ESTIMATION OF MACHINING TIME
INTRODUCTION: Machining time:
It is the time for which the machine works on the component that is from the time when the tool touches the work to when the tool leaves the component after completion of the operations.
Terms used in the study of machining time: 1. Length of cut
Length of cut = approach length + length of work piece to be machined + over travel 2. Feed
3. Depth of cut 4. Cutting speed
Cutting speed = Πdn / 1000m / min
CALCULATION OF MACHINING TIME FOR LATHE OPERATION: 1. Turning operation:
Turning is the process of removing excess material from the work piece by means of a single point cutting tool.
T= L / (F×N)
2. External relief turning:
It is the operation of removal of material from a previously turned surface along the same axis and within the limits of turned area.
T = L/ (F×N) × NUMBER OF CUTS 3. Chamfering:
It is the operation of removing material from the edges of external or internal diameters. T= L / (F×N)
4. Facing:
It is the process of removing material from the ends of the job by moving the tool perpendicular to the axis of the job.
5. Knurling:
It is the operation of upsetting material so as to produced diamond shaped or straight lined patterns on the surface of the material.
T = length of cut / feed
6. Drilling: It is the process of producing holes in a material.
7. Boring: It is the operation of enlarging or finishing an internal hole which has been previously drilled.
8. Reaming: It is the process of removing a small amount of material from a previously drilled hole.
9. Taping: It is the operation of cutting internal threads with the help of a tool called tap.
10. Threading:
It is the process of removing material to produce helix on external or internal circular surface for fastening purposes.
ESTIMATION OF MACHINING TIME FOR SHAPING, PLANNING AND SLOTTING OPERATION:
Formula used:
1. Effective cutting speed (c)
C = L / 1000 × N Where, L = length of forward stroke
N = number of forward stroke per minutes 2. Cutting speed (s)
If k = time for return stroke / time for forward stroke S = L (1+K) / 1000 × N
3. Total time for one cut (T)
T = L / (S×1000) + (L×K) /S ×1000 4. Total time required
If p is the number of cuts required then
T = L (1+K) / S×1000× (W/F) ×P
ESTIMATION OF TIME FOR MILLING OPERATION:
Milling is the operation of removing material from a surface by using a rotary multi point tool called cutter.
Formula used
Milling time: (T = L / F×N)
1. To find feed per revolution
Feed per revolution = feed per tooth × number of cutter teeth 2. To find the time taken per cut
Time taken per cut = (length of job + added table travel) / (feed per revolution × rpm) Added table travel = ½ × (D- (D^2-W^2) ^ (1/2))
ESTIMATION OF TIME FOR GRINDING OPERATIONS: Grinding is the process of metal removal by abrasion.
Formula used
Grinding time per cut = length of cut / (feed per revolution × rpm) Length of cut, L = L0 – w + 5
Total grinding time = (L0 – w +5) / (w/2or w/4) × N ×Number of cut required Rpm is given by N = 1000×S / 3.14 × D
Number of passes, n = d/t
PROBLEMS: Example: 1
Estimate the machining time to turn a 4 cm diameter MS bar 10cm long , down to 3.5cm diameter in a single cut, using high speed steel tool. Assume the cutting speed of the tool to be 30m/min and a feed of 0.30mm/rev.
Solution:
N = 1000s / 3.14 D = 1000 ×30 / 3.14 × 40 = 238.7rpm T = L / F×N = 1000 / .3 × 238.7 = 1.39min
Example: 2
An operator is required to tap a hole with 20mm dia tap to length of 35mm having 3 threads per cm. The cutting speed is 10m/min and the number of cut required is 4. The return speed of the tap is to be 3 times the cutting speed. How much time will be taken to cut these internal threads?
Solution:
N = 1000s / 3.14 D = 1000 × 10 / 3.14 ×20 = 159.15rpm Pitch = 1/ threads per cm = 1/ 3 = 3.33mm
Time for taping per cut, T1 = L + (D/2) / (Pitch × rpm) = 0.27min Return speed = 3 × cutting speed
Return time, T2 = 1 / 3 tapping time = 1/3 × 0.27 = 0.092min Total time for one cut = T1 + T2 = 0.27 + 0.092 = 0.369min
Total time for tapping = 0.369 × 4 = 1.47min
Example: 3
A MS bar of 120mm long and 40mm diameter is turned to 38mm diameter and was again turned to a diameter if 35mm over a length of 50mm as shown in figure. The bar was chamfered at both the ends to give a chamfer of 45 degree×4mm after facing. Calculate the machining time. Assume cutting speed of 50m per min and feed 0.3mm/rev the depth of cut is not to exceed 3mm in any operation.
FIGURE 11.16 Solution:
1. First operation: turning from 40mm diameter to 38mm diameter and 120mm long N = 1000 × 50 / 3.14 × 40 = 397.89rpm
Time taken for turning, T1 = 120/.3 × 397.89 = 1.005min
2. Second operation: external relief from 38mm diameter to 35mm diameter and 50mm long N = 1000×50 /3.14 × 38 =418.83 rpm
Time taken for external relief, T2 = 50/.3 × 418.83 = 0.4min
3. Third operation: facing of both ends
L = 38/2 = 19mm
N = 1000 × 50 / 3.14 × 38 = 418.83rpm Time for facing one end = 19 /0 .3× 418.83 = 0.15min
Time for facing both ends, T3 = 0.15 × 2 = 0.30min 4. Fourth operation: chamfering 45degree × 4mm on both sides
N = 1000 × 50 / 3.14 × 38 = 418.83rpm Time for chamfering on both sides, T4 = 0.032 × 2 = 0.064min
Machining time (or)
Example: 4
Find the time required on the shaper to complete one cut on a plate 600 × 900mm, if the cutting speed is 6m/min. The return time to cutting time ratio is 1:4 and the feed is 2mm per stroke. The clearance at each end along the length is 75mm.
Solution:
Length of stroke = length of plate + clearance on both sides = 900 + 2 × 75 =1050mm
T = L (1+k) / s×1000 × w/f = 1050 × 1.25 / 6000 ×300 = 65.62min
Example: 5
A face milling cutter of 150mm diameter is used to give a cut on a block
500mm×250mm. The cutting speed is 50m /min and feed 0.2mm/rev. Calculate the time required to complete one cut.
Solution:
N = 1000 × 50 / 3.14 × 150 =106.1rpm L = 500mm, w = 250mm
Approach = 0.5D = 0.5 × 150 = 75mm Over travel = 7mm
Added table travel = approach +over travel =75 + 7 =82mm
Milling time per cut = (length of the job + added table travel) / feed per revolution = 582 / 0.2 × 106.1 = 27.43min
Example: 6
Find the time required for doing rough grinding of a 15cm long steel shaft to reduce its diameter from 4 to 3.8cm with the grinding wheel of 2cm face with assume cutting speed as 15m/min and depth of cut as 0.25min.
Solution:
N =1000 × 15 / 3.14 × 40 = 119.37rpm
Number of passes required = depth of stroke to be removed / depth of cut = ((40-38)/2) / 0.25 = 4
Length of cut, L =L0 – w + 5
= 150 – 20 + 5 =135mm
Total grinding time = length of cut / (feed per revolution × rpm) number of passes = 135 / (10×119.37) ×4 = 0.452min
