Assignment [4] machining with solutions
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1. The document contains 5 problems related to machining operations including turning, drilling, and tool life calculations. 2. Problem 1 involves determining the cutting time and metal removal rate for turning a cylindrical workpiece. 3. Problem 2 calculates the required cutting speed to complete a turning operation in 5 minutes given specifications for the workpiece, feed rate, and depth of cut.
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![ASSIENMENT [4]
MACHINING
PROBLEMS
1- A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine
lathe. Cutting conditions are as follows: cutting speed is 2.30 m/s, feed is 0.32 mm/rev, and
depth of cut is 1.80 mm. Determine (a) cutting time, and (b) metal removal rate.
Solution: (a) N = v/(D) = (2.30 m/s)/0.200 = 3.66 rev/s
fr = Nf = 6.366(.3) = 1.17 mm/s
Tm = L/fr = 700/1.17 = 598 s = 9.96 min
Alternative calculation using Eq. (22.5), Tm = 200(700)/(2,300 x 0.32) = 597.6 sec = 9.96
min
(b) RMR = vfd = (2.30 m/s)(103)(0.32 mm)(1.80 mm) = 1320 mm3/s .
2- In a production turning operation, the foreman has decreed that a single pass must be
completed on the cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in
diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must
be used to meet this machining time requirement?
Solution: Starting with Eq. (22.5): Tm = DoL/vf.
Rearranging to determine cutting speed: v = DoL/fTm
v = (0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min
3- A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel
workpart. The hole is a blind hole at a depth of 60 mm and the point angle is 118属. The
cutting speed is 25 m/min and the feed is 0.30 mm/rev. Determine (a) the cutting time to
complete the drilling operation, and (b) metal removal rate during the operation, after the drill
bit reaches full diameter.
Solution: (a) N = v/D = 25(103) / (12.7) = 626.6 rev/min
fr = Nf = 626.6(0.30) = 188 mm/min
Tm = L/fr = 60/188 = 0.319 min
(b) RMR = 0.25D2fr = 0.25(12.7)2(188) = 23,800 mm3/min
14- Tool life tests in turning yield the following data: (1) when cutting speed is 100 m/min,
tool life is 10 min; (2) when cutting speed is 75 m/min, tool life is 30 min. (a) Determine the
n and C values in the Taylor tool life equation. Based on your equation, compute (b) the tool
life for a speed of 110 m/min, and (c) the speed corresponding to a tool life of 15 min.
Solution: (a) Two equations: (1) 120(7)n = C and (2) 80(28)n = C.
120(7)n = 80(28)n
ln 120 + n ln 7 = ln 80 + n ln 28
4.7875 + 1.9459 n = 4.3820 + 3.3322 n
4.7875 - 4.3820 = (3.3322 1.9459) n
0.4055 = 1.3863 n n = 0.2925
C = 120(7)0.2925 = 120(1.7668) C = 212.0
Check: C = 80(28)0.2925 = 80(2.6503) = 212.0
(b) 110 T0.2925 = 212.0](https://image.slidesharecdn.com/assignment4machining-withsolutions-121213110841-phpapp02/85/Assignment-4-machining-with-solutions-1-320.jpg)
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This document describes the design and analysis of an automatic stirrup making machine. Some key points:
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The document discusses the selection of a ball screw for a machine tool based on its operating conditions and design specifications. It outlines the steps to determine the key parameters for the ball screw like lead accuracy, axial clearance, screw length and diameter, support method, permissible axial load and rotational speed, nut model, rigidity, positioning accuracy, torques required, and motor specifications. An example selection process is provided based on the given design data and machine specifications for a lathe machine. Key factors like buckling load, tensile strength and critical speeds are examined to ensure safe design of the ball screw.Blow time current function(Device Modeling of Fuse)
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The document provides information about static shear testing, including direct shear testing, torsion testing, and calculations related to shear stress, shear strain, shear modulus, and other properties. It includes examples of calculations for shear stress, shear modulus, angle of twist, and other values using data from torsion tests on steel specimens of various dimensions under increasing torque loads. Diagrams are presented showing typical shear stress distributions and failure shapes for ductile and brittle materials in torsion testing.theory of metal cutting assignment problems
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The document contains 7 questions related to metal cutting practices and calculations involving cutting speed, feed rate, shear angle, chip thickness, cutting forces, friction forces, power requirements, and production time. Specific calculations are asked to determine motor power, cutting time, drilling power, milling cutter offset, and total production time for machining various metals like mild steel, C-20 steel, and AISI-304 steel. Shear angle, chip reduction coefficient, forces, and energy are to be calculated using given tool geometries and machining conditions.Harmonic rh dc_servo_specsheet
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This document provides specifications for the RH Mini Series DC Servo Actuators from D C Servo Systems. It includes:
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- Graphs showing the torque and current curves for the RH-8D-3006 and RH-5A-5502 models.
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This document discusses compression testing and summarizes:
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1) The document provides three examples of determining the moment capacity of reinforced concrete beams using the strength design method. The examples calculate steel ratios, check code requirements, and determine moment capacities.
2) Design examples are also provided, including calculating reinforcement needed to resist a given moment and designing a beam to support specific service loads. Optimal dimensions are selected to maximize steel ratio within code limits.
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This document contains information about a 6th semester B.E. degree examination in Design of Machine Elements including:
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- Part B includes topics like bevel gears, worm gears, cone clutches, band brakes, and lubrication
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6th Semester Mechanical Engineering (June/July-2015) Question PapersBGS Institute of Technology, Adichunchanagiri University (ACU)
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6th Semester Mechanical Engineering (2013-December) Question Papers
6th Semester Mechanical Engineering (2013-December) Question PapersBGS Institute of Technology, Adichunchanagiri University (ACU)
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Assignment [4] machining with solutions
- 1. ASSIENMENT [4] MACHINING PROBLEMS 1- A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine lathe. Cutting conditions are as follows: cutting speed is 2.30 m/s, feed is 0.32 mm/rev, and depth of cut is 1.80 mm. Determine (a) cutting time, and (b) metal removal rate. Solution: (a) N = v/(D) = (2.30 m/s)/0.200 = 3.66 rev/s fr = Nf = 6.366(.3) = 1.17 mm/s Tm = L/fr = 700/1.17 = 598 s = 9.96 min Alternative calculation using Eq. (22.5), Tm = 200(700)/(2,300 x 0.32) = 597.6 sec = 9.96 min (b) RMR = vfd = (2.30 m/s)(103)(0.32 mm)(1.80 mm) = 1320 mm3/s . 2- In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in 5.0 min. The piece is 400 mm long and 150 mm in diameter. Using a feed = 0.30 mm/rev and a depth of cut = 4.0 mm, what cutting speed must be used to meet this machining time requirement? Solution: Starting with Eq. (22.5): Tm = DoL/vf. Rearranging to determine cutting speed: v = DoL/fTm v = (0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min 3- A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel workpart. The hole is a blind hole at a depth of 60 mm and the point angle is 118属. The cutting speed is 25 m/min and the feed is 0.30 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal removal rate during the operation, after the drill bit reaches full diameter. Solution: (a) N = v/D = 25(103) / (12.7) = 626.6 rev/min fr = Nf = 626.6(0.30) = 188 mm/min Tm = L/fr = 60/188 = 0.319 min (b) RMR = 0.25D2fr = 0.25(12.7)2(188) = 23,800 mm3/min 14- Tool life tests in turning yield the following data: (1) when cutting speed is 100 m/min, tool life is 10 min; (2) when cutting speed is 75 m/min, tool life is 30 min. (a) Determine the n and C values in the Taylor tool life equation. Based on your equation, compute (b) the tool life for a speed of 110 m/min, and (c) the speed corresponding to a tool life of 15 min. Solution: (a) Two equations: (1) 120(7)n = C and (2) 80(28)n = C. 120(7)n = 80(28)n ln 120 + n ln 7 = ln 80 + n ln 28 4.7875 + 1.9459 n = 4.3820 + 3.3322 n 4.7875 - 4.3820 = (3.3322 1.9459) n 0.4055 = 1.3863 n n = 0.2925 C = 120(7)0.2925 = 120(1.7668) C = 212.0 Check: C = 80(28)0.2925 = 80(2.6503) = 212.0 (b) 110 T0.2925 = 212.0
- 2. T0.2925 = 212.0/110 = 1.927 T = 1.9271/0.2925 = 1.9273.419 = 9.42 min (c) v (15)0.2925 = 212.0 v = 212.0/(15)0.2925 = 212.0/2.2080 = 96.0 m/min . 5- Turning tests have resulted in 1-min tool life at a cutting speed = 4.0 m/s and a 20-min tool life at a speed = 2.0 m/s. (a) Find the n and C values in the Taylor tool life equation. (b) Project how long the tool would last at a speed of 1.0 m/s. Solution: (a) For data (1) T = 1.0 min, then C = 4.0 m/s = 240 m/min For data (2) v = 2 m/s = 120 m/min 120(20)n = 240 20n = 240/120 = 2.0 n ln 20 = ln 2.0 2.9957 n = 0.6931 n = 0.2314 (b) At v = 1.0 m/s = 60 m/min 60(T)0.2314 = 240 (T)0.2314 = 240/60 = 4.0 T = (4.0)1/0.2314 = (4)4.3215 = 400 min