Crane Runway Girder

Dr. Ibrahim Fahdah

Damascus University

2011-2012

https://sites.google.com/site/ifahdah/home/lectures

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Components of Crane system

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The Crane Runway Girder and the Structure

Issue1: Vertical Load TransformationThe 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

Dangerous details for lateral forces

Issue3: Transverse Load TransformationFigure (b) illustrates the reversible strain to which the girder web is subjected - an action leading to the result shown in Figure (c)

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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

Wh2

Wh1FR Wheel

Rail

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. Major axis bending

2. Lateral-torsional buckling

3. Horizontal moment capacity

4. Consider combined vertical and horizontal moments

5. Web shear at supports

6. Local compression under wheels

7. Web bearing and buckling under the wheel

8. 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 5950-1-20004.2.5.1

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.

BS 5950-2000 4.11.3

• 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 5950-20004.3.6.3 ,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 CapacityHorizontal 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

BS 5950-1-2000 4.8.3.2

1-Section Capacity:

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

BS 5950-1-2000 4.11.1

The local compressive stress in the web due to a crane wheel load may be obtained by distributing it overa length xR given by:

The stress (fw) obtained by dispersing the wheel load over the length xR should not be greater than py for the web.

2(HR+T)

HR

TplateTflange

45

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Web bearing and buckling under the wheel/supports

BS 5950-1-2000 4.5.2.1

Bearing capacity of web for unstiffened web

Buckling resistance of the unstiffened web BS 5950-1-2000 4.5.31.

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Deflections

BS 5950-1-2000 2.5.2 Table 8(c)

Vertical deflection due to static vertical wheel loads from overheadtravelling 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.

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