1 1 fundamental physical and technical...
TRANSCRIPT
1
1
1 Fundamental Physical and Technical Terms
1.1 Units of physical quantities
1.1.1 The International System of Units (Sl)
The statutory units of measurement are1)
1. the basic units of the International System of Units (Sl units) for the basic quantitieslength, mass, time, electric current, thermodynamic temperature and luminous in-tensity,
2. the units defined for the atomic quantities of quantity of substance, atomic massand energy,
3. the derived units obtained as products of powers of the basic units and atomicunits through multiplication with a defined numerical factor,
4. the decimal multiples and sub-multiples of the units stated under 1-3.
Table 1-2DecimalsMultiples and sub-multiples of units
Decimal power Prefix Symbol
1018 Exa E 10–1 Dezi d1015 Peta P 10–2 Zenti c1012 Tera T 10–3 Milli m109 Giga G 10–6 Mikro µ106 Mega M 10–9 Nano n103 Kilo k 10–12 Piko p102 Hekto h 10–15 Femto f101 Deka da 10–18 Atto a
Table 1-1
Basic SI units
Quantity Units UnitsSymbol Name
Length m metreMass kg kilogrammeTime s secondElectric current A ampereThermodynamic temperature K kelvinLuminous intensity cd candelaQuantity of substance mol mole
1)DIN 1301
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arks
Nam
eS
ymb
olN
ame
Sym
bol
1 L
eng
th, a
rea,
vo
lum
e
1.1
Leng
thm
etre
m
1.2
Are
asq
uare
met
rem
2
are
a1
a =
102
m2
for
land
mea
sure
men
the
ctar
eha
1 ha
= 1
04m
2on
ly
1.3
Volu
me
cub
ic m
etre
m3
litre
l1
l =
1 d
m3
= 1
0–3m
3
1.4
Rec
ipro
cal
reci
pro
cal m
etre
1/m
leng
thd
iop
tre
dp
t1
dp
t =
1/m
1.5
Elo
ngat
ion
met
re p
erm
/mm
etre
(con
tinue
d)
Num
eric
al
valu
e o
fel
onga
tion
ofte
n ex
pre
ssed
in p
er c
ent
only
fo
r re
frac
tive
ind
ex
ofop
tical
sys
tem
s
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hip
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arks
Nam
eS
ymb
olN
ame
Sym
bol
2 A
ngle
2.1
Pla
ne a
ngle
rad
ian
rad
(ang
le)
1 ra
d =
1 m
/m
full
angl
e1
full
angl
e =
2 π
rad
right
ang
le1
= (π
/2) r
ad
deg
ree
°1
°=
(π/1
80) r
adm
inut
e’
1’=
1°/
60se
cond
”1”
= 1
’/60
gon
gon
1 go
n=
(π/2
00) r
ad
2.2
Sol
id a
ngle
ster
adia
nsr
1 sr
= 1
m2 /
m2
see
DIN
131
5
(con
tinue
d)
see
DIN
131
5
In c
alcu
latio
n th
e un
it ra
d a
sa
fact
or c
an b
e re
pla
ced
by
num
eric
al 1
.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
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uant
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ions
hip
Rem
arks
Nam
eS
ymb
olN
ame
Sym
bol
3 M
ass
3.1
Mas
ski
logr
amm
ekg
gram
me
g1
g=
10–3
kgto
nne
t1
t=
103
kgat
omic
u1
u=
1.6
6056
55 ·
10–2
7kg
mas
s un
it
met
ric c
arat
Kt
1 K
t=
0.2
· 10
–3kg
3.2
Mas
s p
er u
nit
kilo
gram
me
kg/m
leng
thp
er m
etre
Tex
tex
1 te
x=
10–6
kg/m
=
1 g
/km
(con
tinue
d)
Uni
ts
of
mas
s ar
e al
sous
ed t
o d
esig
nate
wei
ght
as t
he r
esul
t of
wei
ghin
gq
uant
ities
of
good
s (D
IN13
05)
only
for
gem
s
only
for
tex
tile
fibre
s an
dya
rns,
see
ISO
114
4
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No.
Q
uant
ityR
elat
ions
hip
Rem
arks
Nam
eS
ymb
olN
ame
Sym
bol
3.3
Den
sity
kilo
gram
me
kg/m
3se
e D
IN 1
306
per
cub
ic m
etre
3.4
Sp
ecifi
ccu
bic
met
rem
3 /kg
see
DIN
130
6vo
lum
ep
erki
logr
amm
e
3.5
Mas
s m
omen
tki
logr
amm
e-kg
· m
2
of in
ertia
1)sq
uare
met
re
1)S
ee a
lso
note
s on
pag
e 15
.
( con
tinue
d)
see
DIN
130
4-1
and
N
ote
to N
o. 3
.5
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No.
Qua
ntity
Rel
atio
nshi
pR
emar
ksN
ame
Sym
bol
Nam
eS
ymb
ol
4 T
ime
4.1
Tim
ese
cond
sm
inut
em
in1
min
= 6
0 s
hour
h1
h =
60
min
day
d1
d
= 2
4 h
year
a
4.2
Freq
uenc
yhe
rtz
Hz
1 H
z =
1/s
4.3
Rev
olut
ions
reci
pro
cal
1/s
per
sec
ond
seco
ndre
cip
roca
lm
inut
e1/
min
1/m
in =
1/(
60 s
)
(con
tinue
d)
In t
he p
ower
ind
ustr
y a
year
is t
aken
as
8760
hou
rs.
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uant
ityR
elat
ions
hip
Rem
arks
Nam
eS
ymb
olN
ame
Sym
bol
4.4
Cyc
licre
cip
roca
l1/
sfr
eque
ncy
seco
nd
4.5
Velo
city
met
re p
erm
/sse
cond
1ki
lom
etre
km/h
1 km
/h =
—
—m
/sp
er h
our
3.6
4.6
Acc
eler
atio
nm
etre
per
m/s
2
seco
nd s
qua
red
4.7
Ang
ular
rad
ian
per
rad
/sve
loci
tyse
cond
4.8
Ang
ular
rad
ian
per
rad
/s2
acce
lera
tion
seco
nd s
qua
re
(con
tinue
d)
The
2πfo
ld o
f th
e pe
riod
frequ
ency
is
calle
d an
gula
rfre
quen
cy
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uant
ityR
elat
ions
hip
1)R
emar
ksN
ame
Sym
bol
Nam
eS
ymb
ol
5 F
orc
e, e
nerg
y, p
ow
er
5.1
Forc
ene
wto
nN
1 N
= 1
kg
m/s
2
5.2
Mom
entu
mne
wto
n-se
cond
N ·
s1
N·s
= 1
· kg
m/s
5.3
Pre
ssur
ep
asca
lP
a1
Pa
= 1
N/m
2
bar
bar
1 b
ar
= 1
05P
a
(con
tinue
d)
Als
o ca
lled
wei
ght
see
DIN
130
5.
see
DIN
131
4
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uant
ityR
elat
ions
hip
1)R
emar
ksN
ame
Sym
bol
Nam
eS
ymb
ol
5.4
Mec
hani
cal
new
ton
per
N/m
2 , P
a1
Pa
= 1
N/m
2
stre
sssq
uare
met
re,
1 M
Pa
= 1
N/m
m2
pas
cal
1 G
Pa
= 1
kN
/mm
2
5.5
Ene
rgy,
wor
k,jo
ule
J1
J=
1 N
m =
1 W
sq
uant
ity o
f hea
t=
1 k
g m
2 /s2
kilo
wat
t-ho
urkW
h1
kWh
= 3
.6 M
Jel
ectr
on v
olt
eV1
eV=
1.6
0219
·10
–19J
5.6
Torq
ue
new
ton-
met
reN
· m
1 N
· m
=
1 J
= 1
W ·
s
5.7
Ang
ular
new
ton-
seco
nd-
N ·
s · m
1 N
· s
· m =
1 k
g · m
2 /s
mom
entu
mm
etre
(con
tinue
d)
see
DIN
134
5
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uant
ityR
elat
ions
hip
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arks
Nam
eS
ymb
olN
ame
Sym
bol
5.8
Pow
erw
att
W1
W =
1 J
/s
ener
gy fl
ow,
=1
N ·
m/s
heat
flow
= 1
V ·
A
6 V
isco
met
ric
qua
ntit
ies
6.1
Dyn
amic
pas
cal-
seco
ndP
a · s
1 P
a · s
= 1
N ·
s/m
2se
e D
IN 1
342
visc
osity
= 1
kg/
(s ·
m)
6.2
Kin
emat
icsq
uare
met
rem
2 /s
see
DIN
134
2vi
scos
ityp
er s
econ
d
(con
tinue
d)
app
aren
t p
ower
(V ·
A)
reac
tive
pow
er (v
ar)
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uant
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ions
hip
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arks
Nam
eS
ymb
olN
ame
Sym
bol
7 T
emp
erat
ure
and
hea
t
7.1
Tem
per
atur
e1)ke
lvin
K
deg
ree
Cel
sius
°C
The
deg
ree
Cel
sius
is(c
entig
rad
e)th
e sp
ecia
l nam
e fo
rke
lvin
whe
n ex
pre
ssin
gC
elsi
us t
emp
erat
ures
.
7.2
Ther
mal
squa
re m
etre
m2 /
sse
e D
IN 1
341
diff
usiv
ityp
er s
econ
d
7.3
Ent
rop
y, t
herm
aljo
ule
J/K
see
DIN
134
5ca
pac
ityp
er k
elvi
n
7.4
Ther
mal
w
att
per
W/(
K ·
m)
see
DIN
134
1co
nduc
tivity
kelv
in-m
etre
1)S
ee a
lso
note
s on
pag
e 15
.(c
ontin
ued
)
The
rmo
dyn
amic
te
mp
era-
ture
; see
Not
e to
No.
7.1
and
DIN
134
5.K
elvi
n is
al
so
the
unit
for
tem
per
atur
e d
iffer
ence
s an
din
terv
als.
Exp
ress
ion
of
Cel
sius
te
m-
per
atur
es a
nd C
elsi
us t
em-
per
atur
e d
iffer
ence
s,
see
Not
e to
No
7.1.
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uant
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arks
Nam
eS
ymb
olN
ame
Sym
bol
7.5
Hea
t tr
ansf
erw
att
per
W/(
K ·
m2 )
see
DIN
134
1co
effic
ient
kelv
in-s
qua
rem
etre
8 E
lect
rica
l and
mag
neti
c q
uant
itie
s
8.1
Ele
ctric
cur
rent
,am
per
eA
see
DIN
132
4m
agne
tic
pot
entia
ld
iffer
ence
8.2
Ele
ctric
vol
tage
,vo
ltV
1 V
=
1 W
/Ase
e D
IN 1
324
elec
tric
pot
entia
l
8 3
Ele
ctric
siem
ens
S1
S
= A
/Vse
e N
ote
to c
olum
ns 3
and
cond
ucta
nce
4 an
d a
lso
DIN
132
4
8.4
Ele
ctric
ohm
Ω1 Ω
= 1
/Sse
e D
IN 1
324
resi
stan
ce
(con
tinue
d)
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uant
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elat
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hip
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arks
Nam
eS
ymb
olN
ame
Sym
bol
8.5
Qua
ntity
of
coul
omb
C1
C=
1 A
· s
see
DIN
132
4el
ectr
icity
, ele
ctric
amp
ere-
hour
Ah
1 A
· h
= 3
600
A ·
sch
arge
8.6
Ele
ctric
fara
dF
1 F
= 1
C/V
see
DIN
130
4ca
pac
itanc
e
8.7
Ele
ctric
flux
co
ulom
b p
erC
/m2
see
DIN
132
4d
ensi
tysq
uare
met
re
8.8
Ele
ctric
fiel
dvo
lt p
er m
etre
V/m
see
DIN
132
4st
reng
th
8.9
Mag
netic
flux
web
erW
b1
Wb
= 1
V ·
sse
e D
IN 1
324
8.10
Mag
netic
flux
te
sla
T1
T =
1 W
b/m
2se
e D
IN 1
324
dens
ity, (
indu
ctio
n)
8.11
Ind
ucta
nce
henr
yH
1 H
= 1
Wb
/Ase
e D
IN 1
324
(per
mea
nce)
(con
tinue
d)
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uant
ityR
elat
ions
hip
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arks
Nam
eS
ymb
olN
ame
Sym
bol
8.12
Mag
netic
fiel
dam
per
eA
/mse
e D
IN 1
324
inte
nsity
per
met
re
9 P
hoto
met
ric
qua
ntit
ies
9.1
Lum
inou
sca
ndel
acd
inte
nsity
9.2
Lum
inan
ceca
ndel
a p
ercd
/m2
see
DIN
503
1 P
art
3sq
uare
met
re
9.3
Lum
inou
s flu
xlu
men
Im1
Im=
1 c
d ·
srse
e D
IN 5
031
Par
t 3
9.4
Illum
inat
ion
lux
Ix1
lx=
1 Im
/m2
see
DIN
503
1 P
art
3
see
DIN
503
1 P
art
3.
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 14
15
1
Not
es t
o Ta
ble
1-3
To
No
. 3.5
:
Whe
n co
nver
ting
the
so-c
alle
d “
flyw
heel
ine
rtia
GD
2 ” i
nto
a m
ass
mom
ent
ofin
ertia
J, n
ote
that
the
num
eric
al v
alue
of G
D2
in k
p m
2is
eq
ual t
o fo
ur t
imes
the
num
eric
al v
alue
of t
he m
ass
mom
ent
of in
ertia
J in
kg
m2 .
To
No
. 7.1
:
The
(ther
mod
ynam
ic) t
emp
erat
ure
(T),
also
kno
wn
as “
abso
lute
tem
per
atur
e”, i
sth
e p
hysi
cal q
uant
ity o
n w
hich
the
law
s of
the
rmod
ynam
ics
are
bas
ed.
For
this
reas
on,
only
thi
s te
mp
erat
ure
shou
ld b
e us
ed i
n p
hysi
cal
equa
tions
. Th
e un
itke
lvin
can
als
o b
e us
ed t
o ex
pre
ss t
emp
erat
ure
diff
eren
ces.
Cel
sius
(cen
tigra
de)
tem
per
atur
e(t)
( ) i
s th
e sp
ecia
l diff
eren
ce b
etw
een
a gi
ven
ther
mod
ynam
ic t
emp
erat
ure
Tan
d a
tem
per
atur
e of
T0
= 2
73.1
5 K
.
Thus
, t =
T –
T 0=
T –
273.
15 K
.(1
)
Whe
n ex
pre
ssin
g C
elsi
us t
emp
erat
ures
, the
sta
ndar
d s
ymb
ol °
C is
to
be
used
.
The
diff
eren
ce ∆
t b
etw
een
two
Cel
sius
tem
per
atur
es,
e. g
. th
e te
mp
erat
ures
t 1=
T1
– T 0
and
t2
= T
2–
T 0, i
s
∆t
=t 1
–t2
=T 1
– T 2
=∆
T(2
)
A t
emp
erat
ure
diff
eren
ce o
f th
is n
atur
e is
no
long
er r
efer
red
to
the
ther
mo-
dyn
amic
tem
per
atur
e T 0
, an
d h
ence
is n
ot a
Cel
sius
tem
per
atur
e ac
cord
ing
toth
e d
efin
ition
of E
q. (
1).
How
ever
, th
e d
iffer
ence
bet
wee
n tw
o C
elsi
us t
emp
erat
ures
may
be
exp
ress
edei
ther
in
kelv
in o
r in
deg
rees
Cel
sius
, in
par
ticul
ar w
hen
stat
ing
a ra
nge
ofte
mp
erat
ures
, e. g
. (20
±2)
°C
Ther
mod
ynam
ic t
emp
erat
ures
are
oft
en e
xpre
ssed
as
the
sum
of
T 0an
d a
Cel
sius
tem
per
atur
e t,
i. e
. fol
low
ing
Eq
. (1)
T =
T0
+ t
(3
)
and
so
the
rele
vant
Cel
sius
tem
per
atur
es c
an b
e p
ut i
n th
e eq
uatio
n st
raig
htaw
ay. I
n th
is c
ase
the
kelv
in u
nit s
houl
d a
lso
be
used
for t
he C
elsi
us te
mp
erat
ure
(i. e
. fo
r th
e “s
pec
ial
ther
mod
ynam
ic t
emp
erat
ure
diff
eren
ce”)
. Fo
r a
Cel
sius
tem
per
atur
e of
20
°C, t
here
fore
, one
sho
uld
writ
e th
e su
m t
emp
erat
ure
as
T =
T0
+ t
= 2
73.1
5 K
+ 2
0 K
= 2
93.1
5 K
(4)
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 15
16
1.1.
2 O
ther
uni
ts s
till
in c
om
mo
n us
e; m
etri
c, B
riti
sh a
nd U
S m
easu
res
Som
e of
the
uni
ts li
sted
bel
ow m
ay b
e us
ed f
or a
lim
ited
tra
nsiti
on p
erio
d a
nd in
cer
tain
exc
eptio
nal c
ases
. The
sta
tuto
ry r
equi
rem
ents
var
yfr
om c
ount
ry t
o co
untr
y.
ångs
tröm
Åle
ngth
1 Å
= 0
.1 n
m =
10–1
0 mA
stro
nom
ie u
nit
AE
leng
th1
AE
= 1
49,5
9787
· 10
–9 m
atm
osp
here
phy
sica
lat
mp
ress
ure
1 at
m =
1,0
1325
bar
atm
osp
here
tec
hnic
alat
, ata
pre
ssur
e1
at =
0,9
8066
5 b
arb
arre
lb
bl
volu
me
1 b
bl
= 1
58,9
88 l
Brit
ish
ther
mal
uni
tB
tuq
uant
ity o
f hea
t1
Btu
≈10
55.0
56 J
calo
rieca
lq
uant
ity o
f hea
t1
cal
= 4
.186
8 J
cent
igon
cp
lane
ang
le1
c =
1 c
gon
= 5
π· 1
0–5ra
dd
egre
ed
egte
mp
erat
ure
diff
eren
ce1
deg
= 1
Kd
egre
e fa
hren
heit
°Fte
mp
erat
ure
T K=
273
.15
+ (5
/9) ·
(tF
– 32
)d
ynd
ynfo
rce
1 d
yn =
10–5
Ner
ger
gen
ergy
1 er
g =
10–7
Jfo
otft
leng
th1
ft =
0.3
048
mga
llon
(UK
)ga
l (U
K)
volu
me
1 ga
l (U
K) ≈
4.54
609
· 10–3
m3
gallo
n (U
S)
gal (
US
)liq
uid
vol
ume
1 ga
l (U
S) ≈
3.78
541
· 10–3
m3
gaus
sG
mag
netic
flux
den
sity
1 G
= 1
0–4T
gilb
ert
Gb
mag
netic
pot
entia
l diff
eren
ce1
Gb
= (
10/4
π) A
gon
gp
lane
ang
le1
g =
1 g
on =
5 π
· 10–3
rad
hors
epow
erhp
pow
er1
hp ≈
745.
700
Whu
ndre
dw
eigh
t (lo
ng)
cwt
mas
s1
cwt
≈50
.802
3 kg
inch
(inc
hes)
in, "
leng
th1
in =
25.
4 m
m =
254
· 10
–4m
kilo
gram
me-
forc
e, k
ilop
ond
kp, k
gffo
rce
1 kp
= 9
.806
65 N
≈10
Nkn
otkn
time
1 kn
= 1
sm
/h =
0,5
144
m/s
Ligh
t-ye
arlj
leng
ht1
Lj =
9,4
6053
· 10
15 m
= 6
3240
AE
Uni
t of
mas
sM
Em
ass
1 M
E =
9.8
0665
kg
max
wel
lM
, Mx
mag
netic
flux
1 M
= 1
0 nW
b =
10–8
Wb
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 16
17
1
met
re w
ater
col
umn
mW
Sp
ress
ure
1 m
WS
= 9
806,
65 P
a ≈
98,0
665
mb
arm
icro
nµ
leng
th1
µ=
1 µ
m =
10–6
mm
ilem
ilele
nght
1 m
ile =
160
9,34
4 m
mill
imet
res
of m
ercu
rym
m H
gp
ress
ure
1 m
m H
g ≈
133.
322
Pa
mill
igon
ccp
lane
ang
le1
cc =
0.1
mgo
n =
5 π
· 10–7
rad
oers
ted
Oe
mag
netic
fiel
d s
tren
gth
1 oe
= (
250/π)
A/m
≈80
A/m
ounc
eoz
mas
s1
oz =
28.
3495
gP
ferd
estä
rke,
che
val-
vap
eur
PS
, CV
pow
er1
PS
= 7
35.4
9875
WP
fund
Pfd
mas
s1
Pfd
= 0
.5 k
gp
ieze
pz
pre
ssur
e1
pz
= 1
mP
a =
10–3
Pa
pin
tp
tvo
lum
e1
pt
= 5
68,2
62 m
lp
oise
Pd
ynam
ic v
isco
sity
1 P
= 0
.1 P
a · s
pon
d, g
ram
-for
cep
, gf
forc
e1
p =
9.8
0665
· 10
–3N
≈10
mN
pou
nd1)
Ibm
ass
1 Ib
= 0
,453
5923
7 kg
pou
ndal
pd
lfo
rce
1 p
dl ≈
0.13
8255
Np
ound
forc
eIb
ffo
rce
1 Ib
f ≈
4.44
822
Nse
a m
ile, i
nter
natio
nal
smle
ngth
(mar
ine)
1 sm
= 1
852
msh
ort
hund
red
wei
ght
sh c
wt
mas
s1
sh c
wt
≈45
.359
2 kg
stilb
sblu
min
ance
1 sb
= 1
04cd
/m2
stok
esS
tki
nem
atic
vis
cosi
ty1
St
= 1
0–4m
2 /s
torr
Torr
pre
ssur
e1
Torr
= 1
.333
224
mb
arty
pog
rap
hica
l poi
ntp
leng
th (p
rintin
g)1
p =
(1.
0033
3/26
60) m
≈0.
4 m
mya
rdyd
leng
th1
yd =
0.9
144
mZ
entn
erZ
trm
ass
1 Z
tr =
50
kg
1)U
K a
nd U
S p
ound
s av
oird
upoi
s di
ffer o
nly
afte
r the
six
th d
ecim
al p
lace
.
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 17
18
Tab
le 1
-4
Met
ric, B
ritis
h an
d U
S li
near
mea
sure
Met
ric u
nits
of l
engt
hB
ritis
h an
d U
S u
nits
of l
engt
h
Kilo
met
reM
etre
Mill
imet
reM
ileYa
rdFo
otIn
chM
il
kmm
mm
mile
ydft
in o
r "
mil
11
000
1 00
0 00
00.
6213
1 09
3.7
3 28
139
370
3 93
7 · 1
04
0.00
11
1 00
00.
6213
· 10
–31.
0937
3.28
139
.370
39 3
700.
0000
010.
001
10.
6213
· 10
–60.
0010
940.
0032
810.
0393
739
.37
1.60
953
1 60
9.53
1 60
9 52
81
1 76
05
280
63 3
606
336
· 104
0.00
0914
0.91
4391
4.32
0.56
82 ·
10–3
13
3636
000
0.30
5 · 1
0–30.
3047
930
4.79
0.18
94 ·
10–3
0.33
331
1212
000
0.25
4 · 1
0–40.
0253
925
.399
70.
158
· 10–4
0.02
777
0.08
331
1 00
00.
254
· 10–7
0.25
4 · 1
0–40.
0253
90.
158
· 10–7
0.02
77·1
0–30.
0833
· 10
–30.
001
1
Sp
ecia
l mea
sure
s:
1 m
etric
nau
tical
mile
= 1
852
m1
Brit
. or
US
nau
tical
mile
= 1
855
m1
met
ric la
nd m
ile =
750
0 m
1 m
icro
n (µ
) = 1
/100
0 m
m =
10
000
Å
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 18
19
1
Tab
le 1
-5
Met
ric, B
ritis
h an
d U
S s
qua
re m
easu
re
Met
ric u
nits
of a
rea
Brit
ish
and
US
uni
ts o
f are
a
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Sq
uare
Circ
ular
kilo
met
res
met
red
ecim
.ce
ntim
.m
illim
.m
ileya
rdfo
otin
chm
ils
km2
m2
dm
2cm
2m
m2
sq.m
ilesq
.yd
sq.ft
sq.in
cir.m
ils
11
· 106
100
· 106
100
· 108
100
· 1010
0.38
6013
1 19
6 · 1
0310
76 ·
104
1 55
0 · 1
0619
7.3
· 1013
1 · 1
0–61
100
10 0
001
000
000
0.38
6 · 1
0–61.
1959
10.7
641
550
197.
3 · 1
07
1 · 1
0–81
· 10–2
110
010
000
0.38
6 · 1
0–80.
0119
60.
1076
415
.50
197.
3 · 1
05
1 · 1
0–10
1 · 1
0–41
· 10–2
110
00.
386
· 10–1
00.
1196
· 10
–30.
1076
· 10
–20.
1550
197.
3 · 1
03
1 · 1
0–12
1 · 1
0–61
· 10–4
1 · 1
0–21
0.38
6 · 1
0–12
0.11
96 ·
10–5
0.10
76 ·
10–4
0.00
155
1 97
32.
5899
92
589
999
259
· 106
259
· 108
259
· 1010
130
976
· 10
227
878
· 10
340
145
· 10
55
098
· 1012
0.83
61 ·
10–6
0.83
6130
83.6
130
8 36
1.30
783
6 13
0.7
0.32
28 ·
10–6
19
1296
1 64
6 · 1
06
9.29
0 · 1
0–89.
290
· 10–2
9.29
034
929.
034
92 9
03.4
0.03
58 ·
10–6
0.11
111
114
418
3 · 1
06
6.45
2 · 1
0–10
6.45
2 · 1
0–46.
452
·10–2
6.45
162
645.
162
0.23
96 ·
10–9
0.77
16 ·
10–3
0.00
6940
11.
27 ·
106
506.
7 · 1
0–18
506.
7 · 1
0–12
506.
7·1
0–10
506.
7 · 1
0–850
6.7
· 10–6
0.19
6 · 1
0–15
0.60
7 · 1
0–90.
0054
7·1
0–60.
785
· 10–6
1
Sp
ecia
l mea
sure
s:1
hect
are
(ha)
= 1
00 a
re (a
)1
sect
ion
(sq
.mile
) =
64
acre
s =
2,5
89 k
m2
1ar
e (a
) =
100
m2
1 ac
re =
484
0 sq
.yd
s =
40.
468
aU
SA
1 B
ad. m
orge
n =
56
a =
1.3
8 ac
re1
sq. p
ole
= 3
0.25
sq
.yd
s =
25.
29 m
2
1 P
russ
ian
mor
gen
= 2
5.53
a =
0.6
3 ac
re1
acre
= 1
60 s
q.p
oles
= 4
840
sq.y
ds
= 4
0.46
8 a
1 W
uert
tem
ber
g m
orge
n =
31.
52 a
= 0
.78
acre
1 ya
rd o
f lan
d =
30
acre
s =
121
4.05
a
Brit
.1
Hes
se m
orge
n =
25.
0 a
= 0
.62
acre
1 m
ile o
f lan
d =
640
acr
es =
2.5
89 k
m2
1 Ta
gwer
k (B
avar
ia) =
34.
07 a
= 0
.84
acre
1sh
eet
of p
aper
=
86
x 61
cm
give
s 8
pie
ces
size
A4
or 1
6 p
iece
s A
5or
32
pie
ces
A6
⎫ ⎬ ⎭⎫ ⎬ ⎭
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 19
20
Tab
le 1
-6
Met
ric, B
ritis
h an
d U
S c
ubic
mea
sure
s
Met
ric u
nits
of v
olum
eB
ritis
h an
d U
S u
nits
of v
olum
eU
S li
qui
d m
easu
re
Cub
icC
ubic
Cub
icC
ubic
Cub
icC
ubic
Cub
icG
allo
nQ
uart
Pin
tm
etre
dec
imet
rece
ntim
etre
mill
imet
reya
rdfo
otin
ch
m3
dm
3cm
3m
m3
cu.y
dcu
.ftcu
.inga
lq
uart
pin
t
11
000
1 00
0 · 1
031
000
· 106
1.30
7935
.32
61 ·
103
264.
21
056.
82
113.
61
· 10–3
11
000
1 00
0 · 1
031.
3079
·10
–30.
0353
261
.023
0.26
421.
0568
2.11
361
· 10–6
1 · 1
0–31
1 00
01.
3079
·10
–60.
3532
· 10
–40.
0610
230.
2642
· 10
–31.
0568
· 10
–32.
1136
· 10
–3
1 · 1
0–91
· 10–6
1 · 1
0–31
1.30
79 ·
10–9
0.35
32 ·
10–7
0.61
0 · 1
0–40.
2642
· 10
–61.
0568
· 10
–62.
1136
· 10
–6
0.76
4573
764.
573
764
573
764
573
·10
31
2746
656
202
808
1 61
60.
0283
170
28.3
1701
28 3
17.0
128
317
013
0.03
7037
11
728
7.48
224
29.9
2896
59.8
5792
0.16
38 ·
10–4
0.01
6387
116
.387
1616
387.
160.
2143
· 10
–40.
5787
· 10
–31
0.00
433
0.01
732
0.03
464
3.78
5 · 1
0–33.
7854
423
785.
442
3 78
5 44
20.
0049
457
0.13
3679
723
11
48
0.94
63 ·
10–3
0.94
6360
594
6.36
0594
6 36
0.5
0.00
1236
40.
0334
199
57.7
50.
250
12
0.47
32 ·
10–3
0.47
3180
247
3.18
0247
3 18
0.2
0.00
0618
20.
0167
099
28.8
750.
125
0.50
01
Con
vers
ion
: 1 in
ch =
0,0
3937
x l/
m
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 20
21
1
1.1.3 Fundamental physical constants
Universell gas constant: R = 8.314472 J K–1 mol–1
is the work done by one mole of an ideal gas under constant pressure (1013 hPa) whenits temperature rises from 0 °C to 1 °C.
Avogadro’s constant: NA (Loschmidt’s number NL): NA = 6.0221415 · 1023 mol–1
number of molecules of an ideal gas in one mole.
Base of natural logarithms: e = 2.718282
Bohr’s radius: ao = 0.5291772108 · 10–10 mradius of the innermost electron orbit in Bohr’s atomic model
Boltzmann’s constant: k = 1.3806505 · 10–23 J · K–1
is the mean energy gain of a molecule or atom when heated by 1 K.
Elementary charge: eo = F/NA = 1.60217653 · 10–19 As is the smallest possible charge a charge carrier (e.g. electron or proton) can have.
Electron-volt: eV = 1.60217653 · 10–19 J
Energy mass equivalent: mec2 = 8.1871047 · 10–14 J = 0.510998918 MeVaccording to Einstein, following E = m · c2, the mathematical basis for all observedtransformation processes in sub-atomic ranges.
Faraday’s constant: F = 96485.3383 C · mol–1
is the quantity of current transported by one mole of univalent ions.
Field constant, electrical: ε0 = 0.885418781762 · 10–12 F · m–1
a proportionality factor relating charge density to electric field strength.
Field constant, magnetic: µ0 = 4 · π · 10–7 H · m–1
a proportionality factor relating magnetic flux density to magnetic field strength.
Gravitational constant: γ = 6.6742 · 10–11 m3 · kg–1 · s–2
is the attractive force in N acting between two masses each of 1 kg weight separatedby a distance of 1 m.
Velocity of light in vacuo: c = 2.99792485 · 108 m · s–1
maximum possible velocity. Speed of propagation of electro-magnetic waves.
Mole volume: Vm = 22.710981 · 10–3 · m3 · mol–1
the volume occupied by one mole of an ideal gas at 0 °C and 1013 mbar. A mole is thatquantity (mass) of a substance which is numerically equal in grammes to the molecularweight (1 mol H2 = 2 g H2)
Planck’s constant: h = 6.6260693 · 10–34 J · s a proportionality factor relating energy and frequency of a light quantum (photon).
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 21
22
Stefan Boltzmann’s constant: σ = 5.6704 · 10–8 W · m–2 · K–4 relates radiant energyto the temperature of a radiant body. Radiation coefficient of a black body.
Temperature of absolute zero: T0 = –273.16 °C = 0 K.
Wave impedance of space: Z0 = 376.73031346152 Ωcoefficient for the H/E distribution with electromagnetic wave propagation.
Z0 = µ0 /ε0 = µ0 · c = 1/ (ε0 · c)
Weston standard cadmium cell: E0 = 1.0186 V at 20 °C.
Wien’s displacement constant: b = 2.8977685 · 10–3 m · Kenables the temperature of a light source to be calculated from its spectrum.
1.2 Physical, chemical and technical values
1.2.1 Electrochemical series
If different metals are joined together in a manner permitting conduction, and both arewetted by a liquid such as water, acids, etc., an electrolytic cell is formed which givesrise to corrosion. The amount of corrosion increases with the differences in potential.If such conducting joints cannot be avoided, the two metals must be insulated fromeach other by protective coatings or by constructional means. In outdoor installations,therefore, aluminium/copper connectors or washers of copper-plated aluminium sheetare used to join aluminium and copper, while in dry indoor installations aluminium andcopper may be joined without the need for special protective measures.
Table 1-7
Electrochemical series, normal potentials in volts, at 25 ºC.
Li/LI+ – 3.05 Mn/Mn2+ – 1.18 Fe/Fe3+ – 0.04K/K+ – 2.93 Zn/Zn2+ – 0.76 H2/H+ 0.00Ba/Ba2+ – 2.91 Cr/Cr3+ – 0.74 Cu/Cu2+ + 0.34Ca/Ca2+ – 2.87 Fe/Fe2+ – 0.44 Cu/Cu+ + 0.52Na/Na+ – 2.71 Cd/Cd2+ – 0.40 Hg/Hg2
2+ + 0.79Mg/Mg2+ – 2.37 In/In3+ – 0.34 Ag/Ag+ + 0.80Be/Be2+ – 1.85 Co/Co2+ – 0.28 Pd/Pd2+ + 0.99Al/Al3+ – 1.66 Ni/Ni2+ – 0.25 Pt/Pt+ + 1.20Ti/Ti2+ – 1.63 Sn/Sn2+ – 0.14 Au/Au3+ + 1.50Zr/Zr4+ – 1.53 Pb/Pb2+ – 0.13 Au/Au+ + 1.70
Note: The electrode designation to the left of the slash indicates the electrode material and to theright of the slash to ion layer which forms in front of the electrode surface (dependent on variousinfluences).
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1
If two metals included in this table come into contact, the metal mentioned first willcorrode.
The less noble metal becomes the anode and the more noble acts as the cathode. Asa result, the less noble metal corrodes and the more noble metal is protected.
Metallic oxides are always less strongly electronegative, i. e. nobler in the electrolyticsense, than the pure metals. Electrolytic potential differences can therefore also occurbetween metal surfaces which to the engineer appear very little different. Even thoughthe potential differences for cast iron and steel, for example, with clean and rustysurfaces are small, as shown in Table 1-8, under suitable circumstances these smalldifferences can nevertheless give rise to significant direct currents, and hence corrosiveattack.
Table 1-8
Standard potentials of different types of iron against hydrogen, in volts
SM steel, clean surface approx. – 0.40 cast iron, rusty approx. – 0.30cast iron, clean surface approx. – 0.38 SM steel, rusty approx. – 0.25
1.2.2 Faraday’s law
1. The amount m (mass) of the substances deposited or converted at an electrode isproportional to the quantity of electricity Q = l · t.
m ~ l · t
2. The amounts m (masses) of the substances converted from different electrolytes byequal quantities of electricity Q = l · t behave as their electrochemical equivalentmasses M*. The equivalent mass M* is the molar mass M divided by theelectrochemical valency n (a number). The quantities M and M* can be stated ing/mol.
M*m = — l · t
F
If during electroysis the current I is not constant, the product
l · t must be represented by the integral l dt.
The quantity of electricity per mole necessary to deposit or convert the equivalentmass of 1 g/mol of a substance (both by oxidation at the anode and by reduction atthe cathode) is equal in magnitude to Faraday's constant (F = 96480 As/mol).
t2
t1
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Table 1-9
Electrochemical equivalents1)
Valency Equivalent Quantity Approximaten mass2) precipitated, optimum current
g/mol theoretical efficiencyg/Ah %
Aluminium 3 8.9935 0.33558 85 … 98Cadmium 2 56.20 2.0970 95 … 95Caustic potash 1 56.10937 2.0036 95Caustic soda 1 30.09717 1.49243 95Chlorine 1 35.453 1.32287 95Chromium 3 17.332 0.64672 —Chromium 6 8.666 0.32336 10 … 18Copper 1 63.54 2.37090 65 … 98Copper 2 31.77 1.18545 97 … 100Gold 3 65.6376 2.44884 —Hydrogen 1 1.00797 0.037610 100Iron 2 27.9235 1.04190 95 … 100Iron 3 18.6156 0.69461 —Lead 2 103.595 3.80543 95 … 100Magnesium 2 12.156 0.45358 —Nickel 2 29.355 1.09534 95 … 98Nickel 3 19.57 0.73022 —Oxygen 2 7.9997 0.29850 100Silver 1 107.870 4.02500 98 … 100Tin 2 59.345 2.21437 70 … 95Tin 4 29.6725 1.10718 70 … 95Zinc 2 32.685 1.21959 85 … 93
1) Relative to the carbon-12 isotope = 12.000.2) Chemical equivalent mass is molar mass/valency in g/mol.
Example:
Copper and iron earthing electrodes connected to each other by way of the neutralconductor form a galvanic cell with a potential difference of about 0.7 V (see Table 1-7). These cells are short-circuited via the neutral conductor. Their internal resistance isde-termined by the earth resistance of the two earth electrodes. Let us say the sum of allthese resistances is 10 Ω. Thus, if the drop in “short-circuit emf” relative to the “open-circuit emf” is estimated to be 50 % approximately, a continuous corrosion current of35 mA will flow, causing the iron electrode to decompose. In a year this will give anelectrolytically active quantity of electricity of
h Ah35 mA · 8760 — = 306 —– .a a
Since the equivalent mass of bivalent iron is 27.93 g/mol, the annual loss of weightfromthe iron electrode will be
27.93 g/mol 3600 sm = ————————— · 306 Ah/a · ————— = 320 g/a.
96480 As/mol h
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1
1.2.3 Thermoelectric series
If two wires of two different metals or semiconductors are joined together at their endsand the two junctions are exposed to different temperatures, a thermoelectric currentflows in the wire loop (Seebeck effect, thermocouple). Conversely, a temperaturedifference between the two junctions occurs if an electric current is passed throughthe wire loop (Peltier effect).
The thermoelectric voltage is the difference between the values, in millivolts, stated inTable 1-10. These relate to a reference wire of platinum and a temperature differenceof 100 K.
Table 1-10
Thermoelectric series, values in mV, for platinum as reference and temperaturedifference of 100 K
Bismut ll axis –7.7 Rhodium 0.65Bismut ⊥ axis –5.2 Silver 0.67 … 0.79Constantan –3.37 … –3.4 Copper 0.72 … 0.77Cobalt –1.99 … –1.52 Steel (V2A) 0.77Nickel –1.94 … –1.2 Zinc 0.6 … 0.79Mercury –0.07 … +0.04 Manganin 0.57 … 0.82Platinum ± 0 Irdium 0.65 … 0.68Graphite 0.22 Gold 0.56 … 0.8Carbon 0.25 … 0.30 Cadmium 0.85 … 0.92Tantalum 0.34 … 0.51 Molybdenum 1.16 … 1.31Tin 0.4 … 0.44 Iron 1.87 … 1.89Lead 0.41 … 0.46 Chrome nickel 2.2Magnesium 0.4 … 0.43 Antimony 4.7 … 4.86Aluminium 0.37 … 0.41 Silicon 44.8Tungsten 0.65 … 0.9 Tellurium 50Common thermocouplesCopper/constantan Nickel chromium/nickel(Cu/const) up to 500 °C (NiCr/Ni) up to 1 000 °CIron/constantan Platinum rhodium/(Fe/const) up to 700 °C platinum up to 1 600 °CNickel chromium/ Platinum rhodium/constantan up to 800 °C platinum rhodium up to 1 800 °C
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1.2.4 pH value
The pH value is a measure of the “acidity” of aqueous solutions. It is defined as thelogarithm to base 10 of the reciprocal of the hydrogen ion concentration CH3O1).
c(H+)pH ≡ –log –––––––
mol · L–1
pH scale1 m = 1 mol/ l hydrochloric acid (3.6 % HCl —– 0
0.1 m hydrochloric acid (0.36 % HCl)—–—–—–—–—–—–—–gastric acid—–—–—–—–—–—–—– —– 1
—– 2vinegar ( ≈ 5 % CH3 COOH)—–—–—–—–—–—–—–
—– 3
acidmarsh water—–—–—–—–—–—–—– —– 4
—– 5
—– 6
river water—–—–—–—–—–—–—– —– 7
tap water 20 Ωm—–—–—–—–—–—–—– neutral
—– 8see water 0.15 Ωm (4 % NaCl)—–—–—–—–—–—–—–—– 9
—– 10
0.1 m ammonia water (0.17 % NH3)—–—–—–—–—–—–—– —– 11alkaline
saturated lime-water (0.17 % CaOH2)—–—–—–—–—–—–—– —– 12
0.1 m caustic soda solution (0.4 % NaOH)—–—–—–—–—–—–—– —– 13
⎧⎨⎩
⎧⎨⎩
⎧⎨⎩
Fig. 1-1
pH value of some solutions
1.2.5 Heat transfer
Heat content (enthalpy) of a body: Q = V · ρ · c · ∆ϑV volume, ρ density, c specific heat, ∆ϑ temperature difference
Heat flow is equal to enthalpy per unit time: Φ = Q/t
Heat flow is therefore measured in watts (1 W = 1 J/s).
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1
Specific heat (specific thermal capacity) of a substance is the quantity of heat requiredto raise the temperature of 1 kg of this substance by 1 °C. Mean specific heat relatesto a temperature range, which must be stated. For values of c and λ, see Section1.2.7.
Thermal conductivity is the quantity of heat flowing per unit time through a wall 1 m2 inarea and 1 m thick when the temperatures of the two surfaces differ by 1 °C. Withmany materials it increases with rising temperature, with magnetic materials (iron,nickel) it first falls to the Curie point, and only then rises (Curie point = temperatureat which a ferro-magnetic material becomes non-magnetic, e. g. about 800 °C forAlnico). With solids, thermal conductivity generally does not vary much (invariable onlywith pure metals); in the case of liquids and gases, on the other hand, it is oftenstrongly influenced by temperature.
Heat can be transferred from a place of higher temperature to a place of lowertemperature by
– conduction (heat transmission between touching particles in solid, liquid orgaseous bodies).
– convection (circulation of warm and cool liquid or gas particles).
– radiation (heat transmission by electromagnetic waves, even if there is no matterbetween the bodies).
The three forms of heat transfer usually occur together.
Heat flow with conduction through a wall:
λΦ = — · A · ∆ϑs
A transfer area, λ thermal conductivity, s wall thickness, ∆ϑ temperature difference.
Heat flow in the case of transfer by convection between a solid wall and a flowingmedium:
Φ = α · A · ∆ϑ
α heat transfer coefficient, A transfer area, ∆ϑ temperature difference.
Heat flow between two flowing media of constant temperature separated by a solidwall:
Φ = k · A · ∆ϑ
k thermal conductance, A transfer area, ∆ϑ temperature difference.
In the case of plane layered walls perpendicular to the heat flow, the thermal conduct-ance coefficient k is obtained from the equation
1 1 sn 1— = —— + —— + –—k α 1
∑ λn α 2
Here, α1 and α2 are the heat transfer coefficients at either side of a wall consisting ofn layers of thicknesses sn and thermal conductivities λn.
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Thermal radiation
For two parallel black surfaces of equal size the heat flow exchanged by radiation is
Φ12 = σ · A (T14–T2
4)
With grey radiating surfaces having emissivities of ε1 and ε2, it is
Φ12 = C12 · A (T14–T2
4)
σ = 5.6704 · 10–8 W · m–2 · K–4 radiation coefficient of a black body (StefanBoltzmann’s constant), A radiating area, T absolute temperature.
Index 1 refers to the radiating surface, Index 2 to the radiated surface.
C12 is the effective radiation transfer coefficient. It is determined by the geometry andemissivity ε of the surface (table 1-12).
Special cases: A1 A2 C12 = σ · ε1
σA1 ≈ A2 C12 = —————
1 1– + – – 1ε1 ε2
Table 1-11
Emissivity ε (average values ϑ < 200 °C)
Black body 1 Oil 0.82Aluminium, polished 0.038 (230 °C) Glass 0,94 (22 °C)Aluminium, raw 0.079 (26 °C) Porcelain, glazed 0,92 (22 °C)Copper, polished 0.049 (23 °C) Water 0.96Copper, oxidized 0.639 (600 °C) Wood (oak) 0,89 (21 °C)Brass, polished 0.059 (19 °C) Roofing felt 0,91 (21 °C)Brass, dull 0.229 (56-338 °C) enamel varnish 0,91 (24 °C)Steel, dull, oxidized 0.969 (26-356 °C) spirit varnish 0,82 (25 °C)Steel, polished 0.29 Soot 0.93
σA2 includes A1 C12 = ————————
1 A1 1– + — · — – 1ε1 A2 ε2
Table 1-12
Heat transfer coefficients α in W/(m2 · K) (average values)
Natural air movement in a closed spaceWall surfaces 10Floors, ceilings: in upward direction 7
in downward direction 5Force-circulated airMean air velocity w = 2 m/s 20Mean air velocity w > 5 m/s 6.4 · w0.75
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1
1.2.6 Acoustics, noise measurement and noise abatement
Audible sound comprises the mechanical oscillations and waves of an elastic mediumin the frequency range of the human ear between 16 Hz and 20.000 Hz.
Oscillations below 16 Hz are termed infrasound and above 20.000 Hz ultrasound.Sound waves can occur not only in air but also in liquids (water-borne sound) and insolid bodies (solid-borne sound). Solid-borne sound is partly converted into audibleair-borne sound at the boundary surfaces of the oscillating body. The frequency ofoscillation determines the pitch of the sound. The sound generally propagatesspherically from the sound source, as longitudinal waves in gases and liquids and aslongitudinal and transverse waves in solids.
A sound source is characterized by its sound power W. The sound power is the soundenergy radiated by a sound source in unit time. Its unit of measurement is the watt.Sound propagation gives rise to an alternating pressure, the root-mean-square valueof which is termed the sound pressure p. It decreases approximately in proportion tothe square of the distance from the sound source. The sound intensity I is the soundenergy flowing perpendicularly through a surface in unit time; it is therefore a vectorialvalue. Its unit of measurement is the watt/m2.
Since the sensitivity of the human ear is proportional to the logarithm of the soundpressure, a logarithmic scale is used to represent the various types of sound levels.
The sound power level Lw is defined as
Lw = 20 lg W/Wo in dB.
Here: W Sound power radiated from the sound source
Wo Reference power 10–12 W
The sound power level is determined indirectly by measuring sound pressure or soundintensity levels at an enveloping surface surrounding the source. ISO 3740, 3744 to3748 or DIN 45635-1 standards can be applied as general stipulations for determiningthe sound power of machinery, and modified methods have also been developed for aseries of machines. Power transformers are governed by IEC 610076-10 with typicalsound power levels to VDI 3739.
The sound intensity level LI is defined as the logarithm of the ratio of the soundintensity at the measuring point to the reference intensity Io. Measurement isperformed with a sound intensity probe in which two microphones are located at ashort distance opposite each other. The sound intensity in the direction of themicrophone axes is calculated using the sound pressure and the gradients of thesound pressure between the microphones.
LI = 10 lg I/Io in dB
Io The reference intensity 10–12 W/m2
I Measured intensity
The sound intensity measuring method permits determination of the sound powerunder non-ideal conditions, for instance in the presence of interference noise orreflections. Further stipulations on determination of the sound power by the soundintensity method can be found in ISO 9614, Parts 1 and 2.
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The sound pressure level Lp is measured with a sound level meter as the logarithm ofthe ratio of the sound pressure to the reference pressure po. Measurement can beperformed, for example, to IEC 61672-1 and -2.
Lp = 20 lg p/po in dB
Here: po Reference pressure, approximately corresponding to the audible thresholdat 1000 Hz
po 2 · 10–5 N/m2 = 2 · 10–5 Pa
p = Root-mean-square sound pressure
Example:
Sound pressure p = 2 · 10–3 N/m2, measured with a sound pressure meter
Sound level = 20 lg (2 · 10–3) / (2 · 10–5) = 40 dB.
The volume of a noise can be stated as a linear sound pressure level (to DIN 45630,Sheets 1 and 2) or as a frequency-dependent weighted sound pressure level (to DIN45631, E. Zwicker method). The weighted sound pressure levels LA, LB and LC,whichare obtained by switching in defined weighting networks in the sound level meter, arestated in the unit dB with a suffix (A), (B) or (C).
The total sound power level of several sound sources is obtained by adding theirsound powers, i.e. the individual levels are delogarithmized, added and the sumlogarithmized again. Addition of two sound sources of equal strength increases thesound level by 3 dB (example: 2 sound sources of 85 dB together have 88 dB). Withseveral sound sources with different volumes, the volume of the loudest sound sourceis dominant. (Example: 2 sound sources with 80 and 86 dB have a total volume of 87dB.) It follows from this that with two equally loud sound sources both of themattenuate, and with sound sources of different loudness only the louder ones causeattenuation. Every increase in the level by 10 dB causes the perceived loudness todouble, and every reduction by 10 dB causes it to halve.
There are legal limits to permissible noise immissions; in Germany, according to theTechnical Directive on Protection against Noise (TAL) of 26.08.1998, the loudness ofnoises on average must not exceed the following values at the point of immission:
Area Day (6 a.m. - 10 p.m.) Night (10 p.m. - 6 a.m.)dB (A) dB (A)
Industrial 70 70Commercial 65 50Composite 60 45generally residential 55 40Purely residential 50 35Therapy (hospitals, etc.) 45 35
Short-lived, isolated noise peaks can be disregarded (ISO 1996-1, -2, DIN 45641,45645-1). Many standards and regulations, including the TAL, make further allowancesof between 3 dB and 6 dB deducted from the measured sound pressure level fornoises containing tones or pulses. Other limits apply in many other countries. Thereare various methods for assessing the tone content of a noise: one example is that ofDIN 45681.
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1
Disturbing noise is propagated as air-borne and solid-borne sound. When air-bornesound waves strike a wall, some are thrown back by reflection and others areabsorbed by the wall. If air-borne noise striking a wall causes it to vibrate, the wallstransmit the sound into the adjacent space. Solid-borne sound is converted intoaudible air-borne sound by radiation from the boundary surfaces. Ducts, air-shafts,piping systems and the like can transmit sound waves to other rooms. Specialattention must therefore be paid to this when buildings are designed.
Sound pressure and sound intensity decrease with increasing distance from the soundsource. A rough estimate of the sound level Lp(r) to be expected at distance r from asound source with sound power LW can be calculated as follows:
Lp(r) = Lw –10 lg (2π r2)
It is assumed here that the sound source is mounted on a reflecting level surface andthe propagation conditions are otherwise homogeneous.
In the open air, sound propagation is not only affected by distance, but also byreflection and absorption on buildings and plant components, and by the acousticproperties of the ground, by plants and by meteorological influences such as wind andtemperature gradients. ISO 9613-2 and VDI 2714 are often used as guidelines inforecasting sound propagation.
In general, noise emissions must be kept as low as possible at their point of origin. Thiscan often only be achieved by enclosing the noise sources. VDI 2711 provides aguideline for the practical implementation of noise reducing enclosures. The acousticeffect of an enclosure results from a combination of sound reflection from the walls onthe inside, and sound absorption. Without sound absorption, there would be anincrease in sound pressure inside the enclosure as a result of the reflected sound,which would reduce the effectiveness of the enclosure. The most commonly usedsound-absorbent materials are porous substances, plastics, cork, glass fibre andmineral wool, etc. Higher frequency noise components (> approx. 250 Hz) are easier tocombat than lower frequency noises. For noise reduction by more than around 10 dB,not only the choice of materials for the enclosure walls, but also the careful sealing ofall openings, is important. Openings to channel heat and gas are to be fitted withacoustic dampers. A further method of reducing noise immission is to erect anacoustic screen. The effect of an acoustic screen depends on its dimensions, positionand distance from the sound source, distance of the receiver from the edge of thescreen, and the absorption capacity of the wall (see VDI 2720).
When testing walls and ceilings for their behaviour regarding air-borne sound (DIN52210, EN 20140-3, ISO 140-4), one determines the difference D in sound level L forthe frequency range from 100 Hz to 3150 Hz.
D = L1 – L2 in dB where L = 20 lg p/po dB
Here: L1 = Sound level in room containing the sound source
L2 = Sound level in room receiving the sound
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Table 1-13
Noise attenuation figures of various construction materials in the range from 100 to3200 Hz
Structural component Attenuation Structural AttenuationdB dB
Brickwork, rendered, 45 Single door without up to 2012 cm thick extra sealingBrickwork, rendered, 50 Single door with good seal 3025 cm thickConcrete wall, 10 cm thick 42 Double door without seal 30Concrete wall, 20 cm thick 42 Double door without extra sealing 40Wood wool mat, 8 cm thick 50 Single window without sealing 15Straw mat, 5 cm thick 38 Spaced double window with seal 30
The reduction in level ∆L, obtainable in a room or enclosure by means of sound-absorbing materials or structures is:
∆L = 10 lg A2/A1 = 10 lg T1/T2 in dB
Here: A Equivalent sound absorption area in the room concerned (frommultiplication of the geometrical areas with their corresponding degrees ofsound absorption )
T Reverberation time of the room in s (Index 1 applies to the untreated room,and index 2 to the room treated with sound-absorbing materials)
Typical degrees of sound absorption a :
Material
Smooth concrete, tiles, masonry 0.05Room with furniture, rectangular machine room 0.15Irregularly shaped room with furniture, machine room 0.2Room with upholstered furniture, room with insulating material on parts of ceiling and walls 0.25Room with insulating material on ceiling and walls 0.35Room with large quantities of insulating material on ceiling and walls 0.5
The equivalent degree of sound absorption of a room can be determined experimen-tally using the reverberation time T:
A = 0,163 V/T in m2
Here: V Room volume in m3
T Reverberation time in s, in which the sound level L falls by 60 dB after soundemission ceases.
When a reduction in noise of approx. 10 dB and more is required, it should also beexamined whether sound can be radiated by other components excited by thetransmission of structure-borne noise. If necessary, the sound source is to be mountedon anti-vibration bearings. These can be simple rubber springs or dissipative steelsprings, depending on the requirements. For high demands, an intermediatefoundation which is also installed on anti-vibration mountings may also be necessary– see also VDI 2062.
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1
1.2.
7 Te
chni
cal v
alue
s o
f so
lids,
liq
uid
s an
d g
ases
Tab
le 1
-14
Tech
nica
l val
ues
of s
olid
s
Mat
eria
lD
ensi
tyM
eltin
gB
oilin
gLi
near
Ther
mal
Mea
nS
pec
ific
Tem
per
atur
eρ
orp
oint
ther
mal
cond
ucti-
spec
.el
ectr
ical
coef
ficie
nt α
free
zing
exp
ansi
onvi
ty λ
athe
at c
at
resi
stan
ce ρ
of e
lect
rical
poi
ntα
20 °
C0
..10
0 °C
at 2
0 °C
resi
stan
ceat
20
°Ckg
/dm
3°C
°C1)
W/(
m ·
K)
J/(k
g · K
)Ω
· mm
2 /m
1/K
Alu
min
ium
e.g
. EA
L 99
.5(A
)2.
7065
822
7024
220
920
0.02
876
0.00
42A
l-al
loy
e.g.
EA
I MgS
i(B)
2.70
≈63
023
190
920
0.03
330.
0036
Lead
11.3
432
71
730
2834
130
0.21
0.00
43
Bro
nze
e.g.
CuS
n4P
b4Z
n48.
9≈
930
≈17
.387
377
≈0.
090
0.00
07C
adm
ium
8.64
321
767
31.6
9223
40.
762
0.00
42C
hrom
ium
6.92
180
02
400
8.5
452
0.02
8
Iron,
pur
e7.
881
530
2 50
012
.371
464
0.10
0.00
58Iro
n, s
teel
≈7.
8≈
135
0≈
11.5
4648
50.
25..
0.10
≈0.
005
Iron,
cas
t≈
7.25
≈1
200
≈11
4654
00.
6..
10.
0045
Gol
d19
.29
106
32
700
14.2
309
130
0.02
20
0038
Con
stan
tan
Cu
+ N
i8
..8.
91
600
16.8
2241
00.
48..
0.50
≈0.
0000
5C
arb
on d
iam
ond
3.51
≈3
600
4 20
01.
350
2C
arb
on g
rap
hite
2.25
7.86
571
1
Cop
per
e.g
. Cu-
ETP
R20
08.
921
083
2 33
017
385
393
0.01
754
0.00
392
Mag
nesi
um1.
7465
011
1025
.016
710
340.
0455
0.00
4
1)b
etw
een
0 °C
and
100
°C
(con
tinue
d)
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 33
34
Tab
le 1
-14
(con
tinue
d)
Tech
nica
l val
ues
of s
olid
s
Mat
eria
lD
ensi
tyM
eltin
gB
oilin
gLi
near
Ther
mal
Mea
nS
pec
ific
Tem
per
atur
eρ
orp
oint
ther
mal
cond
ucti-
spec
.el
ectr
ical
coef
ficie
nt α
free
zing
exp
ansi
onvi
ty λ
athe
at c
at
resi
stan
ce ρ
of e
lect
rical
poi
ntα
20 °
C0
..10
0 °C
at 2
0 °C
resi
stan
ceat
20
°Ckg
/dm
3°C
°C1)
W/(
m ·
K)
J/(k
g · K
)Ω
· mm
2 /m
1/K
Bra
ss e
.g. C
uZn3
78.
591
218
120
377
≈0.
0555
0.00
24N
icke
l8.
91
455
3 00
013
8345
2≈
0.12
0.00
46P
latin
um21
.45
177
33
800
8.99
7113
4≈
0.11
0.00
39
Mer
cury
13.5
4638
.83
357
618.
313
90.
698
0.00
08S
ulp
hur
(rho
mb
ic)
2.07
113
445
900.
272
0S
elen
ium
(met
allic
)4.
2622
068
866
351
Silv
er10
.50
960
195
019
.542
123
30.
0165
0.00
36Tu
ngst
en19
.33
380
6 00
04.
5016
713
40.
060.
0046
Zin
c7.
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990
716
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387
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0037
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7.28
232
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026
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230
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90.
004
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etw
een
0 °C
and
100
°C
ABB_11_E_01 13.08.2007 8:16 Uhr Seite 34
35
1
Tab
le 1
-15
Tech
nica
l val
ues
of li
qui
ds
Mat
eria
lC
hem
ical
Den
sity
Mel
ting
Boi
ling
Exp
ansi
onTh
erm
alS
pec
ific
Rel
ativ
efo
rmul
aρ
orp
oint
at
coef
ficie
ntco
nduc
tivity
heat
cp
die
lect
ricfr
eezi
ng76
0 To
rrx
10–3
λat
20
°Cat
0 °
Cco
nsta
nt ε
r
poi
ntat
180
°C
kg/d
m3
°C°C
at 1
8 °C
W/(
m ·
K)
J/(k
g · K
)
Ace
tone
C3H
6O0.
791
–95
56.3
1.43
2 16
021
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thyl
alc
ohol
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789
–11
478
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100.
22
554
25.8
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yl e
ther
C4H
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0.71
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124
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1.62
0.14
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84.
3
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mon
iaN
H3
0.77
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77.8
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187
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ABB_11_E_01 13.08.2007 8:16 Uhr Seite 35
36
Tab
le 1
- 16
Tech
nica
l val
ues
of g
ases
Mat
eria
lC
hem
ical
Den
sity
Mel
ting
Boi
ling
Ther
mal
Sp
ecifi
cR
elat
ive1)
form
ula
ρ1)
poi
ntp
oint
cond
uctiv
ity λ
heat
cp
at 0
°C
die
lect
ricco
nsta
nt ε
rkg
/m3
°C°C
10–2
W/(
m ·
K)
J/(k
g · K
)
Am
mon
iaN
H3
0.77
1–
77.7
–33
.42.
172
060
1.00
72E
thyl
ene
C2H
41.
260
–16
9.4
–10
3.5
1.67
1 61
11.
0014
56A
rgon
Ar
1.78
4–
189.
3–
185.
91.
7552
31.
0005
6
Ace
tyle
neC
2H2
1.17
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81–
83.6
1.84
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97
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272
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ABB_11_E_01 13.08.2007 8:16 Uhr Seite 36
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1
1.3 Strength of materials
1.3.1 Fundamentals and definitions
External forces F acting on a cross-section A of a structural element can give rise totensile stresses (σz), compressive stresses (σd), bending stresses (σb), shear stresses(τs) or torsional stresses (τt). If a number of stresses are applied simultaneously to acomponent, i. e. compound stresses, this component must be designed according tothe formulae for compound strength. In this case the following rule must be observed:
Normal stresses σz. σd. σb,Tangential stresses (shear and torsional stresses) τs, τt.
are to be added arithmetically;
Normal stresses σb with shear stresses τs,Normal stresses σb with torsional stresses τt,
are to be added geometrically.
Fig. 1-2
Stress-strain diagram, a) Tensile test with pronounced yield point, material = structuralsteel; b) Tensile test without pronounced yield point, material = Cu/Al, ε Elongation,σ Tensile stress, σs Stress at yield point, σE Stress at proportionality limit, Rp02 Stresswith permanent elongation less than 0.2 %, σB Breaking stress.
Elongation ε = ∆ l/l0 (or compression in the case of the compression test) is foundfrom the measured length l0 of a bar test specimen and its change in length ∆ l = l – l0
in relation to the tensile stress σz, applied by an external force F. With stresses belowthe proportionality limit σE elongation increases in direct proportion to the stress σ(Hooke’s law).
Stress σE σEThe ratio ———————— = —– = E is termed the elasticity modulus.Elongation ε εE
E is an imagined stress serving as a measure of the resistance of a material todeformation due to tensile or compressive stresses; it is valid only for the elasticregion.
E is determined in terms of the load σ0.01, i.e. the stress at which the permanentelongation is 0.01 % of the measured length of the test specimen.
a) b)
σE
⎪↓
⎪↓
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If the stresses exceed the yield point σs, materials such as steel undergo permanentelongation. The ultimate strength, or breaking stress, is denoted by σB, although a bardoes not break until the stress is again being reduced. Breaking stress σB is related tothe elongation on fracture εB of a test bar. Materials having no marked proportionallimit or elastic limit, such as copper and aluminium, are defined in terms of theso-called Rp0.2-limit, which is that stress at which the permanent elongation is 0.2 %after the external force has been withdrawn.
For reasons of safety, the maximum permissible stresses, σmax or τmax in the materialmust be below the proportional limit so that no permanent deformation, such aselongation or deflection, persists in the structural component after the external forceceases to be applied.
Table 1-17
Material Elasticitymodulus E1)
kN/mm2 2)
Steel 191-224Cast iron 110-140Bronze, CuAl5 123Copper e.g. Cu-ETP 110Al-alloy EAl MgSi(B) 70Aluminium EAl 99,5(A) 65Magnesium alloy MgMn2 45Lead 16
1) Typical values 2) 1 kN/mm2 = 1 GPa
Fatigue strength (endurance limit) is present when the maximum variation of a stressoscillating about a mean stress is applied “infinitely often” to a loaded material (at least107 load reversals in the case of steel) without giving rise to excessive deformation orfracture.
Cyclic stresses can occur in the form of a stress varying between positive and negativevalues of equal amplitude, or as a stress varying between zero and a certain maximumvalue. Cyclic loading of the latter kind can occur only in compression or only in tension.
Depending on the manner of loading, fatigue strength can be considered as bendingfatigue strength, tension-compression fatigue strength or torsional fatigue strength.Structural elements which have to withstand only a limited number of load reversalscan be subjected to correspondingly higher loads. The resulting stress is termed thefatigue limit.
One speaks of creep strength when a steady load with uniform stress is applied,usually at elevated temperatures.
1.3.2 Tensile and compressive strength
If the line of application of a force F coincides with the centroidal axis of a prismatic barof cross section A (Fig.1-3), the normal stress uniformly distributed over the cross-
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1
section area and acting perpendicular to it is
Fσ = — .A
With the maximum permissible stress σmax for a given material and a given loading, therequired cross section or the maximum permissible force, is therefore:
FA = ——— or F = σmax · A.σmax
Example:
A drawbar is to be stressed with a steadyload of F = 180 000 N.
The chosen material is structural steel St 37 with σmax = 120 N/mm2.
Required cross section of bar:
E 180 000 NA = ——— = ———————— = 1500 mm2.
σmax 120 N/mm2
Round bar of d = 45 mm chosen.
1.3.3 Bending strength
The greatest bending action of an external force, or its greatest bending moment M,occurs at the point of fixing a in the case of a simple cantilever, and at point c in thecase of a centrally loaded beam on two supports.
Fig. 1-4
Maximum bending moment at a: M = F · l; bei c: M = F · l/4
In position a and c, assuming the beams to be of constant cross section, the bendingstresses σb are greatest in the filaments furthermost from the neutral axis. M may begreater, the greater is σmax and the “more resistant” is the cross-section. The followingcross sections have moments of resistance W in cm, if a, b, h and d are stated in cm.
The maximum permissible bending moment is M = W · σmax and the required momentof resistance
MW = —–— .σmax
Fig. 1-3
l
F
l/2 l/2
l
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Example:
A mild-steel stud σmax = 70 N/mm2 with an unsupported length of
l = 60 mm is to be loaded in the middle with a force F = 30 000 N. Required momentof resistance is:
M F · l 30 000 N · 60 mmW = ————— = ————— = ——————————— = 6.4 · 103 mm3.
σmax 4 · σmax 4 · 70 N/mm2
According to Table 1-21, the moment of resistance W with bending is W ≈ 0.1 · d 3.
The diameter of the stud will be: d =310 W, d =
364 000 mm = 40 mm.
1.3.4 Loadings on beams
Table 1-18
Bending load
Case Reaction force A, B max. Deflection fBending permissible moment M load F, Q
F l F l3A = F W = ——— f = ———
σzul 3 E J
σzul WMmax = F l F = ———— —
l
Q l Q l3A = Q W = ———— f = ———
2 σzul 8 E J
Q l 2 σzul WMmax = —— Q = ————— —
2 l
F F l F l3A = B = — W = ——— f = ————
2 4 σzul 48 E J
F l 4 σzul WMmax = —— F = ————— —
4 l
Q Q l 5 Q l3A = B = — W = ———— f = —— · ——
2 8 σzul 384 E J
Q l 8 σzul WMmax = —— Q = ————— —
8 l
l
l
l
l
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1
Table 1-18 (continued)
Bending load
Case Reaction force A, B max. Deflection fBending permissible moment M load F, Q
F b F a 2 b 2A = —— f = —————
l 3 E J l
F a σzul W lB = —— F = —————
l a b
F · a · bMmax = ————l
F aA = B = F f = ———— · [3 (l + 2 a) 2 – 4 a2]24 E J
σzul WMmax = F a F = ————
a
F1 e + F2 c A a F1 a2 e2+ F2l2 d2
A = ——————— W1 = ——— f = ————————l σzul 3 E J l
F1 a + F2 d B cB = ——————— W2 = ———
l σzul
determine beam forgreatest “W“
W selection modulus (bending)
F Single point load, Q Uniformly distributed load, J axial angular impulse
E elasticity modulus of material
Q Q l Q l3A = B = — W = ———— f = —— — · ——
2 12 σzul E J 384
Q l 12 σzul WMmax = —— Q = ——————
12 l
l
l
l
l
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1.3.5 Buckling strength
Thin bars loaded in compression are liable to buckle. Such bars must be checked bothfor compression and for buckling strength. Buckling strength is calculated with Euler'sformula, a distinction being drawn between four cases.
Table 1-19
Buckling
Case I
One end fixed, other end free
Case II
Both ends free to move along bar axis
Case III
One end fixed, other end free to movealong bar axis
Case IV
Both ends fixed, movement alongbar axis
E Elasticity modulus of material s Factor of safety:J Minimum axial angular impulse for cast iron 8,F Maximum permissible force for mild carbon steel 5,I Length of bar for wood 10.
l l
l l
π2 E JF = ————
4 s l2
4 π2 E JF = ————
s l2
( π )2E J
F = ——————s l2
π2 E JF = ————
s l2
0.7
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1
1.3.6 Maximum permissible buckling and tensile stress for tubular rods
Threaded steel tube (gas pipe) or seamless steel tube
10 E 10 E D4 – d4 D 4 – d 4
Fbuck = ——— · J = ——— · ————— where J ≈ ———— from Table 1-22s l2 s l2 20 20
Ften = A · σmax
in which F ForceE Elasticity modulus (= 210 kN/mm2
J Axial angular impulse in cm4
s Factor of safety (= 5)σmax Max. permissible stress (= 350 N/mm2)A Cross-section areaD Outside diameterd Inside diameterl Length
Fig. 1-5
Table 1-20
Nomi- Dimensions Cross- Moment Weight Fbuck for tube length l ≈ Ften
nal dia- sec- of ofmeter tions inertia tube
D D a A J 0.5 m 1 m 1.5 m 2 m 2.5 m 3 minch mm mm mm2 cm4 kg/m kN kN kN kN kN kN kN
10 ³⁄8 17.2 2.35 109.6 0.32 0.85 5.26 1.31 0.58 0.33 0.21 0.15 7.6715 ¹⁄₂ 21.3 2.65 155.3 0.70 1.22 11.69 2.92 1.30 0.73 0.47 0.32 10.8720 ³⁄₄ 26.9 2.65 201.9 1.53 1.58 25.48 6.37 2.83 1.59 1.02 0.71 14.13
25 1 33.7 3.25 310.9 3.71 2.44 61.84 15.46 6.87 3.87 2.47 1.72 21.760.8 25 2 144.5 0.98 1.13 16.34 4.09 1.82 1.02 0.65 0.45 10.120.10431.8 2.6 238.5 2.61 1.88 43.48 10.87 4.83 2.72 1.74 1.21 16.70
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1.3.7 Shear strength
Two equal and opposite forces F acting perpendicular to the axis of a bar stress thissection of the bar in shear.
FThe stress is τs = —;
A
For for given values of F and τs max, the required cross section is A = –——–
τs zul
Fig. 1-6
Pull-rod coupling
Stresses in shear are always combined with a bending stress, and therefore thebending stress σb has to be calculated subsequently in accordance with the followingexample.
Rivets, short bolts and the like need only be calculated for shear stress.
Example:
Calculate the cross section of a shackle pin of structural steel, with Rp 0.2 min = 300 N/mm2 and τs max = 0.8 Rp 0.2 min, for the pull-rod coupling shown in Fig. 1-6.
1. Calculation for shear force:
F 150 000 NA = ———— = ————————————— = 312 mm2
2 τs max 2 · (0.8 · 300) N/mm2
yields a pin diameter of d ≈ 20 mm, with W = 0.8 · 103 mm3 (from W ≈ 0.1 · d 3, seeTable 1-21).
F = 15 000 kp ≈ 1.5 · 105 N
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1
2. Verification of bending stress:
The bending moment for the pin if F l/4 with a singlepoint load, and F l/8 for a uniformly distributed load. The average value is
F l F l—— + ——
4 8 3Mb = ———————— = — F l
2 16
when F = 1.5 · 105 N, l = 75 mm becomes:
3Mb = —— · 1.5 · 105 N · 75 mm ≈ 21 · 105 N · mm;
16
Mb 21 · 105 N · mm NσB = –— = —————————— ≈ 2,63 · 103 ———W 0,8 · 103 mm3 mm2
i. e. a pin calculated in terms of shear with d = 20 mm will be too weak. The requiredpin diameter d calculated in terms of bending is
Mb 21 · 105 N · mmW= ——— = —————————— = 7 · 103 mm3 = 0.7 cm3
σmax 300 N/mm2
d ≈310 · W =
310 · 7 · 103 mm3 = 41,4 mm ≈ 42 mm.
i. e. in view of the bending stress, the pin must have a diameter of 42 mm instead of20 mm.
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1.3.8 Moments of resistance and moments of inertia
Table 1-21
Cross- Moment of resistance Moment of inertiasection torsion bending1) polar1) axial2)
W 4) W 4) Jp J
0.196 d 3 0.098 d 3 0.098 d 4 0.049 d 4
≈ 0.2 d 3 ≈ 0.1 d 3 ≈ 0.1 d 4 ≈ 0.05 d 4
D 4 – d 4 D 4 – d 4 0.098 (D4– d4) 0.049 (D 4 – d 4)0.196 ———— 0.098 ————
D D
0.208 a3 0.118 a3 0.141 a 4 0.083 a 4
0.208 k1 b 2 h 3) 0.167 b h2 0.141 · k2 b4 0.083 b h 3
B H 3 – b h 3 B H 3 – b h 3——————— ———————
6 H 12
B H 3 – b h 3 B H 3 – b h 3——————— ———————
6 H 12
B H 3 – b h 3 B H 3 – b h 3——————— ———————
6 H 12
b h 3 + bo ho3 b h 3 + bo ho
3——————— ———————
6 h 12
1) Referred to CG of area.2) Referred to plotted axis.3) Values for k: if h : b = 1 1.5 2 3 4
———————————————————— ———-----then k1 = 1 1,11 1,18 1,28 1,36then k2 = 1 1,39 1,62 1,87 1,99
4) Symbol Z is also applicable, see DIN VDE 0103
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1
1.4 Geometry, calculation of areas and solid bodies
1.4.1 Area of polygons
Regular polygons (n angles)The area A, length of sides S and radii of the outerand inner circles can be calculated using the angle αand the number of sides n. (α = 360 ˚/n).
Irregular polygons
g1 h1 g 2 h 2A = ——— + ——— + …2 2
1= – (g1 h1 + g 2 h 2 + …)
2
Pythagoras theorem
c 2 = a 2 + b 2; c = a 2 + b 2
a 2 = c 2 – b 2; a = c 2 – b 2
b 2 = c 2 – a 2; b = c 2 – a 2
nA/S2 = – cot (α/2)
4
A/r2 = n tan (α/2)
nA/R2 = –– sin α
2
S/R = 2 sin (α/2)
R/r = cos (α/2)
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1Triangle A = – a h U = a + b + c
2 1e = – h3
a + bTrapezium A = ——— · h U = a + b + c + d
2 h a + 2 be = – · ————
3 a + b
Rectangle A = a b U = 2 (a + b)
b r αCircle A = —— = —— r2 π U = 2 r + bsegment 2 180
α 2 sin α 180b = r π —— e = – r ——— · ———
90 3 α π
1A = – π r2 U = r (2 + π) ≈ 5,14 r
2 4 re = – · – = 0,425 r
3 π
48
1.4.2 Areas and centres of gravity
Table 1-22
Shape of surface A = area U perimeterS centre of gravity (cg)e distance of cg
Semicircle
Circle
Annularsegment
Semi-annulus
Annulus
Circularsegment
Ellipse
d 2
A = r 2 π = π — U = 2 π r = π d4
π αA = —— α (R 2 – r 2) U = 2 (R – r) + π · —– (R + r)
180 90
2 R 2 – r 2 sin α 180e = – · ———— · ——— · ——
3 R2 – r 2 α π
π 4 R2 + R · r + r2A = —— α (R 2 – r 2) e = — · ————————
2 3 π R + r
A = π (R 2 – r 2) U = 2 π (R + r)
r2 π · α π r αA = —— ––––– - sin 2α U = 2 r 2 – h 2 + ————
2 90 90
s = 2 r 2 – h 2s 2
e = ————12 · A
a b πA = —— π U = – 1,5 (a + b) – ab 4 2
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1
1.4.3 Volumes and surface areas of solid bodies
Table 1-23
Shape V volume O Surfaceof body A base surface
M Nappe
Solid V = a b c O = 2 (a b + a c + b c)rectangle
d 3Cube V = a3 = ——— O = 6 a2 = 3 d 2
2.828
Prism V = A h O = M + 2 A
Pyramid 1V = – A h O = A + M
3
1Cone V = – A h O = π r (r + s)
3 s = h2 + r2
Truncated π h O = (R + r) π s + π (R2 + r2)cone V = (R 2 + r 2 + R r) · —–
3 s = h2 + (R – r)2
Truncated 1pyramid V = – h (A + A1 + AA1) O = A + A1 + M
3
4Sphere V = – π r 3 O = 4 π r 2
3
2Hemisphere V = – π r 3 O = 3 π r 2
3
1Spherical V = π h 2 r — – h O = 2 π r h + π (2 r h – h 2) =segment 3 π h (4 r – h)
2Spherical V = – π r 2 h π rsector 3 O = —— (4 h + s)
2(continued)
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Table 1-25 (continued)
Shape V Volume O Surfaceof body A Area
Zone π h O = π (2 r h + a 2 + b 2)of sphere V = —— (3a 2 + 3b 2 + h 2)
6
h + h1Obliquely V = π r 2 ———— O = π r (h + h1) + A + A1cut cylinder 2
2 πCylindrical V = – r 2 h 0 = 2rh + – r 2 + Awedge 3 2
Cylinder V = π r 2h O = 2 π r h + 2 π r2
Hollow V = π h (R 2 – r 2) O = 2π h (R + r) + 2π (R2 – r2)cylinder
π D + d πBarrel V = — l · O = ————π d + – d 2
15 2 2(2 D 2 + Dd + 0.75 d 2) (approximate)
Body of ro- V = 2 π A O = circumference of cross-tation (ring) A = cross-section section x 2 π
50
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