0
1.2kviews
Broach Design :
1 Answer
0
25views

Design a circular broach pull type to machine a hole of diameter 5048. The length of workpiece is 100mm. The broach is of HSS material. Permissible strength is not to exceed 25 $Kgf/mm^2$.

Solution Given data:

Hole diameter =(D)= 50mm

Length of workpiece = (L)=100mm

Permissible strength = $(\sigma_s) = 25Kgf/mm^2$

1] Selection of rise per tooth or cut tooth$(S_2)$

$S_z = 0.02 \ to \ 0.05 mm $ General value of $S_z$

Selecting $S_z =0.04 \ mm$

2] Broaching Allowance(A)

Allowance implies the amount of material to be removed i.e. to determine the size of pre-machined hole

$A=0.05D + (0.1 \ to \ 0.2)\sqrt L$

$A=0.005D +0.1\sqrt{100} =1.25$

$A=0.005(50) +0.2\sqrt{100} =2.25 $

Selecting A = 2.0 mm

Diameter of pre-machined hole = 50 - 2

= 48 mm

3] Number of cutting teeth $(Z_c)$

Cutting teeth comprises of roughing teeth and semi finishing teeth

$Z_c \frac{a}{2S_Z}$ (2 to 3) teeth

$Z_c = \frac{2}{2 * 0.04} +3 = 28$ teeth

No. of rough teeth $Z_r =28-5 =23$ teeth

1] Design of tooth profile

  • Gullet design

enter image description here

i) K=filling factor or chip space ratio

K = 3 to 6

Selecting K=5

ii) Area of underformed chip $= S_zL$

Area of gulled $=\pi / 4h^2$

= $\frac{\pi}{4}h^2 = K S_zL$

= $\frac{\pi}{4}h^2 = 5 * 0.042188$

h = 511 mm

ii) Pitch of rough died teeth (K)

p = (2.5 to 2.8)h

p = 2.5 * 5.1 = 12.75 (min)

p = 2.8 * 5.1 = 14.28 mm (max)

p = 13 mm

4] Land width (i.e. width of teeth) (f)

f = (0.3 to 0.)p

f = 0.3 x 13 =3.9 mm (min)

f = 0.4 x 13 = 5.2 mm (max)

f = 5 mm

5] Sullet ratius(r)

r = (0.2 to 0.25)p

r = 0.2 x13 = 2.6 mm (min)

r = 0.25 x 13 = 3.25 mm (max)

r = 3mm

6] Rake angle (r)

r = 6 to $10^o$

Selecting $r=8^o$

7] Clearance or relief angle $(\alpha)$

$\alpha = 3^o$ for rough teeth

$\alpha = 2^o$ for semi finish teeth

$\alpha = 1^o$ for finish teeth

8] No. of finishing teeth

$n_f = $ 3 to 6

Selecting n = 3

5] Cutting teeth diameter

Total No. of teeth $ = n_f+n_f+n_f$

z = 23+5+3 =31

Diameter of 1st tooth $(D_1)$

$D_1 = D_0 -overcut$

$D_0 = 48mm$

$D_1 =48-0.005$

= 47.995 mm

In case of broaching overcut allowance varies from $5 \mu m \ to \ 10\mu m$

6] Diameter of last cutting teeth diameter cutting teeth includes only rough and S.F. teeth & finish teeth

$D_31 = 50H_8 -(0.005 \ to \ 0.01)$

= 50.027 - 0.005

= 50.022 mm

$D_1 = 47.995$

$D_2 = D_1+2S_z = 47.995 + 2 * 0.04 = 48.075 mm$

$D_2 = D_2 +2S_z = 48.075 + 2 * 0.04 = 48.155 mm$

7] Cutting length of broach

= P x no. of teeths

= 13 x 28 = 364 mm

8] Details of broach finishing teeth

Np. of finish teeth = 3

Length of finish teeth

= 3 x 13

= 39 mm

9] Design of breakers $(n_c{}_b)$

No. of chip breaker = 20 to 30

Increase in diameter increases the no. of chip breaker

Let $n_cb$ =24

Pitch of chip breaker

= $\frac {\pi * D_avg}{n_cb}$

= $\frac{\pi 49}{24}$

= 6.4 mm

$h_c{}_b = $ 2 to 3 mm

$r_c{}_b= $ 1 to 2 mm

10] Stress induced

i) No. of teeth taking the cut simultaneously

n = Length of hole to be machined / pitch of broach teeth

=$\frac{100}{13} = 7.69 = 8 $ (Always round to next number)

ii) Weakest section

= Cross section area of 1st cutting tooth root diameter

=$(D_1 - 2h)^2\pi /4$

=$[48-2(5)]^2 * \pi/4$

=$38^2 * \pi / 4$

=$1134.11 mm^2$

iii) Broaching force

= Q x 2l

Q = cutting force per unit length = 20 Kgf/mm

Broach force = $20 * n * \pi * D_avg$

=$20 * 8 * \pi * 49$

= 24630 Kgf

Stress induced = broaching force / Weakest section

= $\frac{24630}{1134.11}4$

= 21.71 Kgf/mm <permissible stress, hence soft</p>

11) Dimensions of other portion of broach

i) pull end diameter = pilot hole diameter - (1 to 1.5 mm)

= 48 - 1.5

= 46.5 mm

ii) Neck diameter = Pull end dia -(1 to 2 mm)

= 46.5 - 1

= 45.5 mm

iii) length of neck = diameter of neck

= 45mm

iv) Front pilot diameter = (Diameter before broaching)Hs

= 48 Hs

v) Front pilot length = length of w/p

= 100mm

vi) Rear pilot length = $\ \phi \ (Finished \ hole) \ = \ \phi \ 50H_8$

vii) length of rear pilot = (0.6 to 0.8 )L

= 0.8 x 100

= 80 mm

12] Total length on pull broach

Shank length = 200 mm (Assuming)

Front pilot length = 100 mm

cutting teeth length = 364 mm

Finish teeth length = 39 mm

Rear pilot length = 50 mm

Total length = 783 mm

13] Find the permissible length of broach

Length of broach must not exceed 40 times the diameter of hole to bebroached

L permissible <= 4 x dia of hole

783 <= 40 x S

783 < 2000

The broach is sufficiently rigid

Please log in to add an answer.