gantry crane - baixardoc

SUMITOMO CORPORATION

PROJECT : SKS, Prai - 350MW CCGT Power Plant Checked By Page of

DOC.TITLE: Design of Super structure-Design of Crane-girder Area: Turbine build.

DOC. NO : CGPR1-100-5-011 Rev 0 Dept Structural

DESIGN CALCULATIONS REFERENCES /

REMARKS

6.4.a DESIGN OF CRANE GANTRY GIRDER 11M span All below

references are

a) INPUT DATA :- BS 5950,

(Refer Appendix-E, for EOT drawing) part-1, UNO

Crane Capacity = 1050 kN

Weight of Crab = 320 kN

Weight of Crane Bridge = 780 kN

Self weight of the Rail = 2 kN/m

Width of Walk way = 0.6 m

Dead Load of the Walkway = 1.5 kN/m²

Live Load of the Walkway = 5 kN/m²

Height of the Crane Rail = 65 mm

Span of the Crane Girder, Lg = 11 m

Centre to centre distance of , Lc = 32 m

Rail (i.e. Span of Crane Bridge)

Mini. approach of crane hook to the gantry = 1.800 m

No. of Wheels = 4

Wheel Spacing1 = 1.40 m

Wheel Spacing2 = 4.70 m

C.G of loading from left load = 3.75 m 1.40 4.70 1.40

Impact Factor : Vertical = 30 %

Horizontal = 10 %

(Transverse to rail)

Deflection Factor Vertical = 600 Table:5

Horizontal = 500

Load Factor : Imposed load vertical -g = 1.6

Imposed load Horiz.gIhf = 1.6

Dead load gdf = 1.4

Design strength of steel, py = 265.0 N/mm2

Table:6

Maximum unsupported length Top Flange = 2.60 m

146905687.xls.ms_office .xls REF gvrs/ST






REMARKS

Depth of the surge girder = 0.60 m

Maximum unsupported length Bottom Flange = 2.60 m

1.80m (1050+320)kN 780 kN

Kicker

RL 32.00m RR

RL = (1370 x 30.20 + 780 x 32.00/2)/32.00= 1682.938 kN

Wheel Load by calculation 420.73 kN/wheel

b) LOAD CALCULATIONS:

b.1) Vertical Loads

b.1.a) Conc. Loads

Max. static Wheel Load say Wm = 421 kN

875.7 875.7

Load due to Impact = 0.30 x 421 = 126.3 kN

Total load = 547 kN

Factored Load Wmf = 1.60 x 547.30 = 875.68 kN 1.40 4.70 1.40

b.1.b) Uniform Dirstributed Load

Self weight of rail = 2.00 kN/m

Walkway Dead Load = 0.45 kN/m

Walkway Live Load = 1.50 kN/m

Self weight of girder = 4.66 kN/m

8.61 kN/m

Factored load Wdf = 1.40 x 8.61 12.06 kN/m

b.2) Horizontal Loads

Maximum lateral load per wheel is equal to 10% Static vertical wheel load,

l = 0.1 from Fig-1

Max. Lateral load WH = 0.10(421*4) = 168.4 kN BS:2573,part-1

4 wheels are resisting the total lateral load

Factored lateral load Wdf = 1.60 x 168.40 / 4 67.36 kN/wheel







REMARKS

c) MAXIMUM BENDING MOMENT AND SHEAR FORCE:

c.1) For vertical loads

c.1.a) Bending Moment :-

The maximum Bending moment under moving loads occurs when line of

action of one load and centre of gravity of the loads are at equal distance

from the centre of span.

875.68kN 875.68kN 875.68kN 875.68kN

12.06kN/m

C

RA RB

11.00m

Reactions :-

Ra = 4x875.68x(11 - 11*0.5 - 0.25*4.7)/ = 1443.525 kN

+ 12.06 x 11 /2

Rb = 4x875.68+12.06x11- 1,443.525 = 2191.834 kN

Maximum Bending moment occurs at C. =

Mux1 = (1443.53 x 4.33) -875.68 x 1.4 - (12.06 x 4.33²/2)

= 4904.517 kN.m

c.1.b) Shear Force:-

875.68kN 875.68kN 12.06kN/m

RA 11.00m

Reactions:

RA = 4 x 875.7 x [11.0-3.8] /11+ (12.1 x 11.0/2) 2374.930 kN

RB = (4 x 875.7) + (12.1 x 11.0) - 2374.93 1260.428 kN

Max. Reaction = 2374.930 kN

c.2) For Horizontal loads :-

CG. OF LOADS

Mid Span of Crane Girder

= =

CG. OF GANTRY







REMARKS

67.36kN

C

c.2.a) Local Bending Moment at C,

Crane Girder is laterally bending between Node points of surge Girder

Muy = 67.360 x 2.6 /4 43.784 kN.m

c.2.b) Axial Force:

Because of Lateral force, the Crane Girder is subjected to axial force.

Max lateral bending Moment 4904.5 x 67.36 / 875.68 377.27 kN-m

F=Axial force in the surge girder 377.27 / 0.6 628.78 kN

c.2.c) Shear force :-

67.36kN 67.36kN

RA 3.75m 11.00m RB

Reactions :-

RA = 4x 67.4[11.0 - 3.8]11.00 = 177.585 kN

RB = 4 x 67.360 - 177.585 = 91.855 kN

Max. Horzontal reaction RH = 177.585 kN

d) DESIGN OF GANTRY GIRDER: y







REMARKS

Depth 1250 mm

Width 450 mm 20

t = 20 mm x x 1250

T = 40 mm

40

450

Properties :-

Depth of the section, D = 1250 mm

Width of the section, B = 450 mm

Thickness of web, t = 20 mm

Thickness of flange, T = 40 mm

Effective depth of web, d = 1170 mmSecond moment of inertia, Ixx = 1.59E+10 mm

4

Second moment of inertia, Iyy = 6.08E+08 mm4

rmin = 101.19 mm

Section modulus, Zxx = 2.54E+07 mm3

Section modulus, Zyy = 2.70E+06 mm3

Plastic modulus, Sxx = 2.96E+07 mm3

Plastic modulus, Syy = 4.28E+06 mm3

Buckling parameter, u = 1 conservatively

Torsional index, x : D/T = 31.25 as per Cl.4.3.7.5

Sectional Area, A = 59400 mm2

Flange Area on one side, Ag = 18000 mm2

Out stand width of panel, b = 215 mm

Constant, e, = sqrt(275/py) = 1.02

Outstand element of compression flange, b/T = 5.38 Plastic Cl.3.5.2 and

Web slenderness, d/t = 58.50 Plastic Table:7

d.1) Shear Capacity

Web slenderness, d/t = 58.50 < 63*1.02 Cl.4.4.4.1

Satisfactory

Shear area parallel to the web, Avx=t*d = 23400 mm2 Cl.4.2.3,

Critical Shear strength, qcr for t/d =58.50 = 159 N/mm2 Table:21,

Shear Capacity, Vcr=qcr*Avx = 3720.6 kN Cl.4.4.5.3

>2,374.93 kN Satisfactory

d.2) Moment capacity, Mb







REMARKS

d.2.a) Lateral-torsional buckling moment, Mb:

( as per clause 4.3.7.3 of BS 5950, part-1)

Effective length factor = 1.00 Table:9

( Destabilizing condition)

(As per table:9,BS 5950,part-1: Beam partial restrained against rotation)

Effective length, LE = 2.60 m

Slenderness, l = LE/rmin = 25.69

Equivalent slenderness, lLT = nunl Cl.4.3.7.5

Slenderness correction factor, n = 1.0 conservatively

Uniform moment factor, m = 1.0 conservatively

Buckling parameter, u = 1.000

l/x = 0.822

N = 0.50

Slenderness factor, n = 1.00 Table:14

lLT = 25.69

pb = 265.00 N/mm2 Table:12

Buckling resistance, Mb = pb*Sxx

= 7843.23 kN.m Satisfactory

>4904.52 kN.m Cl.4.3.7.2

> m*Mux1

e) CHECK FOR COMBINED BENDING COMPRESSIVE STRESS

IN EXTREME FIBRE (FOR VERTICAL PLUS LATERAL)

e.1) Compressive strength pc :-

Slenderness, l = LE/rmin = 25.69

Reduced design strength, py = 245.00 N/mm2 Cl.4.7.5

pc = 240.00 N/mm2 Table:27c

e.2) Overall buckling check

(As per Clause 4.8.3.3.1, BS 5950: part-1)

F/Ag*pc + mMux1/Mb + mMuy/py*Zyy = 0.832 Satisfactory

< 1.000

f) CHECK FOR LONGITUDINAL STRESS:







REMARKS

Height of rail = 65 mm

5% of the static wheel load = 5/100 x4x 875.7 175.14 kN

Bending moment in the longitudinal direction is equal to Longitudinal Force into

Crane Rail Depth plus half of Crane Girder depth

Mux2 = 175136 x (65 + 625.0) 120.84 kN.m

CHECK FOR COMBINED BENDING COMPRESSIVE STRESS

IN EXTREME FIBRE (FOR VERTICAL PLUS LONGITUDINAL)

F/Ag*pc + m(Mux1+Mux2)/Mb = 0.681 Satisfactory

g) CHECK FOR DEFLECTION:

Allowable deflection for vertical loads

d lim, v = Span / 600 =11,000.0 / 600.0 = 18.33 mm

Allowable deflection for horizontal loads

d lim, h = Span / 500 = 11,000.0 /500 = 22.00 mm

Vertical Deflection:-

3.15

1.75

547.3kN 547.3kN 8.61kN/m

c

RA 11.00 RB

d v =

= #VALUE!

{( 2 x 547300 x 11000³)/( 48 x 205000 x 1.59E+10)} x

{[3 x 1.75/11 - 4 x (1.75/11)³] + [3 x 3.15/11 - 4 x (3.15/11)³]}

= 11.960 mm

CHECK dv < Allowable Deflection 11.960 < 18.3 HENCE SAFE

h) Crane Girder Welding Calculation

Top Flange & Web is welded by full Penetration Butt weld.

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úúû

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-´+-´+´3333

L

a23a2

48EI

PL

L

a13a1

48EI

PL

EI

WL

384

5

LL44

4

CG. OF GANTRY

CG OF LOADS

==







REMARKS

Bottom Flange Weld.

Horizontal Shear = FAy/ Ixx

A- Area of the Bottom Flange = 18000 mm2

y - C.G of flange Plate from C.G of section = 605 mm

Ixx of the section = 1.59E+10 mm4

Maximum vertical shear = 2374.930 kN

Horizontal Shear 2,374.9 x 1000 x 18000x605 / 15851055 1631.626 N/mm

Size of the weld on each side 1,631.6/ ( 2 x 215x 0.707) 5.421 mm

Provide weld as = 12 mm

i) DESIGN OF BEARING STIFFENER

Bearing check:

Minimum area of stiffener in contact with the flange = 0.8*Fx/pys Cl.4.5.4.2

Fx = External reaction

pys = Design strength of stiffener

Minimum Area of stiffener required = 7169.60 mm2

Conside Thk. Of Stiffener , ts = 25.00 mm

Width of the stiffener, bs = 450.00 mm

Area of the stiffener = 11250.00 mm2 Satisfactory

Check for outstands

Outstand from the face of the web = bs/2-web thickness

= 215.00 mm

Outstand of web stiffeners, as per Cl.4.5.1.2 of BS5950: Limits:

19tse = 483.88 mm

13tse = 331.08 mm Satisfactory

Bearing resistance of the stiffener

Bearing Stress in member = 211.10 N/mm2







REMARKS

< 265 N/mm2 Satisfactory

Buckling resistance of the stiffner

(as per Cl.4.5.1.5 of BS5950,part-1)

Design strength of the stiffner in buckling = py-20 Cl.4.5.1.5

= 245.0 N/mm2

Buckling resistance check as a column:

Area of combined section 450 x25 + 20 x 20 x 20 19250.00 mm2

Ixx = 1.90E+08 mm4

= 99.38 mm

l = l / Rmin =1250x 1000 / 99.4 = 12.58

Compressive strength, pc = 245.00 N/mm2 Tb.27c,

Buckling resistance of the stiffener = 4716.25 kN

> 2374.93 kN Satistactory

Weld between Stiffener & web

Vetical Height avilable for Welding = 1170.00 mm

Thk. of weld reqd =2,374.9 x1000/(1170x2x0.7*215) 6.74 mm

Provide weld thickness = 12.00 mm

j) Shear buckling of Web under Wheel load

Web bearing under wheel load

(as per Cl.4.11.4,BS 5950, part-1)

Load dispersion under wheel,lw= 2(Height of the wheel + Thickness of the flange)

= 210 mm

Bearing Capacity = lw*py*t = 1113 kN

> 875.68 kN Satisfactory

Web buckling under wheel load

(as per Cl.4.5.2.1, BS 5950,part-1)

AI /Rmin =







REMARKS

b1 = Stiff bearing length = 2(Height of the crane rail)

= 130.00 mm

n1 = Dispersion at 45degrees through half the depth of the section

= (depth of the web + 2*thickness of the flange)

= 1250 mm

d = Depth of the web

= 1170 mm

Web slenderness, l = 2.5*depth of the web/thickness of the web Cl.4.5.2.1

= 146.25

Compressive resistance, pc = 70 N/mm2 Table 27c

Buckling resistance, Pw = (b1+n1)*t*pc

= 1932.00 kN

> 875.68 kN Satisfactory

k) Connection for Longitudinal Force

Longitudinal Force = 175.14 kN

Dia of bolt provided = 24.00 mm

No. of bolts provided = 4.00

Stress in Bolts = 96.78 N/mm2

< 160 N/mm2

l) Design of Surge Girder

Design of bracing members

Maximum Horizontal force = 177.585 kN

Max Force in diagonal = 335.1 kN

Angles provided = 100X100X8 RSC

Area of the Section = 15.60 cm2

Rmin of the section = 3.07 cm

Length of diagonal = 1.50 m

Inclination of diagonal w.r.t Horizontal = 32.00

Stress in member = 214.82 N/mm2

(No.bays are

not to count in

the sketch)

Allowable Stress in member

l=1.5 *100 / 3.07 = 48.86


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