'HVLJQH[DPSOHIRUWKHDSSOLFDWLRQRI
(852&2'(3DUW $FWLRQVLQGXFHGE\FUDQHV
DQGPDFKLQHU\
DQG
(852&2'(3DUW &UDQHVXSSRUWLQJVWUXFWXUHV
QGW'5$)7
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Aachen, October19th, 2003
Prof. Dr.-Ing. G. Sedlacek
Dipl.-Ing. R. Schneider
Dipl.-Ing. N. Schäfer
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-2-
7DEOHRIFRQWHQWV
Page
)25(:25' 3DUW$'HVLJQH[DPSOHIRU(XURFRGH3DUW
'$7$2)7+(&5$1( 1.1 GENERAL ............................................................................................................................................ 6
1.2 GEOMETRIC PROPERTIES ..................................................................................................................... 6
1.3 MECHANICAL PROPERTIES .................................................................................................................. 6
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
'<1$0,&0$*1,),&$7,21)$&7256MM 2.1 GENERAL ............................................................................................................................................ 6
2.2 DYNAMIC MAGNIFICATION FACTOR ϕ1 ................................................................................................ 6
2.3 DYNAMIC MAGNIFICATION FACTOR ϕ2 ................................................................................................ 7
2.4 DYNAMIC MAGNIFICATION FACTOR ϕ3 ................................................................................................ 7
2.5 DYNAMIC MAGNIFICATION FACTOR ϕ4 ................................................................................................ 7
2.6 DYNAMIC MAGNIFICATION FACTOR ϕ5 ................................................................................................ 7
'(7(50,1$7,212)7+(9(57,&$/:+((//2$'6 3.1 GENERAL ............................................................................................................................................ 8
3.2 UNLOADED CRANE .............................................................................................................................. 9
3.3 LOADED CRANE .................................................................................................................................. 9
'(7(50,1$7,212)7+(+25,=217$//2$'6 4.1 GENERAL .......................................................................................................................................... 11
4.2 CAUSED BY ACCELERATION AND DECELERATION OF THE CRANE ...................................................... 11
'ULYHIRUFH. /RQJLWXGLQDOORDGV 7UDQVYHUVHORDGV+7 4.3 CAUSED BY SKEWING OF THE CRANE ................................................................................................ 12
6NHZLQJDQJOH 1RQSRVLWLYHIDFWRU )RUFHIDFWRUV /RQJLWXGLQDOIRUFHV 7UDQVYHUVHIRUFHV 4.4 CAUSED BY ACCELERATION OR BRAKING OF THE CRAB .................................................................... 15
(&&(175,&,7<2)9(57,&$/:+((//2$'6 )$7,*8(/2$'6 6800$5<2)7+(&5$1($&7,216 Design example for Eurocode 3 – Part 6: Cranes supporting structures
-3-
'HVLJQH[DPSOHIRU(XURFRGH3DUW
'$7$2)7+(&5$1(581:$<*,5'(5 1.1 SYSTEM............................................................................................................................................. 18
1.2 CROSS-SECTION PROPERTIES ............................................................................................................. 18
*HQHUDO &URVVVHFWLRQFODVVLILFDWLRQ ,17(51$/)25&(6$1'020(1762)7+(&5$1(581:$<*,5'(5 2.1 GENERAL .......................................................................................................................................... 18
2.2 INTERNAL FORCES AND MOMENTS AT POINT 2.875 ........................................................................... 19
6HOIZHLJKWRIWKHFUDQHUXQZD\JLUGHU 9HUWLFDOZKHHOORDGVRIWKHFUDQH $FFHOHUDWLRQDQGGHFHOHUDWLRQ 7RUVLRQGXHWRYHUWLFDODQGKRUL]RQWDOORDGV 2.3 INTERNAL FORCES AND MOMENTS AT SUPPORT................................................................................. 22
6HOIZHLJKWRIWKHFUDQHUXQZD\JLUGHU 9HUWLFDOZKHHOORDGVRIWKHFUDQH $FFHOHUDWLRQDQGGHFHOHUDWLRQ 7RUVLRQGXHWRYHUWLFDODQGKRUL]RQWDOORDGV s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
&52666(&7,215(6,67$1&(2)7+(&5$1(581:$<*,5'(5 3.1 POINT 2.875 ...................................................................................................................................... 24
6KHDUUHVLVWDQFHRIWKHZHE ]D[LV 6KHDUUHVLVWDQFHRIWKHWRSIODQJH \D[LV 6KHDUUHVLVWDQFHGXHWRWRUVLRQ ,QWHUDFWLRQEHWZHHQQRUPDODQGVKHDUIRUFHV %HQGLQJDQGD[LDOIRUFHV 3.2 SUPPORT ........................................................................................................................................... 26
6KHDUUHVLVWDQFHRIWKHZHE ]D[LV 6KHDUUHVLVWDQFHRIWKHWRSIODQJH \D[LV 6KHDUUHVLVWDQFHGXHWRWRUVLRQ $QQH[*RI(XURFRGH3DUW ,QWHUDFWLRQEHWZHHQQRUPDODQGVKHDUIRUFHV $[LDOIRUFHV 5(6,67$1&(2)7+(:(%7275$169(56()25&(6 )$7,*8( 5.1 GENERAL .......................................................................................................................................... 29
5.2 DETAIL CATEGORIES ......................................................................................................................... 30
5.3 POINT 2.785 ...................................................................................................................................... 31
9HULILFDWLRQRIWKHFURVVVHFWLRQ 9HULILFDWLRQRIWKHZHE 6KHDUVWUHVVHV 'LUHFWVWUHVVHV ,QWHUDFWLRQEHWZHHQGLUHFWDQGVKHDUVWUHVVHVLQWKHZHE 5.4 SUPPORT ........................................................................................................................................... 35
9HULILFDWLRQRIWKHZHE 6KHDUVWUHVVHV 'LUHFWVWUHVVHV ,QWHUDFWLRQEHWZHHQGLUHFWDQGVKHDUVWUHVVHVLQWKHZHE Design example for Eurocode 3 – Part 6: Cranes supporting structures
-4-
)RUHZRUG
This report demonstrates the application of Eurocode 1 - Part 3: “Actions induced by
cranes and machinery” and the application of Eurocode 3 - Part 6: “Crane supporting
structures” for a top mounted crane.
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-5-
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
3DUW$
'HVLJQH[DPSOHIRU(XURFRGH3DUW
$FWLRQVLQGXFHGE\FUDQHVDQGPDFKLQHU\
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-6-
'DWDRIWKHFUDQH
*HQHUDO
The geometric properties which are assumed in the design example are summarized in
section 1.2 and the mechanical details of the crane are defined in section 1.3. Further
assumptions for the crane are given where they are necessary.
*HRPHWULFSURSHUWLHV
The following geometric properties are assumed in the design example for the crane:
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Span length of the crane bridge:
15,00 m
Wheel spacing a:
2,50 m
Min. spacing between crab and supports emin: 0,00 m
0HFKDQLFDOSURSHUWLHV
The following mechanical properties are defined for the crane:
)LJXUH'HILQLWLRQRIWKHKRLVWORDGDQGWKHVHOIZHLJKWRIDFUDQH
Self-weight of the crane Qc1 : 60,0 kN
Self-weight of the crab Qc2 :
10,0 kN
Hoistload Qh,nom :
100,0 kN
'\QDPLFPDJQLILFDWLRQIDFWRUVMM
*HQHUDO
The dynamic effects of a crane structure are taken into account by magnification factors
which are defined in Eurocode 1 - Part 3.
'\QDPLFPDJQLILFDWLRQIDFWRUM
The magnification factor ϕ1 takes into account vibrational excitation of the crane
structure due to lifting the hoist load off the ground and is to be applied to the selfweight of the crane.
ϕ1 = 1,1 (upper value of the vibrational pulses)
(EC 1- P 3: Table 2.4)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-7-
'\QDPLFPDJQLILFDWLRQIDFWRUM
The magnification factor ϕ2 is only to be applied to the hoistload and takes into account
the dynamical effects when the hoistload is transferred from the ground to the crane.
The magnification factor depends on the hoisting class of the crane. It is assumed that
the crane is classified as HC 3. Recommendations about the classification of cranes are
given in Annex B of Eurocode 1 - Part 3.
Assumption:
Hoisting class of the crane: HC 3
vh = 6 m/min
6
ϕ 2 = ϕ 2,min + β 2 v h = 1,15 + 0,51 ⋅
= 1,20
60
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
(EC 1- P 3: Table 2.4)
The parameters ϕ2,min and β2 were obtained from table 2.5 of EC 1- Part 3.
'\QDPLFPDJQLILFDWLRQIDFWRUM
The magnification factor ϕ3 considers the dynamical effects when a payload is sudden
released. These dynamic effects occur at cranes which use magnets as hoist tools. In the
design example it is assumed that no part of the payload is able to sudden release.
Assumption:
No sudden release or dropped part of the load.
ϕ3 = 1,00
(EC 1- P 3: Table 2.4)
'\QDPLFPDJQLILFDWLRQIDFWRUM
This magnification factor is to be applied to the self-weight of the crane and to the
payload, if the rail track observes not the tolerances specified in ENV 1993 - 6.
Assumption:
The tolerances for rail tracks are observed as specified in ENV 1993 - 6.
ϕ4 = 1,00
(EC 1- P 3: Table 2.4)
'\QDPLFPDJQLILFDWLRQIDFWRUM
The magnification factor ϕ5 takes into account the dynamic effects caused by drive
forces and depends on the characteristic of the drive forces.
Assumption:
The drive force change smoothly.
ϕ5 = 1,50
(EC 1- P 3: Table 2.6)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-8-
'HWHUPLQDWLRQRIWKHYHUWLFDOZKHHOORDGV
*HQHUDO
In this section the minimum and the maximum vertical wheel loads of the crane are calculated according to table 3.1 which was obtained from Eurocode 1 - Part 3 (Table 2.2).
Table 3.1 defines the groups of loads which are to be considered as one characteristic
crane load, when additional actions apply at the structure (for example: self-weight,
wind action, snow). With the definition of the groups of loads the relevant combinations
of the magnification factors are given.
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
7DEOH *URXSV RI ORDGV DQG G\QDPLF IDFWRUV WR EH FRQVLGHUHG DV RQH
FKDUDFWHULVWLFFUDQHDFWLRQ
Groups of loads
Symbol
Section
ULS
SLS
Acci-
dental
1
2
3
4
5
6
7
8
9
10
2.6
Q
Q
1
Q
Q
Q
1
Q1
1
1
Qh
2.6
Q
Q
-
Q
Q
Q
-
1
1
HL, HT
2.7
Q
Q
Q
Q
-
-
-
Q5
-
-
4 Skewing of crane bridge
HS
2.7
-
-
-
-
1
-
-
-
-
-
5 Acceleration or braking of
crab or hoist block
HT3
2.7
-
-
-
-
-
1
-
-
-
-
6 In service wind
FW*
Annex A
1
1
1
1
1
-
-
1
-
-
7 Test load
QT
2.10
-
-
-
-
-
-
-
Q6
-
-
8 Buffer force
HB
2.11
-
-
-
-
-
-
-
-
Q7
-
9 Tilting force
HTA
2.11
-
-
-
-
-
-
-
-
-
1
1 Self-weight of crane
2 Hoist load
3 Acceleration of crane
bridge
Qc
is the part of the hoist load that remains when the payload is removed, but is not included in the selfweight of the crane.
1)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-9-
8QORDGHGFUDQH
The minimum vertical wheel load apply at a crane runway girder when the crane is
unloaded.
Qr,min
Qr,min
ΣQr,min
ΣQr; (min)
Qr,´(min)
Qr,(min)
"
)LJXUH/RDGDUUDQJHPHQWRIWKHXQORDGHGFUDQHWRREWDLQWKHPLQLPXP
ORDGLQJRQWKHUXQZD\EHDP
a
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
a) Load group 1,2
ϕ = 1,1:
⇒ Q C1,k = 1,1 ⋅ 60,0 = 66,0 kN
⇒ Q C 2,k = 1,1 ⋅ 10,0 = 11,0 kN
∑Q
∑Q
r ,(min)
r , min
=
=
1
⋅ 66,0 + 11,0 = 44,0 kN ⇒ Q r ,(min) = 22,0 kN
2
1
⋅ 66,0 = 33,0 kN
2
⇒ Q r , min = 16,5 kN
b) Load group 3,4,5,6
ϕ = 1,0:
⇒ Q c1,k = 1,0 ⋅ 60,0 = 60,0 kN
⇒ Q c 2,k = 1,0 ⋅10,0 = 10,0 kN
∑Q
∑Q
r ,(min)
r , min
=
=
1
⋅ 60,0 + 10,0 = 40,0 kN ⇒ Q r ,(min) = 20,0 kN
2
1
⋅ 60,0 = 30,0 kN
2
⇒ Q r ,min = 15,0 kN
/RDGHGFUDQH
The maximum vertical wheel loads apply at a crane runway girder when the crane is
loaded.
Qr,max
Qr,max
ΣQr,max
Crab
ΣQr ,(max)
Qr, (max)
Qr, (max)
Qh,nom = nominal hoist load
emin
"
)LJXUH/RDGDUUDQJHPHQWRIWKHORDGHGFUDQHWRREWDLQWKHPD[LPXP
ORDGLQJRQWKHUXQZD\EHDP
a) Load group 1
ϕ = 1,1:
⇒ Q c1,k = 1,1 ⋅ 60,0 = 66,0 kN
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 10 -
⇒ Q c 2,k = 1,1 ⋅10,0 = 11,0 kN
ϕ = 1,2:
∑Q
∑Q
⇒ Q h ,k = 1,2 ⋅100,0 = 120,0 kN
r ,(max)
r , max
=
=
1
⋅ 66,0 = 33,0 kN
2
⇒ Q r ,(max) = 16,5 kN
1
⋅ 66,0 + 11,0 + 120,0 = 164,0 kN ⇒ Q r ,max = 82,0 kN
2
b) Load group 2
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
ϕ = 1,1:
⇒ Q c1,k = 1,1 ⋅ 60,0 = 66,0 kN
⇒ Q c 2,k = 1,1 ⋅10,0 = 11,0 kN
ϕ = 1,0:
∑Q
∑Q
⇒ Q h ,k = 1,0 ⋅100,0 = 100,0 kN
r ,(max)
r , max
=
=
1
⋅ 66,0 = 33,0 kN
2
⇒ Q r ,(max)
= 16,5 kN
1
⋅ 66,0 + 11,0 + 100,0 = 144,0 kN ⇒ Q r ,max = 72,0 kN
2
c) Load group 4,5,6
ϕ = 1,0:
⇒ Q c1,k = 1,0 ⋅ 60,0 = 60,0 kN
⇒ Q c 2,k = 1,0 ⋅10,0 = 10,0 kN
ϕ = 1,0:
∑Q
∑Q
⇒ Q h ,k = 1,0 ⋅100,0 = 100,0 kN
r ,(max)
r , max
=
=
1
⋅ 60,0 = 30,0 kN
2
⇒ Q r ,(max) = 15,0 kN
1
⋅ 60,0 + 10,0 + 100,0 = 140,0 kN ⇒ Q r ,max = 70,0 kN
2
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 11 -
'HWHUPLQDWLRQRIWKHKRUL]RQWDOORDGV
*HQHUDO
In this section the following horizontal loads are calculated:
-horizontal loads caused by acceleration and deceleration of the crane bridge, see 4.2;
-horizontal loads caused by skewing of the crane bridge, see 4.3;
-horizontal loads caused by acceleration or braking of the crab, see 4.4;
&DXVHGE\DFFHOHUDWLRQDQGGHFHOHUDWLRQRIWKHFUDQH
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
'ULYHIRUFH.
Rail i = 1
Rail i = 2
)LJXUH'HILQLWLRQRIWKHGULYHIRUFH
K
K2
Friction factor:
Number of single wheel drivers:
∑Q
*
r , min
= 0,2
(EC 1- P 3: 2.7.3(4))
mw = 2
= m w ⋅ Q r ,min = 2 ⋅ 15,0 = 30,0 kN
(EC 1- P 3: 2.7.3(3))
K = µ ⋅ ∑ Q *r ,min = 0,2 ⋅ 30,0 = 6,0 kN
(EC 1- P 3: 2.7.3(3))
/RQJLWXGLQDOORDGV
Rail i = 1
Rail i = 2
H L ,2
H L ,1
)LJXUH/RQJLWXGLQDOKRUL]RQWDOORDGV+
/L
Number of runway beams:
H L,1 = H L,2 = ϕ5 ⋅
K
nr
= 1,5 ⋅
nR = 2
6,0
= 4,5 kN
2
(EC 1- P 3: 2.7.2(2))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 12 -
7UDQVYHUVHORDGV+
7
Rail i = 1
Rail i = 2
H T ,2
H T ,1
M
S
a
HT ,1
H T ,2
O
K1
s
K = K 1+
ξ
K2
K2
ξ
O
O
)LJXUH'HILQLWLRQRIWKHWUDQVYHUVHORDGV+ 1
2
O
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
7L
ξ1
=
∑ Qr,max
∑ Q =∑ Q
r
(EC 1- P 3: 2.7.2(3))
∑ Qr
r , max
+ ∑ Q r ,(max) = 140,0 + 30,0 = 170,0 kN
(EC 1- P 3: 2.7.2(3))
ξ1
=
140,0
= 0,82
170,0
(EC 1- P 3: 2.7.2(3))
ξ2
= 1 − ξ1 = 0,18
(EC 1- P 3: 2.7.2(3))
lS
= (ξ1 − 0,5)⋅ l = (0,83 − 0,5)⋅ 15,0 = 4,95 m
(EC 1- P 3: 2.7.2(3))
M
= K ⋅ l S = 6,0 ⋅ 4,95 = 29,7 kNm
(EC 1- P 3: 2.7.2(3))
M
29,7
= 1,5 ⋅ 0,18 ⋅
= 3,2 kN
a
2,5
M
29,7
= ϕ 5 ⋅ ξ1 ⋅ = 1,5 ⋅ 0,82 ⋅
= 14,6 kN
a
2,5
H T ,1 = ϕ 5 ⋅ ξ2 ⋅
(EC 1- P 3: 2.7.2(3))
H T,2
(EC 1- P 3: 2.7.2(3))
&DXVHGE\VNHZLQJRIWKHFUDQH
6NHZLQJDQJOH
0,75 x
10
=
= 0,004 rad
a
2500
y
0,1 ⋅ 50
αV =
=
= 0,002 rad
a
2500
α0 =
= 0,001 rad
-------------α = α F + α V + α 0 = 0,007 rad
αF =
(EC 1- P 3: Table 2.7)
(EC 1- P 3: Table 2.7)
(EC 1- P 3: Table 2.7)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 13 -
1RQSRVLWLYHIDFWRU
f = 0,3 (1 − exp (− 250 α )) = 0,3 (1 − exp (− 250 ⋅ 0,007 )) = 0,248
)RUFHIDFWRUV
(EC 1- P 3: 2.7.4(2))
(a) Distance ei of the wheel pair i from the guidance means
e1 = 0 as flanged wheels are used
e2 = a = 2,50 m
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
(b) Combination of wheel pairs: IFF
m=0
(c) Distance h:
h=
mξ1ξ 2 l 2 + ∑ e 2j
∑e
=
j
0 + 2,50 2
= 2,50 m
2,50
(EC 1- P 3: Table 2.8)
n=2
λS = 1 −
∑e
j
n⋅h
= 1−
2,50
= 0,5
2 ⋅ 2,50
λ S,1,L = λ S, 2,L = 0
(EC 1- P 3: Table 2.9)
(EC 1- P 3: Table 2.9)
for wheel pair 1:
ξ2
n
 e1 
1 −  =
h

ξ  e 
= 1 1 − 1  =
n
h
0,18
(1 − 0) = 0,09
2
(EC 1- P 3: Table 2.9)
0,82
(1 − 0) = 0,41
2
(EC 1- P 3: Table 2.9)
λ S,1, 2,T =
0,18  2,50 
1 −
=0
2  2,50 
(EC 1- P 3: Table 2.9)
λ S, 2 , 2 , T
0,82  2,50 
1 −
=0
2  2,50 
(EC 1- P 3: Table 2.9)
λ S,1,1,T =
λ S, 2,1,T
for wheel pair 2:
ξ2  e2 
1 −  =
n 
h
ξ  e 
= 1 1 − 2  =
n
h
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 14 -
/RQJLWXGLQDOIRUFHV
Rail i = 1
Rail i = 2
H L ,2
H L ,1
)LJXUH/RQJLWXGLQDOKRUL]RQWDOORDGV+
/L
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
H S,1,L = f ⋅ λS,1,L ⋅ ∑ Q r = 0
(EC 1- P 3: 2.7.4(1))
H S, 2,L = f ⋅ λS, 2,L ⋅ ∑ Q r = 0
(EC 1- P 3: 2.7.4(1))
7UDQVYHUVHIRUFHV
Rail i = 1
Directon of motion
Rail i = 2
α
H
S
Wheel pair j = 1
HS ,2,1,T
67
Wheel pair j = 2
)LJXUH/RQJLWXGLQDOKRUL]RQWDOORDGV+
/L
Guide force S:
S = f ⋅ λS ⋅ ∑ Q r = 0,248 ⋅ 0,5 ⋅ 170,0 = 21,1 kN
(EC 1- P 3: 2.7.4(1))
for wheel pair 1:
H S,1,1,T = f ⋅ λS,1,1,T ⋅ ∑ Q r = 0,248 ⋅ 0,09 ⋅ 170,0 = 3,8 kN
H S, 2,1,T = f ⋅ λS, 2,1,T ⋅ ∑ Q r = 0,248 ⋅ 0,41 ⋅ 170,0 = 17,3 kN
(EC 1- P 3: 2.7.4(1))
(EC 1- P 3: 2.7.4(1))
⇒ H S,1,T = S − H S,1,1,T = 17,3 kN
⇒ H S, 2,T = H S, 2,1,T
= 17,3 kN
for wheel pair 2:
H S,1, 2,T = f ⋅ λS,1, 2,T ⋅ ∑ Q r = 0,248 ⋅ 0 ⋅ 170,0 = 0 kN
H S, 2, 2,T = f ⋅ λS, 2, 2,T ⋅ ∑ Q r = 0,248 ⋅ 0 ⋅ 170,0 = 0 kN
(EC 1- P 3: 2.7.4(1))
(EC 1- P 3: 2.7.4(1))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 15 -
&DXVHGE\DFFHOHUDWLRQRUEUDNLQJRIWKHFUDE
H T ,3 = 0,1 ⋅ (10,0 + 100,0 ) = 11,0 kN
(EC 1- P 3: 2.7.5)
(EC 1- P 3: 2.11.2)
(FFHQWULFLW\RIYHUWLFDOZKHHOORDGV
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
)LJXUH(FFHQWULFLW\RIWKHZKHHOORDG
e=
1
1
⋅ b r = ⋅ 55 = 13,75 mm
4
4
(EC 1- P 3: 2.5.3(2))
)DWLJXHORDGV
Q e ,i = ϕ fat ⋅ λ i ⋅ Q max,i
1 + ϕ1 1 + 1,1
=
= 1,05
2
2
1 + ϕ 2 1 + 1,2
=
=
= 1,10
2
2
(EC 1- P 3: 2.12.1(4))
ϕ fat ,1 =
(EC 1- P 3: 2.12.1(7))
ϕ fat , 2
(EC 1- P 3: 2.12.1(7))
Assumption: crane is classified in class S6:
λ i = 0,794
λ i = 0,871
for normal stresses
for shear stresses
(EC 1- P 3: Table 2.12)
(EC 1- P 3: Table 2.12)
For normal stresses:
Q e ,i = ϕ fat ⋅ λ i ⋅ Q max,i = 1,1 ⋅ 0,794 ⋅ 70,0 = 61,1 kN
(EC 1- P 3: 2.12.1(4))
For shear stresses:
Q e ,i = ϕ fat ⋅ λ i ⋅ Q max,i = 1,1 ⋅ 0,871 ⋅ 70,0 = 67,1 kN
(EC 1- P 3: 2.12.1(4))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 16 -
6XPPDU\RIWKHFUDQHDFWLRQV
For the XOWLPDWHOLPLWVWDWH the results are summarised in the following table according
to the groups of loads.
7DEOH 6XPPDU\ RI WKH YHUWLFDO DQG KRUL]RQWDO ORDGV IRU WKH FUDQH UXQZD\
JLUGHU
Groups of loads
Magnification factor which are
considered for the group of load
1
2
3
4
5
6
Q = 1,10
Q = 1,10
Q = 1,00
Q = 1,00
Q = 1,00
Q = 1,00
Q = 1,20
Q = 1,00
Q = 1,50
Q = 1,50
Q = 1,50
Q = 1,50
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Vertical
loads
Self-weight of the
crane
Qr,(min)
22,0 kN
22,0 kN
20,0 kN
20,0 kN
20,0 kN
20,0 kN
Qr,min
16,5 kN
16,5 kN
15,0 kN
15,0 kN
15,0 kN
15,0 kN
Self-weight of the
crane and hoistload
Qr,(max)
16,5 kN
16,5 kN
-
15,0 kN
15,0 kN
15,0 kN
Qr,max
82,0 kN
72,0 kN
-
70,0 kN
70,0 kN
70,0 kN
HL,1
4,5 kN
4,5 kN
4,5 kN
4,5 kN
-
-
HL,2
4,5 kN
4,5 kN
4,5 kN
4,5 kN
-
-
HT,1
3,2 kN
3,2 kN
3,2 kN
3,2 kN
-
-
HT,2
14,6 kN
14,6 kN
14,6 kN
14,6 kN
-
-
HS1,L
-
-
-
-
0
-
HS2,L
-
-
-
-
0
-
HS1,T
-
-
-
-
17,3 kN
-
HS2,T
-
-
-
-
17,3 kN
-
HT,3
-
-
-
-
-
11,0 kN
Horizontal Acceleration of the
loads
crane
Skewing of the
crane
Acceleration of the
crab
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 17 -
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
3DUW%
'HVLJQH[DPSOHIRU(XURFRGH3DUW
&UDQHVVXSSRUWLQJVWUXFWXUHV
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 18 -
'DWDRIWKHFUDQHUXQZD\JLUGHU
6\VWHP
Single-span girder with fork-support, length: l = 7,00 m
&URVVVHFWLRQSURSHUWLHV
*HQHUDO
In the design example it is assumed that the rail is rigid fixed with clamps on the crane
runway girder.
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
The benefit effects of the rigid fixed rail on the design resistance are not taken into
account in the design example (see 5.3.3 (2) of EC 3 - Part 6)
Cross-section properties of the crane runway girder (without rail) HE-B 500:
2
4
A [cm ]
239,0
Iy [cm ]
107200
Area of the flange:
Area of the web:
4
Iz [cm ]
12620
3
Wel,y [cm ]
4290
3
Wel,z [cm ]
842
AF = 300$28,0 = 84,0 cm2
AW = 444$14,5 = 64,4 cm2
Cross-section properties of the rail A55:
A [cm2]
40,5
Iy [cm4]
178
Iz [cm4]
337
Material S235
&URVVVHFWLRQFODVVLILFDWLRQ
The cross-section is classified into class 1.
,QWHUQDOIRUFHVDQGPRPHQWVRIWKHFUDQHUXQZD\JLUGHU
*HQHUDO
For the verification of the crane runway girder the internal forces and moments are
calculated with influence lines for the following points:
Point 2.875: Maximum bending moment of the crane runway girder (in field)
Support:
Maximum shear forces of the crane runway girder (at support)
The design example is carried out for load group 1, see table 7.1.
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 19 -
,QWHUQDOIRUFHVDQGPRPHQWVDWSRLQW
Load position for the maximum bending moment:
Qr,max
2,50
Qr,max
x
7,00
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
A=
11,5 − 2 ⋅ x
7,0
11,5 ⋅ x − 2 ⋅ x 2
M( x) = A ⋅ x =
7,0
11,5 − 4 ⋅ x
M ’( x ) =
= 0 for max M ⇒ x = 2,875
7,0
6HOIZHLJKWRIWKHFUDQHUXQZD\JLUGHU
gk = 1,873 + 0,318 = 2,2 kN
2,2 ⋅ 7,0
A (g k ) =
= 7,7 kN
2
2,875 2 ⋅ 2,2
M y, k = 7,7 ⋅ 2,875 −
= 13,0 kNm
2
Vz , k = 7,7 − 2,875 ⋅ 2,2 = 1,4 kN
9HUWLFDOZKHHOORDGVRIWKHFUDQH
a) Bending moment
Qr,max
2,50
Qr,max
0,095
0,242
max M y, k = Q r , max ⋅ (η1 + η 2 ) ⋅ l = 82,0 ⋅ (0,242 + 0,095) ⋅ 7,0 = 193,7 kNm
min M y ,k = 0 kNm
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 20 -
b) Shear force
Qr,max
2,50
Qr,max
0,411
0,232
0,589
max Vz ,k = Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (0,589 + 0,232) = 67,3 kN
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Qr,max
2,50
Qr,max
0,411
0,054
min Vz ,k = Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (−0,054 − 0,411) = −38,1 kN
$FFHOHUDWLRQDQGGHFHOHUDWLRQ
a) Bending moment
HT,2
2,50
HT,2
0,095
0,242
min M z , k = (H T , 2 ⋅ η1 + H T , 2 ⋅ η 2 ) ⋅ l = (−14,6 ⋅ 0,242 + 14,6 ⋅ 0,095) ⋅ 7,0 = −15,0 kNm
HT,2
2,50
HT,2
0,032
0,242
max M z , k = (H T , 2 ⋅ η1 + H T , 2 ⋅ η 2 ) ⋅ l = ( −14,6 ⋅ 0,0316 + 14,6 ⋅ 0,242) ⋅ 7,0 = 21,5 kNm
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 21 -
b) Shear force
HT,2
2,50
HT,2
0,411
0,232
0,589
max Vy ,k = H T , 2 ⋅ η1 + H T , 2 ⋅ η 2 = (−14,6) ⋅ (−0,411) + 14,6 ⋅ 0,232 = 9,4 kN
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
HT,2
2,50
HT,2
0,411
0,054
min Vy,k = H T , 2 ⋅ η1 + H T , 2 ⋅ η 2 = (−14,6) ⋅ (−0,054) + 14,6 ⋅ (−0,411) = −5,2 kN
N k = − 4,5 kN
7RUVLRQGXHWRYHUWLFDODQGKRUL]RQWDOORDGV
Rail A 55:
br =55 mm
h1 =65 mm
Wheel loads: Qr,max = 82,0 kN
e y = 0,25 ⋅ b r = 13,75 mm
(EC 1-P 3: 2.5.3 (2))
Horizontal loads due to acceleration and deceleration: H T = ±14,6 kN
e z = 0,5 ⋅ h + h 1 = 0,5 ⋅ 500 + 65 = 315 mm
M t1 = 82,0 ⋅ 0,01375 + 14,6 ⋅ 0,315 = 5,7 kNm
M t 2 = 82,0 ⋅ 0,01375 − 14,6 ⋅ 0,315 = −3,5 kNm
2,50
Mt2
Mt1
0,411
0,054
0,589
+
max M t ,k = 5,7 ⋅ 0,589 − 3,5 ⋅ (−0,054) = 3,5 kNm
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 22 -
,QWHUQDOIRUFHVDQGPRPHQWVDWVXSSRUW
There are no bending moments at the support (Single-span girder)
6HOIZHLJKWRIWKHFUDQHUXQZD\JLUGHU
Vz , k = 7,7 kN
9HUWLFDOZKHHOORDGVRIWKHFUDQH
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
max Vz , k = Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (0,0 + 0,0) = 0 kN
Qr,max
2,50
0,643
Qr,max
1,0
min Vz ,k = Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (−1,0 − 0,6428) = −134,7 kN
$FFHOHUDWLRQDQGGHFHOHUDWLRQ
HT,2
2,50
HT,2
1,0
max Vy ,k = H T , 2 ⋅ η1 + H T , 2 ⋅ η 2 = (−14,6) ⋅ (−1,0) + 14,6 ⋅ 0,0 = 14,6 kN
HT,2
HT,2
2,50
0,357
min Vy ,k = H T , 2 ⋅ η1 + H T , 2 ⋅ η2 = (−14,6) ⋅ 0,0 + 14,6 ⋅ (−0,357) = −5,2 kN
N k = − 4,5 kN
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 23 -
7RUVLRQGXHWRYHUWLFDODQGKRUL]RQWDOORDGV
Rail A 55:
br =55 mm
h1 =65 mm
Wheel loads: Qr,max = 82,0 kN
e y = 0,25 ⋅ b r = 13,75 mm
(EC 1- P 3: 2.5.3 (2))
Horizontal loads due to acceleration and deceleration: H T = ±14,6 kN
e z = 0,5 ⋅ h + h 1 = 0,5 ⋅ 500 + 65 = 315 mm
M t1 = 82,0 ⋅ 0,01375 + 14,6 ⋅ 0,315 = 5,7 kNm
M t 2 = 82,0 ⋅ 0,01375 − 14,6 ⋅ 0,315 = −3,5 kNm
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
2,50
Mt2
Mt1
1,0
+
0,643
max M t ,k = 5,7 ⋅ 1,0 − 3,5 ⋅ 0,643 = 3,4 kNm
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 24 -
&URVVVHFWLRQUHVLVWDQFHRIWKHFUDQHUXQZD\JLUGHU
3RLQW
6KHDUUHVLVWDQFHRIWKHZHE ]D[LV d/tw = 390/14,5 = 26,9 < 60 ⇒ Verification for shear buckling is not necessary
(EC 3- P 1: 5.1)
max Vz ,Sd = γ G G k + γ Q Q k = 1,35 ⋅ 1,4 + 1,35 ⋅ 67,3 = 92,75 kN
A V = 390 ⋅ 14,5 = 56,55 cm 2
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Vz , Rd = A V ⋅
fy / 3
= 56,55 ⋅
γM
235 / 3
= 697,5 kN > 92,75 kN
1,1
(EC 3- P 1: 6.2.6)
6KHDUUHVLVWDQFHRIWKHWRSIODQJH \D[LV It is assumed that the horizontal loads are resisted by the top flange of the girder.
max Vy ,Sd = γ Q Q k = 1,35 ⋅ 9,4 = 12,7 kN
A V = A TV = 300 ⋅ 28 = 84,0 cm 2
Vy ,Rd = A V ⋅
fy / 3
= 84,0 ⋅
γM
235 / 3
= 1036,1 kN > 12,7 kN
1,1
(EC 3- P 1: 6.2.6)
6KHDUUHVLVWDQFHGXHWRWRUVLRQ
max M t ,Sd = 1,35 ⋅ 3,5 = 4,7 kN
τ V ,Ed =
M t ,Sd ⋅ t
It
=
kN f y / 3
kN
4,7 ⋅ 2,8 ⋅ 100
= 2,45 2 <
= 12,3 2
538
γM
cm
cm
(EC 3- P 1: 6.2.6)
,QWHUDFWLRQEHWZHHQQRUPDODQGVKHDUIRUFHV
Vpl,Rd =
(
Av ⋅ fy / 3
γM
Vpl,T ,Rd = 1 −
) = 697,5 kN
(
τ t ,Ed
)
1,25 ⋅ f y / 3 / γ M 0
(EC 3- P 1: 6.2.6 (2))
⋅Vpl,Rd = 1 −
2,45
⋅ 697,5 = 639,5 kN
1,25 ⋅ 12,3
(EC 3- P 1: 6.2.7)
VEd = 92,75 kN ≤ 319,8 kN = 0,5 ⋅ Vpl,T ,Rd
⇒ no interaction between shear and normal stresses necessary
(EC 3- P 1: 6.2.8)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 25 -
%HQGLQJDQGD[LDOIRUFHV
It is assumed that the horizontal loads are resisted by the top flange.
a) Verification for max My,Sd:
N Sd = −1,35 ⋅ 4,5 = 6,1 kN
max M y ,Sd = 1,35 ⋅ 13,0 + 1,35 ⋅ 193,7 = 279,0 kNm
M z ,Sd = 1,35 ⋅ 15,0 = 20,3 kNm
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
ATF
Wel,y
Wel,z
=
=
=
84
4290
842
cm2
cm3
cm3
M y ,Sd
M z ,Sd
N Sd
+
+
≤ 1,0
A TF ⋅ f y,d Wel, y ⋅ f y ,d Wel,z ⋅ f y ,d
(EC 3- P 1: 6.2.1)
6,1
279,0 ⋅ 100
20,3 ⋅ 100
+
+
= 0,42 ≤ 1,0
84 ⋅ 23,5 / 1,1 4290 ⋅ 23,5 / 1,1 842 ⋅ 23,5 / 1,1
b) Verification for max Mz,Sd:
N Sd = −1,35 ⋅ 4,5 = 6,1 kN
M y, k = 82,0 ⋅ (0,0316 + 0,2420) ⋅ 7,0 = 157,0 kNm
M y,Sd = 1,35 ⋅ 13,0 + 1,35 ⋅ 157,0 = 229,5 kNm
max M z ,Sd = 1,35 ⋅ 21,5 = 29,03 kNm
84
cm2
ATF =
4290
cm3
Wel,y =
Wel,z =
842
cm3
M y ,Sd
M z ,Sd
N Sd
+
+
≤ 1,0
A TF ⋅ f y,d Wel, y ⋅ f y ,d Wel,z ⋅ f y ,d
(EC 3- P 1: 6.2.1)
6,1
229,5 ⋅ 100
29,03 ⋅ 100
+
+
= 0,42 ≤ 1,0
84 ⋅ 23,5 / 1,1 4290 ⋅ 23,5 / 1,1 842 ⋅ 23,5 / 1,1
1RWH
The cross-section properties of the rail are not taken into account though the
rail is rigid fixed.
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 26 -
6XSSRUW
6KHDUUHVLVWDQFHRIWKHZHE ]D[LV d/tw = 390/14,5 = 26,9 < 60 ⇒ Verification for shear buckling is not necessary
(EC 3- P 1: 5.1)
max Vz ,Sd = γ G G k + γ Q Q k = 1,35 ⋅ 7,7 + 1,35 ⋅ 134,7 = 192,2 kN
A V = 390 ⋅ 14,5 = 56,55 cm 2
Vz , Rd = A V ⋅
fy / 3
= 56,55 ⋅
γM
235 / 3
= 697,5 kN > 192,2 kN
1,1
(EC 3- P 1: 6.2.6)
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
6KHDUUHVLVWDQFHRIWKHWRSIODQJH \D[LV It is assumed that the horizontal loads are resisted by the top flange of the girder.
max Vy ,Sd = γ Q Q k = 1,35 ⋅ 14,6 = 19,7 kN
A V = A TV = 300 ⋅ 28 = 84,0 cm 2
Vy ,Rd = A V ⋅
fy / 3
= 84,0 ⋅
γM
235 / 3
= 1036,1 kN > 19,7 kN
1,1
(EC 3- P 1: 6.2.6)
6KHDUUHVLVWDQFHGXHWRWRUVLRQ $QQH[*RI(XURFRGH3DUW max M t ,Sd = 1,35 ⋅ 3,4 = 4,6 kN
τ V ,Ed =
M t ,Sd ⋅ t
It
=
kN f y / 3
kN
4,6 ⋅ 2,8 ⋅ 100
= 2,39
<
= 12,3 2
2
538
γM
cm
cm
(EC 3- P 1: 6.2.6)
,QWHUDFWLRQEHWZHHQQRUPDODQGVKHDUIRUFHV
Vpl,Rd =
(
Av ⋅ fy / 3
γM
Vpl,T ,Rd = 1 −
) = 697,5 kN
(
τ t ,Ed
)
1,25 ⋅ f y / 3 / γ M 0
(EC 3- P 1: 6.2.6 (2))
⋅Vpl,Rd = 1 −
2,39
⋅ 697,5 = 641,0 kN
1,25 ⋅ 12,3
(EC 3- P 1: 6.2.7)
VEd = 192,2 kN ≤ 321,0 = 0,5 ⋅ Vpl,T ,Rd
⇒ no interaction between shear and normal stresses necessary
0
(EC 3- P 1: 6.2.8)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 27 -
$[LDOIRUFHV
1RWH
It is assumed that the horizontal loads are resisted by the top flange. The rail is
rigid fixed with clamps on the top flange. Therefore the net section properties
of the crane runway girder are considered. The cross-section properties of the
rail are not taken into account though the rail is rigid fixed.
N Sd = −1,35 ⋅ 4,5 = −6,1 kN
∆A = 2 ⋅ 21 ⋅ 28 = 11,8 cm 2
A TF,net = A TF − ∆A = 84,0 − 11,8 = 72,2 cm 2
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
N Sd
≤ 1,0
A TF,net ⋅ f y,d
6,1
= 0,004 ≤ 1,0
72,2 ⋅ 23,5 / 1,1
(EC 3- P 1: 6.2.4)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 28 -
5HVLVWDQFHRIWKHZHEWRWUDQVYHUVHIRUFHV
The resistance of the web to transverse forces is determined according to section 4.4 of
the draft of Eurocode 3 - Part 1.5: „Supplementary rules for planar plated structures
without transverse loading“.
s s = 2 ⋅ h + 50 = 2 ⋅ (0,75 ⋅ 65) + 50 = 14,75 cm
h w = 500 − 2 ⋅ 28 = 444 mm
k f = 6,0 + 2,0 ⋅ (h w / a ) 2 = 6,0 + 2 ⋅ (44,4 / 700) 2 = 6,0
(EC 3- P 5: 6.1 (4))
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
3
0,9 ⋅ k f ⋅ E ⋅ t 3w 0,9 ⋅ 6,0 ⋅ 21000 ⋅ 1,45
Fcr =
=
= 7786,4 kN
hw
44,4
m1 =
f yf ⋅ b f
f yw ⋅ t w
=
(EC 3- P 5: 6.4 (1))
235 ⋅ 300
= 20,7
235 ⋅ 14,5
2
(EC 3- P 5: 6.5 (1))
h 
 444 
m 2 = 0,02 ⋅  w  = 0,02 ⋅ 
 = 5,0
28
t

 f 
2
[
]
(EC 3- P 5: 6.5 (1))
[
]
l y = s s + 2 ⋅ t f ⋅ 1 + m 1 + m 2 = 14,75 + 2 ⋅ 1,45 ⋅ 1 + 20,7 + 5,0 = 32,4 cm
(EC 3- P 5: 6.5 (2))
λF =
l y ⋅ t w ⋅ f yw
Fcr
=
32,4 ⋅ 1,45 ⋅ 23,5
= 0,38 < 0,5 ⇒ κ F = 1
7786,4
(EC 3- P 5: 6.4 (1))
⇒ l eff = κ F ⋅ l y = 1,0 ⋅ 32,4 = 32,4 cm
(EC 3- P 5: 6.2)
FRd = l eff ⋅ t w ⋅ f yw = 32,4 ⋅ 1,45 ⋅ 23,5 = 1104,0 kN
(EC 3- P 5: 6.2)
FSd = γ Q ⋅ Q r ,max = 1,35 ⋅ 82,0 = 110,7 kN
FSd < FRd
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 29 -
)DWLJXH
*HQHUDO
According to 9.1.4 of Eurocode 3 - Part 6 no fatigue assessment is necessary, if the
number of load cycles with more than 50 % of the full payload is smaller than 10000
cycles.
In the design example this condition is not fulfilled, so that a fatigue check is necessary.
The fatigue assessment is carried out for the crane runway girder on the basis of
nominal stress ranges.
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
∆σ c
γ Mf
∆σ E 2 = λ ⋅ Φ fat ⋅ ∆σ P
γ Ff = 1,0
γ Mf = 1,15
γ Ff ∆σ E 2 ≤
(EC 3- P 9: 8 (2))
(EC 3- P 6: 9.4.1 (4))
(EC 3- P 6: 9.3 (1))
(EC 3- P 9: Table 3.1)
Provided that the crane is classified into loading class S6 the following values are
obtained from Eurocode 1 – Part 3:
λ = 0,794 for normal stresses
λ = 0,871 for shear stresses
Φ fat = 1,1
(EC 1- P 3: Table 2.12)
(EC 1- P 3: Table 2.12)
(EC 1- P 3: 2.12.1 (7))
In the design example the stresses ∆σ E 2 are direct calculated with the following fatigue
loads:
for normal stresses:
Q e ,i = λ ⋅ Φ fat ⋅ Q max,i = 0,794 ⋅ 1,1 ⋅ 70,0 = 61,1 kN
(EC 1- P 3: 2.12.1 (4))
for shear stresses:
Q e ,i = λ ⋅ Φ fat ⋅ Q max,i = 0,871 ⋅ 1,1 ⋅ 70,0 = 67,1 kN
(EC 1- P 3: 2.12.1 (4))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 30 -
'HWDLOFDWHJRULHV
The runway beam is checked for the following detail categories which were obtained
from Eurocode 3 - Part 9 (Tab. 8.1, 8.2, 8.10).
Detail category
125
Constructional detail
Amendments
Verification of normal
stresses in the runway
beam.
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
80
Verification of normal
stresses in the runway
beam.
80
Verification of shear
stresses in the web.
160
Verification of vertical
stresses in the web due
to wheel loads.
(Eurocode 3 - Part 6)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 31 -
3RLQW
9HULILFDWLRQRIWKHFURVVVHFWLRQ
a) Selfweight
M y = 13,0 kNm
b) Wheel loads
max M y = Q e ,i ⋅ (η1 + η 2 ) ⋅ l = 61,1 ⋅ (0,242 + 0,0952) ⋅ 7,0 = 144,2 kNm
min M y = 0,0 kNm
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Normal stresses at the top flange
Detail category 80 (due to the net section properties by clamps)
144,2 + 13,0
kN
= 3,7
4290,0
cm 2
0,0 + 13,0
kN
min σ x =
= 0,3 2
4290,0
cm
kN
∆σ E 2 = 3,7 − 0,3 = 3,4
cm 2
kN
8,0
∆σ c =
= 7,0 2
1,15
cm
max σ x =
∆σ E 2 < ∆σ c
Normal stresses at the lower flange
Detail category 125
144,2 + 13,0
kN
= 3,7
4290,0
cm 2
0,0 + 13,0
kN
min σ x =
= 0,3 2
4290,0
cm
kN
∆σ E 2 = 3,7 − 0,3 = 3,4
cm 2
12,5
kN
∆σ c =
= 10,9
1,15
cm 2
max σ x =
∆σ E 2 < ∆σ c
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 32 -
9HULILFDWLRQRIWKHZHE
6KHDUVWUHVVHV
a) Selfweight
Vz = 1,4 kN
kN
τ xz ≈ 0
cm 2
b) Wheel loads
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
max Vz = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (0,589 + 0,232) = 55,1 kN
min Vz = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (−0,054 − 0,411) = −31,2 kN
55,1
kN
= 0,9 2
44,4 ⋅ 1,45
cm
− 31,2
kN
=
= −0,5 2
44,4 ⋅ 1,45
cm
max τ xz =
min τ xz
c) Local shear stresses in the web due to wheel loads
d r = 0,75 ⋅ h r + t f + r = 0,75 ⋅ 65 + 28 + 27 = 104 mm
b eff = b fr + d r = 150 + 104 = 254 mm < b = 300 mm
3
t 3f ⋅ b eff 2,8 ⋅ 25,4
I f ,eff =
=
= 46,5 cm 4
12
12
I r = 136 cm 4 (25 % wear, see “Petersen Stahlbau”, page 1360)
I rf = I r + I f ,eff = 136 + 46,5 = 182,5 cm
1
4
Fz
67,1
kN
=
= 2,8
l eff ⋅ t w 16,3 ⋅ 1,45
cm 2
kN
τ || = 0,2 ⋅ σ ⊥ = 0,2 ⋅ 2,8 = 0,6
cm 2
kN
max τ || = 0,9 + 0,6 = 1,5
cm 2
kN
min τ || = −0,5 − 0,6 = −1,1
cm 2
kN
∆τ E 2 = 1,5 + 1,1 = 2,6
cm 2
8,0
kN
∆τ c =
= 6,4
1,25
cm 2
∆τ E 2 < ∆τ c
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 6.2.1 (13))
(EC 3- P 6: 7.5.2 (2))
1
l eff = 3,25 ⋅ [I rf / t w ] 3 = 3,25 ⋅ [182,5 / 1,45] 3 = 16,3 cm
σ⊥ =
(EC 3- P 6: 7.5.2 (1))
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 7.5.2 (1))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 33 -
'LUHFWVWUHVVHV
a) Local stresses in the web due to wheel loads
l eff = 16,3 cm
σ⊥ =
(EC 3- P 6: 7.5.2 (2))
Fz
61,1
kN
=
= 2,6
l eff ⋅ t w 16,3 ⋅ 1,45
cm 2
(EC 3- P 6: 7.5.2 (1))
b) Local stresses in the web due to bending
TSd = Fz ,d ⋅ e y = 61,1 ⋅ 0,01375 = 0,84 kNm
(EC 3- P 6: 9.4.2.2 (1))
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
a = 700,0 cm
d w = 50,0 − 2 ⋅ 2,8 = 44,4 cm
t w = 1,45 cm
1
I t ≈ ⋅ 30,0 ⋅ 2,8 3 = 220 cm 4
3
 0,75 a t 3w

sinh 2 (π d w / a )
η=
⋅

sinh(2 π d w / a ) − 2 π d w / a 
 It
0,5
 0,75 ⋅ 700 ⋅ 1,45 3

sinh 2 (π ⋅ 44,4 / 700)
=
⋅

220
sinh(2 ⋅ π ⋅ 44,4 / 700) − 2 ⋅ π ⋅ 44,4 / 700 

6 TSd
⋅ η ⋅ tanh (η)
a t 2w
kN
6 ⋅ 0,84 ⋅ 100
⋅ 5,247 ⋅ tanh (5,247) = 1,8
=
2
700 ⋅ 1,45
cm 2
σ T ,Ed =
kN
cm 2
kN
= 1,8 − 1,8 = 0
cm 2
max σ T ,Sd = 1,8 + 1,8 = 3,6
min σ T ,Sd
⇒ max ∆ σ E = 3,6
∆ σc =
kN
cm 2
16,0
kN
= 12,8
1,25
cm 2
∆σ E < ∆σ c
(EC 3- P 6: 9.4.2.2 (1))
0, 5
= 5,247
(EC 3- P 6: 9.4.2.2 (1))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 34 -
,QWHUDFWLRQEHWZHHQGLUHFWDQGVKHDUVWUHVVHVLQWKHZHE
3
5

 

 γ ⋅ ∆σ   γ ⋅ ∆τ 
E2 
E2 
 Ff
+  Ff
≤ 1,0
 ∆σ c   ∆τ c 
 γ
  γ

Mf
Mf

 

3
(EC 3- P 9: 8 (3))
5

 

1,0 ⋅ 3,6  1,0 ⋅ 2,6 

 +
 = 0,033 ≤ 1,0
 16,0   8,0 
 1,25   1,25 
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 35 -
6XSSRUW
9HULILFDWLRQRIWKHZHE
6KHDUVWUHVVHV
a) Selfweight
Vz = −7,7 kN
− 7,7
kN
τ xz =
= −0,1 2
44,4 ⋅ 1,45
cm
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
b) Wheel loads
max Vz = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ 0,0 = 0,0 kN
min Vz = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (−1,0 − 0,6428) = −110,2 kN
0,0
kN
max τ xz =
= 0,0
44,4 ⋅ 1,45
cm 2
kN
− 110,2
min τ xz =
= −1,7
44,4 ⋅ 1,45
cm 2
c) Local shear stresses in the web due to wheel loads
1
3
1
3
l eff = 3,25 ⋅ [I rf / t w ] = 3,25 ⋅ [182,5 / 1,45] = 16,3 cm
Fz
67,1
kN
=
= 2,8
l eff ⋅ t w 16,3 ⋅ 1,45
cm 2
kN
τ xz = 0,2 ⋅ σ ⊥ = 0,2 ⋅ 2,8 = 0,6
cm 2
kN
max τ xz = 0,0 + 0,6 = 0,6
cm 2
kN
min τ xz = −1,7 − 0,6 = −2,3
cm 2
kN
∆τ E 2 = 0,6 + 2,3 = 2,9
cm 2
8,0
kN
∆τ c =
= 6,4
1,25
cm 2
σ⊥ =
∆τ E 2 < ∆τ c
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 7.5.2 (1))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 36 -
'LUHFWVWUHVVHV
a) Local stresses in the web due to wheel loads
l eff = 16,3 cm
kN
, see 6.3.2.2 (a)
σ ⊥ = 2,6
cm 2
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 7.5.2 (1))
b) Local stresses in the web due to bending
kN
, see 6.3.2.2 (b)
cm 2
kN
= 1,8
cm 2
σ T ,Ed = 1,8
(EC 3- P 6: 9.4.2.2 (1))
s
e
I
d
B
o
c
A
F uro 11
0
E
2
e
l
0
c
1
y
0
C 2
σ T ,Ed
kN
cm 2
kN
= 1,8 − 1,8 = 0
cm 2
max σ T ,Sd = 1,8 + 1,8 = 3,6
min σ T ,Sd
⇒ max ∆ σ E = 3,6
∆ σc =
kN
cm 2
kN
16,0
= 12,8
1,25
cm 2
∆σ E < ∆σ c
,QWHUDFWLRQEHWZHHQGLUHFWDQGVKHDUVWUHVVHVLQWKHZHE
3
5

 

 γ ⋅ ∆σ   γ ⋅ ∆τ 
E2 
E2 
 Ff
+  Ff
≤ 1,0
∆
σ
∆
τ

 

c
c
 γ
  γ

Mf
Mf

 

3
5

 

1,0 ⋅ 3,6  1,0 ⋅ 2,9 

 +
 = 0,041 ≤ 1,0
 16,0   8,0 
 1,25   1,25 
(EC 3- P 9: 8 (3))