
n<t
+ n2
+
.
.
.
ip\
-
of a mixture
'" n
n
Vi =
o 49
•
•
+ n n = ^rtj
i
Mole fractions
•
.
= n
i
X^i
and
7^
=
1
Equivalent molecular mass A/ of a mixture
For the molecular mass the following formulae apply:
"i-7?
50
M
and
where the equivalent molecular mass
;V/
=
g
of the mixture
can be
calculated as follows:
M
o 51
=
i^Mi-Vi)
i=1
and
-j-
M
=
'i?(^)
i-
\A/|/
1
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
Volume
O
8
fractions n of a mixture
o 54
JE
and
Vi
r,
V\ we mean the volume the component
the temperature T and the total pressure p of
the mixture. For ideal gases the following formulae apply:
Here, by partial
would occupy
volume
at
m.R.T
n-,R m
P
P
o 55
Internal
energy u and enthalpy h
T
^
and
u
=
n
i: (£rui)
1
=
V
=
of a mixture
n
o 56
Vi
=
h
;
1.
i:
=
1
(£r/zi)
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 +
v
o 58
enthalpy
where the
specific
CP m 71
i
m
1
+ c
cD
heats
of
the
»mn
-/
n
•
+
.
t
Wn'«n
m
mixture are determined
o 59
^.(*r«bJ
™n
m
+C Pm2^2m 2
follows:
o 60
.
energy
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:
Am
=
ji
dm L
A
*w1
=
<P =
Q
= X
^w2
S
~ *w2
plane wall
pipe
is:
d a -d,
where
;
^w1
Q
The mean logarithmic area
o 63
XA
<P =
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
vection
the flow originates by itself, the conis called natural convection. The
convection taking place
forced convection.
o 64
in
a flow
=
Q
is
called
= <xA(t-t w )
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:
k
A
«
:
:
A
«1 ^1
thermal conductivity
(for
heat transfer coefficient
values see Z
(for
1
.
.
.
calculation see
Z
O
5)
12)
HEAT
11
Heat transmission
Heat exchanger
A heat exchanger transmits heat from one
heat flow
may be
fluid to another.
The
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
O
used on page
enveloped surface
C-,
12:
(A,
enveloping surface J
inside diameter of pipe
outer diameter of pipe
,C 2
radiation constants
:
< A2
Grashof number
Nusselt' number
Gr
)
Nu
H
height of plate
length of plate
L
of
the
surfaces
8
2
ra-
4
)
o 70
exchanging
values see Z 1 2)
radiation constant of the black body
C s = 5.67*1(T W/(m K
values
see Z 14)
Pr
Prandtl-Number Pr =(r/c p )/A (for
At = \t
absolute temperature difference between wal
and fluid in the thermally not affected region
ambient temperature (7^ = t^ + 273.15°)
V
kinematic viscosity (v = rj /g)
(for values see Z 14)
dynamic viscosity
V
(for
diation
:
:
'
:
|
dynamic viscosity
dynamic viscosity
at
mean temperature
at wall
A
thermal conductivity
y
volume expansion coefficient
temperature factor
of fluid
temperature
(for
of fluid
|
values see Z 5, Z 6)
and o 77)
(see Z 11
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
= 0.13
$GrPr
gyAtm
plate
Gr
o 73
o 74
o 76
o 77
o 78
for
,
for
1700
g Y At q?
< GrPr <
10
GrPr>
10
A/u
horizon-
= 0.41
YCr-Pr'
GrPr<
10
for
,
8
8
//3
on a
WuA
tal
o 75
_
,
D
5
3
plate
Fluid properties
must be related
The expansion
coefficient of
to reference temp.:
gases
=
tB
= MTc.
y
is:
fc
SL+
x
A/V
-
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
4
RePr^>
U
>
10
,
Re =
where
^4^
7/
turbulent
Re>2Z2Q
6
/?<? < 10
< L/d< oo
2320 <
if
1
0.6
;
<
Pr
<
500;
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:
o 82
«Str
o 84
o 85
c
parallel
o 83
4
sur-
between
enveloping
orstr)
0*^1,2
=
7i
faces
i,
q
-72 4
h-h
Ci,
J_ +
Q
')a in J/(m 2 s K) or
W/(m 2
K)
For explanation of symbols see
O
11
Al
>4 2
/J_ _ J_\
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.
o
p1
In
or
N/mm 2
f
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
(e.g. mild steel)
(e.g.
plastic yield
aluminium alloy)
a
*eH
"
1/
TW^
-
RP
-
Jf
i
r
/
:
=
tt-;
\o B =
jo
L
F
—A J
:
'
'
'
'
'
eP
€
'
'
/
A
'
*
'
'
'
/
/
Notation The standard symbols
Rm
'
/
/
€p
P3
#m
are from
A
BS
tensile stress,
e
18 and DIN 50145.
where
tensile force
:
s o> [A
]
:
original cross section (of
Al
p4
1
Lo
AL;
;
[lo]
[Al]
:
:
00
%
unloaded specimen)
strain,
where
unloaded specimen)
length of loaded specimen
original length (of
change
in
continued on P 2
STRENGTH
General terms
continued from P
ReH> [^Sol
(stress-strain
1
diagram)
Proof stress or yield strength (offset)
:
The limit of proportionality is sometimes
known as the elastic limit.
[rj
ep = 0.01 % => /? p0-01
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™
P5
1^
"
4L
p6
I
°b "
R pQ2
=>
^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
Rm
v
:
:
is:
'
p
=
x
Km
—
yield strength of material
safety factor, always greater than
Ultimate safety factor
against fracture)
1.
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
(see Z 16/17 for values of £). £ is known as "Young's
law)
is
limit
modulus ".
P 7
o
E
=
•
E
e =
Al
E
Olr
~ e ~
Tensile and compressive stresses
^l
a and o7Cc
iF,
t
f/
Fx
P 8
^U
^n
Oc
Pi
Strain e under tension
P 9
I- h
Al
=
e
°1 =
E
Compressive
strain f c
Ft
"i±
E A
Fc
p10
£
/
$
under compression
E
E A
•
A
fi
= tensile or
com-
pressive stiffness
Transverse contraction under tension (Poisson
s ratio)
For circular cross section
—"«
where e along
P 11
along
"
/-
/o
For most metals Poisson's ratio can be
P12
%
expansion (see also o
assumed
to
stress
caused by
is
be
\i
ath
=
E
is
the temperature difference between the unstressed
E-a-At
original state
= 0.3.
re-
13/14):
At
-
d
UQ
dn
Thermal stresses: Tensile or compressive
stricting thermal
dp-
_
and e
/<,
(£, h
= a At)
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
Tensile and compressive stresses
in thin-wall
ee
;
=
,
FI(E-A)
cylinders
(boiler formula):
Hoop
p14
p15
o
o
o
stress
Tensile stress
= p d I (2 s)
= + p, 4 / (2 s)
| vg|jd
= -p a d a /(2s) )
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
:
t
M
5.
da
d
x
^
1
2
STRENGTH
Tension, Compression
Tensile stress
in
a shrunk-on ring (approximate formulae)
Shrunk-on
Fu
ring on a rotating shaft:
The shrinkage force Fh of the ring
must be at least twice the centripj_
petal force Fq.
p 16
Fh
^
2
=
4n 2 m-y s -n'
Fc
P 17
Fc
p 18
>'s
p 19
Cross section
p 20
Shrinkage allowance A =
m-y s
=
-
a)
2
•
R3 R2 -
4
3ji
r3
A
=
_Fh_
2-Pt
(A
= outside diameter
Shrunk-on ring
Split, rotating
Fc
1
of shaft
n
- inside diameter
clamping
for
clamped
of ring)
Clamped
parts
parts.
comprises:
Centripetal force
Centripetal force
FC r
FCM
for ring
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
U = w- V where
p 22
is:
;
-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
:
centre of
D m mean
:
mass
of half ring (see
diameter (D m = R +
r)
K
7)
STRENGTH
Loads
in
beams
Explanation
external loads on a beam (including support reactions and
weight) produce internal forces and moments which stress
All
its
own
the material. By taking a section through the beam at a point x
it
is possible to show the internal
loads: Vertical shear for-
ces
V and bending moments M.
End loads P and torsion Tare considered
separately.
[I
y
B
r ^f=i«^ttt:
(f
Referring to the x-y plane
Forces
(z
axis
is
at right angles):
end loads
P
r-axis
shear forces
bending moments
M
x-axis
torsion
T
.v-axis
in
direction of
v-axis
Moments
about the
produce
Always consider the left-hand side
P
V
of the section.
each part of the beam there must be equilibrium between all external and internal forces and moments:
Considered separaterly:
In
V +
p 25/26
M
p 27/28
Method
1.
2.
^L
V;
hi.
=0
P +
T +
of calculation
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.
4.
Find the forces and moments on the left hand side of the
section as in p 25 ... p 28
Plot shear force and bending moment diagrams.
Relations between h
m
p 29/30
,
_
V and
M at any
dV
v
dx
point x
dx
Rules:
M
is
In
sections with no loads
a
maximum when
V =
V =
constant
Example: Simply supported beam
The reactions
/?
with end load. (Fixed at A)
are:
A = 2.5kN;
Pa = 3 kN;
RB =
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
Loads
in
D
IT 7
beams
continued from P 6
c)
<x <
1
from
equation p
27
26
29
1
m <
27
26
x < 4
from
equation p
m
x < 6 m
from
equation p ...
m
.
.
25
!
29
m <
4
27
25
*
26; 29
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
ft?
II
*
ft?
II
^
E
|
cvj
1
i
II
*H
+
Z
E
ft,
e"
:
i
*
i
II
a?
5
^
i
1
CSJ
H
.
,
in
7'°
H
+
E
II e
»-
d
i
i
z
ii
a.
i
-o|-3
o
to
d
'
ii
ii
+
H
^
11
+
in
Z
E"
*~
in
,*
+
^
+
ft?
ft?
^
^
aight line
+
ft,
i
z
.*
CO
z
"
* !S
1
1
if)
ii
!
+
^i E
°
z
_*
I
8?
1
qT
ft,
ii
i
E
z
i"
I
s
i
7
o
II
+
ftT
'
i
i
k
ii
o
o
CO
ftT
+
a?
'
CM
5
H
"
1
H
k
E
H
CO
o
II
^
I
,—
o
«q
i
in
O
|
jl ;II e
CM
Z
in
*
CNJ
E
1
ii
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
p 31
p 32
Bending moment:
+F| -r (1 -cos
M
M
= —
+
F-|
At the section
q>,
F-|
r
•
•
^ y^
90°
o^j^rj
or:
qp)
r
+ F2 r-sin
cos cp - F2
=
cp
•
•
r
•
sin
cp
F and F2
:
are resolved into tangential
and
radial
components.
Shear force
p 33
Fq +
p 34
Fq = -
p 35
Fq
F-|
(radial):
sin cp
•
F-\
•
sin
dM =
= 6s
—
p 36
+
cp
F2
-
F2
cos
•
<p
cos
cp
— -6M
1
=
;
or from p 30:
,.
(because
•
s
6<p
r
1
d(-Fi-r+Fi •/••cosy-^-r-sincp)
r
6<p
=
r
•
cp
^
F-,
Normal force
p 37
Fn - F-i
•
p 38
Fn
•
-
jF,
;
j =
6s
r
j
x
dcp)
^'
sin^-Z^cosg?
(tangential):
cos
cos
g?
+
F2
•
sin
<p
<p
-
F2
•
sin
<p
For a graphical method of determining bending
moment
see K 4
STRENGTH
9
Bending
Maximum bending
p 39
_
u
btmax
stress
A*->W
_
,
^
'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
Second moment
:
neutral axis
of area
about S z or about plane of neutral
axis.
Bending stress
p 41
Section modulus
p 42
from the neutral axis
Yy
=
°\>
at distance y
M
Z mm
= -i^max
Z
/_
Second moments
of
area
second moment of area see
Polar second moment of area see
Axial
Product moment
Principal
The
see
I
I
I
17 and table P 10
17
17
second moments of area and
second moments of
axes
principal
principal
\
area
h =
to
Anax and
FH
= Anm are applicable
h
A
asymmetric sections, when the
axes are rotated through
^
^^
principal
the angle
p 43
*1
2
p 44
Anax
-.mm
cp
-
.
o U'y
*
x
)±|y(/y
-/x )2 + 4 /xy 2
2/xv
tan 2<p
7/
For calculations of
The
The
7 xy
see
117/18.
principal axes are always perpendicular to each other.
axis of
axis e.g.
I-\
symmetry
=
7X
.
of a
symmetrical section
is
one
principal
STRENGTH
10
Bending
Values
of
/
and Z
some common
for
sections
(see p 41 and p 42)
For position of centroid 5 (or neutral axis) see K 7
and
/x
/
Z x and Z
y
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
32
zx
/x
=
4 -d 4
£(£>
64
=
/'y
y
)
Jt
,
32
=
= 0.06014
= 0.5412
/
y
p 51
p 52
p 53
/x
•
RA
3
•
b
7
^•y
p 56
24
62
-h
-ft
24
48
p 57
a+6
36
h2
12
2a+b
h
3
p 58
Z>
6-ft 2
36
b3
2
4
2
= jt-a
4
b-h 3
p 55
fc
Z, =
4
-a
zy
0.1203- j 3
0.6250 R 3
0.1042 -s 3
0.5413 R 3
-s A
3
= K-a-b
jt
p 54
10
P 4 -rf 4 _, fl 4 -<i 4
10£>
D
p 49
p 50
d3
K-d 3
n-d'
64
/v«/«
p 48
M
a
+
b
(a+b) 2 +2ab
2a+&
a+2b
ft
p 59
centroid axis
Steiner's theorem
(Parallel axis
moments
p 60
I
JB
-b =
theorem
of area).
/x
+ ^-a 2
for
second
1
I
STRENGTH
Deflection of
Beams
beams
in
11
bending
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
EI-y"
p 62
^M
±
R
-M
=
£./^=
EI-V = -[Mdx + C,
dX
Ely = -JftJ M 6x dx + C, x +
p 63
p 64
•
R
p 65
_M_
„
y
dx 2
y'
:
•
C2
radius of curvature of the elastic curve at point x.
= tan
q>
inclination of the tangent to the elastic curve
:
at point x.
v
C,
= deflection of
beam
at point x.
and C2 are constants of integration and are determined 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
For a
p 66
U
beam
=
due
JE-J
~2
to
bending U:
of length
I:
6x
A beam
uous
with discontinloads may be
divided into n length:
p 67
"m-JMlIt*
STRENGTH
Deflection of
beams
+
12
bending
in
Tote.
co
IK
*_
p 68
cnjIco CCCNJIcM
ICO
ii
6
-
^ -O ^ -C
— !•©
+
+
—I
cc
+
*'*
+
,
p 69
*'IcN
>lcO >lcO
II
^N
or
^.
Ico
Ico
II
II
II
y-
co
CD
3
|
CM
col"""
co
II
r-
C\J
C\J
O<
O
CD
££
p 70
c
§2
<
^
-C|^ CCS
i
P 71
CD
s:
«S
o ££
o *T
o
.
„
ii
-o
l
<
ll
*?
II
<
a:
CM
I
%jTih
Q-O
II
I
£
£ ^ ^
OS
££
II
00
az
i«
i"
£
1
JW
-* ^
5 5
*
*
STRENGTH
Deflection of
«5
P 72
Uj
~
£
I
«
in
P13
bending
OcM
*
|S
"
beams
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
CO
CD
<D
t0|— v^H,
*
coioo
n
*
*
mioo t-ico ^ ^ £ ^
ii
<
cd
as os
*4m
-
<
tn
^ ^
.
STRENGTH
beams
Deflection of
in
14
bending
Mohr's analogy
Graphical method
1. Determine the bending
polygon (see also K 4).
moment
curve by constructing a
Original
Position diagram
beam
1
2.
Fig.
1
link
beam
Equivalent force diagram
Fig.
beam
Deflection of the original
y
=
/»'
Slope
tan
P 77
Force diagram
polygon as the equivalent distributed load w*
on the "equivalent beam". Another link polygon will give the
tangents to the elastic curve.
Construct a
Equivalent
p 76
link
at
HH* m
EI
...
m A m L3
w
F
•
2
at point x:
.
•
support A and B:
yA = Rff-pr—;
m F m A m L2
•
resp. tan
.
H
EI
<p B
rrip
•
mA
•
rri\_
.
Mathematical method
1
p 78
2.
p 79
Calculate the equivalent support reaction /? A * of the
"equivalent beam" carrying the equivalent distributed load
w* = A-i +... + A n (see fig. 2).
Calculate the equivalent bending moment
equivalent shear force V* at the point x:
V* = R,
M* = /? A*-2-A-z A
;
3.
and the
(see
fig. 1
+2)
Distance between center of gravity of equivalent distributed load A and the section x.
Slope y = V*/E-I
Deflection
y = M*IEI\
xA
p 80
M*
:
continued on P 15
STRENGTH
Deflection of
beams
in
Pl5
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.
Original
Simple beam
Cantilever
A
beam
A
Equivalent
A
A
beam
A
A8
A %
%
D
A
.
beam
V
Beam
of varying cross section
Fig. 1
Original beam
e. g. shaft
z
I
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 the
Plot
w*(z) according
to the ratio
/xmax
.
p 81
Fig.
2
Equivalent beam
of the shaft
in Fig.
1
J1
Then calculate according
to
P 14 (items 2 and
I
3)
or p 78
...
p 80.
STRENGTH
Beams
maximum
p 83
t bp
SWl
6W-z
h 2 PbX
3w/
V
ft-Pbt
6
W(
\
2_wl 2
bl- PbX
•
V^p
t
3
/
V
32
W
V
)
•
J_3jWi/
W
v Ab Pm
V
I
E\h
3vv / z 2
h 2 -l- Pbi
~hTp^
p 87
V
I
2
b
p 86
I
b-E\h
lwl(
p 85
beam
8W( l\
bE\hl
-
yl
type of
/=
bt
h2 Pbt
'
p 84
deflection
6W-z
6W-I
V b-Pbx
aI
p 82
maximum
typical
dimension
x = resp. v =
section
dimension
P16
of uniform strength
•
/
^-Pbt
3
/
32
V
n
,
2
l
W
*</'
4z^\
1
bt
/
•
z
'Pb\
W
point load
w
uniformly distributed load
Pbt
permissible bending stress
/
64-
192_
5
EI
W-/ 3
'End*
kN
kNm
N mm 2
(see
Z
17)
*
|
STRENGTH
Statically indeterminate
17
beams
Fig. 1
Convert
beam
a
(fig.
into a statically deter-
1)
minate one (fig. 2) by replacing one
support by its support reaction {R c in
I
^
I
F,9-
Divide into two separate beams or subsystems. Determine the deflections at
the point of the statically indeter-
I
I
/[
|
V
A
i
/*
*
J
fig- 2).
———— c
I
indeterminate
statically
I
I
I
IcKI
I
^
I.
I
A
Wl
/W
1s
stt su
subsystem
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
2nd subsystem
A^
- ycz
Hence, calculate the support force
support reactions.
Method
at
at C,
R c and then
the remaining
of solution for simple statically indeterminate
Statically in-
beam determinate beam
determinate
w,
—
2nd
subsystem
subsystem
Rc
]%] w
yC 2
\m
c
i
B
beams
1st
Statically
IV,
4
JV2
^r
5&
*»-~±^" e
s
J/?C
2
W;
^2
c
^Xt 4
I
V\\C\
r
I
I
I
l'T™
'
l"I)
4
2
^~
--*^
indeterminate support reactions and
rf
ffA
g"
rc\
2
statically
fr^-
^
l/? A
^3
*C ^S
A
V4
Ma
moments
J±Sn
C|
Mb
STRENGTH
18
Shear
Hooke's law
p 89
shear stress
G
T =
or
q
for
y
G: shear modulus
y shear strain
:
Relation between shear modulus and modulus of elasticity or
Young's modulus
G
p 90
=
—— £
2
(1
Mean shear
p 91
q
f
or
- 0.385
+
•
E
;
correspond to P 3 with u = 0.3
A*)
stress
</ f
or
rf
T,
Permissible shear strength p q (values for
dead
type of load
(see P 2)
92
Q
guillotine
cross cutting
undulating
alternating
^p0.2 /3.0
applied by
shears
(punch
1°
-!°-j
Q
1-2 9a
-
G-
2 tan a
1-2
<o
i
^/-.v
< 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.
tan a = 5//;
p 94
q =
q
:
q
:
/
q'
transverse shear stress (transverse to
axis) resulting from shear forces Q
axial shear stress (parallel to beam
axis) "complementary shear"
beam
0.8
/?„
cutting tools
parallel cutting
^\-\\\\\v \\^\
p 93
16)
^P0.2 /1.5
^p0.2 /2.2
Ultimate shear stress q £
for ductile metals
_ vmax
for cast iron:
A
Shearing force
R p02 see Z
.
.
.)
STRENGTH
19
Shear
Axial shear stress
p 95
1 =
QM
TT
p 96
A/=
AA-y s
q =
q max
due
to
shear forces
" 1
for
ab = abmax
occurs when ab = 0,
e. on neutral axis.
.
i.
Max. shear stress
for different
<?max
cross sections
= q
Ta
4
3
'
d a 2 + d a -dj +
d 2 + d2
dj
2
p 97
for thin-wall tubes: 2
(<*a«4)
Strain energy u
due
to
shear
1
p 98
Shear deflection
of a
q-y =
2G
beam
x -pr—i +
p 99
G
C
Determine the constant
-A
C
K^+C
£v4
from known factors, e.g.
v
=
at
the
supports.
p100
The factor
»-*/[£
d/l
allows for the form of cross
(A)
section. For
H =
examp ie:
1.2
®
I 80
1.1
2.4
I 240
I 500
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
5 AA
:
the z axis
width of section at point y
centre of area of section A
A about
STRENGTH
20
Torsion
General
T a <
Tfmax
—
•
p 101
p 102
Shear stress due torsion r
t
=
=
T -
Torque
p«
t
=
2-Ji-n
CO
power
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
p 104
<P
of twist
!
-G
p
Stepped
u
k
shafts:
T
p 105
"
i
-
l\
1
'pi
"
T
180°
p 106
polar
7-/
m 180/
III
-
(see e 5)
(f
moment
of
inertia / n
p 107
32
P 108
32
Angle
-^ 4
"
)
Tl
of twist
<p
Ix
180°
-G
Jt
magnitude
constant J
in
1
Tt
:
C1
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
G
cross
section
of rt
K
- Ttmax c2-
Tt2 = c3 .Ttmax
rt3 =
in 3:
=
•
C^h-b 2
in 2:
'6
-|
=
c^h-b 3
p 110
T-l
/t
'
position and
torsion
/i
D*-d 4
"
'
of non-circular, solid or thin-wall hollow section
Bars
p 109
(D 4
Vi"
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
J
at
1
46.19
m
26
- r max
20 T ^ 13 T
T
:
a4
p 111
t
a
p 113
at
0.1154-j4
0.0649 d A
p 114
t i
1
3
h?
T t2 =
at 2:
p 112
p 115
cross
section
=
T t1
:
T tmax
~~
= 5.297
-8.157
p 116
at 1: Tt1
_k_
16
p 117
D 3 -d3
'
*
D^d
=
2
T
at 2:
n2 +
16
p 120
=
n^
,
--^r-i
tmax
-
7-d
5.1
n
D/d = DJd =
2
r"t max
-
t
at 1: Tt1
p 118
p 119
'tmax
Dd
5.1
(rf
4
-
4
4
)
at 2: r, 2
1
with varying wall thickness:
at 1: Tt1 =
p 121
Ttmax
7
p 122
2/1 m 'mm
at
p 123
Tt2 -
r
p
*
with thin, uniform wall:
AA
p 124
/4 m
(
i
r
2-f
2 A,
"imax
J|n=3
M
[""
_[_ rp
1.0
1.12
1
/4
5
t
s
m
:
:
<
7;
+
n=3
<
1.3
the
n=2
1.17
rectangular
tion of
ness b max
sition
sec-
-§*
max. thick1
(e. g.
po-
in
the
y?7?7Z
I-JT
sketch).
area enclosed by the median
line
length of median line
wall thickness, (min. wall thickness)
):
part length of median line when wall thickness
(f min
:
of
long side h of the
n=2
|
«1.3
up
rectangular
cross
sections
r
midway along
Foppl's factor:
p 125
profiles built
T-b,
2>i 3 '*i
t,
= const.
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=
p 128
Permissible working load
F
p 129
Slenderness
W-'V^
2-
/
Limit
p 130
/ic
=
L
/
ratio
=
based on R pO.01
1
,m
L
= 0.707-/
= 0.5-/
P e /v
on R p02
Limit 2 based
^lim
-"V^
*
V* P «
Tetmajer formula
Valid
in
the range
p0.01
Material of strut fails
p 131
Px
due
=
to
=a
A
— A ~
^P°- 2
bending and compression
- b-k + C A 2 = P'V k
Material
a
US:ASTM
GB-Standard
\
b
Calculation method
First
mum
determine
area using
mula:
p 132
_
i/
min ~
*Vfk
2
the
Euler
the
valid
for
/.
=
80... 100
60... 100
0.053
5... 80
2. .100
0...100
material failure
caused by buckling
mini-
moment
second
c
N/mm 2
BS 970, 050 A20 A283Gr.C 289 0.818
BS970,080M30 A 440
589 3.918
BS 1452 -220
A 48 A 258 776 12000
timber
30 020
beech or oak
38 025
mild steel
mild steel
cast iron
»
\
of
forffpO.01
-
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
vk = 4
s
vk = 6
.
.
.
5
.
.
.
6
H
8
...
|
in
the Tetmajer range
.
,
in
_
—
larae
kfor
for small
,
|
the Euler range
M
1
structures
,,
|
|
|
Present
1
slenderness
ratio
A
A| im0 01
.
|
j
1
Limiting
|
and
p 129
p 130
calculate from
A| im0 -2
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
U)
A 283 Grade C
c
USA
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
1.41
1.79
2.53
3.65
4.96
6.48
8.21
10.31
cast iron
AA2017 ASTM-A48A25B
1.03
1.39
1.99
3.36
5.25
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
and
appropriate new
values are identical.
the
final
value
of
co,
until
the
1
STRENGTH
Combination
24
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 £
Fz = F- cos y
p 137
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
w '" D © in
functions see E 2.
and y
Long beams
in
Neutral axis of
is
y
Xq
Either
in
Fx
for buckling.
Fy
x
Fy A-
/
y
at:
Fz
h
7y
Fx 'Al
With asymmetrical cross section
the principal axes (see P 9).
Bending
is
the straight line:
which intercepts the axes
p 141
If
compression should be examined
o =
p 140
Fz
a compressive force, a, (3
different quadrants. For the sign of cosine
y.
>'o
F
resolve
in
the directions of
one axis with end load
or
Fy
in
formulae p 139
tension
Bending with
...
p 141
is
zero
displaces the
compression
neutral axis
compression
towards the
tension
-
,
STRENGTH
Combination
Stress
The
in
25
of stresses
curved beams (R
<
5 h)
Fn and bending
direct force
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
C
the
at
R+
y
—
and
inner
outer surfaces are:
Fn
p 143
A/x
.
M
p 144
C
=
R+\e<i\
for coefficient
b-R 3
X
R
\e 2
+
In
\
C
d
2R
d
A.
R
2R
p 146
C
=
*Pt
\
R-\e 2 ^Px
C
1
p 145
Icil
C
M
x
AR
Formulae
Mx-R
j
VMi
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
must be added vectorially.
The maximum shear
A complementary shear
it.
Maximum
torsional stress r res
point
in
r
5.1
1.7/
~
~d 2
where T =
F|
d*
5.17D
D 4 -d*
,
,
n
_
'
D + Dd +
2
T7
t
p 152
d
D2 +
-\.
f|
7
Pqt
2
PqX
D*~^d*
where T =
F 4.244
Pqt
:
4.244
p 150
cross section
1
p 149
p 151
:
-d(D +
d)
Pqt
D*-d*
For thin wall tubes:
5.1
p 153
TD
D A -d A
2.55 F-
p 154
2.55-F
D 2 -d 2
2D 2 + d 2
d4
Pqt
Pqt
T + 1.5F
<
~ p«
2h^>)
PqX
££
p 155
c,'b 2~h
~b~h
where T
p 156
p q{
t
rq
F
T
for
1
and acts in the
acts perpendicular
stress rres occurs at point
cross sectional plane.
to
torsion at any cross section
permissible shear stress (see Z 16)
:
shear stress
:
:
:
calculated maximum torsional sheat stress
force producing torsion
:
c-,
torque produced by
and c 2 see P 20
F
STRENGTH
Combination
Combination
27
of stresses
and shear stresses
of direct
Material strength values can only be determined for single-axis
stress
to
Therefore,
conditions.
multi-axis
av
single-axis equivalent stresses
are
stresses
P
(see
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
ov
in v
z direction
in
direction
y-z 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
0.5
Direction
rotation
the
of
az
ov
+
)
highest
±
0.5 l/(az -
2r
=
The
,
T™
a-i
+
ki 1
at
angle of
is:
*>
(where the shear stress
Rotating the element through the angle
T max
2
)
is
zero).
Oz
Maximum shear
p 159
oy
stress
principal
Uorw the original position
(p
tan 2 <po
p 158
(
=
±
tp T
gives the
stresses
0.5
]/(
2
oz
av )
+4r
0.5 (a, - o 7 )
2
direct stresses act simultaneously:
p 160
OM
=
0.5 (C7Z +
Direction of the
Oy
maximum
)
=
0.5(O, +
C7
2 )
shear stress r max
is
oz
p 161
tan 2
The
<p
r
Principal stresses
2r
and Maximum shear stresses
lie
at
45
to
each other.
*
}
The solution gives 2 angles. It means that both the
stresses and the Maximum shear stresses occur in
ions at right angles.
Principal
2 direct-
STRENGTH
Combination
Stress
The
in
28
of stresses
three dimensions
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 +
p 163
where
So-T=
R
=
ax
+
+
<7
y
ax ay
oz
p 164
S =
p 165
T = ox oy az + 2
•
+
ay
•
<jz
+
oz -ox -
rxy Ty.T^ -
cr
rxy
x
•yz
Tyz 2 -
o^zj -
cr
z
Txy 2
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
According
in
p 166
Equivalent stress
p 167
Equivalent
moment
To find the diameter
modulus Z from:
P
shafts of circular cross section
to the theory of
maximum
strain energy:
2
oE
= \jfbx + 3(a
ME
=
yM b
2
-/qt ) 2
+ 0.75(a
-
< p bt
T) 2
of the shaft, determine the necessary section
z
I'
=Mi
Pbt
due to bending
shear stress due to torsion
bending moment
tensile stress
torque
according to P 29
n
r 29
STRENGTH
Combination
;
N
1
co
w
b
u
1
£
CO
=
= =
tT
b~
s
00
j?
.CM
+
CO
^-~
in
b
1
of stresses
CD
?
E
CM
+
CO
>
3
CD
.^
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
+
CNJ
°-2w.E
CX
^>
o~
•
00
Q.
-
' ?«
11
+
CM
c
CO
CM
3§
CM
CM
c
E
—
+
0.0
a
*-
c
fj
+
='
- = =
,-
2.
co
5
E
CO CO
E
11
5
B
CM
13
-
8
b
C
O
b
«
E
0)
CO
CO
6
CO
11
.J
b
1
1
CO
b
b
11
b
11
H
z
b
z
tr
CD
V
b
£?
c
J
.2
Q.
E
55
~§
1
*
-<r
"D
£
CD
C
T3
O
in
f£
b"
O
-*
+
CM
= = = =
— =£ — =
CM
tT
II
E
lr
b
11
E
Is
1
CM
b
c
tf
:
01
in
Q.
i
c
°
s
II
<
+
•
CD
O
D)
00
3
O
0)
CO
3
c
O
c
O
3
O
w
*
E
-
to
"ro
c
J>
-E
O
-—
Cfl
c
O
C
Q)
2
~ £
to
S3
E
to
CO
CO
at
>
E
"co
2
a
01 >^
,
O W
TO
C 3
3 cr
3
»»!
t.
a)
II
ra
00
w to
S
CP
£
g g
<D
co
(D
0)
a3
9 «
c
E
CO
a;
*"
1 E
m
°
^
CO
£ «
X!
ro
0)
| 5CO
a
+
»
is
-%%%%
SiSS
.
N
fc
OS
„_
-2.
II
b
cm
b
s
«=
W
Cfl
C -P
S
(-
2
A
b
*=
-
IS)
CM
CD
A
0,-°
woD
E-o c c 3
C <D .2
O CO rC
c°
11
+
..
C/3
CD
X
11
-
to
c
cm
l
CM
T3
to
cm
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
Fmax =
q 2
shear load Fs
calculation for friction effect:
Ao ~ A<
1
A3 « As
FA
(1.3... 1.6)
Fc
=
see P
q 3
= (0.25
Pi
(allowing
for
/?
p
.
2
1
.
.
.
v-Fs
0.5) /? p02
fere
(U-
and
torsion
factor)
safety
req
(0.25... 0.5)
(using load-extension diagram,
(for
values of
m -w
[/.
see Z
7)
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
^
&
tation,
-0-
^
&
O-
a
e. g.
q 4
is
the point of ro-
For a
F
q 5
q 6
A1
rigid
FA1
=
/
•
«
/
\
h/A.
-
i
^>
attachment:
•
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
.
.
.
m
1
= 3;n = 2
(2)]
outside diameter of bolt
core diameter of bolt
MACHINE PARTS
Axies and Shafts
Axles and shafts (approximate
Stability
Axis
7
q
8
q
9
fixed 2
of circular
modulus
cross section
bending
bending
stress 3
(Z^^/10)
'
Qbtu
M
=
Pbt
'
m a/ 10-M
V Pbt
Pbt
rotating
(3...
Q'bt
Pbt
(3 ... 5)
torsional
solid shaft
stress 3
pure torsion
P^
V
torsion and bending
Pqt
5)
A
permissible
diameter for
Shafts
q 11
q 12
'
permissible
'
Z
q 10
1
solid axle
required
section
for
q
calculation)
n
Pqx
'
(3 ... 5)
-
TtU
(10... 15)
Bearing stress
on shaft \
q 13
extension]
F
r
/bm = d-b
(Pb see
Shear due
/
/
>
>
Z
'
18)
.
for
325 h
all
shafts
|
... 8iU
with
2)
M
circular
cross
rectangular
sections
6.
For precise calculation see DIN 15017
(P 2)
Formulae are restricted to load classes +
and pqt allow for stress concentration-, roughness-,
15017), safety-factor and combined stresses
I
3)
|
for fixed axles
Deflection due to bending see P 12
due to torsion see P 20
Vibrations
see
1)
unnecessary when
to lateral load: Calculation
d/A
II
pbt
size-, (see
DIN
arm of force F
M, T bending moment, torque
/bm (Pb) mean (permissible) bearing stress [see Z18)
/
:
:
/bmax see
Obt
d.
47 f° r hydrodynamically lubricated plain bearings,
values see Z16.
[other cases see Z18.
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
Clamped
see DIN 7190 (graphical method).
T- v
q 14
fid
Taper
q 15
fits
joint
D-d
a =
Taper from
tan
For
shaft
I
tapered
extensions
DIN 1448, 1449.
Approximate formula for
on the
q 16
imaginary joint,
not to stiff
joint
Fa
see
axial force
FA
nut:
=
tan
-'&
D
q 17
+ *)
+ d
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
The bearing length
q
U
/
is
t2
.
t
to transmit
a torque T:
2 T
j
t
r77777%777777A
((^L^^
K^
I
j
-
\^ x^
SV*±-
d-h'Pb
Dimensions
fillets
For
to
DIN
6885,
preferably
Sheet
1.
Allowance
for
with form A.
precise
nigung
calculations refer to
Antriebstechnik
e.V.
heft 26, 1975.
For symbols see
Q
Mielitzer,
ForschungsvereiForschungs-
Frankfurt/M.,
continued on
4
Q
4
MACHINE PARTS
Shaft-hub joints
continued from
Q
3
Splined shaft
Shaft
-
q 19
/
q 20
dm =
q 21
h
=
•
2
The load
~
2
not shared equally between the splines so allowance
for unequal bearing is 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
needed to transmit a torque T of 3000 N m, made in cast
steel fitted with a plain key.
*
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
CI
CS
(gray cast iron)
(cast steel), St (steel)
[p^
in
N/mm 2
50
100
/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
ASTM A 572 Grade 42 - resp. to BS 4360 43 B - but not
(such as high centrifugal force, etc.). Increase
there are other forces or moments being carried.
cases
cial
when
u
CD
a
o
<D
CL
w
®
JD -*
c
CO
QC
to
for spe-
steel
£
CO
X
-
LU
o
—
en
o
^
£en O
.
C/D
c
-Era
"O
*C
-D
(j)
©
C
.C
CD
Q.
'
-
\
tarQ
\\
\
\
\\ \ \
L
MACHINE PARTS
Springs
R and
Spring rate
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
=
J,
F2
+
F,
R tot
q 26
=
=
S3 ...
52
/?!
F3
+
+
Springs
in
+
...
F 2 + R3
.
.
F2
F,
.
+
S2
S;
R
+
53
+
= F3
s
i
=
F\
1
R
t
:
R2
'
R,
tension and compression
e.g. ring spring (Belleville spring)
Springs
bending
in
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:
bQ =
q 30
z-b
no. of leaves.
Then
(as q 28):
b-h 2
F =
q 31
6
-
Pbx
/
leaves 1 and 2 are the same
length (as in the sketch):
If
q 32
= 2b
b,
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
n-FB
q 34
'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 2
oscillating (500
± 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
/
i
F-r-l
IE
= Dm
-i
no. of coils.
:
f
coil length
—-^
(Additional correction
is
needed
for deflection of long arms).
For precise calculation, see DIN 2088.
Springs
in
torsion
Torsion bar spring
Shear stress
T
q 38
Angle of twist
d3
#
d3
spring length as
q 39
Torque
shown
:
rA
:
lor-/s
G-d 4
sketch.
N/mm 2
Tf in
static
oscillating 2
preloaded
rm
Tls
GI D
Stress p qt and fatigue strength
not preloaded
Pqt
in
=
|
700
1020
Tf
=
Tm ± TA
d=
d=
20mm
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)
*
Normal
s
Tension spring
f
Did = 5
coil ratio:
Static stress:
D
... 15;
> /8
= (D
D unknown
d ^l/ 8 ^ctheor
+ £>j)/2
d
> j/ QFn -D
d
> l/
r
q 41
estimate Did
q 42
max. perm, deflection
q 43
sum
D
^-Pqt
8
Fn
P'
3A
x-dn
of min. distances
between
•;
tension spring
fctheoiH
d
-|
without
preload
compression spring
D known
—
f
3
q 40
Fn
Fy
,*ith
<-.
Compr. spring s
% Fb
coils
sA =
with x ={0.21.
10.7
D/d=l 4
120
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
t
/? n
diagram x 0.8
For higher relaxation requirement see DIN 2089.
Cold-wound compr. springs
Theoretical spring i
deflec-l
iperm.
f
,orce
tion
I
when
Istress
~1000
just solid
(compression spring)
*c
Circ. spring steel wire
|^ctheor| Pqtc
ASTM-
1200
spring wire
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
and
Length/diameter
e* = b/d
ratio
0.5
fete^
*-B*
1.5
1.0
in transverse
longitudinal sections
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
therefore
adequate
Excellent
good
oil
at
each end, Small pressure drop
cooling
with
high
load
at
low rotational speeds.
rotational
speeds.
Low
each end,
flow.
Good
for
at
therefore high load capacity
Poor cooling
capacity
tational speeds.
Symbols see Q 14
at
low
facilities.
ro-
Danger
of
edge loading.
continued on
Q
1
MACHINE PARTS
Q11
Bearings
Q
continued from
Bearing pressure p, p
q 46
q 47
r
F
p
bearing
pressure
10 (journal bearings)
D B
sf*dF
Pmax
<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
mean bearing pressure
to the
thickness of the
(According to
relative
lubrication
to
relation
in
film.
2
0.3
0.4
0.5
Bauer, VDI 2204).
Bearing clearance
DR -
q 48
y>
is
ing
relative bearing clearance
C,
y
CID
D,
basically the
operation
relative
(including
bearing clearance established durthermal expansion and* elastic de-
formation).
q 49
Typical values
y=
(0.3
... 1 ... 3)
Criteria for the choice of
10~ 3
1)
\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
length/diameter ratio
B*
support
self-aligning
Minimum
<
0.8
values for plastics
q 51
q 52
Upper value
bearing material
sintered metals
1)
Grease-lubricated plain bearings
Symbols see Q 14
relatively
B*
>
low
0.8
rigid
4)10" 3
2)10" 3
3)
10" 3
continued on
Q
12
MACHINE PARTS
Q 12
Bearings
continued from
Minimum permissible
during operation h Um in
Theoretical
q 54
>
+
Q
11 (journal bearings)
lubricant film thickness
urn.
Actual
shaft deflexion
hv
,
bearing
distortion
+
*zj)
l*
_
mm
"min
C/2
_
special
cases,
U*
some
J
crankshaft
2
i
I
J
^
Relative minimum
lubricant film thickness //* min
q 55
(15)]
+ sum
of peak-to-valley heights.
(*z.B
10...
3.5
[(D
e.g.
small-
automobile
bearings
1
large
shaft
diameters
Relative eccentricity e
AZmin
V D
•
For statically loaded bearings:
otherwise instability
Sommerfeld
number So
(Dimensionless)
q 56
So =
PV2
V
Qetf
Nr.
BID
1
1/1
2
3
4
5
6
1/2
1/3
1/4
1/6
1/8
So
Inserting
the adjacent
diagram gives
/i* min and therein
fore also h min
To a
mate
sity is
first
?/ eff
.
esti-
visco-
based on
mean temperature of bearing.
A
better
mate
7"
ef
esti-
is:
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
e3 =
0.5
Rules:
hydrodynamic
flow rate required to maintain
theoretical
lubrication
(exactly values see DIN 31652):
is
5u(C-2A mln
Oil inlet to the
)
expanding part
of the bearing.
Oil velocity u:
q 58
In
supply
q 59
In
return lines:
Oilbags,
oil
lines: v
slots
o
= 2 m/s
= 05 m/s;
;
(depth
=
p
p
= 0.05
= 0.
2 C) never
in
...
0.2
MPa
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
Friction
power P = f-Fu = P th (rate of heat removal)
/by the use of the following diagrams and So by q 56)
f
(Calculate
500
-I
0.01
0.02
0.05.
10
0,
So
Ah, amb
k
W/(m^K)
100
So
Heat removal by convection
q 61
50
20
_ k'A(Ts
= 7+12
at the
Tamb)
Wamb
tm
s-1
200
1000
housing surface A.
with empirical formula for k:
(equation scaled\
{
for units
continued on
Symbols see Q 14
500
*
/
Q
14
MACHINE PARTS
14
Bearings
Q
continued from
area
13 (journal bearings)
the heat removal surface
approximate value for step-bearings:
the
If
of
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
Pth amb = P
until, according to q 60,
Since
f
Heat removed by lubricant
P th
if
necessary with
conduction are neglected):
Oil circulation,
q 69
q 70
c
min
k
P>
minimum
:
Ten)
(1.6... 1.8)
Z
*10 6
Jm_3 K"
1
5)
Z
7)
lubricant film thickness during operation
lubricant feed pressure
:
mean, max. bearing pressure
Pmax
u
-
heat transmission coefficient
:
p en
.p«
coefficient of friction (values see
:
/i
D
specific heat (values see
.
p
cooler (convection, heat
oil
for simple calculations for mineral oils:
c
/
.
:
L= QCpQiTex
^th,
Guide values
L
peripheral speed of bearing journal
:
warnb
:
»
velocity of air surrounding the bearing housing
external width of bearing housing
in axial
direction
C
nominal clearance, concentration, chamfer
nominal bearing diameter
D
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
H
I
:
:
:
Q
:
lubricant flow rate,
mean
^z.B
J-
am
7en>
Tq
r]
rj
p
Od f
rate
average peak-to-valley height of bearing sliding surface
ambient temperature
Tex lubricant temperature at bearing entrance, exit
bearing temperature
dynamic viscosity of the lubricant (values see Z 14)
effective dynamic viscosity of the lubricant
:
:
:
e ff
volume flow
value of surface finish C.L.A. of shaft mating surface
:
:
density (val.s. Z 5)
:
crushing yield point
I
\p
:
co
:
relative bearing
angular velocity
clearance (val.s. Q11)
(co
= 2n
\)
MACHINE PARTS
15
Cross head guide, Clutches
Crosshead guide
Crosshead guide
operate smoothly only
will
when
L
q 71
tan
a <
(2-h +
the length ratio
l_
q 72
F cos a
is
2 n
,
h
'////////////////A
l)-fi
Y////////////////A
tana
- n tana
1
If
the above conditions for tan a are not satisfied
a danger of tilting and jamming.
there
is
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
approximate calculation:
to 102 = &>i,
Acceleration of driven side from C02 =
= const.; T\_ = const.; 7c = const. > T\_. Then, per
a>i
simplified
for
operation:
q 73
energy loss
q 74
slip
W,
h-
(**Y&d
2
time
Calculating the area of the friction surface
flat
single-
twin-
|
cylin-
multi-plate
plate
drical
clutches
clutches
III!
TTT
ii
i
Number and
size of friction surfaces
/^r
^
depend on the permissible
contact pressure p^ and the permissible thermal capacity
per unit area q p
continued on Q 16
.
Symbols see Q 17
MACHINE PARTS
16
Clutches
Q
continued from
15 (friction clutches)
Calculation of contact pressure
(for values see Z 19)
For
all
/»
b
types of friction surfaces:
i-A
q 75
Pb ^dyn
'
q 76
R a3
R a2
2
where
3
'
'
^m
-
/?j
-
R?
3
conical
flat
cylindrical
friction surfaces
q 77
operating
Fa = A-p
Fa
Ap
=
sin
a2
force
(axial)
q 78
to prevent
for multiplate
clutches usually:
q 79
IT = 0,6.
.
.0,8
tan
R\ =
Km
a>
/i stat
^stat
T
Calculation for a shaft:
«a =
locking:
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
=
i-A >
Wr z
Wr z
MACHINE PARTS
Friction clutches
17
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
B
:
2pF6 jr
TB =
q 82
7"
n
Expanding-shoe drum brakes
(Drawing of simplex brake showing, simplified, the forces acting on
the shoes).
Leading
|
Trailing
shoes
Fm
Fs-l
=
a- fx-r
(Servoaction)
Braking torque T&:
TB =
q 85
(Fn1 +
Fn2
)
ii
Leading
r
Band brakes see K
shoe
13
Notation for friction clutches and brakes (Q15
A
Trailing
shoe
.
.
.
Q17)
area of friction surface
radius of friction surfaces
:
R
R m R a /?, mean, outside,
operating torque
7C
:
,
,
:
inside radius of friction surface
:
TL
TM
T
t
W,
load torque
:
motor torque
:
transfer
:
i
:
/
:
z
of clutch
no. of friction surface
no. of calipers on a disc brake
operating frequency
:
fx,
nd
co
:
(for
moment
energy loss per operation
:
fx sXaX
:
(EU: s"\ h
_1
)
friction-, sliding friction-, static coefficient of friction
angular velocity
properties of friction materials see Z19)
MACHINE PARTS
Q 18
Involute-tooth gears
Involute-tooth gears
Spur gears, geometry
q 86
Gear
q 87
Transmission
ratio
na
ratio
0) b
Transmission
ratio of multi-stage gearing:
q 88
=
q 89
invor =
i,
•
•
/„
•
;„,
tana — a
Showing
the transverse path of contact
(see ISO
Active
flank area
of wheell
R
1
122)
^&\°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
helical
=
m
•
=
Pi
~ cos
ji
z
" cos
<
m
m
h a = h aP =
q 97
addendum
dedendum
q 98
bottom clearance
ht =
c
=
/j
|8
m « =2i =
mn
m
= d
s
m a -K
m n -n
Po
f
p
(0.1
=
.
.
J-cos/8
_ d
z
j8
+ c
0.4)
m «
0.2
-m
continued on
For symbols see
Q
29, suffixes
see
Q
23
Q
19
:•
MACHINE PARTS
19
Involute-tooth gears
Q
continu 3d from
18 (spur gears)
Standard gears
spur
q 99/100
reference diameter
helical
m-z
=
d
d =
tip
q 102
root diameter
q 103
pressure angle
d
f
-z
d-2-hf
=
a = a n = a = ap
On = «p
t
a
tan
q 104
q 107
x
d a = d + 2-h a
diameter
q 101
q 105/106
m
=
cos pc
dcos
db =
base diameter
=
t
1
a?n
tan
cos p
d b = d- cos a
a
t
1
equivalent no.
of teeth
"cos 2 /3b -cos/3
3960
table see DIN
z
q 108
cos 3
/?
min. no. of teeth
q 109
2
to avoid
theory
sm 2 a
9
(undercut)
~
'
17- cos 3 p
a P = 20°
for
when produced
z gs
^
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\
36
9a=
length of path
of contact
(total length)
I
116/117
+ Zl
centre distance
,
q 115
helical
l\
'" n
2
r
2-cosi3
.
^V^1 -^1
2
2
+ V<*a2
2
-4>2 2
- Wbi + d b 2) -tan a x )\
transverse
contact ratio
ea -
9«
p cosa
9a
^
q 118
overlap ratio
to
q 119
contact
e
p
ratic )
Y
p
—
•
t
cos a
t
fc-sinl/Sl
m n -Jt
'
= e a + ep
continued on
For symbols see
'-
Q
29, suffixes
see
Q
23
Q
20
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)
14-z
a wt calculated from
17
* 1 + X2 ~
cos a,
(z :
+Z2)-('nvaw t-invat )
(z-i
+
2-tanan
-
invot +
centre distance
q 129
addendum
q 130
zo)
2-a
invawt
or
modifica-
mn=
k*
a-
q 131
addendum
h a = h aP +
q 132
dedendum
hf
= h iP
a6
q 133
outside diameter
d a = d + 2-h a
root diameter
d<
q 135
length of path of
x
2-f
COSOft
"
i
^
-m
n
+
{x
-
2 9a = -2-[V^1
(d b1
e a = 9a'(P
+
•
rf
rf
2
b1
'+
£a =
£p
contact ratio
£y =
tool data
unknown
take ap
=
Q
29, suffixes
2)
2
2
V d a2 -^2
'-
9jiP\ -cos at)
-jc)
ea + £p
20°.
Note the sign. With external gears k x
modification can often be avoided
For symbols see
+ x2 )
= ft-sin|/j|/(w n
overlap ratio
If
x
b2 )-tana wt J
cosa)
q 139
1)
-tana n
k*-m n
q 138
2)
2
d-2-hf
=
-
transverse
contact ratio
•
x-m n
x-m n
-
q 134
contact
—m
C0S at
a,
3d COS
Owt
a
9
tion coefficient
)
mm
^ 14-(z/cos 3 ff)
(total)
q 128
•
17
centre distance
q 127
2
to 0.17
'
to give a specific
q 126
t
+
COS fi
/iao-paoC -sin gn
+
can be up
ditto 1
_ z-sin2g
*min -
p
see
Q
mn <
23
0!
When
k
<
0.1
addendum
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
Safety factor
q 140
.S>
against fatigue failure of tooth root:
^Flim^ST ^NT'
_
c
of tooth (approximate calculation)
m
b
^SrelT' ^RrelT
•
^X
^
st
K.K,
•
r
Using the simplification:
{KFa 'YB 'Y$~V,
(^fcrelT' ^RrelT" ^x)
*t
q 141
'
^Fg
•
Y Fs
•
=
2;
Ym ~
1;
Sf min
2-o> lim
b
FFS
q 142
(^A ^v)
•
.
Y ST
~ 1
tooth form factor for external gearing (see diagram)
:
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
For symbols see
Q
29, suffixes
see
Q
23
Q
22
|
MACHINE PARTS
Q 22
Involute-tooth gears
Q
continued from
Load capacity
21 (spur gears, design)
of tooth flank (approximate calculation)
Safety factor 5 H against pitting:
q 145
°H lim Znt
_
c
'
•
V^
Zy Zp)
{Z\.-
'
-Zw Zx
*Sl
Zh 2 E Z z Zp y^A ^v Ky
•
'
•
•
"
For metals the factor of elasticity
ZE
where
£
VO-175
q 146
is
simplified to:
2
•
E
E2
y
Ei
Therefore, the approximate formula becomes:
*
q
147
>
d
fe,!^ 0175
,
£
g,'
,
z
cos
=1
q 151
only valid for a n
s
Appr oxima te values for strength
(D/a grams in DIN 3990 part 5)
Sp
Mat.
scification to
Norm
6
-
N/mm 2
A 48-50 B
80
360
4
ASTM A 536-20-90-02 230 560
A 572
cs
AS
200
400
3
-
1064
220
620
2
-
SAE 4140
290
670
Gr. 65
20°
7
2.5CI
=
28
^liml^lim
Grade
1
H^E'V^'V^'V% ,SHp
—^\0
h**,v\\\
.^^^S^nN
Crts"^****^.
\ \
^
\ \ ^
\\
v
jSw
1
ASCH
500 1630
3240
CS:
carbon steel
alloy steel
O.l^"" "*-*-"^
1
8
ASCH: case hardened
_ __
continued on Q 23
K A Ky
-
5 Hmi n
q 148
q 149
q 150
q 151
a H iim
ZH
:
(Z L
•
:
-
^^S^,
0.08
9
AS:
^^**s^S
0.05
2.0cast iron
CI:
I
I
alloy steel
ZH
1.7
-
J
J skew angle
for
see load capacity
for helical gears \
)
(pitch cylinder)
a n = 20°
a
P ~*
of
tooth root (q 142)
1.2 (guide value)
guide values see table
:
zone factor (see diagram)
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:
^shaft
q 152
q 153
pinion integral with shaft
1.2
pinion free to turn on shaft
...
from gear ratio i
and a specified
centre distance a
(seeq 113-114-129)
1
1.5
2
Tooth width ratios
Tooth- and bearing-quality
q 154
teeth smoothly cast or flame cut
6
b_
b
m
dt
...
10
teeth machined; bearings supported each
q 155
q 156
on
overhung
side
steel
construction
pinion
(6) ...
10... 15
teeth well machined; bearings supported
each side
in
gear casing
teeth precision machined;
q 157
or
each side and lubrication
<
«1
50 s-
overhung gearwheel
fully
m
n
:
:
:
:
,
:
v
:
1
:
2
:
20
...
40
good bearings
in
gear casing:
Q
18
...
25
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
£
0.7
$ 1.5
supported
Suffixes for
b
25
.
q 158
:
...
1
q 159
a
15
Q
29
MACHINE PARTS
24
Bevel gears
Bevel gears
Bevel gears, geometry
Equations q 86... q 88
are applicable and also:
cone angle
q 160
q 161
q 162
q 163
tan
(5-,
d:
sin I
cos I + u
I = 90°
tan 6
=> tan 61
sin I
cos I + 1/w
(I = 90° => tan 6 2
angle between
q 164
q 165
shafts
I
y
z
Only the axial and radial
forces acting on mesh
wheel 1 are shown
J
external pitch
cone distance
}
*e = 2
-sin
Y
6
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
dm
cos (5
^v1
Formulae q 92, q 95 ... q 100 are also applicable
back cone (suffix "e").
to the surface
of the
Bevel gears, design
The design
(suffix
referred
is
to
the
MID-POINT OF THE WIDTH
b
"m") with the values:
dm
-
169/170
Re
171/172
2-fl,
z
sin
2-T
<5
dm
continued on
For symbols see
Q
29, for suffixes
see
Q
23
Q
25
MACHINE PARTS
25
Bevel gears
Q
continued from
and
Axial
24
mesh
= Fmt -tan a sin 6
= Fmt -tan a cos <5
radial forces in
q 173
axial force
Fa
q 174
radial force
F,
Load capacity
•
•
of tooth root (approx. calculation)
Safety factor 5 F against fatigue failure of tooth root:
q 175
°F
_
c
b ef
-m
lim
^ST ^6
'
'
^FS
rel
T
'
^R
rel
T
'
^X
_
Y E' Y K' (^A K v K Fa
'
*
r
With the exception of YSJ the factors
substitute spur gears (index „v").
Y
,
Giving the approximate formula:
FmX
are determined for
...
^•y s-*WWY
q 176
Sfi
K -(KA .tfv)tf F|r
F
^ST (^6
"
*
0.85-6
Y FS
Substitute the
:
*
'
^
T ^R
rel
number
rel
"
T ^x)
'
aF
I
T1
2
of teeth of the
complementary spur
gear z v The graph for spur gears on page
applicable to bevel gears.
For all other data see q 142, q 143 and q 144.
.
Load capacity
S Fl
A" F p)
•
Q
21
is
then also
of tooth flank (approximate calculation)
Safety factor 5 H against pitting of tooth surface.
q 177
OH
Sh
•
(
ZL
~b
\
'
ZE
=
Yo.175 E
ZE
is
with
Giving the approx. formula:
2-
<mi^V
Ty cosd] M v +
Kh
0.85
q 180
'
2 H ^E Z e Z K
For metals the factor
q 179
Zy Zr)
•
'
-^X
KH $
=
~
«
*
"
"
V-^A
'
-^v
"
*
^H a
*
-^H
c
^H
min
f
simplified to:
E
2
=
•
y
E2
=1=1
s.q181
1
E
E,+E 2
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
supported pinion and wheel
1 .88 for fully supported and overhung
V
2.25 for overhung pinion and wheel
Z H 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.
For all other data see q 148 ... q 151.
A" Fp
1.65 for
fully
:
)
q 181
^
—
.
*
V:^
q 178
lim
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
Worm
Worm
27
gearing
gearing, geometry
(Cylindricalworm gear-
normal module
ing,
BS
axial section,
angle between
I=
in
2519,
shafts
90°).
Drive
worm
All
the
ing
on the teeth
mesh
forces
are
the three
F and F
r
x
In
actin
shown by
arrows Fa
,
.
the example:
z-i
=
2,
right-hand
helix.
Gear tooth
ratio
and transmission
q 185
module
q 186
pitch
q 187
mean diameter
(free to
rf
1
...
88
Worm
|
wheel, suffix 2
= p 2 = d2 nl z 2
=
m1
2-r m1
chose, for normal values see DIN 3976)
form factor
q 189
centre helix angle
q 190
pitch diameter
q 191
q 192
addendum
dedendum
q 193
tip
q 194
outside diameter
q 195
tip
q 196
tooth width
b^
q 197
root diameter
rffi
q = d my
tany
=
=
m
a1
hu
clearance factor
Profile offset factor
otherwise x =
mm
»
j
_ ii
0.
Q
+0
= m(1
cf = (0.167
4n
=
groove radius
For symbols see
lm
^
<*mi
.
.
.
m
=
d2
/i
centre distance
'>
suffix
m-n
q 188
q 198
as q 86
ratio
Worm,
•
22
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
= a - d a2 /2
rk
W
" d22
b2
«
= d m1 - 2 /z f1 rff2 =
a = (</ m1 + d 2 )/2 +
0,9-d m1
f/2
— ^
xm
-2m
^f2
1
>
x for check of a pre-set centre distance,
continued on Q 28
29, for suffixes
see
Q
23
>
MACHINE PARTS
Worm
28
gearing
Q
continued from
Worm
27
gearing, design (worm driving)
Worm
q 199
peripheral force
Worm
*"»
"ml
q 200
q 201
q 202
axial force
Fn
*.i
*
tan (y+p)
cosptana p
Fr1
radial force
1
,1
(y+p)
sin
rubbing speed
wheel
Fx2
=
Fm
Fa2
=
^t1
=
*"r2
- Ft
i
0>!
V9
cos y m
2
Efficiency
Worm
tan ym
q 203
/
Worm
driving
tan (ym + p)
(7m
Coefficient of friction (typical values)
vg
worm
«
teeth hardened
and ground
teeth tempered
and machine cut
wheel driving
= tan (ym - p) / tan yn
77'
<?)==>
=
\i
1
self-locking!
tan
p
m/s
~
vg
10 m/s
0.04
0.02
0.08
»0.05
worm
worm
For calculation of
Calculation of module
Load capacity
are combined
q 204
Fx2 =
Cb
2
-i£
in
and flanks and temperature rise
the approximate formula:
where b 2 ~0.8 d m1
p 2 = m-n.
•
P2
;
;
'
q
*
2 T2 /d2 = 2 T2 /(m z 2 )
10 for i = 10,20, 40
9
»
17 for
Fx2 =
.8-7-2
CpernVtf
m
of teeth root
q 205
q 206
shaft deflection see P 12
22
•
•
•
= 80, self-locking
1
for normal, naturally-cooled worm gears
ground steel, worm wheel of bronze):
Assumed values
hardened and
v9
Cperm
When
cooling
is
adequate
m
I
|
N
For
all
symbols see
Q
-
8 N
29, for suffixes
I
1
|
8
2
this value
Cnerm
q 207
s~ 1
mm
I
2
I
5
I
can be used for
mm ~ 2
see
Q
10
3.5
23
all
I
(worm
15
2.4
I
20
2.2
speeds:
MACHINE PARTS
29
Gears, Gearings
Notation for
Q
18
...
Q
28 (suffixes see
Q
23)
centre distance (a d standard centre distance)
facewidth
eH effective facewidth (root /flank) for bevel gears
:
b ef I
fe
:
addendum
addendum
dedendum
change
of cutting tool
of reference profile
of reference profile
of addendum factor
(e. g.
DIN 867)
no. of teeth
z
Fx
*A
K Fa /
K Ha /
*e
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.)
transverse load- /face load factor (root stress)
Kf$
K H a transverse load- /face load factor (contact stress)
total pitch cone length (bevel gears)
mean pitch cone length (bevel gears)
:
:
T
torque
tooth form factor (force applied at the
tooth form factor for external gearing
»s
tip of
a tooth)
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
:
/3 b
:
g
p a0
<7FM
normal pressure angle
:
:
:
:
m
CTHMm
:
:
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
DIN 3971
bevel gears and gearing
definitions for
DIN 3975
straight worm gearing
]
I
BS 2519
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
tables, work spindles with bearings) are designed to give high
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
approximately 0.03 mm. For spindle deflection refer to formula P13 and for cutting forces see r 4.
of chip formation)
Cutting
is
(main
drives
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
The progressive
ratio
cp
n1
<pk-i
speeds
for the
n k with k number
n-\
output speeds
r
2
are calculated by:
<P
and the preferred series
is
Standardized progressive
selected.
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
Cutting
Pc
power
General
Fc
Pc =
*?mech
r
4
Cutting force
Fc =
-
'
Drilling
Fc {D +
v
2
7?electr
'
d)
^mech
1 - mC
'
b, h,
ze
(/c
-mc
1
c
see Z
°«-
5
I
o = 7
O
1
.
2e
7)
0Q
w
O
O
<N
•
mm
^•*c,,*( 4)
i
Tables of values for K,
n n
*?electr
I
i-
NO
ffl'c
"5 *C
6
E
CO
<o
B-
fc>|CM
o x
^ ^
X
x
t-
-CO
t-
x
\i
^
I
J^&
en
»-
—
E S o
=
r
"°
£
in
o
}no-dn pue jno-u/v\op
BujHjuj
8UB|d
For explanation or symbols refer to R 5
PU3
1
PRODUCTION ENGINEERING
Machining
m
o
o
d
CD
c o
Wo
TD
O)
c
jz
,
1
eg
o
o
d
CO
Q)
O
5°
Q.CVJ
x:
o
ai
§H*
1
co
CO
Qlcsj
^m
m
oo en en co
co
in <d cd
*
C
CO
,
O)
1
O
"* CD
r»
CO -^ CO CD CD
<J>
CD
>
o
o
en in -* rt- CVJ CO
CD
*
CO t00 o
ico * •* m
-5J"
cvj
-<i-
<*•
cvj
o>o
=3
o ,_
OC
h- in oo c\j -<^
co cd
co
O
c
o
Tlcvj
d
CM
co
CO CD 00
Om
in m
-<3-
9ZIS UIBJ6
9|qei
2
%
-4
<*
o
o
if
o
in co
co cvj
CO CO CO cvj CVJ
2
O
O O
° c
CO
1
.2
O
2
Q>;z
"D
'»
CD
-
~C
o
C
azis uibj6
Q)
"•*-
o
t- in
't c\i
in -<fr <* co co co
"O
For explanation or symbols refer to R 5
*
cvj
o
O C
£:
Cvi
CO O) CD
OO CD
^J- CO CO CO Cvj CM*
8 o
d 2
o
5
o
d
C
o
in evi oo co
Tt co co cvi cvi
CO
CVJ
cvj
*•
I
aiqei
PRODUCTION ENGINEERING
FU
Machining
Feed drives
Feeds
in
geometrical progression with progressive ratio
q>
Feed
= 1.12- 1.25-1.4-1.6-2.0.
rate
Method
Feed
Turning, longitudinal
15
r
(external
and
internal)
u = n-sz -z s
Drilling
16
r
Notes
rate
u = n-s
for twist drills
Ze = 2 S = 2
sz = 0.5 5
u = V
Planing, shaping
17
r
Milling, plane milling
18
r
and end
Cutting times
r
= n-s z -z s
u
milling
tc
where
19
f,
=
/
+
/'
calculating the cycle and machining times for each workand infeed travels and also the lengths covered during non-cutting motions, divided by the corresponding
When
piece, the feed
speeds, must be taken into account.
Feed power P v
r20
Feed power
r21
Feed force
r22
Friction force
u(FR +
rV
^mech
Fv «
0.2
=
rn b
^R
*
Fv
)
V alectr
Fc
;
(Fc from
r
4)
-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
= uE
-m b
\
For explanation or symbols refer to R 5
L-
(fi.g + !^.\
f
b
/
'/mech
H
electr
PRODUCTION ENGINEERING
Machining
Explanation of symbols
used on pages R
infeed
/
K
1
.
R 4
see table 2
effective width
speed
5 S /1.
b w = B s /3
rough plane grind. b w =
smooth grinding
minimum output speed
maximum output speed
milling width
:
.
effective grain distance
:
width of chip
B2
.
power
cutting
milling width
measured
feed power
from tool centre
feed
width of the buffing disk
diameter of pre-drilled hole
acceleration time
feed per cutting edge
working-part - outside,
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
in
slenderness
gravitational acceleration
£s
chip thickness
*?eiectr
number
j:
of output
speeds
basic cutting force related
^mech
x
:
H
to area
method
:
cutting travel
(p
work
<p s
travel
overrun travel
:
o
factor
at
:
:
'
both ends
of cutting
:
1
ratio (e s
number
of cutting
edges
= a Is)
electrical efficiency
mechanical efficiency
setting angle
friction coefficient,
drill tip
see Z 7
angle
progressive ratio
incident angle for milling,
cutting or grinding resp.
with feed rate u
per tool
edges
action
HM
Ihss
carbide
tip
high-speed
tip
PRODUCTION ENGINEERING
Metalworking
Cold working of sheet
Deep drawing
Initial
blank diameter
w
°-
r24
Am
D
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
Am
=
±[2nd,r z + 4(K-2)r z
Example: (assume
D
r26
= }Jd 4 2 + d 6 2
rs
-d 5
2
=
+
rz
/L
=
f
r)
fl
2
zd
and 2nd stages
r27
ft
r28
I
r29
-7
«1
02
'<*1
0.001
FD1 =nd^skim ^q>^
ftmax
=
r31
r32
r33
<p<i
=
In
FD2 =
2
\/0-6^i -0.4
(p2
without
37"
=^00 + 0.1-^-0.001
*7ei
r30
1
-<*,!•
cf,
max = 0100 + 0-1
*"
5
2nd stage
1st stage
—
^d 4 2
(27td 4 + 8rs )r s +
4d^
+ 2nr(di + d 4 )+4nr
1st
m =
~2~ + red 2 s ktm2<P2-rf-
=
In
I
V0-6^2
2
_
inter-
-0.4'
frl
+
fc f2
mediate'
with
annealing
"fm2 -
g)2
continued on
R7
PRODUCTION ENGINEERING
R
Metalworking
continued from
The work
7
R6
volume and the
w, related to the
yield strength k f is
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!&[<*Bottom tearing occurs
r35
r36
Maximum drawing
Material
_ F D2
conditions
(3
Specification to
U.S.A.
619-75
steel
620-75
]~
o>
-*
*
283 Gr. C
Stainless steel
(18%
Cr;
9%
Ni)
SAE 3310
AA 6004
AIMgSi
soft
soft
/?100
BS Grade
366-79
Carbon
1490
0.1
FB2
without] with
intermediate
annealing
& max
-
S20
2.0
1.2
1.8
600
LM
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
-
1.7
321
5
*
:
'•
:
:
:
:
:
!
;
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
N/mm 2
390
360
350
340
410
15
1
970
+
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 = —
:
f]
and R m
- Sheet metal
^
H
CO
<
+
if
_ F D1 +0.1 FB1
„
i )2
PRODUCTION ENGINEERING
8
Metalworking
Extrusion
r37
Remodel
F
force, pressure force
r38
Remodel work
r39
Mean
strength for deformation
=
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
4
r43
r/
F
^
d?
"o
-'o
= 0.7
..
.
0.8
V
=
T]
= 0.6
F
Jh
{d Q
.
.
.
2
-d
logarithmic ratio of deformation
without intermediate annealing
AI99.5
methoa\
AIMgSi
soft
3.9
3.0
backward
4.5
40
1.2
:
r]
<p
F
= 0.5
low
:
:
:
alloyed
1.2
0.9
0.8
0.7
1.1
1.1
0.95
0.8
used area
logarithmic deformation ratio
deformation efficiency
V remodelled volume
w volume related remodel-work according to curves Z 20
Ah: depth of stroke
7/ F
0.6
Amax
alloyed
forward
<pA
fdo 2 h
)
C<0.1% C<0.15% 00.15%
1.4
:
do 2
d^l?
Steel
"^material
A
,
2
0.7
Maximum
=ln
ELECTRICAL ENGINEERING
General terms
The most important electrical quantities
and their units. - Basic rules
Note regarding capital and small
letters used as symbols
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
Electrical
Examples: formulae
s 8, s 9, s 13
Exceptions:
PFeio
Electrical
/, a),
i,
v,
work W
work W
W
is equivalent to mechanical work
as explained
Energy conversion, however, is subject to losses.
Units: J; Ws (wattsecond); kW h; MW h; see also A 3, A 5)
Electrical
on
M
1.
1
Further the following
plained on S1 and S2:
relation
Ws
W
s 2
power P
Electrical power P
=
1
applies,
=
Joule
using
J
1
=
1
quantities
N
m
ex-
2
=
I-V-t =
¥-t
=
I2
Rt
Electrical
explained on
M
power P, as
is equivalent to mechanical
Energy conversion, however, is subject to
1.
losses.
Units:
W
(Watt);
MW
kW;
;
see also
A
3,
A
5)
1W
Further the following
plained on S1 and S2:
relation
applies,
see L
1
Period T
see L
1
Angular frequency
co,
_
P
s 3
Frequency/
=
using
1
±s
= 1*LS1
quantities
Yl
s
ex-
2d
(T=1//)
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
s
General terms
continued from S
2
1
Current density J
S 4
-i
>
Applicable only where
over cross section A.
Units: A/m 2 A/mm 2
distribution
of
current
/
uniform
is
;
V
Potential difference
v-i
s 5
Units: V
(volt);
Where
mV; kV
1 A through a conductor converts energy
W, the voltage across this conductor is 1 V.
a direct current of
a rate of
1
IV
-
1$
1^
-
1AO
-
-
at
,${.
Resistance R
R -
s 6
Units:
Q
Where
(ohm); kQ;
a voltage of
the flow through
Conductance G
Conductance G
is
(Ohm's law)
j
MQ
it,
1
V across a conductor causes a current
is 1
Q.
= J_V =
1 A
1
resistance
W
=
A2
1
_W_» =
s
A2
of
1
A
to
_Nm_
2
s A
the reciprocal of resistance R.
G
=
1/Q
=
s 7
MR
Unit: 1/Q
Quantity of electricity, charge
s 8
Mho]
[1
Q
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
electricity
Q stored
of a capacitor
in
it
s 10
C
Units: F
Where
of
1
capacitance
F =
C.
1
Magnetic
C
1
to
be charged to a voltage
is 1 F.
=
<
A2
As
1
_
s
tf_£ =
1
W
'"
V
V
s
V
a capacitor requires a charge of
1
it:
pF
(farad); jxF; nF;
V, its
the ratio of quantity of
is
and voltage V across
J
A2
1
s
2
Nm
flux
11
<P
=
X
jj
u df
f
(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.
1
Wb (weber) = Vs = 10 8 M (maxwell)
Wb is the magnetic flux which, linking a circuit
it
a voltage of
Units:
1
V as
it
is
reduced to zero
Magnetic induction (flux density) B
The magnetic induction in a cross section A
B
s 12
of 1 turn, induces
uniform rate in 1 s.
at a
is:
&
~
A
Here A is the cross-sectional area traversed
by the homogeneous magnetic flux <P.
Units: T
(tesla); nT;
1T =
Where
a
1
nT;
;
homogeneous magnetic
ses an area of
1
m2
,
perpendicularly
V s/m 2 G (gauss)
= 10- 4 -Vs =
V-f-
in
its
[
10 4 G
flux of
1
=
Wb
magnetic induction
1Q
4_M,
perpendicularly traver-
is 1
T.
continued on S 4
1
ELECTRICAL ENGINEERING
General terms
continued from S 3
Inductance L
Ny
L =
s 13
N
=
1
(seesl)
-j
Here / is the current flowing through a
the magnetic flux linked with this coil.
Units: H (henry);
coil of
N
turns and
<P
mH
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.
Vs.
=
1
1H =
A
A
1
iM
Magnetic
S
field
strength
H
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
s
F\ in
the
16
f-th
Fj
here
l\
is
section of a magnetic circuit:
-
Hv
-
F
2>i
s 17
i
Reluctance S
of a
-
homogeneous
section of a magnetic circuit:
„
s
S =
18
Units: 1/H; = A/V
s;
symbols see S 16
/
(Ampere Turn/Wb)
equivalent to
Ohm's law
-£-
<P
for
l\
the length of this section.
\
\
for
magnetic circuits/
ELECTRICAL ENGINEERING
Electric circuits
Basic properties of electric circuits
s 20
s 21
Directions of currents, voltages, arrows representing them
Direction of the current and of arrows
generator --» +
+-*representing positive currents in
load
I
I
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
where function
of element
determine
(generator or
load) as well
as polarity is
directions
as stated
s
23
known
s
24
unknown
arrows
of
in a
negative
value, direction with respect
to arrow of current
or voltage is
positive
above
at
random
—
—
equal
opposite
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
=
Resistance R
V
of
El
(see also s 6)
R
conductor
/
A
Y-A
Resistance R of conductor
(in degrees centigrade)
S
R
27
=
Vl-t-7]
a
:
Y
g
=
temperature
mass m
c-m-A&
temperature coefficient
(s.
conductivity
(s.
:
resitivity
c
:
specific heat
rj
:
efficiency
&
/?2o[l+a(0-2O°C)]
Electric heating of a
s 28
at
(s.
(s.
Z1
...
Z
Z
Z
21)
21)
21)
Z4 and 02)
A&:
t
/?2o
:
:
temperature change
time
resistance at
# = 20°C
continued on S 6
ELECTRICAL ENGINEERING
Resistor combinations
continued from S 5
Law
The algebraic sum
1st Kirchhoff
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
total current and partial
currents are inversely proportional to
their respective resistances.
in
parallel,
,. J
1
/1
'
30
.
*
j
/z
.
*
_
,
"
/3
1
1
.
R
1
.
R,
'
1
.
R2
R3
Current division
Partial currents of 2 resistors
ed
s 31
h
G
+
:
G2
Ri
2nd Kirchhoff Law
The algebraic sum of all
a closed mesh (loop) is
s
connect-
in parallel:
IV
32
=
voltages around
zero.
0.
Here voltages traversed in accordance
with (opposite to) direction of arrow are
considered positive (negative).
Ratio of voltages
Where
in
ages
s
33
several resistors are connected
the ratio of partial voltequal to the ratio of the re-
series,
is
spective resistances.
V,
V2
V3 =
:
:
^
:
R2
:
r,i
/?,;
R3
Voltage divider
Partial
voltages
connected
s
34
v,
=
<^2
v
Gi
across
2
resistors
in series:
,
= TV-
#1
R2
rS5n
ELECTRICAL ENGINEERING
Combination
of resistances
Series connection
R s (according
Total resistance
to s 26)
generally:
s
35
s
36
Rs
=
R2 + R3 +
R) +
.
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;
for 3
for 2
for
39
R,R 2
+ R2
Rp
/?!
s
R-\
P
/?1 /?2
G, +
« equal
"*"
R] R3
R
G2
G-j
-
£
1
1
1
40
R 2 /?3
"2 "3
"*"
-^^p
resistances
several resistances
s
ft*
nG
+ G2 + G3
Multiple connection
A
multiple connection of several
known
resistances
is
subdivided
and series connections, proceeding outwards. These
are separately converted as to be conveniently combined again,
into parallel
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^
G-| + G2 + G3
G-\
A3 + R 2 R 3
R2R3
R-\
y
=
V
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
• The remaining
The complete
voltage sources are short-circuited.
current sources are open-circuited.
solution
the
is
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
V
r
x
= a
r
-V + a -V +... + a v -V v +
1
^xaO+^xal
^xbO+^xbl
Computation
S46
S47
\X
|^
,
|P^
xbn
where
/
.
.
.
Vy
=
1^
=0, with
0,
with
Vq *
0,
/
0,
q
4=
and
and
/
V
V +b -I
VX a0+^xa1+^xbO
a -V + a :
:
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
R< Ro
S51
;
of the partial solut.
S48
S49
,
1
Required
(Cf.
voltage
V, -
s 48)
Explanations
cf.
S9
(£
+
£
7
>
.
^^
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
»
Produce
to
compute R and V
Q|*
:
x
t
Remove
is
/?i
V\ is
Note:
the branch AA' from the network.
the resistance between A and A'.
the voltage at AA'.
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
Computation
i//?i
+
1//?
Computation
of /S(
Ri
of
R
ft
4
rf
Equivalent circuit
"
*.
A
-$ + £ + <
Hence: V
s53
Using
V
s52:Kx = (-^ +
,
V
-
/?
1
1
1/^ +
4c «i
cf.
1
/o)
1//?!
+
1/^2
UR 2 + MR
of
result
S8, s51.
Explanations:
ap-ay
'
2
'
coefficients
voltages
of
currents
voltage source
current source
with internal
resistance
which are determined by the
resistors in the network
R\
ft
=
{
ELECTRICAL ENGINEERING
10
Resistor combinations
Transformation of a
delta to a star-circuit
^10 ^20 + ^10 ^30 + ^20 ^30
^30
•
S
s
54
i? 12
-
'
*10 *20 +
55
and vice versa
/?irj-
R,o-Ri
'
R30 +
10
R 20 #30
^20
R23' R-\2
R23 + R^2 + R^3
30
R23' R-\3
#23 + #12 + ^13
^10 ^20 + ^10 ^30 + R20 R30
56
l
R-\o
R-\3
120
'
s
+ R)2 +
/?23
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
-
RM
=
s 59
1
£&
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
:
riation
resistor,
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
Electrical
work
Wc
Capacitors connected
Where capacitors
C
=
Where capacitors
field
C
in parallel,
C
increases.
+
C3
in
C3
series
are added
the total capacitance
67
i<
in parallel
C2
Ci +
Capacitors connected
s
an electric
are added
the total capacitance
66
in
cv
Wc
65
stored
—
8)
C
c2
c3
C3
C2
1
in series,
decreases.
Capacitance of two coaxial cylinders
V o
s 68
s
69
2
£r
£
'•
:
A
:
a
:
r-[
:
r2
:
/
:
•
Jt
•
£o
-r
r2
relative permittivity
absolute permittivity
plate area (one side)
thickness of dielectric
radius of inner cylinder
radius of outer cylinder
length of cylinders
Z 22)
As/(Vm)
(see
£ Q = 8.85 x 10" 12
ELECTRICAL ENGINEERING
13
Electro-magnetic rules
s
Deflection of a magnetic needle
70
The N-pole of a magnetic needle is
and repelled by a magnetic N-pole.
attracted by a magnetic S-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
the current through the
of
the direction of its
axial motion would indicate the direction of the lines
of magnetic flux through the coil.
coil,
Movable conductors and
73
^
—
"Z-
~
N
coils
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
J^l
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
t^. s
**c^-i
J^~f
Machines
75
Hand Rule (generator)
Where the thumb points in
Right
the direction of the
and the middle finger
mag-
the direction of
motion, the index finger indicates the direction of
current flow.
netic flux
S 76
Hand Rule (motor)
Where the thumb points
in
Left
the direction of the magin the direction of
current flow, the middle finger indicates the direction of motion.
netic flux
N
t=J-S
other.
S
N
in
and the index finger
r?i
ELECTRICAL ENGINEERING
Magnetic
field
S 14
Quantities of magnetic circuits
Magnetic
s
flux
NI
77
(see also s 11)
S
Magnetic induction
(flux density)
R
s 78
H
T
B
H
li
(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
s
80
strength
field
//
-»-
=
(Magnetising force)
//
=
(see also s 14)
t*rt*o
Magnetomotive force
/
n
F
s 81
;
NI
-
Magnetomotive force
s
2
=
(see also s 15)
F;
F\
82
(see also s 16)
Reluctance S
s
S
83
-^
=
I
=
(see also s 18)
H H A
r
Energy
s
Wm stored in a magnetic field
W m = 1 ^/0 = ±LI
2
85
Leakage
flux
L
Part of the total
is
thus
ful flux
s
86
lost
<V Hence
coefficient
flux <P leaks through the air and
desired effect. &\_ is related to the use-
magnetic
for the
the leakage
is:
j
total flux
_
<p u
For symbols see S 18
useful flux
(1.15... 1.25)
ELECTRICAL ENGINEERING
Magnetic
15
field
The magnetic
field
and
forces
its
Force F m acting between magnetic poles
In the direction of the magnetic flux
Fm
a tensile force
occurs:
B2 A
1
fm
2"
'
Mo
Forces F\ acting on a current-carrying conductor
A conductor carrying a current / encounters
force
over
F|
magnetic
s
its
length
/
perpendicular
a transverse
the lines of
to
flux:
89
Bl
=
F,
I
When
applied to the armature of a DCmachine, the moment is:
s
90
Af
j
1
^-0
2
=
I
Jt
£z
a
/
(p
Induced voltage
(induction law)
V\
N turns and resistance
magnetic flux <P that varies with time,
an open-circuit voltage
Where
s
a coil of
92
V,
:
flux per pole
:
=
N -ry1
R\
is
threaded by a
(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
conductor loop
erator armature
in magnetic field
(p
V
s 93
{
= (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
Sense
of
to alternating-current circuit
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\-
\
Peak values
Current
(see also s
=
360° =
=
<?2
1)
and voltage
i
<Pz
<Pi
v of
an alternating
current vary
with
time
t,
periodically
usually
sinusoidally.
maximum
are
s
The
and v
angular
co
= 2
ed
in
,-r
/ the angle cover-
time
t is:
98
a =
Hence
3
i
peak values. At
frequency
called
an
s
values
at this
to t
=
2nft
time
99
the current
is
=
100
the voltage
is
= 6-sin
i- sin (cot)
(cot)
Root-mean-square (rms) values
These are used for practical
i
•
sin
calculations
indicated by meters
generally
s 101
/
=
/eff
-
102
V
=
veff
-Vi/w
s 103
Vt/'-«
a
v-sin a
and
for sine
7
=
v
=
With these values the relation P = V I
ternating current, if cos cp = 1 (see s 115).
are
usually
waves
=
/eff
vM
V?
=
^
also applies for
al-
continued on S 17
/
ELECTRICAL ENGINEERING
17
Alternating current
continued from S 16
Phase shift, phase angle
Where different kinds
q>
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
Q
factor,
The
s
Q
damping
factor tan
(p)
loss angle S
factor of a circuit has been defined by:
Q
106
Here
6,
-
s\n(o)t
w
is
^**
=
the peak value of the energy stored
loss energy dissipated in one period.
in
the circuit
and VVyp the
The reciprocal
s
Q
of
106
factor
called
is
damping
tan <5= 1/(2
For a choke
combination
(s
(s
125 and s 128)
126 and s 129)
factor
((5 is
and
this
for a
the loss angle)
capacitor-resistor
definition
results
in
the
simple relations:
s 107
Q=
s 108
6 = 90° -
ltan<pl
I
tan 6
\q>\
s 109
For formulae regarding tan
Formulae
s
138 and s 139
not so simple.
MQ
=
=
=
= 1/ltan
IVw /V b
I
cp
7W /
b
l
l
<p\
(for series
connection)
connection)
(for parallel
see S 19 and S20.
applicable for resonant circuits are
B
ELECTRICAL ENGINEERING
18
Alternating current
Basic equations for single phase alternating current
Impedance
Admittance
Z
Y
see
s 110
=
1/Z
s 111
Voltage across impedance
V
=
IZ
s 112
Current through impedance
/
s 113
Apparent power
Reactance
S
=
VI
s 114
X
=
Zsin<p
s 115
Active power
P
=
s 116
Reactive power
Q
=
VI -cos q>
VI -sin cp
S
117
s 118
Power
cos
factor
Alternating magnetic
1
flux in a coil
J
cp
^
S19 and S20
V
z
P_ =
)2
^/p^+
=
=
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
for
for
(fi
=
Anx
10" 7
vacuum, gases, fluids and most solids:
magnetic materials take \i from Z23
t
number
of parallel paths through winding
length of magnetic circuit
number of turns of coil
number of pole pairs
number
of
conductors
resistance
series
in
parallel
inductance
equivalent circuit
of
in
parallel
choke
Vs/Am)
n,
c 19
W
ELECTRICAL ENGINEERING
.
Alternating current
Components,
and
series-
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
C
^o
<B
°-
co
a
—
CO
^
to
^
^
o
CO
>«
.a
^
JO
a>
V** I**-/
*
^
.^
^/
2±
Ix,
i^
1
i.
H
1
%
lit-
v° =
r
S
4I
i.
*r
:».
^|,
—
<D
c
c
°
>
O i
E
o
o
No.
V
OJ
I
I
D
S
_
o
c
CO
CO
§!
"I
120
3
S?
~>
> >
cd
S
8
Q.
Jo
o
S 121
Oc-E.*
A
r-
co
CD
co
C
a)
-l°
'3
-i°
'a
O)
=5
V-
S
o
22
si?
3
a>
*•
£
Q)
I s
il||l
sii>3
1-,
1
^
^nt^
-C
—
-
-|
*ft*
•U>
co
n
1
-A
-1
1
O
a.
8
>*
-°
^r
\V
^
°*
8,
J5
i^
rsT
-A
c
a>
c
o
(0°O
CO
"O O)
i-.
CO
CO
8-
o
--.
2
E
V
ov
o
o
CO
&8
CO
?J
&
^
8-
8-
o
8-
o
II
CO
o
o
CD
ov
D)
C
- E
2
o
°o
o
o
_0>
Q.
3
C
C
O
o
a:
ii
N
CD
CO
CO
CD
+
ii
Q.
o
""J
i
ii
<D
CNJ
r«
'"la
I^J
-J
o
Ik
3
—c
w
OJ
>
o
"
s€|i
5
l»
124
CO
<0
CD
5r
CD
o
> o
°
3 CD
"O '^
—C CD
*-
co
!
S125
I
^
ELECTRICAL ENGINEERING
20
Alternating current
continued from S 19
1
^
8
re
oc
c
o
T3
CO
0)
—
o
—<D
C
>.
i3
^
J5
re
i_
a
©
£*
^
"o
to
a>
re|
<» ° „> c
^ „ 9 o
o
re *£
sz —
CO
c
g>
^o
<K
ro
.E
E
re
5 $
Jrexo
X3
re
.:=
g
|S.2
*~
^
re
° 852
-r,
V
*'5
_
o
_
ore
re
ra
©
> a
£ ore
o.
w
£
—
."K
re
c 5 &
re 2 3
OT
0)
S126
^
re
a.
.re
+
a
° c © c
<d
.E
c
S128
n
1
o
> $ re
.1
*1
o 3
^ o
eg
'5
ro
O
.Q
i
-o
*-
Q.
>
S127
re
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
resonance
at
Wc
I
=
Io
k
resonance
condition
co r
a) r
resonant
frequency
2
LH C =
2ji
where
(o r
Lc
C
wfLpC
1
yL R C
2k V l p c
,
current at
It
Vb
(p
Q
factor
loss angle 6
from
Qr =
tan
<5
KL
=
=
-
Rr
R
Vc
(o r
J_ =
Gr
=
CR R
R*
co L R
c_
/r
resonant
period
s 141
Tank
2it
at
=
QP
7C =
/b
=
(O r
tan St
RP
CR P
(D r
Lf
J_
0p
r
wavelength
s 140
1
frequency/ = /r resonance occurs
line
resonance
at
=
300-10 6
(O r
CR
r
m
frS
V^r C
2ji
V^p C
Circuit
resonant circuit has its maximum impedance Z max at
resonant frequency. Therefore it acts as a rejector for
currents of this frequency.
A
parallel
its
142
RR C
for
symbols see S 18
and current
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
—
Vr
e
A/vz
z
h©
-4
select auxiliary resistor
R such
unknown capacitance
unknown inductance
unknown resistance of
that
I
|
coil or
6
V R « Vz
I
<5
X
R2
:
&4
I
S17
known resistances
loss angle, see
.
a-
capacitor
calibrated adjustable capacitance
unknown impedance (inductive or capacitive)
ELECTRICAL ENGINEERING
23
Alternating current
Inductance L from impedance and resistance
Calculating
s
150
/. from impedance and resistance
Pass an alternating current (J = I/A « 3 A/mm 2 through a
)
measure the terminal voltage
s 151
Z
impedance
Calculating
s
/.
current
=
L-
s 152
V,
/,
active
power
resistance
±^*-W
for a toroidal coil
153
2:r
L
Calculating
for a
square
r.
coil
armatures must be circular
D
s 154
S
s
155
156
<
>
£
inductance
1.05
1
1.05
1
3
values
m
V
2
^/V
m
become
HH
\u
V (-)
\u
V
^H
'
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
157
circumference of
ratio a
s
158
armature winding
mean length of armature winding
number of turns
degree
:
coil
(/
m =
+
/
cross section
b
of loosing of turns
(/3
=
ab
.
jra)
coil
P:
and
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
©*«"
-
Low frequency
Assuming
2VA^
4d
+ «)
Vf[
coils
that
D
s 161
-
s 162
V -
975^
s 163
a
i.±V?
and
1
Calculation of
number
=
u,
then
of turns \ of a coil
From cross section
s 164
TV
« ab
Ar
« RA
From resistance
s 165
Using reference
Position
coil
unknown
coil
of
Nx
turns and reference coil of
turns at short distance on closed
iron core. Magnetize core by alternating voltage Ve applied to
magnetizing
voltages
coil
Ne
Vx and V
.
Measure
using high
impedance voltmeter. Then
s 166
«.£
For explanation of symbols see S 23
separation gap
N
lrcn core
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
Hc
The coercive force
reduce the
Hc
has to be applied to
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
addition for punching ridges etc.
+
x)
(0.1 ... 1.0)
ELECTRICAL ENGINEERING
26
Alternating current
Choke
Choke
coil
used
in
to a value
Vv
impedance
of
It
is
coil
used as a dropping impedance
an ac
circuit to
reduce the
S 170
S 171
voltage
line
minimum
for a restitive load with
choke
Z D -y]R* 2 +
(coL R
total circut
z
/? v )
-
-
-
V down
losses.
V(Rr +
2
*-±V(l-VPl
f
+ (a >L R
f
-(^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„ =
and repeat procedure,
if
VRr
Z
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).
case of unfavourable results repeat procedure. Determine
resistance of choke according to s 26.
In
Choke
The
coil of
constant inductance with iron core
core essentially serves
for guiding the magnetic flux and
should incorporate as many single
air gaps 6^ as possible. These
should be filled with insulating
layers and should not exceed 1 cm
iron
in length.
The m.m.f. required
inductance
The
gaps
to
magnetize the core is neglected.
Peak values of H and B are used
for calculations.
air
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
current-depending variation
relative
I^Rtot
--^r
.
—
1
1
.
(
^R
A
—
fe
=
H Fe
gL
l
fi fe
fe
ti
d
+
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
175
s
176
effective
A F e=
cross section
of core
with
VK
VLe „'
VLe „
/ ef(
-
I eU
2ji/Lr
number
s 177
V
of turns
of air
s 179
gap
length
total
of air
s 180
gap
single
s 181
Ai= ab
by
cm
of wire
Choke
+
s
[A L= ab +
5{a + b)6A
=
nL B -5N 2
<
d'/n
1
cm
(5,
to
= din
<
2
n
n
{a
+
b)
cm
1
Use next standard values for
d a including insulation
d,
Aw
l
b)
u
cross section
of winding
length of limb
(a
abnN
=
6
+ 5
'•^
diameter
s 182
^Lefl
.
4A4f8 Fe A Fe
cross section
s 178
take AfQ from standards
or determine a and b by
\A Fe = 0.9 a b ~ A F e
I
=
1.12 da
2
N
%
determined from dimensions of core
section and
Aw
current-depending inductance
This type of choke employs an iron core without an air gap.
is used only for special purposes, e. g. as a magnetic amplifier.
coil of
power
coefficient of
»
cm 4 /VA
0.24
sb 0. 15
for
k
c
|a o|
choke
chokes
core section see S 26
/VA for oil chokes
increase
values by 75%
core section
m
for air-cooled
4
preliminary current density for air-cooled choke
2
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
It
flFe
t0
be
J'
=
2
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
A/mm^
)
Z
23
N
ELECTRICAL ENGINEERING
28
Alternating current
Transformer
Designation of windings
distinction by
function
nominal voltages
winding with
higher
|
input
lower
low-end
winding
183
s 184
rated
primary (index
|
Nominal values (index
s
power
nominal transformation ratio
transfer)
output
winding
nominal voltage
high-end
in circuit
power
(direction of
secondary (index
winding
1)
|
2)
n)
5N
1
=
=
V 1N
^1 N
-/ 1N
=
^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
transformer at rated currents. P-|«
total
power dissipation and
ing operational
s
186
"
^1 K
The values measured are used
is
required for calculat-
efficiency.
^Cu N
for calculating the relative short-
voltage v^, which, for bigger transformers,
indicated on the name plate:
= 100(V 1K /V 1N )%.
circuit
s
187
The following
s 188
may be determined
quantities
diagram:
^Cu = ^r/^1 N
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-
mer
s 189
ratio of
Vo = i-V
\
i
=
1
U
(index
=
hn
_/
cos
;
i'«
simplified
Rv
L
Vrv
Vl
'
I
1
>
of
V£
= 4%)
s 191
K 1K -(cos<p 1K
Jl
'):
« V 1K -COS(g?iK-<P2 )-/2/^2N
Secondary voltage V2
V2 = V, - AV
VR /Vi
vector diagram
Vi
AV =
using the vector
simplified
1^.
s 190
always
equivalent circuit
Load-dependent variation AV
(approximation for
=
<p 1K
is
cos<p2 + sin<p 1K -sin<p2)
/ 2 //2n
V9 '/i
J
ELECTRICAL ENGINEERING
30
Three-phase current
Basic connections
Star
VD
s 192
^
uW
^M?
*—
1=1
s 193
\v
.
v,
Ph
^Y
1
f
Delta
S 194
V -
s 195
/
V,P h
1
=
V3
/ Bhh
P
Measuring three-phase power
Load balanced
with neutral point
(star
connected)
—•-
from
mains
rr\
v
conn< sction
without neutral point
(delta connected)
-*- I
I
L\
to
from
.
L2
load
mains
^s5i_l
L3
196
total
to
'° ad
3
{c
power
h
.
JLJl
"1
N
s
ȣi
(w)
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
,
L2 L3
,
:
outer conductors
neutral conductor
Fwph
:
active
power
l3
P2
yph
V
line current
:
:
P, +
power
phase current
:
^=^r
of
one phase
:
phase voltage
:
line
voltage
ELECTRICAL ENGINEERING
31
Three-phase current
Reactive and active power, power factor
(for
reactive
s 199
active
power
200
power
factor
s
symmetrical load)
power
s 198
q
= y/2'v-I-s\x\
p
=
<p
y/3V-l-cosq)
P
V3V-/
Power
factor correction
consumers)
(for inductive
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
inherent consumption of condenser
Pc
«
0.003
- tan
q>,
q> 2 )
P
Q
Table (numerical)
cos
q
tan
cos
q
q
tan
q)
cos
q
tan
q
062
1
265
0.81
0724
cos
q
tan
q
0.44
2.041
0.64
1.201
0.82
0.698
92
456
426
0.46
1.930
1.138
83
0.672
0.93
0395
0.48
1.828
066
068
078
0.84
0.646
0.94
0.363
0.50
1.732
70
1.020
0.85
0620
0.95
0.329
0.52
1.643
0.72
0.964
0.86
0.593
0.96
0.292
0.42
2.161
1
91
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
can be calculated from the above table, cos q-\
representing the required power factor and cos q- 2 the consumer
tan
<pi
or tan
power
factor.
q) 2
ELECTRICAL ENGINEERING
32
Motors
Direct-current machine
(motor and generator)
General
203
moment constant
s
204
rotational
s
205
torque
s
206
armature current
s
207
terminal voltage
s
s
208
speed
s
209
internal
c
--fh
CM
source voltage
<Pa> =
2nC M 0n
c M <Pia
± (V ~
Vq)
K,±'aKa
V+
/a
2jrC M
power
to
generator
*'
**'
<*>
= M; m =
Pi
s 210
Ra
*»
Vq
Pa =
/a
v/,,
mechanical power supplied
by motor
s 211
Shunt motor
Easy
(for circuit
speed
starting,
r)Vl
x
diagram see S33)
%
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. Main circuit winding
ensures a powerful starting torque.
number
number
of
magnetic
*'
armature pairs
of pole pairs
2
:
Ra
:
number of conductors
armature resistance
flue
+ motor
- generator
")
- 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-sp
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
s
218
V
=
400 V
in star,
delta,
meaning
Vph
=
V
meaning
Vph
-
77=;
= 230 V
V_ m 400 V
=
.)i,
= 230
V3"
VS
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
IV
s
220
s
221
s
222
s
223
W^iw
D
n/-^iw
X
IV
IW
2V
IU
IV
IW
l^2V
X
W^-^IW
z
IU
2U /^-2W
2W-{
d
5|
IU
224
s
225
IV
IW
iu^iw
2V»
N2
£A?3
m
?>N2
2U2V2W
urn
2W^2U
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
11
2W~^j2U
lU^IW
*2V
\—2W
iu^Aiw
r
V?N2
Yin,
N2
2N,
YJN2
s2
6
1l
111
iu-^ IW
D d
AN
2
2U 2V 2H
2U2V2W
IV
s
1/1/1
2
IV
IV
LU
2U 2V
IU
IV
IW
y/j
IU
IV
IW
IV
IW
LU
IU
ri/i/i
M/3
2U 2V
2W
3
2W
2U 2V
jW
2V
JL
N2
L
Wn
2
IV
11
\>2W
w^^iw
2U 2V
IV
z
2W
*V2W
lUy^lW
N2
2N<
Single phase-output transformers
1
i
1.2
Key
numbers
(= key
\
i
*
2.2
1.2
2.2
1
PV: pri nary voltage
SV: sec ondary voltage
number x
2.1
2.1
;.;
1
s 231
D
|Y|
star
dSlta
d
N2
|
|y|
zig-zag
1
z
used to calculate the phase angle
are
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
1
2 «
3 CO
£
n
i
CO
11
^8
O
o
co
kJ-
to
to
CO
CD
CO
>
co
co
*l
1
Si™**
^
CO
(D
i_
O —5 x_S ~c
>.-
2-
CO
co
~l
1
o
l|
CO
!
o
c
CO
CD
CD
E
E
2
To
"co~
c
-3
CO O"
.2 -a
CO
CO
"
CD
*-
co
>
&o
o
Q>
o
E
£
co
CO
a)
a3
"°
0>
.
©
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
O N
of
b
mov-
«
CO
fixed
contact
springs
tension
h- O)
permanent
current
feeds
c
o
for
O
of
other
2
b
a
of
instruments
leads
moment
magnet.
in
to
coil
in
current
and
fixed
coil
non-uniform
2
a
bo
en
c
or
o>£ E
1
o
E
E
CO
5 o
o
o
co
co
o
o
o
O
i.
CD
£Q-
Q.
O
O)
uniform
magnetic
c
'>
O
E
in
counter
spiral
-o
c
coil
leads,
ffl
in
c
o
b
CO
V
^
CO
—
E c
o
with
co
moment
springs
CO
CD
co
2
'>
magnet
X
°8
CO
moving
as
-o
c
i
iS
°b>
bands
coil,
CO
in
b
'>
O
E
o
2
moving
screen
c£
heater.
Thermocouple
coils
o
o
and
fc
ing
permanent
moment
<o
©
O c
T3 O 3
c
o
CO-D O
T-
CO
or
field
Thermocouple
field
tension
.—
Q.
close
each
—
*l|
o
x
application
uniform
of
Q.
CO
thermal
without
serve
bands
Q.
°.E
moving
spiral
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
F
Tl
CO
c
o
||
2 -g"
|
2
.2
b
"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
PVC
/
z
unburied copper conductors including overload
protection devices at an ambient temperature of 30 °C 1)
insulation,
Nominal
cross section
Group
7 Z in
1
/ n in
7 Z in
Group 2
7 n in
Group
7Z in
3
1.5
mm 2
7 n in
A
A
A
A
A
A
Group
mm,
1
:
2:
3:
approx.
16
25
18 26
10 2 20
108 135 168
80 100 125
>
80
1.1
1.4
1.8
50
35
103 132
80 100
50
Nominal Cu-wire
diam.
10
2.5
2.3
multiwire
3.6
2.8
129 158 198
100 125 160
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
value decreases (increases) by about
increase (-decrease).
50 °C should not be exceeded!
2)
7Z
7%
per 5 °C temperature
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
overload protection devices of type B,
Live
*)
N:
Neutral
maximum
C
or
K
(for
PE:
each additional switch requires an extra cross-over
allowed value of
/
n iz ).
Earth
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
follow those given
in
of the
the
following terms very strongly
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
be influenced only by itself or its input
of a system
variables.
Delay time T, time constant T
The P-Tj element, the first order delay element,
which can
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
(1/a)
2
)
v (t) =
K P u(t)
the characteristic angular frequency and
ration. (For the solution of this differential
(o
is
damping
tion
+ (20/a>o) Ht) +
see J
#
the
equa-
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
t3
ration
&
are used.
a> d
-
aJo-Vl-fl 2
Explanation of the symbols see
T35
CONTROL ENGINEERING
Control engineering terms
Step response
The step response
(see
Steady
Tu
1
J>7
is
Tg
build up time
intersection of
with
\l
Tu
Tg
f
\
Trise
Settling time
Fig. 1
k
Step response of a
transfer element
fig. 1).
Build up time
of
Step response v
Step function u
defined
as the time between t
and the point of intersection
the
of
first inflexion of the
step response with
the x-axis
The
te/^TX
/
/
Tu
The equivalent dead
(see
stc
value
fig. 1).
Equivalent dead time
time
Transient value
\
tolerance
0\ ^ershoot
u,v
progress in
time of the output
variable of a transfer element when a
step function is used
input
variable
as
the
is
the
Tc
,s
the
and
x-axis
defined as the time between the point
inflexion of the step response
first
the point when this first inflexion
reaches the steady state value.
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
able when a ramp function with given rate
of the output variof change is used
as input variable (ramp generation).
'<i
MO
ik
Fig.
r(t)
2
T
t
4
t
5
r
H
e(t)dt
Ramp
T
r
is
the
ramp
£{t)
te(t)
generation
time, e
=
C/q
(t)
/
Ofor
I
1
for
the unit step generation:
t
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
(t)
characterises the dynamic behaviour of the transfer
element. For unit step responses for the most important transelements see T 14 to T 17.
h
fer
Transfer Function F(s)
The transfer function F(s)
is the ratio of the Laplace Transthe 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
form
v(s) of
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
tions
in
symbolic
is the
a considered system.
illustration
of
all
opera-
Elements of the action diagram
Elements of the action diagram are the action line, the functional block, the addition and the branching point.
Fig.
Functional
block
Action
line
u,
w-i,
u 2 input stimuli
:
Addition
= ± u 1 ± u2
v
v
:
output result
T
3
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.
Components
Fig.
trol
4
To achieve
this,
equa-
of the control loop and its quantities
typical action diagram of a closed loop con-
shows a
system 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
final controlled variable x A 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*.
The
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
Example:
is
not possible.
Temp, of the contents of a pot.
Temp, of the hotplate.
The final controlled variable *A can also occur within the controlled system, that means within the control loop, m this case
the controlled variable is formed via the final controlled
variable; influencing disturbance variables can be controlled.
Example: Final controlled variable: Mixing ratio of two liquids.
Final controll. variable:
Controlled variable:
Specific resistance.
Controlled variable:
Measuring equipment, feedback variable r
The measuring equipment is the sum total of all the funcrecording,
transferring,
for
determined
elements
tional
adapting and distributing the variables (see fig. 4). The
feedback variable r is the variable which results from the
measurement of the controlled variable x.
The reference variable adjuster, reference variable w
The reference variable adjuster
outputs a reference variable
w
is
a functional unit which
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
The
equipment
controlling equipment consists of the actuator
final controlling element.
final
and
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
transfer function F (s)
transfer function F {s) is the product of
transfer functions serial of a control loop or a loop.
Open loop
The open loop
al
Example:
u(s)
:
9
^
D
y
t
F{
' 1
M
(W
F2 (s)
1
"«
.
F
(5)
= F,(5)
-F2
(j)
*J
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.
Open loop
I
Control factor
The control
1
H F (0)
factor
10
F F (0) is
R F (0)
given by
= 1/(1 + V
Gain crossover angular frequency
The gain crossover angular
frequency co D is the open „, a
loop frequency where the | £
absolute value (amplitude)
of the open control loop
is equal to 1.
co
)
D
H 8.
<»o
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.
Diagram of absolute value and
phase response (not logarithmic)
of an open controlled loop
5
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
Tstart-
Agreed tolerance
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
t
3
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.
is
built
by using each individual
Series combination
U(S)
^(S)
t11
Parallel
u(s)
1
12
—
-^
v(s)
F2 (S)
-
F(s) = F,(s)-F2
(s)
combination
—
—
Feedback
*(s) -Zl
I
..
v(s)
F{s) =
F2 (S)
—I
Fi(s)+F2
(s)
rule
u(s)
,(s)
K(s)
1
13
F^(s)
F(s)
1+F,(s)-F2
F2 (s)
(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.
If F-i (s) and/or F2 (s) contain sign changes, then negative (positive)
feedback will result when the number of sign changes in the whole
loop
is
odd
(even).
Special case:
u(s)
~-\
p_
»
*
F2
{s)
=
1
(direct feedback).
V(SJ
/"-,
' 1
/c)
(S)
F^{s)
I
1
14
F(s)
1
+ F<
(s)
CONTROL ENGINEERING
10
Rules
Extended feedback
rule
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:
If
1
^w-44--^^
+XF
15
u(s)
1
F0i
means
oi
with
FVre8 (j)
= IlFv
(s)
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
(s)
loops
the
or
.
m
1
16
^Vres( 5 ) =
n FVk
k -
is
the product of
all
transfer functions of the
1
transfer elements which
lie in the forward path.
The overlapping of action lines does not affect the application of the extended feedback rule.
Example:
u(s)
kH f
vi (s)
r^9
HF
v2 (s)
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
to 6
respective time functions are determined at the points
the following example:
in
X;(S) =X${S)+X2(S) =
xm (s)
v{s);
FV2 FV3
v(s)
Fv
F\j9
^V3
•
v(s)
Fw(S) L^QJ^Fy 2 (s)--Y^Fyz(s)
Fr,(s)
<f
1
19
xs
(s)
=
FR2 (s)
FH2
xz
-v(s)
xi(s)_
t20
(s)
=
1
^-v(s)
^
F
+
1
FV2 (s)-Fsj 3
*V1 (s)
(s)
li£)l
v(s)
V; lis)}
At the reference addition point B there
is
the following
rela-
tionship:
t
u
21
(s)
- x®
(s)
= x%
(s)
for
;
xs
(s)
and x t
(s)
use the
values obtained above:
u(s)-FR2
t22
The solution Fres
found
t23
in t
Fresis)
=
1
-v(s)
(s)
FV i(*) I^V2(^)-^V3(^)
of this equation
is
^R1 (5)
+
Fv3
same as
the
V(s)
(s)
that already
17:
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
nominator
b)
c)
of s occur (meaning that neither de-
nominator of the transfer function
contain a fraction where s occurs in the denominator).
the lowest power of s has a coefficient of 1 and
there is no polynomial common factor.
nor
may
Example:
t24
v{s) m
F:(s)
Exception:
a
If
+ ajS + a 2 s 2 +
+
fc s + &
2 *2
(1
+
° (1
u{s)
PI
1
.
.
.) (1
+c,s+c 2 s 2 +
-
•
•)
+<M + d2
or PID factor occurs,
F2
t25
(s)
vis± =
1/(Tn s) +
„
p
u{s)
1
1
+
_
+ Ts
..)(..
.
still
in which
remains,
KP
^+Tn s
Tn s
1
From the table showing the most important
+ Ts
transfer elements
the following type can be found for this equation:
(PI )
_T ^|-(PD)-T
1
1
.
The following table shows
Type
of the
normalized form
Product
normalized form
Sum
normalized form
different types of
Representation
denominator and
nominator are
factorized form
denominator and
nominator are
in
written
in
normalized form:
Application
Bode diagram and
series stabilization
of control loops
Hurwitz criterium
sum
denominator as
Mixed
far as
possible divided into
Determination of
the step
Nominator in
and ramp response
in the form of a sum
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.
normalized form
factors.
.)
..)( )
.
another form
+ 1 in the example]
an approved normalized form.
this factor [\/(T„s)
is
2
.y
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)
= b qR (s)/F k (s)
= b R (s)/F k (s)
F2 (s)
and F3 (s)
are determined; then
F(s) =
t30
t
31
^1 (s) =
Fns(s)
1/(1
A+M{nB)-M{rms 2
b qR (s)
t32
F2 (s)
_
B + F2 (s) + F3 (s)
M(qRs)
b q R(s)
fkW
qRs
+
nrmBs 2
1
_
F, (s)
1
)
q Rs + ^1
W
?#*
1
+
nrmBs 2
2
t33
qRs
t34
t35
+
bn(s)
F3 (s)
F(s)
[1
nrmBs
nrmBs2 + rmsl[qR
Fk ($;
=
2?
1
M(ms)
+R/{ms)
1
s
/?5
1
+
[1+mj/R]
+
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
I
order delay element
dent if[er
Symbol
in
the
in
Structure
Equation
the time domain
Examples
action diagram
Kp
KP
v =
u
•
Proportional element
v=
=
K,
fudt
K Ju dt+
{
v =
Kr
v(0)
u
Intergral
M&
V =
Kn
=
KD
fvdt
•
— L^—
U
u
-u
Derivative element
-M^
i
v[t)-u(t-T
t )
^cn^
5
Dead time element
KP
P-Ti
v +
KP
element
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]
= Unit of
•
t
•
6
t
h)
KD
(t)
"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
Symbol in the
action diagram
in
Structure
Equation
the time domai
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
Parallel
15
combinational element PI
Unit step response
Equation h{t) =
Transfer
function F{s) =
diagram
*p[1
"^e^'-cos^ /-©)]; <od
&2
#
=a>Q \j\ -
= Arcsin
<6><
90°
KP
KP (l+e"'rtanC:')
+
'con
+
s
bi
\a)r
< #<1
a: p
Kp
+
(1
7, *)
(1
+
T2
n+20
3ti+20
5n+20
2<y<j
2£Dd
2wd
- [T,e^-T2
1-
e^
s)
i
1
IW)
9>
l
Kp
=
^1,2
V^-D
JC^
.1 -
K
i
,/ xf
J"
>
*
1
7 lnK
Ks^/r,
1
tf,f
Kp
^7^
Kp
Tn -s
(1
+
+ Kp =
KP
K-1
(1
T
1
InK
K-1
-
|T +7 2
'
+=r)
W)
+1
V*)
Explanation of the symbols see
T35
CONTROL ENGINEERING
Tie
|
ce~:'e-
S.-rc
= ::
z-"
z
Eca:
:-
« :-e
aran
"
v
-1—
-
=
KP
=
k p (ii
=
^i
•
u + Kq
-
u
+ r„ u)
•
*
—-
—
rlt—
-—
.
-&
f r
LjLa
= :
lit
-
hP^
—3^—1
—
r
l-T,
-ZZ-
-^—^—
D-T.
-
-~ZrExr
="=
z~
z~
e
:
•
-
-
s
see ~
2 r
/
CONTROL ENGINEERING
T16
combinational elements PD, PID
Parallel
Series combinational elements
and
I-T-,
D-T-,
Unit step response
Equation h(t) =
Transfer
function F{s) =
diagram
*P
+
a: d<5(0
i
Kp + Kq
KP
1
•
s
+ T^-s)
(1
-
~^0
K^+ K
P
+
KD
-s
K,f +
KP
for
<
KD
+
—<
<5(0 =
oo
KP
\jr +
l
+
Tv
r
6(t)}
;
1
*Pk
I
jnk
'*
hjnk
^nk
^vk
-5)(i +
rvk
-5)
=
V
=
^r (i-Vi-4rv /rn \
*
«
n
*Pk ~ *,
•
)
'o
-Tn
f
7nk
A-,(f-r+ r-e~' /T
i I
)
W)
K\
s(1 + 7-5)
37K|TN
*~
f
7
^D
KD
Kd-s
(1/7-)e-'
/T
im
.
T
1+7-5
0.37^
~^0
7-
"
f
Explanation of the symbols see T 35
CONTROL ENGINEERING
Series combinational element
17
Group combinational elements (PD)-"^ and
Identifier
Symbol
in
Structure
Equation
the
in
(PID)-T.,
the time domain
Examples
action diagram
-/w
D-T5
v+
2
Kp
(PD)-l,
KP
=
u +
j
rv
1
r¥ u
•
Kp T,,T
Kp TnTv
v +
7
v =
AT|
J
u
I"
1
dJ -
+ 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
Group combinational elements (PD)-Ti and
17
(PID)-T-)
Unit step response
Equation h(t) =
Transfer
function F(s)
diagram
Mo
-0O)
"" 0,
of
,
^Dag--e
-sinfi\ J
ATd-5
6)
+
Kp + *p
„
#2
5<
*
u
2
,|o_^j e-;
=
a
2<u d
2<y d
^^^-iJeT
Kp7,/r
7V -J
1
+
1
+ r-5
T
1
Kp
v
K
+
rv
=
-
5
(7*
KP
a> \/l
+ T'S
1
=
=
= Arcsin
2<yd
•
o) d
;
Q_
rt(t)
1+2^5+,,
r
r
-
- 7)
1
+
=
rv
-5
r-j
+ 0.377v /r)
*
Kp-
Ts
+
\
KP (0.63
,''(t)
KpTv /7
KJ+ K^+ \KJ-
KP
+
KD
j
TTT^
"(t)A
'
7-
n
rv -7Tn«
7"n
Kp[l+— ln(1--^ +
,
rn
(i
+ r-5)
rn
= Xp /K,
r*
=Tn -T
;
'n
7V = Xq
,
Kp [0.37(^-+-k)+0.63]
7"nV
'
//Cp
Explanation of the symbols see
T35
CONTROL ENGINEERING
Methods
18
determining stability
for
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
1.
transfer function of the
open control loop
is
known.
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 +
118
a-i
s
+ a 2 s* +
.
.
.
+ a n sn =
satisfy the following conditions:
(see also D 9)
must be >
The coefficients themselves have to fulfil special depend-
• All coefficients a v
*
encies.
The conditions
for
Equation
1st degree
2nddegree
a
a
3rd degree
4th degree
a-\
5th degree
A
a-|
,
equations up to degree 5:
coefficient conditions
and
a^
a-|,
a2
>
>
>
a2 - a3 a
2
a 2 a 3 - a3 a
-
a*
a4
>
= a a 2 a 3 a 4 + a a-i 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
D n _! =
-
1
A-B>0
--
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
Advantage: The method
statement concerning the
19
leads to a swift and accurate
a given control loop.
stability of
Disadvantage: The Hurwitz
criterium gives neither a stateto 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.
ment as
2.
Reduction
into single
Transform
the
polynomials
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.
of how close the system
given by two characteristic values:
A measure
is
is
to
becoming unstable
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
The determination of the actual
J*lmF
Recommended
Recommended
Re F
Ga>)
Fig.
8
values for the phase margin <5: 30°
values for the gain margin
e: 8 dB
(corresponds to the factors 2.5
Advantages: Examination
8).
(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.
60°
16 dB
..
.
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
,
is an influence of disturbance variables applied
controlling
Point of
element and point of
measurement
u
Controlled
ps. Controlling
there
between
measurement
finite
only
accuracy
a
is
This
accuracy is given by
the
of
the
value
control factor R F (0).
possible.
"""
_
"T^T
element
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.
Only factors of the frequency response of type I, P, D,
P-T 1( PD and P-T 2 with attenuation # < 1 are permitted
t
(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.
•All the serially occuring Integral - (K,), Proportional-(Kp)
*
T
,
I
or
Derivative (K D ) action coefficients
single action coefficient.
are summarized
into
one
« Representation of amplitude response.
phase response.
Completion by the controlling element.
The tables on pages T 22 and T 23 show the Bode diagrams for
P-T 1t P-T 2 and PD elements. These diagrams are used
P, I, D, T
for the' determination of the amplitude and phase responses.
* Representation of
*
t
,
CONTROL ENGINEERING
Bode diagrams
22
for basic
elements
and P-Ti elements
Amplitude response F(co) =
Phase response Arc F{\o)) =
q>
=
Symbol
Diagram (Ampl. logarithm.)
Diagram (Phase
linear)
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
<i
F ldB
-oo
<
<
n
<n <
-oo
Kr M(D
-90
+oo, integer
n <p/1°
10
dB/decade
n
10
n+1
10
-oo
<
10
n+3
10
n 4
l
,„
<
+oo, integer
<
+oo, integer
n
-90"
<
n
n+2
<
-oo
Kn-O)
n F ldB
+oo, integer
c
+oo, integer
<
-oo
10
n
10
n+ 1
10
n
n+2
1Q
n +3
1Q
n+4
\co\
io7io n+1
1/K
D
=(o
+2
io
i(f
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/ ~¥+ *
lO^IO'^
H
i-
"
\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.
20 dB/d«
Tto
/2
10°
/2
1Q
1
W
2
^
CONTROL ENGINEERING
Bode diagrams
23
for P-T2
and PD element
Amplitude response F((o) =
Phase response Arc F(\(o) =
<p
Symbol
Diagram (Ampl. logarithm.)
Diagram (Phase
< 0<1
(*)'
Arctan
KP
i
V[ -(*)T
linear)
90
20%+
"
c
,
(*)'
< #<
co
1
>
P-T;
Kp
1
I
(T
V
P dB
Vl-2 9
yKP 2
=
+ (K D -cof
2
V1 + (rv w)
KP
with
2
Tv =
Arctan (Tv w)
/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
<*>/
wn
CONTROL ENGINEERING
24
Bode diagrams
and
for (PD)-"^
(PID)-"^ element
Amplitude response F(co) =
Phase response Arc F[]co) =
cp
=
Symbol
Diagram (Ampl. logarithm.)
KP2
V
Diagram (Phase
+ (K D o)) 2
+ (Tea) 2
1
linear)
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
10' 1
io
0/
io
'
I
—
—71
1
1
10
T<T,
^^^
10" 2 10' 1
2
7-v*
10°
10
TJT
1
\10 2
TJT
Arctan 7> —— ]- Arctan (To)
Tn
w>o
o<r /r <oo
-
KP2
+ (KpCD- KJco) 2
1
gP
_
+ (Tco) 2
Wi*P>-i /(7»r
V
1
idBi
(PIDJ-T,
/
V
K|_
i
1
_J
_
,
2
- Arctan (7a)
r k >7
^) K ldB/7
7>-i/(7»
Vl+[7>-1/(7»]
+ (To;
r>4r
v
= Arcsin
o
F
co\
L
n
1_
< rn /
r„
<
°°
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 # <
P-T-, factors.
1,
P-T 2 factors with $
>
are divided into
1
two
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
omitted
here
physical
and
units
of
K
it
KD
or
KP
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
gradient
break
is
the PD factor (inverse to the P-T factor) is used the
corresponding corrections have to be made at the breakpoint o) E = 1/TV upwards (i.e. + 3 dB)!
When
1
of a P-T 2 factor with &
2
corrected by -10 Ig [1 - ((o/co ) +
Around the break point
totic
curve
is
<
(2
1
the
dw/(o
asymp2
)
].
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
P-Ti
-Arctan
t
PD
=
A
(Tco)
Arctan (Tv co)
P-T2 with # <
PI
Table
(o
I
- PD
Comment:
1
- 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
P-Ti
-Arctan {Toy)
+ Arctan (Tv co)
I
x
PD
P-T 2 with
PI
=
I
#<1
- PD
Comment:
t
A
co
"-"(t^S 2 -"-)
- 90° + Arctan (Tn (o)
whilst
factor has a
creases with a).
T
Table
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 /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)*)
controlled system and measuring
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.
value of
equipment
of the
at the
.
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
F
i
4
= rlA =
5)
(1+10sec-s)[l+0.8/(5sec)s + 1/(25sec
y(s)
Solution: According
T13
T17
2
2
)5
l
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
for determination are given as numbers (in
These step
the left margin on T27...T29.
numbers are also plotted in the Bode diagram on T30 at
the corresponding positions. The following results are found
at the step numbers:
The next steps
circles)
at
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
from
this
data:
K PR
list
the required controller has the characteristic
Tn = 2.08 sec.
= 2.82;
The speed by which the controlled loop using
reaches
values
its final
is
characterized by
cj d
this controller
= 1.2 sec
-1
.
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
11
1
==
6.7
<pt
= -180°
1
= 5 sec"
16 /Cp RE indB = -10^K PRE = 3.16
12
o) Db
a»j,
E
continued on T 33
CONTROL ENGINEERING
32
Examples
o
1
o
CM
O
CO
1
o
"aI
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
)
in the cases of a Pl-controlling element.
Determination of a (PID)-T., controlling element
Problem: For a control loop with the same series connection as in example 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
example 1.
sated as
Example
2:
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 =
11
<p e
(2
+ 0.67) sec = 2.67 sec;
=-180 o
-<p R1
=-242°
7"
v6
= 2 0.67 sec/2.67 = 0.5 sec
•
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
value
is
for the desired control loop to reach the
-1
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:
C
Table
con-
Ty
^PR
troller
TV
P
K Py
n n
°J
PI
•
Tty
Ty
3.3
AV Py T
lty
.
-
1"
PID
^
A'py
•
rty
2 7\
y
Tty
q.5
rty
Characteristic data of the controlled system are unknown:
Table
con-
K PR
troller
P
0.5 A'pRc-jt*
PI
0.45 tfPRcrK*>
0.83
PID
0.6 KpRcrit*
0.5
•'A'pRcrit
,)7
crit
:
:
D
Tn
7crit **)
7crit ">
K PR value when
the control loop.
permanent
Oscillation period
in
tion.
rv
0.125
oscillation
rcrit ">
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
l-T
t
Integral
:
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
element with 1st order delay
(PD)-T, element:
PD
elem. with 1st order delay
P element: Proportional element
(PIDJ-T, elem.:
PD
T element: Dead time element
element: Proportional derivative element
List of
Error variable
:
Incline of the
r
:
Bode diagram
Feedback variable
u
:
mF
v
:
vm
:
t
symbols for Control engineering terms
:
e
Tu
Equivalent dead time
r¥
Rate time
rnk ,(ryk ): Reset time (rate
:
amplitude response
in
the
:
Input variable
Overshoot of the
w
ries
representation
with
Tn > 4 rv
Tnkb>(TrtbY- Rese '
Output variable
unit
step function of
ries
a transfer element
with
:
Reference variable
:
Target variable
:
Controlled variable
ries
xA
:
Final controlled variable
with
xm
iv*
x
PID elem. with 1st order delay
time)
the
of
in
the
time
^e
ra te time ) in
(
representation
the
of
to the
phase requirement
Reset time
7n >
4
time)
(rate
representation
rv
of the
the se-
in
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
led system,
F((o)
:
Amplitude response
the gain crossover angular frequency
Frequ. response of the open contr. loop
and meeting the phase
ip
:
Transfer fund, of the open contr. loop
:
Ampl. response of the open contr. loop
:
Ampl. response of the controllg. elem.
Ampl. response of the series connec-
:
tion of the controlled
system and the
KD
:
Derivative action coefficient
:
Integral action coefficient
K P Proportional action coefficient
Control factor
R F (Q)
K Pk (a>): Proport. act. coeff. in the
Phase
:
8
<p e
Phase
:
KPR
series re-
(o
coefficient
Proportional
:
Tg
> 4 Tv
a)
of
the
co
of
the
7"
n
Th
Tn
Tend
b
c>
E
Damping frequency
:
Angular frequency
:
Characteristic angular frequency
:
:
:
w D6
:
(i)
:
:
Build
:
Half-life
:
Reset time
u) x
Time
to reach
o)
Time
to reach lower tolerance
:
7,,,,,:
Eigen angular frequency
Gain crossover angular frequency
Corner angular frequency
Gain crossover angular frequency when
realizing the
up time
D(
phase margin condition
Gain crossover angular frequency when
realizing the gain
period
steady state
at the
equipment
:
Delay time
:
control-
:
<y D
coefficient
controlled element
T
-
measuring equipment -
:
%
uring
controlled element
action
at
a>
6
phase crossover angular frequency wn
Phase response
<p(a))
Phase response of the open control loop
((o)
i?
present, of the PID elem. with
action
njargin
(a)):
:
Integral
control-
measuring equipment -
of the series connection
led system,
:
Km (o)):
of the series connection
Phase response of the controlling elem.
qk
%[(o): Phase response of the series connection
of the controlled system and the meas-
measuring equipment
K,
se-
PID element
Tn > 4 Tv determined according
rnk£1 (Tvke):
>
F (jw):
FQ (s)
F ((o)
Fn (w)
Fy (o)
se-
PID element
.
nt
:
:
margin condition
Phase crossover angular frequency
Phase crossover angular frequency when
the gain margin e
is
met.
CHEMISTRY
Ui
Elements
atomic
element
in
element
antimony
argon
Ar
39.948
As
Ba
Be
137.34
bismuth
boron
bromine
Bi
208.980
cadmium
Cd
112.40
caesium
Cs
Ca
132.905
barium
beryllium
calcium
carbon
cerium
26.9815 neodymium
144.240
neon
Nd
Ne
nickel
Ni
58.71
74.9216 niobium
Nb
92.906
nitrogen
osmium
N
Os
121.75
9.0122
B
Br
O
palladium
106.4
79.909
phosphorus
Pd
P
platinum
Pt
195.09
potassium
K
praseodymium
Pr
140.907
226.04
150.35
chlorine
CI
35.453
rubidium
chromium
Cr
51.996
ruthenium
cobalt
Co
Cu
58.9332 samarium
Sm
63.54
copper
erbium
Er
12.0112 radium
rhodium
140.12
scandium
selenium
167.26
18.9984 silicon
fluorine
F
gadolinium
Gd
Ga
Ge
157.25
Au
He
H
196.967
In
114.82
gallium
germanium
gold
helium
hydrogen
indium
iodine
1
14.0067
190.2
oxygen
Ra
Rh
Rb
Ru
Ce
20.183
10.811
40.08
C
mass
in u
Al
arsenic
symbol
u
aluminum
Sb
atomic
mass
symbol
15.9994
30.9738
39.102
102.905
85.47
101.07
Sc
Se
44.956
78.96
28.086
Si
69.72
sodium
Ag
Na
72.59
strontium
Sr
87.62
sulfur
180-948
tellurium
S
Ta
Te
thallium
Tl
204.37
Th
232.038
Tm
168.934
118 69
silver
4.0026 tantalum
1.008
126.9044 thorium
thulium
192.2
107.870
*
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
magnesium
manganese
Mg
Mn
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
:
atomic mass unit
6.939
24.312
54.9381 yttrium
(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
(CH 3
C2H2
ammonia
ammonia
ammonium
(hydroxide
of)
ammonium
)
CO
•
2
NH 3
NH4OH
hydroxide
C6 H 5
NH 2
aniline
aniline
bauxite
hydrated aluminium oxides Al 2 3 2 H 2
calcium hypochlorite
CaCI (OCI)
•
•
bleaching powder
blue vitriol
copper sulfate
CuS0 4
sodium tetraborate
Na 2 B 4
butter of zinc
zinc chloride
ZnCI 2
cadmium
cadmium
CdS0 4
borax
sulfate
sulfate
5
•
7
H2
3
•
H2
10 H 2
•
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
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
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
fixing salt or
hypo
vitriol
C0 2
SiC
HgS
•
7
•
•
2
H2
H2
4 H2
PbS
<
continued on U 3
CHEMISTRY
Us
Chemicals
continued from U 2
chemical
trade
chemical
formula
name
limestone
magnesia
marsh gas
minimum
or red lead
calcium carbonate
magnesium oxide
CaC0 3
methane
CH 4
2 PbO Pb0 2
plumbate
MgO
•
phosphoric acid
potash
potassium bromide
ortho phosphoric acid
HNO3
H 3 P0 4
potassium carbonate
potassium bromide
KBr
potassium
potassium
potassium
potassium
potassium
potassium
potassium
potassium
nitric
acid
nitric
chlorate
chloride
chromate
cyanide
potassium dichromate
potassium iodide
acid
K 2 C0 3
chlorate
KCIO3
chloride
KCI
chromate
K 2 Cr0 4
cyanide
KCN
prussic acid
potassium dichromate
potassium iodide
hydrogen cyanide
pyrolusite
manganese dioxide
Mn0 2
quicklime
calcium monoxide
potassium ferrocyan.
CaO
red prussiate of potassium
K 2 Cr 2
7
Kl
HCN
K 3 Fe(CN) 6
NH4CI %
salammoniac
silver bromide
ammonium
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
white
zinc sulphate
ZnS0 4
vitriol
silver
chloride
bromide
AgBr
Na 2 C0 3
Na 2
10 H 2
2
•
H2
)
•
7
yellow prussiate of potass.
potass, ferrocyanide
K 4 Fe(CN) 6
zinc blende
zinc sulphide
ZnS
zinc or Chinese white
zink oxide
ZnO
(OH) 2
H2
•
3
H2
CHEMISTRY
U
II 4
A
Acids, Bases
pH
The negative
pH
values
log of the hydrogen-ion-concentration c H - indicates
its
value:
pH
cH +
10"'
1
=
10" 2
-log c H -
10- 7
u
|io-
12
|io- 13
1fJ
-14
{
pH-y a\ue
2
1
7
|
1
-
_
"
..
-
nnirt
13
|
14
neutral
1
Establishing
12
/?//
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
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
6.0...7.6
yellow
phenol red
6.4. .8.2
red
.80
.88
neutral red
6.4..
cresol red
7
meta cresol purple
7.4.. .9.0
thymol b lue
phenolpl ital
8.0. .96
alizarin
>
ellc
82.9.8
Bin
w
0.
66
(orange)-yellow
blue
purple
yell.
purple
blue
yellow
red
(blue)-red
orange-yellow
purple
purple
yellow
yellow
yellow
colourless
100. .12.1 light-yellow
blue
red-violet
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
-2NH
hydrogen
hydrogen sulfide
CO(NH 2 2 + H 2
NH 4 OH + HCI
+ H2
NH 3
CdS0 4 +H 2 S
CaC0 3 + 2 HC
CaOCI 2 + 2 HC
+ Zn
H 2 S0 4
+ 2 HC
FeS
H2 S
+
lead sulfic ie
Pb(N0 3
-*PbS
oxygen
sodium
2 KCIO3
+
+
ammonia
)
ammoniur n chloride
ammoniur n hydroxide
cadmium sulfide
carbon
di
Dxide
chlorine
)2
Na 2
h) 'droxide
ZnS0 4
zinc sulfic e
H2
—
--Cds
+
-*C0 2
+
H 2 S0 4
CaCI 2 + H 2
+ CaCI 2 + H 2
-*CI 2
^H
—
+ H 2S
+
+
co 2
+
3
-* NH 4 CI +
NH4OH
+
2
—>3
—
2
ZnS0 4
%
FeCI 2
2
2 KCI
NaOH
+
-*ZnS
H2
H 2S
HN0 3
2
H 2 S0 4
Freezing mixtures
Drope
from
°C
+
+
10
10
+
8
+
15
in
t<
Mixture
jmperature
(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)
1
1
H2
H2
1-4 ice
1
+
+
+
1
1
1
KCI
NH 4 NO
NaN0 3
(crushed)
methyl alcohol
c
+
+
+
+
+
-
1
1
NH 4 CI
NACI
2CaCI 2
•
2
CaCI 2
•
1
C0 2
6 H?
6 H2
solid
CHEMISTRY
u
Moisture, Drying agents, Water-hardness
Atmospheric relative humidity
closed containers
in
Relative humidity above
the solution (%) 20°C
Supersaturated
= 65°F
aqueous solution
92
86
80
76
Na 2 C0 3
63
55
45
35
NH4NO3
Ca(N0 3 2
•
10
H2
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
25°C
(77°F),
g/m 3
name
air
1.4
copper
0.8
zinc chloride
calcium chloride
0.14 ... 0.25
0.16
0.008
formula
CuS0 4
dehydrated
ZnCI 2
CaCI 2
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
0.000025
phosphorus pentoxide
P2O5
Hardness
1
sulfate,
German hardness
Vd
=
water
mg MgO
7.19
water
1.25° English hardness = 1.78° French hardness
17.8 American hardness (1.00 ppm CaC0 3
1
1°d
water
mg CaO
of a
10
I
1
I
)
Classification of hardness
4°d
8°d
12°d
12
18
very soft
soft
slightly hard
...
18 d
...
30°d
above 30°d
rather hard
hard
very hard
Mixture rule for fluids (mixture cross)
capa-
in
mixed
of the
admixture
for
water
Example:
I"
starting
city
is
b =
fluid
weight-%
0.
a = 54%; b = 92%; c shall
One should mix
become 62%.
thus 30 weight-sharings of a with 8 of
3
KOH
b.
)
x
RADIATION PHYSICS
V
Photometry and Optics
General
For every photometric quantity there
same
physical quantity and the
Photometry
Symbol
candela
radiant
cd
intensity
'v
luminous
4> v
flux
lumen
Im = cd
= fi-/ v
radiant
power
sr
lumensecond
quantity
of light,
Gv=<V'
luminous
energy
They
and e for
Radia tion physic s
Quantity
Symbol
Units
Units
luminous
intensity
a corresponding radiation-
by different suffixes, v for visual
are differentiated
energy.
Quantity
is
relationships apply to both.
Im
sr
*,-fl-/.
W
Q e =4>*t
J
quantity
radiance
m2
A^ COS
e.
sr
lx=
light
ff v
exposure
= £v
Definition of the
The luminous
a black body
—
m^
Im
irradiance
radiant
•
Ixs
t
base
unit
exposure
"candela"
A
=
Ee
2
-t
m2
Ws
m2
(cd)
intensity of a surface of 1/600000
at a
W
Ee
He
m'
£f
lux
illuminance
Ws
=
W
cd
/v
A, -cos
J/s
of radiation
Lv =
luminance
=
radiant
energy,
s,
also Im h
W
U
temperature of 2042
m2
(= 1%
mm
2
)of
K.
Photometric radiation equivalent
A = 555 nm.
1 watt = 680 Im for wavelength
Luminous
A
flux
surface
A
Z 25)
require a luminous
consumption
for lighting(values see
to an
Ev
lit
illumination
flux of
Z<Py
For symbols see T 2
A-Ey
will
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
sources produce equal illumination of a surface,
of the squares of their distances from the surface is
the ratio
equal to
light
the
of
ratio
their
/
_
..
surface
(
luminous
intensities:
t=;
hi
/V 2
Light refraction
np
sin a
n a ~ sin /?
= const, for
Where
sin
fi
^
—?-
\>-~i
\*
all
angles.
total reflection
dense
oc ' urs
|
in relation to
fluid
1.49 water
quartz
1.54 alcohol
crown glass
diamond
2.41
A-\
:
A2
'•
benzol
in
relation to
vacuum
hydrogen
1.36 oxygen
1.47 atmosphere
1.000292
1.000271
1.50 nitrogen
1.000297
1.33
1.000292
area of radiating surface
area of illuminated or irradiated surface
projection of the radiating surface
cos £^
A-i
1.56 glycerine
= 589.3 nm
gasous matter
matter
atmosphere
plexiglas
|\
medium ft\
Refractive index for yellow sodium ligh tning A
solid matter
thin
medium
:
A<\
perpendicular to the
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
2
2
unit sr = m /m
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
Type
v16
atmosphere)
(in
Wavelength A =
of radiation
X-rays
hard
0.0057
soft
ultra-soft
0.08
nm
nm
2.0
nm..
UV-C
UV-C
UV-B
UV-A
optical radiation
ultra-violet radiation
...
nm
nm
nm
nm
380 nm
420 nm
490 nm
530 nm
650 nm
780 nm
1.4 nm
3.0 nm
100
280
315
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
IR-C
elf
0.08
.
280
315
380
420
490
530
650
780
nm
nm
nm
nm
nm
nm
nm
nm
Mirrors
Plane mirrors
The image
at the
is
same
dis-
tance behind the mirror as the
object
is in
u =
v17
front of
it:
-v
Concave mirrors
i.i
v18
i
+
v
u
f
Depending upon the position of
object, the image will be real
or virtual:
u
v
00
f
>2/
f<v<2f
2/
2/
2/
>u > f
>2/
00
f
negative
<f
at focal point
real, inverted,
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
« 0.3 x 10 9 m/s (velocity of light)
For explanation of symbols for mirrors refer to V4
•)c = 299 792 458 m/s
|
/:
frequency 1/s
RADIATION PHYSICS
V
Lenses
Lenses
Refraction
D
of a lens
Unit:
v20
Lens equation
1
dpt
=
1
dioptrics
(thin lenses only)
v21
'>£ + t8
v22
5
v23
Where two
diately
lenses with focal depths /1 and /2 are placed immeother, the equivalent focal length/, is given
one behind the
11^1
by
v24
/i
Magnifying lens
._ _ object
where object
general
is in
m
v25
=
-j
focus
j+-\
Microscope
total
h
&
magnification
*
§
j
m
v26
t-s
fvh
v27
v28
Macro photograpy
camera extension
v29
a = f(m +
distance of object
B
F
/
G
:
:
size of
n^-k)
image
focus
n
:
r
:
:
focal length
t
:
size of object
m
range of vision
1)
(=
25
cm
for
:
:
refractive index (see
V2)
radius of curvature
optical length of tube
magnification factor
normal vision)
RADIATION PHYSICS
Ionizing radiation
Ionizing radiation
Ionizing
radiation
any
is
which causes direct or
permanent gas.
Accumulated
radiation
indirect
of
As
absorbed
energy
(measured
kg
v30
value)
particles
excitation
rate of
258^
R
J
t
kg
kg
m
absorbed
absorbed
dose
dose,
1
gy_
*-?-£m
D-f.J
=
vv
"
s
v33
a
Units
absorbing
energy
kg
v31
v32
of
rontgen
1
1
or
time rate
values
Units
values
amount
charged
of
ionization
kg
31.56X10 6
i
kg a
W
m
kg
6.242
dose
1
equivalent
=
v35
H=D
1
^y
0.01
sv
dose
VAs
Ws
equivalent,
kg
kg
-rate
sievert
1
(theoretical
value)
v34
x10
1
w
gy
kg
= q-.
q
[100 rem
1
sv]
=
= q-f-J
Ionization current
and a voltage
is
/;
When
air
q-D
molecules are ionized by radiation
applied, an ionization current / flows.
(Instrument: the ionization chamber).
Charge Q: When an
duces a charge
ionization current / flows for a time
Q
v36
Units
in
(
)
=
t
it
pro-
It
are earlier units
continued on T 6
RADIATION PHYSICS
V
Ionizing radiation
Dose
J:
Dose /
is
mass m,
a value related to
W
Radiation energy W:
e.g.
J =
Q/m.
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"
is
W
The
Activity A:
activity
A
number
the
is
substance that disintegrates per
of
atoms
of a radioactive
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
The
X:
=
X
half
life is
Ty2
In
2/Ty2
mass
the time taken for one half of the radioactive
to decay.
Units: s~ 1 min
-1
,
Half lives of
h~ 1 d
,
,
some
_1
,
a
-1
and
natural
.
artificial
isotopes
relative
atomic
1
number
'
Z
Element
atomic 2 >
mass
potassium
cobalt
strontium
Symbols used
mass (base
N
q
/
:
:
:
:
number
number
:
mass
caesium
caesium
5.3a
thorium
29 a
uranium
unit)
Plutonium
|
(3
Ty2
half
life
FA
2.1a
30 a
1600 a
14xi0 9
a
4.5 x I0 9 a
24000 a
life
atoms
y - und X-rays
for other radiation
ionization constant for tissue
quality factor for
half
>
Ar
134
137
226
232
238
239
radium
8.0d
of radioactive
atomic 2
Element
12a
12.4 h
=
q =
-,
for
fL
Z
1.3xl0 9 a
131
iodine
1>
life
Ar
3
40
42
60
90
tritium
m
half
atomic
<7
1
1
...
20
/-/l
/=(1 ...4)/L
bone
WL/e
fL =
ionization constant for air
= 33.7 V)
Notation of units used
A:
ampere
|
C:
coulomb
J:
|
joule
a:
annum
(1
J
I
d: dies
(1
annum =
1
=
1
31.56x 10 6 s)
86 400 s)
1991 the average person
dies
Exposure to radiation (dose equivalent): In
the Federal Republic of Germany would have been exposed
follwing radiation:
h
Type
msv
from natural sources
for medical reasons
other
artificial
<2.4
<
t
<0.1
radiation
<
*permitted by law
')
number
of protons
2)
I
0.5
number
of protons
0.3
and neutrons
r
in
[m rem]
240
50
<
<
in
to the
10
30
TABLES
Zi
Properties of solids
Reference conditions
Density pat
= 20°
point
Boiling
i.e.
t
t:
C
The values
direct transition
in
brackets
Thermal conductivity A
at
t
9
t
t
kg/dm 3
°C
°C
1600
1040
658
658
300
2600
2300
2200
2200
630
815
1440
2.6
aluminum bronze
aluminum cast
2.6
aluminum,
2.7
rolled
7.7
amber
1.0
antimony
6.67
arsenic
5.72
artificial
wool
2.5
1300
704
3.59
4.5
1.85
bismuth
9.8
271
boiler scale
2.5
borax
1.72
1200
740
brass, cast
8.4
brass, rolled
8.5
brick
1.8
bromine
bronce (Cu Sn
3.14
6)
iron ore
321
1.55
3.51
cast iron
7.25
850
3600
1200
cerium
6.77
630
chalk
127.9
209.4
209.4
22.53
1100
1100
165
= 0.858 9
kcal/(h
K)
1.2. ..3.5
113
113
»
0.80
0.436
0.904
0.904
0.209
0.348
1.357
0.816
0.29
0.80
0.996
0.385
0.385
0.92
63
2300
64
0.37
0.58
765
1439
92.1
2500
58
8.9
2>
|
kJ/(kg K) 2
1.02
1
kJ/(kg K)
=
0.67
0.234
0.63
0.854
0.532
0.92
rr
»
0.13
8.1
1.8
K)
cp
1
10.89
1570
5.1
864
W/(m
k)
1.0
8.83
heat
0.46
2970
1560
2800
-7.3
910
calcium
carbon
1
W/(m
specific
A
tivity
0.17
900
900
cadmium
»
100
1700
1580
1280
barytes
beryllium
1
o<r<
1.5
asbestos
barium
brown
state.
meltinglboiling thermal
point
conduc-
density
agate
sublimation,
to
= 20° C
Specific heat c p for the temperature range
andp = 1.0132 bar.
Substance
refer
from the solid to the gaseous
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
kg/dm 3
°C
°C
chromium
1800
1600
1490
7.1
1.8. ..2.1
cobalt
8.8
coke
1.4
concrete reinforce
constantan
copper, cast
copper, rolled
cork
2.4
diamond
1
2700
2980
3100
69
1
69.4
0.8. ..1.7
1600
1083
1083
8 89
8.8
8.9
2400
2500
2500
23.3
384
384
0.05
0.2... 0.3
650
350
2000
129.1
40. ..50
2.8
>
kJ/(kg K) 2
0.84
0.452
0.88
0435
0.84
0.88
0.41
0.394
0394
2.0
0.52
[3540]
0.9. ..1.0
duralium
ebonite
K)
0.184
3.5
dripping, beef
W/(m
A
0.084
0.4
clay
tivity
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
brick
fire
glass,
1.8. ..2.2
window
2.5
glass-wool
gold
0.15
0.47
0.88
0.81
0.84
0.04
0.84
19.29
1063
2700
310
0.130
graphite
2.24
3800
168
0.71
ice
0.92
ingot iron
7.9
1460
iodine
4.95
113.5
4200
100
2500
184
4800
iridium
iron,
7.25
forged
7.8
iron-oxide
5.1
lead
11.3
leather
09.
limestone
lithium
3.2.
magnesium
2)
1
W/(m
1
kJ/(kg K)
K)
2500
327.4
1740
:
=
59.3
58
0.58
179
1372
657
650
1110
1500
34.7
2.09
0.49
0.218
0.134
0.532
0.461
0.67
0.130
0.15
1.5
2.2
0.909
301.2
0.36
157
1.05
70. ..145
1.01
.3.6
1.74
alloy
0.44
46. ..58
..1.0
0.53
magnesium,
1200
1200
1570
2.6
magnesia
')
2450
22.5
iron, cast
2.3
47. ..58
1.8
0.3598 kcal/ (h
m
K)
2388 kcal (kgK)
>
TABLES
zs
Properties of solids
melting boiling
thermal
specific
po int
conduc-
heat
density
Substance
Q
kg/dm
manganese
t
3
7.43
marble
mica
2.0... 2.8
molybdenum
10.2
°C
°C
1221
2150
2.8
nickel
8.9
osmium
22.48
oxide of chrom
palladium
12.0
paper
0.7. ..1.1
5.21
paraffin
2600
1452
5500
2730
2500
2300
1552
5300
52
0.9
peat
0.2
phosphorbronce
phosphorus
8.8
pig iron, white
1.82
8.65
pitch
1.25
coal
21.5
porcelain
2.2... 2.5
potassium
0.86
quartz
2.5
radium
2930
300
280
2500
1300
0.381
0.14
88
0.24
0.95
125
39
1550
1.52
1500
220
1420
4.4
2.33
W/(m
1
kJ/(kg
K)=
0.2: $88
960
kcal/(h
kcal/(
m
K)
<gK)
1
0.92
0.32
*
1
1.30
0.2.0.35
700
2230
688
2600
3.12
105
K)
1.02
0.13
70
71
rubber, raw
1
0.38
0.092
•
2>
0.80
0.54
52.3
159
0.7
100... 300
= 0.8598
1.9
0.36
127.9
1.07
1)
3.26
2300
5500
rosin
silver
1.336
0.26
0.80
2500
carbide
0.24
0.14
9.9
1960
silicon,
70.9
2230
1140
12.3
silicon
0.130
075
1470
960
rhodium
2.1. ..2.5
0.272
0.461
0.8... 1.0
762-2
21.4
sandstone
selenium
0.87
145
59
4400
rhenium
1.4. ..1.6
0.84
1770
1650
63
950
3175
rubidium
0.46
2.8
0.35
0.24
8.8
sand, dry
kJ/(kg K) 2
»
110
8.6. ..9.1
red metal
K)
1
0.13
5
red lead
W/(m
cp
A
0.42
1.35
platinum
tivity
0.08
900
44
1560
1000
7.0. ..7.8
pinchbeck
pit
t
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
density
Substance
Q
t
t
kg/dm 3
°C
°c
slate
2.6. ..2.7
2000
snow
0.1
sodium
97.5
0.98
soot
1.6... 1.7
steatite
2.6. .2.7
7.85
steel
sulfur, cryst.
2.0
tantalum
16.6
tar
1.2
tellurium
6.25
thorium
- 15
455
1800
11 7
timber, alder
"
,
ash
,
birch
,
larch
"
"
0.49
0.70
0.138
0.19
4.9
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
7.5. ..10
300 .400
titanium
4.5
tungsten
19.2
uranium
vanadium
19.1
6.1
zinc, cast
6.86
zinc, die-cast
6.8
zinc, rolled
1
0.20
54
2.4
1.28
,
1
0.83
0.17
pockwood
"
»
2
47. .58
0.14
,
"
1
1.26
0.84
1.6
0.85
0.75
"
2>
0.07
0.16
pitchpine
"
0.76
4.187
0.75
,
"
4000
kJ/(kg K) 2
0.17
0.16
0.14
0.12
,
,
2500
445
4100
300
1300
126
>
0.55
0.75
0.65
0.75
maple
oak
"
K) 1
0.5
100
880
1600
1460
115
2990
W/(m
cp
A
tivity
W/(m
K)
kJ/(kg K)
419
393
419
7.15
=
-
0.85 98 kcal/(h
0.23 88
m
K)
kcal/(kc JK)
0.47
0.13
31.4
0.084
54.7
35.
1.4
0.24
0.24
15.5
110
140
110
1.3
.70
0.117
0.50
3.43
0.515
0.147
0.38
0.38
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
< t < 100°C and
p = 1.0132 bar.
melting| boiling
thermal
specific
point
conduc-
heat
density
Substance
Q
t
kg/dm 3
acetic acid
1.08
acetone
0.79
alcohol
benzene
0.79
0.89
benzine
0.7
chloroform
1.53
diesel
°C
°C
16.8
118
-95
-130
5.4
-150
0.88
-70
-5
ether
0.73
117
gas
0.86
oil
oil
glycerine
1.27
heating oil
f 10%
hydrochlor
acid
1 40%
hydrofluoric acid
0.92
linseed
1.05
nitric
3
1.56
0.94
of turpentine
perchlor ethylene
0.87
petroleum
petroleum ether
0.80
0.67
1.84
sulfuric acid
50%
1.49
trichlor ethylene
water
>
2»
1
W/(m
1
kJ/(kg K)
K)
78.4
80
50.. .200
2.43
1.80
0.17.. .0.2
0.137
0.16
2.1
61
175
35
200... 300
290
0.13
175.. .350
0.12
102
0.50
0.14
2.26
0.15
0.29
2.43
19.5
3.14
316
-5
380... 400
-41
357
66
86
-20
150. ..300
-10
-20
-70
-160
-10
160
119
2.33
0.15
2.09
0.126
1.68
8.4
0.138
2.51
0.26
0.15
1.72
0.10
1.80
150.300
0.159
0.905
2.09
40. ..70
0.14
1.76
0.5
1.38
0.14
1.59
338
1.40
sulfurus acid
toluene
>
1.62
sulfuric acid cone.
kJ/(kg K) 2
>
56.1
-38.9
-98
0.8
acid
oil
1
-20
13.6
of resin
K) 3
W/(m
cp
A
»
-92.5
0.91
oil
methyl alcohol
oil
-5
-14
1.20
0.99
0.96
oil
machine
mercury
-30
-20
3>
tivity
/
4'
0.88
1.47
4°
1 at
= 0.8598
- 0.23JJ8
kcal/(h
-73
-10
-94.5
110
87
100
-^86
m
K)
1.34
0.16
1.30
0.58
4.183
4
kcal/(k
JK)
at
t
=
'at
t
=
3)
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
melting|boiling
thermal
point
conduc-
density
9
kg/m 3
acetylene
tivity
t
I
°C
°c
1.17
- 83
1.293
-213
-
81
ammonia
0.77
argon
1.78
-192.3
- 77.9 - 33.4
-189.3 -185.9
air,
atmosphere
cp
1
K)
cv
|
kJ/(kg K) 2
'
0.018
>
1.616 1.300
1.005 0.718
0.02454
0.022
0.016
0.02
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
0.58
-230
1.26
-169.3 -103.7
- 270.7 - 268.9
-111.2 - 84.8
0.017
0.143
0.013
0.171
1.28
-210
-170
2.67
-145
-135
-
carbon disulfide
carbon monoxide
3.40
chlorine
coal gas
ethylene
helium
hydrochlor acid
0.18
hydrogen
hydrogen sulfide
0.09
gas
heat
A
2.056 1.568
0.52 0.312
1.05 0.75
butane, isobutane, ncarbon di-oxide
blast furnace
W/(m
specific
2.70
1.97
1.63
10
1
-210
2.14
0.36
1.59
1.47
1.173
5.20
3.121
0.795 0.567
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
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
1
W/(m
1
kJ/(kg K)
1
1.54
K)
=
0.8598 kcal/(h
=
02 388
0.00
-246.1
m
K)
14.05 9.934
0.992 0.748
151
0.25
2.19 1.672
1.03 0.618
0.024
0.024
1.038 0.741
0.909 0.649
-112
100 00
-111.9 -108.0
kcal/(k gK)
0.013
0.0088
0.030
0.046
3)
at
t
--
=
100°C
TABLES
numbers
Friction
Coefficients of sliding
and
sliding friction
static friction
static friction
\x
\i
on
material
material
-Q
bronze
bronze
0.20
0.18
0.18
cast iron
steel
oak
oak
rubber
hemp
0.20. ..0.40
0.15. ..0.35
||
oakf
cast iron
rope
cast iron
steel
0.17. ..0.24
asphalt
0.50
0.60
concrete
belt
cast iron
PTFE
POM
W
0.18
031
0.16
0.10
0.10
0.02. ..0.05 0.18. ..0.24
0.30
0.50
0.40
0.20. ..0.50
ice
0.014
steel
0.10. ..0-30
PE-W
0.20
0.30
0.50
0.40
1
»
0.40. ..0.50
PTFE
PA 66
2
)
0.03. ..0.05
3
>
0.30. ..0.50
POM
4
>
0.35... 0.45
PTFE
2
)
0.035. ..0.055
POM
4>
PE-W
0.2. ..0.7
0.11
0.02. ..0.08 0.15. ..0.30
0.10
0.10
0.50. ..0.70
0.40.
..(
Rolling friction
K 12 and L
9)
arm /
of
f rictional
force
in
mm
lever
material on material
0.10
0.15
0.50
rubber on asphalt
rubber on concrete
lignum vitae on lignum vitae
on steel (hard:
on steel (soft)
elm on lignum vitae
steel
steel
:
-H-:
ball
0.05
0.8
movement with grain of both materials
movement perpendicular to grain of sliding body
polyethylene with plasticizer
2)
polytetrafluorethylene
4)
0005. ..0.01
bearing)
1)
3)
0.50 0.12
0.26 0.02. ..0.10 0.50. ..0.60
0.027
(for section
II
0.10
0.19
0.10 0.05. ..0.15 0.40. ..0.60
0.50
0.08 0.04. ..0.12
0.40
wood
PE-W
0.11
0.50
oak
1)
0.06
0.08
0.07
timber
leather
steel
0.10
polyamide
polyoxymethylene
from BASF)
126 from Dupont)
CA from BASF)
Hostaflon C 2520 from Hoechst)
(e.g. Lupolen
(e.g. Teflon C
(e.g. Ultramit
(e.g.
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)
%
'/a
19
14
Vb
Threads per inch
28
19
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
Flow area
(mm 2 ) 30
119 201
581 1012 1371 2206
2
Ration of flow area (mm
to nominal bore (inches)
)
242
243
317 402
488
581
810 914 1103
Roughness k
(according to Richter, Hydraulics of Pipes)
Material and
kind of pipe
new seamless
rolled or
steel
pipes
cast iron
pipes
pipes folded
and
riveted of
sheet steel
k
in
mm
0.02. .0.06
typical rolled finish
drawn pickled
(commercial)
steel pipes
used
Condition of pipe
»
cleanly galvanized (dipping process)
0.03. ..0.04
0.07.. .0.10
commercial galvanized
0.10. ..0.16
uniform corrosion
about 0.15
pits
medium
medium
corrosion, light incrustation
incrustation
heavy incrustation
cleaned after long use
new, typical cast finish
new, bituminized
used, corroded
0.15. .0.4
about
1.5
2. ..4
0.15. ..0.20
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
new, depending on kind and quality
about 0.15
...4
...1.5
1.2
4.5
of riveting
light riveting
about
heavy riveting
to 9
25 years old, heavily incrusted, riveted pipe
12.5
1
TABLES
Zio
Heat values
Latent heat of fusion per unit
kJ
material
material
kg
aluminum
377
164
168
46
126
134
243
172
113
antimony
brass
cadmium
cast iron
chromium
cobalt
copper
ethyl ether
mass
kJ
glycerine
gold
kg
176 paraffin
147
67 phenol
109
manganese
335 platinium
205 potassium
23 silver
155 sulfur
mercury
11.7 tin
naphthaline
151
nickel
234 zinc
ice
iron
lead
kJ
alcohol
59
109
38
33.5
alloy
117
mass
/d
kJ
kg
material
kg
503 oxygen
88C hydrogen
)
ammonia
113
59
Wood's
kJ
material
kg
kJ
material
kg
Latent heat of evaporation per unit
at 101.32 kN/m 2 (= 760 torr)
material
/.<
141C mercury
281
carbon dioxide
59J methyl chloride
chlorine
29C nitrogen
214
402
sulfur dioxide
406 toluene
201 water
Calorific value
365
2250
Hu
(average values)
Hu
MJ
Solids
Hu
MJ
Liquids
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
brown coal
42.1
coal gas
furnace coke
30.1
gasoline
42.5
hydrogen
gas coke
29.2
heating
41.8
non coking coal
31.0
methyl alcohol
19.5
peat, dry
14.6
methylspirit (95%)
25.0
wood, dry
13.3
petroleum
40.8
9.6
Diesel fuel
oil
oil
4.1
5.2
119.9
10.8
methane
50.0
36.0
municipial gas
18.3
11.3
natural gas, dry*
43.9
39.0
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
MJ
(Cf.
i
K3)
56.4
TABLES
7
£m 11
Heat values
-inear coefficient of expansion i
=
at/
a/10
material
-6
...
material
aluminium
bismuth
23.8
German
13.5
brass
18.5
a/10
-6
porcelain
gold
14.2
quartz glass
ead
29.0
silver
bronze
17.5
molybdenum
30.0
nickel
5.2
cast iron
10.5
nickel steel
15.2
=
copper
16.5
platinum
Ni
steel,
8.5
mild
12.0
23.0
ungsten
1.5
material
4.5
line
9.0
Cubic coefficient of expansion
at/ = 15°C
y/lQ-z
0.5
19.7
in
36%
Invar
4.0
steatite
13.0
constantan
a/10- 6
material
18.0
silver
cadmium
material
1/K
in
100°C
)
y/10~ 3
in
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)
thiol*
material
3
10
ness of in sula ing lc iyer i mrr
20
50 100 120 250 380 510
i
reinforced concrete
4.3
3.7
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 =
ac =
ac =
2.45
4.90
7.35
N/mm 2
N/mm 2
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
material
acetylene
air
ammonia
carbonic acid
carbon monoxide
and molecular mass
A'
R
M
J
kg
kg K
kmol
319
287
488
189
297
26
29
hydrogen
material
17
oxygen
44
sulfuric acid
28
water vapor
C
C
W/(m 2 K 4
R
M
J
kg
kgK
kmol
4124
297
260
130
462
nitrogen
Radiation constant
material
M
2
28
32
64
18
C
at 20
C
material
W/(m 2 K 4
)
)
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
soot,
aluminum, unpolished
0.40 x 10~ 8
nickel, polished
0.40 x 10
brass, unpolished
8
1.25 x 10"
absolutely
ice
3.60 x 10" 8
black surface
silver,
iron,
tin,
polished
polished
Dynamic
/
in
r\
N s/m 2 =
5.30 x 10"
8
5.30 x 10~
zinc,
unpolished
8
5.30 x 10~
iron,
unpolished
5.40 x 10
motor
of
°C
SAE
1
viscosity
-8
8
smooth
brickwork
oils in
5.67 x 10
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
1
kg/(m
s)
=
1
Pa
s
- 1000 cP
-8
-8
TABLES
z 13
Heat values
Mean
in
t
°c
specific heat c pm
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,
Got tingen Heidelt erg: S pringer 1975.
I
11.
Ai iflage. Berlin/
TABLES
Zl4
Heat values
Liquids*)
t
C
Q
Substance
°C
water
octane C 8 H 18
ethane
C 2 H 5 OH
toluene
oil
*) Explanation of the
s
1791.8
1002.6
4.181
864-7
4.215
4.494
-25
738
719
0.144
0.137
1020
714
14.62
2.131
2093
2232
0.183
0.177
3241
1786
1201
701
3707
2.064
20
50
100
806
789
763
716
20
50
100
879
847
793
200
661
20
50
100
200
885
867
839
793
672
1435
1383
1296
1.33
20
50
695
636
609
4.45
0173
0.165
0.152
547.1
281.7
134.6
22.52
16.63
11.90
7.41
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
1821
1.968
1.612
1.717
1.800
1.968
2.617
137
1.48
185
2.06
2.19
20
60
100
866
842
818
2.09
0.141
0521
261
113
229
0.124
0.122
0.119
13546
0.139
9.304
1558
1260
2.366
0.286
1.5*10 6
11
11.11
326
871
Cf.
13.44
6.99
3.57
1.75
0.90
0.144
0.134
0.127
0.108
1.729
852
820
5
06685
2.801
20
50
100
ymbol.
0.5620
0.5996
0.6405
0.6803
2.395
3.454
561
20
3
-
988.1
958.1
mercury Hg
C3H8
Pa S
200
-50
spindle-oil
Pr
4.217
4.182
20
50
glycerine
kgK
10 6 *7
999.8
998.3
S0 2
ammonia NH 3
insulating
m3
A
W
mK
20
50
100
C7H8
sulfur dioxide
kJ
-25
benzene C 6 H 6
P
kg
2.31
196
0.02
1.24x10"
TABLES
Z
Heat values
Gases
(at
t
Substance
°C
air,
-20
dry
carbon dioxide
C0
1000 mbar)*)
Q
C
kg
kJ
m3
kg K
20
100
200
400
1.006
1.006
1.007
1.012
1.026
1.069
-30
2.199
0.013
0.015
0.016
0.022
0.030
12.28
13.75
14.98
18.59
26.02
078
0.70
0.70
0.69
0.68
0.66
0.0081
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
1.41
0.903
0.909
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
1.038
1.038
1.038
1.047
1
2.88
2.64
2.11
nitrogen N 2
1.23
1.13
0.90
25
100
200
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
tr ie
0.71
0.473
0.477
0.494
25
100
*) Explanation of
16.15
17.10
17.98
21,60
25.70
32.55
3.13
2.87
2.29
S0 2
saturation)
s
25
100
25
100
200
(at
Pa
Pr
1.784
1.422
1.120
ammonia NH 3
hydrogen H 2
0.023
0.025
0.026
0.032
0.039
0.053
10 6 r?
25
100
200
chlorine CI
sulfur dioxide
A
W
mK
0.800
0.827
0.850
0.919
0.997
2
2
P
1.377
1.275
1.188
0.933
0.736
0.517
1,951
oxygen
15
symt )OlS
cf
.
011
13.50
14.05
14.34
14.41
14.41
1.864
1.907
2.034
2.883
6.144
-
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
0.024
0.026
16.6
17.8
20.9
24.7
0.031
0.037
0.141
0.171
0.181
0.211
0.249
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
V|
Vj
d
CD
cb
T-
d
*
co
T- ft
A A
A
A A
o
CM
CM CM
CO
o CM
o o CD
o O O O
O
^
O
O
O
O o
O
o
3 o, O)
W t- s
CO CM
CM CO
"*
in
in in
< O O O O o o o
1^ O "* CO
iCO
m
< 1 iCM CM CM
CO
CO CO
m
m
in in
o
o
co
CD
o) co
en m
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
ia.
s
•^-
I
CM
ci.
•<*
c
©
d)
VI
"tf
£
-a
£
vj
o o
o o o o o
CD - O)
3
->
C
o
Vj
r*»
CM CM CM CO
-o
o
CO
m m m s o
o
CO 1^
CO
<t CO
o
CO
CO
-i-
CM CO CO
cd
JS
o
o
in in
130
180
210
250
380
440
«>
Jfi
reat
for
heat-1
f- CO
18)
<«•
E
co
._
o
o
o o Ev.
o
CO CM 00
CM CO CO
CD
p
o o
o
o
he
t
actor
o
O iO O
O o o
co
h- r^
m
co -t -t m
CD
CO CO
strength
elasticity
CO
<D
C\J
i
Tensile
of
o
o
o
in
o o
o o o
o
S m
co t- in
o o
en 00
a < ^ t- i- CM CM CO CM CM
Q-cvj t?
Modulus
CO
in
in
CO Is- CJ> CO
CM CM CM CO
>
o
en
CO CO
CO
2
<J
CM
210000
210000
210000
210000
C D
55 65
Grade
Grade
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
in
o
CM
o
o
o
m
o
CM
/vith
P
400
600
700
safety
and
a
2
F
specially
Allow
165000
170000
es
(see
185000
ameter,
*)
d
the
Grade
Grade
CO
Material
A-Standard
283
3
IUOJJ SUOI}
-BOjjpads
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
CM
536
ASTM
A
2
CO CM
oo
*
in
seeP2
igth
^-^2
O 0)0) to
S .E .= CD
DOS * £2 H
CD
o
WOJJ SUOjJ
-BOjjpads
•<>
;s
CO
Jj
co
on
O
Alter
CO CM
CO
<J>
CO
o
en
U:Undi
BS
2789
Note:
A:
TABLES
17
Strength and Machining values
Allowable bending and torsional stresses;
for elastic materials in
Modulus
Material
of
elasticity
Type
1
'
A
210000
1000
750
500
Yellow Brass
ASTM-B 134(274)
CDA-419 HV190
of
Pqt
650
500
350
120
100
80
200
180
150
80000
42000
142000
55000
110000
200 100
150
100 50
42000
120
100
80
117000
300
220
150
45000
200
180
150
Phosphor Bronze
CDA-529
G
300
250
200
ASTM-B 122(752)
Tin Bronze
rigidity
B
200 100
150
100 50
Nickel Silver
HV160
Pbt
110000
HV150
65-18
moduli
Modulus
of
loading
E
Spring Steel
SAE1078;
hard. + temp.
E and G
N/mm 2
HV190
for simple springs
for bent and shaped springs
for springs with no hysteresis effect
(safety factor
(
(
v For explanation refer to P 1.
For cylindrical helical springs use diagram on page
2)
Q
— 1-5)
-
"
-
"
"
^10)
3)
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
Permissible contact pressure p b
N/mm 2
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
hardened and ground: 1)2)
lubrication, shaft
m
s
Material
Pb
m
8. .12
Leaded
(836)
j.
20 3)
Red Brass
Leaded
(938)
Tin
Bronze
sintered iron
Pz
m/s
Cast Tin Bronze CDA90I
gray cast iron
3
15 3
grease lubrication
quality bearings
PA66
4. ..12
6G
^0
(polyamide)
>
15
009
dry
...1
grease lubrication
<1
5
0-35
'
HDPE
2. ..4
(high-density
sintered iron
with copper
002
polyethylene)
PTFE
12
sintered
bronze
(polytetra-
-*0
30
fluorethylene
006
enclosed)
PTFE - lead
+ bronze
20
tin-bronze
graphite
(DEVA
metal)
90 4
(GLACIER-DU)
>
<0005
0-5.
1404)
.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.
:
—
Normal values of p b under
shock load
undulating load
dead load
Material
bronze
cast iron
gunmetal
10
20
..
8
..
20
30
malleable iron
steel
..
..
..
15
30
12
30
50
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
is
lubrication
possible
3)
Limited
41
Specially developed metals
life
(wearing parts).
5)
I
For shell thickness
1
mm
TABLES
2:i9
Data for clutches and brakes
Properties of friction materials
(Q 15
"CO
...
Q
17)
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
«-
O
_
Q)
CO
CVJ
-
Pb
-
o
c
CO
c
'c
c
B
-
c
N/mm
in
d
d
T3
oo
CO
CD
o
O
o
°
T
(O X
°C
transient
o
o
o
o
perature
ten
Max.
o
o
CO
o
m
CO
o
o
o
m
o
m
O
w
*O
CO
1
o
°
co
-r
1-
in
*
CO
°C
continuous
o o
o
o
CD
CD
o
m
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.
CO
E
3
Q)
*
5
friction
a)
Q.
CD
o
E
o
en
CO
CD
general
lining/steel
organic
ll
|
CO
a
or
£
2
c
CD
O
C
o
o
en
0)
0)
C
-Q
Q)
O
CO
5?
o
CD
0)
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
c
o
1
75
w
II
II
=
0)
CO
03
o
/
N
C
o
m o
-
£
to
"CD
a)
CO
00
(9^ JM
CO
^
^
=L 3.
~
ST
«
Qco
m
TABLES
20
Work w and
0,5
1,0
yield strength kf
2,0
1,5
SAE3310
0,25
<p
:
0,50
0,75
•
1,00
0,5
logarithmic deformation ratio
|
w
energy per unit volume
For other materials see VDI 3200
:
strain
1,0
1,5
<P
Af
:
yield strength
2,0,
TABLES
Z21
Electrical properties
and
Electrical specific resistance g
conductance y of conductors at
specific
Q
material
aluminium
antimony
brass - 58% Cu
brass - 63% Cu
cadmium
m
m
Q mm
36
iron (pure)
0.417
2.4
lead
0.059
17
0.071
14
magnesium
manganese
cast iron
chromium-Ni-Fe
constantan
copper
silver
gold
graphite
13.1
40
0.025
1
1
= 20 C
y
Q mm 2
m
m
Q mm 2
material
2
0.0278
0.076
carbon
German
Q mm 2
/
9
Y
mercury
mild steel
nickel
0.10
10
0.208
4.8
0.0435
23
0.423
2.37
0.941
1.063
013
7.7
0.087
11.5
2.0
0.10
10
nickeline
0.5
0.48
2.08
platinum
0.111
9
0.0172
58
silver
0.016
62.5
0.12
8.3
0.369
2.71
0.0222
45
8.00
0.125
tin
tungsten
0.059
17
zinc
0.061
16.5
Electrical resistance g of insulators
material
9
14
baskelite
10
glass
10 15
marble
mica
10
10
paraffin
oil
paraffin
wax
(pure)
Electric
materia
Q cm
plexiglass
aluminium
brass
carbon
constantan
copper
German
silver
graphite
manganese
10
15
polystyrene
10 18
10
porcelain
10
14
17
pressed amber
10
18
10
18
vulcanite
10
10
18
water, distilled
10
temperature coefficient « 2 o
at
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
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
4
3
soft
25
olive
oil
rubber
4.7
paper
25
paper,
glass
5
guttapercha
hard paper
4
paraffin
paraffin
(laminated)
insulation of high
voltage cables|
insulation of telephone cables
4.5
petroleum
phenolic resin
1.5
plexiglass
polystyrene
porcelain
4.4
water
marble
8
pressed board
4
oil
1
|
impregnated
4.2
6
2.3
sulfur
teflon
5
8
transformer
mineral
transformer
vegetable
turpentine
3.2
vulcanised fibres
3
vulcanite
oil
2.2
wax
2.2
El ectro-motive
2.5
steatite
ebonite
caster
£<
6
oil
compound
£r
mica
2.2
3.5
2
oil
2.2
oil
2.5
2.2
2.5
80
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
Magnetic
strength H and relative
as a function of induction B
field
permeability
// r
steel casting
cast ron
T.34
m
2
tesla
0.1
0.2
0.3
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
-
H
B
G
[
[gauss
A/m
Mr
-
dynamo
|
allow sheet and strips steel
P1.0
induction
alloyed
i
and dynamo
=
H
P1.0 = 1.3*
3.6^
kg
kg
H
Mr
A/m
-
Ur
A/m
-
I
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
practica limit
181
215
243
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
power
T in W/kg
low
medium
3.0
2.3
high
stray
at 1:0
thickness
3.6
mm
density kg/dm
1.3
1.5
0.35
0.5
3
7.8
7.75
7.65
7.6
core
losses
per unit
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
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
fl 2 5 = 1-53 Vs/m
2
indicates that a minimum induction of 1.53 Vs/m
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.
[or
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
Type
for
il
lumination
Es/ in Ix
of establishment
lighting
or location
lighting
bench
only
rough
workshops
to
work done
precise
very precise
normal
offices
medium
lighting
streets
and
20
50
100
light
medium
traffic
heavy
factory yards
light
with traffic
heavy
200
500
1000
1500
200
500
bright
squares with
general
50
100
200
300
500
750
open
rooms,
living
100
200
300
500
medium
according
= lm/m 2
General and spec.
General
20
50
Lumin ous efficac
Type
of
Colour o f illuminated surface
medium
dark
ighting
light
direct
0.60
0.45
indirect
0.35
0.25
street
0.30
0.15
deep bowl
and
widespread
reflector
0.40
0.4 5
Lum inoui
Standard lamps
Pel
with single coiled
filament
(at operating voltage)
<P V
i
flu: (
w
V
40
60
75
100
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
<Z> V
mps
0.23
klm 0.12
Pel
of la
25
15
40.0
tubular
Fluorescent lamps
values for
diameter
26
mm
Pe\
V
'Warmwhite'
'Daylight'
38
mm
Pel
V
High-pressure
lamps filled with
mercury vapour
Pel
</>
v
|
W
18
36
58
klm
1.45
3.47
5.4
W
15
klm 0.59
W
125
klm
6.5
20
40
65
1.20
3.1
5.0
250
400
700
24
42
1
4
1000 2000
60
125
TABLES
26
—
Statistics
1
<p(x)
,
I
0-398 942
0398 922
0398 862
0.398 763
0398 623
0398 444
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
0.392 531
0.391 806
0.391 043
0.390
0.389
0.388
0.387
0.386
0.385
0384
0.383
0.382
0.381
0.380
242
404
529
617
668
683
663
606
515
388
226
*oW
erf (x)
A
cp(x)
0000 000
0007 979
0.000 000
0.50
0352 065
0011 283
051
350 292
0.022 565
0.033 841
0.045 111
0.52
0.348 493
0.346 668
0.344 818
0.056 372
0.55
0.56
015 957
0.023 933
0.031 907
0.039 878
0047 845
0.055 806
0.063 763
0.071 713
0079 656
691
0.211
519
332
129
908
670
413
136
840
522
184
823
439
032
600
143
0.222
0.233
0.244
379
389
0.181
0.189
0.197
0.205
0.212
0220
228
0.235
0.243
0.258
0.266
0.273 661
0.281 153
0.288 617
0.296 054
0.303 463
0.310 843
0.318 194
366 782
263
714
135
527
890
225
533
812
0.60
0.167
0.179
0.189
0.200
0.375 240
373 911
0.372 548
0.365
0.363
0.362
0.360
0.358
0.357
0.355
0.353
463
623
758
867
947
996
012
992
936
840
702
522
296
022
700
326
900
418
880
283
627
908
126
279
365
382
330
206
009
739
392
969
468
887
225
482
656
745
750
668
235
119
990
847
0.377 801
0.376 537
0.368 270
057
058
0.111 340
0.119
0.127
0.134
0.142
0.150
0.158
0.166
0.174
325 514
0.332 804
0.340 063
0347 290
0.354 484
0.361 645
0.368 773
0.375 866
0.54
0.078 858
0.090 078
0.101 281
0.112
0.123
134
0.145
0.156
0.087 591
0.095 517
103 434
0.251
0.369 728
0067 622
053
255
0.265
0276
0.286
0.297
0307
0.318
0.328
0338
0.349
359
0369
0.399
0.409
0.418
0-428
0.437
0.447
0.456
0.466
0.475
0.484
0.493
0.502
0.511
0.59
0.61
0.62
0.63
064
0.65
0.66
0.67
0.68
0.69
0.70
0.71
0.72
073
0.74
0.75
0.76
0.77
£K-
erf(x) =
$oW
379 031
0.371 154
df;
e
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
0.307 851
0.305 627
0.303 389
0.301 137
0298 872
0.296 595
o (x)
erf (x)
382 925
0.389 949
0.396 936
0.403 888
0.410 803
417 681
0.520
0.529
0.537
0.546
0.554
0.424 521
0.571
0.431 322
0.579
0.438 085
0.444 809
0.595
0.451 494
0.458 138
0.464 742
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
0.563
587
603
500
244
899
464
939
323
616
816
923
937
856
0.611 681
0.619 412
0.627 047
0.634 586
0642 029
0.649 377
656 628
0.663 782
670 840
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
078
294 305
0.79
0.292 004
0.80
0.289 692
0.287 369
0.576 289
582 060
748 003
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
0587 784
0.753 811
0.759 524
0.765 143
0.81
0.82
0.83
0.84
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
0.258 881
0.256 471
0.254 059
0.251 644
0.249 228
0.246 809
0.244 390
0.593 461
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
178
427
629
782
888
945
954
914
826
0.801 883
0.637
0.642
0.647
0.652
0.657
0.662
0.667
0.672
0.677
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
1.00
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
<p(x)
0.241 971
551
0.239
0.237
0.234
0.232
0.229
0.227
0.225
0.222
0.220
0.217
0.215
0.213
0.210
0.208
0.205
0.203
132
714
297
882
470
060
653
251
852
458
069
686
308
936
571
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
0.187
0.184
0.182
0.180
0.178
0.175
0.173
1.30
0.171
1.31
0.169
0.166
0.164
0.162
0.160
0.158
0.156
0.153
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
1.45
1.46
1.47
1.48
1
.49
543
235
937
649
371
<Po(x)
0842
701
150
0.846
0.850
0.854
0.858
810
838
784
650
436
144
773
326
803
205
533
788
1.51
0-701 660
0.706
0.710
0.715
0.719
0.724
282
855
381
858
287
0.728 668
0866
0.869
0.873
0.876
0.880
0.883
0.886
0.741 524
0889
0.745 714
0.749
0.753
0.757
0.762
0.765
856
951
999
000
953
971
0.893 082
0.896 124
0.899 096
0.902 000
0.904 837
0.907 608
152
1.53
1.54
1.60
1.61
1.62
1.63
1.64
165
166
1.67
1.68
1.69
0915 534
0.781 303
0.918 050
0.920 505
1.72
1.73
1.74
700
331
727
639
564
505
0.141 460
0844 392
0.806
0.809
0.813
0.816
0.819
0822
0826
0.838 487
0.841 460
0.847 283
0.850 133
0.852 941
0.855 710
0.858 438
0.861 127
0.863 776
0.922
0.925
0.927
0.929
0.931
0.934
0.936
0.938
0.940
0941
0.943
0.945
0947
949
950
0.952
0.953
0.955
0.956
0.958
0.959
0.961
0.962
0.963
0.964
900
236
514
734
899
008
063
065
015
914
762
562
313
016
673
285
853
376
857
297
695
054
373
654
898
<f(x)
0.129
0.127
0.125
0.123
0.121
118
0.116
0.114
0.112
0.110
0.109
107
0.105
103
0.102
0.100
0.098
0.097
0.095
0.094
0.092
0.090
0.089
0.087
1.57
1.58
1.59
1.71
1.75
1.76
1.77
1.78
1.79
0086
1.80
0.078
0.077
0.076
0.074
0.073
1.81
1.82
1.83
1.84
1.85
186
1.87
1.88
189
1
90
1.91
1.92
1.93
1.94
195
1.96
1.97
1.98
1
99
=
-im V'.d,
V>T.
56
1
1.70
0.788
0.792
0.795
0.799
0.802
erf(jc)
0120
0.910 314
0.912 956
0.785 024
df;
1.55
0.769 861
0.773 721
0.777 535
471
417
418
435
0.131 468
0.862
0.733 001
0.737 286
0835
431
X
erf (x)
682 689
0.151 831
0.139
0.137
0.135
0.133
-
0.687 505
0.692 272
0.696 990
0.829
0.832
0.149
0.147
0.145
0.143
f
--JU fe" 2
V2FJ
915
455
949
399
804
165
482
755
984
170
313
413
104
847
602
369
147
937
740
555
383
225
080
948
2
*
fW-^yi*
0.084
0.083
0.081
0080
0.072
0.070
0.069
0.068
518
583
665
763
878
009
157
323
505
704
921
155
406
675
961
265
586
925
282
657
049
459
887
333
796
277
776
293
828
380
950
538
143
766
407
065
740
433
144
066 871
0.065 616
0.064 378
0.063 157
0.061 952
0.060 765
0.059 595
0.058
0.057
0.056
0.055
441
304
183
079
<P (x)
erf (x)
0.866 336
0.868 957
0966 105
0.871 489
0.967 277
0.968 414
0.873 983
0.876 440
0969 516
0.878 858
0.971 623
0.881 240
0972 628
0.883
0.885
0.888
0.890
0.892
585
893
165
401
602
0894 768
0896 899
0.970 586
0.973
0.974
0.975
0.976
0.977
0.898 995
0978
0978
0979
901 057
0.980
903 086
0.981
0.981
0.905 081
907 043
0.908 972
0.910
0.912
0.914
0.916
869
734
568
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
0.983
0.983
0.984
0.985
0.985
0.986
0.986
0.987
0.987
0.988
0.988
0.989
603
547
462
348
207
038
843
622
376
105
810
493
153
790
407
003
578
135
672
190
691
174
641
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
,
,
acid-base-indicators
action diagram
basic structures
- - rule for addition
,
,
-
line
active
11
U
T
T
T
T
4
3
4
4
3
9
7
S 18
D 9
D 10
,
relationship
6
Q20
18
algebraic equation, definition
general solution
between
D
D
D
zeros and coefficients
of any degree
- expressions
D
alternating current
S
9
9
9
.D 12
solution
zeros, roots
,
G
T
Q
addendum
admittance
,
2
2
S29
power
actuator
,
L
L
L
L
9.
2 D21
16. ..S29
S22
bridge
,
angle of twist
angular frequency
annuity interest calculation
annulus
apparent power
A.Q.L.
arc differential
- length
- of circle
Archimedes principle
area expansion
,
units of
arithmetic mean
- series
asymptote
atmospheric
axial
I
relative humidity
shear stress
P20
S16
D25
K13
belt drive
P 9 P 10
bending
- in one axis
P24
P28
P24
shafts
D
American hardness
U 6
T 3
amplitude response
for the whole circuit
T24
P 5, P 6 P 7
analysis of forces
A 1
Angstrom
-
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
two planes
- moment
- of uniform beams
- stress
Bernoulli Differential
Bernoulli's theorem
bevel gears
K
P
2
.P 13
1 1
P 9 Z 17
Equation
J 10
,
N
Q
4
Q25
O 3
24
bimetallic strips
binomial series
D 18
B 3
S 18
- theorem
D 4
- Bode diagram
T3, T22.T23, T 29 T31
boiling point
Z 1. ..Z 6
G
11
bolted joints
14
14
bolts
Q
Q
boring
R
bottom clear
- tearing
Bow diagram
bracket attachment
brakes
Q18
I
I
K
N
A
D
D
7
3
3
1
17
17
F 3
U 6
P 19
,
R
K
Q
Q
build
up time
axiomes to the probability
G
1
axles
Q
2
band brakes
K 13
bankholder forces
R
7
calculation for a controller
calculation of module
calorific value
barrel
C
4
candela
buoyancy
1
7
6
1
17 Z 19
Q17
brake torque
branching point
buckling
1
2
T
P 22
3
P23
T
N
2
3
T18
Q28
Z10
V 1
S3
capacitance
S 12
S 12
- of coaxial cylinders
capacitor power
S31
A 3
carat
Cardan
L 10
joint
D23
Cartesian coordinate system
Cavalieri principle
center of gravity
14,
central limit theorem
centrifugal force
changes of state of ideal
1
C
K
15
I
G
M
gases
1
8
3
5
5
characteristic angular
frequency
charge
chemical elements
- terms
choke
U
,
.
.
2
B
3
arc of
B3
B3
sector of
segment
of
circumcircle
Clairaut' Differential equation
clamped
1
2
1
joint
Q
,
F
2
7
7
7
E 6
J 11
Q 3
K
K
K
Q
16 Z 19
coaxial cylinders, capac. of
S 12
P 2
Z 7
Z 11
clutches
15
coeff. of elongation
friction
Q28
- - heat transmission
M
restitution
Z
Z
Z
rolling friction
- -
sliding
static friction
non-magnetic
cold working of sheet
combination of stresses
combinations
comparator
T
6
D30
4
2
G
conductance
S
2
1. ..Z
10
6
Z
7
list of symbols
control factor
control loop, rules to determine
the transfer function
control loops, components
quantities and
functions to describe the
1
T34
T
7
T
T
9
4
,
dynamic behaviour
T
T
T
T
T
controlled system
- variable
,
- -
overshoot
point of measurement
controller
- output variable
controlling element
.
,
,
,
T
T
T
setting rules for P-, PI-
7
4
5
8
5
6
6
6
T20
,T26
choice of the type
determinations
T 25
proved
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
P
7
conditional probability
conduction of heat
conductivity, thermal
10
M
P
M
P3
2
G
conical pendulum
control engineering terms,
S32
4
D 12
interest
2
15
confidence statement
crosshead guide
cross section, nominal
- switch
cube
cubic coeff. of expansion
cuboid
T
loop
- pendulum
- wound motor
compression
compressive stress
7
R 6
P24. .P29
D 5 D 6
complex numbers
D29
components of the control
compound
7
S25
S24
S24
Q 8
S24
coercive force
coil, high frequency
- low frequency
coiled spring
coils,
8
7
C
Q
C
,
3
S26 S27
coil
circle
-
T
S
U
U
cone
- clutch
- frustum
function
current
- density
current rating
- source
2
11
K
M
6
6
10
15
Q
Q
S37
S37
C
1
Z
11
C
G
G
9
S
S
1
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
dynamo sheet
Q18 Q20
dedendum
deep drawing
R
P 12
deflection
deformation efficiency
1,
Z
1
..
,
6
P20
,
R
R
T
.Z
N
8
8
1
6
3
D25
H 4
H
H
..
D
Descartes theorem
desiccants
determinants
determination of o
U
D7 D
G
3
6
9
6
8
6
a (PID) -T, controlling
T31 ..T33
a PI- or P-controlling
element (example)
diagram-hub geometry
diameter, base
-
,
..T32
Q 5
Q 19
Q 19
Q20
Q19
root
•
tip
Q
19
Z
H
H
dielectric constant
difference coefficient
differential coefficient
- equations
,
order
- - ,2 nd order
th
n order
,
,
- -
,
1
.
.
definition
st
,
J
1
1
J 12
J
,
.
.
.
.
reduction of order
with constant
J 5
coefficients
D 17
DIN-Series
dioptrics
direct-current machine
1
1
J 10
J 4, J 9,
J 4, J 5, J 11
J 5
J 2
linear
partial solution
2
J 12
J
J
J
J
,
.
.
,
R
V
4
1
S32
T
R
6
3
U
I
N
1
6
18
,Z 12
Z23
properties
Eigenangular frequency
T
elastic curve
electric circuits
P 11
-
S
measurements
S8 ,S
S8 S
- networks
calculation
,
-
,
9
9
S
specific resistance
temperature coefficient
Z21
Z21
work
S
work
in
S13
Z22
B3
,
English hardness
U
1
F
4
U
6
4
4
enthalpy
entropy
epicyclic gearing
equations of any degree, approximate solutions D 13 ..
equilateral triangle
B2
equilibrium
equivalent beam
- dead time
Q26
.
D 16
D26
K 4
P 15
T
2
G
G
8
8
8
6
- molecular mass
error curve
equation
expansion, area
differential
,
linear
of gases
of liquid bodies
of solid bodies
,
volume
1
S 12
electro-magnetic rules
-motive series
elements, chemical
-
1
an
electric field
-
5
power
electrical
-
1
S12
S11
field
electrical
Euler formula
7
3
eddy currents
7
J
1
L
Q 12
S25
8
,
V
Q27
eccentricity, relative
function
variable
3
8
,
inertia
ellipse
T30
reference
,
R2
drilling
dynamic moment of
dynamic viscosity
S33
element (example)
7
1
D 17
derivative
derivatives
9
F
Q17
Q 7
4
decimal-geometric series
- ratio, logarithmic
delay time
density
N
- determination of
deposit calculation
T
disc brakes
- springs
distance law, optical
- -time curve
disturbance variable
drive worm
drying agents for diseccators
damping ratio
D.C. machines with
commutating poles
T
direct feedback
direction cosines of vectors
T
P22
J 11
3
3
4
3
3
3
G
G
expected mean
experimental probability
exponential curve
- equation
- functions
extended feedback
3
1
F
4
D 4
H 5
T 10
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
D23
L
U
.
15 Q
K9 ,Z
Q
N
6
,
for fluids
K
frustrum of cone
C
C
pyramid
functional block
functions, exponential
-
,
,
,
,
7
3
K13
rope
sliding
of
17
Z 8
Z 19
Z 15
,
-
3
3
K13
,
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
12
9
Z14
gases, tables
laws
O
S
gauss
Gaussian curve
G
Q
18
..
general equation of state
generators
Q29
Q 5
4
- series
D 17
D31
application of
German hardness
golden section
gradient of a curve
H
graphical determination of a
T 21
linear controller
4
3
8
S33
D17
geometric mean
U
6
D32
1
..
H
2
.T32
Grashof number
group combinational elements
(PD)-T, and (PID)-!",
012
gravitational force
K 1
N 2
V 5
R 3
- pressure
grinding
1
2
A
A
04 ,Z
08
,
gray
L
S
,
- United States
gas constant
- mixtures
8
5
,
T28
T8 T20
realization
-
gallon, Imperial
,
forming device for the referFourier series
Q
locked joints
losses in pipe flow
materials, properties
numbers
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
M
oscillations
M
6
heat conduction
-
latent
of sublimation
radiation
,
removal
by lubricant
transfer
coeff.
10,
-
transmission
values
helical spring
henry
7
integral, definite
O 11
O 7
O 2
O 2
O 10
Q13
Q 14
O 10
flow
-
involute toothing
10
10
exchange
exchanger
,
6
6
4
Z 14
11
11
10,0
Z 10
..
012
.Z13
Q 9
S
S
-
indefinite
-
,
1
D 10
D 12
..
Hurwitz criterium
hydrodynamics
hydrodynamics values
N4
..
hydrostatics
N
..
1
N
Z
N
hyperbola
F
F 5, F6
hypergeometric distribution
H
hyperbolic functions
- power
- work
9
9
E
F
- elastic
- plastic
impedance
M
M
M
,
,
S 19
,
5
5
5
5
5
5
Johnson formula
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
E
inclined plane
independent events
induced voltage
inductance
induction motor
inertia,
dynamic moment
K 10 L
,
G
S
moment of
inhomogeneous differential
equation
4
,
of
correlation-table
16,
1
variable transf.
02 ,Z
lattice girder
K5
1
K
10
6
S14
hand
S 13
A 1
V 4
V 4
V 2
length, units of
lens equation
J
,
leakage flux
S14
S34
I
D27
D26
D26
latent heat
left
19
19
T 10
Laplace Transform
Laplace transformation
differential equations
- - translation
S 15
I
I
6
9
6
7
D28
D26
D26
D28
..
calculation rules
4
8
8
8
1
%
S20
,
incircle
6
V5 V 6
S25
ionizing radiation
7
,
impact
7
D23
8
V1 Z21
illumination
2
,
convolution
N
ideal fluid
15
P
S
G
S25
S25
S25
hysteresis
2
2
joule
Q 4
T12 ,T18
hub dimensions
19
D 17 Z22
- hyperbolic functions
isothermal state
isentropic
isobare
isochore
isothermal
2
I
- enthalpy of a mixture
International E-series
inverse circular functions
- Fourier transformation
isentropic
J
I
I
iron losses
differential
I
I
1
equation
Horner method
..
energy of a mixture
2
homogeneous
1
numerical
4
P 18
1
I
rules
L
S24
14
1
,
internal
6
I
11,12
integration
- application of
- by parts
- by substitution
B
C
Q23
..
S37
hexagon
Hooke's law
hollow cylinder
18
instruments, electrical
hertz
high frequency coil
Q
installation
rule
lenses
light refraction
lighting values
linear coefficient of
Z25
expansion
Z
11
measuring instruments
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
P
H
D
flux
of lamps
- intensity
- quantity
Mac
V
V
field
- strength
S
- flux
- induction
magnetomotive force
4,
S 14,
S 14,
S 3.
S 3.
S4.
magnifying lens
manipulated variable
manipulating point
,
S 15
Z23
S 14
S 14
S 14
V 4
V
R2 R
V2 V
mirror
,
units of
maximum
Q
Q
module, circular
- normal
P 18 Z
of elastic ty P 3.
Z
P
rigidity
Mohr's analogy
- stress circle
molecular volume
- mole fraction
moment
1
8
K
of a force
inertia
I
16
.
- theorem
rolling
,
rotational
,
simple harmonic
,
sliding
18,
M2, M 3
A 2
P27
S 3
P 18
maxwell
mean shear stress
- specific heat of various
gases
- value
mechanical oscillation
mechanisms
meaning of derivative
measure of plane angle
L
L
L
L
L
L4
,
,
,
,
,
S32
multi plate clutch
switch
13,
M
2,
1
5
9
6
7
9
density
- distribution
- probability curve
G
8,
D 14
S 24
G
G
G
3
6
L 10
H 3
E 1
T 5
7
7
7
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
H
1
19
S32
S32
S34
S32
S34
S34
S34
Q 15
S37
compound wound
- direct current
- induction
- shunt
- synchronous
- three phase
motors
D
shear stresses
I
L4
,
motor,
.
K
motion, linear
-
4
4
3
6
6
18
18
17
17
14
P13
non-magnetic coils
normal curve for probability
of inertia
measuring equipment
U
U
ds
equivalent
moment
T18
control loop for stability
milling
flu
7
P23
microscope
mixture rule for
- cross
6
9
Newton's approximation method
1,
fraction
,
I
-
1
Z
S
R
..
method of bucklinc coefficient
methods for check ng the
circuit
M
mass
-
1
D 18
V 4
Laurin series
macro photography
magnetic
4
S22
Q 12
Q 13
Z25
V1, Z25
Z25
lubrication film
lubricant flow rate
luminous efficacy
-
8
6
1
S8
R6
metalworking
Q21, Q22, Q24, Q25
loads in beams
logarithmic functions
logarithms
low frequency coils
Z
mesh-work
modulus
load capacity of tooth
S36
electr.
melting point
,
I
oblique angle triangle
octagon
S2
Ohm's law
,
bags
- slots
open loop gain
oil
transfer function
operating characteristic
optical distance law
optics
5
13
13
T
T
7
7
11
2
G
3 10
V
V
ordinary differential equat on
harmonic
mechanical
overshoot
- of the controlled variable
oscillation,
-
S
Q
Q
,
1
(PID)-^ controlling
element, characteristics
of a control loop
PI- or
-
normal
,
Q
plain key
plane, inclined
-
J
1
L
4
mirrors
planing
M
6
2
8
plastic yield
point of disturbance
Poisson distribution
T
T
T20
Q23
Q18
Q18
pinion, dimensions
pitch, circular
,
R2
G4 G
,
P
T
- second moment
T20
Pappus theorems
I
parabola
parallel combination
- combinational element
elements PD, PID
F
2
T
9
T15
T16
PI
C
parallelepiped
parallel-flow-exchanger
- resistances
- resonance
6
S21
B
partial fraction
differential
D 3
expansion
equation
particular
1
J
1
,
J
1
J
2
D
4
of area
I
17
polygon
B
2
polytropic state
potential difference
- divider
S
S
6
2
8
M
1
power
- active
- factor
S31
S31
S31
S31
A3 A 5
,
correction
1
11
S
parallelogram
-
15
1
6
10
D30
polar coordinate system
P- or (PD)-"^ controlling
element, characteristics
of a control loop
3
K 10
V 3
R 4
- reactive
- units of
powers
Prandtl-number
,
,
,
D
1
12 ,Z 15
U 5
11,
preparation of chemicals
pressure
- hydrostatic
- in a fluid
N
1
N
1
1
,
N
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
G
G
D
,
,
,
,
P2, P
18,
Z 16
.
.
permittivity, absolute
-
,
Z22
relative
permutations
phase crossover angular
frequency
- margin condition,
realization
- margin
- response
Z 18
S 12
D5
,
D
6
T
7
T27
T7 ,T20
T
3
representation
T25
- shift
photometric radiation
S 17
,
equivalent
photometry
pH-values
V 1
V 1
U 4
,
*
probability density function
- distribution
-
,
horizontal
,
vertical
P
L
L
L
properties of gases
liquids
solids
Z1
:
Z
Z
.Z
- ,4th
- mean
pyramid
- frustrum
,
C
C
,
factor
quadratic equation
quantity of electricity
4
2
2
17
2
8
8
8
6
5
4
D32
D32
D32
proportional, 3rd
Q
2
1
1
S 17 S21
D 1
S 2
,
radians
E
radiation
10
Z 12
- constant
radius of gyration
M
ramp response
random variable
range of an ammeter
- - a voltmeter
T
G
2
1
2
2
G
3
S
S
11
11
ratio of
slenderness
reactance
Z 17
S 18
reactive power
reagents, chemical
S31
U
N
B
T
T
V
real fluid
rectangle
reference variable
- - adjuster
refractive index
Regula
falsi
related step response
relative
- eccentricity
S4
reluctance
rem
remanent
-
,
moment
,
rolling
in
5
5
2
S2
of
S6
S6
series
- parallel
resisting moment
resistor combinations
,
7
S
S
5
5
7
P
K 12
S 7
S 7
11
S21
S21
restitution, coefficient of
M
reversible processes
8
7
6
2
Reynolds' number
right angled triangle
N
- hand rule
method
Rontgen
- unit
S 13
K 5
V 5
E
A
1
12 Z
7
rolling friction coefficient
L9 K
root diameter
roots
K 12
S 16
Q
19,
rotation
rotational motion
rough plane
roughness in pipes
Q20
D
1
K 13
rope friction
- operated machines
K 14
L7,
M
L
R
Z
4
4
3
9
,
B3
B3
circle
sphere
self-induction
series
D 17
- arithmetic
series combination
series combinational element
..
,
,
,
P
2
D
7
9
8
F
F
K
11
Q
1
D 15
P
9
19
T 15
P 9
I
,
K
K
K
7
7
7
C
3
S 15
D 19
D 17
T 9
T 17
geometric
T16
D17
Mac
D 18
and D-T,
I-T,
-
9
L10
I
12
S 14
V 5
S 10 S
- resistance
root-mean-square
- product
Scotch-Yoke mechanism
screws
- and bolts
secant approximation method
second moment of area
19
- - - volume
- order delay element
T 1
section modulus
- - circle
segment of a
1
resonant circuits
- frequency
Ritter
F8
scalar
sector of an annulus
2
P20 P21
S6, S
Sarrus rule
T
S25
flux density
resistance
- of a conductor
resistances
1
T12
T
safety factor
D 16
G
Q
frequency
5
4
rules for the normalized form
of the transfer function
- to determine the
transfer function
Laurin
S32
motor
resonance
S21
D 18 D 19
Taylor
shaft-hub joints
shafts
,
Q
..
Q
Q
5
2
R 4
P 18 P 19
P 18
P 1
- modulus
- stress
due torsion
P20
shearing force
- deflection
of a
beam
shrinkage allowance
shrunk-on ring
shunt motor
side-rake angle
Simpson's
3
R2
shaping
shear
rule
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
Sommerfeld number
G
special distributions
spectral energy
specific conductance
- heat
- heat, liquids
12
9
D23
2,0
Z
Z
.Z
1
6
-
,
P
compressive
in
,
,
- heat of various gases
- lighting
- resistance
speed diagram
- rational
Z15
Z 17
stresses
sphere
C
C
C
C
Z14
,
spread
..
spring, coiled
spring, disc
-
,
helical
,
laminated leaf
,
leaf
,
torsion-bar
rate
Q
spur gears
square
-
1
L
1
2
3
3
3
4
19
9
8
6
9
Q
Q
Q
Q
Q
Q
Q 7
Q 7
Q 6
Q 8
Q21
splined shaft
3 6
R
18
B
1
S23
coil
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
- moment
of a
body
statistic-tables
Steiner's theorem
stiffness
steel tubes
,
M
2
,
S34
tangent rule
taper joint
Q
target variable
Taylor series
D
6
3
6
8
A
3
- stress
tension
P
theorem
,
,
shear stresses
Tetmajer formula
- stresses
of
rel.
P 1
P 3
P 4
S 3
P 18
P22
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
time, units of
,
tonne
A
A
tooth width ratios
Q23
toroidal coil
1
torque
torsion
P 4 P 11 ,Z20
Z 16 .Z 18
P 1
P3 Z 17
,
1
P3
tesla
1
bending
1
,
3
,
T
- units of
tensile force
F
-
E
temperature
P
.
2
8
S
synchronous motor
straight line
strain
- energy
I
superposition
Stokes
strength values
stress
5
Q 2
P27
two dimensions
substitution method
in
Z
T
N
step response
M
6
9
2
Q
14
14
9
Z26 ,Z27
16
P25
curved beams
rotating bodies
shafts
-
9
P 1
P 3
Z 17
P 10
I
1
,
M
I
a curve
- stress
in
1
threephase current
S30 S31
- motors
S34
- output transformers
S35
- power, measurement
S30
T 8
time to reach lower tolerance
steady state
T 8
Z20
static friction coefficient
Q
P
,
Z21
- with conical boring
cylindrical - zone of a
springs
-
3
P 2
P 18
static
strain diagram
tensile
stress, torsional
..
P
Q 9
P28
three dimensions
9
1
,
permissible
shearing
specific heat of a mixture
7,
1
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
T 10 ,T 11
T 12
normalized form
determination
T12
T 12
rules
T12
type
transformation of a delta to a
S 10
star-circuit
-
of a star to a delta-circuit
transformer
-
phase output
switch groups
threephase output
transmissions ratios
transverse contraction
trapezium
,
single
,
- rule
triangle
-
,
centroid
,
equilateral
,
1
- angular
- of light
- time curve
Venn diagram
P
3
vibrations
viscosity,
1
K
7
I
15
K
7
E
F
6
1
B2 D32
,
turbulent flow
turning
two coils facing each other
types of transfer elements
R2
F
F
F7, F
F
F
L
L
vertex radius
V
L3, L
G
F2
...
,
- divider
- source
S
voltmeter
volume expansion
T34
water pipes
-
fraction
-
per unit
1 1
1
S
2
wavelengths
A1,V
3
wave
wedges
T
F
2
7
Wheatstone bridge
work
- units of
Z 9
S 3
K11
S11
A
A
A
A 2 .. A
A3 A
A
A
A2 A
4
2
4
4
5
3
2
4
power
pressure
time
volume
1
.
,
,
,
worm
2
S21
trap
4
3
1
,
N 1
S 2
S 7
S 9
S 36
3
9
watt
P 18
A
6
A
C
A
A
4
units of
,
ungula
- step response
- vectors
1
F
M
mass
ultimate shear stress
unit of fineness
6
5
9
8
8
9
8
2
2
3
7
N1.Z16
dynamic
- kinematic
voltage
weber
mass
3
F8, F
7,
,
6
6
6
2
5
N 6
Z 17
S 13
E
E
E
E
E
circum circle
right angled
trigonometric conversions
force
length
G
gases
- product
- sum
S28
S35
S35
S35
M 4
1
,
real
velocity
F
oblique angle
radius of incircle
units of area
5
2
gases
vectors
vector difference
- equation
S 10
B
acute angle
,
,
B
T
variations of state of deal
method
,
A3, A
work
- ramp response
variances
,
- -
units of
gearing
Q
27,
M
1
A
5
Q 28
X-rays
V
3
yield point
P
1
Young's modulus
Z20
Z16
zone
C
- strength
of a
sphere
3
McGraw-Hill
\
/
Hvision ol The McGraw-Hill
gg
(
AMpames
ISBN 0-07-024572-X