Introduction Load acting on Gantry girder Types of Gantry Girders Limiting Deflection of Gantry Girders Maximum Load Effects Design Procedure Examples

1. Subject : Design of steel structures(2180610)
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2. Contents
 Introduction
 Load acting on Gantry girder
 Types of Gantry Girders
 Limiting Deflection of Gantry Girders
 Maximum Load Effects
 Design Procedure
 Examples
2
Guided by : Prof. Pritesh Rathod
Prof. Sunil Jaganiya
Prepared By: kajol Panchal

3. Introduction
• Gantry girders are laterally unsupported beams to carry
heavy loads from place to place at the construction sites,
mostly these are of steel material.
• A girder is a support beam used in construction.
• It is the main horizontal support of a structure which supports
smaller beams. Girders often have an I-beam cross section
composed of two load-bearing flanges separated by a
stabilizing web, but may also have a box shape, Z shape and other
forms.
• Gantry girders or crane girders carry hand operated or electrically
operated overhead cranes in industrial building such as factories,
workshops, steel work , etc. to lift heavy materials, equipment etc.
and to carry them form one location to the other, within the building .
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4. Components
The essential component
of crane system are :
1. crane girder or crane
bridge or cross girder
2. Trolley or crab
3. Gantry Girder
4. Crane rails
4

5. Crane Notations
5

6. Loads on Gantry Girder
1) Vertical Loads
from crane
2) Impact load
from crane
3) Longitudinal
horizontal
force(Drag
force)
4) Lateral
load(surge load)
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7. Additional impact loads on cranes (IS:875)
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8. Types of Gantry Girders
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9. Limiting Deflection
9

10. Maximum Load Effects
10

11. Max. S.F., B.M., and Deflection
11

12. Two Cranes at the Same Span
12

13. Max. BM for Two Cranes At the Same
Span
13

14. Design Procedure
 Assume that the lateral load is resisted entirely by the top flange of
the beam plus any reinforcing plates, channels etc. and the vertical
load is resisted by the combined beam.
1. Find the maximum wheel load: This load is maximum when the
trolley is closest to the gantry girder. Increase it for the impact
2. Calculate the maximum bending moment in the gantry girder due
to vertical loads.
3. To simplify the calculations, add the maximum bending moment
due to dead load to the maximum wheel load moment.
4. The maximum shear force is calculated. When the gantry is not
laterally supported, the following may be used to select a trail
section.
Zp = Mu / fy Zp (trial) = k Zp (k = 1.40-1.50)
Economic depth ≈ 1/12th of the span.
Width of flange ≈ 1/40 to 1/30th of the span
14

15. 5.The plastic section modulus of the assumed combined section
Mp = 2 fy A / 2 = A fy
where A is called the plastic modulus Zp
6. Check for moment capacity of the whole section (as lateral support
is provided at the compression flange)
Mcz = βb Zp fy ≤ 1.2 Ze fy / γm0 <Mu
7.Check top flange for bending in both the axes using the interaction
equation
(My / Mndy)+ (M2/Mndz) ≤ 1.0
8. If the top (compression) flange is not supported, Check for buckling
resistance in the same way as in step 6 but replacing fy with the
design bending compressive stress fbd.
9. Check web of the girder at points of concentrated load for local
buckling or local crushing, and provide load carrying/ bearing
stiffeners, if necessary.
10. Check for deflection under working loads
15

16. Design Check
• Check for moment capacity
• Check for shear capacity
• Check for Buckling Resistance
• Check for Local Buckling
• Check for Deflection
16

17. Design a gantry girder to be used in an industrial
building carrying a manually operated overhead traveling
crane for the following data:
• Crane capacity - 200kN
• Self-weight of the crane girder excluding trolley - 200kN
• Self-weight of the trolley, electric motor, hook, etc. - 40kN
• Approximate minimum approach of the crane hook to the
gantry girder - 1.20m
• Wheel base - 3.5m
• c/c distance between gantry rails - 16m
• c/c distance between columns (span of gantry girder) - 8m
• Self weight of rail section - 300N/m
• Diameter of crane wheels - 150mm
• Steel is of grade Fe 410.
Design also the field welded connection if required. The
support bracket connection need to designed.
17

18. SOLUTION :
For Fe410 grade of steel:
 fu = 410 Mpa
 fy = fyw = fyf = 410 MPa
For hand operated OT crane:
 Lateral loads = 5% of maximum static wheel load
 Longitudinal loads = 5% of weight of crab and weight lifted
 Maximum permissible deflection = L/500
Partial safety factors
 ‫ע‬m0 = 1.10
 ‫ע‬mw = 1.50 (for site welds)
Load factor
 ‫ע‬m1 = 1.50
ξ = ξw = ξf = √ (250/ fy) = √ (250/ 250) = 1.0
18

19. STEP 1: CALCULATION OF DESIGN FORCES
Maximum wheel load:
 Maximum concentrated load on crane = 200 + 40 = 240kN
 Maximum factored load on crane = 1.5 x 240 = 360kN
 The crane will carry the self-weight as a UDL = 200/16 = 12.5 kN/m
 Factored uniform load = 1.5 x 12.5 = 18.75kN/m
For maximum reaction on the gantry girder the loads are placed on the
crane girder as shown in the figure below,
19

20. Taking moment about B,
RA x 16 = 360 x (16 - 1.2) + (18.75 x 16 x 16) / 2
RA = 483kN
RB = 177kN
The reaction from the crane girder is distributed equally on the
two wheels at the end of the crane girder.
Therefore, Maximum wheel load on each = 483/2 = 241.5 kN
wheel of the crane
STEP 2: CALCULATION OF MAXIMUM BM & SF:
Maximum Bending Moment:
 It consists of maximum moments caused by the moving wheel
loads on the gantry girder and self weight of the gantry girder.
20

21. For Maximum Bending Moment, the wheel loads shall be placed as
shown in figure.
The calculation of maximum bending moments due to wheel loads and
self weight of gantry girder has been done separately because
calculation of impact load and bending moment due to it involve live
load only,
Assume, Self Weight of the Gantry Girder as 2 kN/m.
Total Dead Load, w = 2000 + 300 = 2300 N/m = 2.3 kN/m
Factored Dead Load = 1.5 x 2.3 = 3.45kN/m
21

22. The position of one wheel load from the mid point of span
= Wheel Base/4
= 3.5/4 = 0.875 m
Bending moment due to live load only,
 Taking moment about D,
Rc x 8 = 241.5 x (8 - 1.375) + 241.5 x 3.125
Rc = 294.33kN
 Taking moment about C,
RD x 8 = 241.5 x 1.375 + 241.5 x 4.875
RD = 188.67kN
Maximum Bending Moment due to Live Load
= 188.67 x 3.125 = 589.6 kNm
Bending Moment due to Impact
= 0.10 x 589.6 = 58.96 kNm
22

23. Total Bending Moment due to live load and impact loads
= 589.6 + 58.96 = 648.56 kNm
Bending Moment due to Dead Load
= wl2/8 = 3.45 x (82/8) = 27.6 kNm
Therefore,
Maximum B.M = 648.56 + 27.6 = 676.16 kNm
= 676.16 x 106 N_mm
Maximum Shear Force:
 It consists of maximum shear due to moving wheel loads and self
weight of the girder.
For maximum shear force the wheel loads shall be placed as shown in
figure, i.e one of the wheel loads should at the support.
23

24. Taking Moment about D,
Rc x 8 = (241.5 x 8) + (241.5 x 4.5)
Rc = 377.34 kN
Hence,
Maximum Shear Force due to wheel loads = 377.34 kN
Lateral force transverse to the rails
= 5% of weight of crab and weight lifted = 0.05 x 240 = 12kN
Factored lateral force = 1.5 x 12 = 18 kN
STEP 3: CALCULATION OF LATERAL FORCES:
24

25. Maximum horizontal reaction due to lateral force by proportion at C,
= (lateral force x Reaction at c due to vertical load)
(maximum wheel load due to vertical load)
= (9 x 294.33) / 241.5
= 10.97 kN
Horizontal reaction due to lateral force at D
= 18 - 10.97
= 7.03 kN
Maximum bending moment due to lateral load by proportion
= (9.0 / 241.5) x 589.6
= 21.97 kNm
Maximum shear force due to lateral load by proportion
= (377.34 / 241.5) x 9.0
= 14.06 kN
25

26. Approximate depth of section = L / 12 = (8 x 103) / 12
= 666.66mm ≈ 600mm
Approximate width of the flange = L / 30 = (8 x 103) / 30
= 266.66mm ≈ 300 mm
Approximate section modulus required, Zpz = 1.4 (Mz / fy )
= (1.4 x 676.16 x 106) / 250 = 3786.5 x 103 mm3
STEP 4: SELECTION OF PRELIMINARY TRIAL SECTION :
Let us try ISWB 600 @ 1311.6 N/m
with ISMC300 @ 351.2 N/m on its
top flange as shown in figure,
26

27. Properties Notation
I-Section
ISWB 600
Channel Section
ISMC 300
Area A 17038 mm2 4564 mm2
Thickness of Flange tf 21.3 mm 13.6 mm
Thickness of Web tw 11.2 mm 7.6 mm
Width of Flange bf 250 mm 90 mm
Moment of Inertia Iz 106198.5 x 104 mm4 6362.6 x 104 mm4
Iy 4702.5 x 104 mm4 310.8 x 104 mm4
Depth of Section h 600 mm 300 mm
Radius at root Rl 17 mm
Cyy 23.6 mm
27

28. The distance of NA of built up section from the extreme fibre of
compression flange,
ỹ = (∑AY) / ∑A = {17038 X (300 +7.6)} + {4564 X 23.6}
(4564 + 17038)
= 247.59 mm
Gross Moment of Inertia of the Built-up section
Iz gross = Iz beam + Iz channel
(The channel is placed over the I-section in such a manner that its yy-
axis becomes the zz-axis. Therefore, in calculating Iz of the total section,
Iy of the channel will become Iz of the channel.)
Iz gross = [(106198.5 x 104) + (17038 x (307.6 - 247.59)2)]
+
[(310.8 x 104) + (4564 x (247.59 -23.6)2)]
= 135,543x104 mm4
STEP 5: MOMENT OF INERTIA OF GANTRY GIRDER :
28

29. Iy gross = Iy beam + Iy channel
= (4702.5 x 104) + (6362.6 x 104)
= 11,065.1x104 mm4
Zez = (Iy /y)
= (135,543 x 104) / (600 + 7.6 - 247.59)
= 3764.98x103 mm3
Plastic Modulus of Section
(Ignoring the Fillets)
Equal area axis (Refer to Fig)
= 4564 + (250 X 21.3) + (ỹ1 tw)
= 250 X 21.3 + (600-2 X 21.3 ỹ1 ) X 11.2
Z = 74.95 mm
(From Lower surface of top flange of I – Section)
29

30. Plastic section modulus of the section above equal axis,
Zpz1 = [300 X 7.6 X (74.95 + 21.3 + (7.6 / 2))]
+ [2 X (90-7.6)x13.6(74.95 + 21.3 – (90-7.6/2))]
+ [250 X 21.3 x (74.95 + (21.3/2))]
+ [74.95 X 11.2 X (74.95/2)]
Zpz1 = 838.775 X 103 mm3
Plastic section modulus of the section below equal axis,
Zpz2 = [250 X 21.3 X ( 600 - 21.3 - 74.95 - ( 21.3/2))]
+ [((600 – 2 X 21.3 – 74.95)2 / 2)]
Zpz2 = 3929.202 X 103 mm3
Zpz = Zpz1 + Zpz2 = (838.775 X 103) + (3929.202 X 103)
Zpz = 4767.977 X 103 mm3
30

31. Plastic section modulus of compression flange about yy – axis,
Zpfy = [((250 X 21.3 X 250) /4)] + [((2 X (300 – (13.6 X 2))2 X 7.6) / 8)]
+ [2 x (13.6 x 90 x ( 300 – 13.6) / 2)]
= 824.763 X 103 mm3
STEP 6: CLASSIFICATION OF SECTION:
Outstand of flange of I-Section,
b = bf/2 = 250/2 = 125 mm.
b/tf of flange of I-section
= 125/21.3 = 5.86 < 8.4 (8.4ε = 8.4 x 1 = 8.4)
Outstand of flange of Channel Section,
b = bf - tw = 90 - 7.6 = 82.4 mm.
b/tf of flange of Channel section
= 82.4/13.6 = 6.05 < 8.4 (8.4 ε = 8.4 x 1 = 8.4)
31

32. d/tw of web of I-Section = ((h – 2tf) / tw)
= ((600 – 2 x 21.3) / 11.2)
= 49.76 < 84 (84 ε = 84 x 1 = 84)
Hence the entire section is plastic, (βb = 1.0)
STEP 7: CHECK FOR MOMENT CAPACITY:
Local Moment capacity, Mdz = βb Zpz (fy / ‫ע‬m0 ) ≤ 1.2 x Ze (fy / ‫ע‬m0 )
Mdz = 1.0 x 4767.977 x 103 x (250 x 10-6 / 1.10)
= 1083.63 kNm ≤ 1.2 x 3764.98 x 103 x (250 x 10-6 / 1.10)
= 1026.81 kNm
Hence, Moment capacity of the section,
Mdz = 1026.81 kNm > 676.16 kNm
which is safe.
(IS :800-2007,P,53)
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33. Moment capacity of compression flange about y-axis,
Mdy,f = βb Zpyf (fy / ‫ע‬m0 ) ≤ 1.2 x Zey,f (fy / ‫ע‬m0 )
= 1.0 x 824.76 x 103 x (250 x 10-6 / 1.10)
= 187.44 kNm ≤ 1.2 x 580.92 x 103 x (250 x 10-6 / 1.10)
= 158.43 kNm
Hence moment capacity of flange,
Mdy,f = 158.43 kNm
Combined check for local moment capacity,
(Mz / Mdz ) + (My,f / Mdy,f) ≤ 1.0
(676.16 / 1026.81) + (21.97 / 158.43) = 0.797 < 1.0
Which is safe.
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34. The elastic lateral buckling moment,
Mcr = c1(π2EIYhf/2L2
LT)[1+1/20((LLT/ry)/(hf/tf))2]0.5
Overall depth of the section,
hf = h = 600 + 7.6 = 607.6 mm
Note: hf is the centre to centre distance between flanges. In the
calculation, value of hf has been taken equal to overall depth of the
section to avoid cumbersome calculations. Further, this approximation
will result in conservative design.
Effective length, LLT = 8 x 102 mm
Thickness of flange, tf = 21.3 + 7.6 = 28.9 mm
Radius of gyration, ry = √(IY/A)
= √((11065.1x104)/(17038+4564))
= 71.57mm
STEP 8: BUCKLING RESISTANCE IN BENDING CHECK:
(IS :800-2007,CL.8.2.2, P .54)
34

35. The coefficient, c1 = 1.132 (From Appendix XIV, assuming uniform
load condition)
Mcr = [(1.132x2x2x105x11065.1x104x607.6) / (2x(8x103)2)]
x [1+1/20((8x103/71.57)/(607.6/28.9))2]0.5
= 1823.24 x 106 Nmm
Non dimensional slenderness ratio,
λLTz = √(βb Zpz fy / Mcr)
= √((1 x 4783.594 x 103 x 250) / (1823.24 x 106))
= 0.809
ΦLTz = 0.5 [1 + αLT (λLTz – 0.2) + λLTz2]
αLT = 0.21 (assuming connection of channel with I-section flange
by intermittent fillet welds. For uniform weld, αLT = 0.49)
ΦLTz = 0.5 x [1+0.21 x (0.809 – 0.2) + 0.8092]
= 0.891
35

36. χLTz = [1/( ΦLTz + ΦLTz
2 - λLTz
2)0.5]
= [1/(0.891 + 0.8912 – 0.8092)0.5]
= 0.791
Design bending compressive stress,
fbd = χLTz (fy / ‫ע‬m0 )
= 0.791 (250/1.10)
= 179.77 N/mm2
The design bending strength,
Mdz = βb Zpz fbd
= 1.0 x 4767.977 x 103 x 179.77 x 10-6
= 857.14 kNm > 676.16 kNm (which is alright)
Hence, the beam is safe in bending under vertical load.
36

37. Since lateral forces are also acting, the beam must be checked for bi-
axial bending. The bending strength about y-axis will be provided by the
top flange only as the lateral loads are applied there only.
Mdy =Zyt (fy/γm0)
Zyt = section modulus of top flange about yy-axis.
= (4702.5x104/2 +6362.6x104)/(300/2) = 580.92 x 103 mm3
(Assuming the moment of inertia of top flange to be half of the moment
of inertia of I-section)
Mdy = 580.92 x 103 x (250/1.1) x (10-6) = 132.02kNm.
(Mz/Mdz) + (My/Mdy) < 1.0
= (676.16/859.946) + (21.97/132.02) = 0.952 < 1.0
Hence, the section is safe.
STEP 9: CHECK FOR BIAXIAL BENDING
37

38. Maximum shear force due to wheel load = 377.34 kN
Impact load = 0.1 x 377.34
= 37.73 kN
Design shear force = 377.34 + 37.73
= 415.05 kN
Shear capacity = Av fyw / (√3 x γmo)
= [((600x11.2) x 250x10-3) / (√3x1.1) ]
= 881.77 kN > 415.07kN, which is safe.
Maximum shear,
V = 415.07 kN < 529 kN
(i.e. 0.6*Vd = 0.6 x 881.77 = 529kN)
Since V < 0.6*Vd, the case is of low shear.
No reduction will be therefore there in the moment capacity.
STEP 10: CHECK FOR SHEAR CAPACITY :
(IS :800-2007,CL.8.4.1, P.59)
38

39. We should check for buckling under the wheel load.
Buckling Resistance = (b1+n1)tw*fcd
Where, b1 = bearing length=150mm(diameter of wheel)
n1 = (600/2) +2x7.6 = 315.2 mm
Slenderness ratio of the web,
λw = 2.45 (d1/tw)
= 2.45(600 -2x(21.3 + 17))/11.2 = 114.49
For λw=114.49, fy=250N/mm2 and buckling curve c, the design
compressive stress from Table 7.6,
fcd = 89.64 N/mm2
Therefore, Buckling Resistance = (150 + 315.2) x 7.6 x 89.64 x 10-3
= 316.9 kN > 241.5kN
(which is safe)
STEP 11: WEB-BUCKING CHECK: (IS :800-2007,CL.8.7.3.1,P.67)
39

40. δ = wL3x(3a/4L –a3/L3)
6EI
W = Maximum Static Wheel Load = (241.5 / 1.5)
= 161 kN
a = (L - c) /2 = [(8x103) – (3.5x103)] / 2
= 2.25 x 103 mm
Therefore, Vertical deflection
= (161x103)x(8x103)3x{(3x2.25x103)/(4x8x103)-(2.25x103)3/(8x103)3)
[6 x (2x105) x 135,543x104]
= 9.56 mm
Permissible maximum deflection, = L / 500
= 8000 / 500
= 16 mm > 9.56 mm, which is safe.
STEP 12: DEFLECTION CHECK:
(IS :800-2007)
40

41. The required shear capacity of the weld,
qw = Vaỹ / Iz
V = 377.34 kN
A = 4564 mm2 (area above the section)
ỹ = (247.59 -7.6) = 239.9 mm
Iz = 135,543x104 mm4
Therefore, qw = [(377.34 x 103) x 4564 x 239.9] / [135,543x104]
= 304.82 N/mm
Let us provide 3 mm weld to connect channel with flange of I-section.
Strength of the weld provided, = (0.7x3x410)/(√3x1.50)
= 331.39 N/mm > 304.82 N/mm
Hence, provide 3 mm size intermittent fillet welds for making the
connection.
STEP 13: DESIGN OF WELD:
(IS :800-2007,CL.10.5.7)
41

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