th
Aachen, October19 , 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-
Page
1.1 GENERAL ......................................................... ............................................................ ....................... 6 1.2 GEOMETRIC PROPERTIES ...................................................... ............................................................ ... 6 1.3 MECHANICAL PROPERTIES ................................................... ............................................................ ... 6
2.1 GENERAL ......................................................... ............................................................ ....................... 6 2.2 DYNAMIC MAGNIFICATION FACTOR ϕ 1 ..................................................... ........................................... 6 2.3 DYNAMIC MAGNIFICATION FACTOR ϕ 2 ..................................................... ........................................... 7 2.4 DYNAMIC MAGNIFICATION FACTOR ϕ 3 ..................................................... ........................................... 7 2.5 DYNAMIC MAGNIFICATION FACTOR ϕ 4 ..................................................... ........................................... 7 2.6 DYNAMIC MAGNIFICATION FACTOR ϕ 5 ..................................................... ........................................... 7
3.1 GENERAL ......................................................... ............................................................ ....................... 8 3.2 UNLOADED CRANE ..................................................... ............................................................ ............. 9 3.3 LOADED CRANE ......................................................... ............................................................ ............. 9
4.1 GENERAL ......................................................... ............................................................ ..................... 11 4.2 CAUSED BY ACCELERATION AND DECELERATION OF THE CRANE ................................................. ..... 11 4.3 CAUSED BY SKEWING OF THE CRANE ................................................... ............................................. 12 4.4 CAUSED BY ACCELERATION OR BRAKING OF THE CRAB ..................................................... ............... 15
Design example for Eurocode 3 – Part 6: Cranes supporting structures
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Page
1.1 GENERAL ......................................................... ............................................................ ....................... 6 1.2 GEOMETRIC PROPERTIES ...................................................... ............................................................ ... 6 1.3 MECHANICAL PROPERTIES ................................................... ............................................................ ... 6
2.1 GENERAL ......................................................... ............................................................ ....................... 6 2.2 DYNAMIC MAGNIFICATION FACTOR ϕ 1 ..................................................... ........................................... 6 2.3 DYNAMIC MAGNIFICATION FACTOR ϕ 2 ..................................................... ........................................... 7 2.4 DYNAMIC MAGNIFICATION FACTOR ϕ 3 ..................................................... ........................................... 7 2.5 DYNAMIC MAGNIFICATION FACTOR ϕ 4 ..................................................... ........................................... 7 2.6 DYNAMIC MAGNIFICATION FACTOR ϕ 5 ..................................................... ........................................... 7
3.1 GENERAL ......................................................... ............................................................ ....................... 8 3.2 UNLOADED CRANE ..................................................... ............................................................ ............. 9 3.3 LOADED CRANE ......................................................... ............................................................ ............. 9
4.1 GENERAL ......................................................... ............................................................ ..................... 11 4.2 CAUSED BY ACCELERATION AND DECELERATION OF THE CRANE ................................................. ..... 11 4.3 CAUSED BY SKEWING OF THE CRANE ................................................... ............................................. 12 4.4 CAUSED BY ACCELERATION OR BRAKING OF THE CRAB ..................................................... ............... 15
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-3-
1.1 SYSTEM.................................................. ............................................................ ............................... 18 1.2 CROSS-SECTION PROPERTIES ...................................................... ....................................................... 18
2.1 GENERAL ......................................................... ............................................................ ..................... 18 2.2 INTERNAL FORCES AND MOMENTS AT POINT 2.875 .................................................... ....................... 19 2.3 INTERNAL FORCES AND MOMENTS AT SUPPORT ........................................................ ......................... 22
3.1 POINT 2.875 ........................................................ ........................................................ ...................... 24 3.2 SUPPORT .......................................................... ............................................................ ..................... 26
5.1 GENERAL ......................................................... ............................................................ ..................... 29 5.2 DETAIL CATEGORIES ............................................................ ............................................................ . 30 5.3 POINT 2.785 ........................................................ ........................................................ ...................... 31 5.4 SUPPORT .......................................................... ............................................................ ..................... 35
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-4-
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.
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-5-
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-6-
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.
The following geometric properties are assumed in the design example for the crane: Span length of the crane bridge: 15,00 m Wheel spacing a: 2,50 m Min. spacing between crab and supports e min: 0,00 m
The following mechanical properties are defined for the crane:
Self-weight of the crane Q c1 : 60,0 kN Self-weight of the crab Q c2 : 10,0 kN Hoistload Qh,nom : 100,0 kN
The dynamic effects of a crane structure are taken into account by magnification factors which are defined in Eurocode 1 - Part 3.
The magnification factor ϕ 1 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.
ϕ 1 = 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-
The magnification factor ϕ 2 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 ⋅ 60
= 1,20
(EC 1- P 3: Table 2.4)
The parameters ϕ 2,min and β 2 were obtained from table 2.5 of EC 1- Part 3.
The magnification factor ϕ 3 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.
ϕ 3 = 1,00
(EC 1- P 3: Table 2.4)
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.
ϕ 4 = 1,00
(EC 1- P 3: Table 2.4)
The magnification factor ϕ 5 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.
ϕ 5 = 1,50
(EC 1- P 3: Table 2.6)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-8-
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. Groups of loads Symbol
Section
ULS
SLS
Accidental
1
2
3
4
5
6
7
8
9
10
1 Self-weight of crane
Qc
2.6
1
1
1
1
1
2 Hoist load
Qh
2.6
-
-
1
1
HL, HT
2.7
-
-
-
5
-
-
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
-
-
-
-
-
-
-
6
-
-
8 Buffer force
HB
2.11
-
-
-
-
-
-
-
-
7
-
9 Tilting force
HTA
2.11
-
-
-
-
-
-
-
-
-
1
3 Acceleration of crane bridge 4
1)
Skewing of crane bridge
is
*
the part of the hoist load that remains when the payload is removed, but is not included in the selfweight of the crane.
Design example for Eurocode 3 – Part 6: Cranes supporting structures
-9-
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)
a
a) Load group 1,2
⇒ Q C1,k = 1,1 ⋅ 60,0 = 66,0 kN ⇒ Q C 2,k = 1,1 ⋅ 10,0 = 11,0 kN
ϕ = 1,1:
∑Q ∑Q
1
r ,(min)
= ⋅ 66,0 + 11,0 = 44,0 kN ⇒ 2
1
r , min
= ⋅ 66,0 = 33,0 kN
⇒
2
Q r ,(min)
Q r , min
= 22,0 kN
= 16,5 kN
b) Load group 3,4,5,6
⇒ 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
1
r ,(min)
= ⋅ 60,0 + 10,0 = 40,0 kN ⇒ 2
1
r , min
= ⋅ 60,0 = 30,0 kN
⇒
2
Q r ,(min) Q r ,min
= 20,0 kN
= 15,0 kN
The maximum vertical wheel loads apply at a crane runway girder when the crane is loaded. Qr,max
Qr,max
ΣQr,max
ΣQr ,(max)
Crab
Qr, (max)
Qr, (max)
Qh,nom = nominal hoist load emin
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
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⇒ Q c 2,k = 1,1⋅10,0 = 11,0 kN ⇒ Q h ,k = 1,2 ⋅100,0 = 120,0 kN
ϕ = 1,2:
∑Q ∑Q
1
r ,(max)
= ⋅ 66,0 = 33,0 kN 2
⇒
1
⋅ 66,0 + 11,0 + 120,0 = 164,0 kN ⇒ r , max = 2
Q r ,(max)
Q r ,max
= 16,5 kN
= 82,0 kN
b) Load group 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 h ,k = 1,0 ⋅100,0 = 100,0 kN
∑ ∑Q
Q r ,(max)
1
= ⋅ 66,0 = 33,0 kN 2
⇒
1
r , max
= ⋅ 66,0 + 11,0 + 100,0 = 144,0 kN ⇒ 2
Q r ,(max)
Q r ,max
= 16,5 kN
= 72,0 kN
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 h ,k = 1,0 ⋅100,0 = 100,0 kN
∑Q ∑Q
1
r ,(max)
= ⋅ 60,0 = 30,0 kN 2
1
r , max
⇒
= ⋅ 60,0 + 10,0 + 100,0 = 140,0 kN ⇒ 2
Q r ,(max)
Q r , max
= 15,0 kN
= 70,0 kN
Design example for Eurocode 3 – Part 6: Cranes supporting structures
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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; Rail i = 1
Rail i = 2
K
K2
Friction factor:
Number of single wheel drivers:
∑Q K
* 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))
= µ ⋅ ∑ Q *r ,min = 0,2 ⋅ 30,0 = 6,0 kN
(EC 1- P 3: 2.7.3(3))
Rail i = 1
H
L,1
Rail i = 2
H
L,2
Number of runway beams: H L,1 = H L,2 = ϕ5 ⋅
K nr
= 1,5 ⋅
nR = 2
6,0 2
= 4,5 kN
(EC 1- P 3: 2.7.2(2))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
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Rail i = 1
ail i =
H T ,2
H T ,1 M
S
a
HT ,1
H T ,2
s
K
K = K 1+
1
ξ
1
K2
K2
ξ
2
ξ1
=
∑ Qr,max ∑ Qr
∑ Q =∑ Q r
140,0
(EC 1- P 3: 2.7.2(3))
r , max
+ ∑ Q r ,(max) = 140,0 + 30,0 = 170,0 kN
(EC 1- P 3: 2.7.2(3))
= 0,82
(EC 1- P 3: 2.7.2(3))
= 0,18
(EC 1- P 3: 2.7.2(3))
ξ1
=
ξ2
= 1 − ξ1
lS
= (ξ1 − 0,5)⋅ l = (0,83 − 0,5)⋅ 15,0 = 4,95 m
(EC 1- P 3: 2.7.2(3))
M
= K ⋅ lS
(EC 1- P 3: 2.7.2(3))
170,0
= 6,0 ⋅ 4,95 = 29,7 kNm
H T ,1 = ϕ 5
⋅ ξ2 ⋅
H T,2 = ϕ5
⋅ ξ1 ⋅
M a M a
= 1,5 ⋅ 0,18 ⋅ = 1,5 ⋅ 0,82 ⋅
29,7 2,5 29,7 2,5
= 3,2 kN
(EC 1- P 3: 2.7.2(3))
= 14,6 kN
(EC 1- P 3: 2.7.2(3))
αF =
0,75 x
αV =
y
10
= 0,004 rad
(EC 1- P 3: Table 2.7)
= 0,002 rad
(EC 1- P 3: Table 2.7)
α0 =
= 0,001 rad
(EC 1- P 3: Table 2.7)
α = αF + α V + α 0
-------------= 0,007 rad
a a
= =
2500 0,1 ⋅ 50 2500
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 13 -
f = 0,3 (1 − exp (− 250 α )) = 0,3 (1 − exp (− 250 ⋅ 0,007 )) = 0,248
(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 (b) Combination of wheel pairs: IFF m=0 (c) Distance h:
h
=
mξ1ξ 2 l 2
+ ∑ e j2
∑e
j
=
0 + 2,50 2 2,50
= 2,50 m
(EC 1- P 3: Table 2.8)
n=2
λ S = 1 − ∑ j = 1 − 2,50 = 0,5 n⋅h 2 ⋅ 2,50
(EC 1- P 3: Table 2.9)
λ S,1,L = λ S, 2,L = 0
(EC 1- P 3: Table 2.9)
e
for wheel pair 1:
ξ 2 e1 0,18 (1 − 0) = 0,09 1 − = n h 2 ξ e 0,82 (1 − 0) = 0,41 = 1 1 − 1 = n h 2
λ S,1,1,T =
(EC 1- P 3: Table 2.9)
λ S, 2,1,T
(EC 1- P 3: Table 2.9)
for wheel pair 2:
ξ 2 e 2 0,18 2,50 1 − = 0 1 − = n h 2 2,50 ξ e 0,82 2,50 = 1 1 − = 0 1 − 2 = n h 2 2,50
λ S,1, 2,T =
(EC 1- P 3: Table 2.9)
λ S, 2, 2,T
(EC 1- P 3: Table 2.9)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 14 -
Rail i = 1
H
Rail i = 2
H
L,1
L,2
= f ⋅ λ S,1,L ⋅ ∑ Q r = 0 H S, 2,L = f ⋅ λ S, 2,L ⋅ ∑ Q r = 0 H S,1,L
(EC 1- P 3: 2.7.4(1)) (EC 1- P 3: 2.7.4(1))
Rail i = 1
Directon of motion
Rail i = 2
α H
S
Wheel pair j = 1
HS,2,1,T
Wheel pair j = 2
Guide force S: S = f ⋅ λ S ⋅
∑Q
r
= 0,248 ⋅ 0,5 ⋅170,0 = 21,1 kN
(EC 1- P 3: 2.7.4(1))
for wheel pair 1:
= f ⋅ λ S,1,1,T ⋅ ∑ Q r = 0,248 ⋅ 0,09 ⋅ 170,0 = 3,8 kN H S, 2,1,T = f ⋅ λ S, 2,1,T ⋅ ∑ Q r = 0,248 ⋅ 0,41 ⋅ 170,0 = 17,3 kN H S,1,1,T
(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:
= f ⋅ λ S,1, 2,T ⋅ ∑ Q r = 0,248 ⋅ 0 ⋅ 170,0 = 0 kN H S, 2, 2,T = f ⋅ λ S, 2, 2,T ⋅ ∑ Q r = 0,248 ⋅ 0 ⋅ 170,0 = 0 kN H S,1, 2,T
(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 -
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)
e=
1 4
1
⋅ b r = ⋅ 55 = 13,75 mm 4
(EC 1- P 3: 2.5.3(2))
Q e ,i
= ϕ fat ⋅ λ i ⋅ Q max,i
ϕ fat ,1 = ϕ fat , 2 =
1 + ϕ1 2 1 + ϕ2 2
= =
(EC 1- P 3: 2.12.1(4))
1 + 1,1 2 1 + 1,2 2
= 1,05
(EC 1- P 3: 2.12.1(7))
= 1,10
(EC 1- P 3: 2.12.1(7))
Assumption: crane is classified in class S 6:
λ i = 0,794 λ i = 0,871
for normal stresses
(EC 1- P 3: Table 2.12)
for shear stresses
(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 -
For the the results are summarised in the following table according to the groups of loads.
Groups of loads
1
Magnification factor which are considered for the group of load
Vertical loads
2
3
4
=
1,10
=
1,10
=
1,00
=
1,00
=
1,20
=
1,00
=
1,50
=
1,50
=
1,50
=
1,50
5 =
1,00
6 =
1,00
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 -
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 18 -
Single-span girder with fork-support, length: l = 7,00 m
In the design example it is assumed that the rail is rigid fixed with clamps on the crane runway girder. 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
3
Iz [cm ] 12620
Wel,y [cm ] 4290
3
Wel,z [cm ] 842
2
AF = 30028,0 = 84,0 cm 2 AW = 44414,5 = 64,4 cm
Cross-section properties of the rail A55: 2
4
A [cm ] 40,5
Iy [cm ] 178
4
Iz [cm ] 337
Material S235
The cross-section is classified into class 1.
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
Load position for the maximum bending moment: Qr,max
2,50
Qr,max
x
7,00
A=
11,5 − 2 ⋅ x 7,0
M( x ) = A ⋅ x M’( x ) =
=
11,5 ⋅ x − 2 ⋅ x 2
11,5 − 4 ⋅ x 7,0
7,0
=0
for max M
⇒ x = 2,875
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 − 2 Vz , k = 7,7 − 2,875 ⋅ 2,2 = 1,4 kN
= 13,0 kNm
a) Bending moment Qr,max
2,50
Qr,max
0,095 0,242
= Q r ,max ⋅ (η1 + η2 ) ⋅ l = 82,0 ⋅ (0,242 + 0,095) ⋅ 7,0 = 193,7 kNm min M y ,k = 0 kNm max M y , k
- 19 -
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
Qr,max
Qr,max
2,50
0,411
0,054
min Vz ,k
= Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (−0,054 − 0,411) = −38,1 kN
a) Bending moment HT,2
2,50
HT,2
0,095 0,242
min M z , k = ( H T , 2 HT,2
2,50
⋅ η1 + H T, 2 ⋅ η 2 ) ⋅ l = (−14,6 ⋅ 0,242 + 14,6 ⋅ 0,095) ⋅ 7,0 = −15,0 kNm 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 HT,2
⋅ η1 + H T, 2 ⋅ η 2 = (−14,6) ⋅ (−0,411) + 14,6 ⋅ 0,232 = 9,4 kN
2,50
0,411
0,054
min Vy ,k N k
HT,2
= H T , 2 ⋅ η1 + H T ,2 ⋅ η 2 = (−14,6) ⋅ (−0,054) + 14,6 ⋅ (−0,411) = −5,2 kN
= − 4,5 kN
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 Horizontal loads due to acceleration and deceleration: H T ez
= 0,5 ⋅ h + h 1 = 0,5 ⋅ 500 + 65 = 315 mm
= 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 M t1
2,50 Mt2
Mt1
0,054
0,411 0,589
max M t ,k
+
= 5,7 ⋅ 0,589 − 3,5 ⋅ (−0,054) = 3,5 kNm
(EC 1-P 3: 2.5.3 (2))
= ±14,6 kN
Design example for Eurocode 3 – Part 6: Cranes supporting structures
There are no bending moments at the support (Single-span girder)
Vz , k
= 7,7 kN
max Vz , k
= Q r ,max ⋅ (η1 + η2 ) = 82,0 ⋅ (0,0 + 0,0) = 0 kN Qr,max
2,50
0,643
min Vz ,k
Qr,max
1,0
= Q r ,max ⋅ (η1 + η 2 ) = 82,0 ⋅ (−1,0 − 0,6428) = −134,7 kN
HT,2
2,50
HT,2
1,0
max Vy , k = H T , 2 HT,2
⋅ η1 + H T , 2 ⋅ η 2 = (−14,6) ⋅ (−1,0) + 14,6 ⋅ 0,0 = 14,6 kN HT,2
2,50
0,357
min Vy , k = H T , 2 ⋅ η1 + H T , 2 ⋅ η2
N k
= − 4,5 kN
= (−14,6) ⋅ 0,0 + 14,6 ⋅ (−0,357) = −5,2 kN
- 22 -
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 23 -
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 Horizontal loads due to acceleration and deceleration: H T ez
= 0,5 ⋅ h + h 1 = 0,5 ⋅ 500 + 65 = 315 mm
= 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 M t1
2,50 Mt1
1,0
Mt2
+ 0,643
max M t ,k
= 5,7 ⋅1,0 − 3,5 ⋅ 0,643 = 3,4 kNm
(EC 1- P 3: 2.5.3 (2))
= ±14,6 kN
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 24 -
⇒ Verification for shear buckling is not necessary
d/tw = 390/14,5 = 26,9 < 60
(EC 3- P 1: 5.1)
= G G k + Q Q k = 1,35 ⋅ 1,4 + 1,35 ⋅ 67,3 = 92,75 kN = 390 ⋅ 14,5 = 56,55 cm 2
max Vz ,Sd AV
Vz , Rd
= AV ⋅
f y / 3
= 56,55 ⋅
γ M
235 / 3 1,1
= 697,5 kN > 92,75 kN
(EC 3- P 1: 6.2.6)
It is assumed that the horizontal loads are resisted by the top flange of the girder.
= Q Q k = 1,35 ⋅ 9,4 = 12,7 kN = A TV = 300 ⋅ 28 = 84,0 cm 2
max Vy ,Sd AV
Vy , Rd
= AV ⋅
f y / 3
= 84,0 ⋅
γ M
235 / 3 1,1
= 1036,1 kN > 12,7 kN
(EC 3- P 1: 6.2.6)
max M t ,Sd
τ V ,Ed
=
= 1,35 ⋅ 3,5 = 4,7 kN
M t ,Sd ⋅ t It
=
4,7 ⋅ 2,8 ⋅ 100 538
= 2,45
kN cm 2
<
f y / 3
γ M
= 12,3
kN cm 2
(EC 3- P 1: 6.2.6)
Vpl, Rd
=
Vpl,T ,Rd
(
A v ⋅ f y / 3
)
γ M =
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 1,25 ⋅ 12,3
⋅ 697,5 = 639,5 kN (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 -
It is assumed that the horizontal loads are resisted by the top flange.
a) Verification for max My,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 N Sd
ATF Wel,y Wel,z
= = =
N Sd A TF ⋅ f y,d
2
84 4290 842
+
M y ,Sd Wel, y ⋅ f y ,d
6,1 84 ⋅ 23,5 / 1,1
+
cm 3 cm 3 cm
+
M z,Sd Wel,z ⋅ f y ,d
279,0 ⋅ 100 4290 ⋅ 23,5 / 1,1
+
≤ 1,0
20,3 ⋅ 100 842 ⋅ 23,5 / 1,1
(EC 3- P 1: 6.2.1)
= 0,42 ≤ 1,0
b) Verification for max Mz,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 N Sd
ATF Wel,y Wel,z
= = =
N Sd A TF ⋅ f y,d
+
M y ,Sd Wel, y ⋅ f y ,d
6,1 84 ⋅ 23,5 / 1,1
2
84 4290 842
+
cm 3 cm 3 cm
+
M z,Sd Wel,z ⋅ f y ,d
229,5 ⋅ 100 4290 ⋅ 23,5 / 1,1
+
≤ 1,0
29,03 ⋅ 100 842 ⋅ 23,5 / 1,1
(EC 3- P 1: 6.2.1)
= 0,42 ≤ 1,0
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 -
⇒ Verification for shear buckling is not necessary
d/tw = 390/14,5 = 26,9 < 60
(EC 3- P 1: 5.1)
= G G k + Q Q k = 1,35 ⋅ 7,7 + 1,35 ⋅ 134,7 = 192,2 kN = 390 ⋅ 14,5 = 56,55 cm 2
max Vz ,Sd AV
Vz , Rd
= AV ⋅
f y / 3
= 56,55 ⋅
γ M
235 / 3 1,1
= 697,5 kN > 192,2 kN
(EC 3- P 1: 6.2.6)
It is assumed that the horizontal loads are resisted by the top flange of the girder.
= Q Q k = 1,35 ⋅14,6 = 19,7 kN = A TV = 300 ⋅ 28 = 84,0 cm 2
max Vy ,Sd AV
Vy , Rd
= AV ⋅
f y / 3
= 84,0 ⋅
γ M
235 / 3 1,1
= 1036,1 kN > 19,7 kN
(EC 3- P 1: 6.2.6)
= 1,35 ⋅ 3,4 = 4,6 kN ⋅ t 4,6 ⋅ 2,8 ⋅ 100 M = t ,Sd = = 2,39
max M t ,Sd
τ V ,Ed
It
538
kN cm 2
<
f y / 3
γ M
= 12,3
kN cm 2
(EC 3- P 1: 6.2.6)
Vpl, Rd
=
Vpl,T ,Rd
(
A v ⋅ f y / 3
)
γ M =
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 1,25 ⋅ 12,3
⋅ 697,5 = 641,0 kN (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 -
N Sd
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.
= −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 N Sd A TF, net ⋅ f y,d
≤ 1,0
6,1 72,2 ⋅ 23,5 / 1,1
= 0,004 ≤ 1,0
(EC 3- P 1: 6.2.4)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 28 -
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“.
= 2 ⋅ h + 50 = 2 ⋅ (0,75 ⋅ 65) + 50 = 14,75 cm h w = 500 − 2 ⋅ 28 = 444 mm
ss
= 6,0 + 2,0 ⋅ (h w / a ) 2 = 6,0 + 2 ⋅ (44,4 / 700) 2 = 6,0
k f
0,9 ⋅ k f ⋅ E ⋅ t 3w
Fcr
=
m1
=
m2
2 hw 444 = 0,02 ⋅ = 0,02 ⋅ = 5,0 28 t f
hw f yf ⋅ b f f yw ⋅ t w
=
=
0,9 ⋅ 6,0 ⋅ 21000 ⋅ 1,453 44,4
235 ⋅ 300 235 ⋅ 14,5
(EC 3- P 5: 6.1 (4))
= 7786,4 kN
= 20,7
(EC 3- P 5: 6.4 (1))
(EC 3- P 5: 6.5 (1))
2
ly
= s s + 2 ⋅ t f ⋅ [1 +
m1
+ m 2 ]= 14,75 + 2 ⋅ 1,45 ⋅ [1 +
(EC 3- P 5: 6.5 (1))
20,7 + 5,0
]= 32,4 cm (EC 3- P 5: 6.5 (2))
λ F =
l y ⋅ t w ⋅ f yw Fcr
=
32,4 ⋅ 1,45 ⋅ 23,5 7786,4
= 0,38 < 0,5 ⇒ κ F = 1
⇒ l eff = κ F ⋅ l y = 1,0 ⋅ 32,4 = 32,4 cm
FSd
= l eff ⋅ t w ⋅ f yw = 32,4 ⋅ 1,45 ⋅ 23,5 = 1104,0 kN = Q ⋅ Q r ,max = 1,35 ⋅ 82,0 = 110,7 kN
FSd
< FRd
FRd
(EC 3- P 5: 6.4 (1))
(EC 3- P 5: 6.2) (EC 3- P 5: 6.2)
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 29 -
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 f atigue check is necessary. The fatigue assessment is carried out for the crane runway girder on the basis of nominal stress ranges.
∆σ c γ Mf ∆σ E 2 = λ ⋅ Φ fat ⋅ ∆σ P Ff = 1,0 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 λ = 0,871 Φ fat = 1,1
for normal stresses for shear stresses
(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 loads:
∆σ E 2 are direct calculated with the following fatigue
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 -
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
Constructional detail
Amendments
125
Verification of normal stresses in the runway beam.
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
a) Selfweight My
= 13,0 kNm
b) Wheel loads max M y min M y
= Q e,i ⋅ (η1 + η 2 ) ⋅ l = 61,1 ⋅ (0,242 + 0,0952) ⋅ 7,0 = 144,2 kNm = 0,0 kNm
Normal stresses at the top flange Detail category 80 (due to the net section properties by clamps) max σ x
=
min σ x
=
144,2 + 13,0 4290,0 0,0 + 13,0 4290,0
= 0,3
∆σ E 2 = 3,7 − 0,3 = 3,4 ∆σ c =
8,0 1,15
= 7,0
= 3,7
kN
cm 2 kN
cm 2 kN
cm 2
kN cm 2
∆σ E 2 < ∆σ c Normal stresses at the lower flange Detail category 125 max σ x
=
min σ x
=
144,2 + 13,0 4290,0 0,0 + 13,0 4290,0
= 0,3
∆σ E 2 = 3,7 − 0,3 = 3,4 ∆σ c =
12,5 1,15
∆σ E 2 < ∆σ c
= 10,9
= 3,7
kN
cm 2 kN
cm 2 kN
cm 2
kN cm 2
- 31 -
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 32 -
a) Selfweight Vz
= 1,4 kN
τ xz ≈ 0
kN cm 2
b) Wheel loads
min Vz
= Q e,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (0,589 + 0,232) = 55,1 kN = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (−0,054 − 0,411) = −31,2 kN
max τ xz
=
min τ xz
=
max Vz
55,1 44,4 ⋅ 1,45 − 31,2
= 0,9
kN
cm 2 kN = −0,5 2 44,4 ⋅ 1,45 cm
c) Local shear stresses in the web due to wheel loads
= 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 = = 46,5 cm 4 I f ,eff =
dr
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 4 1
l eff
l eff ⋅ t w
=
67,1 16,3 ⋅ 1,45
= 2,8
τ|| = 0,2 ⋅ σ ⊥ = 0,2 ⋅ 2,8 = 0,6 max τ||
= 0,9 + 0,6 = 1,5
kN cm 2
kN cm 2
kN
cm 2 kN min τ || = −0,5 − 0,6 = −1,1 cm 2 kN ∆τ E 2 = 1,5 + 1,1 = 2,6 2 cm 8,0 kN ∆τ c = = 6,4 2 1,25 cm
∆τ E 2 < ∆τ c
(EC 3- P 6: 7.5.2 (2)) (EC 3- P 6: 6.2.1 (13))
1
3
Fz
(EC 3- P 6: 7.5.2 (2))
(EC 3- P 6: 7.5.2 (2))
= 3,25 ⋅ [I rf / t w ] = 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 (1))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 33 -
a) Local stresses in the web due to wheel loads l eff
= 16,3 cm
σ⊥ =
Fz
(EC 3- P 6: 7.5.2 (2))
=
l eff ⋅ t w
61,1 16,3 ⋅ 1,45
= 2,6
kN
(EC 3- P 6: 7.5.2 (1))
cm 2
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))
a = 700,0 cm tw
= 50,0 − 2 ⋅ 2,8 = 44,4 cm = 1,45 cm
It
≈ ⋅ 30,0 ⋅ 2,83 = 220 cm 4
dw
1
3
0,5
0,75 a t 3w sinh 2 ( π d w / a ) η= ⋅ (EC 3- P 6: 9.4.2.2 (1)) π − π I sinh( 2 d / a ) 2 d / a t w w 0,5 0,75 ⋅ 700 ⋅ 1,453 sinh 2 ( π ⋅ 44,4 / 700) = ⋅ = 5,247 220 sinh( 2 44 , 4 / 700 ) 2 44 , 4 / 700 ⋅ π ⋅ − ⋅ π ⋅ σ T,Ed = =
6 TSd a t 2w
⋅ η ⋅ tanh (η)
6 ⋅ 0,84 ⋅ 100 700 ⋅ 1,45
2
⋅ 5,247 ⋅ tanh (5,247) = 1,8
max σ T ,Sd
= 1,8 + 1,8 = 3,6
min σ T ,Sd
= 1,8 − 1,8 = 0
⇒ max ∆ σ E = 3,6 ∆ σc =
16,0 1,25
∆σ E < ∆σ c
= 12,8
(EC 3- P 6: 9.4.2.2 (1))
kN cm 2 kN cm 2
kN
cm 2 kN
cm 2
kN cm 2
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 34 -
3
5
γ ⋅ ∆σ γ ⋅ ∆τ E2 Ff + Ff E 2 ≤ 1,0 ∆σ c ∆τ c γ Mf γ Mf 3
5
1,0 ⋅ 3,6 1,0 ⋅ 2,6 + = 0,033 ≤ 1,0 16,0 8,0 1,25 1,25
(EC 3- P 9: 8 (3))
Design example for Eurocode 3 – Part 6: Cranes supporting structures
- 35 -
a) Selfweight
= −7,7 kN kN − 7,7 τ xz = = −0,1 2 44,4 ⋅ 1,45 cm Vz
b) Wheel loads
min Vz
= Q e,i ⋅ (η1 + η2 ) = 67,1 ⋅ 0,0 = 0,0 kN = Q e ,i ⋅ (η1 + η 2 ) = 67,1 ⋅ (−1,0 − 0,6428) = −110,2 kN
max τ xz
=
min τ xz
=
max Vz
0,0 44,4 ⋅ 1,45 − 110,2
= 0,0
kN
cm 2 kN = −1,7 2 44,4 ⋅ 1,45 cm
c) Local shear stresses in the web due to wheel loads 1
l eff
1
= 3,25 ⋅ [I rf / t w ] = 3,25 ⋅ [182,5 / 1,45] 3 = 16,3 cm
σ⊥ =
3
Fz l eff ⋅ t w
=
67,1 16,3 ⋅ 1,45
= 2,8
τ xz = 0,2 ⋅ σ ⊥ = 0,2 ⋅ 2,8 = 0,6 max τ xz
= 0,0 + 0,6 = 0,6
cm 2
kN cm 2
kN
cm 2 kN min τ xz = −1,7 − 0,6 = −2,3 cm 2 kN ∆τ E 2 = 0,6 + 2,3 = 2,9 2 cm 8,0 kN ∆τ c = = 6,4 2 1,25 cm
∆τ E 2 < ∆τ c
kN
(EC 3- P 6: 7.5.2 (2)) (EC 3- P 6: 7.5.2 (1))
