1. Calculation of Machining Time for Drilling Operations

2. Drilling Operation: Producing holes in the material – drill tool

3. Calculation of Machining Time for Drilling Operations
T = time required for drilling (min) f = feed per revolution (mm) L = Depth of hole to be produced (mm) N = Revolutions of the job per minute (rpm) = 1000 × 𝑆 𝜋 𝐷 S = Cutting speed (m/min) D = diameter of the hole to be drilled (mm)

4. T = 𝐿 𝑓𝑁 Time taken to drill unit length = 1 𝐹𝑒𝑒𝑑/𝑚𝑖𝑛
Time taken to drill L mm length = 𝐿 𝐹𝑒𝑒𝑑/𝑚𝑖𝑛
T = 𝐿 𝑓𝑁
Time required for drilling = T = 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 ℎ𝑜𝑙𝑒 𝑡𝑜 𝑏𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑓×𝑁

5. Sample problem
A 10 cm thick laminated plate used in pressure vessels consists of 8 cm thick steel plate cladded with titanium plate of thickness 2 cm. A 1 cm diameter hole is to be drilled through this composite plate. Estimate the time taken for drilling this hole if, cutting speed of steel and titanium are 2.5cm/min and 1cm/min respectively. Also the feed of drill for steel and titanium are cm/rev and 0.02 cm/rev respectively. Given data L = 100mm; L1 = 80mm; L2 = 20 mm; D = 10 mm; Ssteel = 25 m/min; Stitanium = 10 m/min; fsteel = 0.25 mm/rev; ftitanium = 0.2 mm/rev

6. To find: total time taken for drilling Solution rpm of steel Nsteel = 1000 × 𝑆𝑠𝑡𝑒𝑒𝑙 𝜋 𝐷 = 1000 ×25 𝜋×10 = rpm rpm of steel Ntitanium= 1000 × 𝑆𝑡𝑖𝑡𝑎𝑛𝑖𝑢𝑚 𝜋 𝐷 = 1000 ×10 𝜋×10 = rpm At rpm drilling operations should be carried out T1= 𝐿1 𝑓𝑠𝑡𝑒𝑒𝑙×𝑁 = × = min

7. T2= 𝐿2 𝑓𝑡𝑖𝑡𝑎𝑛𝑖𝑢𝑚×𝑁 = × = min Total time for drilling operation = T1 + T2 = = min

8. Calculation of Machining Time for Boring Operation
T = time required for boring (min) f = feed per revolution (mm) L = Depth of bore to be produced (mm) N = Revolutions of the job per minute (rpm) = 1000 × 𝑆 𝜋 𝐷 S = Cutting speed (m/min) D = diameter of the hole to be bored (mm)

9. T = 𝐿 𝑓𝑁 Time taken to bore unit length = 1 𝐹𝑒𝑒𝑑/𝑚𝑖𝑛
Time taken to bore L mm length = 𝐿 𝐹𝑒𝑒𝑑/𝑚𝑖𝑛
T = 𝐿 𝑓𝑁
Time required for boring = T = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑟 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑏𝑜𝑟𝑒 𝑡𝑜 𝑏𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑓×𝑁

10. Sample problem
What will be the machine time required to rough bore, to second bore and to finish bore a soft cast iron cylinder whose diameter is 220 mm and the length of bore is 260 mm. (i) rough bore 39 rpm, mm feed, (ii) second rough bore 45 rpm, mm feed and (iii) finish bore 25 rpm, 5.75 mm feed. Given data Diameter (D) = 220 mm; Length of the bore (L) = 260 mm To find Time required to rough, second and finish bores

11. Diameter (D) = 220 mm; Length of the bore (L) = 260 mm To find Time required to rough, second and finish bores Solution (i) Rough bore time (T1) f = mm ; N = 39 rpm T1 = 𝐿 𝑓𝑁 = ×39 = min

12. (ii) Second Rough bore time (T2) f = 0
(ii) Second Rough bore time (T2) f = mm ; N = 45 rpm T2 = 𝐿 𝑓𝑁 = ×45 = min (iii) Finish bore time (T3) f = 5.75 mm ; N = 25 rpm T3= 𝐿 𝑓𝑁 = ×25 = min Total time to rough, second and finish bore = T1 + T2 + T3 = = min

13. Reaming – removing a small amount of material from a previously drilled or bored hole for obtaining the perfect hole – tool name – reamer Tapping– cutting internal threads – tool name – tap Actual tapped length = L + 𝐷 2 Time taken for tapping = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑡𝑎𝑝 𝑃𝑖𝑡𝑐ℎ (𝑜𝑟) 𝑓𝑒𝑒𝑑 𝑝𝑒𝑟 𝑟𝑒𝑣 ×𝑟𝑝𝑚 = L + 𝐷 2 𝑃𝑖𝑡𝑐ℎ×𝑟𝑝𝑚 Returning the tap ½ of the time required for pushing it in the workpiece Total Time taken for tapping = 3 2 (L + 𝐷 2 ) 𝑃𝑖𝑡𝑐ℎ×𝑟𝑝𝑚 Pitch = 1 𝑇.𝑃.𝐶 (T.P.C = Threads per centimeter) l

14. Sample problem
Calculate the drilling and tapping time for producing threads in a mild steel sheet of 2.5 cm thickness. The size of H.S.S drill to be used is cm and the number of threads to be cut is 3 per cm. Taking cutting speed and feed for drill as 20 m/min and cm/revolution respectively, tapping speed as 5 m/min. Neglect the time taken for setting up and approaching and over travelling of tools.
Solution
To find drilling time; t = 25mm; D = 20mm; S = 20 m/min;
f = 0.25mm/rev
N = × 𝑆 𝜋 𝐷 = ×20 𝜋×20 = rpm l

15. T1 = 𝐿 𝑓𝑁 = ×318.3 = min (ii) To find tapping time; L = 25 mm ;D = 20mm; S = 5 m/min; Threads per centimeter = 3 N = 1000 × 𝑆 𝜋 𝐷 = 1000 ×5 𝜋×20 = rpm Pitch = 1 𝑇.𝑃.𝐶 = 1 3 = Time for tapping T2= L + 𝐷 2 𝑃𝑖𝑡𝑐ℎ×𝑟𝑝𝑚 = ×79.58 = min l

16. Thread cutting – removal of material to produce helix on external or internal circular surface for fastening Time required for threading = T = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑡 𝑓×𝑁 × no. of cuts Pitch for a single start thread = Pitch of the thread Pitch for a multi-start thread = Pitch × number of starts For external threads, number of cuts = 25 𝑡ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚 For internal threads, number of cuts = 32 𝑡ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚 l

17. Sample problem
Estimate the time required for cutting 3 mm pitch threads on a mild steel bar of 2.8 cm diameter and 8 cm long. Assume the cutting speed for threading as 15 m/min. Solution N = 1000 × 𝑆 𝜋 𝐷 = 1000 ×15 𝜋×28 = rpm Threads per cm = 1 𝑝𝑖𝑡𝑐ℎ = = Number of cuts = 25 𝑡ℎ𝑟𝑒𝑎𝑑𝑠 𝑝𝑒𝑟 𝑐𝑚 (for external threads) l

18. = = 7.5 or 8 approximately Threading time / cut = 𝐿 𝑝𝑖𝑡𝑐ℎ × 𝑁 = × = min Time for 8 cuts = × 8 = 1.25 min l

19. References
T. R. RusselBanga and S. C. Sharma., Mechanical Estimating and Costing, Khanna Publishers, New Delhi, 2007.
Sinha.B.P., "Mechanical Estimating and Costing", Tata McGraw-Hill, Publishing Co.,
Russell.R.S and Tailor, B.W, "Operations Management", PHI, 4th Edition,
Chitale.A.V. and Gupta.R.C., "Product Design and Manufacturing", PHI, 2nd Edition,2002.
Gideon Halevi and Roland Weill, “Principles of Process Planning – A Logical Approach” Chapman and Hall, London,  2014

20. References
Mikell.P.Groover “Automation, Production Systems and Computer Integrated Manufacturing”, Prentice Hall of India, New Delhi, 2008.
S. K. Mukhopadhyay, Production Planning and Control-Text and cases, PHI Pvt. Ltd 2007.
Kesavan. R, Elanchezhian.C, Vijaya Ramnath. B, Process Planning and Cost Estimation, New Age International Publishers Ltd., 2009
Jayakumar. V, Process Planning and Cost Estimation, Lakshmi Publications, Chennai, 2016