Haider Crane Co.

112/25/07

Hoist Crane Single Girder

Haider Crane Co.

212/25/07

Hoist Crane Single Girder

Haider Crane Co.

312/25/07

Hoist Crane Single Girder

Haider Crane Co.

412/25/07

Hoist Crane Double Girder

Haider Crane Co.

512/25/07

Ho

ist

Cra

ne

Do

ub

le G

ird

er

Haider Crane Co.

612/25/07

Hoist Crane Single Girder Gantry crane

Haider Crane Co.

712/25/07

Ho

ist

Cra

ne

Sin

gle

Gir

de

r G

antr

y cr

ane

Haider Crane Co.

812/25/07

Hoist Crane Gantry crane

Haider Crane Co.

912/25/07

Ho

ist

Cra

ne

Gan

try

cran

e

Haider Crane Co.

1012/25/07

Suspension Crane

Haider Crane Co.

1112/25/07

Haider Crane Co.

1212/25/07

Suspension Crane

Haider Crane Co.

1312/25/07

Jib Crane

Haider Crane Co.

1412/25/07

Car

go L

ift

Haider Crane Co.

1512/25/07

Setting Crane Girder

Haider Crane Co.

1612/25/07

Crane Design Basics

Haider Crane Co.

1712/25/07

Span

PDead Load Bending

Max Moment =

l Pl2

8 4

P = Center drive + Controls

w = Footwalk + Beam + Lineshaft

Dynamic Mx = Max Moment x factor

Dynamic My = Max Moment x factor

Span

P1 P2

x y

Live Load BendingL = Span

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.

Compute Moments

Haider Crane Co.

1812/25/07

Compute Stress

a

b

DL – Dead load, LL = Live Load, Fy = Yield normally @ 36 ksi for A36 material, Fb = Allowable Stress

Dynamic Mx = LL Mx + DL Mx

My = LL My + DL My

fM

SFb

x

bott

yxT

0 6.

fM

S

fM

S

f

F

f

F

b

x

Top

b

y

Top

b

b

b

y

xC

yC

xC

x

yC

0 6

10.

.

Girder Beam

Fabricated/Box Beam

Available to 60 ft max. length

L/h should not exceed 25

L/b should not exceed 65

h

b

Haider Crane Co.

1912/25/07

Compute Deflection

5

384

48

4

3

wl

EI

Pl

EI

For Uniform Load

For Conc. Load

For Trolley:

PLL TW

Pa

EIl a

2

243 42 2

Haider Crane Co.

2012/25/07

12000

Ld

Af

0 6. y

Allowable Compressive Stress Fb per CMAA 74

1/600

1/888

Use when the flanges

are not welded on the

top and bottom

Haider Crane Co.

2112/25/07

Allowable Compressive Stress

yb

f

b

t

b

y

t

y

b

ytt

F

CMAAper

A

dL

F

r

LFOtherwise

Fr

LF

FFr

Land

yFr

L

6.0

)(12000

170000,

15300003

2,

510000102000

1

3 2

2

2

Select Allowable Stress which is the Greatest of all.

Then check for the following:

16.0

6.0

y

ycomp

b

xcomp

ytensile

f

Haider Crane Co.

2212/25/07

Lower Flange loading per CMAA 74

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.

This is a empirical formula

Haider Crane Co.

2312/25/07

Lower Flange loading per CMAA 74

Haider Crane Co.

2412/25/07

Lower Flange loading per CMAA 74

Haider Crane Co.

2512/25/07

Lower Flange loading - Alternate procedure I

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 :

f

P

A

P

t

P

tt

w w

2 2 35 7.

30 deg

P/2

3.5k

tfk

Bottom Flange

e

ePoint of Load

tw

ek1

P/4

tf

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.

Tegangan lentur sebesar ;

fM

S

Pe

bd

Pe

et

P

tb

f f

4

6

4

6

2

0 752 2 2

.

Refer to Engineering Journal, 4th quarter, 1982, Tips for avoiding Crane Runway Problems by David T. Ricker

Haider Crane Co.

2612/25/07

Lower Flange loading - Alternate procedure II

Now, the angle is changed from 30 degree to 45 degrees.tw

ek1

P/4

tf

45 deg

b=2e

Load

Capacity = 6000 lb (daya angkut crane)

Hoist, Wt = 1000 lb (berat crane)

Load = 6000 +1000 = 7000 lb

Wheel load = 7000/4 = 1750 lb

With 15% impact = 1750(1.15) = 2013 lb

Lebar flens, b =11.5 inc, sehingga e = b/2 = 5.75 inc

Tebal flens, tf = 0.875 inc

Momen pada flens, M = 2013 (5.75) = 11574.75 lb-inc

Teg arah-y, σy = M/Sy = 11574.75. (6)/(11.5)(0.875)2 = 7890.15 lb/in2

Moment = 7000(1.15)(30)(12)/4 + 110 (12x30)2/(8 x12) = 873.000 lb-in

30 ft crane

Span

110 lb/ftTeg arah-x, σx = M/Sx = 873.000/280 = 3117.8 lb/in2

15,78908,31172

15,78902

8,311722

yxyx

= 9.827 << 0.6 σy = 21.600 lb/in2 OK

Haider Crane Co.

2712/25/07

EXAMPLE – Simple Approach

Capacity: 2 Ton (=4000 lb), Span: 20 Ft (=240 in)

Hoist Wt: 200 lb, Hoist W.B: 12 in

Vertical Impact factor = 15%, Hor. Impact = 10%

Solution:

inL

in

EE

EI

aLPa

EI

wL

inlbS

M

inlbS

M

inlbMx

aL

L

PwLMx

y

y

x

x

xx

4.0600

237.0

21824

)11442403(11415.12100

21812384

240)8.31(5

24

)43(

384

5

/5.303915.127.9

1.01.11.294571

/6.80924.36

1.294571

1.2945712

12240

2402

15.12100

8

240

12

8.31

228

224

224

2

2

22

22

240

P1 P2

114 12P=2100lb

OK

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

P=2100 lb, w = (31.8/12) lb/in

.

2

21

2

2

2

1

2

/8.337.11,min_

/8.337.1141.4240

1200012000

/000.216366.06.0

/1.132.115.30396.8092

allcomb

allallall

f

all

yall

comb

inlbof

inlb

AdL

inlb

inlb

OK

Beam must be checked for

Lower flange load, if the

trolley is under running

Haider Crane Co.

2812/25/07

EXAMPLE – Conservative Approach

11 1052000

12. ..

. Br speed

015 0005 05. . ( ) . HoistSpeed

0078 0025. ( _ .) .Bridge acc

C T WB X X

T WBHLF

T WtDLF

2 2

1 2_

_

_

P P MP P

LX X

L WB P

P POR M

PL1 2

1 2 2 2

1 2

054

,

( ), . , ,

wL PL2

8 4

MxSxb

DLF =

HLF =

IFD =

Wheel Ld (P1/P2) =

= 1.10

= 1.15

= 0.39

= 2410 240

P1 P2

114 12 P=2100lb

Moment A = HLF x M (whichever is greater) = = 166290

Moment B = IFD x M (whichever is greater) = 56394

Static Moment = = 19080

Moment C = DLF x Static Moment

Moment D = IFD x Static Moment

= 20988

= 7441.2

Moment Mx = A+C = 187278

Moment My = B+D = 63835.2

Tensile Stress = = 5.15 < 0.6(36) OK

Comp. Stress X = MxSxt

= 5.15

= 17.07Comp. Stress Y = My

Sy

C

f

C

F

x

b

y

y

0 6

1.

515

216

17 07

216103 1

.

.

.

.. ERR

dwL

EI1

5

3840 0181

4

.

dPL

EI2

30

3

d

P Pa

L a

EI3

2

3 4

240 2188

1 22 2

.

dP L

EI4

4801098

1

3

.

Total Deflection = d1+d2+(Greater of d3 and d4)

Deflection = 0.0181+0+0.2188 = 0.2369 in

Deflection 0.2369 < L/600 (0.4) OK

Above calculation is for S12x31.8 Beam

Beam must be

checked for

Lower flange

load, if the trolley

is under running