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dokumen.tips crane-beam-design-guide

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2011-2012
Crane Runway Girder
Dr. Ibrahim Fahdah
Damascus University
https://sites.google.com/site/ifahdah/home/lectures
2011-2012
2011-2012
Components of Crane system
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The Crane Runway Girder and the
Structure
Issue1: Vertical Load Transformation
The support method of the crane
runway girder depends on the
magnitude of the reactions being
transmitted. Some typical
arrangements ranging from the lightest
to the heaviest are shown
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The Crane Runway Girder and the
Structure
Issue2: Free Rotation at the Supports
Free rotation at the supports of crane
runway girders is important in order to
prevent bending and torsional moments
in the columns.
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The Crane Runway Girder and the
Structure
Issue3: Transverse Load Transformation
Figure (b) illustrates the reversible strain
to which the girder web is subjected - an
action leading to the result shown in
Figure (c)
Dangerous details for lateral
forces
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The Crane Runway Girder and the
Structure
Issue3 could easily be prevented by
simply connecting the top flange directly
to the column, as shown. The top flange
acts as a horizontal beam delivering its
reaction to the column.
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Typical Section of Crane Girders
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Rail Fastenings
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Design Procedure (BS5950 Code)
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Classification of Cranes
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Step1: Calculate the maximum vertical
Loads
• The weight of the trolley (carriage) + Lifted
Load (Rh)
• The weight of the crane bridge (Rs)
• The self weight of the crane girder & Rails (Rg)
Note: The load to the crane girder will be maximum
when trolley wheels are closest to the girder.
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Step1: Calculate the maximum vertical
Loads (cont.)
For Warehouse or workshop F=1.3 => the load combinations below
Conservatively we can simplify the calculation , a factor of 1.3 can be applied
simultaneously to both the lifted load and to the self-weight of the crane.
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Step1: Calculate the maximum vertical
Loads (cont.)
So the maximum unfactored static point load per wheel, assuming there are
two wheels on each side, is:
Rw=1.3*0.5*(Rs/2+Rh*(Lc-ah)/Lc)
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Step2: Calculate the Horizontal Loads
Plan View
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Step2: Calculate the Horizontal Loads
• Inertia forces produced by the motion drives
or brakes. Referred to as the surge load.
(clause 3.1.5.1 of BS 2573-1:1983[4]).
• Skew loads due to travelling referred to as the
crabbing force. (clause 3.1.5.2, BS 2573: Part
1:1983 [4])
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Step2: Calculate the Horizontal Loads
• Transverse Surge load is taken as 10% of the combined
weight of the crab and the lifted load.
• Longitudinal Surge load of 5% of the static vertical
reactions. (i.e. from the weight of the crab, crane
bridge and lifted load).
• Crabbing forces are obtained from clause 4.11.2 (BS
5950-1:2000). If the crane is class Q1 or Q2, then the
crabbing forces would not need to be considered.
Note : Horizontal loads need not to be combined together.
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Step3: Load Combinations
Wv
FR
Wh1
Rail
Wheel
Wh2
Load combination according to BS 5950-1:2000 (Table 2) are:
• LC1 =1.4 DL + 1.6 Wv
• LC2 =1.4 DL + 1.6 (Wh1 or Wh2 or FR)
• LC3 =1.4 DL + 1.4 Wv + 1.4 (Wh1 or Wh2 or FR)
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Step4: Design Checks
1.
2.
3.
4.
5.
6.
7.
8.
Major axis bending
Lateral-torsional buckling
Horizontal moment capacity
Consider combined vertical and horizontal
moments
Web shear at supports
Local compression under wheels
Web bearing and buckling under the wheel
Deflection
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Major Axis Bending
For plastic section:
Note : Sx is for the whole section
BS 5950-1-2000
4.2.5
Check limit to avoid irreversible deformation under serviceability
loads.
BS 5950-1-2000
4.2.5.1
Note: for section classification of compound I- or H-sections, see BS
5950-1-2000 : 3.5.3 & Table 11.
Note: Moment capacity should be reduced in case of high shear
according to BS 5950-1-2000 : 4.2.5.3
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Lateral-Torsional Buckling
• Check gantry girder as an unrestrained member for vertical loads.
• Due to interaction between crane wheels and crane rails, crane
loads need not be treated as destabilizing, assuming that the rails
are not mounted on resilient pads.
• No account should be taken of the effect of moment gradient i.e.
mLT (lateral-torsional buckling factor) should be taken as 1.0.
BS 5950-2000
4.11.3
BS 5950-2000
4.11.3
BS 5950-2000
4.3.6.3 ,4.3.6.2, and
4.3.6.4
Pb is the bending strength and is dependent on the design strength
py and the equivalent slenderness λLT.
BS 5950-2000
4.3.6.7(a)
For compound section (Rolled section + plate ), use I and H with
unequal flanges to calculate λLT.
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Horizontal Moment Capacity
Horizontal loads are assumed to be carried by the top flange plate only.
Moment capacity of the top flange plate, Mc,plate is equal to the
lesser of 1.2py Zplate and py*Splate.
BS 5950-1-2000
4.2.5
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Consider Combined Vertical and
Horizontal
Moments
1-Section Capacity:
BS 5950-1-2000
4.8.3.2
2-Buckling Capacity: “simplified method”
BS 5950-1-2000
4.8.3.3.1
For simplicity take maximum M x and M y (rather than coexistent M x
and M y) and assume that the minor axis loads are carried by the plate
only.
M LT is the maximum major axis moment in the segment.
Note : mx, my factors can be taken as 1.0 for simplicity.
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Web Shear at Supports
BS 5950-1-2000
4.8.3
Note: It is ok to assume that the sear is resisted by the UB section =>
Av = tD (for rolled I-sections, load parallel to web)
BS 5950-1-2000
4.8.3 (a)
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Local Compression under Wheels
The local compressive stress in the web due to a crane wheel load
may be obtained by distributing it over
a length xR given by:
45
HR
Tplate
Tflange
2(HR+T)
The stress (fw) obtained by dispersing the wheel load over the length
xR should not be greater than py for the web.
BS 5950-1-2000
4.11.1
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Web bearing and buckling under the
wheel/supports
Bearing capacity of web for unstiffened web
BS 5950-1-2000
4.5.2.1
Buckling resistance of the unstiffened web
BS 5950-1-2000
4.5.31.
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Deflections
Vertical deflection due to static vertical wheel loads from overhead
travelling cranes
Horizontal deflection (calculated on the top flange properties
alone) due to horizontal crane loads
Note : The deflection of crane beams can be important and the exact
calculations can be complex with a system of rolling loads. However,
For two equal loads, a useful assumption is that the maximum
deflection occurs at the centre of the span when the loads are
positioned equidistant about the centre.
BS 5950-1-2000
2.5.2 Table 8(c)
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