Design of Gantry Girders 1

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

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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%( 200+454 ) =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

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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 MMax =WL4

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

BM due to dead load =WL2/8 = 27.6 KNm

Moment due to lateral force

For this also the wheels are to be placed as earlier

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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 of 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

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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 ISMC 400@ 49.4 kg/m

A=62.93 x102 A=184.86 x 102 mm2

Tt=23.6 mm Tt=15.3 mm

Tw = 11.8 mm Tw = 8.6 mm

B=250mm B=100 mm

Izz=1.15x109 mm Izz=1.5x108 mm4

IYY=5.29 X 107 mm4 IYY=5.0 X 106 mm4

Zzz=3.85 x 106 mm3 Zzz=7.54 x 105 mm3

ZYY= 4.23 x 105 mm3 ZYY= 6.7 x 104 mm3

Cy = 24.2 mm

Section classification

t=√250/ fy =√250/250 =1

(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

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Hence the section is plastic (from table 2 , P-18, IS 800)

Elastic properties of combined section

Total area = 184.86 x 102 + 62.93 x 102 =247.79x102

Distance of NA from tension fibre

y ̅̅̅̅ = 184.86∗102∗( 6002 )+62.93∗102∗(600+8.6−24.2)

247.79∗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+ I YYC + A2Y2^2

=(1.15x109)+(184.86x102x72.232)+(5x106 +62.93 x102x212.172)

= 1.534 x109 mm4

Zzz = (Izz/ y ̅̅̅̅) = 1.534∗109

372.23 = 4.12 x106 mm3

IYY = Iyy(1) +IZZ(C)

=(5.29x107 + 1.5 x108)mm4 = 2.03 x108 mm4

= 2.03x108 mm4

IYY of compression flange

=(Izz)channel +((Iyy/2)) I Section

=(1.5x 108)+(5.29∗107

2¿ = 1.76 x 108 mm4

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ZY for top flange alone

=1.76∗108

200 = 8.82 x105 mm3

Calculation of plastic section modulus

Total area = 247.79 x 102 mm2

Let dp be the distance between the centre of I Section to equal area axis

dp =Ach /(2 twi) = 62932∗11.8 =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

∑Ay = (23.6X250)(566.5-(23.6/2)) + ((566.5-23.6)x11.8x((566.5-23.6)/2)

=5.01x106

Above equal area axis

∑Ay = (6293(41.95-24.2) + (250x23.6) (41.95-8.6 –(23.6/2))

+( (41.95 -8.6 -23.6 )x 11.8) x ((41.95 – 8.6 -23.6)/2)

=2.39x105 mm3

Zpz = 5.01x106 + 2.39x105

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= 5.25 x 106 mm3

For the top flange only

Zpy = 2(2502

x23.6)(2504

) + (2(400−2∗15.3

2)x8.6 x(

400−2∗15.34

) +(2(100x 15.3 x(200−15.32

))

=1.25 x 106 mm4

Check

b/t of the flange of beam = (250−11.8

2)/23.6 = 5.04 < 9.4t

b/t of the flange of channel = (100−8.615.3

) = 5.97 < 9.4t

d/t of the web of I section = 600−2∗23.6

11.8 =46.84 < 84

hence the section is plastic

Local moment capacity

M dz =Bb∗Zp∗fyΓ mo

Bp = 1 for plastic section (clauses 2.1.2 of IS 800)

=(1∗5.25∗106∗250

1.1) = 1.193 x 10 9 Nmm

1.2∗Ze∗fyΓmo

=1.2∗4.12∗106∗250

1.1 = 1.123 x 10 9 Nmm

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Moment due to vertical load = 758.19 KNm

Factored moment due to self weight = 1.5 x 27.6 = 41.4 KNm

Moment due to horizontal force parallel to rail = (1.5x16.848x103x236.37) = 6KNm

Total moment = 805.59 KN

Hence take M dz = 1123 KNm

Mdy =Zpy∗fyΓmo

= 1.25∗106∗250

1.1 = 284 KNm

1.2∗Zey∗fyΓmo

= 1.2∗8.82∗105∗250

1.1 = 2.4 *108 Nmm = 240 KNm

ie Mdy = 240 KNm

combined load capacity check

MzMdz

+ MyMdy

<= 1.0

805.591123

+ 31240

= 0.846 < 1.0

Hence safe

Shear capacity

For vertical load Vz =519.26 KN

Shear capacity

Vp = Av∗t yw√3∗1.1 (clause 8.4.1 page 59)

Av = 600x11.8 = 7080 mm2

Yp = 7080∗250√3∗1.1 = 9.29 x 105 N = 929 KN > 519.26 KN

Hence safe

Buckling resistance (clause 8.2.2 )

Md = Bb x Zp xfbd

Bb = 1.0 (plastic section)

H = 600+8.6 =608.6mm

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L=8000mm

E=2x105 N/mm2

Tt = 23.6 + 8.6 = 32.2mm

Iyy = (Izz)channel + (Iyy) I

=(1.5x108 + 5.29 x 107) = 2.03 x107 mm4

Γyy = √ Iyy / A = √(2.03∗108)/24779 = 90.5

From table 14 of IS 800 for L/r = 90.5

fcr=402N/mm2 ht/tt = 18.03

From table 13 (a) –Is 800 ht = (608.6 –(23.6/2) –((23.6+8.6)/2))

αlt = 0.21 = 580.7mm

fbd = 184.1 N/mm2 tt = 23.6 + 8.6 =32.2 mm

Mdz = BxZpxf bd hf/tt =580.7/ 32.2 = 18.03

=1.0 x 5.25 x106 x 184.1

=966KNm >805.59 KNm hence safe

Check for biaxial bending

(Mz/Mdz) + (My/Mdy) <=1.0

(805.54/966) + (31/240) =0.96

Hence safe

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Design of Gantry Girders 1

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