Crane Design and Calculation

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

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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.

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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

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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