The
design of the crane runway beams
The loads on crane runway beams are determined in accordance
with BS EN 1991-3[2]. This code sets out the groups of loads and dynamic
factors to be considered as a single characteristic crane action. The relevant
partial factors are set out in Table A.1 in Annex A of the code. At ultimate
limit state for the design of the crane and its supporting structures, the
characteristic crane action being considered is combined with simultaneously
occurring actions (eg wind load) in accordance with BS EN 1990. The final
ultimate design loads from the crane end carriage which are supported by the
runway beam can thus be determined. The groups of loads are identified in Table
2.2 of BS EN 1991-3 and include the actions listed in the table below. Several
of the loads have a dynamic factor associated with them which depend on the
class and function of the crane.
Unfavorable crane actions have a γQ value of 1.35, not the
usual value of 1.5. Fatigue assessment is regarded as a serviceability limit
state with a partial factor of 1.0.
Fatigue affects
Is the occurrence of a collapse as a result of the exposure
of some part of the repeated cycles of loads for example when holding an iron
ruler and bending it to the top it does not break and then bend it down it does
not break but with the repetition of this process several times it will break.
For example, in the example of the ruler depends on the
number of times the ruler is bent up and down and depends on (fatigue) where
the ruler is moved up and down.
BS EN 1991-3 provides
a simplified approach to designing crane runway beams (gantry girders) for
fatigue loads to comply with incomplete information during the design stage, when full details of the crane may not be
available. The crane fatigue loads are given in terms of fatigue damage
equivalent loads Qe that are taken as constant for all crane positions. The
fatigue load may be specified as follows:
Qe = φfat λi Qmax
,i where, as stated by the code, Qmax,i is the maximum value
of the characteristic vertical wheel load, i and λi = λ1,i λ2,i is the damage
equivalent factor to make allowance for the relevant standardized fatigue load
spectrum and absolute number of load cycles in relation to N = 2.0 × 106
cycles. This concept was discussed in reference [1]. The damage equivalent dynamic
impact factor φfat for normal conditions may be taken as:
Φfat1=
The factors φfat,1 and φfat,2 apply to the self-weight of the
crane and the hoist load respectively. In BS EN 1991-3, Annex B Table B.1 gives
recommendations for loading classes S in accordance with the type of crane and
Table 2.12 gives a single value of λ for each of normal and shear stresses
according to the crane classification. Overhead travelling cranes are in either
S-class S6 or S7 so that, having selected an S class, the corresponding λ value
is determined. (The classes Si correspond to a stress history parameter s defined
in BS EN 13001-1[3] but the details are not required for this example). The
method for carrying out the fatigue assessment is set out in section 9 of BS EN
1993-6[4]. Once the fatigue loads are determined, the stress ranges (denoted
ΔσE,2 ) for the critical details of the crane runway beam can be calculated.
These are the damage equivalent stress ranges related to 2 million cycles. The
fatigue stress range is multiplied by the partial factor for fatigue loads γFf
stated in BS EN 1993-6 section 9.2 which is equal to 1.0. The critical details
must be categorized according to Tables 8.1 to
γFf ∆σE,2
A common practice in industrial buildings is to weld a channel, open
side down, to the top flange of a standard rolled beam for use as a crane
runway. In many cases, it is not possible to brace a crane runway laterally
between columns, so the channel provides additional lateral stiffness. There
are several interesting structural questions associated with the practice,
like, what should be the welding pattern? How are the residual stresses
affected? What if the channel has yield strength different from the beam?
However, the primary question addressed in this paper is, how does one check
such a beam for lateral-torsional buckling?
The location of the maximum design moments and shears due to the
crane traveling along
The idea of moving loads, so we need to know where you are,
giving (maximum moment) and likewise where you are maximum shear
The (moving load) wheels will be crane
To find the maximum moment for grouo of moving load
1. We calculate where the outcome is for all the strong
2. We describe the distance between where the outcome is
influenced and the nearest force, for example.(point a )
3- Put (point a) in the middle of the kemra and stack the
loads around it without the result.
4- We calculate the zoom at the nearest force of the middle
of the zoom
To find (maximum shear) for a group of moving load
We put the first big of load on up support and count the reaction for this support , so it's him
the . maximum shear
Impact factor
In
the case of the (moving load ) dynamic effect of the movement resulting from
the shock, which is expressed by the impact factor and will high live load
I------- 25% ------ In case of
Electrical operation
I------- 10% ------ In case of manual
operation
If
not specified, we assume that it is 25% and the added (impact factor )only is
to (live load)
mLL+I=mLL
*(1+I) QLL+I= mLL
*(1+I)
Crippling
checks
We have to check crippling if there is direct loading on the
flange , so we have to check crane girder against crippling .(Assume contact
length after the rail=10 cm if not given ) the contact length in ECP is (N)
which is number of cycle
As a result of the presence of (concentrated load) high,
(web) is exposed to (crippling) the occurrence of (deformation) the occurrence
of(deformation) in (web)
We study the first part (straight) in where it is the first
weak part of the exhibition of the highest stress and the division of the load
of the wheel on the area exposed to stress we get (web) and should not exceed
the value (crippling stress) present in the code
Fcrp =
N=contact
length, 10cm if not given k=tf+r=2tf
And if the result is unsafe that we do
one of the two solutions
a) Choose bigger section in order to increase tw
b) Use stiffeners to strengthen the web
Reference:
1.
Lectures of Design of steel
structures by: Dr. Eng.
Medhat.M. Momtaz
2. Egyptian Code of Practice LRFD, 2005
3 .Steel Structures,
design & Behavior, Charles G. Salmon & John E. Johnson, Fifth Edition,
2009..
4. WWWEB
Enhanced Teaching of Structural Steel Design, AISC.
5. Euro code 3:
Design of steel structures - Part 1-1: General rules and rules for buildings.
.
6. Euro code 3:
Design of steel structures - Part 1-6: Strength and stability of shell
structures
7.BS EN 1993-6: 2007
Euro code 3 Design of steel structures – Part 6: Crane supporting structures
8. BS EN
1993-1-9:2005 Euro code 3 Design of steel structures – Part 1-9 Fatigue
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