Transcript
Project: 35 Ton hydraulic mooring winch
1 Duty pull F= 35 Ton @ 1st Layer
Drum Pipe D= 0.762 m Ratio: D/d= 19.1
Wire Length L= 500 m Drum Length Ld= 1229.6
wire dia. d= 0.040 m
Colum X= 29
Layer 0.802 L/LAYER L/TOTAL FORCE SPEED
1 0.802 73.067 73.1 75 5 m/min
2 0.871 76.642 149.7 69.0 5.4 m/min
3 0.941 85.691 235.4 64.0 5.9 m/min
4 1.010 88.830 324.2 59.6 6.3 m/min
5 1.079 98.315 422.5 55.7 6.7 m/min
6 1.148 101.019 523.6 52.4 7.2 m/min
7 1.218 110.938 634.5 49.4 7.6 m/min
8 1.287 113.207 747.7 46.7 8.0 m/min
9 1.356 123.562 871.3 44.4 8.5 m/min
10 1.426 125.396 996.7 42.2 8.9 m/min
11 1.495 136.186 1132.9 40.2 9.3 m/min
12 1.564 137.584 1270.4 38.5 9.8 m/min
13 1.633 148.810 1419.2 36.8 10.2 m/min
14 1.703 149.772 1569.0 35.3 10.6 m/min
15 1.772 161.433 1730.5 33.9 11.0 m/min
16 1.841 161.961 1892.4 32.7 11.5 m/min
17 1.910 174.057 2066.5 31.5 11.9 m/min
DYNAMICS CALCULATIONS Title: HYDRAULIC AHTW C/W HYD. POWER PACK
Model No. ME75THAHTW Dwg. Number 75THAHTW-100
SHEET: 1/2 Proj. No. J070029E Rev. No. 0 Date: 20/04/07
a. Winch
Drum pull Fd = 75 Ton
Wire rope speed V = 5.0 m/min (1st Layer, LOW SPEED)
Wire rope dim. d = 0.052 m
Drum Dia. D = 0.762 m
PCD at 1st Layer 0.814 m
Gear ratio i = 29.05 (113/26*127/19)
Shaft revolution speed n = 1.96 rev/min
1. Hyd. motor output torque required
Td= Fd x 9810 x (D+d) / (2i)*(1+η)
= 13400 Nm
Mechanical loses: η= 30%
2. Hyd. motor output speed required
n xi
= 56.80 rev/min
3. Choose hyd. Motor
Type: HMC270 280
Displacement: Vm Vm= 4588 cc/rev
p= 69.4 Nm/bar
Mechanical loses: 92%
Volumetric efficiency: 97%
4. Working flow
Qm= Vm x nm/ηv
= 268.7 l/min
5. Max. load pressure (Winch pull)
209.9 bar
6. Hyd. Motor output power
Nm= 94.0 KW
D1=
nm=
ηm=
ηv=
PLt= Td/p/ηm
DYNAMICS CALCULATIONS Title: HYD. TOWING WINCH C/W HYD. POWER PACK
Model No. ME75THAHTW Dwg. Number 75THAHTW-100
SHEET: 2/2 Proj. No. J070029E Rev. No. 0 Date: 20/04/07
b. Power pack
1. Hyd. Pump flow required Qm= 268.7 l/min
Displacement:
= 186.49 cc/rev
Hyd. System volum efficency 0.98
2. Choose hyd. Pump
Pump Type:
Qty. of pump n= 2
Displacement: 93.2 cc/rev
Duty output: P = 219.9 bar
Q =
= 137.1 lpm
Input power: N =
= 58.21 kw
Elec. Motor safty factor: 0.94
Hyd. pump overall efficency: 0.9
3. Choose elec. Motor: n= 2
415 v 1470 rpm
3 PH
50 Hz W= 90 kw
qp = Qm/ηv/ne
ηv =
qp =
neqpne
PQ/ηp/ηe/612
ηe =
ηp =
ne=
DYNAMICS CALCULATIONS Title: HYDRAULIC AHTW C/W HYD. POWER PACK
Model No. ME75THAHTW Dwg. Number 75THAHTW-100
SHEET: 1/2 Proj. No. J070029E Rev. No. 0 Date: 20/04/07
a. Winch
Drum pull Fd = 12 Ton
Wire rope speed V = 20.0 m/min (1st Layer, HIGH SPEED)
Wire rope dim. d = 0.052 m
Drum Dia. D = 0.762 m
PCD at 1st Layer 0.814 m
Gear ratio i = 29.05 (113/26*127/19)
Shaft revolution speed n = 7.82 rev/min
1. Hyd. motor output torque required
Td= Fd x 9810 x (D+d) / (2i)*(1+η)
= 2144 Nm
Mechanical loses: η= 30%
2. Hyd. motor output speed required
n xi
= 227.20 rev/min
3. Choose hyd. Motor
Type: HMC270 60
Displacement: Vm Vm= 980 cc/rev
p= 12.2 Nm/bar
Mechanical loses: 92%
Volumetric efficiency: 97%
4. Working flow
Qm= Vm x nm/ηv
= 229.5 l/min
5. Max. load pressure (Winch pull)
191.0 bar
6. Hyd. Motor output power
Nm= 73.1 KW
D1=
nm=
ηm=
ηv=
PLt= Td/p/ηm
DYNAMICS CALCULATIONS Title: HYD. TOWING WINCH C/W HYD. POWER PACK
Model No. ME75THAHTW Dwg. Number 75THAHTW-100
SHEET: 2/2 Proj. No. J070029E Rev. No. 0 Date: 20/04/07
b. Power pack
1. Hyd. Pump flow required Qm= 229.5 l/min
Displacement:
= 159.34 cc/rev
Hyd. System volum efficency 0.98
2. Choose hyd. Pump
Pump Type:
Qty. of pump n= 2
Displacement: 79.7 cc/rev
Duty output: P = 201.0 bar
Q =
= 117.1 lpm
Input power: N =
= 45.47 kw
Elec. Motor safty factor: 0.94
Hyd. pump overall efficency: 0.9
3. Choose elec. Motor: n= 2
415 v 1470 rpm
3 PH
50 Hz W= 90 kw
qp = Qm/ηv/ne
ηv =
qp =
neqpne
PQ/ηp/ηe/612
ηe =
ηp =
ne=
MARINE EQUIPMENTS PTE LTD 9/21 WINCH STRESS
CALCULATION (For BV Approval)
75TON ANCHOR HANDLING/TOWING WINCH
The winch duty pull load is: Fd= 75000 kg x 5 m/min
Max. braking force is: 150000 kg
Wire PCD @ 1st Layer: 814.0 mm
1. SHAFT STRESS CALCULATION (neglet wire angle)
Duty pull capacity: 30525.0 kgm
The Max. bending torque applied on the shaft:
46875.0 kgm
Where a= 1250.0 mm
b= 1250.0 mm
Based on the ASME code Max. shear equation:
T=sqrt[(Kb*Mb)^2+(Kt*Mt)^2]= 76652.6 kgm
Where 1.5
1.0
Ss=16T/(πd^3)= 13.10 d= 310 mm
Shaft material ST52.3, Yielding-point> 355 N/mm2
Safty-factor= 2.71 >1.4
Winch brake holding load 150000 kg
The Max. bending torque applied on the shaft:
94500.0 kgm
Where a= 1260.0 mm
b= 1260.0 mm
Max. tensile stress on shaft=32Mb/(πd^3) 323.3 d= 310 mm
Safty-factor over yield strength= 1.10 >1.4
The shaft mainly under shear force, The min. shaft diameter is: 260.0 mm
27.7
Fb=
D1=
TD=FdxD1/2=
Mb=Fd*ab/l=
Kb=
Kt=
kg/mm2
Fb=
Mb=Fd*ab/l=
N/mm2
Sshear=Fb/A=4Fb/(3.14d2)= N/mm2
MARINE EQUIPMENTS PTE LTD 10/21 WINCH STRESS
Safty-factor over yield strength= 12.80 >2
2. PIN STRESS CALCULATION
There are 12 pin in the gear: dia= 30 mm
PCD= 1730 MM
Force on pin Fp= 346185260 N
The shear stress acting on the pin:
S=Fp/n/A= 61250.05 N/mm^2
Safty-factor over yield strength> 4.0
3. KEY STRESS CALCULATION (2nd STAGE)
There are two keyway in the gear: n= 2
The key section dimension: bxH 45 (mm) x 25 (mm)
The max torque is caused by winch:Td= 7627.5 kgm
The squeeze stress acting on the key:
24.91
Where h= 12.5 mm
L= 140 mm
d= 175 mm
The shear stress acting on the key:
6.92
The Gear material is ASTM A148 grade80/40,
Safty-factor over yield strength= 1.35 >1.3
4. BEARING & BEARING HOUSING
1). BEARING CAPACITY: spherical roller bearing
The bearing selected is: 23052MB
Basic dynamic load rating: 1500 KN
Pin material is AISI1045, Tensile stress>600 Mpa, yielding-point >350 Mpa,
S1=2TD/hLdn= kg/mm2
S2=2TD/bLdn= kg/mm2
Key & shaft material is AISI1045, Yielding-point >355 Mpa,
Tensile stress >620 Mpa, Yielding-point>335 Mpa
MARINE EQUIPMENTS PTE LTD 11/21 WINCH STRESS
Basic static load rating: 2800 KN
See attached FAG bearing catalogue.
2). Bearinghousing material for winch is mild steel: ASTM-A36
5. FOUNTATION BOLTS
1). REVERSAL MOMENT: 238500000 kg.mm
Where: 150000 kg (Winch brake holding)
distance between wire and foundation: H= 1590 mm
2). MAXIMUM TENSION FORCE OF BOLT:
13140.5 kg
MAXIMUN SHEAR FORCE OF BOLT:
2542.4 kg
Last row number of bolts: 15
Totally Number of bolts: 59
1210 mm
3). MAXIMUM TENSION & SHEARING STRESS OF BOLT:
We use M30 ISO G 8.8 bolts, Tensile stress= 80.0
Yielding point = 64.0
The bolt stress area is: A= 561
23.4
4.5
Safty-factor over yield strength= 2.73 >4
6. DRUM BRAKE SYSTEM CALCULATION (TWO STAGE)
Max. static brake holding: 150000 kg
Brake Race Dia.: D= 1834 mm
wire P.C.D. 814.0 mm
Brake Lap: α= 275 deg.
Coefficient of friction: μ= 0.35
Dictance: A= 782 mm
B= 450 mm 600 mm
M=Fb*H=
proof load: Fb=
Fr=M/(R2*z1)=
Fs=Fb/z2=
z1=
z2=
R2=
kg/mm2
kg/mm2
mm2
Stension=Fr/A= kg/mm2
Sshear=Fs/A= kg/mm2
Fb=
D1=
L1=
MARINE EQUIPMENTS PTE LTD 12/21 WINCH STRESS
C= 782 mm 150 mm
b= 65 mm
s= 8 mm
2). Force analysis & Calculation
Max. torque @ brake holding: M= 598900500 N/mm
Brake force on brake drum: P= 653109 N
Moving pin load(loose side):
149589 N
Tau side force: 802697 N
Spindle load: 37397 N
Anchor pin load: 802697 N
3). Brake band:
Material is ASTM-A36
250.0
Permissible design stress(Tensile) = 162.5
Brake band size: w= 300 mm t= 25.0 mm
Max. tensile stress: 107
4). Pin selection:
Material is AISI1045, 110.0 mm
350
Permissible design stress(Shear) = 122.5
a.Tau side pin:
84.5
50.0
Shear stress: 76.2
5). Anchor plate, bar & brake lug:
L2=
T2=P/(e^(μα)-1)=
T1=T2*e^(μα)=
T=T2/L2*L1=
R=T1=
Yield stress = N/mm2,
N/mm2,
σ=T1/(wt)= (N/mm2) < 162.5 N/mm2,
d1=
Yield stress = N/mm2,
N/mm2,
Max. Shear stress: t=R/(1/4*3.14*d12)= N/mm2, < 122.5 N/mm2
b. Moving pin (loose side)@T2
d2=
t=T2/(1/4*3.14*d22)= N/mm2, < 122.5 N/mm2
MARINE EQUIPMENTS PTE LTD 13/21 WINCH STRESS
Material is ASTM-A36
250.0
Permissible design stress(Tensile) = 162.5
Anchor plate thickness: 38 mm
Number of anchor plate lug: 2
Outer radius: 110 mm
Max. tension stress=T1/[nxtx(2r-d1)]= 96
Link plate @ R
Bar thickness: 38 mm
Number of anchor bar: 2
Outer radius: 110 mm
96
Link plate lug @ T2
Moving pin (loose side)@T2 30 mm
Number of anchor bar: 2
Outer radius: 50.0 mm
50
6). Brake Cylinder & hand wheel:
Required Brake force: T= 37397 N = 3808 KG
Disc spring: AM2008210 ID82XOD200X9.5THKX15.5
Stroke: 180 mm
PRE-PRESS: 1.38mm FORCE= 5304 KG > 3808 KG
Number of spring: 70
Yield stress = N/mm2,
N/mm2,
Anchor plate @ T1
N/mm2, <162.5 N/mm2
Max. tension stress=R/[nxtx(2r1-d1)]= N/mm2, <162.5 N/mm2
Max. tension stress=T2/[nxtx(2r2-d2)]= N/mm2, <162.5 N/mm2
MARINE EQUIPMENTS PTE LTD 14/21 1st STAGE GEAR
SPUR GEAR CONTACT STRESS CALCULATION (BS436)
FOR 75 TON ANCHOR HANDLING/TOWING WINCH (WIRE Φ52mm )
(first stage)
PINION SPUR GEAR (UNIT)
center distance: a= 1251 1251 mm
mormal module: 18 18 mm
pressure angle 20 20 deg
tooth number: 26 113
face width: b= 270 250 mm
reference diameter: 468 2034 mm
tip diameter: 504 2070 mm
tooth depth: 40.5 40.5 mm
number of revolution 8.50 1.96 rev/min
construction: solid solid
mean roughness: 3.2 3.2 μm
meterial type: AISI 4140 ASTM A148 grade80/40
tensile stress: 900 620
yielding point: 700 335
300 Cst
application factor: 1.0
required life time: 10000 hours
1.0
The winch overload : F= 75000 kg (BS MA 35 : 1975)
Norminal tangential force for the contact stress:
Wire PCD. 814 mm
30015 kg
mn=
αn=
z1 & z2
d1, d2
da1, da2
Rα(root)
σb=
σs=
lubricant viscosity at 40˚C:
minimun demanded safety factor for contact stress: SHMin=
Dwpcd=
FHT=F*Dwpcd/d2=
MARINE EQUIPMENTS PTE LTD 15/21 1st STAGE GEAR
Calculation for the contact stress:
(equation 2)
4.35
439.8 mm
1911.3 mm
65 (B.11)
2.49 (Equation 10)
53.1 mm (B.3)
92.6 (B.2)
1.74 (B.8)
0.867 (Equation 11)
180.1 (Equation 14)
Poisson's ratio 0.3
Steel: 21000 kg/mm^2
Cast steel: 20600 kg/mm^2
1.0
1.1 (from 16.3)
v= 12.49 m/min= 0.21 m/sec
5.96
1.0 (spur gear from fig. 11)
X= 0.8687 (grade 9 see table 7)
0.35 (Equation 22)
1.0 (Equation 20)
250 (Equation 35)
7164.4 kg (Equation 37)
σH=ZH*ZE*Zε*SQRT(FHT*(u+1)*KA*KV*KHα*KHβ/u/b/d1)
u=Z2/Z1=
db1=d1*cosαn=
db2=d2*cosαn=
Prel=(d1/2)*(u/(u+1))*(cosαt*tanαtw/cosβt)=
ZH=2*SQRT(cosβt/sin(2*αt))=
Pbt=mn*π*cosαt/cosβ=
ga=0.5*(SQRT(da1^2-db1^2)+SQRT(da2^2-db2^2))-a*sinαtw=
εa=ga/Pbt=
Zε=SQRT((4-εa)/3)=
ZE=SQRT{1/[π*((1-v1^2)/E1+(1-v2^2)/E2)]}=
v1=v2=
E1=
E2=
KA=
QV=
Qv v z1=
Kv350=
B=(350/(FtKA/b))^X=
Kv=1+(Kv350-1)/B=
beff=b-lc/2= mm lc=0
Fmest=bd1u/(u+1)*[σHP/(ZH*ZE*Zε)]^2=
MARINE EQUIPMENTS PTE LTD 16/21 1st STAGE GEAR
281.1 N/mm (Equation 38)
K= 0.48 (Fig. 12)
A= 0.023 μm (Table 8)
120.06 kg/mm= 1178 N/mm
210 mm
l= 770 mm
231 mm
9.13 μm (Equation 40)
12.9 μm (Equation 42)
25.04 μm (Equation 45)
0.45 (from fig 13. 700MPa)
11.27 μm (Equation 47)
20 N /mm.μm (from 17.2.5 steel to steel)
1.1 (Equation 49)
1.74
12.2 μm (BS436 part 2. Table 3)
0.915 μm (Equation 54)
0.85 (Equation 50)
1.0
The contact stress:
718 MPa
900 MPa (Fig. 2)
750 MPa (Fig. 2)
Zc= 1.0
0.85 (Equation 16)
0.94
Wmest=Fmest/beff=
Wm=FtKAKv/beff=
dsh=
ls=0.3l=
fsh=WmA[(1+K*(ls/d1^2)*(d1/dsh)^4-0.3)+0.3]*(b/d1)^2=
fma=fHβ=
Fβx=/1.33fsh+fma/=
qy=
Fβy=qyFβx=
cγ=
KHβ=1+cγ*Fβy/(2Wm)=
εγ=εβ+εα=εα=
fpe=
γα=0.075fpe=
KHα=εγ*[0.9+0.4*cγ*(fpe-γα)/(Wm*KHβ)]/2=
take KHα=
σH=ZH*ZE*Zε*SQRT(FHT*(u+1)*KA*KV*KHα*KHβ/u/b/d1)=
σHD1=
σHD2=
ZG1=0.9*ZG2=
ZG2= (Fig 3, Prel/mn=3.4 σB=800MPa)
MARINE EQUIPMENTS PTE LTD 17/21 1st STAGE GEAR
761 MPa (Equation 15)
705 MPa (Equation 15)
1.02 (Fig. 5)
0.9 (Fig. 6)
1.06 (Fig. 7)
1.0
1173130
1.2 (BS436 Fig. 8 curve 1)
5098601
1.1 (BS436 Fig. 8 curve 3)
815 MPa (Equation 1)
823 MPa (Equation 1)
1.14 >1
1.15 >1
σHlim1=σHD1*Zc*ZG1=
σHlim2=σHD2*Zc*ZG2=
ZLZv=
ZR=
ZW=
Zx=
N2=
ZN2=
N1=
ZN1=
σHP1=σlim1*ZL*Zv*ZR*Zw*Zx*ZN1=
σHP2=σlim2*ZL*Zv*ZR*Zw*Zx*ZN2=
Spur gear safety factor for contact stress SHmin:
SHlim1=σHP1/σH=
SHlim2=σHP2/σH=
MARINE EQUIPMENTS PTE LTD 18/21 2nd STAGE GEAR
SPUR GEAR CONTACT STRESS CALCULATION (BS436)
FOR 75TON ANCHOR HANDING/TOWING WINCH (WIRE Φ52mm )
(second stage)
PINION SPUR GEAR (UNIT)
center distance: a= 876 876 mm
mormal module: 12 12 mm
pressure angle 20 20 deg
tooth number: 19 127
face width: b= 145 125 mm
reference diameter: 228 1524 mm
tip diameter: 252 1548 mm
tooth depth: 27 27 mm
number of revolution 56.66 8.48 rev/min
construction: solid solid
mean roughness: 3.2 3.2 μm
meterial type: AISI 4140 ASTM A148 grade80/40
tensile stress: 900 620
yielding point: 700 335
300 Cst
application factor: 1.0
required life time: 10000 hours
1.0
The overload : F= 30015 kg (BS MA 35 : 1975)
Norminal tangential force for the contact stress:
Load PCD. 468 mm
9217 kg
mn=
αn=
z1 & z2
d1, d2
da1, da2
Rα(root)
σb=
σs=
lubricant viscosity at 40˚C:
minimun demanded safety factor for contact stress: SHMin=
Dwpcd=
FHT=F*Dwpcd/d2=
MARINE EQUIPMENTS PTE LTD 19/21 2nd STAGE GEAR
Calculation for the contact stress:
(equation 2)
6.68
214.3 mm
1432.1 mm
34 (B.11)
2.49 (Equation 10)
35.4 mm (B.3)
60.6 (B.2)
1.71 (B.8)
0.874 (Equation 11)
180.1 (Equation 14)
Poisson's ratio 0.3
Steel: 21000 kg/mm^2
Cast steel: 20600 kg/mm^2
1.0
1.1 (from 16.3)
v= 40.59 m/min= 0.68 m/sec
14.14
1.0 (spur gear from fig. 11)
X= 0.8687 (grade 9 see table 7)
0.53 (Equation 22)
1.0 (Equation 20)
125 (Equation 35)
1840.5 kg (Equation 37)
144.4 N/mm (Equation 38)
K= 0.48 (Fig. 12)
σH=ZH*ZE*Zε*SQRT(FHT*(u+1)*KA*KV*KHα*KHβ/u/b/d1)
u=Z2/Z1=
db1=d1*cosαn=
db2=d2*cosαn=
Prel=(d1/2)*(u/(u+1))*(cosαt*tanαtw/cosβt)=
ZH=2*SQRT(cosβt/sin(2*αt))=
Pbt=mn*π*cosαt/cosβ=
ga=0.5*(SQRT(da1^2-db1^2)+SQRT(da2^2-db2^2))-a*sinαtw=
εa=ga/Pbt=
Zε=SQRT((4-εa)/3)=
ZE=SQRT{1/[π*((1-v1^2)/E1+(1-v2^2)/E2)]}=
v1=v2=
E1=
E2=
KA=
QV=
Qv v z1=
Kv350=
B=(350/(FtKA/b))^X=
Kv=1+(Kv350-1)/B=
beff=b-lc/2= mm lc=0
Fmest=bd1u/(u+1)*[σHP/(ZH*ZE*Zε)]^2=
Wmest=Fmest/beff=
MARINE EQUIPMENTS PTE LTD 20/21 2nd STAGE GEAR
A= 0.023 μm (Table 8)
73.74 kg/mm= 723 N/mm
150 mm
l= 500 mm
150 mm
6.78 μm (Equation 40)
12.9 μm (Equation 42)
21.92 μm (Equation 45)
0.45 (from fig 13. 700MPa)
9.86 μm (Equation 47)
20 N /mm.μm (from 17.2.5 steel to steel)
1.1 (Equation 49)
1.71
12.2 μm (BS436 part 2. Table 3)
0.915 μm (Equation 54)
0.86 (Equation 50)
1.0
The contact stress:
799 MPa
900 MPa (Fig. 2)
750 MPa (Fig. 2)
Zc= 1.0
0.85 (Equation 16)
0.94
761 MPa (Equation 15)
705 MPa (Equation 15)
1.02 (Fig. 5)
Wm=FtKAKv/beff=
dsh=
ls=0.3l=
fsh=WmA[(1+K*(ls/d1^2)*(d1/dsh)^4-0.3)+0.3]*(b/d1)^2=
fma=fHβ=
Fβx=/1.33fsh+fma/=
qy=
Fβy=qyFβx=
cγ=
KHβ=1+cγ*Fβy/(2Wm)=
εγ=εβ+εα=εα=
fpe=
γα=0.075fpe=
KHα=εγ*[0.9+0.4*cγ*(fpe-γα)/(Wm*KHβ)]/2=
take KHα=
σH=ZH*ZE*Zε*SQRT(FHT*(u+1)*KA*KV*KHα*KHβ/u/b/d1)=
σHD1=
σHD2=
ZG1=0.9*ZG2=
ZG2= (Fig 3, Prel/mn=3.4 σB=800MPa)
σHlim1=σHD1*Zc*ZG1=
σHlim2=σHD2*Zc*ZG2=
ZLZv=
MARINE EQUIPMENTS PTE LTD 21/21 2nd STAGE GEAR
1.05 (Fig. 6)
1.06 (Fig. 7)
1.0
5086105
1.2 (BS436 Fig. 8 curve 1)
33996595
1.1 (BS436 Fig. 8 curve 3)
951 MPa (Equation 1)
960 MPa (Equation 1)
1.19 >1
1.20 >1
ZR=
ZW=
Zx=
N2=
ZN2=
N1=
ZN1=
σHP1=σlim1*ZL*Zv*ZR*Zw*Zx*ZN1=
σHP2=σlim2*ZL*Zv*ZR*Zw*Zx*ZN2=
Spur gear safety factor for contact stress SHmin:
SHlim1=σHP1/σH=
SHlim2=σHP2/σH=
top related