beam model calculation
DESCRIPTION
Calculation of beams accordig to EC2TRANSCRIPT
EEUU
RROO
CCOO
DDEE
22
WWOO
RRKK
EEDD
EEXX
AAMM
PPLL EE
SS
EEUU
RROO
CCOO
DDEE
22
WWOO
RRKK
EEDD
EEXX
AAMM
PPLL EE
SS
Cop
yrig
ht:
Euro
pean
Con
cret
eP
latf
orm
ASB
L,M
ay20
08.
Allr
ight
sre
serv
ed.N
opa
rtof
this
publ
icat
ion
may
bere
prod
uced
,sto
red
ina
retr
ieva
lsys
tem
ortr
ansm
itted
inan
yfo
rmor
byan
ym
eans
,el
ectr
onic
,m
echa
nica
l,ph
otoc
opyi
ng,
reco
rdin
gor
othe
rwis
e,w
ithou
tth
epr
ior
writ
ten
perm
issi
onof
the
Euro
pean
Conc
rete
Plat
form
ASBL
.
Publ
ishe
dby
the
Euro
pean
Conc
rete
Plat
form
ASBL
Edito
r:Je
an-P
ierr
eJa
cobs
8ru
eVo
lta10
50Br
usse
ls,B
elgi
um
Layo
ut&
Prin
ting
byth
eEu
rope
anCo
ncre
tePl
atfo
rm
Alli
nfor
mat
ion
inth
isdo
cum
ent
isde
emed
tobe
accu
rate
byth
eEu
rope
anCo
ncre
tePl
atfo
rmAS
BLat
the
time
ofgo
ing
into
pres
s.It
isgi
ven
ingo
odfa
ith.
Info
rmat
ion
onEu
rope
anCo
ncre
tePl
atfo
rmdo
cum
ents
does
not
crea
tean
ylia
bilit
yfo
rits
Mem
bers
.W
hile
the
goal
isto
keep
this
info
rmat
ion
timel
yan
dac
cura
te,
the
Euro
pean
Conc
rete
Plat
form
ASBL
cann
otgu
aran
tee
eith
er.
Ifer
rors
are
brou
ght
toits
atte
ntio
n,th
eyw
illbe
corr
ecte
d.
The
opin
ions
refle
cted
inth
isdo
cum
ent
are
thos
eof
the
auth
ors
and
the
Euro
pean
Conc
rete
Plat
form
ASBL
cann
otbe
held
liabl
efo
ran
yvi
ewex
pres
sed
ther
ein.
All
advi
ceor
info
rmat
ion
from
the
Euro
pean
Conc
rete
Plat
form
ASBL
isin
tend
edfo
rth
ose
who
will
eval
uate
the
sign
ifica
nce
and
limita
tions
ofits
cont
ents
and
take
resp
onsi
bilit
yfo
rits
use
and
appl
icat
ion.
No
liabi
lity
(incl
udin
gfo
rne
glig
ence
)fo
ran
ylo
ssre
sulti
ngfr
omsu
chad
vice
orin
form
atio
nis
acce
pted
.
Read
ers
shou
ldno
teth
atal
lEur
opea
nCo
ncre
tePl
atfo
rmpu
blic
atio
nsar
esu
bjec
tto
revi
sion
from
time
totim
ean
dth
eref
ore
ensu
reth
atth
eyar
ein
poss
essi
onof
the
late
stve
rsio
n.
This
publ
icat
ion
isba
sed
onth
epu
blic
atio
n:"G
uida
all'u
sode
ll'eu
roco
dice
2"pr
epar
edby
AICA
P;th
eIt
alia
nAs
soci
atio
nfo
rRe
info
rced
and
Pres
tres
sed
Conc
rete
,on
beha
lfof
the
the
Ital
ian
Cem
ent
Org
anzi
atio
nAI
TEC,
and
onba
ckgr
ound
docu
men
tspr
epar
edby
the
Euro
code
2Pr
ojec
tTe
ams
Mem
bers
,du
ring
the
prep
arat
ion
ofth
eEN
vers
ion
ofEu
roco
de2
(pro
fA.
W.
Beeb
y,pr
ofH
.Co
rres
Peire
tti,
prof
J.W
alra
ven,
prof
B.W
este
rber
g,pr
ofR.
V.W
hitm
an).
Auth
oriz
atio
nha
sbe
enre
ceiv
edor
ispe
ndin
gfr
omor
gani
satio
nsor
indi
vidu
als
for
thei
rsp
ecifi
cco
ntrib
utio
ns.
Att
ribu
tabl
eFo
rew
ord
toth
eC
omm
enta
ryan
dW
orke
dE
xam
ples
toE
C2
Euro
code
sar
eon
eof
the
mos
tadv
ance
dsu
iteof
stru
ctur
alco
des
inth
ew
orld
.The
yem
body
the
colle
ctiv
eex
perie
nce
and
know
ledg
eof
who
leof
Euro
pe.T
hey
are
born
out
ofan
ambi
tious
prog
ram
me
initi
ated
byth
eEu
rope
anU
nion
.With
aw
ealth
ofco
dew
ritin
gex
perie
nce
inEu
rope
,itw
aspo
ssib
leto
appr
oach
the
task
ina
ratio
nal
and
logi
calm
anne
r.Eu
roco
des
refle
ctth
ere
sults
ofre
sear
chin
mat
eria
ltec
hnol
ogy
and
stru
ctur
albe
havi
our
inth
ela
stfif
tyye
ars
and
they
inco
rpor
ate
allm
oder
ntre
nds
inst
ruct
ural
desi
gn.
Like
man
ycu
rren
tna
tiona
lco
des
inEu
rope
,Eu
roco
de2
(EC
2)fo
rco
ncre
test
ruct
ures
draw
she
avily
onth
eC
EBM
odel
Cod
e.A
ndye
tth
epr
esen
tatio
nan
dte
rmin
olog
y,co
nditi
oned
byth
eag
reed
form
atfo
rEu
roco
des,
mig
htob
scur
eth
esi
mila
ritie
sto
man
yna
tiona
lco
des.
Als
oEC
2in
com
mon
with
othe
rEu
roco
des,
tend
sto
bege
nera
lin
char
acte
rand
this
mig
htpr
esen
tdiff
icul
tyto
som
ede
sign
ers
atle
ast
initi
ally
.Th
epr
oble
ms
ofco
min
gto
term
sw
itha
new
set
ofco
des
bybu
sypr
actis
ing
engi
neer
sca
nnot
beun
dere
stim
ated
.Thi
sis
the
back
drop
toth
epu
blic
atio
nof
‘Com
men
tary
and
Wor
ked
Exam
ples
toEC
2’by
Prof
esso
rM
anci
nian
dhi
sco
lleag
ues.
Com
mis
sion
edby
CEM
BU
REA
U,
BIB
M,
EFC
Aan
dER
MC
Oth
ispu
blic
atio
nsh
ould
prov
eim
men
sely
valu
able
tode
sign
ers
indi
scov
erin
gth
eba
ckgr
ound
tom
any
ofth
eco
dere
quire
men
ts.T
his
publ
icat
ion
will
assi
stin
build
ing
conf
iden
cein
the
new
code
,w
hich
offe
rsto
ols
for
the
desi
gnof
econ
omic
and
inno
vativ
eco
ncre
test
ruct
ures
.Th
epu
blic
atio
nbr
ings
toge
ther
man
yof
the
docu
men
tspr
oduc
edby
the
Proj
ect
Team
durin
gth
ede
velo
pmen
tof
the
code
.The
docu
men
tis
rich
inth
eore
tical
expl
anat
ions
and
draw
son
muc
hre
cent
rese
arch
.C
ompa
rison
sw
ithth
eEN
Vst
age
ofEC
2ar
eal
sopr
ovid
edin
anu
mbe
rofc
ases
.The
chap
ter
onEN
1990
(Bas
isof
stru
ctur
alde
sign
)is
anad
ded
bonu
san
dw
illbe
appr
ecia
ted
bypr
actio
ners
.Wor
ked
exam
ples
furth
erill
ustra
teth
eap
plic
atio
nof
the
code
and
shou
ldpr
omot
eun
ders
tand
ing.
The
com
men
tary
will
prov
ean
auth
entic
com
pani
onto
EC2
and
dese
rves
ever
ysu
cces
s.
Prof
esso
rRS
Nar
ayan
anC
hairm
anC
EN/T
C25
0/SC
2(2
002
–20
05)
Fore
wor
dto
Com
men
tary
toEu
roco
de2
and
Wor
ked
Exam
ples
Whe
na
new
code
ism
ade,
oran
exis
ting
code
isup
date
d,a
num
bero
fprin
cipl
essh
ould
bere
gard
ed:
1.C
odes
shou
ldbe
base
don
clea
ran
dsc
ient
ifica
llyw
ell
foun
ded
theo
ries,
cons
iste
ntan
dco
here
nt,c
orre
spon
ding
toa
good
repr
esen
tatio
nof
the
stru
ctur
albe
havi
oura
ndof
the
mat
eria
lphy
sics.
2.C
odes
shou
ldbe
trans
pare
nt.T
hatm
eans
that
the
writ
erss
houl
dbe
awar
e,th
atth
eco
deis
notp
repa
red
fort
hose
who
mak
eit,
butf
orth
ose
who
will
use
it.3.
New
deve
lopm
ents
shou
ldbe
reco
gniz
edas
muc
has
poss
ible
,but
nota
tthe
cost
ofto
oco
mpl
exth
eore
tical
form
ulat
ions
.4.
Aco
desh
ould
beop
en-m
inde
d,w
hich
mea
nsth
atit
cann
otbe
base
don
one
certa
inth
eory
,exc
ludi
ngot
hers
.Mod
elsw
ithdi
ffer
entd
egre
esof
com
plex
itym
aybe
offe
red.
5.A
code
shou
ldbe
sim
ple
enou
ghto
beha
ndle
dby
prac
ticin
gen
gine
ers
with
out
cons
ider
able
prob
lem
s.O
nth
eot
herh
and
sim
plic
itysh
ould
notl
ead
tosi
gnifi
cant
lack
ofac
cura
cy.H
ere
the
wor
d“a
ccur
acy”
shou
ldbe
wel
lund
erst
ood.
Ofte
nso
-ca
lled
“acc
urat
e”fo
rmul
atio
ns,d
eriv
edby
scie
ntist
s,ca
nnot
lead
tove
ryac
cura
tere
sults
,bec
ause
the
inpu
tval
uesc
anno
tbe
estim
ated
with
accu
racy
.6.
Aco
dem
ayha
vedi
ffer
entl
evel
sof
soph
istic
atio
n.Fo
rins
tanc
esi
mpl
e,pr
actic
alru
les
can
begi
ven,
lead
ing
toco
nser
vativ
ean
dro
bust
desi
gns.
As
anal
tern
ativ
em
ore
deta
iled
desi
gnru
les
may
beof
fere
d,co
nsum
ing
mor
eca
lcul
atio
ntim
e,bu
tre
sulti
ngin
mor
eac
cura
tean
dec
onom
icre
sults
.
Forw
ritin
ga
Euro
code
,lik
eEC
-2,a
noth
erim
porta
ntco
nditi
onap
plie
s.In
tern
atio
nal
cons
ensu
shad
tobe
reac
hed,
butn
oton
the
cost
ofsi
gnifi
cant
conc
essi
onsw
ithre
gard
toqu
ality
.Alo
tofe
ffor
twas
inve
sted
toac
hiev
eal
ltho
sego
als.
Itis
aru
lefo
rev
ery
proj
ect,
that
itsh
ould
not
beco
nsid
ered
asfin
aliz
edif
impl
emen
tatio
nha
sno
tbe
enta
ken
care
of.
This
book
may
,fu
rther
toco
urse
san
dtra
inin
gson
ana
tiona
lan
din
tern
atio
nal
leve
l,se
rve
asan
esse
ntia
lan
dva
luab
leco
ntrib
utio
nto
this
impl
emen
tatio
n.It
cont
ains
exte
nsiv
eba
ckgr
ound
info
rmat
ion
onth
ere
com
men
datio
nsan
dru
les
foun
din
EC2.
Itis
impo
rtant
that
this
back
grou
ndin
form
atio
nis
wel
ldo
cum
ente
dan
dpr
actic
ally
avai
labl
e,as
such
incr
easi
ngth
etra
nspa
renc
y.Iw
ould
like
toth
ank
my
colle
ague
sof
the
Proj
ectT
eam
,esp
ecia
llyR
obin
Whi
ttle,
BoW
este
rber
g,H
ugo
Cor
res
and
Kon
rad
Zilc
h,fo
rhe
lpin
gin
getti
ngto
geth
eral
lbac
kgro
und
info
rmat
ion.
Als
om
yco
lleag
ueG
iuse
ppe
Man
cini
and
his
Italia
nte
amar
egr
atef
ully
ackn
owle
dged
forp
rovi
ding
ase
tofv
ery
illus
trativ
ean
dpr
actic
alw
orki
ngex
ampl
es.F
inal
lyI
wou
ldlik
eto
than
kCE
MB
UR
AU
,BIB
M,E
FCA
and
ERM
COfo
rth
eiri
nitia
tive,
supp
orta
ndad
vice
tobr
ing
outt
hisp
ublic
atio
n.
Joos
tWal
rave
nC
onve
noro
fPro
ject
Team
forE
C2(1
998
-200
2)
EC2�
wor
ked
exam
ples
sum
mar
y
Tabl
eof
Cont
ent
EU
RO
CO
DE
2-W
OR
KE
DE
XA
MPL
ES
-SU
MM
AR
Y
SEC
TIO
N2.
WO
RK
ED
EXA
MPL
ES–
BA
SIS
OF
DE
SIG
N...
......
......
......
......
......
......
......
......
......
......
..2-
1
EXA
MPL
E2.
1.U
LSCO
MBI
NAT
ION
SO
FAC
TIO
NS
FOR
ACO
NTI
NU
OU
SBE
AM[E
C2
–CL
AUSE
2.4]
......
......
......
......
......
.2-1
EXA
MPL
E2.
2.U
LSCO
MBI
NAT
ION
SO
FAC
TIO
NS
FOR
ACA
NO
PY[E
C2
–CL
AUSE
2.4]
......
......
......
......
......
......
......
......
.2-2
EXA
MPL
E2.
3.U
LSCO
MBI
NAT
ION
OF
ACTI
ON
OF
ARE
SID
ENTI
ALCO
NCR
ETE
FRAM
EDBU
ILD
ING
[EC
2–
CLAU
SE2.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
..2-4
EXA
MPL
E2.
4.U
LSCO
MBI
NAT
ION
SO
FAC
TIO
NS
ON
ARE
INFO
RCED
CON
CRET
ERE
TAIN
ING
WAL
L
[EC
2–
CLAU
SE2.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
...2-
6
EXA
MPL
E2.
5.C
ON
CRET
ERE
TAIN
ING
WAL
L:G
LOBA
LST
ABIL
ITY
AND
GRO
UN
DRE
SIST
ANCE
VERI
FICA
TIO
NS
[EC
2–
CLAU
SE2.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
...2-
9
SEC
TIO
N4.
WO
RK
EDE
XA
MPL
ES–
DU
RA
BIL
ITY
......
......
......
......
......
......
......
......
......
......
......
......
..4-
1
EXA
MPL
E4.
1[E
C2
CLAU
SE4.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
....4
-1
EXA
MPL
E4.
2[E
C2
CLAU
SE4.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
....4
-3
EXA
MPL
E4.
3[E
C2
CLAU
SE4.
4]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
....4
-4
SEC
TIO
N6.
WO
RK
EDE
XA
MPL
ES
–U
LTIM
AT
EL
IMIT
STA
TES
......
......
......
......
......
......
......
......
6-1
EXA
MPL
E6.
1(C
ON
CRET
EC
30/3
7)[E
C2
CLAU
SE6.
1]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.6-1
EXA
MPL
E6.
2(C
ON
CRET
EC
90/1
05)[
EC
2CL
AUSE
6.1]
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
..6-3
EXA
MPL
E6.
3C
ALCU
LATI
ON
OF
VRD
,CFO
RA
PRES
TRES
SED
BEAM
[EC
2CL
AUSE
6.2]
......
......
......
......
......
......
......
......
6-4
EXA
MPL
E6.
4D
ETER
MIN
ATIO
NO
FSH
EAR
RESI
STAN
CEG
IVEN
THE
SECT
ION
GEO
MET
RYAN
DM
ECH
ANIC
S
[EC
2CL
AUSE
6.2]
......
......
......
......
......
......
......
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......
......
......
......
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......
......
...6-
5
EXA
MPL
E6.
4B–
THE
SAM
EAB
OVE
,WIT
HST
EEL
S500
Cf yd
=43
5M
PA.[
EC
2CL
AUSE
6.2]
......
......
......
......
......
......
..6-7
EXA
MPL
E6.
5[E
C2
CLAU
SE6.
2]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
....6
-9
EXA
MPL
E6.
6[E
C2
CLAU
SE6.
3]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
..6-1
0
EXA
MPL
E6.
7SH
EAR
–TO
RSIO
NIN
TERA
CTIO
ND
IAG
RAM
S[E
C2
CLAU
SE6.
3]...
......
......
......
......
......
......
......
......
......
.6-1
2
EXA
MPL
E6.
8.W
ALL
BEAM
[EC
2CL
AUSE
6.5]
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
...6-
15
EC2�
wor
ked
exam
ples
sum
mar
y
Tabl
eof
Cont
ent
EXA
MPL
E6.
9.TH
ICK
SHO
RTC
ORB
EL,a
<Z/2
[EC
2CL
AUSE
6.5]
......
......
......
......
......
......
......
......
......
......
......
......
......
.6-1
8
EXA
MPL
E6.
10TH
ICK
CAN
TILE
VER
BEAM
,A>Z
/2[E
C2
CLAU
SE6.
5]...
......
......
......
......
......
......
......
......
......
......
......
...6-
21
EXA
MPL
E6.
11G
ERBE
RBE
AM[E
C2
CLAU
SE6.
5]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.6-2
4
EXA
MPL
E6.
12PI
LECA
P[E
C2
CLAU
SE6.
5]...
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
....6
-28
EXA
MPL
E6.
13V
ARIA
BLE
HEI
GH
TBE
AM[E
C2
CLAU
SE6.
5]...
......
......
......
......
......
......
......
......
......
......
......
......
......
...6-
32
EXA
MPL
E6.
14.3
500
KNCO
NCE
NTR
ATED
LOAD
[EC
2CL
AUSE
6.5]
......
......
......
......
......
......
......
......
......
......
......
......
6-38
EXA
MPL
E6.
15SL
ABS, [E
C2
CLAU
SE5.
10–
6.1
–6.
2–
7.2
–7.
3–
7.4]
......
......
......
......
......
......
......
......
......
......
......
.6-4
0
SEC
TIO
N7.
SER
VIC
EAB
ILIT
YL
IMIT
STA
TE
S–
WO
RK
ED
EXA
MPL
ES...
......
......
......
......
......
..7-
1
EXA
MPL
E7.
1E
VALU
ATIO
NO
FSE
RVIC
EST
RESS
ES[E
C2
CLAU
SE7.
2]...
......
......
......
......
......
......
......
......
......
......
......
..7-1
EXA
MPL
E7.
2D
ESIG
NO
FM
INIM
UM
REIN
FORC
EMEN
T[E
C2
CLAU
SE7.
3.2]
......
......
......
......
......
......
......
......
......
......
...7-
5
EXA
MPL
E7.
3E
VALU
ATIO
NO
FCR
ACK
AMPL
ITU
DE
[EC
2CL
AUSE
7.3.
4]...
......
......
......
......
......
......
......
......
......
......
.....7
-8
EXA
MPL
E7.
4.D
ESIG
NFO
RMU
LAS
DER
IVAT
ION
FOR
THE
CRAC
KIN
GLI
MIT
STAT
E
[EC
2CL
AUSE
7.4]
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.7-1
0
5B7.
4.2
APP
ROXI
MAT
EDM
ETH
OD
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.7-1
1
EXA
MPL
E7.
5A
PPLI
CATI
ON
OF
THE
APPR
OXI
MAT
EDM
ETH
OD
[EC
2CL
AUSE
7.4]
......
......
......
......
......
......
......
......
....7
-13
EXA
MPL
E7.
6V
ERIF
ICAT
ION
OF
LIM
ITST
ATE
OF
DEF
ORM
ATIO
N...
......
......
......
......
......
......
......
......
......
......
......
......
..7-1
8
SEC
TIO
N11
.LIG
HT
WE
IGH
TC
ON
CR
ETE
–W
OR
KE
DEX
AM
PLE
S....
......
......
......
......
......
......
...11
-1
EXA
MPL
E11
.1[E
C2
CLA
USE
11.3
.1–
11.3
.2]..
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
.11-
1
EXA
MPL
E11
.2[E
C2
CLA
USE
11.3
.1–
11.3
.5–
11.3
.6–
11.4
–11
.6].
......
......
......
......
......
......
......
......
......
......
......
...11
-3
EC2�
wor
ked
exam
ples
2-1
Tabl
eof
Cont
ent
SEC
TIO
N2.
WO
RK
ED
EX
AM
PLE
S–
BA
SIS
OF
DE
SIG
N
EX
AM
PL
E2.
1.U
LS
com
bina
tions
ofac
tions
fora
cont
inuo
usbe
am[E
C2�
clau
se2.
4]
Aco
ntin
uous
beam
onfo
urbe
arin
gsis
subj
ecte
dto
the
follo
win
glo
ads:
Self-
wei
ght
Gk1
Perm
anen
tim
pose
dlo
adG
k2Se
rvice
impo
sed
load
Qk1
Note
.In
this
exam
ple
and
inth
efol
lowin
gon
es,a
single
char
acter
istic
valu
eis
take
nfor
self-w
eight
and
perm
anen
timp
osed
load,
respe
ctive
lyG
k1an
dG
k2,b
ecaus
eoft
heir
smal
lvar
iabi
lity.
EQ
U�
Stat
iceq
uilib
rium
(Set
A)
Fact
ors
ofSe
tAsh
ould
beus
edin
the
verif
icat
ion
ofho
ldin
gdo
wn
devi
ces
for
the
uplif
tof
bear
ings
aten
dsp
an,a
sind
icate
din
Fig.
2.1.
Fig.
2.1.
Loa
dco
mbin
ation
forver
ifica
tion
ofho
lding
down
devic
esat
thee
ndbe
arin
gs.
STR�
Bend
ingm
omen
tveri
ficat
ionat
mid
span
(Set
B)
Unl
ike
inth
eve
rific
atio
nof
stat
iceq
uilib
rium
,the
parti
alsa
fety
fact
orfo
rper
man
entl
oads
inth
eve
rific
atio
nof
bend
ing
mom
enti
nth
em
iddl
eof
the
cent
rals
pan,
isth
esa
me
fora
llsp
ans:
G=
1.35
(Fig
.2.2
). Fig.
2.2.
Loa
dcom
bina
tion
forve
rifica
tion
ofbe
ndin
gmom
enti
nth
eBC
span
.
EC2�
wor
ked
exam
ples
2-2
Tabl
eof
Cont
ent
EX
AM
PL
E2.
2.U
LS
com
bina
tions
ofac
tions
fora
cano
py[E
C2�
clau
se2.
4]
The
cano
pyis
subj
ecte
dto
the
follo
win
glo
ads:
Self-
wei
ght
Gk1
Perm
anen
tim
pose
dlo
adG
k2Sn
owim
pose
dlo
adQ
k1
EQ
U�
Stat
iceq
uilib
rium
(Set
A)
Fact
orst
obe
take
nfo
rthe
verif
icat
ion
ofov
ertu
rnin
gar
eth
ose
ofSe
tA,a
sin
Fig.
2.3.
Fig.
2.3.
Loa
dcom
bina
tion
forve
rifica
tion
ofsta
ticeq
uilib
rium
.
STR�
Veri
ficat
ionof
resist
ance
ofa
colu
mn(
SetB
)
The
parti
alfa
ctor
tobe
take
nfo
rpe
rman
ent
load
sin
the
verif
icatio
nof
max
imum
com
pres
sion
stre
sses
and
ofbe
ndin
gw
ithax
ialfo
rce
inth
eco
lum
nis
the
sam
e(
G=
1.35
)for
allsp
ans.
The
varia
ble
impo
sed
load
isdi
strib
uted
over
the
full
leng
thof
the
cano
pyin
the
first
case
,an
don
lyon
half
ofit
fort
heve
rific
atio
nof
bend
ing
with
axia
lfor
ce.
EC2�
wor
ked
exam
ples
2-3
Tabl
eof
Cont
ent
Fig.
2.4.
Loa
dco
mbin
ation
forth
ecom
press
ionstr
esses
verif
icatio
nof
thec
olum
n.
Fig.
2.5.
Loa
dcom
bina
tion
forth
everi
ficat
ionof
bend
ingw
ithax
ialf
orce
ofth
ecolu
mn.
EC2�
wor
ked
exam
ples
2-4
Tabl
eof
Cont
ent
EX
AM
PL
E2.
3.U
LS
com
bina
tion
ofac
tion
-re
side
ntia
lco
ncre
tefr
amed
build
ing
[EC
2�
clau
se2.
4]
The
perm
anen
tim
pose
dlo
adis
indi
cate
das
Gk.V
ariab
leac
tions
are
liste
din
tabl
e2.
1.T
able
2.1.
Var
iable
actio
nson
ares
iden
tialc
oncre
tebu
ildin
g.V
aria
blea
ction
sse
rvic
eabi
lity
impo
sed
load
snow
onro
ofin
g(fo
rsite
sund
er10
00m
a.s.l
.)w
ind
Cha
racte
ristic
valu
eQk
Qk,
esQ
k,n
F k,w
Com
bina
tion
valu
e0
Qk
0.7
Qk,
es0.
5Q
k,n
0.6
F k,w
N.B
.The
valu
esof
parti
alfa
ctors
aret
hose
recom
mend
edby
EN
1990
,but
they
may
bede
fined
inth
eNat
ional
Ann
ex.
Basic
com
bina
tions
forth
everi
ficat
ionof
thes
upers
tructu
re-S
TR(S
etB)
(eq.6
.10-
EN
1990
)Pr
edom
inan
tact
ion:
win
dfa
vour
able
verti
call
oads
(fig.
2.6,
a)1.
0·G
k+
1.5·
F k,w
unfav
oura
blev
ertica
lloa
ds(fi
g.2.
6,b)
1.35
·Gk
+1.
5·(F
k,w
+0.
5·Q
k,n
+0.
7·Q
k,es
)=1.
35·G
k+
1.5·
F k,w
+0.
75·Q
k,n
+1.
05·Q
k,es
Pred
omin
anta
ctio
n:sn
ow(fi
g.2.
6,c)
1.35
·Gk
+1.
5·(Q
k,n+
0.7·
Qk,
es+
0.6·
F k,w
)=1.
35·G
k+
1.5·
Qk,
n+
1.05
·Qk,
es+
0.9·
F k,w
Pred
omin
anta
ctio
n:se
rvice
load
(fig.
2.6,
d)1.
35·G
k+
1.5·
(Qk,
es+
0.5·
Qk,
n+
0.6·
F k,w
)=1.
35·G
k+
1.5·
Qk,
es+
0.75
·Qk,
n+
0.9·
F k,w
Fig.
2.6.
Basic
comb
inat
ionsf
orth
everi
ficat
ionof
thes
upers
tructu
re(S
etB)
:a)W
ind
pred
omin
ant,
favou
rabl
evert
icall
oads
;b)
Win
dpr
edom
inan
t,un
favo
urab
leve
rtica
lloa
ds;c
)Sno
wloa
dpr
edom
inan
t;d)
servic
eloa
dpr
edom
inan
t.
EC2�
wor
ked
exam
ples
2-5
Tabl
eof
Cont
ent
Basic
com
bina
tions
forth
everi
ficat
ionof
found
ation
sand
grou
ndres
istan
ce�
STR
/GE
O[eq
.6.1
0-E
N19
90]
EN
1990
allo
ws
for
thre
edi
ffer
ent
appr
oach
es;
the
appr
oach
tobe
used
isch
osen
inth
eN
atio
nalA
nnex
.For
com
plet
enes
sand
inor
dert
ocla
rify
wha
tis
indi
cate
din
Tabl
es2.
15an
d2.
16,t
heba
sicco
mbi
natio
nsof
actio
nsfo
rall
the
thre
eap
proa
ches
prov
ided
byE
N19
90ar
egi
ven
belo
w.
App
roac
h1
The
desig
nva
lues
ofSe
tCan
dSe
tBof
geot
echn
ical
actio
nsan
dof
allot
hera
ctio
nsfr
omth
est
ruct
ure,
oron
the
stru
ctur
e,ar
eap
plie
din
sepa
rate
calc
ulat
ions
.Hea
vier
valu
esar
eus
ually
give
nby
SetC
fort
hege
otec
hnic
alve
rific
atio
ns(g
roun
dre
sista
nce
verif
icat
ion)
,and
bySe
tBfo
rthe
verif
icatio
nof
the
conc
rete
stru
ctur
alel
emen
tsof
the
foun
datio
n.
SetC
(geo
tech
nica
lver
ifica
tions
)
Pred
omin
anta
ctio
n:w
ind
(favo
urab
leve
rtica
lloa
ds)(
fig.2
.7,a
)1.
0·G
k+
1.3·
F k,w
Pred
omin
anta
ctio
n:w
ind
(unf
avou
rabl
eve
rtica
lloa
ds)(
fig.2
.7,b
)1.
0·G
k+
1.3·
F k,w
+1.
3·0.
5·Q
k,n+
1.3·
0.7·
Qk,
es=
1.0·
Gk
+1.
3·F k
,w+
0.65
·Qk,
n+
0.91
·Qk,
es
Pred
omin
anta
ctio
n:sn
ow(fi
g.2.
7,c)
1.0·
Gk
+1.
3·Q
k,n+
1.3·
0.7·
Qk,
es+
1.3·
0.6·
F k,w
=1.
0·G
k+
1.3·
Qk,
n+
0.91
·Qk,
es+
0.78
·Fk,
w
Pred
omin
anta
ctio
n:se
rvice
load
(fig.
2.7,
d)1.
0·G
k+
1.3·
Qk,
es+
1.3·
0.5·
Qk,
n+
1.3·
0.6·
F k,w
=1.
0·G
k+
1.3·
Qk,
n+
0.65
·Qk,
es+
0.78
·Fk,
w
Fig.
2.7.
Basic
combi
natio
nsfor
thev
erific
ation
ofth
efou
ndat
ions(
SetC
):a)
Win
dpr
edom
inan
t,fa
vour
able
verti
call
oads
;b)
Win
dpr
edom
inan
t,un
favo
urab
leve
rtica
lloa
ds;c
)Sno
wloa
dpr
edom
inan
t;d)
servic
eloa
dpr
edom
inan
t.
EC2�
wor
ked
exam
ples
2-6
Tabl
eof
Cont
ent
SetB
(ver
ifica
tion
ofco
ncre
test
ruct
ural
elem
ents
offo
unda
tions
)
1.0·
Gk
+1.
5·Q
k,w
1.35
·Gk
+1.
5·F k
,w+
0.75
·Qk,
n+
1.05
·Qk,
es
1.35
·Gk
+1.
5·Q
k,n
+1.
05·Q
k,es
+0.
9·F k
,w
1.35
·Gk
+1.
5·Q
k,es
+0.
75·Q
k,n
+0.
9·F k
,w
App
roac
h2
The
sam
eco
mbi
natio
nsus
edfo
rthe
supe
rstru
ctur
e(i.
e.Se
tB)a
reus
ed.
App
roac
h3
Fact
ors
from
SetC
forg
eote
chni
cala
ctio
nsan
dfr
omSe
tBfo
roth
erac
tions
are
used
inon
eca
lcul
atio
n.Th
isca
se,a
sge
otec
hnica
lact
ions
are
notp
rese
nt,c
anbe
refe
rred
toSe
tB,i
.e.to
appr
oach
2.
EX
AM
PL
E2.
4.U
LS
com
bina
tions
ofac
tions
ona
rein
forc
edco
ncre
tere
tain
ing
wal
l[E
C2�
clau
se2.
4]
Fig.
2.8.
Acti
onso
na
retain
ingw
alli
nrei
nfor
cedcon
crete
EQ
U-(
static
equi
libriu
mof
rigid
body
:veri
ficat
ionof
globa
lsta
bilit
yto
heav
eand
slidi
ng)(
SetA
)O
nly
that
part
ofth
eem
bank
men
tbe
yond
the
foun
datio
nfo
otin
gis
cons
ider
edfo
rth
eve
rific
atio
nof
glob
alst
abili
tyto
heav
ean
dsli
ding
(Fig
.2.9
).
1.1·
S k,te
rr+
0.9·
(Gk,
wal
l+G
k,te
rr)+
1.5·
S k,so
vr
Fig.
2.9.
Acti
onsf
orE
QU
ULS
verif
icatio
nof
aret
ainin
gwal
lin
reinf
orced
conc
rete
EC2�
wor
ked
exam
ples
2-7
Tabl
eof
Cont
ent
STR
/GE
O-(
groun
dpr
essur
eand
verif
icatio
nof
resist
ance
ofwa
llan
dfoo
ting)
App
roac
h1
Des
ign
valu
esfro
mSe
tC
and
from
Set
Bar
eap
plie
din
sepa
rate
calcu
latio
nsto
the
geot
echn
icala
ctio
nsan
dto
allo
ther
actio
nsfr
omth
est
ruct
ure
oron
the
stru
ctur
e.
SetC
1.0·
S k,te
rr+
1.0·
Gk,
wal
l+1.
0·G
k,te
rr+
1.3·
S k,so
vr
SetB
1.35
·Sk,
terr
+1.
0·G
k,w
all+
1.0·
Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
1.35
·Sk,
terr
+1.
35·G
k,w
all+
1.35
·Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
1.35
·Sk,
terr
+1.
0·G
k,w
all+
1.35
·Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
1.35
·Sk,
terr
+1.
35·G
k,w
all+
1.0·
Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
Not
e:Fo
rall
the
abov
e-lis
ted
com
bina
tions
,tw
opo
ssib
ilitie
sm
ustb
eco
nsid
ered
:eith
erth
atth
esu
rcha
rge
conc
erns
only
the
part
ofem
bank
men
tbey
ond
the
foun
datio
nfo
otin
g(F
ig.2
.10a
),or
that
itac
tson
the
who
lesu
rfac
eof
the
emba
nkm
ent(
Fig.
2.10
b).
Fig.
2.10
.Pos
sible
load
cases
ofsu
rchar
geon
thee
mban
kmen
t.
Forb
revi
ty,o
nly
case
sin
relat
ion
with
case
b),i
.e.w
ithsu
rcha
rge
actin
gon
the
who
lesu
rfac
eof
emba
nkm
ent,
are
give
nbe
low
.
The
follo
win
gfig
ures
show
load
sin
relat
ion
toth
eco
mbi
natio
nsob
tain
edw
ithSe
tBpa
rtial
safe
tyfa
ctor
s.
EC2�
wor
ked
exam
ples
2-8
Tabl
eof
Cont
entFi
g.2.
11.A
ction
sfor
GE
O/S
TRU
LSve
rifica
tion
ofa
retai
ning
wall
inrei
nfor
cedco
ncret
e.
EC2�
wor
ked
exam
ples
2-9
Tabl
eof
Cont
ent
App
roac
h2
SetB
isus
ed.
App
roac
h3
Fact
ors
from
SetC
forg
eote
chni
cala
ctio
nsan
dfr
omSe
tBfo
roth
erac
tions
are
used
inon
eca
lcul
atio
n.1.
0·S k
,terr
+1.
0·G
k,w
all+
1.0·
Gk,
terr
+1.
3·Q
k,so
vr+
1.3·
S k,so
vr
1.0·
S k,te
rr+
1.35
·Gk,
wal
l+1.
35·G
k,te
rr+
1.3·
Qk,
sovr
+1.
3·S k
,sovr
1.0·
S k,te
rr+
1.0·
Gk,
wal
l+1.
35·G
k,te
rr+
1.3·
Qk,
sovr
+1.
3·S k
,sovr
1.0·
S k,te
rr+
1.35
·Gk,
wal
l+1.
0·G
k,te
rr+
1.3·
Qk,
sovr
+1.
3·S k
,sovr
Anu
mer
icex
ampl
eis
give
nbe
low
.
EX
AM
PL
E2.
5.C
oncr
ete
reta
inin
gw
all:
glob
alst
abili
tyan
dgr
ound
resi
stan
ceve
rific
atio
ns[E
C2�
clau
se2.
4]
The
assu
mpt
ion
isin
itiall
ym
ade
that
the
surc
harg
eac
tson
lyon
the
part
ofem
bank
men
tbe
yond
the
foun
datio
nfo
otin
g.
Fig.
2.12
.Wall
dime
nsion
sand
actio
nson
thew
all(su
rchar
geou
tside
thef
ound
ation
footin
g).
wei
ghtd
ensit
y:=
18kN
/m3
angl
eof
shea
ring
resis
tanc
e:=
30°
fact
orof
horiz
.act
ive
earth
pres
sure
:K
a=
0.33
wal
l-gro
und
inte
rfac
efr
ictio
nan
gle:
=0°
self-
wei
ghto
fwal
l:P k
,wal
l=
0.30
2.50
25=
18.7
5kN
/mse
lf-w
eigh
toff
ootin
g:P k
,foot
=0.
502.
5025
=31
.25
kN/m
Gk,
wall
=P k
,wal
l+
P k,fo
ot=
18.7
5+
31.2
5=
50kN
/mse
lfw
eigh
tofg
roun
dab
ove
foot
ing:
Gk,
grou
nd=
182.
501.
70=
76.5
kN/m
surc
harg
eon
emba
nkm
ent:
Qk,
surc
h=
10kN
/m2
grou
ndho
rizon
talf
orce
:S k
,gro
und
=26
.73
kN/m
surc
harg
eho
rizon
talf
orce
:S k
,surc
h=
9.9
kN/m
EC2�
wor
ked
exam
ples
2-10
Tabl
eof
Cont
ent
Ver
ifica
tion
tofa
ilure
bysl
idin
g
Slid
efo
rce
Gro
und
horiz
onta
lfor
ce(
G=
1,1)
:S g
roun
d=
1.1
26.7
3=29
.40
kN/m
Surc
harg
eho
rizon
tal(
Q=
1.5)
:Ssu
r=
1.5
9.90
=14
.85
kN/m
Slid
ingf
orce:
F slid
e=
29.4
0+
14.8
5=
44.2
5kN
/m
Resis
tant
forc
e(in
the
assu
mpt
ion
ofgr
ound
-floo
ring
frict
ion
fact
or=
0.57
)w
alls
elf-w
eigh
t(G=
0.9)
:F s
tab,
wall
=0.
9(0
.57
18.7
5)=
9.62
kN/m
foot
ing
self-
wei
ght(
G=
0.9)
:F s
tab,
foot
=0.
9(0
.57
31.2
5)=
16.0
3kN
m/m
grou
ndse
lf-w
eigh
t(G=
0.9)
:F s
tab,
grou
nd=
0.9
(0.5
776
.5)=
39.2
4kN
/mres
istan
tfor
ce:F s
tab=
9.62
+16
.03
+39
.24
=64
.89
kN/m
The
safe
tyfa
ctor
fors
lidin
gis:
FS=
F sta
b/
F rib
=64
.89
/44
.25
=1.
466
Ver
ifica
tion
toO
vert
urni
ng
over
turn
ing
mom
ent
mom
entf
rom
grou
ndla
tera
lfor
ce(
G=
1.1)
:M
S,gr
ound
=1.
1(2
6.73
3.00
/3)=
29.4
0kN
m/m
mom
entf
rom
surc
harg
ela
tera
lfor
ce(
Q=
1.5)
:MS,
surc
h=
1.5
(9.9
01.
50)=
22.2
8kN
m/m
overt
urni
ngm
omen
t:M
rib=
29.4
0+
22.2
8=
51.6
8kN
m/m
stab
ilizin
gm
omen
tm
omen
twall
self-
weig
ht(
G=
0.9)
:M
stab
,wall
=0.
9(1
8.75
0.65
)=10
.97
kNm
/mm
omen
tfoo
ting
self-
weig
ht(
G=
0.9)
:M
stab
,foot
=0.
9(3
1.25
1.25
)=35
.16
kNm
/mm
omen
tgro
und
self-
weig
ht(
G=
0.9)
:M
stab
,gro
und
=0.
9(7
6.5
1.65
)=11
3.60
kNm
/msta
biliz
ingm
omen
t:M
stab=
10.9
7+
35.1
6+
113.
60=
159.
73kN
m/m
safe
tyfa
ctor
togl
obal
stab
ility
FS=
Mst
ab/M
rib=
159.
73/5
1.68
=3.
09
Con
tact
pres
sure
ongr
ound
App
roac
h2,
i.e.S
etB
ifpa
rtial
facto
rs,is
used
.
Byta
king
1.0
and
1.35
asth
epa
rtial
fact
ors
fort
hese
lf-w
eight
ofth
ew
allan
dof
the
grou
ndab
ove
the
foun
datio
nfo
otin
gre
spec
tivel
y,w
eob
tain
four
diff
eren
tco
mbi
natio
nsas
seen
abov
e:
first
com
bina
tion
1.35
·Sk,
terr
+1.
0·G
k,w
all+
1.0·
Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
seco
ndco
mbi
natio
n
1.35
·Sk,
terr
+1.
35·G
k,w
all+
1.35
·Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
third
com
bina
tion
1.35
·Sk,
terr
+1.
0·G
k,w
all+
1.35
·Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
EC2�
wor
ked
exam
ples
2-11
Tabl
eof
Cont
ent
four
thco
mbi
natio
n
1.35
·Sk,
terr
+1.
35·G
k,w
all+
1.0·
Gk,
terr
+1.
5·Q
k,so
vr+
1.5·
S k,so
vr
the
cont
act
pres
sure
ongr
ound
isca
lcul
ated
,for
the
first
ofth
efo
urth
abov
e-m
entio
ned
com
bina
tions
,asf
ollo
ws:
mom
entv
s.cen
treof
mass
ofth
efoo
ting
mom
entf
rom
grou
ndla
tera
lfor
ce(
G=
1.35
):M
S,te
rr=
1.35
(26.
733.
00/3
)=36
.08
kNm
/mm
omen
tfro
msu
rcha
rge
late
ralf
orce
(Q=
1.5)
:M
S,so
vr=
1.5
(9.9
01.
50)=
22.2
8kN
m/m
mom
entf
rom
wall
self-
weig
ht(
G=
1.0)
:M
wall
=1.
0(1
8.75
0.60
)=11
.25
kNm
/mm
omen
tfro
mfo
otin
gse
lf-w
eigh
t(G=
1.0)
:M
foot
=0
kNm
/mm
omen
tfro
mgr
ound
self-
wei
ght(
G=
1.0)
:M
grou
nd=
-1.0
(76.
50.
40)=
-30.
6kN
m/m
Tota
lmom
entM
tot=
36.0
8+
22.2
8+
11.2
5�
30.6
=39
.01
kNm
/m
Vert
icall
oad
Wall
self-
wei
ght(
G=
1.0)
:P w
all=
1.0
(18.
75)=
18.7
5kN
m/m
Foot
ing
self-
wei
ght(
G=
1.0)
:P f
oot=
1.0
(31.
25)=
31.2
5kN
m/m
Gro
und
self-
weig
ht(
G=
1.0)
:P g
roun
d=
1.0
(76.
5)=
76.5
kNm
/mTo
tall
oad
P tot
=18
.75
+31
.25
+76
.5=
126.
5kN
/mE
ccen
trici
tye
=M
tot/
P tot
=39
.01
/12
6.5
=0.
31m
B/6
=2.
50/6
=41
.67
cmM
axpr
essu
reon
grou
nd=
P tot
/2.
50+
Mto
t6
/2.
502
=88
.05
kN/m
2=
0.08
8M
Pa
The
resu
ltsgi
ven
atTa
ble
2.2
are
obta
ined
byre
peat
ing
the
calc
ulat
ion
fort
heth
ree
rem
aini
ngco
mbi
natio
nsof
parti
alfa
ctor
s.Th
em
axim
alpr
essu
reon
grou
ndis
achi
eved
with
the
seco
ndco
mbi
natio
n,i.e
.for
the
one
inw
hich
the
wall
self-
weigh
tand
thes
elf-w
eight
ofth
egro
und
abov
ethe
foot
ing
areb
oth
mul
tiplie
dby
1.35.
For
the
verif
icatio
nof
the
cont
act
pres
sure
,the
poss
ibilit
yth
atth
esu
rcha
rge
acts
onth
ewh
ole
emba
nkm
ent
surfa
cem
ust
beals
oco
nsid
ered
.(F
ig.
2.13
);th
eva
lues
give
nat
Tabl
e2.
3ar
eob
tain
edby
repe
atin
gth
eca
lcula
tion
fort
hiss
ituat
ion.
Fig.
2.13
.Dim
ensio
nsof
ther
etain
ingw
allo
fthe
num
erice
xam
plewi
thsu
rchar
geon
thew
hole
emba
nkm
ent.
EC2�
wor
ked
exam
ples
2-12
Tabl
eof
Cont
entT
able
2.2.
Max
press
uref
orfou
rdiff
erent
combi
natio
nsof
parti
alfac
torso
fperm
anen
tloa
ds(su
rchar
geou
tside
thef
ound
ation
footin
g).Co
mbin
ation
first
seco
ndth
irdfo
urth
MS,
groun
d(k
Nm/
m)
36.0
8(
Q=
1.35
)36
.08
(Q=
1.35
)36
.08
(Q
=1.
35)
36.0
8(
Q=
1.35
)M
S,su
rch(k
Nm/
m)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
Mwa
ll(k
Nm/
m)
11.2
5(
G=
1.0)
15.1
9(
G=
1.35
)11
.25
(G
=1.
0)15
.19
(G=
1.35
)M
groun
d(k
Nm/
m)
-30.
60(
G=
1.0)
-41.
31(
G=
1.35
)-4
1.31
(G
=1.
35)
-30.
60(
G=
1.0)
Mtot
(kN
m/m
)39
.01
32.2
428
.30
42.9
5
P wall
(kN
/m)
18.7
5(
G=
1.0)
25.3
1(
G=
1.35
)18
.75
(G
=1.
0)25
.31
(G
=1.
35)
P foo
t(k
N/m
)31
.25
(G
=1.
0)42
.19
(G=
1.35
)31
.25
(G
=1.
0)42
.19
(G
=1.
35)
P gro
und
(kN
/m)
76.5
0(
G=
1.0)
103.
28(
G=
1.35
)10
3.28
(G
=1.
35)
76.5
0(
G=
1.0)
P tot
(kN
/m)
126.
5017
0.78
153.
2814
4
eccen
tricit
y(m)
0.31
0.19
0.18
0.30
press
ureo
ngr
ound
(kN
/m2 )
88.0
599
.26
88.4
898
.83
Tab
le2.
3.M
axpr
essur
eon
groun
dfor
fourd
iffere
ntcom
binat
ionso
fpar
tialf
actor
sofp
erman
entl
oads
(surch
arge
onth
ewho
lefou
ndat
ionfoo
ting).
Com
binat
ionfir
stse
cond
third
four
thM
S,gro
und
(kN
m/m
)36
.08
(Q=
1.35
)36
.08
(Q
=1.
35)
36.0
8(
Q=
1.35
)36
.08
(Q=
1.35
)M
S,su
rch(k
Nm/
m)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
22.2
8(
Q=
1.5)
Mwa
ll(k
Nm/
m)
11.2
5(
G=
1.0)
15.1
9(
G=
1.35
)11
.25
(G=
1.0)
15.1
9(
G=
1.35
)M
groun
d(k
Nm/
m)
-30.
60(
G=
1.0)
-41.
31(
G=
1.35
)-4
1.31
(G=
1.35
)-3
0.60
(G=
1.0)
Msu
rch
(kN
m/m
)-1
0.20
(Q
=1.
5)-1
0.20
(Q=
1.5)
-10.
20(
Q=
1.5)
-10.
20(
Q=
1.5)
Mtot
(kN
m/m
)28
.81
22.0
418
.10
32.7
5
P wall
(kN
/m)
18.7
5(
G=
1.0)
25.3
1(
G=
1.35
)18
.75
(G=
1.0)
25.3
1(
G=
1.35
)P f
oot
(kN
/m)
31.2
5(
G=
1.0)
42.1
9(
G=
1.35
)31
.25
(G=
1.0)
42.1
9(
G=
1.35
)P t
err(k
N/m
)76
.50
(G
=1.
0)10
3.28
(G
=1.
35)
103.
28(
G=
1.35
)76
.50
(G=
1.0)
Psur
ch(k
N/m
)25
.50
(Q
=1.
5)25
.50
(Q
=1.
5)25
.50
(Q
=1.
5)25
.50
(Q=
1.5)
P tot
(kN
/m)
152.
019
6.28
178.
7816
9.50
eccen
tricit
y(m)
0.19
0.11
0.10
0.19
press
ureo
ngr
ound
(kN
/m2 )
88.4
699
.67
88.8
999
.24
The
two
addi
tiona
llin
es,n
otpr
esen
tin
Tabl
e1.
18an
dhe
rehi
ghlig
hted
inbo
ld,c
orre
spon
dto
the
mom
enta
ndto
the
verti
cal
load
resu
lting
from
the
surc
harg
eabo
veth
efo
otin
g.
The
max
pres
sure
ongr
ound
isac
hiev
edon
ceag
ainfo
rthe
seco
ndco
mbi
natio
nan
dits
valu
eis
here
high
erth
anth
eon
eca
lculat
edin
the
prev
ious
sche
me.
EC2�
wor
ked
exam
ples
4-1
Tabl
eof
Cont
ent
SEC
TIO
N4.
WO
RK
ED
EX
AM
PLE
S–
DU
RA
BIL
ITY
EX
AM
PL
E4.
1[E
C2
clau
se4.
4]
Des
ign
the
conc
rete
cove
rofa
rein
forc
edco
ncre
tebe
amw
ithex
posu
recl
assX
C1.
The
conc
rete
inus
eha
sres
ista
nce
clas
sC
25/3
0.B
otto
mlo
ngitu
dina
lbar
sare
520
;the
stirr
upsa
re8
at10
0m
m.
The
max
aggr
egat
esi
zeis
:dg
=20
mm
(<32
mm
).Th
ede
sign
wor
king
life
ofth
est
ruct
ure
is50
year
s.N
orm
alqu
ality
cont
roli
sput
inpl
ace.
Ref
erto
figur
e4.
1.
Fig.
4.1
From
tabl
eE.
1N-E
C2
we
see
that
,in
orde
rto
obta
inan
adeq
uate
conc
rete
dura
bilit
y,th
ere
fere
nce
(min
.)co
ncre
test
reng
thcl
ass
for
expo
sure
clas
sX
C1
isC
20/2
5;th
ere
sist
ance
clas
sado
pted
(C25
/30)
issu
itabl
eas
itis
high
erth
anth
ere
fere
nce
stre
ngth
clas
s.
The
stru
ctur
alcl
assi
sS4.
Firs
t,th
eco
ncre
teco
verf
orth
est
irrup
sis
calc
ulat
ed.
With
:c m
in,b
=8
mm
We
obta
infr
omta
ble
4.4N
-EC
2:c m
in,d
ur=
15m
m
Mor
eove
r:c d
ur,
=0
;c d
ur,st
=0
;c d
ur,a
dd=
0.
From
rela
tion
(3.2
):c m
in=
max
(cm
in,b
;cm
in,d
ur+
c dur
,-
c dur
,st-
c dur
,add
;10
mm
)=m
ax(8
;15
+0�
0�
0;10
mm
)=15
mm
EC2�
Wor
ked
exam
ples
4-2
Tabl
eof
Cont
ent
Mor
eove
r:de
vc
=10
mm
.
We
obta
infr
omre
latio
n(3
.1):
nom
min
dev
cc
c=
15+
10=
25m
m.
Ifw
eno
wca
lcul
ate
now
the
conc
rete
cove
rfor
long
itudi
nalr
einf
orce
men
tbar
s,
we
have
:m
in,b
c=
20m
m.
We
obta
infr
omta
ble
4.4N
-EC
2:m
in,d
urc
=15
mm
.
Mor
eove
r:du
r,c
=0
;
dur,s
tc
=0
;
dur,a
ddc
=0
.
From
rela
tion
(3.2
):m
inc
=m
ax(2
0;15
+0�
0�
0;10
mm
)=20
mm
.
Mor
eove
r:de
vc
=10
mm
.
We
obta
infr
omre
latio
n(3
.1):
nom
c=
20+
10=
30m
m.
The
conc
rete
cove
rfo
rth
est
irrup
sis�d
omin
ant�
.In
this
case
,th
eco
ncre
teco
ver
for
long
itudi
nalb
arsi
sinc
reas
edto
:25
+8
=33
mm
.
EC2�
wor
ked
exam
ples
4-3
Tabl
eof
Cont
ent
EX
AM
PL
E4.
2[E
C2
clau
se4.
4]
Des
ign
the
conc
rete
cove
rfo
ra
rein
forc
edco
ncre
tebe
ampl
aced
outs
ide
are
side
ntia
lbu
ildin
gsi
tuat
edcl
ose
toth
eco
ast.
The
expo
sure
clas
sisX
S1.
We
orig
inal
lyas
sum
eco
ncre
tew
ithst
reng
thcl
assC
25/3
0.Th
elo
ngitu
dina
lrei
nfor
cem
entb
ars
are
520
;the
stirr
upsa
re8
at10
0m
m.
The
max
imal
aggr
egat
esi
zeis
:dg
=20
mm
(<32
mm
).Th
ede
sign
wor
king
life
ofth
est
ruct
ure
is50
year
s.A
norm
alqu
ality
cont
roli
spu
tin
plac
e.R
efer
tofig
ure
3.2.
From
tabl
eE.
1N-E
C2
we
find
that
,in
orde
rto
obta
inan
adeq
uate
conc
rete
dura
bilit
y,th
ere
fere
nce
(min
.)co
ncre
test
reng
thcl
ass
for
expo
sure
clas
sX
S1is
C30
/37;
the
conc
rete
stre
ngth
clas
sm
ustt
here
fore
bein
crea
sed
from
the
orig
inal
lyas
sum
edC
25/3
0to
C30
/37,
even
ifth
eac
tions
onco
ncre
tew
ere
com
patib
lew
ithst
reng
thcl
assC
25/3
0.
Fig.
4.2
Inac
cord
ance
with
wha
thas
been
stat
edin
exam
ple
3.1,
we
desi
gnth
em
inim
umco
ncre
teco
verw
ithre
fere
nce
tobo
thth
est
irrup
sand
the
long
itudi
nalb
ars.
The
stru
ctur
alcl
assi
sS4
We
obta
in(
min
,dur
c=
35m
m;
dev
c=
10m
m):
-fo
rthe
stirr
ups:
nom
c=
45m
m;
-fo
rthe
long
itudi
nalb
ars:
nom
c=
45m
m.
The
conc
rete
cove
rfo
rth
est
irrup
sis�d
omin
ant�
.In
this
case
,th
eco
ncre
teco
ver
for
long
itudi
nalb
arsi
sinc
reas
edto
:45
+8
=53
mm
.
EC2�
Wor
ked
exam
ples
4-4
Tabl
eof
Cont
ent
EX
AM
PL
E4.
3[E
C2
clau
se4.
4]
Calc
ulat
eth
eco
ncre
teco
vero
faTT
prec
aste
lem
ent,
mad
eof
pres
tress
edre
info
rced
conc
rete
,pl
aced
outsi
dean
indu
stria
lbui
ldin
gsit
uate
dcl
oset
oth
ecoa
st.
Thee
xpos
ure
clas
sisX
S1.
We
use
conc
rete
with
stre
ngth
clas
sC
45/5
5.A
tthe
low
ersi
deof
the
two
ribbi
ngs
ofth
eTT
elem
entw
eha
ve:
long
itudi
nal
12re
info
rcem
entb
ars;
8st
irrup
sat
100
mm
;st
rand
s0,
5�.
The
max
imal
aggr
egat
esi
zeis
:dg
=16
mm
.Th
ede
sign
wor
king
life
ofth
est
ruct
ure
is50
year
s.A
nac
cura
tequ
ality
cont
rolo
fcon
cret
epr
oduc
tion
ispu
tin
plac
e.R
efer
tofig
ure
3.3.
We
find
out
from
tabl
eE.
1N-
EC2
that
for
expo
sure
clas
sX
S1,t
hem
inim
umco
ncre
test
reng
thcl
assi
sC30
/37;
stre
ngth
clas
sC
45/5
5is
ther
efor
ead
equa
te.
The
orig
inal
stru
ctur
alcl
assi
sS4
.In
acco
rdan
cew
ithta
ble
4.3N
:th
est
ruct
ural
clas
sis
redu
ced
by1
asth
eco
ncre
teus
ed(C
45/5
5)is
ofst
reng
thcl
ass
high
erth
anC
40/5
0;th
est
ruct
ural
clas
sis
redu
ced
by1
assp
ecia
lqua
lity
cont
rolo
fthe
conc
rete
prod
uctio
nis
ensu
red
We
then
refe
rto
stru
ctur
alcl
ass
S2.
Cal
cula
ting
first
the
conc
rete
cove
rfor
stirr
ups.
We
have
:m
in,b
c=
8m
m.
We
obta
infr
omta
ble
4.4N
-EC
2:m
in,d
urc
=25
mm
.
Mor
eove
r:du
r,c
=0
;
dur,s
tc
=0
;
dur,a
ddc
=0
.
From
rela
tion
(3.2
):m
inm
in,b
min
,dur
dur,
dur,
stdu
r,ad
dc
max
(c;
cc
cc
;10
mm
)=
=m
ax(8
;25
+0�
0�
0;10
mm
)=25
mm
.
EC2�
wor
ked
exam
ples
4-5
Tabl
eof
Cont
ent
Con
side
ring
that
the
TTel
emen
tis
cast
unde
rpr
oced
ures
subj
ecte
dto
ahi
ghly
effic
ient
qual
ityco
ntro
l,in
whi
chth
eco
ncre
teco
verl
engt
his
also
asse
ssed
,the
valu
eof
c dev
can
beta
ken
as5
mm
.
We
obta
infr
omre
latio
n(3
.1):
nom
min
dev
cc
c=
25+
5=
30m
m.
Cal
cula
ting
now
the
conc
rete
cove
rfor
long
itudi
nalb
ars.
We
have
:m
in,b
c=
12m
m.
We
obta
infr
omta
ble
4.4N
-EC
2:m
in,d
urc
=25
mm
.
Mor
eove
r:du
r,c
=0
;
dur,s
tc
=0
;
dur,a
ddc
=0
.
From
rela
tion
(3.2
):m
inm
in,b
min
,dur
dur,
dur,
stdu
r,ad
dc
max
(c;
cc
cc
;10
mm
)=
=m
ax(1
2;25
+0�
0�
0;10
mm
)=25
mm
.
We
obta
infr
omre
latio
n(3
.1):
nom
min
dev
cc
c=
25+
5=
30m
m.
Not
eth
atfo
rthe
ordi
nary
rein
forc
emen
tbar
s,th
eco
ncre
teco
verf
orst
irrup
sis�d
omin
ant�
.In
this
case
,the
conc
rete
cove
rfor
long
itudi
nalb
arsi
sin
crea
sed
to:3
0+
8=
38m
m.
Fig.
4.3
Cal
cula
ting
now
the
conc
rete
cove
rfor
stra
nds.
EC2�
Wor
ked
exam
ples
4-6
Tabl
eof
Cont
ent
We
have
:m
in,b
c=
1,5
·12,
5=
18,8
mm
.
We
obta
infr
omta
ble
4.5N
-EC
2:m
in,d
urc
=35
mm
.
Mor
eove
r:du
r,c
=0
;
dur,s
tc
=0
;
dur,a
ddc
=0
.
From
rela
tion
(3.2
):m
inc
=m
ax(1
8,8;
35+
0�
0�
0;10
mm
)=35
mm
.
Mor
eove
r:de
vc
=5
mm
.
From
rela
tion
(3.1
):no
mc
=35
+5
=40
mm
.
The
first
stra
nd�s
axis
ispl
aced
at50
mm
from
the
low
eren
dof
the
ribbi
ngof
the
TTel
emen
t.Th
eco
ncre
teco
verf
orth
elo
wer
stra
nds
ofth
eTT
elem
ent(
one
fore
ach
ribbi
ng)
isth
eref
ore
equa
lto
43m
m.
EC2�
wor
ked
exam
ples
6-1
6-1
SEC
TIO
N6.
WO
RK
ED
EX
AM
PLE
S–
ULT
IMAT
EL
IMIT
STAT
ES
GE
NE
RA
LN
OT
E:Eu
roco
de2
perm
itsto
use
ava
rious
stee
lyi
eldi
nggr
ades
rang
ing
from
400
MPa
to60
0M
Pa.I
npa
rticu
lart
heex
ampl
esar
ede
velo
ped
usin
gS4
50st
eelw
ithdu
ctili
tygr
ade
C,w
hich
isus
edin
sout
hern
Euro
pean
dge
nera
llyin
seis
mic
area
s.So
me
exam
ple
isde
velo
ped
usin
gS5
00to
o.
EX
AM
PL
E6.
1(C
oncr
ete
C30
/37)
[EC
2cl
ause
6.1]
Geo
met
rical
data
:b=
500
mm
;h=
1000
mm
;d'=
50m
m;d
=95
0m
m.
Stee
land
conc
rete
resi
stan
ce,
1an
d2
fact
orsa
ndx 1
,x2
valu
esar
esh
own
inta
ble
6.1.
Bas
is:
1m
eans
the
ratio
betw
een
the
area
ofth
epa
rabo
la�
rect
angl
edi
agra
mat
certa
inde
form
atio
nc
and
the
area
ofre
ctan
gle
atth
esa
me
defo
rmat
ion.
2is
the�p
ositi
onfa
ctor�,
the
ratio
betw
een
the
dist
ance
ofth
ere
sulta
ntof
para
bola
�re
ctan
gle
diag
ram
atce
rtain
defo
rmat
ion
cfr
omc
and
the
defo
rmat
ion
cits
elf.
Fig
.6.1
Geo
met
rica
ldat
aan
dPo
ssib
lestr
ain
distr
ibut
ions
atth
eul
timat
elim
itsta
tes
Tabl
e6.
1M
ater
iald
ata,
1an
d2
fact
orsa
ndne
utra
laxi
sdep
th.
Exam
ple
f yk(M
Pa)
f yd(M
Pa)
f ck(M
Pa)
f cd(M
Pa)
12
x 1(m
m)
x 2(m
m)
6.1
450
391
3017
0.80
0.40
113,
560
8,0
6.2
450
391
9051
0.56
0.35
203.
054
1.5
Firs
tth
eN
Rd
valu
esco
rres
pond
ing
toth
e4
conf
igur
atio
nsof
the
plan
ese
ctio
nar
eca
lcul
ated
.
NR
d1=
0.8·
500·
113.
5·17
·10-3
=77
2kN
NR
d2=
0.8·
500·
608.
0·17
·10-3
=41
34kN
.
The
max
imum
mom
entr
esis
tanc
eM
Rd,
max
=28
21.2
kNm
goes
alon
gsid
eit.
NR
d3=
0.8·
500·
950·
17·1
0-3+
5000
·391
·10-3
=64
60+
1955
=84
15kN
NR
d4=
0.8·
500·
1000
·17·
10-3
+50
00·3
91·1
0-3=
8500
+39
10=
1241
0kN
EC2�
wor
ked
exam
ples
6-2
Tabl
eof
Cont
ent
MR
d3m
usta
lso
bekn
own.
This
resu
lts:M
Rd3
=64
60·(5
00�
0,4·
950)
·10-3
=16
55kN
m
Subs
eque
ntly
,for
ach
osen
valu
eof
NEd
inea
chin
terv
albe
twee
ntw
ofo
llow
ing
valu
esof
NR
dw
ritte
nab
ove
and
one
smal
lert
han
NR
d1,t
hene
utra
laxi
sx,M
Rd,
and
the
ecce
ntric
ity
e=
Rd Ed
M Nar
eca
lcul
ated
.The
irva
lues
are
show
nin
Tabl
e6.
2.
Tabl
e6.2
.Exa
mpl
e1:v
alue
sofa
xial
forc
e,de
pth
ofne
utra
laxis
,mom
entr
esist
ance
,ecc
entri
city.
NEd
(kN
)X
(m)
MR
d(k
Nm
)e
(m)
600
0,10
520
313.
3820
000,
294
2524
1.26
5000
0,66
626
060.
5210
000
virtu
alne
utra
laxi
s10
000.
10
As
anex
ampl
eth
eca
lcul
atio
nre
late
dto
NEd
=50
00kN
issh
own.
The
equa
tion
ofeq
uilib
rium
tosh
iftin
gfo
rdet
erm
inat
ion
ofx
isw
ritte
n:
250
0000
050
0039
150
000.
0035
2000
0050
000.
0035
2000
0050
0095
0x
x0
0.80
500
170.
8050
017
Dev
elop
ing,
itre
sults
:
x2+
66.9
1x�
4889
70=
0
whi
chis
satis
fied
forx
=66
6m
m
The
stre
ssin
the
low
erre
info
rcem
enti
s:2
s95
00.
0035
2000
001
297N
/m
m66
6
The
mom
entr
esis
tanc
eis
:
MR
d=
5000
·391
·(500
-50)
+50
00·2
97·(5
00-5
0)+
0.80
·666
·500
·17·
(500
�0.
4066
6)=
2606
·106
Nm
m=
2606
kNm
and
the
ecce
ntric
ity26
06e
0,52
m50
00
EC2�
wor
ked
exam
ples
6-3
Tabl
eof
Cont
ent
EX
AM
PL
E6.
2(C
oncr
ete
C90
/105
)[E
C2
clau
se6.
1]
Forg
eom
etric
alan
dm
echa
nica
ldat
are
fert
oex
ampl
e6.
1.
Val
ueso
fNR
dco
rres
pond
ing
toth
e4
conf
igur
atio
nsof
the
plan
ese
ctio
nan
dof
MR
d3:
NR
d1=
2899
kN
NR
d2=
7732
kN.
MR
d,m
ax=
6948
.7kN
mis
asso
ciat
edto
it.
NR
d3=
1356
6+
3910
=17
476
kN
NR
d4=
1428
0+
7820
=22
100
kN
MR
d3=
1356
6(0
.5�
0.35
·0.6
19)+
3910
·(0.5
0-0.
05)=
4031
kNm
App
lyin
gth
eex
plai
ned
proc
edur
ex,
MR
dan
dth
eec
cent
ricity
ew
ere
calc
ulat
edfo
rth
ech
osen
valu
esof
NEd
.
The
resu
ltsar
esh
own
inTa
ble
6.3
Tabl
e6.
3Va
lues
ofax
iall
oad,
dept
hof
neut
rala
xis,
mom
entr
esis
tanc
e,ec
cent
ricity
NEd
(kN
)x (m
)M
Rd(k
Nm
)e (m
)15
000,
142
4194
2.80
5000
0,35
054
031.
0810
000
0,61
955
140.
5519
000
virtu
alne
utra
laxi
s27
020.
14
EC2�
wor
ked
exam
ples
6-4
Tabl
eof
Cont
ent
EX
AM
PL
E6.
3C
alcu
latio
nof
VR
d,c
fora
pres
tres
sed
beam
[EC
2cl
ause
6.2]
Rec
tang
ular
sect
ion
b w=
100
mm
,h=
200
mm
,d=
175
mm
.No
long
itudi
nalo
rtra
nsve
rse
rein
forc
emen
tbar
sare
pres
ent.
Cla
ssC
40co
ncre
te.
Ave
rage
pres
tress
ing
cp=
5,0
MPa
.
Des
ign
tens
ilere
sist
ance
inac
cord
ance
with
:
f ctd
=ct
f ctk,
0,05
/C
=1·
2,5/
1,5=
1,66
MPa
Cra
cked
sect
ions
subj
ecte
dto
bend
ing
mom
ent.
VR
d,c=
(m
in+
k 1cp
)bwd
whe
rem
in=
0,62
6an
dk 1
=0,
15.I
tres
ults
:
VR
d,c=
(0.6
26+
0.15
5.0)
100
175
=24
.08
kN
Non
-cra
cked
sect
ions
subj
ecte
dto
bend
ing
mom
ent.
With
I=
1it
resu
lts
I=3
64
200
100
66.6
610
mm
12
S=
33
100
100
5050
010
mm
VR
d,c=
62
3
100
66.6
610
1.66
1,66
5.0
44.3
3kN
500
10
EC2�
wor
ked
exam
ples
6-5
Tabl
eof
Cont
ent
EX
AM
PL
E6.
4D
eter
min
atio
nof
shea
rre
sist
ance
give
nth
ese
ctio
nge
omet
ryan
dm
echa
nics
[EC
2cl
ause
6.2]
Rec
tang
ular
orT-
shap
edbe
am,w
ith
b w=
150
mm
,
h=
600
mm
,
d=
550
mm
,
z=
500
mm
;
verti
cals
tirru
psdi
amet
er12
mm
,2le
gs(A
sw=
226
mm
2 ),s=
150
mm
,fyd
=39
1M
Pa.
The
exam
ple
isde
velo
ped
fort
hree
clas
seso
fcon
cret
e.
a)f ck
=30
MPa
;fcd
=17
MPa
;=
0.61
6
swyw
d2
wcd
Af
sinb
sf
obta
ined
from
VR
d,s=
VR
d,m
ax
itre
sults
:2
226
391
sin0.
375
150
150
0.61
617
henc
eco
t=
1,29
Then
3sw
Rd,s
ywd
A22
6V
zf
cot
500
391
1.29
1038
0kN
s15
0
b)Fo
rthe
sam
ese
ctio
nan
dre
info
rcem
ent,
with
f ck=
60M
Pa,f
cd=
34M
Pa;
=0.
532,
proc
eedi
ngas
abov
eit
resu
lts:
222
639
1sin
0.21
7115
015
00.
532
34he
nce
cot
=1,
90
3sw
Rd,s
ywd
A22
6V
zf
cot
500
391
1.90
1056
0kN
s15
0
c)Fo
rth
esa
me
sect
ion
and
rein
forc
emen
t,w
ithf ck
=90
MPa
,fcd
=51
MPa
;=
0.51
2,pr
ocee
ding
asab
ove
itre
sults
:
222
639
1sin
0.15
0415
015
00.
512
51he
nce
cot
=2.
38
3sw
Rd,s
ywd
A22
6V
zf
cot
500
391
2.38
1070
1kN
s15
0
EC2�
wor
ked
exam
ples
6-6
Tabl
eof
Cont
ent
Det
erm
inat
ion
ofre
info
rcem
ent(
vert
ical
stir
rups
)giv
enth
ebe
aman
dsh
ear
actio
nV
Ed
Rec
tang
ular
beam
b w=
200
mm
,h=
800
mm
,d
=75
0m
m,z
=67
5m
m;v
ertic
alst
irrup
sf y
wd
=39
1M
Pa.T
hree
case
sare
show
n,w
ithva
ryin
gva
lues
ofV
Edan
dof
f ck.
VEd
=60
0kN
;fck
=30
MPa
;fcd
=17
MPa
;=
0.61
6
Then
oE
d
cwcd
w
2V1
12
6000
00ar
csin
arcs
in29
.02
(f
)bz
2(1
0.61
617
)20
067
5
henc
eco
t=
1.80
Itre
sults
:2
swE
d
ywd
AV
6000
001.
263
mm
/m
ms
zf
cot
675
391
1.80
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/170
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tne
cess
ary
for
bend
ing
mus
tbe
incr
ease
dby
F td
=0.
5V
Edco
t=
0.5·
6000
00·1
.80
=54
0kN
VEd
=90
0kN
;fck
=60
MPa
;fcd
=34
MPa
;=
0.53
2
oE
d
cwcd
w
2V1
12
9000
00ar
csin
arcs
in23
.74
2(
f)b
z2
(10.
532
34)
200
675
henc
eco
t=
2.27
Then
with
itre
sults
2sw
Ed
ywd
AV
9000
001.
50m
m/
mm
sz
fco
t67
539
12.
27
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/150
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tne
cess
ary
for
bend
ing
mus
tbe
incr
ease
dby
F td
=0.
5V
Edco
t=
0.5·
9000
00·2
.27=
1021
kN
VEd
=12
00kN
;fck
=90
MPa
;fcd
=51
MPa
;=
0.51
2
oE
d
cwcd
w
2V1
12
1200
000
arcs
inar
csin
21.4
52
(f
)bz
20.
512
5120
067
5
As
issm
alle
rtha
n21
.8o
,cot
=2.
50
Hen
ce2
swE
d
ywd
AV
1200
000
1.82
mm
/m
ms
zf
cot
675
391
2.50
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/120
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tne
cess
ary
for
bend
ing
mus
tbe
incr
ease
dby
F td
=0.
5V
Edco
t=
0.5·
1200
000·
2.50
=15
00kN
EC2�
wor
ked
exam
ples
6-7
Tabl
eof
Cont
ent
EX
AM
PL
E6.
4b�
the
sam
eab
ove,
with
stee
lS50
0Cf yd
=43
5M
Pa.
[EC
2cl
ause
6.2]
The
exam
ple
isde
velo
ped
fort
hree
clas
seso
fcon
cret
e.
a)f ck
=30
MPa
;fcd
=17
MPa
;=
0.61
6
swyw
d2
wcd
Af
sinb
sf
obta
ined
forV
Rd,
s=V
Rd,
max
itre
sults
:2
226
435
sin0.
417
150
150
0.61
617
henc
eco
t=
1.18
Then
3sw
Rd,s
ywd
A22
6V
zf
cot
500
435
1.18
1038
7kN
s15
0
b)Fo
rthe
sam
ese
ctio
nan
dre
info
rcem
ent,
with
f ck=
60M
Pa,f
cd=
34M
Pa;
=0.
532,
proc
eedi
ngas
abov
eit
resu
lts:
222
643
5sin
0.24
215
015
00.
532
34he
nce
cot
=1.
77
3sw
Rd,s
ywd
A22
6V
zf
cot
500
435
1.77
1058
0kN
s15
0
c)Fo
rth
esa
me
sect
ion
and
rein
forc
emen
t,w
ithf ck
=90
MPa
,fcd
=51
MPa
;=
0.51
2,pr
ocee
ding
asab
ove
itre
sults
:
222
643
5sin
0.16
715
015
00.
512
51he
nce
cot
=2.
23
3sw
Rd,s
ywd
A22
6V
zf
cot
500
435
2.23
1073
1kN
s15
0
Det
erm
inat
ion
ofre
info
rcem
ent(
vert
ical
stir
rups
)giv
enth
ebe
aman
dsh
ear
actio
nV
Ed
Rec
tang
ular
beam
b w=
200
mm
,h=
800
mm
,d
=75
0m
m,z
=67
5m
m;v
ertic
alst
irrup
sf y
wd
=39
1M
Pa.T
hree
case
sare
show
n,w
ithva
ryin
gva
lues
ofV
Edan
dof
f ck.
VEd
=60
0kN
;fck
=30
MPa
;fcd
=17
MPa
;=
0.61
6th
en
oE
d
cwcd
w
2V1
12
6000
00ar
csin
arcs
in29
.02
(f
)bz
2(1
0.61
617
)20
067
5he
nce
cot
=1.
80
Itre
sults
:2
swE
d
ywd
AV
6000
001.
135
mm
/m
ms
zf
cot
675
435
1.80
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/190
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tnec
essa
ryfo
rbe
ndin
gm
ust
bein
crea
sed
byF t
d=
0.5
VEd
cot
=0.
5·60
0000
·1.8
0=
540
kN
EC2�
wor
ked
exam
ples
6-8
Tabl
eof
Cont
ent
VEd
=90
0kN
;fck
=60
MPa
;fcd
=34
MPa
;=
0.53
2
oE
d
cwcd
w
2V1
12
9000
00ar
csin
arcs
in23
.74
2(
f)b
z2
(10.
532
34)
200
675
henc
eco
t=
2.27
Then
with
itre
sults
2sw
Ed
ywd
AV
9000
001.
35m
m/
mm
sz
fco
t67
543
52.
27
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/160
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tne
cess
ary
for
bend
ing
mus
tbe
incr
ease
dby
F td
=0.
5V
Edco
t=
0.5·
9000
00·2
.27
=10
21kN
VEd
=12
00kN
;fck
=90
MPa
;fcd
=51
MPa
;=
0.51
2
oE
d
cwcd
w
2V1
12
1200
000
arcs
inar
csin
21.4
52
(f
)bz
20.
512
5120
067
5
As
issm
alle
rtha
n21
.8o
,cot
=2.
50
Hen
ce2
swE
d
ywd
AV
1200
000
1.63
mm
/m
ms
zf
cot
675
435
2.50
whi
chis
satis
fied
with
2-le
gst
irrup
s12
/130
mm
.
The
tens
ilefo
rce
inth
ete
nsio
ned
long
itudi
nal
rein
forc
emen
tne
cess
ary
for
bend
ing
mus
tbe
incr
ease
dby
F td
=0.
5V
Edco
t=
0.5·
1200
000·
2.50
=15
00kN
EC2�
wor
ked
exam
ples
6-9
Tabl
eof
Cont
ent
EX
AM
PL
E6.
5[E
C2
clau
se6.
2]
Rec
tang
ular
orT-
shap
edbe
am,w
ith
b w=
150
mm
h=
800
mm
d=
750
mm
z=
675
mm
;
f ck=
30M
Pa;f
cd=
17M
Pa;
=0.
616
Rei
nfor
cem
ent:
incl
ined
stirr
ups
45o
(cot
=1,
0),d
iam
eter
10m
m,2
legs
(Asw
=15
7m
m2 ),
s=
150
mm
,f y
d=
391
MPa
.
Cal
cula
tion
ofsh
earr
esis
tanc
e
Duc
tility
isfir
stve
rifie
dby
sw,m
axyw
dcw
1cd
w
Af
f0.
5b
ssin
And
repl
acin
g15
739
11
0.61
617
0.5
150
150
0.70
7=
2.72
<7.
40
The
angl
eof
sim
ulta
neou
scon
cret
e�
rein
forc
emen
tste
elco
llaps
e
Itre
sults
cd
swyw
d
bsf
cot
1A
fsin
and,
repl
acin
g15
015
00.
616
17co
t1
2.10
157
391
0.70
7
c)C
alcu
latio
nof
VR
d
Itre
sults
:3
Rd,s
157
V67
539
1(2
.10
1.0)
0.70
710
605.
4kN
150
Incr
ease
ofte
nsile
forc
eth
elo
ngitu
dina
lbar
(VEd
=VR
d,s)
F td
=0.
5V
Rd,
s(c
otco
t=
0.5·
605.
4·(2
.10
-1.0
)=33
3kN
EC2�
wor
ked
exam
ples
6-10
Tabl
eof
Cont
ent
EX
AM
PL
E6.
6[E
C2
clau
se6.
3]
Rin
gre
ctan
gula
rse
ctio
n,Fi
g.6.
2,w
ithde
pth
1500
mm
,wid
th10
00m
m,d
=14
50m
m,
with
200
mm
wid
eve
rtica
lmem
bers
and
150
mm
wid
eho
rizon
talm
embe
rs.
Mat
eria
ls:
f ck=
30M
Pa
f yk
=50
0M
Pa
Res
ults
ofac
tions
:
VEd
=13
00kN
(for
cepa
ralle
lto
the
larg
ersi
de)
T Ed=
700
kNm
Des
ign
resi
stan
ces:
f cd=0
.85·
(30/
1.5)
=17
.0M
Pa
=0.
7[1-
30/2
50]=
0.61
6
f cd=
10.5
MPa
f yd
=50
0/1.
15=
435
MPa
Geo
met
ricel
emen
ts:
u k=
2(15
00-1
50)+
2(10
00-2
00)=
4300
mm
Ak
=13
50·8
00=
1080
000
mm
2
Fig
.6.2
Ring
sect
ion
subj
ecte
dto
tors
ion
and
shea
r
The
max
imum
equi
vale
ntsh
eari
nea
chof
the
verti
calm
embe
rsis
(zre
fers
toth
ele
ngth
ofth
eve
rtica
lmem
ber)
:
V* E
d=
VEd
/2+
(TEd
·z)/
2·A
k=
[130
010
3 /2+
(700
106
1350
)/(2
1.08
106 )]
10-3
=10
87kN
Ver
ifica
tion
ofco
mpr
esse
dco
ncre
tew
ithco
t=1
.Itr
esul
ts:
VR
d,m
ax=
tzf cd
sin
cos
=20
013
5010
.50.
707
0.70
7=
1417
kN
>V
* Ed
EC2�
wor
ked
exam
ples
6-11
Tabl
eof
Cont
ent
Det
erm
inat
ion
ofan
gle
:
*o
Ed
cd
2V1
12
1087
000
arcs
inar
csin
25.0
32
ftz
210
.520
013
50he
nce
cot
=2.
14
Rei
nfor
cem
ento
fver
tical
mem
bers
:
(Asw
/s)=
V* E
d/(z
f yd
cot
)=(1
087
103
)/(13
502.
14)=
0.86
5m
m2
/mm
whi
chca
nbe
carr
ied
outw
ith2-
legs
12m
mba
rs,p
itch
200
mm
;pitc
his
inac
cord
ance
with
[9.2
.3(3
)-EC
2].
Rei
nfor
cem
ento
fhor
izon
talm
embe
rs,s
ubje
cted
toto
rsio
non
ly:
(Asw
/s)=
T Ed
/(2A
kf y
dco
t)=
700
106
/(21.
0810
643
52.
14)=
0.34
8m
m2
/mm
whi
chca
nbe
carr
ied
outw
ith8
mm
wid
e,2
legs
stirr
ups,
pitc
h20
0m
m.
Long
itudi
nalr
einf
orce
men
tfor
tors
ion:
Asl
=T E
du k
cot
/(2A
kf y
d)=
700
106
4300
2.14
/(210
8000
043
5)=
6855
mm
2
tobe
dist
ribut
edon
the
sect
ion,
with
parti
cula
ratte
ntio
nto
the
corn
erba
rs.
Long
itudi
nalr
einf
orce
men
tfor
shea
r:
Asl
=V
Edco
t/(
2f y
d)=
1300
000
2.14
/(243
5)=
3198
mm
2
Tobe
plac
edat
the
low
eren
d.
EC2�
wor
ked
exam
ples
6-12
Tabl
eof
Cont
ent
EX
AM
PL
E6.
7Sh
ear�
Tor
sion
inte
ract
ion
diag
ram
s[E
C2
clau
se6.
3]
Fig
.6.3
Rect
angu
lars
ectio
nsu
bjec
ted
tosh
eara
ndto
rsio
n
Exam
ple:
full
rect
angu
lars
ectio
nb
=30
0m
m,h
=50
0m
m,z
=400
mm
(Fig
.6.3
)
Mat
eria
ls:
f ck=
30M
Pa
f cd=
0.85
·(30/
1.5)
=17
.0M
Pa
300.
71
0.61
625
0;
f cd=
10.5
MPa
f yk
=45
0M
Pa;f
yd=
391
MPa
cw=
1
Geo
met
ric
elem
ents
A=
1500
00m
m2
u=
1600
mm
t=A
/u=
94m
m
Ak
=(5
00�
94)
(300
-94)
=83
636
mm
2
Ass
umpt
ion:
=26
.56o
(cot
=2.
0)
Itre
sults
:VR
d,m
ax=
cwb w
zf cd
/(co
t+
tan
)=
10.5
300
400/
(2+0
.5)=
504
kN
and
fort
heta
ken
z=
400
mm
T Rd,
max
=2
10.5
8363
694
0.44
710.
8945
=66
kNm
resi
stan
thol
low
sect
ion
EC2�
wor
ked
exam
ples
6-13
Tabl
eof
Cont
ent
Fig
.6.4
.V-T
inte
ract
ion
diag
ram
forh
ighl
yst
ress
edse
ctio
n
The
diag
ram
issh
own
inFi
g.6.
4.Po
ints
belo
wth
est
raig
htlin
eth
atco
nnec
tsth
ere
sist
ance
valu
eson
the
two
axis
repr
esen
tsaf
ety
situ
atio
ns.F
orin
stan
ce,i
fVEd
=35
0kN
ista
ken,
itre
sults
that
the
max
imum
com
patib
leto
rsio
nm
omen
tis2
0kN
m.
On
the
figur
eot
herd
iagr
ams
inre
latio
nw
ithdi
ffer
ent
valu
esar
esh
own
asdo
tted
lines
.
Seco
ndca
se:l
ight
actio
nef
fect
sSa
me
sect
ion
and
mat
eria
lsas
inth
epr
evio
usca
se.T
hesa
fety
cond
ition
(abs
ence
ofcr
acki
ng)i
sexp
ress
edby
:
T Ed/T
Rd,
c+
VEd
/VR
d,c
1[(
6.31
)-EC
2]
whe
reT R
d,c
isth
eval
ueof
thet
orsio
ncr
acki
ngm
omen
t:
=f ct
d=
f ctk
/c=
2.0/
1.5
=1.
3M
Pa(f
ctkde
duct
edfr
omTa
ble
[3.1
-EC
2]).
Itre
sults
ther
efor
e:
T Rd,
c=
f ctd
t2A
k=
1.3
942
8363
6=
20.4
kNm
VR
d,c
=1/
3Rd
,cl
ckw
Ck
100
fb
d
Inth
isex
pres
sion
,=
0.01
;mor
eove
r,it
resu
lts:
CR
d,c=
0.18
/1.5
=0.
12
200
k1
1.63
500
1/3
1/3
1/3
lck
100
f10
00.
0130
30
EC2�
wor
ked
exam
ples
6-14
Tabl
eof
Cont
ent
Taki
ngd
=45
0m
mit
resu
lts:
VR
d,c=
0,12
1.63
(30)
1/3
300
450
=82
.0kN
The
diag
ram
issh
own
inFi
g.6.
5Th
ese
ctio
n,in
the
rang
eof
actio
nef
fect
sde
fined
byth
ein
tera
ctio
ndi
agra
m,s
houl
dha
vea
min
imal
rein
forc
emen
tin
acco
rdan
cew
ith[9
.2.2
(5)-
EC2]
and
[9.2
.2(6
)-EC
2].N
amel
y,th
em
inim
alqu
antit
yof
stirr
ups
mus
tbe
inac
cord
ance
with
[9.5
N-E
C2]
,whi
chpr
escr
ibes
fors
hear
:
(Asw
/sb w
) min
=(0
.08
f ck)/f
yk=
(0.0
830
)/450
=0.
010
with
snot
larg
erth
an0.
75d
=0,
.75
450
=33
7m
m.
Bec
ause
ofth
eto
rsio
n,st
irrup
sm
ustb
ecl
osed
and
thei
rpitc
hm
ustn
otbe
larg
erth
anu/
8,i.e
.200
mm
.For
inst
ance
,stir
rups
of6
mm
diam
eter
with
180
mm
pitc
hca
nbe
plac
ed.I
tre
sults
:Asw
/s.b
w=
228
/(180
300)
=0.
0010
Fig
.6.5
V-T
inte
ract
ion
diag
ram
forl
ight
lyst
ress
edse
ctio
n
EC2�
wor
ked
exam
ples
6-15
Tabl
eof
Cont
ent
EX
AM
PL
E6.
8.W
allb
eam
[EC
2cl
ause
6.5]
Geo
met
ry:5
400
x30
00m
mbe
am(d
epth
b=
250
mm
),40
0x
250
mm
colu
mns
,col
umns
rein
forc
emen
t620
We
stat
eth
atth
est
rutl
ocat
ion
C 2is
200
cmfr
omth
ebo
ttom
rein
forc
emen
t,so
that
the
inne
rdr
ive
arm
iseq
ualt
oth
eel
astic
solu
tion
inth
eca
seof
aw
allbe
amw
ithra
tio1/
h=2,
that
is0.
67h;
itsu
gges
tsto
use
the
rang
e(0
.60.
7)·l
asva
lues
fort
hele
vera
rm,l
ower
than
the
case
ofa
slend
erbe
amw
ithth
esa
me
span
.
Fig
.6.6
5400
x30
00m
mwa
llbe
am.
Mat
eria
ls:co
ncre
teC2
5/30
f ck=
25M
Pa,s
teel
B450
Cf yk
=45
0M
Pa
2ck
cd0.
85f
0.85
25f
14.1
7N
/mm
1.5
1.5
,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
node
scom
pres
sive
stre
ngth
:
com
pres
sed
node
s ck
21R
d,m
ax1
cd
f1-
2525
0=
kf
=1.
181-
14.1
7=
15N
/mm
0.85
250
node
sten
sione
d�
com
pres
sed
byan
chor
logs
ina
fixed
dire
ctio
n
EC2�
wor
ked
exam
ples
6-16
Tabl
eof
Cont
ent
ck
22R
d,m
ax2
cd
f1-
2525
0=
kf
=1-
14.1
7=
12.7
5N
/mm
0.85
250
node
sten
sione
d�
com
pres
sed
byan
chor
logs
indi
ffere
ntdi
rect
ions
ck
23R
d,m
ax3
cd
f1-
2525
0=
kf
=0.
881-
14.1
7=
11.2
2N
/mm
0.85
250
Act
ions
Dist
ribut
edlo
ad:1
50kN
/mup
pers
urfa
cean
d15
0kN
/mlo
wer
surf
ace
Col
umns
reac
tion
R=
(150
+15
0)5.
40/2
=81
0kN
Eva
luat
ionof
stress
esin
latti
ceba
rs
Equi
libriu
mno
de1
1ql
C40
5kN
2
Equi
libriu
mno
de3
3R
C96
6kN
sen
(whe
re20
00ar
ctg
56.9
813
00)
kN52
6co
sC
T3
1
Equi
libriu
mno
de2
C2
=C
3cos
=T 1
=52
6kN
Equi
libriu
mno
de4
kN40
52lq
T 2
Tens
ionro
ds
The
tens
ion
rod
T 1re
quire
sast
eela
rea
notl
ower
than
:
2s1
5260
00A
1344
mm
391.
,3w
eus
e6
18=
1524
mm
2 ,
the
rein
forc
emen
toft
helo
wer
tens
ion
rod
are
loca
ted
atth
ehe
ight
of0,
12h
=36
0m
m
The
tens
ion
rod
T 2re
quire
sast
eela
rea
notl
ower
than
:
2s1
4050
00A
1035
mm
391.
3W
eus
e4
20=
1257
mm
2
EC2�
wor
ked
exam
ples
6-17
Tabl
eof
Cont
ent
Nod
esve
rifica
tion
Nod
e3
The
node
geom
etry
isun
ambi
guou
slyde
fined
byth
eco
lum
nw
idth
,th
ew
allde
pth
(250
mm
),th
ehe
ight
ofth
esid
eon
whi
chth
elo
wer
bars
are
dist
ribut
edan
dby
the
stru
tC3
fall
(Fig
.6.7
)
Fig
.6.7
Nod
e3,
left
supp
ort.
The
node
3is
aco
mpr
esse
d-st
ress
edno
deby
asin
gle
dire
ctio
nre
info
rcem
enta
ncho
r,th
enit
ism
anda
tory
tove
rify
that
the
max
imal
conc
rete
com
pres
sion
isno
thig
hert
han
the
valu
e:
22R
d,m
ax12
.75
N/m
m
2c1
2Rd,
max
8100
008.
1N
/mm
400
250
Rem
ark
asth
eve
rifica
tion
ofth
eco
lum
nco
ntac
tpr
essur
eis
satis
fied
even
with
out
taki
ngin
toac
coun
tth
elon
gitud
inal
reinf
orcem
ent(
620
)pres
ent
inth
ecolu
mn.
2c2
2Rd,
max
9660
007.
27N
/mm
531.
625
0
EC2�
wor
ked
exam
ples
6-18
Tabl
eof
Cont
ent
EX
AM
PL
E6.
9.T
hick
shor
tcor
bel,
a c<
h c/2
[EC
2cl
ause
6.5]
Geo
met
ry:2
50x
400
mm
cant
ileve
r(w
idth
b=
400
mm
),15
0x
300
load
plat
e,be
amb
xh
=40
0x
400
mm
Fig
.6.8
250
x40
0m
mth
ick
cant
ileve
rbea
m.
Fig
.6.9
Can
tilev
erbe
amS&
Tm
odel
.
Mat
eria
ls:co
ncre
teC3
5/45
f ck=
35M
Pa,s
teel
B450
Cf yk
=45
0M
Pa
2ck
cd0.
85f
0.85
35f
==
=19
.83
N/m
m1.
51.
5,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
node
scom
pres
sive
stre
ngth
:
com
pres
sed
node
s ck
21R
d,m
ax1
cd
f1-
3525
0=
kf
=1.
181-
19.8
3=
20.1
2N
/mm
0.85
250
EC2�
wor
ked
exam
ples
6-19
Tabl
eof
Cont
ent
node
sten
sione
d�
com
pres
sed
byan
chor
logs
ina
fixed
dire
ctio
n
ck
22R
d,m
ax2
cd
f1-
3525
0=
kf
=1-
19.8
3=
17.0
5N
/mm
0.85
250
node
sten
sione
d�
com
pres
sed
byan
chor
logs
indi
ffere
ntdi
rect
ions
ck
23R
d,m
ax3
cd
f1-
3525
0=
kf
=0.
881-
19.8
3=
15N
/mm
0.85
250
Act
ions
F Ed
=70
0kN
Load
ecce
ntric
ityw
ithre
spec
tto
the
colu
mn
side:
e=
125
mm
(Fig
.6.8
)
The
beam
verti
cals
trutw
idth
isev
alua
ted
byse
tting
the
com
pres
sive
stre
sseq
ualt
o1R
d,m
ax:
Ed1
1Rd,
max
F70
0000
xm
mb
20.1
240
087
the
node
1is
loca
ted
x 1/2
44m
mfr
omth
eou
terc
olum
nsid
e(F
ig.6
.9)
We
stat
eth
atth
eup
per
rein
forc
emen
tis
loca
ted
40m
mfr
omth
eup
per
cant
ileve
rsid
e;th
edi
stan
cey 1
ofth
eno
de1
from
the
low
erbo
rder
isev
aluat
edse
tting
the
inte
rnal
driv
ear
mz
equa
lto
0.8
d(z
=0,
836
0=
288
mm
):
y 1=
0.2d
=0.
2·36
0=
72m
m
rota
tiona
lequ
ilibr
ium
:Ed
cF
aF
zc
7000
00(1
2544
)F
288
ct
7000
00(1
2544
)F
F41
0763
N41
1kN
288
node
1veri
ficat
ion:
22
c1R
d,m
ax1
F41
1000
7N
/mm
N/m
mb
2y
400
27
.14
20.1
22
Mai
nup
perr
einfor
cemen
tdesi
gn:
2t
sydF
4110
00A
1050
mm
f39
1.3
we
use
814
(As=
1232
mm
2 )
Seco
ndar
yupp
errei
nfor
cemen
tdesi
gn:
The
beam
prop
osed
inE
C2is
inde
term
inat
e,th
enit
isno
tpos
sible
toev
aluat
eth
est
ress
esfo
rea
chsin
gle
bar
byeq
uilib
rium
equa
tions
only
,but
we
need
tokn
owth
est
iffne
ssof
the
two
elem
enta
rybe
ams
show
nin
Fig.
6.10
inor
der
tom
ake
the
parti
tion
ofth
edi
agon
alst
ress
senF
cosF
FEd
cdi
agbe
twee
nth
em;
EC2�
wor
ked
exam
ples
6-20
Tabl
eof
Cont
ent
Fig
.6.1
0S&
Tm
odel
reso
lutio
nin
two
elem
enta
rybe
amsa
ndpa
rtiti
onof
the
diag
onal
stres
sFdi
ag.
base
don
the
trend
ofm
ainco
mpr
essiv
est
ress
esre
sulti
ngfr
omlin
eare
lastic
analy
sisat
finite
elem
ents
,som
ere
sear
cher
ofSt
uttg
arth
ave
dete
rmin
edth
etw
ora
tesi
nw
hich
F diag
isdi
vide
d,an
dth
eyha
vepr
ovid
edth
efo
llow
ing
expr
essio
nof
stre
ssin
the
seco
ndar
yre
info
rcem
ent
(MC9
0pa
r.6.
8.2.
2.1)
:
wd
cEd
c
z28
82
12
1a
125
44F
F41
121
1kN
3F
/F3
700
/411
22
wd
sw1
sydF
2110
00A
539
mm
kA
0.25
1232
308
mm
f39
1.3
we
use
5st
irrup
s10
,dou
ble
arm
ed(A
sw=
785
mm
2 )
node
2ve
rifica
tion,
below
thel
oad
plat
e:
The
node
2is
atie
d-co
mpr
esse
dno
de,
whe
reth
em
ainre
info
rcem
ent
isan
chor
ed;
the
com
pres
sive
stre
ssbe
low
the
load
plat
eis: 2
2Ed
2Rd,
max
F70
0000
15.5
6N
/mm
17.0
5N
/mm
150
300
4500
0
EC2�
wor
ked
exam
ples
6-21
Tabl
eof
Cont
ent
EX
AM
PL
E6.
10T
hick
cant
ileve
rbea
m,a
c>
h c/2
[EC
2cl
ause
6.5]
Geo
met
ry:3
25x
300
mm
cant
ileve
rbea
m(w
idth
b=
400
mm
),15
0x
220
mm
load
plat
e,40
0x
400
mm
colu
mn
Fig
.6.1
132
5x
300
mm
cant
ileve
r.F
ig.6
.12
Can
tilev
erS&
Tm
odel
.Th
em
odel
prop
osed
inE
C2(F
ig.6
.12)
isin
dete
rmin
ate,
then
asin
the
prev
ious
exam
ple
one
mor
ebo
unda
ryco
nditi
onis
need
edto
evalu
ate
the
stre
sses
valu
esin
the
rods
;Th
est
ress
F wd
inth
eve
rtica
lten
sion
rod
isev
alua
ted
assu
min
ga
linea
rre
latio
nbe
twee
nF w
d
and
the
ava
lue,
inth
era
nge
F wd
=0
whe
na
=z/
2an
dF w
d=
F Ed
whe
na
=2
z.Th
isas
sum
ptio
nco
rres
pond
sto
the
stat
emen
tth
atw
hen
az/
2(a
very
thick
cant
ileve
r),th
ere
sista
ntbe
amis
the
beam
1on
ly(F
ig.6
.13a
)and
whe
na
2z
the
beam
2on
ly(F
ig.6
.13b
).
a)b)
Fig
.6.1
3.El
emen
tary
beam
soft
heS&
Tm
odel
.
EC2�
wor
ked
exam
ples
6-22
Tabl
eof
Cont
ent
Ass
umed
this
stat
emen
t,th
eex
pres
sion
forF
wd
is:
F wd=
F w1
a+
F w2
whe
nth
etw
oco
nditi
ons
wd
zF
(a)
02
and
F wd
(a=
2z)
=F E
dar
eim
pose
d,so
me
trivi
al
algeb
rale
adst
o:
Edw
1F
2F
3z
and
Edw
2F
F3
;
inco
nclu
sion,
the
expr
essio
nfo
rFw
das
afu
nctio
nof
ais
the
follo
win
g:
EdEd
wd
EdF
F2
2a/z
1F
aF
3z
33
.
Mat
eria
ls:co
ncre
teC3
5/45
f ck=
35M
Pa,s
teel
B450
Cf yk
=45
0M
Pa
2ck
cd0.
85f
0,85
35f
19.8
3N
/mm
1.5
1.5
,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
Nod
esco
mpr
essio
nre
sista
nce
(sam
eva
lues
ofth
epr
evio
usex
ampl
e):
Com
pres
sed
node
s2
1Rd,
max
2N
/mm
0.12
tied-
com
pres
sed
node
swith
tens
ion
rods
inon
edi
rect
ion
22R
d,m
axN
/mm
17.0
5
tied-
com
pres
sed
node
swith
tens
ion
rods
indi
ffere
ntdi
rect
ions
23R
d,m
ax1
N/m
m5
Act
ions
:
F Ed
=50
0kN
Load
ecce
ntric
ityw
ithre
spec
tto
the
colu
mn
oute
rsid
e:e
=20
0m
m
The
colu
mn
verti
cals
trutw
idth
isev
aluat
edse
tting
the
com
pres
sive
stre
sseq
ualt
o1R
d,m
ax:
Ed1
1Rd,
max
F50
0000
x62
mm
b20
.12
400
node
1is
loca
ted
x 1/2
=31
mm
from
the
oute
rsid
eof
the
colu
mn;
the
uppe
rre
info
rcem
enti
sst
ated
tobe
40m
mfr
omth
eca
ntile
ver
oute
rsid
e;th
edi
stan
cey 1
ofth
eno
de1
from
the
low
erbo
rder
isca
lcul
ated
setti
ngth
ein
tern
aldr
ive
arm
zto
be0,
8d
(z=
0,8
260
=20
8m
m):
y 1=
0.2d
=0.
2·26
0=
52m
m
EC2�
wor
ked
exam
ples
6-23
Tabl
eof
Cont
ent
rota
tiona
lequ
ilibr
ium
:
1Ed
cx
Fa
Fz
2½
5000
00(2
00+
31)=
F c. 20
8
ct
5000
00(2
0031
)F
F55
5288
N55
6kN
208
node
1ve
rifica
tion
22
c1R
d,m
ax1
F55
6000
==
=13
.37
N/m
m=
20.1
2N
/mm
b2
y40
02
52
Mai
nup
perr
einfor
cemen
tdesi
gn:
2t
sydF
5560
00A
1421
mm
f39
1.3
we
use
816
(As=
1608
mm
2 )
Seco
ndar
yrei
nfor
cemen
tdesi
gn:
(the
expr
essio
nde
duce
dat
the
begi
nnin
gof
this
exam
ple
isus
ed)
wEd
a2
1z
FF
204
kN3
2w
wydF
2040
00A
521
mm
f39
1.3
EC2
sugg
ests
am
inim
umse
cond
ary
rein
forc
emen
tof:
2Ed
w2
ydF50
0000
Ak
0.5
639
mm
f39
1.3
we
use
3st
irrup
s12
(As=
678
mm
2 )
node
2ve
rifica
tion,
below
thel
oad
plat
e:
The
node
2is
aco
mpr
esse
d-st
ress
edno
de,i
nw
hich
the
main
rein
forc
emen
tis
anch
ored
;the
com
pres
sive
stre
ssbe
low
the
load
plat
eis: 2
2Ed
2Rd,
max
F50
0000
15.1
5N
/mm
17.0
5N
/mm
150
220
3300
0
EC2�
wor
ked
exam
ples
6-24
Tabl
eof
Cont
ent
EX
AM
PL
E6.
11G
erbe
rbea
m[E
C2
clau
se6.
5]
Two
diffe
rent
stru
t-tie
truss
esca
nbe
cons
ider
edfo
rthe
desig
nof
aG
erbe
rbea
m,e
vent
ually
ina
com
bine
dco
nfig
urat
ion
[EC2
(10.
9.4.
6)],
(Fig
.6.
14).
Even
ifth
eE
C2all
ows
the
poss
ibili
tyto
use
only
one
stru
tand
then
only
one
rein
forc
emen
tarr
ange
men
t,w
ere
mar
kas
the
sche
me
b)re
sults
tobe
poor
unde
rlo
ad,b
ecau
seof
the
com
plet
elac
kof
rein
forc
emen
tfo
rth
ebo
ttom
bord
erof
the
beam
.It
seem
sto
beop
portu
neto
com
bine
the
type
b)re
info
rcem
entw
ithth
ety
pea)
one,
and
the
latte
rwill
carr
yat
least
half
ofth
ebe
amre
actio
n.
On
the
othe
rhan
d,if
only
the
sche
me
a)is
used
,iti
snec
essa
ryto
cons
ider
alo
ngitu
dina
ltop
rein
forc
emen
tto
anch
orbo
thth
eve
rtica
lstir
rups
and
the
conf
inin
gre
info
rcem
ent
ofth
etil
ted
stru
tC1.
a)b)
Fig
.6.1
4Po
ssib
lestr
utan
dtie
mod
elsf
ora
Ger
berb
eam
.
Her
eafte
rwe
repo
rtth
epa
rtitio
nof
the
supp
ortr
eact
ion
betw
een
the
two
truss
es.
Mat
erial
s:
conc
rete
C25/
30f ck
=25
MPa
,
stee
lB4
50C
f yk=
450
MPa
Es=
2000
00M
Pa[(3
.2.7
(4)-E
C2]
2ck
cd0.
85f
0.85
25f
14.1
7N
/mm
1.5
1.5
,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
Act
ions
:
Dist
ribut
edlo
ad:2
50kN
/m
Beam
spam
:800
0m
m
R Sdu
=10
00kN
Bend
ing
mom
enti
nth
ebe
amm
id-s
pam
:MSd
u=
2000
kNm
Beam
sect
ion:
bx
h=
800
x14
00m
m
Botto
mlo
ngitu
dina
lrein
forc
emen
t(A
s):10
24=
4524
mm
2
EC2�
wor
ked
exam
ples
6-25
Tabl
eof
Cont
ent
Top
long
itudi
nalr
einfo
rcem
ent
(As�)
:820
=25
13m
m2
Trus
sa)R
=R S
du/2
=50
0kN
Defi
nitio
nof
thet
russ
rods
posit
ionTh
eco
mpr
esse
dlo
ngitu
dina
lbar
has
aw
idth
equa
lto
the
dept
hx
ofth
ese
ctio
nne
utra
laxi
san
dth
enit
isx/
2fr
omth
eto
pbo
rder
;the
dept
hof
the
neut
rala
xis
isev
aluat
edfr
omth
ese
ctio
ntra
nslat
ion
equi
libriu
m:
0.8
bx
f cd+
E s� s
A� s
=f y
dA
s
Fig
.6.1
5Tr
ussa
.
'dx
x0,
0035
' sw
here
d�=
50m
mis
the
dist
ance
ofth
eup
pers
urfa
cere
info
rcem
ent
cds
syd
sx
500.
8bxf
E0.
0035
A'
fA
xan
dth
en:
x=
99m
m
yd' s
s
f0.
0035
391.
399
500.
0017
30.
0019
699
E20
0000
then
the
com
pres
sed
stee
lstra
inre
sults
low
erth
anth
estr
ainin
the
elas
ticlim
it,as
stat
edin
the
calcu
latio
n;
the
com
pres
sive
stre
ssin
the
conc
rete
is
C=
0.8
bx
f cd=
0.8·
800·
99·1
4.17
(app
lied
at0.
4x
40m
mfr
omth
eup
pers
urfa
ce)
whi
leth
eto
pre
info
rcem
ents
tress
is:
C�=
E s� s
A� s
=20
0000
·0.0
0173
·251
3
EC2�
wor
ked
exam
ples
6-26
Tabl
eof
Cont
ent
(app
lied
at50
mm
from
the
uppe
rsur
face
)
the
com
pres
sion
net
forc
e(C
+C�
)re
sults
tobe
appl
iedat
45m
mfr
omth
ebe
amup
per
surfa
ce,t
hen
the
horiz
onta
lstru
thas
the
axis
at67
5�
50-4
5=
580
mm
from
the
tens
ion
rod
T2.
Cal
cula
tion
ofth
etru
ssro
dsstr
esses
Nod
e1
equi
libriu
m:
53,7
742
558
0ar
ctg
1R
C62
0kN
sin
kN36
6co
sC
T1
2
Nod
e2
equi
libriu
m:
580
arct
g38
,66
725
23
2T
cos4
5C
cos
C
23
Csi
nC
sin4
5
22
TC
260
kNsi
nco
s
32
sin
CC
230
kNsi
n45
Nod
e3
equi
libriu
m:
T 1=
C1
sin
+C
2si
n=
663
kNTe
nsion
rods
desig
n
the
tens
ion
rod
T 1ne
edsa
stee
lare
ano
tlow
erth
an:
2s1
6630
00A
1694
mm
391.
3
we
use
5st
irrup
s16
doub
lear
m(A
sl=
2000
mm
2 )
the
tens
ion
rod
T 2ne
edsa
stee
lare
ano
tlow
erth
an:
2s1
3660
00A
935
mm
391.
3
we
use
516
(Asl
=10
00m
m2 ).
Trus
sb)R
=R S
du/2
=50
0kN
Fig
.6.1
6C
alcu
latio
nsc
hem
efo
rthe
trus
sbba
rsstr
esse
s.C
alcu
latio
nof
thet
russ
rods
stress
es
EC2�
wor
ked
exam
ples
6-27
Tabl
eof
Cont
ent
node
1eq
uilib
rium
C�1
=50
0kN
node
2eq
uilib
rium
C� 2
=C� 1
=50
0kN
kN70
7C
'2
T'1
1
node
3eq
uilib
rium
C� 3
=T�
1=
707
kN
T�2
=(T� 1
+C� 3
)·cos
45°
=10
00kN
Tens
ionro
ds
fort
ensio
nro
dT�
2it
isne
cess
ary
toad
opta
stee
lare
ano
tlow
erth
an:
2s1
1000
000
A25
56m
m39
1.3
624
=27
12m
m2
are
adop
ted,
alo
wer
rein
forc
emen
tar
eaw
ould
besu
fficie
ntfo
rte
nsio
nro
dT�
1bu
tfo
rqu
estio
nof
bar
anch
orin
gth
esa
me
rein
forc
emen
tasi
nT�
2is
adop
ted.
EC2�
wor
ked
exam
ples
6-28
Tabl
eof
Cont
ent
EX
AM
PL
E6.
12Pi
leca
p[E
C2
clau
se6.
5]
Geo
met
ry:
4500
x45
00m
mpl
inth
(thick
ness
b=15
00m
m),
2000
x70
0m
mco
lum
ns,
diam
eter
800
mm
piles
Fig
.6.1
7Lo
gpl
inth
onpi
lings
.
Mat
eria
ls:co
ncre
teC2
5/30
f ck=
25M
Pa,s
teel
B450
Cf yk
=45
0M
Pa2
ckcd
0.85
f0.
8525
f14
.17
N/m
m1.
51.
5,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
Nod
esco
mpr
essio
nre
sista
nce
(sam
eva
lues
asin
the
exam
ple
6.8)
Com
pres
sed
node
s1R
d,m
ax=
15N
/mm
2
tied-
com
pres
sed
node
swith
tens
ion
rods
inon
edi
rect
ion
2Rd,
max
=12
.75
N/m
m2
tied-
com
pres
sed
node
swith
tens
ion
rods
indi
ffere
ntdi
rect
ions
3Rd,
max
=11
.22
N/m
m2
Pede
stal
pile
NSd
=20
00kN
MSd
=40
00kN
m
EC2�
wor
ked
exam
ples
6-29
Tabl
eof
Cont
ent
Tied
rein
forc
emen
tin
the
pile:
826
(As=
4248
mm
2 )
The
com
pres
sive
stre
ssF c
inth
eco
ncre
tean
dth
est
eelt
ensio
nF s
onth
epe
dest
alpi
lear
eev
aluat
edfr
omth
eU
LSve
rific
atio
nfo
rnor
mal
stre
sses
ofth
ese
ctio
nits
elf:
F s=
f ydA
s=
391.
3·42
48=
1662
242
N=
1662
kN
NSd
=0.
8b
xf cd�
F s·
·x·
x=
462
mm
The
com
pres
sive
stre
ssin
the
conc
rete
is:
C=
0.8
bx
f cd=
0.8·
700·
462·
14.1
7=
3666
062
N=
3666
kN
(app
lied
at0,
4x
185
mm
from
the
uppe
rsur
face
)
piles
stre
ss
pile
stre
sses
are
evalu
ated
cons
ider
ing
the
colu
mn
actio
nstra
nsfe
rin
two
step
s:
inth
efir
stst
ep,
the
trans
fero
fthe
forc
esF c
eF s
happ
ensi
nth
epl
ane
1(F
ig.6
.17)
tillt
oth
eor
thog
onal
plan
es2
and
3,th
enin
the
seco
ndst
epth
etra
nsfe
ris
insid
eth
epl
anes
2an
d3
tillt
oth
epi
les;
the
truss
-tie
beam
inFi
g.6.
18is
relat
ive
toth
etra
nsfe
rin
the
plan
e1:
com
pres
sion:
A�=
(MSd
/3.0
0+
NSd
/2)=
(400
0/3.
00+
2000
/2)=
2333
kN
tens
ion:
B�=
(MSd
/3.0
0-N
Sd/2
)=(4
000/
3.00
-200
0/2)
=33
3kN
fore
ach
com
pres
sed
pile
:A
=A�/
2=
1167
kN
fore
ach
tied
pile:
B=B�
/2=
167
kN
Inth
eev
aluat
ion
ofst
ress
eson
pile
s,th
epl
inth
own
weig
htis
cons
ider
edne
glig
ible
.
Fig
.6.1
8.S&
Tm
odel
inth
epl
ane
1.
EC2�
wor
ked
exam
ples
6-30
Tabl
eof
Cont
ent
11=
arct
g(1
300
/86
0)=
56.5
°12
=ar
ctg
(130
0/
600)
=65
.2°
T 10
=F s
=16
62kN
T 11
=A�c
ot11
=23
33co
t26.
5°=
1544
kNT 1
2=
B�co
t12
=33
3co
t65.
2°=
154
kN
Fig
.6.1
9Tr
usse
sin
plan
2an
din
plan
3.
13=
arct
g(1
300
/13
25)=
44.5
°T 1
3=
A=
1167
kNT 1
4=
Aco
t13
=11
67co
t44.
5°=
1188
kNT 1
5=
Bco
t13
=16
7co
t44.
5°=
170
kNT 1
6=
B=
167
kN
desig
nof
tens
ion
rods
Tabl
e6.
3
Tens
ion
rod
Forc
e(k
N)
Requ
ired
rein
forc
emen
t(m
m2 )
Bars
10(p
linth
tied
rein
forc
emen
t)16
6242
488
2611
1544
3946
924
1215
439
41
12/2
0(6
12)
1311
6729
82st
irrup
s10
2014
1188
3036
724
1517
043
41
12/2
0(5
12)
1616
742
7Pi
lere
info
rcem
ent
EC2�
wor
ked
exam
ples
6-31
Tabl
eof
Cont
ent
Fig
.6.2
0.Sc
hem
atic
plac
emen
tofr
einf
orce
men
ts.
Nod
esve
rific
atio
nCo
ncen
trate
dno
des
are
only
pres
enta
tthe
pede
stal
pile
and
onth
epi
les
top.
Inth
ese
latte
r,th
eco
mpr
essiv
est
ress
esar
eve
rysm
allas
aco
nseq
uenc
eof
the
pile
ssec
tion
large
area
.:
2c
22
A23
3300
04.
64N
mm
r40
0
EC2�
wor
ked
exam
ples
6-32
Tabl
eof
Cont
ent
EX
AM
PL
E6.
13V
aria
ble
heig
htbe
am[E
C2
clau
se6.
5]
Geo
met
ry:l
engt
h22
500
mm
,rec
tang
ular
sect
ion
300
x35
00m
man
d30
0x
2000
mm
Fig
.6.2
1Va
riab
lehe
ight
beam
Mat
eria
ls:co
ncre
teC3
0/37
f ck=
30M
Pa,s
teel
B450
Cf yk
=45
0M
Pa
2ck
cd0.
85f
0.85
3f
17N
/mm
1.5
1,5
0,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
Nod
esco
mpr
essiv
ere
sista
nce:
com
pres
sed
node
s(E
C2
eq.6
.60)
ck
21R
d,m
ax1
cd
f1-
3025
0=
kf
=1.
181-
17=
17.6
5N
/mm
0.85
250
tied-
com
pres
sed
node
swith
tens
ion
rods
inon
edi
rect
ion
ck
22R
d,m
ax2
cd
f1-
3025
0=
kf
=1-
17=
14.9
6N
/mm
0.85
250
tied-
com
pres
sed
node
swith
tens
ion
rods
indi
ffere
ntdi
rect
ions
ck
23R
d,m
ax3
cd
f1-
3025
0=
kf
=0.
881-
17=
13.1
6N
/mm
0.85
250
load
s
F=
1200
kN
(the
own
wei
ghto
fthe
beam
isne
glig
ible)
EC2�
wor
ked
exam
ples
6-33
Tabl
eof
Cont
ent
stru
t&tie
mod
elid
entif
icat
ion
Beam
parti
tioni
ngin
two
regi
onsB
and
D
The
regi
onst
andi
ngon
the
mid
dle
sect
ion
isa
cont
inui
tyre
gion
(B),
whi
leth
ere
main
ing
part
ofth
ebe
amis
com
pose
dof
Dty
pere
gion
s.
The
boun
dary
cond
ition
sfor
the
stre
ssin
the
regi
onB.
Fig
.6.2
2Id
entif
icat
ion
ofB
and
Dre
gion
s.
Stre
sses
eval
uatio
nfo
rthe
bars
ofth
eS&
Tm
odel
T max
=12
00kN
Mm
ax=
1200
3.00
=36
00kN
m=
3.6
109
Nm
m
Fig
.6.2
3Sh
eara
ndbe
ndin
gm
omen
tdia
gram
s.
Calcu
latio
nof
stre
sses
inth
ere
gion
B
The
stre
ss-b
lock
diag
ram
isus
edfo
rthe
conc
rete
com
pres
sive
stre
sses
dist
ribut
ion;
rota
tiona
lequ
ilibr
ium
:
f cd0.
8·x·
b·(d�
0.4
x)=
3.6·
109
17·0
.8·x
·300
·(190
0�
0.4
x)=
3.6·
109
7752
000·
x�
1632
·x2
=3.
610
9x
=52
2m
m
C=
f cd0.
8·x·
b=
17·0
.8·5
22·3
00=
2129
760
N=
2130
kN
EC2�
wor
ked
exam
ples
6-34
Tabl
eof
Cont
ent
Iden
tifica
tion
ofbo
unda
ryst
ress
esin
the
regi
onD
Fig
.6.2
4Re
actio
nsan
dbo
unda
ryst
ress
esin
the
regi
onD
.
stru
t&tie
mod
el
Fig.
6.25
show
sth
elo
adpa
ths
char
acte
rized
bySc
hlaic
hin
the
stru
t&tie
mod
elid
entif
icat
ion,
show
nin
Fig.
6.26
.
Fig
.6.2
5Lo
adpa
ths.
Fig
.6.2
6.St
ruta
ndtie
mod
el.
The
stru
tC2
tiltin
gis
3190
arct
g46
.76
3000
whi
leth
est
rutC
4til
ting
is
116
90ar
ctg
48.4
115
00.
EC2�
wor
ked
exam
ples
6-35
Tabl
eof
Cont
ent
The
follo
win
gta
ble
repo
rtsth
eva
lue
fort
hest
ress
esin
the
diffe
rent
beam
elem
ents
.Ta
ble
6.4
C 1Se
est
ress
esev
aluat
ion
inth
ere
gion
B21
30kN
T 1T 1
=C
121
30kN
C 2C
2=
F/si
n(N
ode
Ave
rtica
lequ
ilibr
ium
)16
47kN
T 3T 3
=C
2co
s(N
ode
Aho
rizon
tale
quili
briu
m.)
1128
kNT 2
T 2=
T 3,b
ecau
seC 5
is45
°til
ted
(nod
eC
equi
l.)11
28kN
C 3C
3=
C2
cos
=T 3
(Nod
eB
horiz
onta
lequ
il.)
1128
kNF l
oop
F loo
p=
C 1�
C3
1002
kNC 4
C4
=F l
oop/c
os15
09kN
C 52
TC
25
(Nod
eC
verti
cale
quil.
)15
95kN
Stee
lten
sion
rods
desig
n
EC2
poin
t9.7
sugg
ests
that
the
min
imum
rein
forc
emen
tfor
the
wall
beam
sis
the
0,10
%of
the
conc
rete
area
,and
notl
ess
than
150
mm
2 /m,a
ndit
has
tobe
disp
osed
onbo
thsid
esof
the
struc
tura
lmem
bera
ndin
both
dire
ctio
ns.B
ars
12/
20�
(=56
5m
m2 /m
>0,
10%
300
1000
=30
0m
m2 /m
edi
150
mm
2 /m)a
reus
ed.
The
follo
win
gta
ble
repo
rtsth
eev
aluat
ion
for
the
rein
forc
emen
tar
eare
quire
dfo
rth
eth
ree
tens
ion
bars
T 1,T
2an
dT 3
.Ta
ble
6.5
T 1A
s=
2.13
·106 /3
91.3
=54
43m
m2
1820
=56
55m
m2
T 2A
s=
1.12
8·10
6 /391
.3=
2883
mm
2
on1,
50m
leng
thst
irrup
s12
/10�
2le
gs=
2260
mm
2 /m(2
260
1,50
=33
90m
m2 )
T 3A
s=
1.12
8·10
6 /391
.3=
2883
mm
2
ontw
olay
ers
1020
=31
42m
m2
Ver
ifica
tion
ofno
des
Nod
eA(le
ftsu
ppor
t)
Fig
.6.2
7N
ode
A.
tied-
com
pres
sed
node
swith
tens
ion
rods
inon
edi
rect
ion
[(6.6
1)-E
C2]
2Rd,
max
=14
.96
N/m
m2
EC2�
wor
ked
exam
ples
6-36
Tabl
eof
Cont
ent
Load
ing
plat
ear
ea:
62
c1
2Rd,
max
F1.
210
A80
214
mm
14.9
6
a30
0x
300
mm
plat
e(A
=90
000
mm
2 )isu
sed
the
rein
forc
emen
tfor
the
tens
ion
rod
T 3is
load
edon
two
layer
s(Fi
g.6.
27):
u=
150
mm
a 1=
300
mm
a 2=
300
sin46
.76°
+15
0co
s46.
76°=
219
+10
3=
322
mm
62
2c2
1.64
710
17.0
5N
/mm
14.9
6N
/mm
300
322
uha
sto
behi
gher
(itis
man
dato
rya
rein
forc
emen
ton
mor
eth
antw
olay
ers,
oran
incr
ease
ofth
epl
ate
leng
th);
this
last
choi
ceis
adop
ted,
and
the
leng
this
incr
ease
dfr
om30
0to
400
mm
:
a 2=
400
sin46
.76°
+15
0co
s46.
76°=
291
+10
3=
394
mm
62
2c2
1.64
710
13.9
3N
/mm
14.9
6N
/mm
300
394
Nod
eB
Com
pres
sed
node
s
1Rd,
max
=17
.65
N/m
m2
Fig
.6.2
8N
ode
B.
a 3=
522
mm
(coi
ncid
entw
ithth
ede
pth
ofth
ene
utra
laxi
sin
the
regi
onB)
3
62
23
cC
1.12
810
7.2
N/m
m17
.65
N/m
m30
052
230
052
2lo
adpl
ate
dim
ensio
ns:
61.
210
a*22
7m
m30
017
.65
a30
0x
300
mm
plat
eis
used
Stru
tver
ifica
tion
EC2�
wor
ked
exam
ples
6-37
Tabl
eof
Cont
ent
The
com
pres
sive
rang
efo
rea
chst
rut
(onl
yex
cept
ion,
the
stru
tC 1
,w
hich
stre
ssha
sbe
enve
rifie
dbe
fore
inth
efo
rces
evalu
atio
nfo
rthe
regi
onB)
can
spre
adbe
twee
nth
etw
oen
ds,i
nth
isw
ayth
em
axim
alst
ress
esar
ein
the
node
s.
The
trans
vers
alst
ress
fort
hesp
litof
the
mos
tstre
ssed
stru
t(C 2
)is:
T s0.
25·C
2=
0.25
1647
=41
2kN
;
and
then
,for
the
rein
forc
emen
treq
uire
dto
carr
yth
isst
ress
:
2s
4120
00A
1053
mm
391.
3,
then
the
min
imum
rein
forc
emen
t(1
12/
20�
onbo
thsid
esan
din
both
dire
ctio
ns,t
hati
sa s
=11
30m
m2 /m
)ise
noug
hto
carr
yth
etra
nsve
rsal
stre
sses
.
EC2�
wor
ked
exam
ples
6-38
Tabl
eof
Cont
ent
EX
AM
PL
E6.
14.
3500
kNco
ncen
trat
edlo
ad[E
C2
clau
se6.
5]
3500
kNlo
adon
a80
0x50
0re
ctan
gula
rcol
umn
bya
300x
250
mm
cush
ion
Mat
eria
ls: conc
rete
C30
/37
f ck=
30M
Pa,
stee
lB4
50C
f yk=
450
MPa
Es=
2000
00M
Pa
2ck
cd0.
85f
0.85
30f
17N
/mm
1.5
1.5
,
yk2
yd
f45
0f
391.
3N
/mm
1.15
1.15
,
load
ing
area
Ac0
=30
0·25
0=
7500
0m
m2
dim
ensio
nsof
the
load
dist
ribut
ion
area
d 23
d 1=
3·30
0=
900
mm
b 23
b 1=
3·25
0=
750
mm
max
imal
load
dist
ribut
ion
area
Ac1
=90
0·75
0=
6750
00m
m2
load
dist
ribut
ion
heig
ht
mm
500
250
750
bb
h1
2
mm
600
300
900
dd
h1
2
h=
600
mm
Ulti
mat
eco
mpr
essiv
est
ress
Rdu
c0cd
c1c0
6cd
c0
FA
fA
/A75
000
1767
5000
/750
0038
25kN
3.0
fA
3.0
1775
000
3.82
510
N
Itis
wor
thto
obse
rve
that
the
F Rdu
uppe
rlim
itco
rres
pond
sto
the
the
max
imal
valu
eA
c1=
3A
c0fo
rthe
load
dist
ribut
ion
area
,jus
tasi
nth
isex
ampl
e;th
e35
00kN
load
resu
ltsto
belo
wer
than
F Rdu
.
Rein
forc
emen
tdes
ign
Poin
t[6
.7(4
)-EC
2]re
com
men
dsth
eus
eof
asu
itabl
ere
info
rcem
ent
capa
ble
tosu
stai
nth
etra
nsve
rsal
shrin
kage
stre
sses
and
poin
t[6
.7(1
)P-E
C2]
send
sth
ere
ader
topa
ragr
aph
[(6.5
)-E
C2]t
oan
alyse
this
topi
c.
Inth
isca
seth
ere
isa
parti
aldi
scon
tinui
ty,b
ecau
seth
est
rutw
idth
(500
mm
)is
low
erth
anth
edi
strib
utio
nhe
ight
(600
mm
),th
en:
a=
250
mm
EC2�
wor
ked
exam
ples
6-39
Tabl
eof
Cont
ent
b=
500
mm
Fb
a35
0050
025
0T
437.
5kN
4b
450
0th
est
eela
rea
requ
ired
toca
rry
Tis:
2s
ydT43
7500
A11
18m
mf
391.
3
usin
g10
mm
diam
eter
bars
,15
bars
are
requ
ired
fora
tota
lare
aof:
As=
1578
.5=
1178
mm
2 .
EC2�
wor
ked
exam
ples
6-40
Tabl
eof
Cont
ent
EX
AM
PL
E6.
15Sl
abs1,
2[E
C2
clau
se5.
10�
6.1�
6.2�
7.2�
7.3�
7.4]
As
two
dim
ensi
onal
mem
bera
pres
tress
edco
ncre
tesl
abis
anal
ysed
:the
actu
alst
ruct
ure
isde
scrib
edin
the
follo
win
gpo
int.
6.15
.1D
escr
iptio
nof
the
stru
ctur
e
The
desi
gnex
ampl
epr
opos
edin
this
sect
ion
isre
late
dto
ara
ilway
brid
gede
ckm
ade
upby
aco
ntin
uous
slab
onth
ree
span
sw
ithtw
oor
ders
ofpr
estre
ssin
gte
ndon
s(lo
ngitu
dina
land
trans
vers
epr
estre
ssin
g).T
hesl
abis
desi
gned
inca
tego
ryA
(see
Euro
code
2,Pa
rt2,
tabl
e4.
118)
forf
atig
uere
ason
s.Th
ede
ckre
sts
onab
utm
ents
and
circ
ular
pier
san
dha
saov
eral
lbr
eadt
hof
13.6
0m
with
two
side
-wal
ksof
1.40
mw
idth
,tw
oba
llast
reta
inin
gw
alls
and,
inth
em
iddl
e,tw
otra
cksp
acin
gof
5.0
m.T
hesl
abpr
esen
tsa
cons
tant
thic
knes
sof
1.50
mfo
race
ntra
lzon
e7.
0m
wid
th,w
hils
tis
tape
red
tow
ards
the
extre
mity
with
afin
alhe
ight
of0.
6m
.Fi
g.s
6.29
and
6.30
repr
esen
tth
epr
inci
pal
geom
etric
dim
ensi
onof
the
slab
brid
gean
dsu
ppor
ts�s
chem
e.
Fig
.6.2
9Pl
anvi
ewof
the
stru
ctur
ean
dsu
ppor
ts�s
chem
e
1Ex
ampl
eta
ken
from
exam
ple
7.2�s
labs�
bypr
of.
Man
cini
,FI
BBu
lletti
nn°
3,�S
truct
ural
Conc
rete
Text
book
onB
ehav
iour
,Des
ign
and
Perfo
rman
ceV
ol.3
:Dur
abili
ty-D
esig
nfo
rFire
Resis
tanc
e-M
embe
rD
esig
n-M
aint
enan
ce,A
sses
smen
tand
Rep
air-
Prac
tical
aspe
cts�
Man
ual-
text
book
(292
page
s,IS
BN
978-
2-88
394-
043-
7,D
ecem
ber1
999)
.
2Se
eto
oEN
1992
-2Eu
roco
de2,
brid
gede
sign.
EC2�
wor
ked
exam
ples
6-41
Tabl
eof
Cont
ent
Fig
.6.3
0G
eom
etric
dim
ensio
nsof
brid
gecr
osss
ectio
n
Mat
eria
lpro
pert
ies
The
follo
win
gm
ater
ials
prop
ertie
sha
vebe
enco
nsid
ered
:
Con
cret
eG
rade
35:
f ck=
35.0
MPa
;co
mpr
essi
vede
sign
stre
ngth
:f cd
=23
.3M
Pa;
com
pres
sive
resi
stan
cefo
runc
rack
edzo
nes:
f cd1
=17
.1M
Pa;
com
pres
sive
resi
stan
cefo
rcra
cked
zone
s:f cd
2=
12.0
MPa
;m
ean
valu
eof
tens
ilest
reng
th:
f ctm
=3.
23M
Pa;
mod
ulus
ofel
astic
ity:
E c=
29.7
·103
MPa
;sh
earm
odul
us:
G=
12.4
·103
MPa
Pois
son
ratio
:=
0.2
Pres
tress
ing
stee
l,(s
trand
s0.
6�):
f ptk
=18
00M
Pa;
0.1%
proo
fstre
ssf p0
.1k
=16
00M
Pato
tale
long
atio
nat
max
imum
load
:pu
>35�
mod
ulus
ofel
astic
ity:
E p=
195.
0·10
3M
Pa;
Rei
nfor
cing
stee
l,G
rade
500:
f yk=
500.
0M
Pa;
desi
gnst
reng
th:
f yd=
434.
8M
Pa;
mod
ulus
ofel
astic
ity:
E s=
200.
0·10
3M
Pa.
Con
cret
eco
ver
As
envi
ronm
enta
lcon
ditio
nan
Expo
sure
Cla
ss2
may
beco
nsid
ered
(Hum
iden
viro
nmen
tw
ithfr
ost:
exte
riorc
ompo
nent
sexp
osed
tofr
ost).
The
min
imum
conc
rete
cove
rfor
Cla
ss2
iseq
ualt
o25
mm
,whi
chsh
ould
bead
ded
toth
eto
lera
nce
valu
eof
10m
m;
asa
cons
eque
nce
the
nom
inal
valu
efo
rco
ncre
teco
ver
resu
lts:
c nom
=c m
in+
10=
25+
10=
35m
m
adop
ted
inth
eca
lcul
atio
ns.
EC2�
wor
ked
exam
ples
6-42
Tabl
eof
Cont
ent
6.15
.2St
ruct
ural
mod
el
Toev
alua
teth
ein
tern
alac
tions
onth
est
ruct
ure
alin
ear
FEM
anal
ysis
has
been
perf
orm
edad
optin
gsh
ell
elem
ents
tore
pres
ent
the
rein
forc
edsl
ab;
this
kind
ofel
emen
tta
kes
acco
unto
fall
the
slab
and
plat
eco
mpo
nent
sas
wel
las
the
out-o
f-pl
ane
shea
rfor
ces.
The
thic
knes
sof
shel
lele
men
tsha
sbe
enas
sum
edco
nsta
ntfo
rth
ein
ner
zone
ofth
esl
aban
dst
eppe
dto
fash
ion
the
tape
red
extre
mity
.In
Fig
6.31
and
6.32
the
FEM
mod
elis
sket
ched
and
the
diff
eren
tthi
ckof
the
elem
enti
srep
orte
dto
o.
Fig
.6.3
1Tr
ansv
erse
view
ofFE
Mm
odel
Fig
.6.3
2Pl
anof
FEM
mod
elan
dco
nsid
ered
elem
ents
The
adop
ted
shel
lele
men
tsar
eor
ient
edw
ithth
efo
llow
ing
guid
elin
es:
loca
laxi
s2
isor
ient
edas
glob
alax
isY
ofth
ede
ck;
loca
laxi
s3is
orie
nted
inth
eop
posi
tedi
rect
ion
ofgl
obal
axis
Xof
the
deck
;
loca
laxi
s1
isor
ient
edas
glob
alax
isZ
ofth
ede
ck.
Posi
tive
forc
esfo
rFEM
prog
ram
outp
utar
ere
porte
din
Fig.
6.33
:
EC2�
wor
ked
exam
ples
6-43
Tabl
eof
Cont
ent
Fig
.6.3
3Po
sitiv
eac
tions
forF
EMel
emen
ts
Rest
rain
ts
The
exte
rnal
rest
rain
tsha
vebe
enin
trodu
ced
inth
eFE
Mm
odel
cons
ider
ing
thei
rre
alge
omet
ricdi
men
sion
s;th
us,f
ewno
des
have
been
rest
rain
edby
mea
nsof
sprin
gel
emen
tsin
orde
rto
repr
esen
ton
lyan
indi
vidu
alre
strai
ntor
supp
ort.
Fig.
6.34
show
sa
sym
bolic
nota
tion
fort
heex
tern
alre
stra
ints
with
the
node
sinv
olve
d.
Fig
.6.3
4Ex
tern
alre
stra
ints
onth
eFE
Mm
odel
The
elas
ticco
nsta
ntof
the
sprin
gre
stra
inin
gel
emen
tsis
calc
ulat
edto
have
the
sam
est
iffne
ssof
the
subs
truct
ure
(abu
tmen
tsor
pier
s)on
whi
chth
esl
abis
rest
ed.
Fort
hex
and
ydi
rect
ions
,it
may
beas
sum
edth
atth
epi
er,o
rth
eab
utm
entf
ront
wal
l,be
have
slik
ea
EC2�
wor
ked
exam
ples
6-44
Tabl
eof
Cont
ent
sing
leco
lum
nfix
edat
the
base
and
free
athi
sto
p,so
that
the
rele
vant
Kx/
yst
iffne
ssis
valu
able
as:
KE
IH
x/y
33
whe
reE
isth
eY
oung
mod
ulus
,I
the
iner
tiaan
dH
the
heig
htof
the
colu
mn.
For
the
verti
cal
dire
ctio
n,th
ein
trins
icst
iffne
ssof
pot-b
earin
gis
assu
med
,co
nsid
erin
gth
esu
bstru
ctur
eve
rtica
lbeh
avio
uras
rigid
.
For
the
sake
ofsi
mpl
icity
the
calc
ulus
ofth
ere
leva
ntst
iffne
ssis
omitt
edan
dth
efin
alva
lues
ofth
esp
ring
cons
tant
sare
repo
rted
inta
ble
6.6
.
Loca
tion
Kx,
tot
Ky,
tot
Kz,
tot
106
kN/m
106
kN/m
106
kN/m
Abu
tmen
tA9.
5517
8.80
10.0
2Pi
erP1
4.74
11.6
1Pi
erP2
2.66
11.6
1A
butm
entB
2.78
10.0
2Ta
ble
6.6
Stiff
ness
forr
estr
aini
ngel
emen
ts
Itca
nbe
notic
edth
atth
epr
evio
usva
lues
are
refe
rred
toth
eov
eral
lst
iffne
ssof
the
rest
rain
t,th
usth
eel
astic
cons
tant
ofan
yin
divi
dual
sprin
gel
emen
tm
aybe
obta
ined
divi
ding
the
Kva
lues
ofta
ble
6.6
byth
enu
mbe
rof
elem
entr
epre
sent
ing
the
rest
rain
tor
the
supp
orts
.
Pres
tres
sing
forc
es
Two
orde
rsof
pres
tress
ing
tend
ons
are
arra
nged
(inlo
ngitu
dina
land
trans
vers
edi
rect
ions
)in
orde
rto
avoi
dan
yte
nsile
stre
ssin
conc
rete
atse
rvic
e(r
equi
red
byra
ilway
code
).Th
ein
itial
tens
ilest
ress
ofte
ndon
is:
po,m
ax=
0.85
f p0.
1k=
0.85
1600
=13
60M
Pa.
The
num
ber
ofte
ndon
sis
39fo
rth
elo
ngitu
dina
ldire
ctio
nan
d64
for
the
trans
vers
eon
e.Ea
chte
ndon
isbu
iltup
with
19st
rand
s0.
6�ha
ving
anar
eaof
1.39
cm2 .F
ig.6
.35
repo
rtste
ndon�s
layo
utfo
rhal
fdec
k,be
ing
sym
met
rical
lydi
spos
ed.
EC2�
wor
ked
exam
ples
6-45
Tabl
eof
Cont
ent
Fig
.6.3
5Pl
anan
dpr
inci
pals
ectio
nof
tend
onla
yout
Imm
edia
telo
sses
ofpr
estre
ssin
gdu
eto
fric
tion
have
been
eval
uate
dby
mea
nsof
the
follo
win
gex
pres
sion
:
po(x
)=po
,max
e-(
+k
x)
with
: =0.
19co
effic
ient
offr
ictio
nbe
twee
nth
ete
ndon
sand
thei
rshe
athi
ng;
k=
0.01
rad/
mun
inte
ntio
nala
ngul
arde
viat
ion.
Pres
tress
ing
has
tobe
intro
duce
din
the
FEM
mod
elin
orde
rto
calc
ulat
edth
ehy
pers
tatic
actio
nsth
atar
ise
inth
est
ruct
ural
sche
me.
Con
side
ring
pres
tress
ing
asan
exte
rnal
load
,iti
spo
ssib
leto
intro
duce
itby
mea
nsof
two
incl
ined
forc
esat
anch
orag
es(r
epre
sent
ing
actio
nsat
the
extre
mity
)an
dof
asy
stem
ofeq
uiva
lent
load
sal
ong
tend
on�s
prof
ile(r
epre
sent
ing
tend
oncu
rvat
ure
and
loss
esdu
eto
fric
tion)
:the
seac
tions
per
tend
on,s
houl
dbe
appl
ied
cons
iste
ntly
atth
eno
deso
fFEM
mod
el.
The
equi
vale
ntlo
ads
may
beca
lcul
ated
subd
ivid
ing
the
tend
onpr
ofile
into
elem
enta
ryse
gmen
tsan
dev
alua
ting
the
inte
rnal
actio
nab
leto
equi
libra
teth
eex
tern
alon
edu
eto
end
actio
nsde
rivin
gby
the
pres
tress
ing.
EC2�
wor
ked
exam
ples
6-46
Tabl
eof
Cont
ent
Fig
.6.3
6Ef
fect
ofpr
estr
essi
ngon
ase
gmen
tand
equi
vale
ntlo
ads
Fig.
6.36
repr
esen
tsth
efo
rces
actin
gon
ase
gmen
tof
conc
rete
due
toa
curv
edpr
estre
ssin
gte
ndon
;ift
hein
clin
atio
nof
the
cabl
ech
ange
sfr
om1
to2
whi
leth
epr
estre
ssfo
rce
chan
ges
from
P 1to
P 2du
eto
fric
tion,
the
equi
libra
ting
verti
cala
ndho
rizon
talf
orce
sin
the
i-seg
men
tres
ult:
F v,i
=P 2
sin
2P 1
sin
1;
F h,i
=P 2
cos
2P 1
cos
1
whi
leth
eba
lanc
ing
mom
entt
urns
out:
Mi=
(P2
cos
2e 2
P 1co
s1
e 1)
(P2
sin
2P 1
sin
1)a
/2
The
abov
epr
oced
ure
shou
ldbe
repe
ated
for
all
the
segm
ents
.It
can
beno
tice
that
the
forc
esat
the
end
ofea
chse
gmen
text
rem
ityar
eth
esa
me
with
oppo
site
sign
s,de
pend
ing
onw
heth
erth
erig
htor
the
left
segm
enti
sco
nsid
ered
;the
sefo
rces
canc
elou
tthe
mse
lves
with
the
exce
ptio
nat
anch
orag
es.F
inal
ly,f
orea
chte
ndon
,th
efo
rces
atth
eex
trem
ityof
the
cabl
epl
usth
eeq
uilib
ratin
gsy
stem
for
each
segm
ent,
shal
lbe
intro
duce
din
the
FEM
mod
el.
The
choi
ceof
the
posi
tion
ofth
eel
emen
tary
segm
ents
isre
lativ
eto
the
kind
ofel
emen
tad
opte
din
the
FEM
mod
el.I
fbe
amel
emen
tsar
eus
ed,i
tis
poss
ible
toin
trodu
cea
poin
tlo
ad(o
rm
omen
t)w
heth
eral
ong
the
elem
entb
ody
orat
node
s,co
nseq
uent
lyth
ese
gmen
tex
trem
ities
may
bepl
aced
indi
ffer
ently
atno
des
orat
the
mid
dle
ofth
eel
emen
t.W
ithsh
ell
elem
ents
,on
lyno
dal
forc
esca
nbe
cons
ider
edso
that
itis
nece
ssar
yto
plac
ese
gmen
tex
trem
ities
with
intw
ose
quen
tialn
odes
;fur
ther
mor
e,du
eto
the
two-
dim
ensi
onal
sche
me,
one
has
toco
nsid
erth
etra
nsve
rse
posi
tion
ofth
ete
ndon
that
,in
gene
ral,
dono
tcoi
ncid
ew
itha
noda
lalig
nmen
t.A
sa
sim
ple
rule
,the
indi
catio
nsof
Fig.
6.37
may
befo
llow
ed.
EC2�
wor
ked
exam
ples
6-47
Tabl
eof
Cont
ent
Fig
.6.3
7Tr
ansv
erse
distr
ibut
ion
ofpr
estre
ssin
g
Tim
e-de
pend
entp
rest
ress
ing
loss
es
Tim
e-de
pend
entl
osse
sof
pres
tress
may
beev
alua
ted
bym
eans
ofth
efo
llow
ing
equa
tion:
pc
sr
css
prcg
cp
p c
c ccp
tt
Et
tA A
A Iz
tt
,
(,
)(
,)(
)
.(
,)
00
0
20
11
10
8
whe
re:
p,c+
s+r:
loss
ofin
itial
tend
onst
ress
due
tocr
eep
and
shrin
kage
ofco
ncre
tean
dre
laxa
tion
ofst
eel,
betw
een
time
t 0an
dtim
et
;
t 0=
28da
ys:
age
ofco
ncre
teat
pres
tress
ing
time;
t=
2555
0ds
.:co
rres
pond
ing
toa
life-
time
of70
year
s;
cs(t
,t 0):
shrin
kage
stra
inat
time
tca
lcul
ated
from
:cs
(t,t 0
)=cs
0s(
t-t
0)=
0.12
710
-3
whe
re:
cso
=s
(fcm
)RH
with
:
s(f
cm)=
[160
+10
sc(9
f cm/f c
mo)
]10
-6=
0.00
0395
f cm=
mea
nco
mpr
essi
vest
reng
htof
conc
rete
at28
days
=f ck
+8
MPa
;
f cmo
=10
MPa
;
sc=
5fo
rrap
idha
rden
ing
cem
ents
;
RH=
155
110
0
3
.R
H=
1.01
8;
RH
=70
%re
lativ
ehu
mid
ityof
the
ambi
enta
tmos
pher
e;
s(t
-t0)
=t
th
tt
02
00
035
.=
0.57
4
EC2�
wor
ked
exam
ples
6-48
Tabl
eof
Cont
ent
h=
(2A
c/u
)=12
17m
mno
tiona
lsiz
eof
mem
ber;
Ac
=17
.43
106
mm
2gr
osss
ectio
nof
the
beam
;
u=
2864
0m
mpe
rimet
erof
the
mem
beri
nco
ntac
twith
the
atm
osph
ere;
(t,t 0
):cr
eep
coef
ficie
ntat
time
tca
lcul
ated
from
:(t
,t 0)=
0c(t
-t0)
=1.
5708
whe
re:
o=
RH(f
cm)
(t 0)=
1.59
8w
ith
RH=
11
100
013
RH
h.
=1.
281;
(fcm
)=53. f
fcm
cmo
=2.
556;
(t 0)=
101
002
..
t=
0.48
8
c(t
-t0)
=t
tt
tH
0
0
03.
=0.
983
with
H=
151
001
225
018
..
RH
h=
2155
>15
0015
00
Ifth
eim
prov
edpr
edic
tion
mod
elof
chap
ter
3is
used
,the
follo
win
gva
lues
for
cs(t
,t0)
and
for
(t,t
0)m
aybe
eval
uate
d:
cs(t
,t0)
=18
2.62
10-6
;(t
,t0)
=1.
5754
ingo
odag
grem
entw
ithth
epr
evio
uson
e,at
leas
tfor
cree
pva
lue.
pr:
loss
ofpr
estre
ssin
gdu
eto
rela
xatio
nof
stee
lcal
cula
ted
for
are
duce
din
itial
tens
ilest
ress
ofp
=pg
o0.
3p,
c+s+
r(w
here
pgo
isth
eef
fect
ive
initi
alst
ress
inte
ndon
sdu
eto
dead
load
and
pres
tress
ing)
and
eval
uate
das
perc
enta
geby
the
follo
win
gfo
rmul
a:
t=
1000
ht
1000
019.
=10
00h
3w
here
t=is
the
rela
xatio
naf
tert
hour
s;fo
rt>
50ye
ars
t.=
1000
h3;
1000
h=
isth
ere
laxa
tion
afte
r100
0ho
urse
valu
ated
from
Fig.
6.38
;
EC2�
wor
ked
exam
ples
6-49
Tabl
eof
Cont
ent
Fig
.6.3
8Re
laxa
tion
loss
esin
%at
1000
hour
sfor
Cla
ss2
c.:
stre
sson
conc
rete
atle
velo
fpre
tens
ione
dst
eeld
ueto
self
wei
ghta
ndpe
rman
entl
oad;
cpo
:st
ress
onco
ncre
teat
leve
lofp
rete
nsio
ned
stee
ldue
topr
estre
ssin
g;
=E s
/Ec:
mod
ulus
ofel
astic
ityra
tio;
Ap
:ar
eaof
pres
tress
ing
stee
latt
heco
nsid
ered
leve
l;
Ac
:ar
eaof
conc
rete
gros
sse
ctio
n;
I c:
iner
tiaof
conc
rete
gros
ssec
tion;
z cp
:le
vera
rmbe
twee
nce
ntro
idof
conc
rete
gros
ssec
tion
and
pres
tress
ing
stee
l.
Tim
e-de
pend
entl
osse
sof
pres
tress
ing
shou
ldbe
calc
ulat
edfo
rea
chte
ndon
alon
ghi
spr
ofile
soth
ata
corr
ectv
alue
may
beus
edfo
reac
hel
emen
t.A
sa
refe
renc
e,th
em
axim
umva
lue
ofpr
estre
ssin
glo
sses
,asp
erce
ntag
eof
initi
alst
eelt
ensi
on,t
urn
out:
long
itudi
nalt
endo
n:19
%at
anch
orag
ean
d14
%at
pier
axis
;
trans
vers
ete
ndon
:18
%at
anch
orag
ean
d12
%at
mid
span
.
The
effe
cts
oflo
sses
are
take
nin
toac
coun
tw
ithth
esa
me
proc
edur
eus
edfo
rth
epr
estre
ssin
g,bu
tasa
ctio
nsof
oppo
site
sign
.
6.15
.2A
ctio
ns
The
exte
rnal
load
sap
plie
don
the
stru
ctur
esh
ould
beev
alua
ted
acco
rdin
gto
the
prov
isio
nsof
Euro
code
1.3
Traf
ficLo
adon
Brid
ges.
As
verti
calt
rain
load
the
load
mod
elLM
71pl
usth
elo
adm
odel
sSW
(SW
/0an
dSW
/2re
spec
tivel
y)ha
vebe
enad
opte
dw
ithan
coef
ficie
ntof
1.1.
For
the
LM71
,th
e4
poin
tlo
ads
have
been
redu
ced
inan
equi
vale
ntun
iform
load
bysm
earin
gth
eirc
hara
cter
istic
valu
eQ
vkal
ong
the
influ
ence
leng
thso
that
aq v
k,1
may
beob
tain
ed:
Qvk
=1.
125
0di
n=
319.
6kN
q vk,
1=
319.
6/1.
6=
199.
75kN
/m
EC2�
wor
ked
exam
ples
6-50
Tabl
eof
Cont
ent
whe
redi
n,b
eing
the
dyna
mic
fact
oreq
ualt
o1.
162,
isev
alua
ted
belo
w.
Fig
.6.3
9Ad
opte
dlo
adar
rang
emen
tfor
LM71
load
mod
el
The
unifo
rmly
dist
ribut
edlo
adq v
kac
cord
ing
toEu
roco
de1.
3is
:
q vk
=1.
180
din
q vk,
2=
102.
3kN
/m
with
outa
nylim
itatio
nin
leng
th.
Fig.
6.39
show
sth
eLM
71ar
rang
emen
tad
opte
din
the
calc
ulat
ions
.
The
load
mod
elSW
/0is
repr
esen
ted
inFi
g.6.
40an
dits
char
acte
ristic
valu
ere
sults
:
q vk
=1.
113
3di
n=
170.
0kN
/m
Fig
.6.4
0Lo
adm
odel
SW/0
The
load
mod
elSW
/2is
repr
esen
ted
inFi
g.6.
41an
dits
char
acte
ristic
valu
ere
sults
:
q vk
=1.
115
0di
n=
174.
3kN
/m
Fig
.6.4
1Lo
adm
odel
SW/2
The
prev
ious
load
mod
elLM
71,S
W/0
and
SW/2
have
been
intro
duce
din
the
FEM
anal
ysis
cons
ider
ing
asp
read
ing
ratio
of4:
1in
the
balla
stan
dof
1:1
inth
eco
ncre
teup
toth
em
iddl
epl
ane
ofth
esl
ab.
Inth
efo
llow
ing
asle
fttra
ckis
deno
ted
the
track
whi
chha
sa
posi
tive
valu
efo
rthe
yco
-ord
inat
e,w
hile
right
truck
the
othe
rone
.Fi
g.6.
42sh
ows
whi
chel
emen
tsar
ein
volv
edby
spre
adin
gef
fect
s,th
eref
ore
subj
ecte
dto
varia
ble
load
.
EC2�
wor
ked
exam
ples
6-51
Tabl
eof
Cont
ent
Fig
.6.4
2Sp
read
ing
effe
ctso
nFE
Mm
odel
and
load
edel
emen
ts
The
dyna
mic
fact
oris
calc
ulat
edby
mea
nsof
the
follo
win
gex
pres
sion
(trac
kw
ithst
anda
rdm
aint
enan
ce):
0.73
0.2
L2.16
31.
162
whe
reL
isth
ede
term
inan
tlen
gth
defin
edin
the
Euro
code
1.3
as:
LL
LL
133
1317
3327
7517
333
2704
12
3.
..
..
.m
Seve
ralo
ther
actio
ns,a
risin
gfr
omva
riabl
elo
ads,
shou
ldbe
cons
ider
edin
the
anal
ysis
(as
tract
ion
and
brak
ing,
cent
rifug
alfo
rces
,de
railm
ent,
win
dpr
essu
re,
diff
eren
tial
tem
pera
ture
varia
tion
etc.
)bu
t,fo
rth
esa
keof
sim
plic
ity,i
nth
ese
calc
ulat
ions
only
the
follo
win
gac
tions
have
been
cons
ider
ed(in
trodu
ced
inth
em
athe
mat
ical
mod
elin
diff
eren
tst
eps)
:
STEP
1:Se
lf-w
eigh
toft
hest
ruct
ure:
adop
ting
aun
itw
eigh
t3 ;
STEP
2:Pr
estre
ssin
gfo
rces
attim
eof
tens
ioni
ng;
STEP
3:Pr
estre
ssin
gfo
rces
afte
rtim
e-de
pend
entl
osse
s:
inth
eca
lcul
atio
ns,a
limit
valu
eof
tens
ilest
ress
inte
ndon
equa
lto
0.6
f ptk
afte
ral
low
ance
for
loss
es(t
),ha
sbe
enco
nsid
ered
,ac
cord
ing
toth
epr
ovis
ions
ofth
eap
plie
dR
ailw
ayC
ode
toav
oid
the
risk
ofbr
ittle
failu
redu
eto
stre
ssco
rros
ion.
STEP
4:Tr
ack
load
com
preh
ensi
veof
;
rails
,sle
eper
san
dba
llast
(wat
erpr
oofin
gin
clud
ed)
eval
uate
das
aco
ver
with
ano
min
alhe
ight
of0.
8m
and
aun
itw
eigh
tof
=18
kN/m
3 ),so
that
fora
wid
thof
9.5
m,a
nun
iform
lydi
strib
uted
load
resu
lts:
g bal
last
=0.
81.
89.
5=
136.
8kN
/m;
STEP
5:O
ther
spe
rman
entl
oads
com
pose
dby
;
trans
vers
egr
adie
ntfo
rdr
ain
wat
er,a
ssum
edas
alo
adof
1.25
kN/m
2it
turn
sou
t:
EC2�
wor
ked
exam
ples
6-52
Tabl
eof
Cont
ent
g dra
in=
1.25
9.5
=11
.875
kN/m
;
balla
stre
tain
ing
wal
ls(w
itha
cros
sse
ctio
nar
eaof
0.25
m2
and
unit
wei
ghto
f25
kN/m
3 ) g wal
ls=
250.
25=
6.25
kN/m
fore
ach;
duct
s:
g duc
ts=
3kN
/mfo
reac
h;
bord
ercu
rbs
(with
acr
oss
sect
ion
area
of0.
1m
2an
dun
itw
eigh
tof
25kN
/m3 ): g r
einf
beam
=25
0.25
=6.
25kN
/mfo
reac
h;
nois
eba
rrie
rs:
g bar
riers
=8.
00kN
/mfo
reac
h;
STEP
6:V
aria
ble
load
sfo
rmax
imum
bend
ing
mom
ento
nfir
stsp
an(x
=6.
18m
);
the
appl
ied
load
isa
LM71
mod
elon
the
left
track
with
the
follo
win
glo
ngitu
dina
larr
ange
men
t:
Fig
.6.4
3LM
71ar
rang
emen
tfor
Load
Step
5
plus
aSW
/2tra
inon
the
right
track
with
the
follo
win
glo
ngitu
dina
lar
rang
emen
t:
Fig
.6.4
4SW
/2ar
rang
emen
tfor
Load
Step
5
STEP
7:V
aria
ble
load
sfo
rmin
imum
bend
ing
mom
enta
tpie
rP1
(x=
18.4
3m
);
the
appl
ied
load
isa
SW/0
mod
elon
the
left
track
with
the
follo
win
glo
ngitu
dina
larr
ange
men
t:
EC2�
wor
ked
exam
ples
6-53
Tabl
eof
Cont
ent
Fig
.6.4
5SW
/0ar
rang
emen
tfor
Load
Step
6
plus
aSW
/2tra
inon
the
right
track
with
the
follo
win
glo
ngitu
dina
lar
rang
emen
t:
Fig
.6.4
6SW
/2ar
rang
emen
tfor
Load
Step
6
STEP
8:V
aria
ble
load
sfo
rmax
bend
ing
mom
ento
nse
cond
span
(x=
32.3
05m
);
the
appl
ied
load
isa
LM71
mod
elon
the
left
track
with
the
follo
win
glo
ngitu
dina
larr
ange
men
t:
Fig
.6.4
7LM
71ar
rang
emen
tfor
Load
Step
7
plus
aSW
/2tra
inon
the
right
track
with
the
follo
win
glo
ngitu
dina
lar
rang
emen
t:
Fig
.6.4
8SW
/2ar
rang
emen
tfor
Load
Step
7
EC2�
wor
ked
exam
ples
6-54
Tabl
eof
Cont
ent
6.15
.3C
ombi
natio
nsof
Act
ions
The
desi
gnva
lues
fort
heex
tern
alac
tions
have
been
calc
ulat
edad
optin
gth
eco
mbi
natio
nsof
load
sspe
cifie
din
the
appl
ied
Cod
eas
follo
win
dica
ted
inth
esy
mbo
licpr
esen
tatio
n:
Ulti
mat
eLi
mit
Stat
e
SS
dG
kG
kp
kQ
koi
iki
GG
PQ
Q1
12
21
1
Serv
icea
bilit
yLi
mit
Stat
e:ra
reco
mbi
natio
n
SS
dG
GP
kk
kk
oiik
i1
21
1
Serv
icea
bilit
yLi
mit
Stat
e:qu
asi-p
erm
anen
tcom
bina
tion
SS
dG
GP
Qk
kk
iik
i1
22
1
whe
re:
G1k
=ch
arac
teris
ticva
lue
ofth
eac
tion
due
tose
lf-w
eigh
tand
perm
anen
tloa
ds,b
alla
stex
clud
ed;
G2k
=ch
arac
teris
ticva
lue
ofac
tion
due
toba
llast
self-
wei
ght;
P k=
char
acte
ristic
valu
eof
actio
ndu
eto
pres
tress
;
Q1k
=ch
arac
teris
ticva
lue
ofac
tion
due
toth
eba
seva
riabl
eac
tion;
Qik
=ch
arac
teris
ticva
lue
actio
ndu
eto
ofth
eot
heri
ndep
ende
ntva
riabl
elo
ads;
1=
parti
alfa
ctor
ofse
lf-w
eigh
tan
dpe
rman
ent
load
s,ba
llast
excl
uded
,equ
alto
1.4
foru
nfav
oura
ble
effe
ctan
d1.
0fo
rfav
oura
ble
effe
ct;
2=
parti
alfa
ctor
ofba
llast
load
equa
lto
1.8
for
unfa
vour
able
effe
ctan
d1.
0fo
rfa
vour
able
effe
ct;
P=
parti
alfa
ctor
ofpr
estre
sslo
adeq
ual
to1.
2fo
run
favo
urab
leef
fect
and
0.9
for
favo
urab
leef
fect
;
Q=
parti
alfa
ctor
ofva
riabl
elo
ads
equa
lto
1.5
for
unfa
vour
able
effe
ctan
d0.
0fo
rfa
vour
able
effe
ct;
0i=
com
bina
tion
fact
orof
varia
ble
load
sequ
alto
0.8;
2i=
com
bina
tion
fact
orof
varia
ble
load
sfo
rqu
asi-p
erm
anen
tcom
bina
tion
atse
rvic
e,eq
ualt
o0.
6.
EC2�
wor
ked
exam
ples
6-55
Tabl
eof
Cont
ent
6.15
.4V
erifi
catio
nat
Serv
icea
bilit
yL
imit
Stat
e
The
verif
icat
ion
atse
rvic
eabi
lity
limit
stat
eis
rela
tive
toth
efo
llow
ing
cond
ition
s:
stre
sslim
itatio
nat
tens
ioni
ng;
stre
sslim
itatio
nat
serv
ice;
crac
kw
idth
s;
defo
rmat
ion.
Veri
ficat
ion
atte
nsio
ning
Att
ime
ofte
nsio
ning
,no
tens
ilest
ress
shou
ldbe
pres
enti
nth
eex
trem
efib
res
ofth
esl
aban
dth
em
axim
umco
mpr
essi
vest
ress
shou
ldno
texc
eed
the
limit
valu
eof
0.6
f ck=
21M
Pa.
For
the
sake
ofsi
mpl
icity
,one
repo
rtsth
eve
rific
atio
nre
late
dto
the
four
elem
ents
show
edin
figii,
assu
bjec
ted
toth
ehi
gher
stre
ssle
vel.
The
exte
rnal
actio
nsar
eca
lcul
ated
adop
ting
the
rare
com
bina
tion
with
only
the
load
step
s1
and
2.Fr
omFE
Man
alys
is,t
heva
lue
ofn 2
2,m
22,n
33,m
33,n
23,m
23ar
eev
alua
ted
soth
atit
resu
lts:
yt
th
h,
,22
2222 2
6n
m;
yb
bh
h,
,22
2222 2
6n
m
xt
th
h,
,33
3333 2
6n
m;
xb
bh
h,
,33
3333 2
6n
m
xyt
th
h,
,23
2323 2
6n
m;
xyb
bh
h,
,23
2323 2
6n
m
whe
reth
esu
bscr
ipts
tan
db
indi
cate
resp
ectiv
ely
top
and
botto
mfib
re.
The
angl
esof
prin
cipa
ldire
ctio
ns(f
orw
hich
isxy
=0)
are:
123
2233
1 22
atan
;2
=1
+90
°
and
the
prin
cipa
lstre
sses
resu
lt:
122
21
332
123
12
,/
,/
,/
,/
cos
()
()
()
tb
tb
tb
tb
sin
sin
222
22
332
223
22
,/
,/
,/
,/
cos
()
()
()
tb
tb
tb
tb
sin
sin
Refe
rring
toth
eel
emen
tsm
arke
din
Fig.
6.32
one
obta
ins:
Tabl
e6.
7
Elem
.h
n22
n33
n23
m22
m33
m23
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
m/m
][k
Nm
/m]
[kN
m/m
]
648
1.5
-309
1-1
3159
6-2
25-2
176
093
0.96
3-7
806
-852
675
743
456
-51
320
1.5
-351
6-1
0418
1-4
5-8
120
589
1.5
-428
0-1
0007
-67
653
1945
20
Tabl
e6.
8
EC2�
wor
ked
exam
ples
6-56
Tabl
eof
Cont
ent
22,b
33,b
23,b
22,t
33,t
23,t
1,b
2,b
1,t
2,t
1,b
2,b
1,t
2,t
[Mpa
][M
pa]
[Mpa
][M
pa]
[Mpa
][M
pa]
[°]
[°]
[°]
[°]
[Mpa
][M
pa]
[Mpa
][M
pa]
-2.6
6-1
4.58
0.00
-1.4
6-2
.97
0.00
0.02
89.9
80.
1589
.85
-2.6
6-1
4.58
-1.4
6-2
.97
-3.3
0-5
.90
-0.2
5-1
2.91
-11.
800.
41-5
.48
95.4
8-1
8.17
108.
17-3
.27
-5.8
3-1
3.05
-12.
15-2
.46
-9.1
10.
00-2
.22
-4.7
80.
000.
0189
.99
0.01
89.9
9-2
.46
-9.1
1-2
.22
-4.7
8-1
.11
-1.4
80.
01-4
.59
-11.
86-0
.10
1.29
88.7
1-0
.77
90.7
7-1
.11
-1.4
8-4
.59
-11.
85
whi
chno
texc
eed
the
limit
one.
Veri
ficat
ion
oflim
itst
ate
ofst
ress
limita
tion
inco
ncre
te
The
serv
icea
bilit
ylim
itsta
tes
chec
ked
inth
isse
ctio
nar
ere
lativ
eon
lyto
stre
sslim
itatio
n,en
surin
gth
at,u
nder
serv
ice
load
cond
ition
s,co
ncre
teex
trem
estr
esse
sdo
note
xcee
dth
eco
rresp
ondi
nglim
it,fo
rthe
quas
i-per
man
enta
ndth
era
reco
mbi
natio
ns.T
helim
itst
ress
esfo
rcon
cret
ear
e:
Qua
si-pe
rman
entc
ombi
natio
n:C
ompr
essiv
est
ress
=0.
4f ck
=14
.00
MPa
Rare
com
bina
tion:
Com
pres
sive
stre
ss=
0.6
f ck=
21.0
0M
PaA
pply
ing
toth
estr
uctu
ral
FEM
mod
elth
eva
riabl
elo
ads
and
com
bini
ngth
emac
cord
ing
toth
era
ilway
code
prov
ision
s,on
eob
tain
the
max
ima
stres
sva
lues
atto
pan
dbo
ttom
fibre
sth
atha
veto
belo
wer
than
the
corre
spon
ding
limit.
One
repo
rtsth
ere
sults
rela
tive
toth
efo
urel
emen
tsin
dica
ted
inFi
g.6.
32.
Qua
si-P
erm
anen
tCom
bina
tion
Tabl
e6.
9
Elem
.h
n22
n33
n23
m22
m33
m23
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
m/m
][k
Nm
/m]
[kN
m/m
]
648
1.5
-242
0-1
0152
4-2
36-1
576
493
0.96
3-6
233
-634
750
589
108
-37
320
1.5
-353
9-7
855
281
233
458
91.
5-2
736
-790
0-3
-151
-396
0
1,b
2,b
1,t
2,t
[Mpa
][M
pa]
[Mpa
][M
pa]
-2.2
4-1
0.97
-0.9
8-2
.57
-2.6
5-5
.86
-10.
31-7
.37
-2.1
4-4
.62
-2.5
8-5
.86
-2.2
3-6
.32
-1.4
2-4
.21
Rare
Com
bina
tion
Tabl
e6.
10
Elem
.h
n22
n33
n23
m22
m33
m23
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
m/m
][k
Nm
/m]
[kN
m/m
]
648
1.5
-223
8-1
0270
3-2
26-6
154
930.
963
-628
4-6
033
457
7-1
33-6
232
01.
5-2
604
-747
97
712
79-9
589
1.5
-379
1-8
243
-55
-689
-127
5-2
6
1,b
2,b
1,t
2,t
[Mpa
][M
pa]
[Mpa
][M
pa]
-2.0
9-8
.49
-0.8
9-5
.21
-2.7
6-7
.02
-10.
29-5
.51
-1.7
2-1
.58
-1.7
5-8
.40
-4.3
6-8
.89
-0.6
9-2
.09
Veri
ficat
ion
ofSe
rvic
eabi
lity
Lim
itSt
ate
ofC
rack
ing
The
char
acte
ristic
crac
kw
idth
shou
ldbe
calc
ulat
edac
cord
ing
toth
epr
ovisi
ons
ofM
odel
EC2�
wor
ked
exam
ples
6-57
Tabl
eof
Cont
ent
Code
90.
Itha
sbe
notic
e,ho
wev
er,t
hat
from
stres
sca
lcul
atio
nne
ither
for
the
quas
i-pe
rman
entc
ombi
natio
nno
rin
the
rare
one,
the
max
imum
stres
sre
sults
tens
ile.T
here
fore
,no
spec
ific
rein
forc
emen
tisr
equi
red
and
itis
suffi
cien
tto
arra
nge
the
min
imum
amou
ntof
rein
forc
ing
steel
,abl
eto
ensu
rea
duct
ilebe
havi
our
inca
seof
corro
sion
ofpr
estre
ssin
gste
el.
Def
orm
atio
n
Def
orm
atio
nlim
itatio
nis
carr
ied
out
toco
ntro
lth
em
axim
umve
rtica
lde
flect
ion
for
pass
enge
rsco
mfo
rt.Th
elim
itva
lues
/L(d
efle
ctio
n/sp
anLe
ngth
)ar
egi
ven
byth
eEu
roco
de1.
3as
afu
nctio
nof
the
span
leng
than
dth
etra
insp
eed.
The
limit
valu
efo
rm
axim
umve
rtica
ldef
lect
ion
isca
lcul
ated
cons
ider
ing
asp
anle
ngth
of27
.75
m(c
entra
lsp
an)a
nda
train
spee
dov
er28
0km
/h;a
ccor
ding
toth
epr
ovisi
onso
fthe
Code
,itr
esul
ts:
L1
1600
that
shou
ldbe
mul
tiplie
dby
afa
ctor
1.1
for
cont
inuo
usstr
uctu
res;
final
ly,t
hefo
llow
ing
limit
may
beac
hiev
ed:
lim.
L11 1600
114
55
As
aco
nseq
uenc
eof
the
trans
ient
natu
reof
this
even
t,th
eel
astic
defle
ctio
n,ca
lcul
ated
byth
eFE
Mm
odel
,is
rela
tive
toth
eon
lyliv
elo
ad;t
hech
eck
shal
lbe
perfo
rmed
load
ing
only
one
track
,re
adin
gth
em
axim
umde
flect
ion
inco
rresp
onde
nce
ofth
etra
ckax
is.H
avin
glo
aded
the
right
track
with
aLM
71lo
adm
odel
plus
dyna
mic
allo
wan
ce,p
lace
din
the
mid
dle
ofth
eof
the
cent
rals
pan,
the
obta
ined
/Lva
lue
is:
effe
ctiv
e
L0
0055
2775
150
45.
.
and
itre
sults
low
erth
anth
eco
rresp
ondi
nglim
it.
Itca
nbe
notic
eth
atno
furth
erca
lcul
atio
nis
requ
este
dbe
caus
e,du
eto
pres
tress
ing
effe
ct,t
hest
ruct
ure
rem
ains
entir
ely
com
pres
sed,
soth
atth
eel
astic
valu
e,ca
lcul
ated
byth
eFE
Man
alys
is,ha
sto
beco
nsid
ered
.
6.15
.5V
erifi
catio
nof
Ulti
mat
eL
imit
Stat
e
Ver
ifica
tion
atU
LSsh
ould
rega
rdth
est
ruct
ure
asa
who
lean
dits
com
pone
ntpa
rts,
anal
ysin
gth
ere
sista
nce
ofth
ecr
itica
lre
gion
s.In
addi
tion
toth
ean
alys
isof
ULS
ofse
vera
lshe
llel
emen
tund
erth
ere
leva
ntco
mbi
natio
nof
inte
rnal
actio
ns,i
nth
isex
ampl
eso
me
case
ofde
taili
ngar
ein
vesti
gate
d,i.e
.:
burs
ting
forc
eat
anch
orag
eof
pres
tress
ing
tend
on;
spal
ling
forc
eat
anch
orag
eof
pres
tress
ing
tend
on;
punc
hing
unde
rsup
port
plat
e.
EC2�
wor
ked
exam
ples
6-58
Tabl
eof
Cont
ent
Slab
ultim
ate
limit
stat
e
Ver
ifica
tion
atU
LSha
sbe
enpe
rform
edad
optin
gth
esa
ndw
ich
mod
elfo
rshe
llel
emen
ts.
The
inte
rnal
actio
nsin
ash
elle
lem
enta
tULS
are
sket
ched
inFi
g.6.
49.
Fig
.649
Inte
rnal
actio
nsat
ULS
ina
shel
lele
men
ts
Let
usco
nsid
erin
this
sect
ion
only
four
elem
ents
onth
ew
hole
(see
Fig.
6.32
).Th
eex
tern
alac
tions
are
deriv
edfr
omFE
Mm
odel
usin
gth
elo
adst
epfo
rtra
ins
whi
chle
ads
toth
em
axim
umva
lues
and
com
bini
ngth
ere
sults
acco
rdin
gto
the
rele
vant
com
bina
tion
form
ula.
Fort
hein
vesti
gate
del
emen
ts,tu
rnso
ut(o
nbr
acke
tsth
eno
tatio
nof
Fig.
6.49
):Ta
ble
6.11
Elem
.h
nSd
,yn
Sd,x
nSd
,xy
mSd
,ym
Sd,x
mSd
,xy
v Sd,
yv S
d,x
(n22
)(n
33)
(n23
)(m
22)
(m33
)(m
23)
(v13
)(v
12)
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
m/m
][k
Nm
/m]
[kN
m/m
][k
N/m
][k
N/m
]
648
1.5
-177
9-9
096
5-2
3947
0-1
4-6
-593
0.96
3-5
746
-461
0-6
349
9-6
71-7
589
-150
320
1.5
-213
0-5
922
1038
3241
-13
2047
589
1.5
-386
5-7
748
-54
-195
0-4
274
-41
-112
4-1
095
Asf
irsts
tep,
one
may
desig
nth
ein
nerl
ayer
chec
king
ifsp
ecifi
csh
earr
einf
orce
men
tis
requ
ired
orno
t.In
fact
,iti
spos
sible
toca
lcul
ate
the
prin
cipa
lshe
arv o
2=
v x2
+v y
2,o
nth
epr
inci
pals
hear
dire
ctio
no
(suc
hth
atta
n0
vv
yx
),an
dto
chec
kth
atit
turn
out:
v<
v0
11
301
210
0R
dckf
d.
whe
rev R
d1is
spec
ified
inch
apte
r6.4
.2.3
ofM
C90
and
xo
yo
sin
cos2
2.
Ifth
eis
nots
atisf
ied,
spec
ific
shea
rrei
nfor
cem
ents
hall
bear
rang
ed(v
ertic
alst
irrup
s)an
ddi
agon
alco
mpr
essiv
efo
rces
inco
ncre
tesh
all
bech
ecke
d.A
ccor
ding
toCE
BBu
lletin
223,
and
havi
ngse
tam
inim
umam
ount
oflo
ngitu
dina
land
trans
vers
ere
info
rcem
enti
nth
ebo
ttom
and
top
laye
rof
As,x
=A
s,y=
22.6
cm2 /m
plac
edat
0.07
mfro
mth
eex
tern
alfa
ce,t
hefo
llow
ing
tabl
em
aybe
calc
ulat
edfo
rthe
elem
ents
cons
ider
ed.
Tabl
e6.
12
EC2�
wor
ked
exam
ples
6-59
Tabl
eof
Cont
ent
Elem
.d
oo
v ov R
d1F
Scw
FR
cwA
s/s2
nxc
nyc
nxy
c
[m]
[°]
[-][k
N/m
][kN
/m]
[°]
[kN
/m]
[kN
/m]
[cm
2 /m2 ]
[kN
/m]
[kN
/m]
[kN
/m]
648
1.43
51.1
80.
0015
87
417
26.5
6-
--
0.0
0.0
0.0
930.
893
-30.
770.
0025
317
432
726
.56
--
-0.
00.
00.
032
01.
4323
.14
0.00
158
5141
726
.56
--
-0.
00.
00.
058
91.
4345
.76
0.00
158
1569
417
26.5
635
0913
860
14.0
763.
980
5.6
784.
5
with
varia
tion
ofsl
abco
mpo
nent
sdue
tov x
and
v y(i.
e.n x
c,n
ycan
dn x
yc)o
nly
fore
lem
ent
num
ber5
89.
The
oute
rlay
ers
shou
ldbe
desig
ned
supp
osin
gan
initi
alth
ickn
ess
forb
oth
laye
rsno
tle
sser
than
twic
eth
eco
ncre
teco
ver
eval
uate
dat
the
cent
roid
ofre
info
rcem
ent.
One
assu
mes
:
t s=
t i=
20.
07=
0.14
mso
that
,int
erna
llev
erar
mz
and
inpl
ane
actio
nsm
aybe
eval
uate
dfo
rthe
oute
rlay
ers
ofea
chel
emen
tref
errin
gto
Fig.
6.50
and
bym
eans
ofth
efo
llow
ing
equa
tions
:
Fig
.6.5
0In
tern
alfo
rces
inth
edi
ffere
ntla
yers
nn
zy z
m zv v
Sdx
sx
sx
,co
t1 2
x2 0
nn
zy z
m zv v
Sdx
ix
ix
,co
t1 2
x2 0
nn
zy z
m zv v
Sdy
sy
sy
,co
t1 2
y2 0
nn
zy z
m zv v
Sdy
iy
iy
,co
t1 2
y2 0
vx
y
0Sd
sxy
sxy
nz
y zm z
vv v
,co
t1 2
vx
y
0Sd
ixy
ixy
nz
y zm z
vv v
,co
t1 2
whe
rete
rms
onbr
acke
tsha
vebe
sum
med
ifsh
earr
einf
orce
men
tis
requ
ired.
Inth
ede
sign
EC2�
wor
ked
exam
ples
6-60
Tabl
eof
Cont
ent
proc
edur
eis
conv
enie
ntto
reac
hth
em
inim
umam
ount
ofre
info
rcem
ent,
soth
ata
valu
eof
45°f
oran
gle
may
bead
opte
d.Fo
rthe
chos
enel
emen
tsit
turn
sout
:Ta
ble
6.13
Elem
.h
t st i
t cy s
y iz
nSd
y,s
nSd
x,s
v Sd,
sn
Sdy,
in
Sdx,
iv S
d,i
[m]
[m]
[m]
[m]
[m]
[m]
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
648
1.5
0.14
00.
140
1.22
00.
680
0.68
01.
360
-713
.5-4
893.
313
.0-1
065.
1-4
202.
2-7
.793
0.96
30.
140
0.14
00.
683
0.41
20.
412
0.82
3-3
479.
6-1
489.
859
.3-2
266.
7-3
120.
5-1
22.3
320
1.5
0.14
00.
140
1.22
00.
680
0.68
01.
360
-109
3.1
-534
4.2
14.8
-103
7.3
-577
.9-4
.458
91.
50.
140
0.14
01.
220
0.68
00.
680
1.36
030
7.0
32.7
787.
8-2
560.
7-6
252.
672
6.9
Att
hiss
tage
each
laye
rmay
bede
signe
dby
appl
ying
the
follo
win
geq
uatio
ns(
=45
°):
ccd
tf
tv
sin
Sd cos
2sa
fety
verif
icat
ion
onco
ncre
tesid
e
n Rdx
=n S
dx+
v Sd
cot
requ
ired
resis
tanc
eal
ong
xdi
rect
ion
nn
Rdy
Sdy
Sdv cot
requ
ired
resi
stanc
eal
ong
ydi
rect
ion
from
whi
ch,i
fres
ults
atis
fied,
the
rein
forc
emen
tare
asm
aybe
calc
ulat
edas
:
Asx
n fRdx
yd
;A
sy
n fRdy
yd
Ifco
ncre
test
reng
thre
quire
men
tis
not
satis
fied,
anin
crea
sela
yer
thic
knes
ssh
allb
epr
ovid
edun
tilve
rific
atio
nis
met
;in
this
case
new
valu
esfo
rth
ela
yer
actio
nha
ving
chan
ged
zval
ue.
Itca
nbe
notic
eth
atif
n Rdx
orn R
dyva
lue
are
nega
tive,
aco
mpr
essio
nfo
rce
ispr
esen
tal
ong
that
dire
ctio
nan
dno
rein
forc
emen
tis
requ
ired;
ifbo
thth
en R
dxan
dn R
dyar
ene
gativ
eit
ispo
ssib
leto
omit
the
rein
forc
emen
tin
both
the
dire
ctio
nsbu
t,in
this
case
the
verif
icat
ion
ispe
rform
edal
ong
the
prin
cipa
lco
mpr
essio
ndi
rect
ion
inth
eco
ncre
tesu
bjec
ted
tobi
axia
lcom
pres
sion
and
the
chec
king
equa
tion
is:
cSd
xSd
ySd
xSd
yt
nn
nn
vt
24
2
Sd2cd
1f
Fort
heco
nsid
ered
elem
ents,
one
obta
ins:
Tabl
e6.
14
EC2�
wor
ked
exam
ples
6-61
Tabl
eof
Cont
ent
Top
Laye
rDes
ign
Bot
tom
Laye
rDes
ign
Elem
.c
f cd1
/2A
syA
sxc
f cd1
/2A
syA
sx
[MPa
][k
N/m
][c
m2 /m
][c
m2 /m
][M
Pa]
[kN
/m]
[cm
2 /m]
[cm
2 /m]
648
-35.
0-1
7.1
0.0
0.0
-30.
0-1
7.1
0.0
0.0
93-2
4.9
-17.
10.
00.
0-2
2.4
-17.
10.
00.
032
0-3
8.2
-17.
10.
00.
0-7
.4-1
7.1
0.0
0.0
589
-11.
3-1
2.0
25.2
18.9
-45.
6-1
7.1
0.0
0.0
Itca
nbe
notic
eth
atve
rific
atio
nfo
rcon
cret
ein
com
pres
sion
isno
tsat
isfie
dfo
rany
laye
rsex
cept
for
elem
ent
589
top
laye
ran
del
emen
t320
botto
mla
yer.
Thus
,an
incr
easin
gof
laye
rthi
ckne
ssis
requ
ired
and
new
valu
esof
plat
eac
tions
are
obta
ined
:Ta
ble
6.15
Elem
.h
t st i
t cy s
y iz
nSd
y,s
nSd
x,s
v Sd,
sn
Sdy,
in
Sdx,
iv S
d,i
[m]
[m]
[m]
[m]
[m]
[m]
[m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
648
1.5
0.30
00.
240
0.96
00.
600
0.63
01.
230
-716
.6-5
040.
814
.1-1
062.
0-4
054.
7-8
.893
0.96
30.
220
0.19
00.
553
0.37
20.
387
0.75
8-3
588.
5-1
465.
566
.5-2
157.
8-3
144.
8-1
29.4
320
1.5
0.34
00.
140
1.02
00.
580
0.68
01.
260
-117
9.8
-576
8.3
16.0
-950
.6-1
53.8
-5.5
589
1.5
0.14
00.
430
0.93
00.
680
0.53
51.
215
708.
787
0.1
794.
7-2
962.
5-7
090.
072
0.0
whi
chle
adto
the
follo
win
gva
lues
:Ta
ble
6.16
Top
Laye
rDes
ign
Bot
tom
Laye
rDes
ign
Elem
.c
f cd1
/2A
syA
sxc
f cd1
/2A
syA
sx
[MPa
][k
N/m
][c
m2 /m
][c
m2 /m
][M
Pa]
[kN
/m]
[cm
2 /m]
[cm
2 /m]
648
-16.
8-1
7.1
0.0
0.0
-16.
9-1
7.1
0.0
0.0
93-1
6.3
-17.
10.
00.
0-1
6.6
-17.
10.
00.
032
0-1
7.0
-17.
10.
00.
0-6
.8-1
7.1
0.0
0.0
589
-11.
4-1
2.0
34.6
38.3
-16.
8-1
7.1
0.0
0.0
Ofc
ours
e,m
inim
ava
lues
shou
ldbe
adop
ted
forA
sxan
dA
syif
nore
info
rcem
enta
reas
are
requ
ired.
For
elem
ent5
89,t
heA
sxan
dA
syva
lue
are
requ
ired
atth
ece
ntro
idof
the
laye
r,w
here
asth
eyar
ear
rang
edat
0.07
mfr
omth
eex
tern
alsu
rfac
eof
the
slab
inan
ecce
ntric
posit
ion
with
resp
ect
tom
iddl
epl
ane
ofth
ela
yer;
so,
the
amou
ntof
rein
forc
emen
tpro
vide
dha
sto
bech
ange
dto
resto
reeq
uilib
rium
cond
ition
s.Th
isva
riatio
nm
aybe
asse
ssed
with
the
aid
ofth
em
echa
nism
pict
ured
inFi
g.6.
51:
EC2�
wor
ked
exam
ples
6-62
Tabl
eof
Cont
ent
Fig
.6.5
1Sh
elle
lem
ente
quili
briu
min
one
dire
ctio
nwi
thtw
ore
info
rcem
entl
ayer
sonl
y
The
new
forc
esac
ting
onth
ere
info
rcem
ents
beco
me:
nn
ht
bn
tb
zs
Sds
si
Sdi
ii
,,
22
n i=
n Sd,
s+
n Sd,
in s
Fort
hein
vesti
gate
del
emen
ts,t
hefo
llow
ing
area
shav
ebe
ende
tect
ed.
Tabl
e6.
17
Forc
esre
ferr
edto
tens
ion
steel
leve
lTo
pla
yerr
einf
Botto
mla
yerr
einf
Elem
.n
s,y
ni,y
ns,
xn
i,xA
syA
sxA
syA
sx
[kN
/m]
[kN
/m]
[kN
/m]
[kN
/m]
[cm
2 /m]
[cm
2 /m]
[cm
2 /m]
[cm
2 /m]
648
-702
.5-1
070.
8-5
026.
6-4
063.
60.
00.
00.
00.
093
-352
2.0
-228
7.2
-139
9.0
-327
4.3
0.0
0.0
0.0
0.0
320
-116
3.8
-956
.1-5
752.
3-1
59.3
0.0
0.0
0.0
0.0
589
1503
.5-2
242.
516
64.8
-637
0.0
34.6
38.3
0.0
0.0
The
prev
ious
proc
edur
esh
ould
bere
peat
edfo
rall
the
elem
ents
ofth
estr
uctu
ralm
odel
findi
ngth
eam
ount
ofre
info
rcem
ent
topr
ovid
ein
the
slab;
itis
usef
ul,
toco
ntro
lth
estr
uctu
ralb
ehav
iour
and
fora
best
fitte
dre
info
rcem
entl
ayou
t,to
sum
mar
iseth
ere
sults
ina
visu
alm
ap.
Veri
ficat
ion
toBu
rstin
gFo
rce
For
the
calc
ulat
ion
ofth
ebu
rstin
gfo
rce
the
sym
met
ricpr
isman
alog
yis
used
,eva
luat
ing
the
heig
htof
the
prism
soth
athi
sce
ntro
idre
sults
coin
cide
ntw
ithth
ece
ntro
idof
pres
tress
ing
tend
ons.
For
the
sake
ofsi
mpl
icity
,on
lyth
elo
ngitu
dina
ldi
rect
ion
ofpr
estre
ssin
gte
ndon
isco
nsid
ered
with
resp
ectt
oth
eve
rtica
lpla
ne,b
uttra
nsve
rse
forc
edu
eto
burs
ting
effe
ctsh
ould
beal
soca
lcul
ated
inth
eho
rizon
tal
plan
ean
dfo
rtra
nsve
rse
pres
tress
ing
too.
EC2�
wor
ked
exam
ples
6-63
Tabl
eof
Cont
ent
Fig
.6.5
2G
eom
etric
dim
ensio
nfo
rbur
stin
gca
lcul
atio
n
Chec
king
situ
atio
nis
repr
esen
ted
inFi
g.6.
52,
and
the
mos
tun
favo
urab
lesit
uatio
noc
curs
whe
na
singl
ete
ndon
iste
nsio
ned;
cons
ider
ing
the
low
erle
vel
ofte
ndon
(firs
tte
nsio
ned)
the
heig
htof
the
prism
resu
lts:
h bs
=2
0.6
=1.
2m
and
hisl
engt
h,fo
rend
anch
ored
tend
on,i
s:l bs
=h b
s=
1.2
mw
hile
the
wid
thfo
llow
sfr
omth
epo
ssib
leen
larg
emen
tof
the
anch
orpl
ate
that
may
beas
sum
edeq
ualt
o0.
43m
,cor
resp
ondi
ngto
the
trans
vers
esp
acin
gof
long
itudi
nall
ower
tend
ons.
The
desig
nfo
rce
pert
endo
nha
sbee
nev
alua
ted
bym
eans
ofth
efo
llow
ing
expr
essio
n:
Ff .
ASd
ptk
sp11
518
0011
513
919
103
.41
34kN
The
burs
ting
forc
efo
llow
sfro
mth
em
omen
tequ
ilibr
ium
alon
gse
ctio
nA
-A:
Nz
Fbs
bsSd
05
12
21
11
.(
)n
nt
nt
=85
2.6
kN
whe
re:
t 1=
0.07
5m
dista
nce
betw
een
the
cent
roid
ofte
ndon
sabo
vese
ctio
nA
-Ato
the
cent
roid
ofth
epr
ism
;
t 2=
0.30
0m
dista
nce
betw
een
the
cent
roid
ofco
ncre
test
ress
bloc
kab
ove
sect
ion
A-A
toth
ece
ntro
idof
the
prism
;
EC2�
wor
ked
exam
ples
6-64
Tabl
eof
Cont
ent
n 1,n
2nu
mbe
rsof
tend
ons
abov
ean
dbe
low
sect
ion
A-A
,re
spec
tivel
y:co
nsid
erin
gth
ean
chor
plat
eas
rigid
ava
lue
of0.
5m
aybe
assu
med
;
1=1
.1su
pple
men
tary
parti
alsa
fety
fact
orag
ains
tove
rstre
ssin
gby
over
pum
ping
.Bu
rstin
gfo
rce
shal
lbe
resis
ted
byan
area
ofre
info
rcem
ents
teel
of:
As,b
s=
Nbs
/fyd
=19
.61
cm2
distr
ibut
edw
ithin
l bs/3
tol bs
,(i.e
.fro
m0.
40m
to1.
20m
)fro
mth
ean
chor
plat
e.Th
usth
eef
fect
ive
area
ona
met
erle
ngth
,may
befo
und
byth
efo
llow
ing:
A ss
A bs
bss
bs,
,.
.43
..4
2 3
1961
012
00
l bs
57.0
cm2 /m
2
that
may
bepr
ovid
edw
ithtie
sha
ving
diam
eter
of22
mm
and
spac
ing
both
trans
vers
ally
and
long
itudi
nally
of25
0m
m(s
eeFi
g.53
);in
fact
22/2
525
corr
espo
nds
to60
.82
cm2 /m
2.
Fig
.53
Burs
ting
rein
forc
emen
tarr
ange
men
t
Veri
ficat
ion
tosp
allin
gfo
rce
The
spal
ling
forc
em
aybe
calc
ulat
edw
ithth
eeq
uiva
lent
pris
man
alog
y.A
sfo
rbu
rstin
gve
rific
atio
n,on
lyth
elo
ngitu
dina
ldi
rect
ion
isco
nsid
ered
;fu
rther
mor
e,sp
allin
gef
fect
sar
iseif
uppe
rten
don
are
tens
ione
dfir
stly
(the
ecce
ntric
ityle
adst
ote
nsio
nst
ress
es).
Thus
,a
sect
ion
with
abr
eadt
hof
0.43
man
da
heig
htof
1.50
mha
sto
beve
rifie
dfo
rone
tend
onte
nsio
ning
.Th
ele
ngth
ofth
epr
ismfo
rend
anch
ored
tend
on,i
seq
ualt
oth
eov
eral
lhei
ghto
fth
ese
ctio
n,i.e
.l sl
=1.
50m
.C
onsid
erin
gan
ecce
ntric
ityfo
rup
per
pres
tress
ing
tend
onof
0.35
m,t
heex
trem
estr
esse
satt
heen
dof
pris
mle
ngth
are
calc
ulat
edby
mea
nsof
the
beam
theo
ry;f
ora
pres
tress
ing
forc
eF S
d=
4134
.0kN
they
resu
lt(n
egat
ive
ifco
mpr
essiv
e):
top
bott
MPa
MPa
om.4
3.
..4
3.
. .F Sd
10
150
035
60
150
1538
256
2
The
sect
ion
alon
gw
hich
nosh
earf
orce
resu
lts,i
spla
ced
at0.
428
mfro
msl
abbo
ttom
fibre
(see
Fig
54)a
ndth
em
omen
tfor
equi
libriu
mtu
rnso
ut:
EC2�
wor
ked
exam
ples
6-65
Tabl
eof
Cont
ent
Msl
bott
2 30
214
010
23
om.
.43
33.6
1kN
m
Fig
.54
Cal
cula
tion
sche
me
fors
palli
ng
Ass
umin
gz s
l=0.
5l sl
and
b sl=
0.43
m,t
hem
axim
umsp
allin
gfo
rce
turn
sout
:N
sl=
Msl
/zsl
=44
.81
kND
isreg
ardi
ngan
yco
ncre
tete
nsile
resis
tanc
e,th
eam
ount
ofre
info
rcem
enti
s:
As=
Nsl
/fyd
=1.
031
cm2
plac
edpa
ralle
lto
the
end
face
inits
clos
evi
cini
ty.
EC2�
wor
ked
exam
ples
6-66
Tabl
eof
Cont
ent
EC2�
wor
ked
exam
ples
7-1
Tabl
eof
cont
ent
SEC
TIO
N7.
SER
VIC
EA
BIL
ITY
LIM
ITST
AT
ES
–W
OR
KE
DE
XA
MPL
ES
EX
AM
PL
E7.
1E
valu
atio
nof
serv
ice
stre
sses
[EC
2cl
ause
7.2]
Eva
luat
eth
eno
rmal
com
pres
sive
forc
ean
dof
the
asso
ciat
edbe
ndin
gm
omen
tin
the
sect
ion
ofFi
gure
7.1,
with
the
boun
dary
cond
ition
s
a)c c
2ck
(yh)
0;(y
0)k
f;
b)c
1ck
s3
sk
(y0)
kf
(yd)
kf
Then
,eva
luat
eth
em
ater
ials
stra
insf
rom
the
stre
sses
c)
N0
=-8
00kN
;M0
=40
0kN
m.
Fina
lly,c
alcu
late
the
coup
lesM
,Nas
soci
ated
toth
eth
ree
path
sd)
M/N
=-0
.5m
;e)N
=N
0=
-800
kN;f
)M=
M0
=40
0kN
m,t
hat,
linea
rlych
angi
ngM
N,o
rM
,N,r
espe
ctiv
elyw
ithco
nsta
ntno
rmal
forc
eor
cons
tant
bend
ing,
keep
the
sect
ion
toth
eul
timat
ete
nsio
nst
ate
unde
rloa
d.
Fig.
7.1.
Rect
angu
lars
ection
,cal
cula
tion
ofser
vices
tresse
s.
The
follo
win
gda
taar
egi
ven:
f ck=
30M
Pa,f
yk=
450
MPa
,e
=15
.
Cons
ider
ing
Fig.
7.1,
we
have
d=
550
mm
;d�=
50m
m;h
=60
0m
mA
s=6
314=
1884
mm
2 ;=
1
The
boun
dary
cond
ition
sfr
omth
efir
stex
ercis
ese
tth
ene
utra
laxi
son
the
bord
erof
the
botto
mse
ctio
n,th
atis
y n=
h.Fo
rthi
svalu
eit
resu
lts3
22
2
2
600
400
400
600
300
1518
8450
550
12e
300
600
400
1518
84(
600)
2
and
then
e=
-120
.65
mm
,*
63
ynS88
.96
10m
m.
The
seco
ndco
nditi
onin
the
first
exer
cise,
assu
min
gk 2
=0.
45,c
anbe
writ
ten
as
6
N(
600)
0.45
3088
.96
10
EC2�
wor
ked
exam
ples
7-2
Tabl
eof
Cont
ent
and
then
N=
-200
1.16
kN,
M=
Ne=
-200
1.16
(-120
.65)
10-3
=24
1.49
kNm
.Th
ete
nsio
nst
ress
post
ulat
edby
the
seco
ndex
erci
segi
ves
the
follo
win
gex
pres
sion
for
the
neut
rala
xis
n3
yk
e2
ck
d55
0y
235.
71m
mk
f0.
845
01
115
0.6
30k
f
and
the
com
pres
sed
stee
lten
sion
and
the
stre
ssco
mpo
nent
sare
'n
ss
ss
n
d'y
5023
5.7
0.59
dy
550
235.
7
N=
-0.6
3040
023
5.7/
2+0.
845
018
84(1
-0.5
9)10
-3=
-570
.48
kN
M=
(-0.6
3040
023
5.7/
2(2
35.7
/3-3
00)+
0.8
450
1884
(1-0
.59)
250)
10-6
=45
7.5
kNm
e=
-457
.5·1
06 /570
.48·
103
=-8
01.9
5m
mCo
nsid
erin
gth
eth
irdex
erci
se
340
0e
1050
0mm
800
and
22
3 nn
n
n2 n
n
400
y15
1884
550
y50
y3
y20
040
0y
1518
8460
02
y2
this
equa
tion
isite
rativ
elyso
lved
:
y n=
272.
3m
m,
*3
3ynS
1326
310
mm
and
then
the
tens
iona
lsta
teis
3
c6
800
10(
272.
3)16
.42M
Pa13
.263
10
s16
.42
(550
272.
3)15
251.
18M
Pa27
2.3
' s16
.42
(50
272.
3)15
201.
07M
Pa27
2.3
Beca
use
the
cond
ition
e=
-500
mm
impl
iest
hatt
hene
utra
laxi
spos
ition
islo
wer
than
the
one
prev
ious
lyev
aluat
edas
sum
ing
the
max
imal
stre
sses
for
both
mat
erial
s,th
eul
timat
ete
nsio
nst
ate
corr
espo
nds
toth
em
axim
alte
nsio
nad
mitt
edfo
rcon
cret
e.If
we
cons
ider
toch
ange
M,
Nke
epin
gco
nsta
ntth
eec
cent
ricity
,the
tens
iona
lsta
tech
ange
prop
ortio
nally
and
we
can
stat
e
ck
00
c
0.6f
NM
1.09
6N
M.
Onc
eth
eco
ncre
teul
timat
eco
mpr
essiv
elim
itsta
teis
reac
hed,
the
stre
ssis
N=
1.09
6N0=
-876
.80
kN;M
=1.
096
M0
=43
8.4
kNm
.
EC2�
wor
ked
exam
ples
7-3
Tabl
eof
Cont
ent
Wor
king
with
cons
tant
norm
alfo
rce
(N=
N0)
the
ultim
ate
limit
stat
efo
rthe
conc
rete
tens
ion
lead
sto
0n
ck* yn
N(
y)
0.6f
S
and
then
n3
2 nn
y0.
630
400
800
10y
1518
8460
02
y2
Solv
ing
with
resp
ectt
oy n
2 nn
2n ' s sy
60y
8479
50
y30
.25
30.2
584
795
262.
51m
m(5
026
2.51
)0.
630
1521
8.57
MPa
262.
51(5
5026
2.51
)0.
630
1529
5.69
MPa
262.
51an
dth
en
640
026
2.51
262.
51M
0.6
3030
018
84(2
95.6
921
8.57
)25
010
442.
56kN
m2
36
3
442.
5610
e=-
553.
2mm
800
10K
eepi
ngco
nsta
ntth
ebe
ndin
gm
omen
t(M
=M
0),th
elim
itst
ate
cond
ition
for
the
conc
rete
stre
ssis n
ck* yn
N(
y)
0.6f
San
dth
en0
0n
*ck
yn
MM
(y)
eN
0.6f
S
and * yn
0n
n*
*yn
ckyn
IM
(y)
hy
S0.
6f
S2
As
66
30
ck
M40
010
22.2
210
mm
0.6
f0.
630
the
prev
ious
num
eric
form
beco
mes
22
36
nn
nn
n2 n
n
400
y15
1884
550
y50
y22
.22
10y
3y
300
400
y15
1884
600
2y
2an
dite
rativ
ely
solv
ing
EC2�
wor
ked
exam
ples
7-4
Tabl
eof
Cont
ent
n ' s sy39
5mm
(50
395)
0.6
3015
235.
82M
Pa39
5(5
5039
5)0.
630
1510
5.95
MPa
395
6
400
395
N0.
630
1884
(105
.95
235.
82)
1666
.67k
N2
400
395
395
M0.
630
300
1884
(105
.95
235.
82)
250
1040
0.34
kNm
23
e=-2
40m
m
Figu
re7.
2re
ports
the
resu
ltsob
tain
edin
the
evalu
atio
nin
term
soff
orce
sand
stres
ses.
Fig.
7.2.
Resu
ltsfor
diffe
rentl
imit
distr
ibut
ionso
fstre
sses.
Asa
rem
ark,
just
inth
eca
sec)
the
conc
rete
tens
ion
limit
stat
eun
derl
oad
isno
trea
ched
whi
lein
the
case
a)(k
1=0.
45)a
ndth
eot
herc
ases
b)d)
e)f)
(k1=
0.65
)res
pect
ivel
yre
ach
the
tens
ion
ultim
ate
stat
esun
der
load
asso
ciat
edto
non
linea
rvi
scos
ityph
enom
ena
and
min
imal
tens
ion
inth
epr
esen
ceof
parti
cular
com
bina
tions
.O
nth
eot
her
hand
,th
ete
nsio
nul
timat
est
ate
unde
rloa
dfo
rtie
dst
eeli
sgot
just
inth
eca
seb)
.2B
EC2�
wor
ked
exam
ples
7-5
Tabl
eof
Cont
ent
EX
AM
PL
E7.
2D
esig
nof
min
imum
rein
forc
emen
t[E
C2
clau
se7.
3.2]
Let�s
cons
ider
the
sect
ion
inFi
gure
7.3
with
the
follo
win
gge
omet
ry:
A=
1.92
5á10
6m
m2
;yG
=80
9m
m;I
=71
.82á
1010
mm
4 ;
Wi=
7.25á1
08m
m3 ;r
2=
I/A
=39
.35á
104
mm
2
Fig.
7.3.
Box
-sect
ion,d
esign
ofm
inim
umrei
nfor
cemen
t.
Eva
luat
eth
em
inim
umre
info
rcem
enti
nto
the
botto
msla
bin
the
follo
win
gca
ses:
App
licat
ion
ofth
efir
stcr
acki
ngm
omen
tMcr
App
licat
ion
ofan
axial
com
pres
sive
forc
eN
=-6
000
kN,a
pplie
din
the
poin
tPat
250
mm
from
the
botto
mbo
rder
ofth
eco
rres
pond
ing
crac
king
mom
ent.
Cons
ider
the
follo
win
gda
ta:
f ck=
45M
Pa;
f ct,ef
f=
3.8
MPa
;s=
200
MPa
;k
=0.
65(h
w>
1m)
The
give
nsta
tem
ents
impl
y:
s25
0/
1800
0.13
88
f30
0/
1800
0.16
67
sf
10.
6945
0 s,m
in0.
653.
8/20
00.
0123
5
Case
a)
The
appl
icat
ion
ofcr
acki
ngm
omen
tisa
ssoc
iate
dto
the
neut
rala
xisp
ositi
ony n
=y G
,and
then
809/
1800
0.44
94.
Itre
sults
also
sf
10.
4860
2an
dfo
rthe
web
s,m
in3
0.69
452(
10.
1667
0.44
94)
0.01
235
0.4
11
0.16
670.
4494
0.00
208
41
0.44
94
EC2�
wor
ked
exam
ples
7-6
Tabl
eof
Cont
ent
2s,
min
A0.
0020
830
018
0011
23m
m
this
rein
forc
emen
thas
tobe
puti
nth
ew
ebtie
dar
eaw
ithhe
ight
over
the
botto
msla
ba
=18
00�
809�
300
=69
1m
m.
We
use
(5+
5)12
mm
equi
vale
ntto
1130
mm
2 .
Refe
rrin
gto
the
botto
msla
bw
ege
t
f9
10.
812
8an
dit
follo
ws:
s,m
in2(
10.
4494
)0.
1667
0.01
235
0,45
0.00
943
10.
4494
2s,
min
A0.
0094
330
015
0042
43m
m
We
use
(14+
14)
14m
meq
uiva
lent
to43
12m
m2 .
The
rein
forc
emen
tsch
eme
isre
port
inFi
gure
7.4
Fig.
7.4.
Min
imum
reinf
orcem
ent,
case
(a).
Case
b)
The
crac
king
mom
enta
ssoc
iated
toth
eax
ialfo
rce
N=
-600
0kN
,with
ecce
ntric
itye N
=18
00-
809-
250
=74
1m
mde
rives
from
the
relat
ion
crN
ct,e
ffi
i
NA
M1
ef
WA
W
and
then
:3
68
6cr
68
6000
1074
11.
825
10M
13.
87.
2510
1095
85kN
m1.
825
107.
2510
the
ecce
ntric
ityof
the
norm
alfo
rce
inth
epr
esen
ceof
Mcr
isth
en:
3e
9585
1060
0074
185
6mm
and
the
neut
rala
xisp
ositi
onre
sults
from
the
rela
tion
EC2�
wor
ked
exam
ples
7-7
Tabl
eof
Cont
ent
24
nG
r39
.35
10y
y80
912
69m
me
856
,=
0.70
50
Cons
ider
ing
the
web
,with
*h1.
8h
we
dedu
ce:
s,m
in0.
6945
2(1
0.16
670.
7050
)0.
0123
50.
41
10.
1667
0.70
500.
0004
63
1.8(
10.
7050
)
2s,
min
A0.
0004
630
018
0024
8mm
We
use
410
equi
valen
tto
314
mm
2 .
The
bars
have
tobe
loca
ted
inth
etie
dpa
rtof
the
web
fora
nex
tens
ion
a=
1800
-126
9-30
0=
231
mm
over
the
botto
msla
b
Inth
ebo
ttom
slab
we
have
:
f9
10.
812
8an
dit
resu
lts
s,m
in2(
10.
705)
0.16
670.
0123
50,
450.
0079
71
0.70
52
s,m
inA
0.00
797
300
1500
3586
mm
We
use
(12+
12)
14m
meq
uiva
lent
to36
92m
m2 .
The
rein
forc
emen
tsch
eme
isre
porte
din
Figu
re7.
5
Fig.
7.5.
Min
imum
reinf
orcem
ent,
case
(b).
3B
EC2�
wor
ked
exam
ples
7-8
Tabl
eof
Cont
ent
EX
AM
PL
E7.
3E
valu
atio
nof
crac
kam
plitu
de[E
C2
clau
se7.
3.4]
The
crac
kw
idth
can
bew
ritte
nas
:
s,cr
sk
31
24
ss
s
w1
kc
kk
kE
(7.1
)
with
ss,
crt
ct,e
ffe
s
kf
1(7
.2a)
11
min
2.5
1;
;3
2(7
.3)
Ass
umin
gth
epr
escr
ibed
valu
esk 3
=3.
4,k 4
=0.
425
and
cons
ider
ing
the
bend
ing
case
(k2=
0.5)
with
impr
oved
boun
dre
info
rcem
ent(
k 1=
0.8)
,the
(7.2
)we
get
s,cr
sk
ss
s
w1
3.4
c0.
17E
(7.4
)
The
(7.4
)can
beim
med
iately
used
asve
rifica
tion
form
ula.
Asa
nex
ampl
ele
t�sco
nsid
erth
ese
ctio
nin
Figu
re7.
6
Fig.
7.6.
Rein
forced
conc
retes
ection
,cra
cksa
mplit
udee
valu
ation
assu
min
ge
=15
,d=
548m
m,
d�=
46.0
mm
,c=
40m
m,
b=40
0mm
,h=
600m
m,
M=
300k
Nm
,A
s=27
12m
m2
(624
),A
s�=45
2mm
2(4
12),
f ck=
30M
Pa,f
ct,ef
f=f ct
m=
2.9M
Pa
Refe
rrin
gto
ash
ortt
ime
actio
n(k
t=0.
6).
Itre
sults
then
=45
2/27
12=
0.16
7,=
548/
600=
0.91
3,�=
460/
600=
0.07
67,
s=27
12/(
400
600)
=0.
0113
And
the
equa
tion
fort
hene
utra
laxi
syn
is
2 nn
n40
0y
1527
1254
8y
0.16
746
y0
2an
dth
en
EC2�
wor
ked
exam
ples
7-9
Tabl
eof
Cont
ent
2 nn
2n
ny23
7.4y
1130
260
y23
7.8
y11
8.7
118.
711
3026
237.
8mm
,0.
3963
h60
0Th
ese
cond
orde
rmom
entr
esul
ts
n
22
*3
94
y40
0I
237.
815
2712
548
237.
80.
167
4623
7.8
5.96
10m
m3
and
we
dedu
ce6
9s
1530
010
548
237.
8/5
.96
1023
4MPa
the
valu
eto
bead
opte
dis
the
low
est
betw
een
2.5(
1-0.
913)
=0.
2175
;(1-
0.39
63)/
3=0.
2012
;0.
5.Th
en=
0.20
12.
The
adop
ted
stat
emen
tsle
adto
s,cr k
5
0.20
120.
0113
0.6
2.9
115
57.0
8MPa
0.01
130.
2012
234
57.0
824
w1
3.4
400.
170.
2012
0.18
4mm
210
234
0.01
13
4B
EC2�
wor
ked
exam
ples
7-10
Tabl
eof
Cont
ent
EX
AM
PL
E7.
4.D
esig
nfo
rmul
asde
riva
tion
fort
hecr
acki
nglim
itst
ate
[EC
2cl
ause
7.4]
8B7.4.
1E
xact
met
hod
Itis
inte
rest
ing
tode
velo
pth
e(7
.4)t
ous
eit
asa
desig
nfo
rmul
a.In
parti
cular
,sta
ted
b,h,
d,d�
,b,a
ndfix
edM
,we
wan
tto
dedu
ceth
em
etal
rein
forc
emen
tam
ount
As
and
itsde
sign
tens
ion
sin
orde
rto
have
acr
ack
ampl
itude
wk
low
erth
anth
efix
edva
lue
kw
.Th
ead
imen
siona
lcalc
ulus
lead
sto
2e
se
s1
1'
02
(7.5
)
ect
mt
s2
23
s
fk
23n
'(7
.6)
setti
ng
20 cr
tct
m
MM b
hM
kf
6(7
.7)
Ded
ucin
gsfr
om(7
.5)a
ndw
ithits
subs
titut
ion
inth
e(7
.6)w
ege
t2
se
21
'(7
.8)
32
e2
2
2p
3p
'1
'(7
.9)
with
p=s/
(ktf c
tm)
(7.1
0)
From
(7.4
),w
here
kw
w,a
fters
ome
calc
ulat
ions
we
dedu
ce0 k
s
s
wp
n3.
4c
0.17
(7.1
1)
setti
ngs
k0k
tct
m
Ew
wk
f(7
.12)
Com
bini
ng(7
.8)a
nd(7
.11)
,the
(7.9
)is
02
ee
ke
22
22
e
23
22
2w
'1
3.4c
0.34
'1
'
32
'1
'
EC2�
wor
ked
exam
ples
7-11
Tabl
eof
Cont
ent
(7.1
3)th
e(7
.13)
,num
eric
ally
solv
ed,a
llow
sth
ede
term
inat
ion
ofth
ene
utra
laxi
spo
sitio
nan
dth
en,
usin
gth
e(7
.11)
(7.8
),th
eev
alua
tion
ofth
ere
info
rcem
entt
ensio
nan
dits
amou
nt..
Ifit
isno
tth
eca
se,i
tis
nece
ssar
yto
set
inth
e(7
.13)
=2.
5(1-
)an
dth
enre
-eva
luat
ing
,bei
ngth
eva
lue
=0.
5pr
actic
ally
impo
ssib
lefo
rben
ding
prob
lem
s.Th
epr
oced
ure,
aimed
toth
ede
term
inat
ion
ofth
ere
info
rcem
ent
amou
ntan
dits
tens
ion
corr
espo
ndin
gto
fixed
crac
kam
plitu
deva
lues
and
stre
ssle
vel,
requ
irest
ose
tbef
ore
the
valu
eof
the
bars
diam
eter
.
Alte
rnat
ively
,it
ispo
ssib
leto
set
the
tens
iona
llev
els,
for
exam
ple
coin
cide
ntw
ithth
epe
rmiss
ible
one,
and
toev
aluat
eth
eco
rres
pond
ing
rein
forc
emen
tam
ount
san
dth
em
axim
alba
rsdi
amet
er.I
nth
isca
se,a
sth
epa
ram
eter
pis
defin
ed,t
hene
utra
laxi
sis
obta
ined
from
(7.9
),s
from
(7.7
)an
dth
em
axim
aldi
amet
erde
rives
from
(7.1
1)so
lved
with
resp
ect
to,
whi
chas
sum
esth
efo
rm:
ss
okm
axe
s
5.88
w2c
(p)
(7.1
4)
5B7.4.
2A
ppro
xim
ated
met
hod
The
appl
icat
ion
ofth
epr
oced
ure
disc
usse
dab
ove
isqu
itelab
orio
usas
itre
quire
sto
itera
tivel
yso
lve
the
(7.1
3).A
nal
tern
ativ
epr
oced
ure,
easie
rto
beap
plied
,con
sisti
nth
est
atem
entt
hat
the
leve
rar
mh 0
isco
nsta
ntan
din
depe
nden
tfro
man
deq
uiva
lentt
o0.
9d.I
nth
isw
ay,
we
have
sAs0
.9d=
Man
dth
en
s=0.
185
/(p
)(7
.15)
the
(7.4
)writ
ten
for
kw
wim
med
iatel
ygi
ves
se
sk
ss
s
w1
13.
4c0.
17E
p(7
.16)
aimin
gto
afu
rther
simpl
ifica
tion
ofth
epr
oble
m,l
et�s
state
=0.
9,=
0.24
3an
das
sum
ing
byde
finiti
on
*1.
181
10.
185
1c
u0k
2w
u(7
.17)
the
(7.1
6)af
ters
ome
algeb
raha
sthe
form
2e
1e
12
p5
*3.
4u0.
20p
*17
u5u
0*
(7.1
8)
the
(11.
67)i
sea
syto
solv
e,an
dto
geth
erw
ithth
e(7
.15)
and
(7.1
1),l
eads
toth
ede
sred
valu
esse
s.
Inth
isca
seto
o,,i
fwe
sett
heva
lue
ofs,
the
solu
tion
for(
7.18
)with
resp
ectt
olea
dsto
the
relat
ion
EC2�
wor
ked
exam
ples
7-12
Tabl
eof
Cont
ent
**
eok
max
*2
e
17c(
p)
5w
pp
(7.1
9)
that
defin
esth
em
axim
alba
rsdi
amet
er,w
hich
,,as
soci
ated
toth
ere
info
rcem
enta
mou
ntgi
ven
byth
e(7
.15)
,allo
wst
osa
tisfy
the
crac
king
ultim
ate
stat
eco
rres
pond
ing
toa
fixed
valu
eof
the
stee
lten
sion.
6B
EC2�
wor
ked
exam
ples
7-13
Tabl
eof
Cont
ent
EX
AM
PL
E7.
5A
pplic
atio
nof
the
appr
oxim
ated
met
hod
[EC
2cl
ause
7.4]
Let�s
use
the
desc
ribed
proc
edur
eto
the
sect
ion
inFi
gure
7.7.
Fig.
7.7.
Rein
forced
conc
reteS
ection
,rein
forcem
entd
esign
forth
ecra
ckin
gulti
mate
state.
Ass
umin
gb=
1000
mm
;h=
500m
m;c
=50
mm
;=
26m
m;f
ck=
33M
Pa;M
=60
0kN
m,d
esig
nth
ese
ctio
nto
have
acr
ack
ampl
itude
kw
0.30
mm
,k
kw
0.20
mm
,w
0.10
mm
.
Itre
sults
f ctm=
0.3. 33
2/3 =
3.08
6MPa
=(5
00-6
3)/5
00=
0.87
4M
0 cr=
0.6. 3.
086. (1
000. 50
02 /6). 10
-6=
77.1
5kN
m(s
eeex
.7.1
)
=60
0/77
.15=
7.77
*=7.
77/(
1-1.
18/7
.77)
=9.
16u 1
=50
/26=
1.92
Def
ined
max
kw
0.30
mm
the
max
imal
ampl
itude
,in
the
thre
eca
ses
unde
rexa
min
atio
nw
eca
nse
tm
axk
kw
ww
kw
here
k w=
1;2/
3;1/
3.Th
enin
age
nera
lfor
m5
0kw
w0.
32
10w
k32
404
k0.
63.
086
2w
w32
404
uk
1246
k26
2w
7.77
0.20
15p
59.
163.
41.
92p
9.16
1715
1.92
512
46k
09.
167.
77
and
then
2w
p23
5.93
p44
8557
067
k0
and
then
ww
pk
117.
965
1840
0.74
5706
7k
Usin
gth
epr
evio
usre
latio
n,to
geth
erw
ithth
e(7
.15)
and
(7.1
0),
inth
eth
ree
case
she
reco
nsid
ered
k w=
1,k
w0.
3m
m,
p(1)
=15
6.75
s0.
185
7.77
10.
0104
915
6.75
0.87
4,
As(1
)=0.
0104
9á50
0á10
00=
5245
mm
2
s(1
)=0.
6á3
.086á1
56.7
5=
290
MPa
k w=
2/3,
kw
0.2
mm
,p
(2/3
)=11
9.62
s0.
185
7.77
23
0.01
375
119.
620.
874
,A
s(2
/3)=
0.01
375á
500á
1000
=68
75m
m2
EC2�
wor
ked
exam
ples
7-14
Tabl
eof
Cont
ent
s(2
/3)=
0.6á
3.08
6á11
9.62
=22
1M
pa
k w=
1/3,
kw
0.1m
m,
p(1
/3)=
75.4
8
s0.
185
7.77
13
0.02
179
75.4
80.
874
,A
s(1
/3)=
0.02
179á
500á
1000
=10
895
mm
2
s(1
/3)=
0.6á3
.086á7
5.48
140
MPa
The
thre
ese
ctio
nsar
ere
porte
din
Figu
re7.
7;th
em
etal
area
sar
eov
eres
timat
ed,a
nd26
mm
diam
eter
bars
are
used
.
Let�s
verif
yth
ead
opte
dde
sign
met
hod
inor
der
toev
aluat
edits
prec
ision
.Th
efo
llow
ing
resu
ltsar
eob
tain
ed:
k w=
1,s=
5310
/(50
0á10
00)=
0.01
062
2 nn
1000
y15
5310
437
y0
22 n
ny
159.
3y69
614
0
2ny
79.6
579
.65
6961
419
5.9
mm
,=
0.39
18
n
32
94
y10
0019
5.9
I15
5310
437
195.
97.
1310
mm
3
6s
9
437
195.
915
600
1030
4M
Pa7.
1310
Fig.
7.8.
Desi
gned
sectio
ns.
k w=
1k
w0.
3m
m
s=29
0M
Pa
k w=
2/3
kw
0.2
mm
s=22
1M
Pa
k w=
1/3
kw
0.1m
m
s=
140
MPa
EC2�
wor
ked
exam
ples
7-15
Tabl
eof
Cont
ent
The
low
estv
alue
for
hast
obe
chos
enbe
twee
n
=2.
5(1�
0.87
4)=
0.31
5;=
(1�
0.39
18)/
3=
0.20
27
Then
s,cr
0.20
270.
0106
20.
63.
086
115
63.1
1M
Pa0.
0106
20.
2027
k5
304
63.1
10.
2027
w1
3.4
500.
1726
0.30
6m
m30
40.
0106
22
10
k w=
2/3
,s=
69.0
3/
(50á1
00)=
0.01
38
2 nn
1000
y15
6903
437
y0
22 n
ny
207.
1y
9049
80
2ny
103.
510
3.5
9049
821
4.6
mm
,=
0.42
92
n
32
94
y10
0021
4.6
I15
6903
437
214.
68.
4110
mm
3
6s
9
437
214.
615
600
1023
8M
Pa8.
4110
=(1�
0.42
92)/
3=
0.19
03
s,cr
0.19
030.
0138
0.6
3.08
61
1553
.31
MPa
0.01
380.
1903
k5
238
53.3
10.
1903
w1
3.4
500.
1726
0.21
3m
m2
1023
80.
0138
k w=
1/3,
s=
111.
51/
(50á1
00)=
0.02
23
2 nn
n10
00y
1595
5843
7y
1593
385
y0
22 n
ny
334.
5y
1437
040
2ny
167.
216
7.2
1437
0424
7.1m
m,
=0.
494
n
32
29
4y
1000
247.
1I
1595
5843
724
7.1
1515
9338
524
7.1
1.06
10m
m3
6s
9
437
247.
115
600
1016
0M
Pa1.
0610
=(1�
0.49
4)/
3=
0.16
87
EC2�
wor
ked
exam
ples
7-16
Tabl
eof
Cont
ent
s,cr
0.16
870.
0223
0.6
3.08
61
1541
.78
MPa
0.02
230.
1687
k5
160
41.7
80.
1687
w1
3.4
500.
1726
0.12
mm
210
160
0.02
23
The
obta
ined
valu
esar
ein
good
agre
emen
twith
thos
eev
alua
ted
with
inth
ede
sign.
The
valu
esfr
omth
eve
rific
atio
nar
esli
ghtly
large
rbe
caus
eof
the
fact
that
inth
eco
nsid
ered
sect
ion
the
inte
rnal
driv
elev
erar
mis
low
erth
anth
eap
prox
imat
edva
lue
0.9d
assu
med
inth
eap
prox
imat
edde
sign
proc
edur
e.In
fact
,bein
gh 0
/dth
ead
imen
siona
llev
erar
min
units
ofef
fect
ive
heig
htd,
inth
eth
ree
case
we
have
k w=
1h 0
/d=
(43.
70�
19.5
9/3)
/43
.70
=0.
85k w
=2/
3h 0
/d=
(43.
70�
21.4
6/3)
/43
.70
=0.
836
k w=
1/3
h 0/d
=[(1
8á16
0á18
.99
+3á
160á
13.7
92 /18.
99)/
(18á
160
+3á
160á
13.7
9/18
.99)
+
+2/
3á24
.71]
/43
.70
0.8
Let�s
rem
ark
that
the
pres
ence
ofa
com
pres
sed
rein
forc
emen
tis
high
lyre
com
men
ded
tom
ake
duct
ileth
ese
ctio
nin
the
ultim
ate
limit
stat
e.Th
ere
info
rcem
enti
ncre
ase
the
leve
rarm
ofth
ese
ctio
nre
duci
ngth
edi
ffere
nce
betw
een
the
appr
oxim
ated
valu
esan
dth
ose
com
ing
from
the
verif
icat
ion.
The
appr
oxim
ated
met
hod
prev
ious
lydi
scus
sed
can
besu
cces
sful
lyap
plie
din
the
desig
nof
the
ultim
ate
crac
kst
ate.
The
obta
ined
resu
ltsar
ere
porte
din
the
Tabl
es7.
1an
d7.
2an
dth
eyar
esh
own
inFi
gure
7.9.
Tabl
e7.
3an
dFi
gure
7.10
repo
rtnu
mer
icalv
alues
and
grap
hsfo
rth
em
axim
aldi
amet
eran
dth
ere
quire
dre
info
rcem
ent
expr
esse
das
afu
nctio
nof
fixed
valu
esfo
rs.
Stat
ing
asu
itabl
epr
ecisi
onfo
rthe
appr
oxim
ated
met
hod,
thos
eva
lues
are
evalu
ated
usin
gth
e(7
.16)
(7.1
4).
Tab
le7.
1.A
ppro
xima
tedm
ethod
.T
able
7.2.
Exa
ctme
thod
.
wk(m
m)
As(m
m2 )
s(M
Pa)
h 0/d
As(m
m2 )
wk
(mm
)s(M
Pa)
h 0/d
0.1
1115
114
00.
911
151
0.12
016
00.
811
0.2
6903
221
0.9
6903
0.21
323
80.
836
0.3
5310
190
0.9
5310
0.30
630
40.
85
Fig.
7.9.
Com
paris
onbe
tween
thee
xact
and
appr
oxim
ated
meth
ods.
EC2�
wor
ked
exam
ples
7-17
Tabl
eof
Cont
ent
Tab
le7.
3.A
ppro
xim
ated
meth
od�
Dete
rmin
ation
ofma
ximu
mdi
amete
r.
wk
=0.
1m
m(A
)w
k=
0.2
mm
(B)
w k=
0.3
mm
(C)
s(M
Pa)
max
(mm
)A
s(m
m2 )
s(M
Pa)
max
(mm
)A
s(m
m2 )
s(M
Pa)
max
(mm
)A
s(m
m2 )
137
3011
111
214
3070
0128
030
5430
140
2611
151
221
2669
0329
026
5310
145
2010
486
233
2065
0830
920
4910
149
1610
205
243
1662
4532
516
4672
156
1097
5026
110
5816
355
1042
82
Fig.
7.10
.Dia
grams
forM
axim
aldi
amete
r(m
ax)�
Meta
larea
(As)�
Steel
tensio
n(
s).
EC2�
wor
ked
exam
ples
7-18
Tabl
eof
Cont
ent
EX
AM
PL
E7.
6V
erifi
catio
nof
limit
stat
eof
defo
rmat
ion
Eva
luat
eth
eve
rtica
ldisp
lacem
enti
nth
em
id-s
pam
ofth
ebe
amin
Figu
re7.
11w
ithco
nsta
nttra
nsve
rsal
sect
ion
repr
esen
ted
inFi
gure
7.12
Fig.
7.11
.defl
ected
beam
,defo
rmat
ionlim
itsta
te.Fi
g.7.
12.T
rans
versa
lsect
ion.
Ass
ume
the
follo
win
gva
lues
fort
hem
ainpa
ram
eter
s
f ck=
30M
Pa;g
+q=
40kN
/m;g
=2q
;l=
10m
;As=
3164
mm
2(7
24);
and
solv
eth
epr
oblem
first
lyin
acu
mul
ativ
ew
ay,s
tatin
ge
=E s
/Ec=
15.
Refe
rrin
gto
the
stag
eI,
asin
dica
ted
inFi
gure
7.13
,*
2
* G
32
*2
94
I
9*
73
i
A70
050
015
3164
3974
60m
m70
050
035
015
3164
650
y38
5.8m
m39
7460
500
700
I50
070
035
.815
3164
650
385.
818
.05
10m
m12 18.0
510
W5.
745
10m
m70
038
5,8
From
Tabl
e[3
.2-E
C2]
we
get
2 3ct
mf
0.30
302.
9MPa
and
then
the
crac
king
mom
entr
esul
ts*
76
crct
mi
Mf
W2.
95.
745
1010
166.
6kN
m
Cons
ider
ing
the
who
leap
plied
load
then
2
max
4010
M50
0kN
m8
max cr
M50
03
M16
6.6
Fig.
7.13
.Se
ction
atsta
geI.
Inth
est
age
II,a
srep
orte
din
Figu
re7.
13,
EC2�
wor
ked
exam
ples
7-19
Tabl
eof
Cont
ent
2 nn
2 nn
2n
32
*10
4II
y50
015
3164
650
y0
2y
189.
84y
1233
960
y94
.92
94.9
212
3396
269m
m
269
I50
015
3164
650
269
1.01
10m
m3
then
c=18
.05/
10.1
3=1.
78
Fig.
7.14
.Sect
ionat
stage
II.
The
evalu
atio
nof
the
mid
dle-
spam
disp
lacem
entc
anbe
easil
yob
tain
edus
ing
the
relat
ion
(7.1
)he
reex
pres
sed
as
II
vl2
vl
2v
l2
1v
l2(7
.20)
whe
rev I
isth
edi
splac
emen
tca
lcul
ated
inth
efir
stst
epan
dv(
l/2)
the
incr
ease
ofth
edi
splac
emen
tits
elfc
ause
dfr
omth
ecr
acki
ng,t
hatc
anbe
expr
esse
dfo
rsym
met
ryre
ason
11
11
22
Mm
axcr
22
M2
cI
max
fM
Mz
v2
c1
fg
dd
,=
2E
IM
gl
(7.2
1)
whe
reE
cis
assu
med
tobe
Ec
=E
s/15
inag
reem
entw
ithth
ein
trodu
ced
stat
emen
tfo
rth
epa
ram
eter
e,.
Def
inin
gth
epa
ram
eter
=M
max
/Mcr
and
cons
ider
ing
that
f M(
)=
/2,g
()
=4(�
2 ),th
eeq
uatio
n(7
.21)
isw
ritte
nas 1
1
11
22
3m
ax2
22
cI
Md
vc
14
d2
EI
41
(7.2
2)
Calcu
latin
gth
ein
tegr
also
nth
erig
htsid
eof
the
equa
tion
we
finall
yob
tain
24
3m
ax1
11
2c
I
M5
4v
c1
ln2
12
EI
483
4(7
.23)
The
absc
issa
1,w
here
the
crac
ked
part
ofth
ebe
amst
art,
isgi
ven
solv
ing
the
equa
tion
2cr
11
max
M1
4M
(7.2
4)
EC2�
wor
ked
exam
ples
7-20
Tabl
eof
Cont
ent
and
then
11
11
2(7
.25)
Fina
lly,c
onsid
erin
gth
at2
max
Ic
I
M5
v48
EI
(7.2
6)
The
(7.2
0)is
expr
esse
das
24
3m
ax1
11
2c
I
M5
484
12v
1c
11
ln2
12
48E
I5
35
(7.2
7)
Ifth
eva
lue
ofc
prev
ious
lyca
lculat
edis
inse
rted
inth
e(7
.27)
stat
ing
=1
and
lettin
gch
angi
ngin
the
rang
e1
,we
obta
inth
ecu
rves
repo
rted
inFi
gure
7.15
,tha
tsho
was
the
incr
ease
ofth
era
tiom
eans
ade
crea
sefo
r1
and
the
incr
ease
ofv(
l/2)
asa
cons
eque
nce
ofa
large
rcra
cked
part
ofth
ebe
am.
Inth
esa
me
way
,aco
ncen
trate
dlo
adQ
=20
0kN
,pro
duci
ngth
esa
me
max
imal
mom
enti
nth
em
id-s
pam
sect
ion,
lead
sto
the
follo
win
gex
pres
sion
fort
hese
ctio
ndi
splac
emen
t2
3m
ax1
1*
2c
I
Ml
3v
1c
11
81
22
12E
I(7
.28)
inth
isca
se2
1m
axc
I
vM
12E
I(7
.29)
1=1/
(2).
(7.3
0)
The
corr
espo
ndin
gcu
rves
are
repo
rted
inFi
gure
7.15
.We
obse
rve
asth
edi
spla
cem
ents
inth
etw
oca
ses
ofdi
strib
uted
and
conc
entra
ted
load
are
resp
ectiv
ely
0.93
and
0.88
ofth
edi
splac
emen
tca
lcul
ated
inth
est
age
II.F
urth
erm
ore,
for
the
sam
eM
max
,the
disp
lacem
enti
nca
seof
conc
entra
ted
load
resu
ltsto
belo
wer
beca
use
the
linea
rtre
ndof
the
relat
ive
bend
ing
mom
ent
isas
soci
ated
toa
small
erre
gion
ofth
ecr
acki
ngbe
amw
ithre
spec
tto
the
case
ofdi
strib
uted
load
,tha
tisc
hara
cter
ized
bya
para
bolic
diag
ram
ofth
ebe
ndin
gm
omen
ts.
Fig.
7.15
.Dia
gram
sfor
v/v 1
,1-
.Th
esa
me
prob
lem
sca
nbe
solv
edin
age
nera
lized
form
eval
uatin
gnu
mer
icall
yth
edi
splac
emen
tfo
llow
ing
the
proc
edur
eex
pres
sed
in(1
1.51
).In
this
way
,it
ispo
ssib
leth
eev
aluat
ion
the
defo
rmat
ion
ofth
ew
hole
beam
,var
ying
z.T
here
sult,
for
adi
strib
uted
load
EC2�
wor
ked
exam
ples
7-21
Tabl
eof
Cont
ent
and
for
=3,
isre
porte
din
Figu
re7.
6,w
here
grap
hsre
fer
toa
20fo
lder
sdi
visio
nfo
rth
ecr
acki
ngpa
rtof
the
beam
.Rem
ark
asth
eco
mm
itted
erro
rin
the
evalu
atio
nof
the
mid
-spa
mde
flect
ion,
asob
tain
edco
mpa
ring
the
valu
esin
Figu
re7.
15an
dFi
gure
7.16
,is
abou
t4%
.In
parti
cular
,int
rodu
cing
the
num
eric
alva
lues
inth
e(7
.26)
(7.2
9)an
dus
ing
the
resu
ltsin
Figu
re11
.25,
we
have
fort
hem
id-s
pam
disp
lacem
ent:
Dist
ribut
edlo
ad6
8
15
9
550
010
1015
v21
.64m
m2
482
1018
.05
10
v1.
6521
.64
35.7
1mm
2
a)Co
ncen
trate
dlo
ad6
8
15
9
150
010
1015
v17
.31m
m2
122
1018
.05
10
v1.
5617
.31
27.0
0mm
2
Fig.
7.16
.Defo
rmat
ionin
thes
tage
I(a)
,disp
lacem
enti
ncrea
seca
used
byth
ecra
ckin
g(b)
And
total
defor
mat
ion(c)
.
EC2�
wor
ked
exam
ples
7-22
Tabl
eof
Cont
ent
EC2�
wor
ked
exam
ples
11-1
Tabl
eof
Cont
ent
SEC
TIO
N11
.LIG
HT
WE
IGH
TC
ON
CR
ET
E–
WO
RK
ED
EX
AM
PLE
S
EX
AM
PL
E11
.1[E
C2
Cla
use
11.3
.1�
11.3
.2]
The
crite
riafo
rdes
ign
ofth
ech
arac
teris
ticte
nsile
stre
ngth
(frac
tile
5%an
d95
%)a
ndof
the
inte
rsec
ting
com
pres
sive
elas
ticm
odul
efo
rlig
htco
ncre
tear
esh
own
belo
w,
inac
cord
ance
with
the
instr
uctio
nsof
para
grap
hs11
.3.1
and
11.3
.2of
Euro
code
2.Te
nsile
stren
gth
The
aver
age
valu
eof
sim
ple
(axi
al)t
ensil
est
reng
th,i
nla
ckof
dire
ctex
perim
enta
tion,
can
beta
ken
equa
lto:
-for
conc
rete
ofcl
ass
LC50
/55
f lctm
=0,
30f lc
k2/3
1
-for
conc
rete
ofcl
ass>
LC50
/55
f lctm
=2,
12ln
[1+(
f lcm/1
0)]
1
Whe
re:
1=
0,40
+0,6
0/2
200
=up
perl
imit
valu
eof
the
conc
rete
dens
ity,f
orth
eco
rresp
ondi
ngde
nsity
clas
sexp
ress
edin
kg/m
3 ;
f lck
=va
lue
ofth
ech
arac
teris
ticcy
lindr
icco
mpr
essiv
estr
engt
hin
MPa
.f lc
m=
valu
eof
the
aver
age
cylin
dric
com
pres
sive
stre
ngth
inM
Pa.
The
char
acte
ristic
valu
esof
sim
ple
tens
ilestr
engt
h,co
rresp
ondi
ngto
fract
iles
0,05
e0,
95,
can
beta
ken
equa
lto:
fract
ile5%
:f lc
tk,0
,05
=0,
7f lc
tm
fract
ile95
%:
f lctk
,0,9
5=
1,3
f lctm
Inte
rsec
ting
com
pres
sive
elas
ticm
odul
eIn
lack
ofdi
rect
expe
rimen
tatio
n,th
ein
ters
ectin
gco
mpr
essiv
eel
astic
mod
ule
at28
days
,w
hich
can
beus
edas
anin
dica
tive
valu
efo
rde
sign
ofth
ede
form
abili
tyof
struc
tura
lm
embe
rs,c
anbe
estim
ated
byth
eex
pres
sion:
0,3
lcm
lcm
Ef
E22
000
10[M
Pa]
whe
re: f lcm
=va
lue
ofth
ecy
lindr
icav
erag
eco
mpr
essi
vest
reng
thin
MPa
;2
E22
00;
=up
perl
imit
valu
eof
the
conc
rete
dens
ity,f
orco
rresp
ondi
ngde
nsity
clas
sin
kg/m
3 .The
resu
ltsof
calc
ulat
ion
ofth
etw
oab
ove-
men
tione
dm
echa
nica
lfe
atur
esar
esh
own
and
com
pare
din
the
follo
win
gta
ble,
for
two
diff
eren
tty
pes
oflig
htco
ncre
tes
and
for
the
corre
spon
ding
ordi
nary
conc
rete
sbel
ongi
ngto
the
sam
estr
engt
hcl
asse
s.
EC2�
wor
ked
exam
ples
11-2
Tabl
eof
Cont
ent
Tabl
e11
.1
Con
cret
ety
pe1
Con
cret
ety
pe2
Lig
htO
rdin
ary
Ligh
tO
rdin
ary
f lck
[MPa
]35
60[k
g/m
3 ]16
5024
0020
5024
00f lc
m[M
Pa]
4368
10,
850
--0,
959
--E
0,56
3--
0,86
8--
f ctm
[MPa
]2,
73,
24,
24,
4f ct
k;0,
05[M
Pa]
1,9
2,2
2,9
3,1
f ctk;
0,95
[MPa
]3,
54,
25,
45,
7E l
cm[M
Pa]
1916
834
077
3395
039
100
EC2�
wor
ked
exam
ples
11-3
Tabl
eof
Cont
ent
EX
AM
PL
E11
.2[E
C2
Cla
use
11.3
.1�
11.3
.5�
11.3
.6�
11.4�
11.6
]
The
max
imum
mom
entt
hatt
here
info
rced
conc
rete
sect
ion
ofgi
ven
dim
ensio
ns,m
ade
ofty
pe1
light
wei
ghtc
oncr
ete,
desc
ribed
inth
epr
evio
usex
ampl
e,is
able
tow
ithst
and
whe
nth
ere
info
rcem
ents
teel
achi
eves
the
desi
gnel
astic
limit.
The
dim
ensi
onso
fthe
sect
ion
are:
b=30
cm,h
=50c
man
dd=
47cm
.Th
ese
ctio
nin
ques
tion
issh
own
inFi
g.11
.1to
geth
erw
ithth
estr
ain
diag
ram
rela
ted
toth
efa
ilure
mod
ere
calle
d,w
hich
impl
ies
the
simul
tane
ous
achi
evem
ent
ofm
axim
umco
ntra
ctio
nsid
eco
ncre
tean
dof
the
stra
inco
rresp
ondi
ngto
the
desi
gnyi
eld
stres
sof
the
tens
ione
dre
info
rcem
ents
teel
.In
case
one
choo
ses,
like
inth
epr
evio
usex
ampl
e,to
use
the
bilin
eard
iagr
amto
calc
ulat
eth
eco
mpr
essiv
est
reng
thon
conc
rete
,the
limits
ofstr
ain
byco
mpr
essi
onha
veva
lues
lc3
=1,
75�
and
lcu3
=3,
51
=2,
98�
.
The
desig
nstr
ain
corre
spon
ding
tost
eely
ield
ing,
forf
yk=
450
MPa
,is
yd=
f yd/(1
,15
xE s
)=
450/
(1,1
5x
2000
00)
=1,
96�
.The
dist
ance
ofth
ene
utra
lax
isfro
mth
eco
mpr
esse
dup
pere
dge
isth
eref
ore
x=
28,3
cm.
Two
area
scan
bedi
sting
uish
edin
the
com
pres
sed
zone
:the
first
one
isco
mpr
ised
betw
een
the
uppe
redg
ean
dth
ech
ord
plac
edat
the
leve
lwhe
reth
eco
ntra
ctio
nis
lc3
=1,
75�
.The
com
pres
sive
stres
sin
itis
cons
tant
and
itis
equa
lto
f lcd
=0,
85f lc
k/c
=19
,8M
Pa;t
hese
cond
rem
aini
ngar
eais
the
one
whe
reco
mpr
essio
non
conc
rete
linea
rlyde
crea
ses
from
the
valu
ef lc
dto
zero
inco
rresp
onde
nce
ofth
ene
utra
laxi
s.Th
ere
sulta
ntof
com
pres
sion
forc
esis
plac
edat
adi
stanc
eof
arou
nd10
,5cm
from
the
com
pres
sed
end
ofth
ese
ctio
nan
dis
equa
lto
C=
1185
kN.
For
the
cond
ition
ofeq
uilib
rium
the
resu
ltant
ofco
mpr
essio
nsC
iseq
ual
toth
ere
sulta
ntof
tract
ions
T,to
whi
chco
rres
pond
sa
steel
sect
ion
As
equa
lto
As
=T/
f yd=
3030
mm
2 .The
arm
ofin
tern
alfo
rces
ish�
=d�
10,5
cm=
36,5
cm,f
rom
whi
chth
eva
lue
ofth
em
omen
tres
istan
ceof
the
sect
ion
can
even
tual
lybe
calc
ulat
edas
MR
d=
1185
x0,
365
=43
2,5
kNm
.
Fig
.11.
1D
efor
mat
ion
and
tens
ion
diag
ram
ofr.c
.sec
tion,
build
upw
ithlig
htwe
ight
conc
rete
(f lck
=35
MPa
,=
1650
kg/m
3 ),fo
rcol
laps
eco
nditi
onin
whi
chm
axim
umre
sist
ing
bend
ing
mom
enti
sre
ache
dwi
thre
info
rcem
enta
tela
stic
desig
nlim
it.
EC2�
wor
ked
exam
ples
11-4
Tabl
eof
Cont
ent