Hello , I typed out the problem. "Materials and processes in manufacturing "
9th edition degarmo. Page 582 case study.
Thanks
Estimating machining time for turning
You are using a 1” micrometer.
Katin would like to know the minimum time required to machine a large forging.
The 8 ft long forging is to be turned down from an original diameter of 10” to a
final diameter of 6”. The forgin has a BHN of 300-400. The turning is to be
performed on a heavy duty lathe, with a 50 hp motor and continuously variable
speed drive on the spindle. The work will be held down between centers , and
the overall efficiency of the late is 75%.
The log is made from medium carbon 4345 alloy steel. The steel manufacturer,
some basic experimentation, nad established knowledge of the product and its
manufacture have provided the following info:
a tool life equation developed for the most suitable type of tool material at a
feed of .02ipr and a rake angle of alpha = 10 degrees. The equation VT^n = C
generall fits the data, with V= cutting speed and T = time in minutes to tool
failure.
Two test cuts were run, one at V = 60sfpm
Where T = 100 min and another at V = 85 sfpm where T = 10 min
The dynamic shear strength of the material is on the order of 125,000 psi. Jay
decides to make two test cuts at the standard feed of .02 ipr. He assumes that
the chip thickness ratio varies almost linearly between the speeds of 20 and 80
fpm, the values being .4 at the speed of 20 fpm and .6 at 80 fpm. The chip
thickness values were determined by micrometer measurements to determine the
value of Rc ( Rc is chip ratio )
The log will be used as a roller in a newspaper press and must be precisely
machined. If the log deflects during the cutting more than .005” the roll will
end up barrel shaped.
How should I proceed to estimate the minimum time required to machine this
forgin, assuming that one finishing pass will be needed when the log has been
reduced to 6” in diameter? The deflection due to cutting forces must be kept
below .005” at the mid log location.
Assume Fc * .5 = Ff and Ft * .5 = Fr and that Fr causes the deflection
9th edition degarmo. Page 582 case study.
Thanks
Estimating machining time for turning
You are using a 1” micrometer.
Katin would like to know the minimum time required to machine a large forging.
The 8 ft long forging is to be turned down from an original diameter of 10” to a
final diameter of 6”. The forgin has a BHN of 300-400. The turning is to be
performed on a heavy duty lathe, with a 50 hp motor and continuously variable
speed drive on the spindle. The work will be held down between centers , and
the overall efficiency of the late is 75%.
The log is made from medium carbon 4345 alloy steel. The steel manufacturer,
some basic experimentation, nad established knowledge of the product and its
manufacture have provided the following info:
a tool life equation developed for the most suitable type of tool material at a
feed of .02ipr and a rake angle of alpha = 10 degrees. The equation VT^n = C
generall fits the data, with V= cutting speed and T = time in minutes to tool
failure.
Two test cuts were run, one at V = 60sfpm
Where T = 100 min and another at V = 85 sfpm where T = 10 min
The dynamic shear strength of the material is on the order of 125,000 psi. Jay
decides to make two test cuts at the standard feed of .02 ipr. He assumes that
the chip thickness ratio varies almost linearly between the speeds of 20 and 80
fpm, the values being .4 at the speed of 20 fpm and .6 at 80 fpm. The chip
thickness values were determined by micrometer measurements to determine the
value of Rc ( Rc is chip ratio )
The log will be used as a roller in a newspaper press and must be precisely
machined. If the log deflects during the cutting more than .005” the roll will
end up barrel shaped.
How should I proceed to estimate the minimum time required to machine this
forgin, assuming that one finishing pass will be needed when the log has been
reduced to 6” in diameter? The deflection due to cutting forces must be kept
below .005” at the mid log location.
Assume Fc * .5 = Ff and Ft * .5 = Fr and that Fr causes the deflection