For a turning operation, you have selected a high-speed steel (hss) tool and turning a hot rolled free machining steel. your depth of cut will be 0.105 in. the diameter of the workpiece is 1.00 in. a. using a speed of 105 sfpm and a feed of 0.015, calculate the metal removal rate (mrr). b. then calculate the cutting time for the operation with a length of cut of 4 in. and 0.10-in. allowance.

a. Material Removal Rate = 1.9845 in3/min, b. Cutting Time = 40.8 seconds

Explanation:

a. Material Removal Rate is defined as MRR = vfd, where 'v' is the cutting speed, 'f' is the feed, 'd' is the depth of the cut. Substituting given values we get

MRR = 105 x 12 x 0.015 x 0.105 = 1.9845in³/min (Multiplying by '12' to give value in inches)

b. Cutting Time is defined as Tm= L/FrNs, where 'L' is the length of the cut including allowances, 'Fr' is the feed rate and 'Ns' is the spindle rotational speed

To find out Ns we use the formula Ns=V/πD₀, where 'V' is the cutting speed and D₀ is the original diameter. Substituting the values given

Ns= 105 x 12/ π x 1 = 401rpm

Using this we get Tm

Tm = (Length of cut + Allowance)/ Feed Rate x Spindle rotational speed

Tm = (4 + 0.1)/ 0.015 x 401 = 0.68 min = 40.8 s

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

glad i could , give me a follow, so if you need with any other question, im here.