Overhead travelling cranes are used in industrial buildings to

lift and transport heavy jobs, machines, and so on, from one

place to another.

The crane system consists of a bridge over which a crab (trolley), and a cabin which houses the operator moves .

The rails on either side of the bridge rest on crane gantry girders. The gantry girders are the girders which support the loads transmitted through the travelling (moving) wheels of the cranes as shown in figure 1.

Gantry is usually laterally unsupported beams which supports crane girder. Is supported on the columns with bracket.

2

4

9

FEATURES

• Design of Gantry Girder is a classic example of

laterally unsupported beam.

• It is subjected to in addition to vertical loads

horizontal loads along and perpendicular to its axis.

• Loads are dynamic which produces vibration.

• Compression flange requires critical attention.

These are subjected to the loads as follows.

1) Vertical load from the crane.

2) Impact load from the crane

3) Longitudinal horizontal force along the crane rail, due to

starting or stopping of crane

4) The lateral thrust, due to starting/stopping of the crab acting

horizontally, normal to the gantry girder.

1) A gantry girder section is subjected to vertical loads and

horizontal thrust (surge loads) simultaneously. Therefore, the

allowable stresses are increased by 10%. This increase in the

allowable stress is not in addition to that allowed for erection

loads with or without wind or seismic forces.

2) Either of the two horizontal forces specified in table should

be considered to act along with the vertical loads at a time.

12

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).

Deflection limitations (Table 6, p. no. 31).

Vertical loads

i) Manually operated… Span/500 ii) Electric operated.... Span/750

up to 50t

iii) Electric operated… Span/1000 over 50t

Lateral Deflection:

Relative between rails 10mm or Span/400

13

To resist heavy moment, I sections are required.

Apart from bending and shear, these girders are

subjected to longitudinal and lateral on compression

flange.

Compression flange needs additional strengthening.

This is achieved by connecting a channel section on

compression flange of I section.

15

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.

STEPS

1. Find the maximum wheel load: This load is maximum when

the trolley is closest to the gantry girder. This distance is the

minimum approach of the crane hook to the gantry. The

vertical reaction of the crane girder is transferred through its

two wheels on to the gantry girder. So maximum wheel load

is half of it. Increase maximum wheel load for the impact.

Lateral thrust due to cross travel of the crab can be neglected.

16

The maximum vertical load on gantry girder is the maximum reaction of crane girder. To get this , crab should be placed as close to gantry girder as possible

Ref fig below. Take moment about B

Lc= span of crane girder

L1= minimum approach of crane hook (distance between

CG of gantry girder and trolley)

W= weight of trolley plus the weight lifted

w= weigth of crane girder per unit length

Reaction from crane girder is distributed equally on the

two wheels at the end of crane girder

Maximum wheel load on each wheel of the crane= RA/2 kn

2. Calculate the maximum bending moment in the gantry girder

due to vertical loads. To simplify the calculations, add the

maximum bending moment due to dead load to the maximum

wheel load moment.

3. If a<0.586L Wheel load bending moment is maximum when

two wheels are in such a position that the center of gravity of

the wheel loads and one of the wheel load are equidistant

from the center of the gantry girder. In other words, the

quarter point of the span of the two wheels must coincide

with the central line of the girder.

4. If a>.586L Wheel load bending moment is maximum when

one wheel is at the center of gantry girder.

19

3. The maximum shear force is calculated. This consists of shear

force due to wheel loads and dead loads from the gantry and

rails.The shear dur to the wheel load is maximum when one of

the wheel is at the support. Fig 4.

21

4. The lateral forces on the girder and the maximum bending

moments and shear due to these are calculated, The position of

the wheel should be same as that in steps 2 and 3 above

Calculate the horizontal surge.

5. Select trial section: 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.

Calculate Izz, Iyy and Zp for a trial section.

22

6. Classify the section: preference to plastic section

b/tf < 8.4

d/tw <84

7. . Check for moment capacity of the whole section (as

lateral support is provided at the compression flange)

Mdz = b Zp fy/ m0 ≤ 1.2 Ze fy / m0 <Mu

When compression flange is laterally unsupported

replace fy by fbd (bending compressive stress) AND find

Moment capacity of compression flange about y axis.

Mdy.f = b Zpyf fy/ m0 ≤ 1.2 Ze fy / m0

8. Check local moment capacity of the girder

23

9. Check for buckling resistance in bending

10. Check for Shear capacity, defelection etc

24

Crane Capacity : 200KN

Self weight of crane girder excluding trolley : 200KN

Self weight of trolley, electric motor, hook etc : 40KN

Minimum hook approach : 1.2m

Wheel Base : 3.5m

c/c distance between gantry rail : 16m

c/c distance between columns ( span of gantry girder) : 8m

Self weight of rail : 300N/m

Diameter of crane wheels : 150mm

yeild stress ratio = ε w= ε f = √ 250/fy = 1

Steel of Grade Fe 410. Support bracket connections need not be

designed

1. Design Forces:

fu=410 Mpa

fy = fyw=fyf=250 Mpa

Load factor = γm1 = 1.50

Partial safety factor = γm0 = 1.1

Maxm permissible deflections = L/500

ε ( yeild stress ratio) = ε w= ε f = √ 250/fy = 1

1. Design Forces:

For manually operated overhead travelling crane:-

Lateral load= 5% of maximum static load (hand operated )

Longitudinal Load= 5% of weight of crab and weight lifted

Maxm concentrated Load on crane= 200+40 = 240 KN

Maxm factored load on crane= 1.5×240= 360 KN

UDL on crane= 200/16= 12.5 KN/m

Factored UDL on crane= 1.5 ×12.5= 18.75 KN/m

For maximum reaction on the gantry girder the loads are placed on

the crane as shown in figure

Taking moment about B

RA ×16= 360×(16-1.2)+ 18.75 ×(16)2/2 = 483 KN

RB = (360 + 16×18.75) – 483 = 177 KN

Reaction from crane girder is distributed equally on the two wheels at

the end of crane girder

Maximum wheel load on each wheel of the crane= 483/2=241.5 kN

since a<0.586L (4.96m) , For maximum BM the wheel loads

should be placed as below

Assume self weight of gantry girder as 2kN/m

Dead Load of gantry girder = w= self wt + wt of rail= 2000 + 300= 2300N/m

Factored Dead Load= 1.5 × 2.3= 3.45 KN/m

Position of one wheel load from the mid point of span= wheel base/4 =

3.5/4= 0.875m

Bending moment due to live load about “C” = RD × 8 = 241.5 ×1.375 + 241.5 × 4.875

or RD = 188.67 KN

Maxm Moment = 188.67 × 3.125= 589.6 KNm

Bending moment due to impact = 0.1× 589.6 = 58.96 KNm

Total Bending moment due to live load= 589.6+ 58.96= 648.56 KNm

Bending moment due to dead load= wl2/8 = 3.45×8×8/8 = 27.6 KNm

Maxm BM= 638.56+ 27.6= 676.16kNm = 676.16 × 106 Nmm

Taking moment about D

RC ×8= 241.5×8 + 241.5 ×4.5

RC= 377.34kN

Hence Shear force due to wheels= 377.34kN

××××

Lateral Force transverse to the rail = 5% of the weight of

crane girder and weight lifted

Or Lateral Load = .05× 240= 12 kN

Factored Lateral Load = 1.5× 12= 18kN

Lateral load on each wheel= 18/2=9kN

For maximum moment in gantry girder the position of loads

is same as above except that they are horizontal

Hence by proportioning we get

Maxm BM due to lateral load by proportion = 9×589.6/241.5

= 21.97KNm

Maxm shear force due to lateral load by proportion

= 377.34× 9/241.5 = 14.06 KN

PREMILANARY TRIAL SECTION

Approximate depth of section= L/12= 8×10 3/12 = 666.6mm ≃600mm

Approximate width of flange= L/30= 8×10 3/30 = 266.6mm

≃300mm

App section modulus required= Zpz = 1.4 Mz/fy =

= 1.4 ×676.16×106 / 250 = 3786.5 ×10 3 mm3

Properties of I section

Area=17038mm2

Depth of section h= 600mm

Width of flange b = 250mm

Thickness of flange t f = 21.3mm

Thickness of web tw = 11.2mm

Ixx= 106198.5 cm4

Iyy= 4702.5 cm4

Properties of Channel section

ISMC 300 @ 351.2 N/m

Area=4564mm2

Depth of section h= 300mm

Width of flange b = 90mm

Thickness of flange t f = 13.6mm

Thickness of web tw = 7.6mm

Center of gravity Cyy = 23.6mm

Ixx= 6362.6×104 mm4

Iyy= 310.8 ×104 mm4

Moment of Inertia and Modulus of combined section

The distance of NA of the built up

section from the extreme fiber of

compression flange ie y.

y = [A beam (h/2 + tf channel)+ Achannel. Cyy] /

Abeam+Achannel

y = [17038 (600/2+7.6)+ 4564×23.6 /

(17038+4564.23)

= 247.59 mm

Gross moment of Inertia of built up

section

I z gross =I z beam +Iz chan

= [106198.5 × 10 4 + 17038 . (307.6-

247.59) 2] + [310.8 × 104 + 4564(247.59-

23.6)2 ] = 135453 × 10 4

I y gross= I y beam + I y chan

= (4702.5 + 6362.6)× 104

= 11065.1 )× 104

Zez= Iz /y

= 135453 × 10 4 / (600+7.6-

247.59)

= 3764.98 × 103 mm3

Equating equal area (ignoring fillet)

A comp = A tenion

or 4564+250×21.3+ y1×11.2=

250×21.3+ 11.2×(600-2×21.3-y1)

Or y1=74.95b(from lower of top flange)

Plastic section modulus of the section

above equal area axis=

[300×7.6×(74.95+21.3+7.6/2) + 2×(90-

7.6)×13.6×(74.95+21.3-(90-7.6)/2]

+250×21.3×(74.95+21.3)

+74.95×11.2 × 74.95/2 = 838.77 ×103

mm3

PLASTIC MODULUS OF SECTION

Plastic section modulus of the

section below equal area axis=

250×21.3×(600-21.3-74.95-21.3/2)

+[11.2 × (600-2×21.3-74.95)]/2 =

3929.202×103 mm3

Total Zpz = 838.77 ×103

+3929.202×103 = 4767.9 ×103

Greater than required

PLASTIC MODULUS OF SECTION

Outstand of flange of I section, b= bf/2= 250/2= 125mm

b/tf of I section= 125/21.3= 5.86<(8.4 ε = 8.4)

Outstand of flange of Channel Section= , b= bf-tw

= 90-7.6= 82.4mm

b/tf of flange of Channel Section= 82.4/13.6= 6.05<8.4

d/tw of web of I section= h-2tf/tw= 600-2×21.3/11.2=

49.76< 84

Entire section in plastic range

( b =1)

Local moment capacity

Mdz = b Zp fy/ m0

Mdz = 1×4767.97×103×250×10-6/ 1.1 =

1083.63KNm

≤ 1.2 Ze fy / m0 or 1.2×3764.98×103×250×10-6/ 1.1 =

1026.81KNm

moment capacity of section= 1026.81>

676.16KNm

Trial section seems to be safe.

Mdy.f = b Zpyf fy/ m0

Mdy.f = 1×824.76×103×250×10-6/ 1.1 = 187.44KNm

≤ 1.2 Ze fy / m0 or 1.2×580.92×103×250×10-6/ 1.1 =

158.43KNm

moment capacity of flange Mdy.f = 158.43 KNm