This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: IE 215 Solutions for Problems due Sep 3, 2008 (Ch 21) 21.6 The cutting force and thrust force have been measured in an orthogonal cutting operation to be 300 lb and 291 lb, respectively. The rake angle = 10 ° , width of cut = 0.200 in, chip thickness before the cut = 0.015, and chip thickness ratio = 0.4. Determine (a) the shear strength of the work material and (b) the coefficient of friction in the operation. Solution : φ = tan1 (0.4 cos 10/(1  0.4 sin 10)) = tan1 (0.4233) = 22.94 ° F s = 300 cos 22.94  291sin 22.94 = 162.9 lb. A s = (0.015)(0.2)/sin 22.94 = 0.0077 in 2 S = 162.9/0.0077 = 21,167 lb/in 2 β = 2(45) + α 2( φ ) = 90 + 10  2(22.94) = 54.1 ° μ = tan 54.1 = 1.38 21.10 The shear strength of a certain work material = 50,000 lb/in 2 . An orthogonal cutting operation is performed using a tool with a rake angle = 20 ° at the following cutting conditions: cutting speed = 100 ft/min, chip thickness before the cut = 0.015 in, and width of cut = 0.150 in. The resulting chip thickness ratio = 0.50. Determine (a) the shear plane angle; (b) shear force; (c) cutting force and thrust force, and (d) ratio = 0....
View
Full
Document
 Fall '08
 Groover
 Shear Stress, Carbon steel, 22.94°, 228 lb

Click to edit the document details