Design of Head Stock: With Diagram | Machine Tools | Industrial Engineering

Now we proceed to the design of all geared head stock. We have seen how minimum, maximum and various other speeds, and number of steps are arrived at.

For obtaining most economical number of gears, the following sub-divisions should be followed for given number of steps.

Many different distributions are possible for obtaining the various sub-divisions. Selection of most suitable layout is determined by economy, simplicity of design and the optimum gear ratio and size.

The most commonly known distributions are:

(i) Unilateral i.e., when speeds of the follower monotonously decrease,

(ii) Bi-lateral when the speed of the input shaft is midway between output speeds,

(iii) Skewed, i.e., when the input shaft speed is off-centered from the middle position.

These distributions are shown in Figs. 11.43 to 11.45. Further depending upon the sub-division of the clusters, number of shafts and sizes of gears, the distribution may be of open form or crossed form.

The design of the all-geared headstock will be very clear from the problem discussed below:

Problem:

Design a kinetic arrangement of head stock and apron mechanism for a lathe having the following speed ranges, feed and screw cutting mechanism.

It is a HMT lathe with following specifications:

(a) Speed min—45 RPM, Speed max—2000 RPM, No. of steps — 18.

(b) Feed: No. of feeds from 0.05 mm/rev. to 1.4 mm/rev., steps: 35.

Solution:

Speed range N_max = 2000, N_min = 45, No. of steps Z = 18

Possible steps for minimum number of gears can be = 2 x 3 x 3 i.e., in the first step we can have two speeds and for each of these two speeds we can have 3 speeds in the second step, i.e., total six speeds. In 3rd step for each of these speeds we can have 3 more speeds, i.e., a total of 18 speeds.

If ɸ is the common ratio of the series of speeds (which is in G.P.)

The various intermediate speeds calculated from this value of ɸ are given below:

Intermediate speeds are calculated as:

As all the speeds and number of steps are known we can draw the ray diagram as shown in Fig. 11.46.

For stable operation, ‘range ratio’ of any stage should not be greater than 8. Synchronous speed of the motor that can be used = 1750 R.P.M.

So for reduction ratio from motor speed to 336, range ratio = 1750/346 = 5.2, which is less than 8.

1st stage range ratio = ɸ⁹ = (1.25)⁹ = 7.45 < 8

2nd stage range ratio = ɸ⁶ = (1.25)⁶ = 3.82

Another condition is that value of ɸ should lie between 1 and 2 which is also satisfied as ɸ = 1.25.

Hence our arrangement is correct.

Calculation for number of teeth:

If a, b, c, d, e, f,………………. etc. are the number of teeth as shown in Fig. 11.47, then speed ratio from these sets of gears is equal to teeth ratio.

These values are within the permissible error, so number of teeth calculated is correct.

As N’ – N = ΔN is very small, this design is correct.

Fig. 11.48 shows the arrangement of lead screw and feed mechanism. Fig. 11.49 shows the arrangement for apron mechanism.