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

1


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


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


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


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


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

 =

y

 (  )  

=

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 9

2

2

6

2

2

=(1.15x10 )+(184.86x10 x72.23 )+(5x10 +62.93 x10 x212.17 ) 9

4

= 1.534 x10 mm

 



Zzz = (Izz/ y) =

= 4.12 x106 mm3

IYY = Iyy(1) +IZZ(C) 7

8

4

8

4

=(5.29x10 + 1.5 x10 )mm = 2.03 x10 mm 8

4

= 2.03x10 mm

IYY of compression flange =(Izz)channel +((Iyy/2)) I Section 8

  

=(1.5x 10 )+(

8

4

= 1.76 x 10 mm

6


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

∑Ay = (23.6X250)(566.5-(23.6/2)) + ((566.5-23.6)x11.8x((566.5-23.6)/2) 6

=5.01x10

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) 5

3

=2.39x10 mm

6

5

Zpz = 5.01x10 + 2.39x10 6

3

= 5.25 x 10 mm

7


For the top flange only Zpy = 2(

     x23.6)( ) + (2( )x8.6 x( ) +(2(100x 15.3 x(      )) 6

4

=1.25 x 10 mm

Check

  )/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


  = 284  = 

Mdy =

KNm

  8 = = 2.4 *10 Nmm = 240 KNm   ie Mdy = 240 KNm combined load capacity check

   +  <= 1.0      = 0.846 < 1.0 Hence safe Shear capacity For vertical load Vz =519.26 KN Shear capacity Vp =

  √  

(clause 8.4.1 page 59)

Av = 600x11.8 = 7080 mm Yp =

 √  

2

5

= 9.29 x 10 N = 929 KN > 519.26 KN

Hence safe Buckling resistance (clause 8.2.2 ) Md = Bb x Zp xf bd Bb = 1.0 (plastic section) H = 600+8.6 =608.6mm L=8000mm 5

2

E=2x10 N/mm

Tt = 23.6 + 8.6 = 32.2mm Iyy = (Izz)channel + (Iyy) I 8 =(1.5x10

Γyy =

7

7

4

+ 5.29 x 10 ) = 2.03 x10 mm

  =    

= 90.5

9


From table 14 of IS 800

for L/r = 90.5

2

fcr=402N/mm

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/mm

2

tt = 23.6 + 8.6 =32.2 mm

Mdz = BxZpxf bd

h f /tt =580.7/ 32.2 = 18.03

6

=1.0 x 5.25 x10 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

10

Design of Gantry Girders 1

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