FEDERATION EUROPEENNE DE LA MANUTENTION
Section IX
SERIES LIFTING EQUIPMENT
Local Girder Stresses
FEM
9.341
'1 st edition
(E)·'10. 1983
2 Determination of the coefficients c
x ' CzThe coefficients established here are based on numerous test results 1). Theoretical investigations of certain test results with the method of finite elements have proved largely concurrent 1). 'he equations listed below, which are the product of test results, are valid for the ascertain-ment of the coefficients cx,Cz. Positive values of cx/z
mean tensile stress on the bottom of the flange. s
2
A= b 2
The variablesR,
t
1, j and A necessary for the stresscomputation have the following meanings:
R represents the maximum wheel load ascertained upon consideration of the dynamic coefficients.
tl is the theoretical thickness of the flange
at
the load positionj (without tolerances and wear). is the distance from the girder edge to the point of load application.A is calculated as the quotient from
j
Figure 1.1 Parallel flange track section
1 Local girder stresses
The flange bending stresses UFx and UFz arise as secon-dary stresses in the vicinity of the place of load appli-cation in a girder, regardless of its supporting structure (figure 1.1 and 1.2).
2.1 Parallel flange track section according to figure 1.1 Cz O ,=005-058.A+0148.e3,015.A (1), , . czl= 2,23 -1,49· A+1,390·e -18,33 . A (2) cz2
=
O.73-1,58.A+2,910.e-6.O' A (3) CxO = -2,11 + 1,977. A+0,0076.e
6,53·A (4) cxl = 10,108 -7,408 'A-l0,108.e-1,364(~
(6) Cx2= 0 Transition web/flange Load appli-cation point Edge of the flange Transition web/flange Load appli-cation point Edge of the flange l - - - -b _ . - ....-,10Figure 1.2 Girder with inclined flanges
2.2 Girder with inclined flanges according to figure 1.2 The stresses are calculated with the help of the equations
R UFx = Cx~ tl R uFz = Cz
2
tl "The factorsCx and c/usedin the equations can be
de-termined separately according to the type of girder (figure 1.1 and 1.2) and the load position j or A for the specially marked points (0), (1), (2) on the flange.
Transition web/flange Load appli-cation point Edge of the flange Transition web/flange CzO = -0,981"':"1,479· A+ 1,120· e1,322 . Am czl = 1.810-1.150.A+l,060.e-7.7OO .A(8) cz2 =1,990-2,810.A+0,840.e-4,690 .A(9) cxo= -1,096+1,095',A+0:192.e-6 •O 'A(tO)
1) Hannover, H.-a. und Reichwald, R.: Lokale Biegebeanspruchung von Triiger-Unterflanschen (Local flexural stressing of girder lower flanges),f+h-fordern und heben 32 (1982) Nr. 6 (Teil 1) und Nr. 8 (Teil 2)
Copyright by FEM Sektion IX . Also available in French and. German . !Sources of supply see back page
Page 2 FEM9.341
Figure2. Lower chord of a box type girder
in welding seams:
°'
cp-
-JO'
x2 2+az - O'x . O'z+2',Txz2<
O'awWhen determining stresses, flange bending stresses should be superimposed with the main stresses from vertical and lateral forces both in the general stress proof and in the service strength proof (e. g. in a box girder lower chord). Thus for O'penn it should be borne in mind that in crane girders there is a principal load picture with vertical load and lateral force. In superimposing flange bending stresses on principal stresses. the former are reduced by a factor
e.
This reduction may be accounted for by two basic facts:Flange bending stress produces a local stress peak only. The flange bending stress is very rapidly attenu-ated in the longitudinal direction of the girder. At a distance of 10mm from the point of the maximum, the flange bending stress is approximately only a half of the maximum.
4 Explanation of the design rules
Design rules are given for local flange bending stresses on rolled sections with inclined and parallel flanges. The design rules are based on the results of tests1). Measure-ments on the following sections were evaluated: 1200, I300as in DIN 1025part1and IPE200. 300, 360as in DIN1025part5.The load was distributed symmetrically along the longitudinal axes of the girders. The lower chord of box girders with an underrunning trolly should also be calculated using geometric characteristics with the equations for the parallel-flange girder.
The wheel load is ideally assumed to be a load point in the middle of the Hertzian surface. Tolerances in the thickness of the flange are not taken into account. Generally, no reduction in the flange thickness as a result of wear is to be taken into account. Results of tests on four overhead travelling cranes with underrunning trol-leys after 14 years of operation have shown wear of less than 1 mm. Only on heavily stressed suspension tracks is it possibly necessary to increase the thickness of the section on account of wear (e. g. by 5 mm for a flange thickness of30mm).
The equations in paragraphs 3.1 and 3.2apply to cranes. Corresponding equations, to be taken from the respec-tive national regulations, apply to crane runways or other steel structures. b* _!. 2 Cxl=3.965 - 4.835· A-3.965·e-2.675 .A (11 ) Cx2
=
0 (121 ~i-R .~<b
(~ ~tf-.!.---, • 2- t -
rr
-II
3.1 General proof of stress
In the case of composite plane stresses, the following must be proven with consideration of the signs in structural parts:
3 Ascertainment of stresses
The flange bending stresses aFz are to be superimposed on the main stresses O'Hz resulting from vertical and lateral forces. The flange bending stresses are diminished by the factor e=0.75.This also holds true for the flange bending stresses to be considered for the proof of service strength. Load
appli-cation point Edge of the flange
2.3 Lower chord of box type girder
The lower chord o'f a box type girder is to be calculated as a parallel flange track section. Figure2represents an analogous depiction.
O'yield point O'penn ' v =
- Taking into account local flange bending stresses increases the accuracy of the carculation. This prevents uncertainties. which would allow a lower safety factor
v.
32 Proo'f of service strength
( O'xmax
r
+(~zmax
)2_O'x~ax
. azmax +(Txzmax)2~
1,0*) O'xa O'za laxal· 100zal Tx:zaDefinitions:
The value € was arrived at by comparing the results of calculations for numerous single-girder overhead travel-ling cranes on the basis of both the traditional and the F EM methods. Consideration was given to craneS-which' had not been damaged by flange bending stresses even after many years of operation.
The flange bending stresses which accur are proved using a calculation example and the superimposition of flange bending stresses,on main stresses demonstrated.
calculated normal stress in the x and z directions
calculated shear stress
admissible normal stress corresponding to O'xmax and O'zmltli stresses
laxal
(UzaI sum total ofaxa and aza
admissible shear stress corresponding to the
Tx z max stress,
.) This inequality represents an unfavourable condition. allowing values slightly above1. If this is so, the following inequality is used for
the c a l c u l a t i o n : ' / 2 2 2
J
(aXmax) +(aZmax) _ ax max . az max+(xz max )
~ 1.05axa aza Iaxal . Iazal Tx za
Txza Txzmax O'xmax O'zmax 'axa O'za Licenced to SCHOSS SA
FEM9.341 Page 3
5.0 Calculation example
The application of the calculation formulae shown for flange bending stresses in the x and z directions is demonstrated in the following calculation example. The stresses in the middle of the main girder (see fig.5.1, intersection b) of a single·girder overhead travelling crane with parallel flange section girder are.to be·calculated. The design rules used are: FEM Rules for the Design of Hoisting Appliances, Section I(2nd edition) ..
5.3 Values incross section of main girdersection:IPBl360 The variables are labelled according to fig.5.2 to fig.5.4.
h 350 mm Wxo WXl = Wx2
=
1890 cm3 b 300 mm Wxow 2100 cm3 s 10mm Wyo 15780 cm3t
17,5 mm Wyl 580 cm3 Ix 33090 cm4 Wy2 526 cm3 I y 7890 cm4 i 14mm5.4 Calculation of the flange bending stresses using equations 1 to 6 (parallel flange section)
5.4.1 Calculation of A
5.1 Technical data of overhead travelling crane Single-girder overhead travelling crane 3,2 t x 11,0 m
SWL GH = 3200 kg
Span lKr
=
11,0 mApproach dimension lan =0,57 m Main girder IPB1360 Material: St 37
(DIN 1025 Part 3) 2 .b-si 300 - 102 . 14 0,0966
5.2 Calculation of the forces and moments in the middle of the main girder (intersectionb)
The proof of stress in the middle of the main girder is carried out with the load case I. Components consisting of the deadweight of the main girder and that of the trol-ley and the load lifted are taken into account in the ver· tical direction (y direction, see Fig. 5.4) and the inertia forces from long travel in the horizontal direction (x direction).
5.2.1 Oscillation coefficient Osci llation coefficient 1JJ
=
1,15End carriage Travel wheel dia. Wheel base
5.2.2 Max. wheel force R R = (G:a + G4H • 1JJ) . g
( 490 + 3200 . 1 15) .' 9 81
4 4 ' ,
10227 N
(6)
has the value
Cx2
=
0,0 (edge of flange)5.4.4 Flange bending stresses UF!(z) and uF!(x)
The quotient, occurring in all the equations 5.4.3 Factors Cxo, Cxl, Cx 2
(equations4to 6)
Cxo
=-
2,110 + 1,977 . 0,0966 + 0,0076 .e6 •53 . 0.0966cxo = - 1,905 (web/flange) (4) cx ) = 10,108 - 7,408·0,0966 - 10,108.e-1,364. 0,0966
cxl =0,532 (load application point) (5) 5.4.2 Factors czo, cZl< c z 2
.(equations 1 to 3)
Czo=0,05 - 0,58·0,0966+0,148. e3 ,015' 0,0966
czo = 0,192 (web/flange) (1 )
Cz 1=2,23 - 1,49 . 0,0966 + 1,390 .e-18 ,33 .0,0966
czl
=
2,323 (load attachment point) (2) Cz 2=0,73 - 1,58 . 0,0966 + 2,910 . e-6,0 . 0,0966 f:z2=2,207 (edge of flange) (3)g=9,81 m/s2 dKT = 0,16 m
eKT = 2,00 m Electric hoist trolley
vKr 31,5 m/min vKa 14,0 m/min VH 6,0 m/min GKa 490 kg eKa
=
0,420 m dKa=
0,140 m is classified as FEM Trolley. max. long travel speed . max. cross travel speed - max. lifting speed Trolley deadweight Trolley wheel base Trolley travel wheel dia
The overhead travelling crane Group 2. ·R R 33,4N/mm2 10227 - 175, 2 R
t.2
1Thus the flange bending stresses are calculated:
uF! (zO) = 0,192 33,4 = 6,4 N/mm2 uF! (zl)
=
2,323 . 33,4=
77,6 N/mm2 UF) (z2)=
2,207 . 33,4 = 73,7 N/mm2 uF! (xO)=
-1,905 33,4 = -63,6 N/mm2 UF! (x l) = 0,532 33,4 = 17,8 N/mm2 UF) (x2) = 0,0 1320 Nm 16614 Nm 12719 Nm 1370 Nm 83063 Nm 95522 Nm 5.2.3.2 ydirectionInertia forces from long travel component:
MY(Kr)
5.2.3 Bending moments in the x and y directions 5.2.3.1 x direction
Main girder deadweight
component Mx(HT)
Trolley deadweight component Mx(GKa)
min Mx(GKa)
Load lifted component Mx(GH)
Mx(GH) .1JJ
P,age 4r, FEM 9.341
5.5 General proof of. stress in middle of main girder (underside of flange)
5.5.1 Bendirig moments
5.5.4 Proof of stress point 0 (transition web/flange) z direction: Mxges ~ az(O) = - - + + € • aFl(zO) Wxo Wyo . Mxges withM= Mxges Mxges Myges Myges (Mx(HT)+Mx(GKa)+Mx(GH) .
1/1) .
M1,0 (increased coefficient Table T - 1,34) (16614
+
12 719+
95 522) . 1,0 124855Nm My(Kr) . M 1320Nm az(O) az(O) 124855 1320 . 1 890 + 15780 + 0,75 . 6,4 66,0 + 0,1 + 4,8 70,9N/mm2 ~ 160,0N/mm25.5.2 Proof of stress point 2 (edge of flange)
Zdirection: Mxges . My aZ(2) = - -W + - - + € . aFl(z2) x2 Wy2 7,7N/mm2 ~ 92,0N/mm2 x direction: ax(O) € . aFl(xO) x direction: aX(2) = 0,0
I
ax(o)1 ~ 160,0N/mm2 0,75 . (-63,6) -47,7 N/mm2 acp(O) ax(O) aep(O) Reference stress:J
a:(O)+a~(O)
- az(O) . ax(O)+ 3 .T to) 66,0 + 2,5 + 55,3 123,8N/mm2 ~ 160,0N/mm2 124855 1320 1 890 +526
+ 0,75 . 73,7 aZ(2) az(2) az(2) acp(O) = 104,2N/mm2~ 160,0N/mm25.5.3 Proof of stress point 1 (load application point) z direction:
Mxges My
az(l) = - -+ - - + € . aFl(zl)
Wxl Wyl
5.6 Welded web plate / flange design
When there is a welded connection between the web plate and flange (fillet welds, see fig. 5.3), the general proof of stress and a proof of service strength is carried out for flange point 0 (top side of flange),
aX(l) € . aFl(xl)
x dii'ection:
az(l) = 66,0 +2,3+ 58,2
az(l) . = 126,5N/mm2 ~ 160,0N/mm2
Reference stress:
5.6.1 General proof of stress for the weld seam point 0 z direction (
Normal stress, longitudinal loading of the weld point 0
az(O)w = 54,8N/mm2~ 160,0N/mm2
Shearing stress, transverse loading of the weld seam (including the stresses resulting from the longitudinel distri-bution of the wheel loads)
T(O)w = 26,7 N/mm2~ 113,0N/mm2
x direction:
Normal stress, transverse loading to the weld seam
ax(O)w= 47,7N/mm2 ~ 113,0N/mm2 0,75 . 17,6 0,75 ' 17,8 13,4N/mm2~ 160,0 N/mm2 124855 1 320 + - - + 1890 580 aX(l) ax(l) aZ(l) withT(l) ~ 0,0 a'cp(l) = 120,4N/mm2 ~ 160,0N/mm2 acp(l) acp(l) ~h26,52+13,42 - 126,5·13,4 2 T(l) Reference; stress : ! 2 2 . 2 acp(O)w=
..J
54,8 +47,7 - 54,8 . 47,7+ 2,26,7 Ucp(O)w= 64,0N/mm2 ~ 160,0N/mm2 Licenced to SCHOSS SAFEM9.341 Page 5 min Uz(O)w = 8,6 N/mm2 16614 + 1 370 min Uz(O)w = 2100 b drive wheel 21
Single-girder overhead travelling crane Figure5.1
t
/~
=:1'"'
:$
, -i_f::J
1ii'tf=leL1 - -1='= = = = = I ---:.j ~ bolted connection d-dwL: ,.'''
11
~
wheel 11 54,8 N/mm2 (tension)...M. -
54,8 - 016, maxuz(O)w = KzlO)5.6.2 Proof of service strength for weld seam point 0
The weld seams are classified -in the K4 case of notch toughness.
5.6.2.1 Limiting stress ratio for normal stress z direction: MXlHTl + minMxlGKal min Uz(O)w = W xOw x direction: min uX(O)w = 0,0 maxuxlO)w= 47,7 N/mm2 (tension) Kx(O) = 0,0
5.6.2.2 Permissible maximum for normal stress
perm UOz(K z) = 160,0 N/mm
2
~
54,8 N/mm2permUOz(Kx) = 160,0 N/mm2 ~ 47,7 N/mm2 Figure5.2 Main girder section, intersection f
5.6.2.3 Limiting stress ratio for shear stress
minT(O)w ==' 0,0
maXT(O)w 26,7 N/mm2
r
5
0,0 x - - - - x
5.6.2.4 Permissible maximum for shear stress
permTO(KT) = 92,4 N/mm 2 ~ 26,7 N/mm2
5.6.2.5 Proof of the combined stress
(UZ(O)w )2+ (1x(O)w )2 _ uz(O)w . (1xlO)w permCTOz(z) permuOz(x) Ipermuoz1·/permUozI
Y11
, f ' - - - - -b - - - (
@
Figure5.3 Main girder section, intersection b welded- design ( TlO)W )2 ~ 1,0 + perm TO 5 x - - - x 0,1173 + 0,0889 - 0,1021 -+ 0,0835 0,1876 ~ 1,0 o
Figure5.4 Main girder section, intersection b
.Erstellt durch den Technischen AusschuB der Sektion IX der Federation Europeenne de la Manutention (FEM) Prepared by the Technical Committee of Section IX of the Federation Europeenne de la Manutention (FEM) Etabli par le Comite Technique de la section IX de la Federation Europeenne de la Manutention (FEM)
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