written 8.3 years ago by | modified 2.8 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Production Process II
Marks: 10M
Year: May 2014, Dec 2014
written 8.3 years ago by | modified 2.8 years ago by |
Mumbai University > Mechanical Engineering > Sem 4 > Production Process II
Marks: 10M
Year: May 2014, Dec 2014
written 8.3 years ago by | • modified 8.3 years ago |
Design Procedure of an Internal broach tool:
a) Details on a internal broach tool:
A typical internal broach is shown in the figure;
b) Determination of maximum and minimum hole sizes;
Assume desired hole 45H7
D=geometric mean diameter whose ranges are 1-3, 3-6, 6-10, 10-18, 18-30, 30-50, 50-80, 80-20.
45 fall in the range of 30-50; $D= \sqrt{30 x 50}$ =38.73mm
i=0.001D + 0.45 $\sqrt[3]{D}=1.56$ microns
tolerance=IT7 (as hole size has 7)
IT7=16i=24.98micron=0.025mm
Now calculate the maximum and minimum size of the hole as basic size of 45 mm is given.
c) Maximum broaching allowance;
It is defined as the total thickness of the metal to be removed by broaching. It is given by:
Ab=0.005Db +(0.1-0.2) √L ; L=length of the hole to be broached.
$\hspace{4.2cm}$ Db= basic diameter of the hole.
d) Rise per teeth;
It is selected from the following table
Let rise per tooth for roughing and semi-finish teeth be 0.03 and 0.015 mm respectively. (Assume)
e) Number of semi finish teeth;
$= \frac{0.1 × A_b}{2 x 0.015}$
round this value to the next integer value if in point form.
f) No of roughing teeth;
=(A_b-4[2 × 0.015])/(2×0.03)
Always round this value to the next higher value.
g) Pitch;
i) for roughing teeth and semi finishing teeth=1.3√L =pr
ii) for finishing teeth=0.6 x 1.3√L = pf
h) No of teeth simultaneously in action;
$Z_{max} = \frac{L}{p_r} +1$
this value should be greater than 3 for proper action of broach.
i) Chip space ratio;
$kc= {\text{Gullet area}}{\text{Area of longitudinal section of chip}}$ kc=6 (Assume)
Area of longitudinal section of chip=L x Max rise per tooth
therefore we get gullet area which is given by the formula
$A_g=0.5 x p_r x h_c$
and hence hc (depth of chip space)can be found out.
j) Land width (W);
i) For roughing an semi finishing teeth=0.3 x pr
ii) for finishing teeth=0.3 x pf
k)Rake angle;
Assume rake angle 15 degree and 5 degree for roughing and semi finish teeth respectively.
i) Primary and secondary clearance angle;
Type of teeth | PCA | SCA |
---|---|---|
Roughing | 3 | 35 |
Semi finish | 2 | 25 |
Finishing | 1 | 15 |
l) Chip Breakers;
Dimensions are 2mm wide and 1.5 mm deep usually.
They are used to break wide chips to several narrow chips.
m) Fillet radius;
i) For roughing an semi finishing teeth=0.3 x pr
ii) ) for finishing teeth=0.3 x pf
n) Force;
$F=e_c x D_b x Z_{max} x K x \text{max rise per tooth} x π \hspace{1cm} ; ec=4250 N/mm^2$ $\hspace{9cm}$ k=1.4
o) Power for Broaching = $\frac{F ×V}{4500}$ ; V for broaching is 8 m/min
$\hspace{9cm}$ F in kgf
p) length of broach;
=length of the toothed portion + length of shank + length of rear pilot length of toothed portion= length of roughing and semi finish teeth + length of finishing teeth = pr x total number of teeth + pf x 5
length of shank = 250mm assume
length of rear pilot= length of hole