Haider Crane Co.
Hoist Crane Single Girder
12/25/07
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Haider Crane Co.
Hoist Crane Single Girder 12/25/07
2
Haider Crane Co.
Hoist Crane Single Girder
12/25/07
3
Haider Crane Co.
Hoist Crane Double Girder
12/25/07
4
Hoist Crane Double Girder
Haider Crane Co.
12/25/07
5
Haider Crane Co.
Hoist Crane Single Girder Gantry crane
12/25/07
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Hoist Crane Single Girder Gantry crane
Haider Crane Co.
12/25/07 7
Haider Crane Co.
Hoist Crane Gantry crane
12/25/07
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Hoist Crane Gantry crane
Haider Crane Co.
12/25/07
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Haider Crane Co.
Suspension Crane
12/25/07
10
Haider Crane Co.
12/25/07
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Haider Crane Co.
Suspension Crane
12/25/07
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Haider Crane Co.
Jib Crane
12/25/07
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Cargo Lift
Haider Crane Co.
12/25/07
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Haider Crane Co.
Setting Crane Girder
12/25/07
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Haider Crane Co.
Crane Design Basics
12/25/07
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Compute Moments Dead Load Bending
Haider Crane Co.
P
L = Span
Span
Live Load Bending
x
P1
P2 y
Span
l 2 Max Moment =
Pl 8 4
P = Center drive + Controls w = Footwalk + Beam + Lineshaft Dynamic Mx = Max Moment x factor Dynamic My = Max Moment x factor
Jika P1=P2, pakai point-41, jika P1≠P2 pakai point-42. Hitung juga moment maksimum memakai rumus PL/4. Dengan kata lain gunakan nilai terbesar yang diperoleh. 12/25/07
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Compute Stress
Haider Crane Co.
Dynamic Mx = LL Mx + DL Mx
a
My = LL My + DL My
f bxT
f bxC
C
Fbx
Girder Beam Available to 60 ft max. length
Mx STop
f byC f bx
b
Mx 0.6 Fy Sbott
My STop f by
C
0.6 Fy
h
10 .
b
Fabricated/Box Beam L/h should not exceed 25 L/b should not exceed 65 DL – Dead load, LL = Live Load, Fy = Yield normally @ 36 ksi for A36 material, Fb = Allowable Stress 12/25/07
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Compute Deflection
5wl 4 384 EI Pl 3 48 EI
12/25/07
Haider Crane Co.
For Trolley: For Uniform Load
For Conc. Load
LL TW P 2 Pa 3l 2 4a 2 24 EI
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Allowable Compressive Stress Fb per CMAA 74
Haider Crane Co.
1/600
1/888
0.6 y
12000 d L Af
12/25/07
Use when the flanges are not welded on the top and bottom
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Allowable Compressive Stress
L rt
Haider Crane Co.
102000 and Fy
Otherwise, Fb3
L rt
510000 , Fb 2 Fy
2 L Fy 2 rt F y 3 1530000
170000 2
L r t 12000 Fb1 d L A f Fb 0.6 y
( per CMAA)
Select Allowable Stress which is the Greatest of all. Then check for the following:
tensile 0.6 y comp x fb 12/25/07
comp y 0.6 y
1 21
Lower Flange loading per CMAA 74
Haider Crane Co.
This is a empirical formula
For a crane where the trolley is running on the bottom flange, it is necessary to check the local bending of flange due to the wheel load. The flange must be OK before a beam selection is made. 12/25/07
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Lower Flange loading per CMAA 74
12/25/07
Haider Crane Co.
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Lower Flange loading per CMAA 74
12/25/07
Haider Crane Co.
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Lower Flange loading - Alternate procedure I
Haider Crane Co.
Flens bagian bawah dari balok crane harus diperiksa terhadap : 1) Tension in the web.
2) Bending of the bottom flange.
Panjang daerah perlawanan diambil 3,5 k (k = jarak yang diukur pada kemiringan 300 dari titik beban crane). Dengan asumsi 4 roda (2 set) pada setiap ujung crane, setiap roda bekerja P/4 yang akan disalurkan ke-perletakan balok crane. Beban as roda sebesar P/2, mengakibatkan tarik pada web.
Tegangan tarik pada web sebesar :
3.5k
P P P ft 2 A 2t w 35 . 7t w
k
tf 30 deg
Lentur pada flens tergantung posisi roda dengan reaksi web balok (e = jarak dari titik beban ke bagian tepi web). Beban roda sebesar P/4. Panjang arah longitudinal flens yang ikut menahan lentur sebesar 2e.
P/2
tw P/4
Tegangan lentur sebesar ;
Bottom Flange
tf
M Pe 6 Pe 6 0.75 P fb S 4 bd 2 4 2et 2f t 2f 12/25/07
e
k1
e Point of Load
e
Refer to Engineering Journal, 4th quarter, 1982, Tips for avoiding Crane Runway Problems by David T. Ricker
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Lower Flange loading - Alternate procedure II
Haider Crane Co.
Now, the angle is changed from 30 degree to 45 degrees.
tw
Capacity = 6000 lb (daya angkut crane)
P/4
Hoist, Wt = 1000 lb (berat crane)
tf
Load = 6000 +1000 = 7000 lb
k1
Wheel load = 7000/4 = 1750 lb
e
With 15% impact = 1750(1.15) = 2013 lb Lebar flens, b =11.5 inc, sehingga e = b/2 = 5.75 inc 45 deg
Tebal flens, tf = 0.875 inc
Momen pada flens, M = 2013 (5.75) = 11574.75 lb-inc
b=2e
Teg arah-y, σy = M/Sy = 11574.75. (6)/(11.5)(0.875)2 = 7890.15 lb/in2 Load Moment = 7000(1.15)(30)(12)/4 + 110 (12x30)2/(8 x12) = 873.000 lb-in
Teg arah-x, σx = M/Sx = 873.000/280 = 3117.8 lb/in2
x2 y2 x y
2 2 3117,8 7890,15 3117,8 7890,15
= 9.827 << 0.6 σy = 21.600 lb/in2
12/25/07
OK
110 lb/ft 30 ft crane Span
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EXAMPLE – Simple Approach
Haider Crane Co.
Capacity: 2 Ton (=4000 lb), Span: 20 Ft (=240 in)
P1 114
Hoist Wt: 200 lb, Hoist W.B: 12 in
P2 P=2100lb 12
Vertical Impact factor = 15%, Hor. Impact = 10% Solution:
Mx
wL P a L 8 2L 2 2
2
Beam must be checked for Lower flange load, if the trolley is under running
240 Say, for example, we select a A36, S beam S-12x31x8, Ix=218 in4, Iy=37.1 in4, Sx=36.4 in3, Sy=9.27 in3, d/Af=4.41
31.8 240 2 2100 1.15 12 Mx 240 294571.1 lb in P=2100 lb, w = (31.8/12) lb/in 12 8 2 240 2 M 294571.1 x x 8092.6 lb / in 2 comb 8092.6 3039.5 11.132.1 lb / in 2 Sx 36.4 2 0 . 6 0 . 6 36 216 . 000 lb / in M y 294571.11.1 0.1 all 1 y x 3039.5 lb / in 2 12000 12000 Sy 9.27 1.15 all 2 11.337.8 lb / in d 240 4.41 4 2 2 L 5wL (3L 4a ) Af Pa 384 EI 24 EI all min_ of all1 , all 2 11.337.8 lb / in 2 5(31.8)240 4 (3 240 2 4 114 2 ) 2100 1.15 114 comb all. OK 384 12 218 E 24 218 E L 0.237 in 0.4 in OK 600 2
12/25/07
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EXAMPLE – Conservative Approach DLF =
11 . 1.05
Br . speed 1.2 2000
Haider Crane Co. P1
= 1.10 114
HLF = 0.15 0.005( HoistSpeed ) 0.5
= 1.15
IFD = 0.078( Bridge_ acc.) 0.025
= 0.39
C T _WB X 1 X 2 T _ Wt HLF DLF = 2410 T _ WB 2
Wheel Ld (P1/P2) = 2
Moment A = HLF x M (whichever is greater) =P1 P2 , M
12
P=2100lb
240
L WB P2 ( P1 P2 ) 2 PL X , X 05 . , OR , M = 166290 L P P 4 1 2
Moment B = IFD x M (whichever is greater) = 56394 Beam must be wL2 PL checked for Static Moment = = 19080 8 4 Lower flange Moment C = DLF x Static Moment = 20988
P2
load, if the trolley is under running
5wL4 d1 0.0181 384 EI PL3 d2 0 3EI
Moment D = IFD x Static Moment = 7441.2 P1 P2 3 L2 4a 2 Moment Mx = A+C = 187278 d3 a 0.2188 Moment My = B+D = 63835.2 2 24 EI P1 L3 Tensile Stress = Mx Sxb = 5.15 < 0.6(36) OK d4 0.1098 48 EI Mx Comp. Stress X = Sxt = 5.15 My Total Deflection = d1+d2+(Greater of d3 and d4) Comp. Stress Y = Sy = 17.07 Deflection = 0.0181+0+0.2188 = 0.2369 in
17.07 Cy Cx 515 . 1 103 . 1 ERR f b 0.6 Fy 216 . 216 . 12/25/07
Deflection 0.2369 < L/600 (0.4)
Above calculation is for S12x31.8 Beam
OK
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