Haider Crane Co.
Crane Design Basics
A brief introduction to Crane Design is presented with the
intent that it will be useful to the users specially those who
are new in this field and have no knowledge of how a crane
is built.
This work is motivated due to the lack of presence of
literature for crane design on the web. The author was
compelled to put up some effort to have at least some stuff
on the web which can at least give a introductory level
information to engineers and students.
Feel free to send me a mail with questions or comments
Beam Calculatorcrane_calc.xls
12/25/07
1
Compute Moments
Dead Load Bending
Haider Crane Co.
P
L = Span
Span
Live Load Bending
x
P1
P2
y
Span
ωl 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
If P1=P2, use the Case 41, otherwise use Case
42. Also calculate the Max moment using the
formula PL/4. In other words, to be
conservative, use the largest value obtained.
12/25/07
2
Compute Stress
Haider Crane Co.
Dynamic Mx = LL Mx + DL Mx
My = LL My + DL My
f bx
Mx
=
≤ 0.6 Fy
Sbott
f bxC
Mx
=
STop
T
f byC =
f bxC
Fbx
+
a
b
Girder Beam
Available to 60 ft max. length
My
STop
f byC
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
3
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
4
Allowable Compressive Stress Fb per CMAA 74
Haider Crane Co.
1/600
1/888
= 0.6σ y
12000
=
d
L
Af
12/25/07
Use when the flanges
are not welded on the
top and bottom
5
Allowable Compressive Stress
L
rt ≥
102000
L
Fy And
rt ≤
Otherwise, Fb3 =
Fb1 =
510000
Fy , Fb 2
Haider Crane Co.
2




L


F
2
y
rt  
 Fy
=  −
3
1530000






170000

L 

 rt 
2
12000
( perCMAA)
 d

L
Af 


Fb = 0.6σ y
Select Allowable Stress which is the Greatest of all. Then check for the following:
σ Tensile < 0.6σ y
σ comp x
fb
12/25/07
+
σ comp y
0.6σ y
<1
6
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
7
Lower Flange loading per CMAA 74
12/25/07
Haider Crane Co.
8
Lower Flange loading per CMAA 74
12/25/07
Haider Crane Co.
9
Lower Flange loading - Alternate procedure I
Haider Crane Co.
The lower flange of the crane beam must be checked for:
1) Tension in the web. 2) Bending of the bottom flange.
Refer to the figure, the length of resistance is seen to be 3.5k. The 30 degree angle is a
consensus figure used for many years. Assuming 4 wheels (2 pair) at each end of the
crane, each wheel will support P/4 delivered to the supporting crane beam. Two wheels
cause the web tension, so the load is P/2. Tensile stress in the web is:
3.5k
P
P
P
=
=
ft =
2 A 2t w ( 35
. ) 7t w
k
tf
30 deg
Flange bending depends upon the location of
the wheels with respect to the beam web. This
dimension is ‘e’ as shown in the figure. The
wheel load is P/4. Longitudinal length of the
flange participating in the bending resistance
is 2e per yield line analysis. Bending stress is:
P/2
tw
P/4
Bottom Flange
tf
M Pe 6
Pe 6
0.75 P
fb =
=
=
=
4 bd 2
S
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
10
Lower Flange loading - Alternate procedure II
Haider Crane Co.
Now, the angle is changed from 30 degree to 45 degrees.
tw
Capacity = 6000 lb
P/4
Hoist wt = 1000 lb
Load = 6000 +1000 = 7000
tf
k1
Wheel load = 7000/4 = 1750
e
With 15% impact = 1750(1.15) = 2013 lb
b =11.5, e = b/2 = 5.75
Tf = 0.875
45 deg
M = 2013(5.75) = 11574.75
Stress = M/S = 11574.75
.
b=2e
(6)/(11.5)(0.875)^2 = 7890.15
Load
Moment = 7000(1.15)(30)(12)/4 + 110((12)(30))^2/(8(12)) = 873000 lb-in
Stress= 873000/280 = 3117.8
. 2 + 3117.8( 789015
. )
σ = σ x2 + σ y2 + σ xσ y = 3117.82 + 789015
Stress = 9827 << 0.6 Sigma y (21600)
12/25/07
OK
110 lb/ft
30 ft Bridge
Span
11
EXAMPLE – Simple Approach
Haider Crane Co.
P1
Capacity: 2 Ton (4000 Lb), Span: 20 Ft (480 in)
114
Hoist Wt: 200 Lb, Hoist W.B: 12 in
P2
12
P=2100lb
Vertical Impact factor = 15%, Hor. Impact = 10%
Solution:
a
wL
P 
+
L− 
8
2L 
2
2
Mx =
2
Beam must be checked for
Lower flange load, if the
trolley is under running
. 2402 2100 × 115
. 
318
12 
Mx =
×
+
.
 240 −  = 2945711
12
8
2 × 240 
2
M
2945711
.
σx = x =
= 8092.6
Sx
36.4
M y 2945711
. × 11
. × 01
.
σx =
=
= 3039.5
Sy
9.27 × 115
.
2
240
Say, for example, we select a A36, S beam
S12x31.8#, Ix=218, Iy=37.1, Sx=36.4,
Sy=9.27, d/Af=4.41
P=2100 lb, w = 31.8/12 lb/in
σ comb = 8092.6 + 3039.5 = 11132.1
σ all1 = 0.6σ y = 0.6 × 36 = 216000
(3 L2 − 4a 2 )
5wL4
12000
12000
δ=
+ Pa
σ
=
=
= 11337.8
all 2
384 EI
24 EI
d
240 × 4.41
L× A
2
2
4
f
5(318
. )240
(3 × 240 − 4 × 114 )
δ=
+ 2100 × 115
. × 114
σ all = Min _ of − σ all1 , σ all 2 = 11337.8
24 × 218 E
384 × 12 × 218 E
L
σ comb < σ all .
= 0.4
δ = 0.237 <
OK
OK
600
12/25/07
12
EXAMPLE – Conservative Approach
DLF =
. ≤ 1.05 +
11
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 
PL
( P1 + P2 ) 2
X , X = 0.5
,
OR
,
M
=
= 166290

P
P
L
+
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
3 EI
Moment D = IFD x Static Moment = 7441.2
P1 + P2 ) [ 3 L2 − 4a 2 ]
(
Moment Mx = A+C = 187278
= 0.2188
d3 =
a
Moment My = B+D = 63835.2
2
24 EI
P1 L3
Tensile Stress = σ = Mx Sxb = 5.15 < 0.6(36) OK
d4 =
= 0.1098
48
EI
Mx
Comp. Stress X = σ =
Sxt = 5.15
My
Total Deflection = d1+d2+(Greater of d3 and d4)
Comp. Stress Y = σ =
Sy = 17.07
Deflection = 0.0181+0+0.2188 = 0.2369 in
17.07
Cy
515
.
Cx
+
<1 =
+
= 103
. > 1 ERR
216
.
216
.
f b 0.6 Fy
12/25/07
Deflection 0.2369 < L/600 (0.4)
Above calculation is for S12x31.8 Beam
OK
13