Shank Size

The main design criterion for shank size is rigidity. The deflection at the cutting edge is restricted to a certain value based on the size of machine, cutting conditions and tool overhung. The tool overhung (L₀) is associated also to the shank size as well as to the end support conditions. Figure depicts graph of the amplitude and frequency of chatter for various overhung values. It is seen from Figure that only below L₀/H = 2, the amplitude is practically zero. The recommended value of (L₀/H) lies between 1.2 and 2. For the given value of chatter frequency f, the shank deflection can be determined from the (Eq. 1) given as follows.

f = (15.76) / √d   c.p.s                              (1)

where, d is deflection in mm.

Now as chatter frequency ranges from 80 to 160 c.p.s.,

Let       f = 100 c.p.s

d = (15.76/100)2 ≅ 0.025 mm                         (2)

The permissible deflection of shanks is from 0.025 mm for finish cuts to 0.9 mm for rough cuts. Assuming shank as a cantilever,

d = F(L₀)³ /3EI

d = F(L₀)³ (12/ BH ³ ) = 4F(L₀)³ /EBH ³            (3)

= 0.025 mm

It can be noted down that the same value of d has been achieved from Eq. (2.2) also.

The shank size can be determined with respect to machine tool size by the following method:

 (a) The force F for specified size of lathe is given by

F = f × t × C

where,  f refer to feed in mm,

t refer to depth of cut in mm, and

C refers to cutting force constant.

(b) Nicolsons Manchester experiments have set a standard area of cut for lathe design given by

A_c = f × t

Let, f = h_c/180 mm and t = h_c/25 mm

A c  =  (h_c /180 )×  ( c/25)

=  (h_c )² /4500  mm²

where, h_c is height of centre in mm,

Let,  σ_ut = 440 N/mm²

C = 4σ_ut

= 4 × 400 = 1760 N/mm²

When F = (h_c)² /4500  mm² × 1760 N/mm²

= 0.4 (h_c)²   N

On substituting the value of F = 0.4(h_c)²    in Eq. (3), we get

 d = 4(0.4 (h _c )² ) (L₀ )³ /EBH ³

0.025 =4(0.4 (h _c )² ) (L₀ )³ /EBH ³       

As, d = 0.025 from Eq. (2.2). Thus,

= 1.6 (h _c )² ) (L₀ )³ /EBH ³

B = 0.6 H for rectangular shanks

Then, (h _c )² /H⁴    = 0.6 ED/(L₀ )³

Let  L₀  = 3 mm, E = 200 kN/mm²

and d = 0.025 mm, (From Eq. (2))

On substituting these values in above equation, i.e. (h _c )² /H⁴    = 0.6 ED/(L₀ )³  we get

(h _c )² /H⁴    = 1000m^-2

Table depicted the standard shank size according to this rule.

                                                                                 Table

Usually the shank size is also verified for strength.

Noting,

FL₀  = (1/6) BH ²  σ₁

∴ σ =   6FL₀ /BH²

When the effect of F_x included,

σ= σ ₁+ σ ₂ = (6FL₀  / BH²  )+ (6 FL₀    / HB² )                  (4)

F_x = Component of force F working in x direction (in Newton)

F_x = 0.3 to 0.40 F

Hence,

σ =       6FL₀  (0.4 /B) + (1/H) < σ_per                                 (5)

This can be expressed as

F = { BH/(0.4/B) + (1/H)} σ_per/6L₀                                     (6)

where, F refer to permissible tangential force during machining. The maximum depth of shank (H_max) has to be less than the value h_k as illustrated in Table                                              

                                                                                             Table