A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to a 30-tooth worm gear. The gear has a transverse diametral pitch of 6 teeth/in and a face width of 1 in. The worm has a pitch diameter of 2 in and a face width of 2\frac{1}{2} in. The normal pressure angle is 14\frac{1}{2} °. The materials and quality of the gearing to be used are such that curve B of Fig. 13–42 should be used to obtain the coefficient of friction.
(a) Find the axial pitch, the center distance, the lead, and the lead angle.
(b) Figure 13–43 is a drawing of the worm gear oriented with respect to the coordinate system described earlier in this section; the gear is supported by bearings A and B.
Find the forces exerted by the bearings against the worm-gear shaft, and the output torque.
Question 13.10: A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to ...
(a) The axial pitch is the same as the transverse circular pitch of the gear, which is
p_x=p_t=\frac{\pi }{p}=\frac{\pi }{6}=0.5236 in
The pitch diameter of the gear is d_G = N_G/P = 30/6 = 5 in. Therefore, the center distance is
C=\frac{d_W+d_G}{2}=\frac{2+5}{2}=3.5 in
From Eq. (13–27), the lead is
L = p_xN_W (13.27)
= (0.5236) (2) = 1.0472 in
Answer Also using Eq. (13–28), we find
\lambda =\tan^{-1}\frac{L}{\pi d_W} (13.28)
=\tan^{-1}\frac{1.0472}{\pi (2)}=9.46°
(b) Using the right-hand rule for the rotation of the worm, you will see that your thumb points in the positive z direction. Now use the bolt-and-nut analogy (the worm is right-handed, as is the screw thread of a bolt), and turn the bolt clockwise with the right hand while preventing nut rotation with the left. The nut will move axially along the bolt toward your right hand. Therefore the surface of the gear (Fig. 13–43) in contact with the worm will move in the negative z direction. Thus, viewing the gear in the negative x direction, the gear rotates clockwise about the x axis
The pitch-line velocity of the worm is
V_W=\frac{\pi d_Wn_W}{12}=\frac{\pi (2)(1200)}{12}=628 ft/min
The speed of the gear is n_G=(\frac{2}{30} ) (1200) = 80 rev/min. Therefore the pitch-line velocity of the gear is
V_G=\frac{\pi d_Gn_G}{12}=\frac{\pi (5)(80)}{12}=105 ft/min
Then, from Eq. (13–47),
V_S=\frac{V_W}{\cos\lambda } (13.47)
the sliding velocity V_S is found to be
V_S=\frac{V_W}{\cos\lambda }=\frac{628}{\cos9.46°}=637 ft/min
Getting to the forces now, we begin with the horsepower formula
W_{Wt}=\frac{33 000H}{V_W}=\frac{(33 000)(1)}{628}=52.5 lbf
This force acts in the negative x direction, the same as in Fig. 13–40. Using Fig. 13–42, we find f = 0.03. Then, the first equation of Eq. (13–43)
W^x=W(\cos\phi_n\sin\lambda +f\cos\lambda )W^y=W\sin\phi_n (13.34)
W^z=W(\cos\phi_n\cos\lambda -f\sin\lambda )
gives
W=\frac{W^x}{\cos\phi_n\sin\lambda +f\cos\lambda }
=\frac{52.5}{\cos14.5°\sin9.46°+0.03\cos9.46°}=278 lbf
Also, from Eq. (13–43),
W^y=W\sin\phi_n=278\sin14.5°=69.6 lbf
W^z=W(\cos\phi_n\cos\lambda -f\sin\lambda )
=278(\cos14.5°\cos9.46°-0.03\sin9.46°)=264 lbf
We now identify the components acting on the gear as
W_{Ga} = -W^x = 52.5 lbf
W_{Gr} = -W^y = -69.6 lbf
W_{Gt} = -W^z = -264 lbf
A free-body diagram showing the forces and torsion acting on the gearshaft is shown in Fig. 13–44.
We shall make B a thrust bearing in order to place the gearshaft in compression. Thus, summing forces in the x direction gives
F^B_x= -52.5 lbf
Taking moments about the z axis, we have
-(52.5) (2.5) 2- (69.6) (1.5) + F^B_y = 0 F^B_y = 58.9 lbf
Taking moments about the y axis,
(264) (1.5) – F^B_z = 0 F^B_z = 99 lbf
Summing forces in the y direction,
-69.6 + 58.9 + F^B_y = 0 F^B_y = 10.7 lbf
Similarly, summing forces in the z direction,
-264 + 99 + F^B_z = 0 F^B_z = 165 lbf
We still have one more equation to write. Summing moments about x,
-(264) (2.5) + T = 0 T = 660 lbf . in
It is because of the frictional loss that this output torque is less than the product of the gear ratio and the input torque.
