Calculate resultant force, metal removal rate and shear strain in an orthogonal cutting the depth ofcut is 8.5 mm, feed is 1cutting speed is 60 m.the back rake angle is 10° , andrevTminchip thickness ratio obtained as 0.33. Shear stress of material at zero compressive stress is 1000kgMerchant's constant is 80.

Answered: Image /qna-images/answer/09c96adf-83bb-4a61-bfab-d29896ddb351.jpg

Calculate resultant force, metal removal rate and shear strain in an orthogonal cutting the depth of cut is 8.5 mm, feed is 1 cutting speed is 60 , the back rake angle is 10° , and ree chip thickness ratio obtained as 0.33. Shear stress of material at zero compressive stress is 1000 kg Merchant's constant is 80.

Calculate resultant force, metal removal rate and shear strain in an orthogonal cutting the depth of cut is 8.5 mm, feed is 1 cutting speed is 60 , the back rake angle is 10° , and ree chip thickness ratio obtained as 0.33. Shear stress of material at zero compressive stress is 1000 kg Merchant's constant is 80.

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

See similar textbooks

Similar questions

In orthogonal turning of a cylindrical tube of wall thickness 5 mm, the axial and the tangential cutting forces were measured as 1259 N and 1601 N, respectively. The measured chip thickness after machining was found to be 0.3 mm. The rake angle was 10° and the axial feed was 100 mm/min. the rotational speed of the spindle was 1000 rpm. Assuming the material to be perfectly and Merchant's first solution, the shear strength of the material is closest to
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.
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.
Q3:- In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation. Suppose the rake angle were changed to a= 0°. Assuming that the friction angle remains the same, determine (a) the shear plane angle, (b) the chip thickness, and (c) the shear strain for the operation.
Low carbon steel having a tensile strength = 300 MPa and a shear strength = 220 MPa is turned at a cutting speed = 2.5 m/s, feed = 0.20 mm/rev and depth of cut = 3.0 mm. The rake angle = 5° in the direction of chip flow. The resulting chip ratio = 0.45. Using the orthogonal model to approximate turning, determine the cutting force and feed force.
The orthogonal cutting of steel is done with 10° rake tool with a depth of cut of 4mm feed rate of 0.4mm/rev. The cutting speed is 200m/min. The chip thickness ratio is 0.35. The vertical cutting force is 1200N and the horizontal cutting force is 700N. Calculate from the Merchant's theory, the various work done in metal cutting and shear stress.
1.10A O50 mild steel bar is being turned at 1.70 revis on a multi-tool center lathe having nine tools. For all tools the depth of cut is 4mm. the feed is 0.255mm/rev, and the specific cutting pressure is 1 545N mm?. Calculate the total power consumed for the operation
In turning of stales steel alloy, 1100 mm length and 400 mm diameter, the Feed was 0.35 mm/rev, and depth of cut = 2.5 mm. The tool used in this cutting is cemented carbide tool where Taylor tool life parameters are n = 0.24 and C = 450 (tool life (min) and cutting speed (m/min). Compute the cutting speed that will allow the tool life to be 10% longer than the machining time for this part.
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.
1.10A O50 mild steel bar is being tumed at 1.70 revis on a multi-tool center lathe having nine tools. For all tools the depth of cut is 4mm, the feed is 0.255mm/rev, and the specific cutting pressure is 1 545N/mm. Calculate the total power consumed for the operation
A student is using a lathe with 80-hp and 80% efficiency to fabricate a copper alloy with Sy = 1200ksi If the width of cut is 0.30 inand the student set a rake angle of 0and a cutting speed of 200fl / min while she assumed a coefficient of friction to be 0.5. What is the maximum depth of cut the student can achieve?
Q1. Calculate the time required to machine a workpiece 170 mm long, 60 mm diameter to 165 mm long 50 mm diameter. The workpiece rotates at 440 rpm, feed is 0.3 mm/rev and maximum depth of cut is 2 mm. Assume total approach and overtravel distance as 5 mm for turning operation. Answer Q2. A gray cast iron surface 280 wide and 540mm long may be machined either on a vertical milling machine, using a 100mm - diameter face mill having eight inserted HSS teeth, or on a horizontal milling machine using an HSS slab mill with eight teeth on a 200-mm. diameter. Which machine has the faster cutting time? The values of feed per tooth and cutting speed for both processes are 0.4mm/tooth and 80m/min, respectively. The depth of cut = 3.0 mm and assume A and O equal to 5. Answer