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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/mm2.
Solution Given data:
Hole diameter =(D)= 50mm
Length of workpiece = (L)=100mm
Permissible strength = (σs)=25Kgf/mm2
1] Selection of rise per tooth or cut tooth(S2)
Sz=0.02 to 0.05mm General value of Sz
Selecting Sz=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)√L
A=0.005D+0.1√100=1.25
A=0.005(50)+0.2√100=2.25
Selecting A = 2.0 mm
Diameter of pre-machined hole = 50 - 2
= 48 mm
3] Number of cutting teeth (Zc)
Cutting teeth comprises of roughing teeth and semi finishing teeth
Zca2SZ (2 to 3) teeth
Zc=22∗0.04+3=28 teeth
No. of rough teeth Zr=28−5=23 teeth
1] Design of tooth profile
- Gullet design
i) K=filling factor or chip space ratio
K = 3 to 6
Selecting K=5
ii) Area of underformed chip =SzL
Area of gulled =π/4h2
= π4h2=KSzL
= π4h2=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 10o
Selecting r=8o
7] Clearance or relief angle (α)
α=3o for rough teeth
α=2o for semi finish teeth
α=1o for finish teeth
8] No. of finishing teeth
nf= 3 to 6
Selecting n = 3
5] Cutting teeth diameter
Total No. of teeth =nf+nf+nf
z = 23+5+3 =31
Diameter of 1st tooth (D1)
D1=D0−overcut
D0=48mm
D1=48−0.005
= 47.995 mm
In case of broaching overcut allowance varies from 5μm to 10μm
6] Diameter of last cutting teeth diameter cutting teeth includes only rough and S.F. teeth & finish teeth
D31=50H8−(0.005 to 0.01)
= 50.027 - 0.005
= 50.022 mm
D1=47.995
D2=D1+2Sz=47.995+2∗0.04=48.075mm
D2=D2+2Sz=48.075+2∗0.04=48.155mm
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 (ncb)
No. of chip breaker = 20 to 30
Increase in diameter increases the no. of chip breaker
Let ncb =24
Pitch of chip breaker
= π∗Davgncb
= π4924
= 6.4 mm
hcb= 2 to 3 mm
rcb= 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
=10013=7.69=8 (Always round to next number)
ii) Weakest section
= Cross section area of 1st cutting tooth root diameter
=(D1−2h)2π/4
=[48−2(5)]2∗π/4
=382∗π/4
=1134.11mm2
iii) Broaching force
= Q x 2l
Q = cutting force per unit length = 20 Kgf/mm
Broach force = 20∗n∗π∗Davg
=20∗8∗π∗49
= 24630 Kgf
Stress induced = broaching force / Weakest section
= 246301134.114
= 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 = ϕ (Finished hole) = ϕ 50H8
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