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