A complete process to design a gantry crane form wheels to girder.
Hormigon preesforzadoFull description
Steps with formulas to design plate girdersFull description
gantry
Gantry Crane Design CalculationsFull description
Descripción: design
designFull description
inspection
Full description
SS 497_Code of Practice for Design, Safe Use and Maintenance of Gantry Cranes
CL
Design of Steel Beams (1)
Descripción completa
gantry
Design of gantry girder
DESIGN OF GANTRY GIRDER
CRANE CAPACITY =200 KN WEIGHT OF CRANE EXCLUDING CRAB = 250 KN c/c distance between columns = 24 m spacing of columns = 8m weight of crab = 45 KN wheel spacing = 4m Distance between centre of column to centre of gantry girder = 0.4 m Crane hook approach distane = 1 m SOLUTION Centre to centre distance between gantry girder =24-(2x0.4)=23.2m CALCULATION OF LOADS
1
Design of gantry girder
Weight of crane= 250 KN Weight of crab= 45 KN Crane capacity=200 KN Maximum static wheel load due to due to self weight of crane=(250/4) =62.5 KN Due to crane load = (200+45)(23.2-1)/(2x23.2)
= 117.22 KN
Total static wheel load = 62.50+117.22 = 179.72 KN Including impact load 25% = 1.25x179.72 = 224.65 KN Factored wheel load on each wheel = 1.5x224.65 = 336.975 KN LATERAL LOAD Lateral load/wheel= 10% ((crane capacity+crab)/4))
( )
=10%
=0.1x(245/4)= 6.125 KN Horizontal load = 5% of wheel load =0.05x 336.975 =16.848 KN BENDING MOMENT CALCULATION Wheel spacing =b=4m Span of gantry = l = 8m b< 0.586 L POSITION OF WHEELS FOR MAXIMUM BENDING MOMENT
Maximum BM will occur under wheel D
2
Design of gantry girder
Taking moment about B 8 RA = 336.975 x 5 +336.975 x1 RA = 252.73 KN Moment at D = 252.73 x 3 = 758.19 KNm NOTE If b > 0. 586 L
Keep one of the wheel loads at centre and M Max =
Assume self weight of the girder = 2KN/m Self weight of rail = 0.3 KN/m Total dead load = 2.3 KN/m Factored dead load = 3.45 KN/m 2
BM due to dead load =WL /8 = 27.6 KNm Moment due to lateral force For this also the wheels are to be placed as earlier
3
Design of gantry girder
8 HA=9.1875(5+1) HA=6.89 KN BM at D = 6.89 x 3 =20.67 KNm Factored moment = 1.5 x 20.67 = 31 KN Shear force Maximum shear force occur when one o f the wheel loads is at support Shear force due to wheel load = 336.975 +(336.975/2) = 505.4625 KN
SHEAR FORCE due to self weight = 3.45 x (8/2) = 13.8 KN Total SF = 519 .26 KN Shear force due to the lateral load
SF = 9.1875 +(9.1875 /2) = 13.78 KN
DESIGN Economic design of girder = (1/12) of span Compression flange width = (1/30) of span L/12 = 8000/12 = 666.667 mm
4
Design of gantry girder
L/ 30 = 8000/ 30 = 266.67 mm Try IS WB 600 @ 145.1 kg/m And ISMC 400 @ 49.4 kg /m
Properties ISWB600 @ 145.1 kg/m A=62.93 x10
2
ISMC 400@ 49.4 kg/m 2
2
A=184.86 x 10 mm
Tt=23.6 mm
T t=15.3 mm
Tw = 11.8 mm
Tw = 8.6 mm
B=250mm
B=100 mm 9
8
Izz=1.15x10 mm
4
I zz=1.5x10 mm
7
4
IYY=5.0 X 10 mm
6
6
3
Zzz=7.54 x 10 mm
IYY=5.29 X 10 mm
Zzz=3.85 x 10 mm 5
3
ZYY= 4.23 x 10 mm
4
5
3
4
3
ZYY= 6.7 x 10 mm Cy = 24.2 mm
Section classification
= =1
t=
(b/t) of I beam = (250 – 11.8)/(2x23.6)) b/t of channel = (100 -8.6 )/15.3 = 5.97 <9.46 d/t of I section = (600-2(23.6))/11.8 =46.84 < 84 t 5
Design of gantry girder
Hence the section is plastic (from table 2 , P-18,
IS 800)
Elastic properties of combined section 2
2
2
Total area = 184.86 x 10 + 62.93 x 10 =247.79x10 Distance of NA from tension fibre
IYY of compression flange =(Izz)channel +((Iyy/2)) I Section 8
=(1.5x 10 )+(
8
4
= 1.76 x 10 mm
6
Design of gantry girder
ZY for top flange alone
=
5
3
= 8.82 x10 mm
Calculation of plastic section modulus
2
2
Total area = 247.79 x 10 mm
Let dp be the distance between the centre of I Section to equal area axis dp =Ach /(2 twi)
=
=266.65 mm
YPT = distance between tension fibre to equal area axis =(300+266.65) =566.65 mm YPC = (600+8.6-566.65) =41.95mm Ignoring the fillets the plastic section modulus below the equal area axis is
)/23.6 = 5.04 < 9.4t b/t of the flange of channel = ( ) = 5.97 < 9.4t d/t of the web of I section = =46.84 < 84 b/t of the flange of beam = (
hence the section is plastic Local moment capacity M dz =
=( )
=
Bp = 1 for plastic section (clauses 2.1.2 of IS 800) 9
= 1.193 x 10 Nmm
9
= 1.123 x 10 Nmm
Moment due to vertical load = 758.19 KNm Factored moment due to self weight = 1.5 x 27.6 = 41.4 KNm 3
Moment due to horizontal force parallel to rail = (1.5x16.848x10 x236.37) = 6KNm Total moment = 805.59 KN Hence take M dz = 1123 KNm 8