Question: In an orthogonal cutting operation shear angle = 11.31°, cutting force = 900 N and thrust force = 810 N. What will be the shear force? [ESE 2014] Solution: With the help of Merchant Circle Diagram (MCD), this problem can be solved easily. A typical MCD for positive rake angle is shown below. Here in the question, the main cutting force (PZ) and thrust force (PXY) for an orthogonal
Question: Mild steel is being machined at a cutting speed of 200 m/min with a tool rake angle of 10°. The width of cut and uncut thickness are 2 mm and 0.2 mm, respectively. If the average value of co‐efficient of friction between the tool and the chip is 0.5 and the shear stress of the work material is 400 N/mm2, then determine: [ESE 2005] (i) The shear angle (ii)
Question: In an orthogonal cutting test, the cutting force and thrust force were observed to be 1000 N and 500 N, respectively. If the rake angle of tool is zero, what will be the coefficient of friction in chip‐tool interface? [ESE 2000] Solution: For an orthogonal machining, the rake angle of the cutting tool is given as zero. The cutting force and thrust force values are also given. Using these
Question: While turning a 60 mm diameter bar, it was observed that the tangential cutting force was 3000 N and the feed force was 1200 N. If the tool rake angle is 32°, then calculate the coefficient of friction. [ESE 2018] Solution: This problem can be solved with the help of Merchant Circle Diagram (MCD). In orthogonal machining of ductile material with a single point turning tool, the relevant forces
Question: In orthogonal cutting test, the main cutting force = 900 N, the thrust force = 600 N and chip shear angle is 30°. Calculate the chip shear force. [ESE 2003] Solution: This problem on cutting force can be solved with the help of Merchant Circle Diagram (MCD). MCD is one vector representation of several cutting forces associated with orthogonal machining with a sharp tool. A typical MCD for positive
Question: The following data from the orthogonal cutting test is available. Rake angle = +10°; Chip thickness ratio = 0.35; Uncut chip thickness = 0.51 mm; Width of cut = 3 mm; Yield shear stress of work material = 285 N/mm2; Mean friction co‐efficient on tool face = 0.65. Determine the following forces: [ESE 2000] Main cutting force Radial force Normal force on tool Shear force Solution: Force calculation in
Question: An orthogonal cutting operation is being carried out under the following conditions: Cutting speed = 2 m/s Depth of cut = 0.5 mm Chip thickness = 0.6 mm Calculate the chip velocity. [ESE 2003] Solution: To solve this question we need to assume two things—(i) orthogonal rake angle (γO) of the cutting tool is 0° as it is the case of orthogonal cutting, and (ii) the given depth of
Question: If α is the rake angle of the cutting tool, φ is the shear angle and V is the cutting velocity, then express the velocity of chip sliding along the shear plane in terms of above three parameters. [ESE 2001] Answer: To solve this question, one need the knowledge of velocity triangle in machining. In an orthogonal machining of ductile material with a sharp cutting tool, three velocity parameters
Question: In a pure orthogonal turning process, the chip thickness is 0.32mm, feed is 0.2 mm/rev. What is the cutting ratio? [ESE 2014] Solution: The ratio between the uncut chip thickness (a1) to the chip thickness (a2) is termed as cutting ratio (rcu). Owing to lamellar shearing and positive strain, the chip thickness increases as compared to the same for uncut chip. Thus cutting ratio is smaller tan 1, especially
Question: In a machining operation, the chip thickness ratio is 0.3 and the rake angle of the tool is 10°. What is the value of the shear strain? [ESE 2004] Solution: In any conventional machining process when sample material is ductile, material removal takes place due to shearing under the action of compressive force exerted by the cutting tool. The assumed 2-D plane along which shearing takes place is termed
