Crane 3hf4x
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3b7i
Overview 3e4r5l
& View Crane as PDF for free.
More details w3441
- Words: 1,501
- Pages: 13
Haider Crane Co.
Crane Design Basics A brief introduction to Crane Design is presented with the intent that it will be useful to the s 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 P/4 delivered to the ing 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
Crane Design Basics A brief introduction to Crane Design is presented with the intent that it will be useful to the s 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 P/4 delivered to the ing 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
Related Documents 3m3m1z
Crane 3hf4x
April 2020 41
Crane 3hf4x
April 2020 30
Sop Crane 5l5z1k
May 2020 70
Crane Simulator x292d
November 2021 0
Makalah Crane 3m523
November 2021 0