2. DESIGN OF GANTRY GIRDER AS PER IS 800 : 1984 Given Data Crane Capacity Self-weight of the crane girder excluding trolly Self-weight of trolly,hook etc. Minimum hook approach Distance between whele centres Span of crane between the rails Span of gantry girder Yield stress of steel
L1
c Lc
fy
=
250 kN
= = = = = = =
200 50 1 3.5 16 6.5 250
E = Load and bending moment calculations
200000
kN kN m m m m N/mm2 N/mm2
Step-1 (1.1) Load (i) Vertical load
250+50 L1=1 m
Wc=200 Lc = 16 m
Maximum static wheel load due to the weight of the crane
weight of trolly and crane capacity
W1 Wt
Maximum static wheel load due to crane load
W1
Total load due to the weight of the crane and the crane load Add for impact @ 25% Factored Wheel Load (ii) Lateral surge load Lateral load (per wheel)
Factored lateral load
Wc
= = = = =
200/4 50 kN Wt Lc L1 / 2Lc 250+50 300 kN
= =
300(16-1)/(2 x 16) 140.63 kN
= = = = =
50+140.63 190.63 kN 190.63 x 1.25 238.2875 kN 357.43125 kN
= = = =
10%(hook+crab load)/4 0.1 X (250+50)/4 7.5 kN 11.25 kN
(iii) Longitudinal braking load Horizontal force along rails
= = = =
Factored longitudinal load
5% of wheel load 0.05 x 238.2875 19.063 kN 28.60 kN
(1.2) Maximum bending moment (i) Vertical maximum bending moment (excluding self weight) 1.75 1.75 w w A D C E 3.25 ME
At center M c M
= = = = =
B
3.25 2Wc ( L / 2 c / 4) 2 / L 2 x 357.43125(6.5/2-3.5/4)^2/6.5 620.35 kN.m 580.83 kN.m 620.35 kN.m
Assume, self weight of gantry girder = 1.6 kN/m self weight of rail = 0.3 kN/m Total factored dead load = 2.85 kN/m Bending at centre moment due to D.L. = 15.06 kN.m (ii) Horizontal bending moment M Ey = 2 x 11.25(6.5/2-3.5/4)^2/6.5 = 19.53 kN.m At center M cy = 18.29 kN.m My = 19.53 kN.m (iii) Bending moment due to drag Assume the rail height = 0.15 m Reaction due to drag force = 28.6 X (0.3+0.15)/6.5 = 1.98 kN M E = R( L / 2 c / 4) / L = 4.71 kN.m Total design bending moment M z = 620.35+15.06+4.71 = 640.12 kN.m (1.3) Shear force (i) Vertical shear force 1.75
1.75
A
B C 3.25
3.25 Wc ( 2 c / L)
o.k.
Reaction at A Shear force due to dead load Maximum ultimate shear force
Wc ( 2 c / L) wl / 2 VZ
= = = =
522.4 kN 9.27 kN 522.4+9.27 531.67 kN
= = = =
11.25 (2 - 3.5/6.5) 16.45 kN 1.98 kN 533.65 kN
6.5/12
b
= = = = =
I
=
(ii) shear force due to surge load Vy
Reaction due to drag force Maximum ultimate reaction Step-2 Preliminary selection of the girder Approximate depth and width of section d = L /12 b = L /30 Provide depth and width of section, d
ISMC 300 = A ch = hc = b = tf = tw = I zz = I yy = Cy = w
0.3512 4564 300 90 13.6 7.6 6362.6 310.8 23.6
(i) Elastic properties of the combined section Total area A = A B + A ch
= 15621+4564 2 = 20185 mm The distance of N.A. of the built-up section from the extreme fibre of tension flange, y = 364.3 mm
kN/m mm2 mm mm mm mm cm4 cm4 mm
Ap
ISMC 300 cyy h2 = 219.7 mm
.N.A dp
h1 = 64.3 mm t
y 364.3
Rolled section As
ISMB 600
h1 h2 Iz zz combinedI yy I y for tension flange about the y-y axis I tf For compression flange about the y-y axis I cf z y (for top flange alone)
= = = = = = = =
y hB / 2
64.3 mm
( hB t ch ) y C y
219.7 mm 2 2 I ZB AB h1 ( I y ) ch Ach * h2 4 1.21E+09 mm 3 3.31E+06 mm 4 9.01E+07 mm
=
4 1.61E+07 mm
= =
4 7.97E+07 mm 3 531189.3333 mm
(2.2) Calculation of plastic modulus The plastic N.A. divides the area into two equal areas 2 = 10092.5 mm = 4564/(2*12) d p Ach / 2t d p 190.17 mm Depth of equal area axis from bottom flange = (600/2)+190.17 = 490.17 mm Depth of equal area axis from top = (600+7.6) - 490.17 = 117.43 mm Ignoring fillets, the plastic section modulus below the equal area axis 3 3.42E+06 mm Ay = Above the equal area axis 3 9.10E+05 mm Ay = 3 z pz = 4.33E+06 mm For the top flange only 3 z py = 7.21E+05 mm
Step-3
Check for moment capacity
Section Classification ε
=
250 fy
= b / t f of the flange of the I-beam b / t f of the flange of the channel-beam d / t w of the web of the of the I-beam The section is classified as a
= = =
4.76 Plastic 6.06 Plastic 46.54 Plastic Plastic
(3.1) Local moment capacity Md
b b Z p f y / m0
1.2 Z e f y / m 0
Design moment Minor axis design moment
1
Mdz
b Z p f y / m0
1.2 Z e f y / m 0
Md y
= = = = = = = =
b Z p f y / m 0 1.2 Z e f y / m 0 1.00 983.54 kN.m 902.95 kN.m 902.95 kN.m 163.93 kN.m 144.87 kN.m 144.87 kN.m
2 1 LLT / ry C1 1 2( LLT ) 2 20 h f / t f
607.6 6500 28.4 66.83 1.132 2065.11
0.5
mm mm mm mm mm kN.m
b Z p f y / M cr
1.2Z e f y / M cr
0.73 0.7 0.7
0.5 [1 LT LT 0.2 LT ] 2
0.80
=
1 [ LTZ (
2
LTZ
2 LTZ ) 0.5 ]
1
LTZ
1
f bd
M dZ
Design strength
833.80
= = = = = >
[ LTZ (
Step-5
LTZ
2 LTZ ) 0.5 ]
b Z p f bd
2 833.80 N/mm 640.12 Required strength safe
f y / m 0 1.2 Z e f y / m 0
=
0.91 safe
67ε
= =
46.53 67 safe
Vd
=
0.6Vd
566.86
= = >
q h2 A q
= VA y / I Z = 219.7 2 = 4564 mm = 442.01 N/mm
Check for shear capacity
Check for Shear Buckling
d / tw 67 d / tw
f y h tw
m0 3
944.76 kN 566.86 kN 531.67 kN safe
(5.1) Weld design y
This shear is taken by the welds. Strength of weld
=
fu 3 mw
size of weld
0 .7 s
connecting the channel to the top flange of the I-beam. Vd y =
3.35 4 mm
f y Av
m0 3
Av
fu 3 mw 2 189 N/mm
= = =
Hence use a minimum weld of
1
0.848 c LT f y /γ m0 2 192.67 N/mm
Thus the beam is satisfactory under vertical loading . Now, it is necessary to check it under biaxial bending. for top flange only M dy = b Z p (4.1) Check for biaxial bending 640.12/833.8 + 19.53/144.87
2
f y Av
m0 3
Av Vd y
523.40 Step-6 Web buckling Assume dispersion length
150 mm 600/2 + 2*7.6 315.2 mm 2.5 d / t 108 2 106.80 N/mm
= =
596.21 kN 357.43125 kN safe
Step-7 Web bearing Load dispersion at support with 1:2.5 dispersion Minimum stiff bearing n2
= R x /(tf yw / 1.1) n2 = (20.8 + 20)*2.5 = 102 mm 533.65 kN Rx = 93.68 mm b1 = Web bearing at support requires a minimum stiff bearing of 93.68 mm Step-8 Check for deflection at working load Serviceability vertical wheel load excluding impact = 190.63 kN Deflection at mid-span = WL 3 [(3a / 4 L) (a 3 / L3 )] /(6 EI ) a = ( L c) / 2 = 1500 (i) Vertical 5.82 mm = Allowable maximum deflection = 8.67 mm safe (ii) Lateral Only the compound top flange will be assumed to resist the applied lateral load as in the bending check 3.46 mm = = 8.67 mm safe Step-9
Fatigue strength
The cranes operates Maximum trips of crane in 1 hours at maximum load level Design life of the building Category classification Number of stress cycles
= =
f fn mft N sc
ff Design fatigue strength Calculation of actual stress range: f min f max f Shear stres at support
Design fatigue strength in shear
f
= = = = =
= = = = = =
200 days/year 8 hrs/day 3 per hrs 50 years 118 1.15 240000 cycles O.K. f fn 3 5 *10 6 / N sc 2 324.69 N/mm 2 282.34 N/mm
100 kN 16 m Given Data Crane Capacity Self-weight of the crane girder excluding trolly Self-weight of trolly,hook etc. L1 Minimum hook approach c Distance between whele centres Span of crane between the rails Lc Span of gantry girder fy Yield stress of steel E
200 kN
=
100
200
= = = = = = =
140 35 1 3.9 16 6 250
200 75 1 4 16 6 250
=
200000
200000
=
100
200
= = = = = = =
165 35 1 4.2 18 6 250
200 75 1 4.1 18 6 250
=
200000
200000
=
100
200
= = = = = = = =
200 40 1.2 3 18 8 250 200000
200 40 1.2 3.5 16 8 250 200000
20 m Given Data Crane Capacity Self-weight of the crane girder excluding trolly Self-weight of trolly,hook etc. L1 Minimum hook approach c Distance between whele centres Span of crane between the rails Lc Span of gantry girder fy Yield stress of steel E negi Crane Capacity Self-weight of the crane girder excluding trolly Self-weight of trolly,hook etc. L1 Minimum hook approach c Distance between whele centres Span of crane between the rails Lc Span of gantry girder fy Yield stress of steel E
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