Gantry Girder
Gantry Girder
Gantry Girder
5.4.1 Features
Partial safety factor for both dead load and crane load is 1.5 (Table 4, p. no. 29).
Partial safety factor for serviceability for both dead load and crane load is 1.0 (Table 4, p.
no. 29).
up to 50t
over 50t
Additional load
Vertical loads
a) EOT crane
b) HOT crane
10% of wt. of
b) HOT crane
05% of wt of
crab & wt. lifted
Note: Gantry Girder & their vertical supports are designed under the assumption that either of
the horizontal forces act at the same time as the vertical load.
5.5 DESIGN OF GANTRY GIRDER
Example 1
Data
a) Wt. of crane girder/truss 180kN
b) Crane capacity
200kN
50kN
16m
1.2m
6m
g) Wt. of rail
0.25kN/m
=160.625 kN
= 1.5 X 200.775
= 301.16 kN
.
yj for lipped flanges of channel section which depends on ratio of f
Where f = Ifc / (Ifc+Ift).
= 0.7
section
yj = 94.055
Iyy = Iyy of ISWB 500 + Ixx of ISLC 350
= 2987.8 + 9312.6 = 12300.4 X 104 mm4
LLT = K . L = 0.8 X 6000 = 4800 mm
Iw = warping constant
= (1- f) f . Iy . (hy)2
= 6.23 X 10 12 mm6
It = Torsion constant
= bt3/3 = 10.86 X 105
G = 0.77 X 105
= 2950 kNm
To find Zp of Gantry Girder section we need to find equal area axis of the section.
This axis is at a depth of 48.74 mm from the top of the section.
Taking moments of areas about equal area axis.
A . y = Zp = 29.334 X 105 mm3
Refering clause 8.2.2 for laterally unsupported beam
(p. no. 54)
= 0.4984
LT = 0.21 for rolled section
= 0.655
= 0.925
Therefore fbd = LT . fy / m0
= 0.925 X 250 / 1.1 = 210.22 N/mm2
MdZ = b . Zp . fbd = 616.66 kNm > Md = 508.21 kNm
OK
Horizontal Action
Total horizontal force perpendicular to span of
Gantry Girder = 10 % (crane capacity + wt. of
crab and motor)
= 10% (200+50) = 25 kN.
As wheels are having double flanges
Horizontal force / wheel = 25/4 = 6.25 kN
Therefore maxm horizontal BM in proportion to vertical bending moment
My = (6.25 /301.16) X 508.21 = 10.546 kNm
This is resisted by ISLC 350 with top flange of ISWB 500
Zpy1y1 = 100 X 12.5 X 337.52 + (1/4) 7.4 X 3252+ (1/4) X 14.7 X 2502
= 8.47 X 105 mm3
Plastic moment capacity about y1y1 axis
Mdy = b . fy . Zp / mo
= 192.5 kNm
Check for biaxial moment
Reffering clause 9.3.1.1 (p. no. 70)
(Mz/Mdz) + (My/Mdy)
= (518.14 / 614.57) + (10.54 / 192.5)
= 0.897 < 1.0 ..
OK
Hence select section for the gantry Girder as ISWB 500 and ISLC 350 over it.
Example:2
LATERAL LOAD
Lateral load/wheel= 10% ((crane capacity+crab)/4))
=10%
(200+ 40)
4
= 0.1x(245/4) = 6.125 KN
NOTE
If b > 0. 586 L
(WL)
4
Shear force
Maximum shear force occur when one of the wheel loads is at support Shear force due to wheel
load = 336.975 +(336.975/2) = 505.4625 KN
L/ 30 = 8000/ 30 = 266.67 mm
Try IS WB 600 @ 145.1 kg/m And
ISMC 400 @ 49.4 kg /m
Properties
A=62.93 x102
Tt =23.6 mm
Tt =15.3 mm
Tw = 11.8 mm
Tw =8.6 mm
B=250mm
B=100 mm
Izz =1.15x109mm
Izz =1.5x108mm4
Section classification
T=
250
fy
250
250
=1
184.86 x 102 x
=
600
+62.93 x 102 x (600+ 8.624.2)
2
247.79 x 102
= 372.23 mm
Y1 = (372.23 300) = 72.23 mm
Y2 = 600+8.6 -372.23 -24.2 =212.17
IZZ = IZZ (1) +A1Y1^2+ IYYC+ A2Y2^2
= (1.15x109) + (184.86x10 2x72.232)+(5x106+62.93 x102x212.172)
= 1.534 x109mm
(1.534 x 103)
Zzz = (Izz/y) =
372.23
= 4.12 x106mm3
IYY= Iyy (1) + IZZ(C) = (5.29x107+ 1.5 x108) mm4 = 2.03 x108mm4
= 2.03x108mm4
IYY of compression flange
= (Izz)channel +((Iyy/2)) I Section
= (1.5x 108) + (
(5.27 x 107)
2
= 1.76 x 108mm4
(1.76 x 108)
200
= 8.82 x105mm3
Calculation of plastic section modulus
Let
Dp be the distance between the centre of I Section to equal area axis
dp = A c h /(2twi) =
(6293)
2 x 11.8
= 266.65 mm
Check