Discussion 1
The corners of the chisel that affect the cutting force is:
A. Free orthogonal angle (α0)
Free angle function is to reduce friction between the main field (Aα) with the transient field of the workpiece. The price of free elections angle determined by the type of the workpiece and cutting conditions. Ate motion (f) will determine the price point free. The greater the motion of eating the cutting force will be greater
2. Growled orthogonal angle (γ0)
Angle affects the formation of furious anger. For a given cutting speed, large angle furious compression ratio will decrease resulting in a thick angry γh increase shear angle Φ. Large shear angle Φ will decrease the cross-cutting shear Ashi so the style will go down. To reduce the large cutting force resulting from the use of negative angle so furiously cutting speeds should be high (in order to decrease the strength of the workpiece).
3. Torsional angle
Because the torsional angle equal to the angle on the outer side growled then the function will be similar to a chisel point on the lathe furious, furious that affect the formation process. The larger the twist angle will mrnurunkan style that cuts torque and compression force will decline.
The corners of the chisel that affect the smoothness of the surface is:
A. Corner radius (re)
Radius corner serves to strengthen the cut end of the meeting between the eyes with the eye's main minor cut S 'and in addition determine the smoothness of the surface of the cutting .. Machining the surface smoothness is determined by the formula:
Rt = (Cr f2) / 8re;
Where:
Rt = total roughness of the surface roughness parameter (μm)
motion f = feed (mm / r)
re = radius corner (mm)
Cr = conversion factor, depending on the nature ketermesinan workpiece, cutting conditions are selected (cutting speed) and the stiffness of the cutting system;
Kaku = 2000
Medium = 2300
Weak = 3000
The formula above states that the greater the corner radius is chosen then the smoothness of the surface may be enhanced.
2. Oblique angle
When the oblique angle value 0, then the furious flow perpendicular to the eye piece.Would make a furious stream of ρc angle to the line perpendicular to the eye piece and according to Stabler furious flow angle is approximately equal to the oblique angle λs.Given the oblique angle, the length of contact between the cutting tool with the workpiece to be more extended and specific cutting energy Esp will not change until reaching 20o oblique angle. The temperature of the contact area will reach a minimum price when λs + 5o valuable for refining process (finishing) and-5o in the hardening process (roughing).
3. Corner of the Bantu (k'r)
Orientation and the auxiliary field A'α cut against the surface of the workpiece has been cut off aid is determined by the angle cut and angle k'r α'o minor. If the minor α'o free angle is large enough to reduce friction, in principle, the corner pieces as small as possible aids k'r chosen because in addition to strengthening the chisel edge (tip angle is enlarged), the fineness of the product can be enhanced, which is a constraint is the stiffness of the workpiece cutting systems, because of the small corner pieces will enhance the force Fx radians as general pointer cutting the rigid system k'f = 5o - 10o and cutting systems are weak, k'f = 10o - 20o to the corner radius r is small, the surface smoothness parameter Rt ( total roughness) were determined simultaneously by k'f, kf and as the following formula:
Rt = fc / (kr + cot cot k'r)
Where:
Rt = total hardness (μm)
The main pieces of Kf = angle (o)
K'f = corner cut aid (o)
F = feed motion (mm / r)
C = conversion factor> 1000, depending on the nature of the workpiece ketermesinan terms of surface roughness and cutting conditions (cutting speed).
Discussion 2
Problem 1
Note:
A shaft of steel material AISI 1020 (200 BHN) has a cutting speed rpm ......
Initial diameter (d0) = 2 in
Final diameter (d1) = 1.9 in
Were:
Rotary speed (S)?
Answer:
Large spindle rotation speed can be obtained by using the formula below:
A. For inch size
Df S = π (rpm) / 12
2. For metric sizes
Df S = π (rpm) / 1000
We chose the first formula, as is known in the matter of using units of inches.
df = (d0 + d1) / 2
= (2 + 1.9) / 2
= 3.9 / 2
= 1.95 in
Df S = π (rpm) / 12
= (3.14) (1.95) (20) / 12
= 122.46 / 12
= 10.205 sfpm (surface feet per minute)
Df S = π (rpm) / 12
= (3.14) (1.95) (176.38) / 12
= 1079.97 / 12
= 89.99 sfpm (surface feet per minute)
Problem 2
Note:
Cutting length (L) = 4 in
Ingestion (S) = 0.001 in / round
Were:
Cutting time (T)?
Answer:
Cutting the time can be obtained by using the formula below:
A. For inch size
T = π df L / (12 FS)
2. For metric sizes
T = π df L / (1000 FS)
We chose the first formula, as is known in the matter of using units of inches.
T = π df L / (12 FS)
= (3.14) (1.95) (4) / (12) (0.001) (10.205)
= 24.492 / 0.12246
= 200 min
No comments:
Post a Comment