Diameter 34mm electric anchor windlass calculation book

By David Zhu
Last updated August 14, 2021

I. Main parameters

Anchor chain diameter: 34 (AM2).

Workload: 49.13kN

Overload pull: 73.7kN

Support load: 294.75 kN

Anchor speed: 12.9m/min

Anchor depth: ≤82.5m

Motor specifications: Type: JZ₂-H51-4/8/16

Power: 16/16/11kW

Speed: 1400/665

Duty: min10/30/5

Power source: AC360V;

anchor windlass

II.Overall calculation

1. Workload calculation

$T_1=42.5d^2=42.5*34^2=49.13KN$

2. Overload pull calculation

$T_2=1.5*T_1=1.5*49=73.7$KN$$

3. Holding load calculation

$T_3=0.45*T_b_r_e_a_k=0.45*655=294.75$KN$$

4. Transmission ratio calculation

Big gear $Z_1=15$ $Z_2=90$

$i_1=\frac{Z_2}{Z_1}=-\frac{90}{15}=6$

Worm Gear $i_2 =11.67$

$i=i_1*i_2=6*11.67=70.02$

5. Speed calculation

$V=\frac{0.04*665*34}{70.02}=12.9(m/min)$

6. Efficiency calculation

There are several factors that affect the efficiency of the anchor windlass

Worm gear efficiency $\eta_1=0.8$

Open Gear drive rate $\eta_2=0.95$

Chain cable lifter engagement rate $\eta_3=0.95$

Sliding bearing carrying efficiency $\eta_4=0.96$

Dog-type clutch meshing efficiency $\eta_5=0.98$

Coupling efficiency $\eta_6=0.98$

windlass efficiency:

$\eta=\eta_1 \times \eta_2 \times \eta_3 \times \eta_4\times\eta_5\times\eta_6=0.67$

7.Power calculation

When workload,

$N=\frac{1000xT_1\timesV}{6120\times9.8\times\eta}=15.77\left(m/min\right)$

Pick motor model: $JZ_2$ -H51-4/8/16 Power: 16/16/11kW

8.Torque calculation of each driven shaft

When workload

Gypsy axle

$M_1=\frac{T_1\timesD_h0}{2\times\eta_3\times\eta_4}=\frac{49.13\times463}{2\times0.95\times0.96}=12471\left(Nm\right)$

$D_h0$ $\mathsf{anchor\ chain\ wheel\ Diameter}$

Worm gear axle

$M_2=\frac{M_1}{i_1\times\eta_2\times\eta_5}=\frac{12471}{6\times0.95\times0.98}=2232.5\left(Nm\right)$

$D_h_0$ $\mathsf{anchor\ chain\ wheel\ Diameter}$

$\mathsf{Worm\ gear\ axle}$

$M_2=\frac{M_1}{i_1\times\eta_2\times\eta_5}=\frac{12471}{6\times0.95\times0.98}=2232.5\left(Nm\right)$

2)The torque of each shaft when the motor is blocked

$\mathsf{Worm\ gear\ axle}$

$M_3=\frac{9550\timesN_m_o_t_o_r\times\eta_6}{n}=\frac{9550\times16\times0.98}{665}=225.2\left(Nm\right)$

When the motor is blocked

$\mathsf{Worm\ gear\ axle}$

$M_max=2.5\timesM_3^' =2.5\times225.2=563\left(Nm\right)$

$\frac{M_max}{M_3}=\frac{563}{239.1}=2.35$

That is: when the motor has blocked the torque that each axis is subjected to increase to 2.35 times on the basis of workload stress on each axis

$P_b_r_a_n_c_h * 1380 = R_A * 1760$

$R_A = \frac{P_b_r_a_n_c_h1380}{1760}=\frac{294.75*1380}{1760}={231.1kN}$

$R_A =\frac{P_b_r_a_n_c_h * 1380}{1760}=\frac{294.75*1380}{1760}=231.1Kn$

$P$ torque

$M_p = 380* R_A = 380*231.1 =87818(Nm)$

III. Worm gear box selection

Select the $WD-type$ normal cylindrical worm reducer according to the $JB/ZQ4390-86$ standard
At workload time, the worm shaft torque is $2232.5 (Nm)$ .
Check the load-bearing capacity table, the transmission ratio $i$ =11.67 ,center distance is 250

$T_p_2=1705 * 1.44 =2455.2(Nm) ,\mathsf{Shaft\ Hardness \ HRC}$ >$ 45$

$T_p_2 $ >$ 2232.5（Nm） ,\mathsf{Shaft \ Hardness \ HRC} $ >$ 45$

therefore selected WD250 gearbox (worm shaft and worm shaft need to be custom-made)

IV. Open gear strength check

$Z_1=15$ m＝10 \#45\n 220～250HB $\delta_s_1=360 \left(MPa\right)$
$Z_2=90$ m＝10 ZG310-570 220～250HB $\delta_s_2=360 \left(MPa\right)$

$Z_1$ =15 $m$ =10 $#45$ $220~250HB$ $360（MPa）$
Small gear $b_1 = 130$ Big gear $b_2=120$
$\mathsf{Proof \ of \ core \ root \ bending \ fatigue \ strength}$

1. Root stress $\delta_F$ calculation
When working, Pinion shaft torque $T_1 =2232.5 \left(Nm\right)$

$\ big\ Gear T_2 = 12471 \left(Nm\right)$

Circumference force

$F_t_1=\frac{2000\timesT_1}{d_1}=\frac{2000\times2232.5}{150}=29766.7\left(N\right)$

$F_t_2=\frac{2000\timesT_2}{d_2}=\frac{2000\times12471}{900}=27713.3\left(N\right)$

Usage factor: $K_A=1. 25$
Dynamic load factor: $K_V=1.14$
Toothed load distribution factor: $K_F_\beta=1.3$
Interdental load distribution factor: $K_F_\alpha=1$
Composite profile coefficient: $K_F_S_1 =4.2$
$Y_F_S_2 =3.95$
Coincidentity and helix angle coefficient: $Y_\varepsilon_\beta=0.67$
If your coefficients are different, the school check will need to be provided
Root stress

$\sigma_F = \frac{F_t \times K_A \times K_V \times K_F_\alpha \times K_F_\alpha \times K_F_S_1 \times Y_\varepsilon_\beta}{b \times m}$

$\sigma_F_1 = \frac{29766.7 \times 1.25 \times 1.14 \times 1.30 \times 1 \times 4.2 \times 0.67}{130 \times 10} = 119.4\left(MPa\right)$

$\sigma_F_2 = \frac{27713.3 \times 1.25 \times 1.14 \times 1.30 \times 1 \times 3.95 \times 0.67}{120 \times 10} = 113.2\left(MPa\right)$

2.Allowed root stress $\sigma_F_P$ calculation
The basic value of bending fatigue strength of the gear material $\sigma_F_E_1 =500 \left(Mpa\right)$
$\sigma_F_E_2 =360 \left(Mpa\right)$
The life factor calculated for bend strength $Y_N_T = 1$
Relative root fillet sensitivity coefficient $Y_\delta_r_e_l_T = 1$
Relative surface condition coefficient $Y_R_r_e_l_T=0.9$
The dimensional factor for bend strength calculation $Y_X = 0.97$
The minimum safety factor for bending strength $S_F_min =1.4$
Allow root stress

$\sigma_F_P = \frac{\sigma_F_t \times Y_N \times \ Y_\delta_r_e_l_T \times Y_R_r_e_l_T \times Y_X \times }{S_F_m_i_n }$

$\sigma_F_P_1 = \frac{500 \times 1 \times 1 \times 0.90 \times 0.97}{1.4} = 311.8\left(MPa\right) >\sigma_F_1 = 119.4 \left( MPa \right)$

$\sigma_F_P_2 = \frac{500 \times 1 \times 1 \times 0.90 \times 0.97} {1.4} = 224.5\left(MPa\right) >\sigma_F_2 = 113.2 \left( MPa \right)$

3. Calculated safety factor of bending strength

$\sigma_F = \frac{\sigma_F_E \times Y_N_T \times \ Y_\delta_r_e_l_T \times Y_R_r_e_l_T \times Y_X }{\sigma_F }$

$\sigma_F_1 = \frac{500 \times 1 \times 1 \times 0.90 \times 0.97}{119.4} = 3.66$

$\sigma_F_P_2 = \frac{360 \times 1 \times 1 \times 0.90 \times 0.97} {113.2} = 2.36$

4 Stress core when the motor is blocked

When the motor is blocked, the stress is 2.35 times that of the workload

When the motor is blocked, the small gear calculates the stress

$\sigma_F_b_l_o_c_k_1=2.35 \times \sigma_F_1=2.35 \times 119.4 = 280.59 \left( MPa \right)$

The large gear calculates the stress

$\sigma_F_b_l_o_c_k_2=2.35 \times \sigma_F_2=2.35 \times 113.2 = 266.02 \left( MPa \right)$

V. Strength check of shaft

Worm Gear Axle

Worm Gear Axle by the reduction box manufacturers, manufacturers in accordance with national standards requirements for manufacturing, to ensure the strength of the shaft requirements, the outstretched end, and pinion link, suspension wall (cantilever) arrangement, in the face of torque at the same time, bear the radial force of the big gear, the current school nuclear worm axle outstretched the end root to withstand the bending stress.

Shaft material 45 steel, tuning treatment HB217 ~255 $\sigma_S =360MPa$

The worm gear axle outstretched end is made by the product sample

1. When the workload is in, the pinion shaft torque

T＝2232.5（Nm）＝2232500（Nm）

Circumference force:

$F_t=\frac{ 2000 \times T_1}{d_1}={2000 \times 2232.5}{150}=29766.7 \left( N \right)$

$F_r= F_t \times Tg \alpha}=29766.7 \times tg 20}=10834.2 \left( N \right)$

The root is subjected to bending moment

$M = F_r \times 105 = 10834.2 \times 105 = 1137591 \left( Nm \right)$

The hazardous cross-sectional stress is calculated by bending synthes

$\sigma =\frac{\sqrt[10]{M^2 + (\alpha \times T)^2}}{d^3} = \frac{\sqrt[10]{10834^2 + (0.7 \times 2232500)^2}}{90^3}=21.4 \left( MPa \right)$

The stress of the promised use $\sigma =0.4 \times \sigma_s=0.4 \times 360=144\left( MPa \right)$

The stress is much greater than the calculated stress $\sigma$ , and the intensity is checked when overloaded.

2 Gypsy strength core

Shaft material $\# 45$ Tuning treatment $HB 217 \slim 255$ $\sigma = 360 \left( MPa \right)$

2.1 a fatiguestrength core

When workloads,

Horizontal force at the cable wheel

$F_1_t = F_1 \times \cos 17^\circ = 49130 \times 0.9563 = 46983 \left( N \right)$

Vertical force at the gypsy

$F_1_r = F_1 \times \sin17^\circ = 49130 \times 0.2924 = 14366 \left( N \right)$

At the time of the workload, the gypsy torque $M=12471$ \left( Nm \right)
Large gear circumference force (i.e. vertical force)

$F_2_t = \frac{2000 \times M}{d}=\frac{2000 \times 12471}{900} = 27713\left( Nm \right)$

Large gear radial force (i.e. horizontal force)

$F_2_r=F_2_t \times tg20^\circ=10087 \left( Nm \right)$

2.1.1 When C-points are subject to workload, ask for A-point and B-point horizontal forces
The level at which point C is forced is tried

Ask for A-point force $R_A$

$1760 \times R_A = \left(795+585\right) \ times F_1_t - 585 \times F_2_r \times R_A =33486 \left(N\right)$

Find the B-point force $R_B$

$1760 \times R_B = 380 \times F_1_t -\left(795+380\right) \times F_2_r \times R_B =3410\left(N\right)$

2.1.2 When C-points are subject to workload, seek vertical force C-Point at points A and B
The vertical force at point C is

Vertical force C-point

$1760 \times \left( R_A^'\right) = (795 +585) \times F_1_r - 585 \times F_2_t \times R_A^' =2053\left(N\right)$

$1760 \times \left(R_B^'\right) =380 \times F_1_r - \left(795+380\right) \times F_2_t \times \left(R_B^'\right) =-15400\left(N\right)$

2.1.3 When C-points are under workload, seek A-point and B-points

$P_A = \sqrt{R_A^2+R_A^'2} =\sqrt{33486^2+2053^2}=33549 \left(N\right)$

$P_B = \sqrt{R_B^2+R_B^'2} =\sqrt{3410^2+14400^2}=15773 \left(N\right)$

2.1.4 When theC point is under the workload, find the C,E point bend moment

$M_C = \frac{P_A \times 380}{1000}=\frac{33549 \times 380}{1000} =12749\left( Nm\right)$

$M_E = \frac{P_B \times 585}{1000}=\frac{15773 \times 585}{1000} =9227\left( Nm\right)$

2.1.5 When the D-point is under the workload, the A-point and B-point levels are forced
Levels are sought

$1760 \times R_A = 380 \times F_1_t - \left(205+380\right)\times F_2_r \times R_A =6791\left(N\right)$

$1760 \times R_B = \left(1175+205\right) \times F_1_t - 1175 \times F_2_r \times R_B =30105\left(N\right)$

2.1.6 When D-points are under the workload, seek vertical force at points A and B

Vertically sought

D-points are under the workload vertical force

$1760 \times R_A = 380 \times F_1_r - \left(205+380\right)\times F_2_r \times R_A^' =6791\left(N\right)$

$1760 \times R_B = \left(1175+205\right) \times F_1_r - 1175 \times F_2_t \times R_B^' =30105\left(N\right)$

2.1.7 When you are under workload at point D, ask for A and B points to work together

$P_A = \sqrt{R_A^2+R_A^'2} = \sqrt{6791^2+6110^2}=9135 \left(N\right)$

$P_B = \sqrt{R_B^2+R_B^'2} =\sqrt{30105^2+7237^2}=30963 \left(N\right)$

2.1.8 When you are under workload at point D, ask for A and B points to work together

$M_D = \frac{P_B \times 380}{1000}=\frac{30963 \times 380}{1000} =11766\left( Nm\right)$

$M_E = \frac{P_B \times 1175}{1000}=\frac{9135 \times 1175}{1000} =10734\left( Nm\right)$

2.1.9 The dangerous cross-section stress is calculated by bending moment synthesis

Comparing $2.1.4 >$ and $2.1.8$ , it can be seen that when $C$ point is forced,the maximum bending moment is $12749$ Nm, at which the stress of the axis can be calculated. As can be seen from the pre-calculation, the torque of the shaft is $M-12471$ Nm when the workload is loaded
Calculate the stress of the axis

$\sigma = \frac{\sqrt[10]{M^2 +\left( \alpha \times T\right)^2}}{d^3}= \frac{\sqrt[10]{12749000^2 +\left( 0.7 \times 12471000 \right)^2 }}{d^3}=41.5\left(MPa\right)$

It is safe to use

$\sigma_stress =0.4 \times \simga_s =0.4 \times 360 = 144 \left(MPa\right)$

2.1.9.2 The core of the shaft strength when the motor is blocked

According to the previous calculation, the stress of the shaft is enlarged $2.35$ times when the motor is blocked.

$\sigma^' = 0.95 \sigma_s = 0.95 \times 360 = 342 \left( MPa \right)$

When the motor is blocked, the shaft is stressed

$\sigma^' = 0.95 \times \sigma_s =0.95 \times 360 = 342 \left( MPa \right)$

2.1.9.3 When under support load, the strength of the axis is checked

The spindle only bears the support load pull and the braking force is stationary, only the bending moment

The gypsy’s support load pull $f_branch$ is 294750 N

Horizontal force at the gypsy

$F_b_r_a_n_c_h_Z= F_branch \times \cos17^\circ=294750 \times 0.9563 = 281869 \left(N\right)$

Vertical force at the gypsy

$F_b_r_a_n_c_h_Y = F_branch \times \sin17^\circ=294750 \times 0.2924 =86185\left(N\right)$

Brake horizontal force $P_1 = 243434 \left(N\right), P_2 =36662 \left(N\right)$

Brake horizontal force

$F_b_r_a_k_e_Z = P_2 \times \cos49^\circ =36662 \times 0.656 =24050 \left(N\right)$

Brake vertical force

$F_b_r_a_k_e_Y =P_1- P_2 \times \sin49^\circ =243434 - 36662 \times 0.7547 =2157765\left(N\right)$

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Diameter 34mm electric anchor windlass calculation book

Table of Contents

I. Main parameters

II.Overall calculation

III. Worm gear box selection

IV. Open gear strength check

V. Strength check of shaft

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