Diameter 34mm electric anchor windlass calculation book

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    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

    1. 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

    torque value when the motor is blocked

        \[ P_b_r_a_n_c_h * 1380 = R_A * 1760 \]

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

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

    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 coefficientY_\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

    1. 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

    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 12749Nm, 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-12471Nm 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 enlarged2.35 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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