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 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 co mments Beam Calculatorcrane_calc.xls
Compute Moments
Haider Crane Co.
P
Dead Load Bending
L = Span
Span
ω l Max Moment =
8
2
+
Live Load Bending
x
P1
P2 y
Span
Pl
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.
Compute Stress
Haider Crane Co.
Dynamic Mx = LL Mx + DL Mx My = LL My + DL My
f b x = T
M x
C
f b y = C
f b x
C
F b x
≤ 0.6 F y
S bott
f b x =
+
a
M x
b
Girder Beam Available to 60 ft max. length
S Top M y S Top f b y
C
0.6 F y
h
≤ 1.0 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
Compute Deflection
∆= ∆=
5wl
4
For Uniform Load
384 EI
Pl
Haider Crane Co.
For Trolley:
P=
3
48 EI
For Conc. Load
∆=
LL + TW
2 Pa 24 EI
2 2 l a − 3 4 [ ]
Allowable Compressive Stress Fb per CMAA 74
Haider Crane Co.
1/600 1/888
= 0.6σ y
=
12000
L
d A f
Use when the flanges are not welded on the top and bottom
Allowable Compressive Stress
L
rt ≥
102000
L F y And rt ≤
Otherwise , F b3 =
F b1 =
12000
L d A f
Haider Crane Co.
510000
F y , F b 2
2 L F 2 y r t F y = − 3 1530000
170000
L r t
2
( perCMAA)
F b = 0.6σ y
Select Allowable Stress which is the Greatest of all. Then check for the following:
σ Tensile σ comp
x
f b
< 0.6σ y +
σ comp
y
0.6σ y
<1
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.
Lower Flange loading per CMAA 74
Haider Crane Co.
Lower Flange loading per CMAA 74
Haider Crane Co.
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
f t =
2 A
=
P
2t w ( 35 . )
=
P
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
f b =
M S
=
Pe 6 2
4 bd
=
Pe
6 2 f
4 2et
=
0.75 P 2 f
t
e
k1
e Point of Load
e
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 45 deg
T f = 0.875
b=2e
M = 2013(5.75) = 11574.75 Stress = M/S = 11574.75 . (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 σ
=
2 x
σ
+ σ y2 + σ xσ y =
3117.82 + 789015 . 2 + 3117.8(789015 . )
Stress = 9827 << 0.6 Sigma y (21600)
OK
110 lb/ft
30 ft Bridge Span
EXAMPLE – Simple Approach
Haider Crane Co.
Capacity: 2 Ton (4000 Lb), Span: 20 Ft (480 in)
P1 114
Hoist Wt: 200 Lb, Hoist W.B: 12 in
P2 12
P=2100lb
Vertical Impact factor = 15%, Hor. Impact = 10% Solution: 2
Mx = Mx =
wL
+
8 318 . 12
M x
×
2402 8
2
2100 × 115 . 12 + . 240 − = 2945711 2 × 240 2 2
=
σ x
2945711 . × 11 . × 01 . = = = 30395 . S y 9.27 × 115 .
δ
δ
= =
384 EI
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
= 8092.6 + 3039.5 = 111321 . σ all1 = 0.6σ y = 0.6 × 36 = 216000 σ comb
M y
5wL4
240
2945711 . = 8092.6 36.4
σ x
S x
=
P a L − 2 L 2
Beam must be checked for Lower flange load, if the trolley is under running
+ Pa
(3 L2 − 4a 2 )
σ all 2
24 EI
5(318 . )2404
384 × 12 × 218 E L δ = 0237 . < = 0.4 600
+ 2100 × 115 . × 114
OK
(3 × 2402 − 4 × 1142 ) 24 × 218 E
=
12000
L × d A f
=
12000 240 × 4.41
= 11337.8
= Min _ of − σ all1 , σ all 2 = 11337.8 σ comb < σ all . σ all
OK
EXAMPLE – Conservative Approach
DLF =
Br . speed
11 . ≤ 1.05 +
2000
≤ 1.2
HLF = 0.15 ≤ 0.005( HoistSpeed ) ≤ 0.5
Haider Crane Co.
P1
= 1.10 114
= 1.15
12
= 0.39 IFD = 0.078( Bridge_ acc. ) ≤ 0.025 C T _ WB − X 1 − X 2 T _ Wt ( HLF ) + ( DLF ) = 2410 Wheel Ld (P1/P2) = 2 2 T _ WB Moment A = HLF x M (whichever is greater) =P1 > P2 , M =
( P1 + P2 ) L
Moment B = IFD x M (whichever is greater) = 56394 2
Static Moment =
wL
PL
+
8
4
= 19080
Moment C = DLF x Static Moment = 20988
Beam must be checked for Lower flange load, if the trolley is under running
Moment D = IFD x Static Moment = 7441.2 Moment Mx = A+C = 187278 Moment My = B+D = 63835.2 Tensile Stress =
σ =
Mx
= 5.15 < 0.6(36) OK = 5.15
Sxb Mx Comp. Stress X = σ = Sxt My Comp. Stress Y = σ = Sy = 17.07
C x f b
+
C y
0.6 F y
<1 =
515 . 21.6
+
17.07 21.6
= 1.03 > 1 ERR
P2 P=2100lb
240
L − WB × P2 PL 2 . , , = X , X = 05 OR M = 166290 4 P1 + P2 4 5wL
d 1 =
384 EI
= 0.0181
3
PL d 2
d 3 =
=
(P
d 4 =
1
3 EI +
2
P2
)a[
=
2
3 L
0 −
4a
2
]
24 EI
=
0.2188
P1 L 3
= 0.1098 48 EI
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
