Crane Runway Girder. Dr. Ibrahim Fahdah Damascus University.

Transcription

1 Crane Runway Girder Dr. Ibrahim Fahdah Damascus University

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3 Components of Crane system

4 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

5 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.

6 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

7 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.

8 Typical Section of Crane Girders

9 Rail Fastenings

10 Design Procedure (BS5950 Code)

11 Classification of Cranes

12 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.

13 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.

14 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)

15 Step2: Calculate the Horizontal Loads Plan View

16 Step2: Calculate the Horizontal Loads Inertia forces produced by the motion drives or brakes. Referred to as the surge load. (clause of BS :1983[4]). Skew loads due to travelling referred to as the crabbing force. (clause , BS 2573: Part 1:1983 [4])

17 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 (BS :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.

18 Step3: Load Combinations Wv Rail FR Wh1 Wheel Wh2 Load combination according to BS :2000 (Table 2) are: LC1 =1.4 DL Wv LC2 =1.4 DL (Wh1 or Wh2 or FR) LC3 =1.4 DL Wv (Wh1 or Wh2 or FR)

19 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

20 Major Axis Bending For plastic section: Note : Sx is for the whole section Check limit to avoid irreversible deformation under serviceability loads. BS BS Note: for section classification of compound I- or H-sections, see BS : & Table 11. Note: Moment capacity should be reduced in case of high shear according to BS :

21 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 BS Pb is the bending strength and is dependent on the design strength py and the equivalent slenderness λlt. For compound section (Rolled section + plate ), use I and H with unequal flanges to calculate λlt. BS , , and BS (a)

22 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

23 1-Section Capacity: Consider Combined Vertical and Horizontal Moments BS Buckling Capacity: simplified method BS 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.

24 Web Shear at Supports BS 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 (a)

25 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: BS (HR+T) 45 HR Tplate Tflange The stress (fw) obtained by dispersing the wheel load over the length xr should not be greater than py for the web.

26 Web bearing and buckling under the Bearing capacity of web for unstiffened web wheel/supports BS Buckling resistance of the unstiffened web BS

27 Deflections Vertical deflection due to static vertical wheel loads from overhead travelling cranes BS Table 8(c) 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.