Question: In plain turning of a brass rod at a feed of 0.4 mm/rev, what will be the theoretical maximum surface roughness, if (i) the tool is relatively sharp with principal and auxiliary cutting edge angles of 90° and 20°, and (ii) the tool has a considerable nose radius of r = 1.0 mm. [Machining and Machine Tools by A. B. Chattopadhyay] This question is related to the calculation of
Question: Using the Taylor equation VTn = C, calculate the percentage increase in tool life when the cutting speed is reduced by 50% (n = 0∙5 and C = 400). [IAS 2002] This is a simple problem based on the Taylor equation of tool life. The Taylor’s Tool Life equation provides an exponential relationship between the cutting velocity and tool life, as given below. The V indicates cutting velocity (m/min),
Question: During plain turning a brass rod by a turning tool whose principal cutting edge angle is 90° and auxiliary cutting edge angle 20°, the maximum surface roughness has been found to be 0.20 mm. At how much feed rate of the tool, the turning was carried out? [Machining and Machine Tools by A. B. Chattopadhyay] The lay marks that are generated owing to the continuous feed of the cutting
Question: While turning a brass rod by a sharp cutting tool in a given condition, how much (in percentage) change will occur in surface roughness if only the depth of cut is changed from 2.0 mm to 2.5 mm. [Machining and Machine Tools by A. B. Chattopadhyay] Conventional machining processes inherently generate lay marks owing to the continuous feed of the tool. This feed marks primarily determine the theoretical surface
Question: In a single point turning operation with cemented carbide tool and steel work piece, it is found that the Taylor’s exponent is 0.25. If the cutting speed is reduced by 50% then what will be the change in the tool life? [GATE 2016] This problem related to the tool life in single point turning operation can be solved using the Taylor’s Tool Life equation. The well-known Taylor’s Tool Life
Question: In a single-point turning operation of steel with a cemented carbide tool, Taylor’s tool life exponent is 0.25. If the cutting speed is halved, how many times the tool life will increase? [ISRO 2013, IAS 1995] This problem can be solved using well-known Taylor’s Tool Life equation which provides the exponential relationship between cutting velocity and tool life. The generalized form of Taylor’s Tool Life equation is shown below.
Question: In the Taylor’s tool life equation, VTn = C, the value of n = 0.5. The tool has a life of 180 minutes at a cutting speed of 18 m/min. If the tool life is reduced to 45 minutes, then what will be the cutting speed? [IAS 1997] This is one simple problem based on Taylor’s Tool Life equation where one combination of velocity and life is given. It
Question: In a machining operation, doubling the cutting speed reduces the tool life to 1/8th of the original value. What is the value of the exponent n in Taylor’s tool life equation? [GATE 2004] This is one simple problem that can be solved using Taylor’s Tool Life equation. In general, the tool life indicates the duration of time within which a cutting tool can provide satisfactory performance in machining. The
Question: A 50 mm diameter steel rod was turned at 284 rpm and tool failure occurred in 10 minutes. The speed was changed to 232 rpm and the tool failed in 60 minutes. Assuming straight line relationship between cutting speed and tool life, calculate the value of Taylorian Exponent (n). [ISRO 2011] Tool life indicates the duration of time within which a cutting tool provides satisfactory performance in machining. Taylor’s
Question: In an orthogonal machining operation, the tool life obtained is 10 min at a cutting speed of 100 m/min, while at 75 m/min cutting speed, the tool life is 30 min. What is the value of index (n) in the Taylor’s tool life equation? [GATE 2009] In this problem, two cutting velocities and corresponding tool life values are given. The task is to calculate the index (or Taylor’s exponent,
Question: In a machining test, a cutting speed of 100 m/min indicated the tool life as 16 min, and a cutting speed of 200 m/min indicated the tool life as 4 min. What are the values of n and C? [ESE 2016] This is a straightforward problem on tool life that can be solved simply by using Taylor’s Tool Life equation. Here, two cutting velocities and the corresponding tool life
Question: A carbide tool while machining a mild steel workpiece was found to have a life of 1 hr and 40 min when cutting at 50 m/min. Find the tool life if the tool is to operate at a speed 30% higher than previous one. Also, calculate the cutting speed if the tool is required to have a life of 2 hr and 45 min. Assume n = 0.28. [BITS
