Question 13.CS.9B: Design of a Wormset Speed Reducer for a Winch Lift Problem S...

See Figures 9-4 and 9-6.

1    A single-start worm will require a 75-tooth worm gear for the desired 75:1 ratio. This number of worm-gear teeth is well above the minimum recommended in Table 13-7 (p. 771).

2    Assume a center distance of 5.5 in for a trial calculation and find a suitable worm diameter based on that assumption from equation 13.16b (p. 771).

d \cong \frac{C^{0.875}}{2.2}      (13.16b)

d \cong \frac{C^{0.875}}{2.2} \cong \frac{5.5^{0.875}}{2.2}=2.02  in         (a)

3    Find a suitable worm-gear diameter from equation 13.17 (p. 771).

d_g=2 C-d      (13.17)

d_g=2 C-d=2(5.5)-2.02=8.98  \text { in }       (b)

4    Find the lead from equation 13.13 (p. 769).

p_x=\frac{L}{N_w}=p_c=\frac{\pi d_g}{N_g}     (13.13)

L=\pi d_g \frac{N_w}{N_g}=\pi(8.98) \frac{1}{75}=0.376  \text { in }           (c)

5    Find the lead angle from equation 13.12 (p. 769).

\tan \lambda=\frac{L}{\pi d}       (13.12)

\lambda=\tan ^{-1} \frac{L}{\pi d}=\tan ^{-1} \frac{0.376}{\pi(2.02)}=3.39^{\circ}          (d)

This is less than 6°, so the wormset will be self-locking, as required.

6    Find the maximum recommended face width from equation 13.19 (p. 771).

F_{\max } \leq 0.67 d     (13.19)

F_{\max } \cong 0.67 d=0.67(2.02)=1.354  \text { in }      (e)

7    Find the materials factor C_{s} from equation 13.24 (p. 772). Since C < 8 in, C_{s} = 1 000.

8    Find the ratio-correction factor C_{m} from equations 13.25 (p. 772). Based on m_{G} = 75, the second of the expressions in that equation set will be used.

C_m=0.0107 \sqrt{-m_G^2+56 m_G+5145}=0.0107 \sqrt{-75^2+56(75)+5145}=0.653        (f)

9    Find the tangential velocity V_{t} from equation 13.27 (p. 773).

V_t=\frac{\pi n d}{12 \cos \lambda} fpm    (13.27)

V_t=\frac{\pi n d}{12 \cos \lambda}=\frac{\pi(1725)(2.02)}{12 \cos \left(3.392^{\circ}\right)}=913.9  fpm        (g)

10    Use this velocity to find the velocity factor C_{\nu} from equations 13.26 (p. 772). For this value of V_{t}, the second of these equations is appropriate.

\begin{array}{l} \text {if}  0<V_t \leq 700 fpm \quad C_{\nu}=0.659 e^{-0.0011 V_t}\\ \text {if}  700<V_t \leq 3000 fpm \quad C_{\nu}=13.31 V_t^{-0.571}\\ \text {if}  3000<V_t \text { fpm } \quad C_{\nu}=65.52 V_t^{-0.774} \end{array}     (13.26)

C_{\nu}=13.31(913.9)^{-0.571}=0.271      (h)

11    Find the tangential load W_{t} from equation 13.23 (p. 772).

W_{t g}=C_s C_m C_{\nu} d_g^{0.8} F    (13.23us)

W_{t g}=C_s C_m C_{\nu} d_g^{0.8} F / 75.948      (13.23si)

W_{t g}=C_s C_m C_{\nu} d_g^{0.8} F=1000(0.653)(0.271)(8.98)^{0.8}(1.354)=1388  lb         (i)

12    Find the coefficient of friction from the third expression in equation 13.29 (p. 773).

\mu=0.103 e^{\left(-0.110 V_t^{0.450}\right)}+0.012=0.103 e^{\left(-0.110[913.9]^{0.450}\right)}+0.012=0.022          (j)

13    Find the friction force W_{f} from equation 13.28 (p. 773).

W_f=\frac{\mu W_{t g}}{\cos \lambda \cos \phi_n}       (13.28)

W_f=\frac{\mu W_{t g}}{\cos \lambda \cos \phi}=\frac{0.022(1388)}{\cos 3.392^{\circ} \cos 20^{\circ}}=32  lb           (k)

14    Find the rated output power from equation 13.21 (p. 772).

\Phi_o=\frac{n W_{t g} d_g}{126000 m_G} hp     (13.21us)

\Phi_o=\frac{n W_{t g} d_g}{1.91 E 7 m_G} kW     (13.21si)

\Phi_o=\frac{n W_{t g} d_g}{126000 m_G}=\frac{1725(1388)(8.98)}{126000(75)}=2.274  hp       (l)

15    Find the power lost in the mesh from equation 13.22 (p. 772).

\Phi_l=\frac{V_t W_f}{33000} hp       (13.22us)

\Phi_l=\frac{V_t W_f}{1000} kW     (13.22si)

\Phi_l=\frac{V_t W_f}{33000}=\frac{913.9(32)}{33000}=0.888  hp       (m)

16    Find the rated input power from equation 13.20 (p. 772).

\Phi=\Phi_o+\Phi_l      (13.20)

\Phi=\Phi_o+\Phi_l=2.274+0.888=3.162  hp          (n)

17    The efficiency of the gearset is

\eta=\frac{\Phi_o}{\Phi}=\frac{2.274}{3.162}=71.9 \%          (o)

18    Find the rated output torque from equation 13.31 (p. 773).

T_g=W_{t g} \frac{d_g}{2}        (13.31)

T_g=W_{t g} \frac{d_g}{2}=1388 \frac{8.98}{2}=6230  lb – in       (p)

19    While the power rating appears to be adequate for this application, the output torque rating falls short of the projected peak torque of 7 800 lb-in modeled in Case Study 9A; thus some redesign is in order.

20    The original assumption for center distance was increased to 6.531 in and the model recalculated. The center distance was also adjusted slightly to give an integer diametral pitch of 7 in^{–1}. This increased the worm-gear diameter to 10.714 in and the output torque rating to 9 131 lb-in. The new input power rating is 4.52 hp and the power loss is 1.18 hp for an efficiency of 73.8%. The output power rating is 3.33 hp. The new lead angle is 3.48°, so the wormset is still self-locking.

21    While this new design appears feasible based on the loading calculations done in the previous case study, one of the original assumptions regarding the electric motor size will need to be revised. The average net power required was estimated to be 0.62 hp. It was hoped that a 1- to 1.25-hp motor would be adequate, which would allow 110-V operation. This now appears impossible due to the 1.18-hp loss in the wormset, which would leave too little power available to lift the load even if a 1.25-hp motor were used.

The flywheel effect of the rotating winch drum can supply bursts of energy to get past the peaks of load oscillation shown in Figure 9-6, but cannot provide a sustained increase in power above the average available. So a 220-V motor of about 2 or 2.25 hp appears to be necessary for this design. The input power rating of the gearset should easily accommodate that level of power with no overheating problems.

22    The files CASE9B-1 and CASE9B-2, respectively, contain the first (unsuccessful) and second (successful) solutions to this problem and are on the CD-ROM.