8- Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density 2.9 g/cm3, and thermal diffusivity = 0.8 cm2/s. The cutting speed is 4.5 m/s, uncut chip thickness (feed) is 0.25 mm, and width of cut (depth) is 2.2 mm. The cutting force is measured at 1170 N. Using Cook's equation, determine the cutting temperature. a) 417.09 b) 707.60 C) 528.03 d) 316.82 Boş bırak

8- Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density 2.9 g/cm3, and thermal diffusivity = 0.8 cm2/s. The cutting speed is 4.5 m/s, uncut chip thickness (feed) is 0.25 mm, and width of cut (depth) is 2.2 mm. The cutting force is measured at 1170 N. Using Cook's equation, determine the cutting temperature. a) 417.09 b) 707.60 C) 528.03 d) 316.82 Boş bırak

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density = 2.9 g/cm, and thermal diffusivity = 0.8 cm2/s. Cutting speed = 3.5 m/s, uncut chip thickness = 0.25 mm, and width of cut = 2.2 mm. Cutting force = 950 N. Determine the cutting temperature if the ambient temperature = 22°C.
(11,00 Puanlar) 39 An orthogonal cutting operation is being carried out under the following conditions: t0=0,38 mm, tc=0,65 mm, width of the cut= 2.5 mm, V= 3.5 m/s, rake angle=D 6 , Fc= 515 N, and Ft= 210 N. Calculate the percentage of the total energy that is dissipated in the shear plane. The power input in cutting= F.V Power for shearing=F,V, t I'c tan ø = Iccos a tc 1-re sin a cin -1-
The shear strength of a certain work material 110000 lb/in?. An orthogonal cutting operation is performed using a tool with a rake angle is equal -12° and chip thickness before the cut is 0.026 in, width of cut is 0.120 in, also the chip thickness after cut is 0.082 in. Determine: (a) the shear plane angle, (b) the shear force, (c) cutting force, (d) thrust force, and (e) friction force.
Note: Read the question carefully and give me right solutions according to the question. In orthogonal cutting of steel tube of 150 mm diameter and 2 mm thick, the cutting force was 130 kg and feed force was 35 kg for chip thickness of 0.3mm. The orthogonal cut was taken at 60 meter per minute with a feed of 0.14 mm/rev. If the back rack angle of the cutting tool was - 8 o (minus 8 degree), then calculate the shear strain and strain energy per unit volume.
In an orthogonal cutting test with a bar of 75 mm diameter is reduced to 73 mm by using a HSS tool with arake angle  = 10o, following observations were made: length of the chip, lc = 69.44 mm, cutting ratio r =0.3, the horizontal component of the cutting force, FH = 1450 N, and the vertical component of the cuttingforce, FV = 850 N. The various parameters recorded in this cutting operation are: depth of cut, d = 2 mm;feed rate, f = 0.3 mm/rev, cutting speed, V = 60 m/min. Using Merchant’s theory calculate the following:1) Friction force along rake face2) Normal force acting on rake face3) Shear force along the shear plane4) Normal force acting on shear plane5) The percentage error in shear angle predicted by Merchant’s theory6) Shear velocity7) Chip velocity8) Total work done9) The shear work proportion out of the total work done10) The friction work proportion out of the total work done
The following data was obtained from an orthogonal cutting test. Rake angle = 20° Depth of cut = 6 mm Feed rate = 0.25 mm/rev Cutting speed = 0.6 m/s Chip length before cutting = 29.4 mm Vertical cutting force = 1050 N Horizontal cutting force = 630 N Chip length after cutting = 12.9 mm Using Merchant's analysis, calculate (a) Magnitude of resultant force, (b) shear plane angle, (c) friction force and friction angle, and (d) various energies consumed.
(b) A 400 mm long, 19.5 mm diameter of 304 stainless steel rod (assume specific energy of steel is 4 W.s/mm³) is being reduced in diameter to 17 mm by turning on a lathe machine. The spindle rotates at N = 700 rpm, and the tool is traveling at an axial speed of 300 mm/min. Calculate: (i) (ii) (iii) (iv) (v) cutting speed material removal rate cutting time power dissipated cutting force
In a turning operation, cutting speed =200 m/min; feed = 0.25mm mm/rev, and depth of cut = 4.00mm Thermal diffusivity of the work material = 20m mm^2/s and volumetric specific heat =3.5(10^ -3 )J/mm^ 3 -C If the temperature increase above ambient temperature (20degreesC) is the angle measured by a tool-chip thermocouple to be 700degreesC, determine the specific energy for the work material in this operation.
The outside diameter of a cylinder made of steel is to be turned. The starting diameter is 120 mm and the length is 1400 mm. The feed is 0.3 mm/rev and the depth of cut is 2.5mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameters are: n= 0.33 and C=500. Units for the Taylor equation are min for tool life and m/min for cutting speed. Compute the cutting speed that will allow the tool life to be just equal to the cutting time required to complete this turning operation.
(b) An orthogonal cutting operation is being carried out under the following conditions: depth of cut, to = 0.1 mm, chip thickness, to = 0.2 mm, width of cut = 4 mm, cutting speed, v = 3 m/s, rake angle, a = 10°, Cutting force, Fc = 500 N, and Thrust force, F1= 200 N. Calculate the percentage of the total energy that is dissipated in the shear plane of cutting process.
A steel rod 250 mm long and 200 mm in diameter is being reduced to 190 mm in diameter all over its length in one travel. The machine spindle rotates at 500 rpm, whereas the tool is moving at an axial feed of 0.5 mm/rev; calculate the following: Material removal rate (mm3/min) Consumed gross power in Nm/s if cutting force is 477.5N and mechanical efficiency is 90% Cutting time plzz complete in 30 minutes
3. Draw the forces and angles involved in the cutting process and calculate shear angle (Ø), friction coefficient and tangential force if ,cutting force = 80 kN , resultant of forces =100kN , friction force=75KN, rake angle =20° undeformed chip thickness = 0.65mm and deformed chip thickness 0.72mm. %3D