A Copper shaft of diameter 145 mm and length 380 mm is to be...

Name: A Copper shaft of diameter 145 mm and length 380 mm is to be turned over the half of its length from its open end. After turning, facing operation is require to be carried out on its open end. Calculate the time required for total machining if cutting speed used is 16 meters/min, longitudinal feed used 1.2 mm/rev and cross feed used is 0.6 mm/rev. Ignore the tool change and setting time.
Uploaded: 2023-07-28T03:16:01-08:00
Duration: 4 min 39 s
Channel: Madhur L
Description: A Copper shaft of diameter 145 mm and length 380 mm is to be turned over the half of its length from its open end. After turning, facing operation is require to be carried out on its open end. Calculate the time required for total machining if cutting speed used is 16 meters/min, longitudinal feed used 1.2 mm/rev and cross feed used is 0.6 mm/rev. Ignore the tool change and setting time.

A Copper shaft of diameter 145 mm and length 380 mm is to be turned over the half of its length from its open end. After turning, facing operation is require to be carried out on its open end. Calculate the time required for total machining if cutting speed used is 16 meters/min, longitudinal feed used 1.2 mm/rev and cross feed used is 0.6 mm/rev. Ignore the tool change and setting time.

Transcript

00:01 Hi, in this question the internal diameter is given as 53 .5 mm, final diameter is given as 47 mm, initial length is given as 154 mm, feed rate is given as 0 .4 mm per revolution and we are given with the depth as 0 .5 mm.

00:21 Total required time for turning operation is equal to 59 .6 mm, tools approach length is 5 mm and we have overall run length is z mm.

00:44 Now we have to find the number of passes to finish the entire turning operation.

00:52 So, there is number of passes to the entire turning operation is given by it is total length of the cut divided by so depth of the cut.

01:10 So, here we have to find so first we have number of passes is given by so let us find total length of the cut.

01:21 So, total length of the cut is given by it is l into d1 minus d2.

01:29 So, on substituting the values we have 154 into 53 .5 minus 47 which is equal to 1001 mm.

01:40 So, now we have this is 1001 divided by 0 .5.

01:46 So, we have n is 2002 turns.

01:51 In the second part of the question we have to find the actual length of the turning operation.

01:56 So, actual length of turning operation.

02:02 So, this is equal to it is the l actual which is equal to total length of the cut plus tools approach length plus overall.

02:14 So, overall run length.

02:18 So, this is equal to it is 1001 plus 5 mm plus z.

02:24 So, now we are given with a time required for the complete operation as 59 .6 minutes...

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