# How to calculate the drum and power and line pull and brake of mooring winch

## 1. Basic parameters

1.1 Dia. of anchor chain：φ44（$$AM_{3}$$）

1.2 working load：F$$_{C}$$=92KN

1.3 over loading pull：F$$_{O}$$=138KN

1.4 unmooring rate：V$$_{C}$$≥10m/min

1.5 number of sprocket： 6

1.6 dia. of chain wheel section：D=$$\phi$$704mm

1.7 computing dia. of chain wheel：D$$_{0}$$=$$\phi$$672mm

1.8 rating pull of fixed cable：F$$_{C1}$$ =60KN

1.9 Polypropylene rope diameter：d=56mm

1.10 reel capacity of rope：200m

1.11 mix supporting load：F$$_{min}$$=691KN

## 2. The Design parameters

### 2.1 Rated output torque

T$$_{C}$$=F$$_{C}$$·D/2=92000×0.704/2=32384N·m

### 2.2 Gypsy speed

n$$_{C}$$≥V$$_{C}$$/πD$$_{0}$$=10000/672π=4.74/min

### 2.3 The total transmission ratio

Three-stage gear reducer

Motor speed：：n$$_{0}$$=980/1430 r/min

The total transmission ratio：i=n$$_{0}$$/n$$_{C}$$=1430/4.74=301

Take302

### 2.4 The total transmission ratio

P$$_{C}$$＝F$$_{C}$$·V$$_{C}$$／60＝92×10／60＝15.3kW

### 2.5 Motor Selection

The total transmission efficiency ηz=0.8

Motor Power P$$_{0}$$= P$$_{C}$$ /ηz=15.3/0.8=19.1 kW

Motor parameters：

Type：QAD200L18.5/22-4/6P

Power：18.5/22 kW

Speed：980/1430 r/min

Output shaft dia.：φ55mm

### 2.6 Drum computing

⑴Grade of reel：Q235A，σs＝235N／mm2

⑵dia.：D＝500mm

⑶length：L$$_{O}$$＝960mm

⑷circle of each layer Z＝L$$_{O}$$／d0＝96$$_{0}$$／56＝17.2(circle) take 17 circle

⑸layer of cable

n=［(D/d)2+4L/Zπd］1/2/2-D/2d

=［(500/56)2+4×200000÷(17×π×56)］1/2/2-500/(2×56)

=4.8(Layers)

take layer 5 outer dia. Of disc φ≈500+5×56×2+3×56=1228 Take φ=1230mm

⑹thickness of reel δ＝22mm

Strength Check：

According to

σ$$_{1}$$= A$$_{1}$$ A$$_{2}$$S$$_{max}$$/δP≤［σy]（N／mm2） σ$$_{1}$$

the max press stress of the inner of reel A$$_{1}$$

A$$_{1}$$－the modulus of reel layer. According to modulus table, A=2

A2 A＝0.75 According the table A=0.75

S$$_{max}$$－，S$$_{max}$$=60000N

P＝56.5mm

P－distance between reels, mm. it is the naked reel. The P is equal to the notch reel with the same dia. Of steel wire, take P＝56.5mm

[σ$$_{y}$$]－Allowable compressive stress，N/mm2

[σ$$_{y}$$]＝σs／1.5＝235÷1.5＝157N/mm2

V$$_{c}$$ the reel speed V$$_{c}$$ (Computing as layer 3) V$$_{c}$$= n$$_{c}$$π(D+d)=4.74×556÷π=8.5m/min

### 2.7 Tc The output torque of reel(computing as layer 1)

T$$_{c}$$=F$$_{c1}$$(D+d)/2=60000×(0.556) ÷2=16680N.m

## 3 Calculation of the gear and smallest shaft

the second level Spur gear power was taken in gearbox，The total transmission ratio=302

i=i$$_{12}$$·i$$_{34}$$ =90×128×129÷17÷17÷17=302

First level power

### 3.1 Design paremeter

Transfer Power P=22(kW)

Transfer Torque T=146.91(N·m)

Speed of Gear 1 n1=1430(r/min)

Speed of Gear 2 n2=188.46(r/min)

transmission rate i=7.588

Load characteristics of the original motivation SF= light librated

Load characteristics of the work machine WF= uniformity calm

designed life H=10000(Hour)

### 3.2 Setting and structure

Structure type ConS= close

Gear 1 setting ConS1=symmetry

Gear 2 setting ConS2=symmetry

### 3.3 material and heat treatment

Mating types of tooth surface GFace=Hard and soft tooth surface

quality level of heat treatment Q=MQ

material and heat treatment of gear 1 Met1=40Cr

hardness range of gear 1 HBSP1=48～55

hardness of gear 1 HBS1=52

material type of gear 1 MetN1=0

limited stress type of gear 1 MetType1=11

material and heat treatment of gear 2 Met2=45

hardness range of gear 2 HBSP2=162～217

hardness of gear 2 HBS2=190

material type of gear 2 MetN2=0

limited stress type of gear 2 MetType2=7

### 3.4 Accuracy of gear

The accuracy of the first group Ⅰof gear 1 JD11=7

The accuracy of the first group Ⅱof gear 1 JD12=7

The accuracy of the first group Ⅲ of gear 1 JD13=7

The up deviation of thickness of gear 1 JDU1=F

The down deviation of thickness of gear 1 JDD1=L

The accuracy of the first group Ⅰof gear 2 JD21=7

The accuracy of the first group Ⅱ of gear 2 JD22=7

The accuracy of the first group Ⅲ of gear 2 JD23=7

The up deviation of thickness of gear 2 JDU2=F

The up deviation of thickness of gear 2 JDD2=L

### 3.5 Basic parameters

Mn=4

End surface modulus M$$_{t}$$=4.00000

helix angleβ=0.000000(Degree)

Base cylinder helix angle βb=0.0000000(Degree)

teeth of Gear1 Z1=17

Coefficient of gear 1 X1=0.00

teeth width of gear 1 B1=70(mm)

coefficient of teeth width of gear 1 Φd1=1.029

teeth of Gear2 Z2=129

Coefficient of gear 2 X2=0.00

teeth width of gear 2 B2=60(mm)

coefficient of teeth width of gear 2 Φd2=0.116

Total coefficient X$$_{sum}$$=0.000

Standard Center Distance A0=292.00000(mm)

Actual Center Distance A=292.00000(mm)

ratio of teeth U=7.58824

Coincidence degree of end surface εα=1.69648

Coincidence degree of longitudinal εβ=0.00000

total Coincidence degree ε=1.69648

Pitch circle dia. Of gear 1 d1=68.00000(mm)

top circle dia. Of gear 1 da1=76.00000(mm)

root circle dia. Of gear 1 df1=58.00000(mm)

top high of teeth of gear 1 ha1=4.00000(mm)

root high of teeth of gear 1 hf1=5.00000(mm)

high of full teeth of gear 1 h1=9.00000(mm)

top teeth press angle of gear 1 αat1=32.777676(Degree)

Pitch circle dia. Of gear 2 d2=516.00000(mm)

top circle dia. Of gear 2 da2=524.00000(mm)

root circle dia. Of gear 2 df2=506.00000(mm)

top high of teeth of gear 2ha2=4.00000(mm)

root high of teeth of gear 2 hf2=5.00000(mm)

high of full teeth of gear 2 h2=9.00000(mm)

top teeth press angle of gear 2αat2=22.279388(Degree)

Pitch circle String teeth thickness of gear 1 sh1=6.27425(mm)

Pitch circle String teeth high of gear 1 hh1=4.14504(mm)

sch1=5.54819(mm)

fixed String teeth high of gear 1 hch1=2.99023(mm)

inter-public law line teeth number of gear 1 K1=2

inter-public law line length of gear 1 Wk1=18.66516(mm)

Pitch circle String teeth thickness of gear 2 sh2=6.28303(mm)

Pitch circle String teeth high of gear 2 hh2=4.01913(mm)

fixed String teeth thickness of gear 2 sch2=5.54819(mm)

fixed String teeth high of gear 2 hch2=2.99023(mm)

inter-public law line teeth number of gear 2 K2=15

inter-public law line length of gear 2 Wk2=178.45048(mm)

coefficient of teeth top high ha*=1.00

coefficient of top gap c*=0.25

press angle α*=20(Deg)

coefficient of teeth top high of end surface ha*t=1.00000

coefficient of top gap of end surface c*t=0.25000

press angle of end surface α*t=20.0000000(Degree)

### 3.6 parameter of Inspecting items

Cumulative pitch tolerance of gear 1 Fp1=0.04569

Gear Radial runout tolerance of gear 1 Fr1=0.03876

alteration tolerance of public law line length of gear 1 Fw1=0.02946

pitch limit tolerance of gear 1 fpt(±)1=0.01666

rofile Tolerance of gear 1ff1=0.01285

tangential comprehensive tolerance of first tooth of gear 1 fi'1=0.01770

Radial comprehensive tolerance of first tooth of gear 1 fi''1=0

tooth direction tolerance of gear 1 Fβ1=0.01676

tangential comprehensive tolerance of gear1 Fi'1=0.05854

radial comprehensive tolerance of gear 1 Fi''1=0.05427

The limit deviation of the base section fpb(±)1=0.01565

spiral wave Tolerance of gear 1 ffβ1=0.01770

Axial limit deviation of pitch of gear 1 Fpx(±)1=0.01676

tooth direction tolerance of gear 1 Fb1=0.01676

axis parallel tolerance of gear 1 fx1=0.01676

axis parallel tolerance of gear 1 fy1=0.00838

up tolerance of tooth thickness of gear 1 Eup1=-0.06662

down tolerance of tooth thickness of gear 2 Edn1=-0.26649

Cumulative pitch tolerance of gear 2 Fp2=0.11007

Gear Radial runout tolerance of gear 2 Fr2=0.06712

alteration tolerance of public law line length of gear 2 Fw2=0.04711

pitch limit tolerance of gear 2 fpt(±)2=0.01991

Profile Tolerance of gear 2 ff2=0.01845

tangential comprehensive tolerance of first tooth of gear 2 fi'2=0.02302

Radial comprehensive tolerance of first tooth of gear 2 fi''2=0

Fβtooth direction tolerance of gear 2 2=0.00630

tangential comprehensive tolerance of gear2 Fi'2=0.12852

radial comprehensive tolerance of gear 2 Fi''2=0.09397

The limit deviation of the base section fpb(±)2=0.01871

spiral wave Tolerance of gear 2 ffβ2=0.02302

Axial limit deviation of pitch of gear 2 Fpx(±)2=0.00630

tooth direction tolerance of gear 2 Fb2=0.00630

axis parallel tolerance of gear 2 fx2=0.00630

axis parallel tolerance of gear 2 fy2=0.00315

up tolerance of tooth thickness of gear 2 Eup2=-0.07964

down tolerance of tooth thickness of gear 2 Edn2=-0.31858

center distance limit deviation fa(±)=0.03891

### 3.7 strength checking

the limit press of contact strength of gear 1 σHlim1=1186.4(MPa)

The basic values of the flexural fatigue of gear 1 σFE1=672.0(MPa)

Contact fatigue strength allowable value of gear 1 [σH]1=1706.5(MPa)

flexural fatigue strength allowable value of gear 1 [σF]1=992.9(MPa)

the limit press of contact strength of gear 2 σHlim2=540.0(MPa)

The basic values of the flexural fatigue of gear 2 σFE2=426.7(MPa)

Contact fatigue strength allowable value of gear 2 [σH]2=776.7(MPa)

flexural fatigue strength allowable value of gear 2 [σF]2=630.5(MPa)

Contact strength with safety factor SHmin=1.00

flexural strength with safety factor SFmin=1.40

calculation stress of contact Strength σH=709.6(MPa)

contact fatigue strength checking σH≤[σH]=Satisfying

calculation stress of flexural fatigue strength of gear 1 σF1=152.9(MPa)

calculation stress of flexural fatigue strength of gear 2 σF2=132.1(MPa)

flexural fatigue strength check of gear 1 σF1≤[σF]1=Satisfying

flexural fatigue strength check of gear 2 σF2≤[σF]2=Satisfying

### 3.8 Strength checking correlation coefficient

Special treating of gear shape Z$$_{ps}$$= Needn’t

surface hardening of gear Zas= surface hardening

gear shape Z$$_{p}$$= Normal

Lubricant viscosity V50=120(mm^2/s)

Us=disallow

Pinion tooth surface roughness Z$$_{1R}$$=R$$_{z}$$＞6μm(R$$_{a}$$≤1μm)

Static Strength Wtype=Degree

surface roughness of gear Root Z$$_{FR=Rz}$$≤16μm (R$$_{a}$$≤2.6μm)

The basic outline of the tool size

Circumferential force F$$_{t}$$=4320.882(N)

Degreeline speed of gear V=5.091(m/s)

coefficient K$$_{a}$$=1.100

Dynamic Load coefficient K$$_{v=1.307 gear direction load distributing coefficient K\(_{Hβ}$$=1.303

the effect of comprehensive deforming to load distribute K$$_{βs}$$=1.191

the effect of setting precision to load distribute K$$_{βm}$$=0.112

tooth internal load distribute KHα=1.303

Coefficient of the node region Zh=2.495

Material elasticity Z$$_{E}$$=189.800

Degree Contact strength coincidence coefficient Z$$_{ε}$$=0.876

Helix angle coefficient of Contact strength Z$$_{β}$$=1.000

coincidence and Helix angle coefficient Z$$_{εβ}$$=0.876

Contact fatigue life factor Zn=1.30000

The impact coefficients of oil film Z$$_{lvr}$$=0.95000

Work hardening coefficient Z$$_{w=1.16471 sizing coefficient of contact strength Z\(_{x}$$=1.00000

load distribution coefficient of tooth direction K$$_{Fβ}$$=1.303

load distribution coefficient of inter-teeth K$$_{Fα}$$=1.445

coincidence coefficient of tensile strength $$_{Yε}$$=0.692

Helix angle coefficient of tensile strength Y$$_{β}$$=1.000

coincidence and Helix angle coefficient of tensile strength Y$$_{εβ}$$=0.692

life coefficient Y$$_{n}$$=2.06859

Sensitivity coefficient of the tooth root fillet Y$$_{dr}$$=1.00000

surface conditions Coefficient of Dedendum Y$$_{rr}$$=1.00000

size coefficient Y$$_{x=1.00000 Compositive gear profile coefficient of gear 1 Y\(_{fs1}$$=4.53369

stress revising coefficient of gear 1 Y$$_{sa1}$$=1.51921

Compositive gear profile coefficient of gear 2 Y$$_{fs2}$$=3.91637

stress revising coefficient of gear 2 Y$$_{sa2}$$=1.79327

## 4 Winch brake holding capacity calculation

calculation by the supporting load

The max braking torque T$$_{max}$$=691000×0.672/2=232176N·m

Dia. of breaking ring：D$$_{Z}$$＝φ800

raking circle power F$$_{t}$$=2T$$_{max}$$/D$$_{Z}$$ =2×232176×1000/800=580440N

two side pull of strip：

F$$_{1}$$=F$$_{t}$$·e$$^{μθ}$$/(e$$^{μθ}$$-1)

F$$_{1}$$— tight side pull。N

e—fundus of Natural logarithm e=2.718

μ—coefficient of friction if the covering material is non asbestos friction strip, take 0.50

θ— angle of breaking strip。 Take 5π/3

=580440×2.72$$^{0.50×5π}$$/3/(2.72$$^{0.50×5π/3}$$-1)

=580440×2.72$$^{0.50×5π}$$/3/(2.72$$^{0.50×5π}$$/3}\)-1)

=624412N

F$$_{2}$$=F$$_{t}$$/(e$$^{μθ}$$-1)

=580440/(2.72$$^{0.50×5π/3}$$-1)

=43973N

of steel strip according to it as follows:

width120，thickness20，grade 16Mn，σs=345N/mm2

tensile stress σ$$_{L}$$=624412/(120×20)=260N/mm2

σ$$_{L}$$ <σ$$_{s}$$

Strength checking passed

## 5. Calculation clutch of mooring winch and Gypsy

Out Dia of jaw clutch D=330mm

Inside Dia of jaw clutch D0=230mm.

Numbers of Jaw clutch z=3.

Height of Jaw clutch h=50mm.

Width of jaw clutch b=25mm

### 5.1 the pressure condition of anchor chain wheel clutch teeth working side

p=2T$$_{C}$$/D$$_{m}$$A$$_{p}$$Z$$_{j}$$≤[p]

p— pressure of tooth side Mpa

D$$_{m}$$— average dia. Of circle of tooth

A$$_{p}$$— area of supporting working side of tooth mm2}\)

Z$$_{j}$$— tooth number computing， Z$$_{j}$$＝（1/2～1/3）Z

[p]—quiet combining [p]＝100～120MPa

=2×32384×1000/280×2×50×25=92.5MPa≤[p] Can meet the requirements

### 5.2 anti-bend strength condition of tooth root of anchor chain wheel clutch

σ＝6h T$$_{C}$$/D$$_{m}$$L$$^{2}$$mbZ$$_{j}$$≤[σ]

Lm— thickness of tooth root of center dia.

6×50×32384×1000/（280×140$$^{2}$$×25×2）=35.4 N/mm$$^{2}$$≤[σ]310 N/mm$$^{2}$$

## 6. principal axis strength computing

### 6.1 strength computing of min principal axis the steel grade of transmission shaft is 40Cr

a. transfer torque TC=32384 N•m

b. d$$_{2}$$=17.2（T$$_{Ⅰ}$$/[τ]）$$^{1/3}$$

=17.2（32384$$_{Ⅰ}$$/52）$$^{1/3}$$

=144.6mm

After Rounding d$$_{2}$$=150mm

The max load of Spindle is supporting load and the clutch should be in a detached state at this time, the only power is bending torque. The distance between the two sides fulcrum of chain wheel is 700mm, type tank center line of chain wheel to a supporting point is 410mm. On the working of supporting load, the max bending torque can be get from the type tank center line（the load map, shearing force diagram and the bending torque figure see the fig 1）

R$$_{A}$$= F$$_{Surport}$$×410÷700=691000×410÷700 =404728N

R$$_{B}$$= F$$_{Surport}$$×390÷700=691000×290÷700

=286271N M$$_{max}$$= F$$_{Surport}$$×410×(700-410)÷700

=691000×410×(700-410) ÷700

=117371286N.mm

The shaft dia. is 165mm, and the bending load modulus is:

W=πd3/32=3.14×1653÷32=441013mm$$^{2}$$

AS σ= M$$_{max}$$ / W

=117371286÷441013=266N/mm$$^{2}$$

40Cr，σ$$_{s}$$=540 N/mm$$^{2}$$

Grade is 40Cr，σ$$_{s}$$=540 N/mm$$^{2}$$

σ<σ$$_{s}$$

Strength checking passed

## 7. Base structure maximum stress check

The bottom of the base is like to the rectangular loop .The size of inner frame is975×810mm.The outer frame size is 1980×1350, 1360×490. Take it as a ring, and then the R=1055,r=501. For the load way is outter circle fixed and inner circle be taken a torque M, the max stress shall be get from the inner circle. According to the machinery design manual (Chemistry, the third version, part 1, P1-148),

σ$$_{max}$$=A$$_{19}$$M/Rh$$^{2}$$

h=59mm h is the thickness of plate(Equivalent plate thickness of base),h=59mm

A$$_{19}$$=0.685,A$$_{19j}$$ is the coefficient. Thorough Mapping we can get A$$_{19}$$=0.685,then

σ$$_{max}$$=0.685×950000×832/(1331×76$$^{2}$$)

=70.5N/mm$$^{2}$$

[σ]=σ$$_{s}$$/n=235/1.5=157N/mm$$^{2}$$

σ<[σ], The strength of base can meet the requirements.

## 8. The strength checking of connect bolts of base

According to the the strength requirements of deck windlass of front hull, in Rules for Classification and Construction(2006) 3.2.5, the computing area and the pressure on the windlass(Combined type) are as follows:

### 8.1 The Power P$$_{x}$$ perpendicular to the axis direction from the top to the end (See fig2):

P$$_{x}$$=200×A$$_{x}$$

A$$_{x}$$——The area perpendicular to the axis direction from the top to the end

Through mapping can get conclusion: Ax=3.36m2

And then：P$$_{x}$$=200×3.36=672 kN

### 8.2 The force P$$_{y}$$: Parallel to the axis, acting on the side and lateral side of respectively

P$$_{y}$$=150×F×A$$_{y}$$

Area A$$_{y}$$: Parallel to the axis, acting on the side and lateral side of respectively

Through mapping can get conclusion: A$$_{y}$$=2.43m$$^{2}$$

F=1+B/H，but less than 2.5

B--The computing width of windlass Parallel to the axis,m. From the fig,the B=1.817 m；

H—the most height, From the fig, H=1.621

f=1+B/H=1+1.817÷1.621=1+1.12=2.12≤2.5

Take 2.12

then ：P$$_{y}$$=150×2.12×2.43=773 kN

### 8.3 The design computing bolts of fixed the windlass on the deck

#### 8.3.1 Ri Axial force Ri

The windlass was supported by 27 bolts which type is M30. The Axial force i of each bolt was computed as follows:

R xi=(P$$_{x}$$×h×xi×Ai)/i x

R yi=(P$$_{y}$$×h×yi×Ai)/i y

R i= R xi + R yi - R si

P$$_{x}$$——The force perpendicular to the axis.kN；

P$$_{y}$$——The force paralleled to the axis, kN. Take the max from in or out of the shipboard

h——Height between the axis line of windlass and the fixed plane, cm. According to fig1, h=90cm.

xi ,yi——the coordinate of the I bolt ,cm. The force in right-about of action force is positive.

Ai——The Cross-sectional area of bolts, cm$$^{2}$$. Choosing the bolt M30, the Ai= 5.61cm$$^{2}$$

i x=∑Ai×xi2，cm$$^{2}$$，from fig 3 can get ix=333317cm$$^{2}$$；

i y=∑Ai×yi2，cm$$^{2}$$，from fig 3 can get iy=151844cm$$^{2}$$；

R $$_{si}$$——the the anti-static force of the weight of windlass pose on the bolt I ,KN. Weight of windlass is W=6.5t，R $$_{si}$$=6.5÷27×9.81=2.36kN.

Number of Srews 1 2 3 4 5 6 7 8 9
xi(cm) -68.8 -38.8 -8.8 21.2 51.2 81.2 128.2 170.7 191.2
yi(cm) 55.8 55.8 55.8 55.8 55.8 55.8 55.8 55.8 55.8
xi2(cm2) 4733 1505 77 450 2621 6593 16435 29138 36557
yi2(cm2) 3113 3113 3113 3113 3113 3113 3113 3113 3113
Rsi (KN) 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36
ai（cm2） 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61
IX（cm4） 333317 333317 333317 333317 333317 333317 333317 333317 333317
iy(cm4) 151844 151844 151844 151844 151844 151844 151844 151844 151844
rxi(kn) -67.47 -34.13 -8.88 21.41 51.71 82.01 129.48 172.4 193.11
ryi(kn) 136.15 136.15 136.15 136.15 136.15 -136.15 136.15 136.15 136.15
ri(kn) 201.26 167.92 142.7 155.2 185.5 215.8 263.27 306.2 326.9
δ(mpa) 35.88 29.93 25.43 27.67 33.06 38.47 46.93 54.58 58.27

xi(cm) 191.2 191.2 191.2 170.7 81.2 128.2 81.2 59.2 59.2
yi(cm) -13.2 -43.2 -71.2 -71.2 55.8 -71.2 -71.2 -103.2 -120.2
xi2(cm2) 36557 36557 36557 29138 6593 16434 6593 3504 3504
yi2(cm2) 174 1866 5069 5069 3113 5069 5069 10650 14448
Rsi (KN) 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36
ai（cm2） 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61
IX（cm4） 333317 333317 333317 333317 333317 333317 333317 333317 333317
iy(cm4) 151844 151844 151844 151844 151844 151844 151844 151844 151844
rxi(kn) 193.11 193.11 193.1 172.4 82.01 129.48 82.01 59.79 59.79
ryi(kn) -32.21 -105.41 -173.73 -173.73 -136.15 -173.72 -173.72 -251.81 293.28
ri(kn) 222.96 296.16 364.5 343.78 215.8 300.85 253.38 309.24 350.72
δ(mpa) 39.74 52.79 64.97 61.28 38.47 53.63 45.17 55.12 62.52

xi(cm) 21.2 -8.8 -38.8 -68.8 -68.8 -68.8 -68.8 -68.8 -68.8
yi(cm) -120.2 -120.2 -120 -120.2 -103 -73.2 -43.2 -13.2 16.8
xi2(cm2) 449 77 1505 4733 4733 4733 4733 4733 4733
yi2(cm2) 14448 14448 14448 14448 10650 5358 1866 174 282
Rsi (KN) 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36 2.36
ai（cm2） 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61 5.61
IX（cm4） 333317 333317 333317 333317 333317 333317 333317 333317 333317
iy(cm4) 151844 151844 151844 151844 151844 151844 151844 151844 151844
rxi(kn) 21.41 -8.88 -39.18 -69.48 -69.48 -69.48 -69.48 -69.48 -69.48
ryi(kn) 293.28 293.28 293.28 293.28 -251.81 -178.61 -105.41 -32.21 40.99
ri(kn) 312.34 299.2 330.12 360.42 318.94 245.74 172.53 99.34 108.12
δ(mpa) 55.68 53.44 58.84 64.25 56.85 43.8 30.76 17.71 19.27

From the table we can know, the max absolute value is the 12th bolt. Checking it can get the follows:

Rx1=(P$$_{x}$$×h×xi×Ai)/ix

=（672×90×191.2×5.61）/333317

=193.11 kN

Ry2=(P$$_{y}$$×h×yi×Ai)/i y

=（773×90×71.2×5.61）/151844

=-173.73 kN

R15= R xi + R yi - R si

=193.11+173.73 -2.36

=364.48KN

σ= R$$_{i}$$/S=364.48÷5.61=64.97Mpa≤[σ]=332MPa

The tensile strength of bolts meet the requirements

#### 8.3.2 Fi shearing force

From analysis the shearing force F xi, F yi and the compose force Fi of the ist bolt, it is know that the total tipping moment is the max than others. Computing as the follows:

Fxi=(P$$_{x}$$-a×g×W)/N

=（672-0.5×9.81×6.5）÷27

=23.7Kn

Fyi=(P$$_{y}$$-a×g×W)/N

=(773-0.5 ×9.81×6.5）÷27

=27.4Kn

Fi=$$\sqrt{Fxi2+ Fyi2}$$

=$$\sqrt{23.72+27.42}$$

=36.2 Kn

a——Fabrication coefficient, take a=5.

W——Weight of windlass

N——Number of bolt, 27.

g——The acceleration due to gravity, take 9.98 m/s2

#### 8.3.3 Checking the diameter of the bolt according to the max axial and shearing force of single bolt.

According to the fourth strength theory in the Machinery Design Manual,2nd edition, Volume 3,page 25-10, the computing stress of pretightening bolts which load be posed on axial and horizontal can be as follows:

σca=4/(πd2) ×$$\sqrt{（R12/6）2+ 3（Fi+0.5 R12/6）2}$$

=4÷(π×1.52) ×$$\sqrt{（364÷6）2+ 3（36.2+0.5× 364÷6）2}$$

=73.7≤[σ]

[σ]——allowable pull stress. Checking the 25.1-11 can know：level 8.8. σs=830N/mm2. If the safety coefficient n=2.5, the allowable force [σ]=σs/n=830/2.5=332N/mm2

And then, σca=73.7 N/mm2

That is σca≤[σ].It can meet the requirements.