Answered: D.= 0.250" D, = 0.125" %3D L-6.250" If…

Engineering

Mechanical Engineering

D.= 0.250" D, = 0.125" %3D L-6.250" If Cutting speed v=211 FPM (feet per minute) and f= 0.0018 in/rev; Find (a) N=?

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

Related questions

Q: Machining time required reduce a mild steel rod of diameter 60mm, initial length (L) 125mm to final…

Q: . In an orthogonal cutting operationn, uncut chip thickness t is = 0.127 mm chip thickness t is =…

Q: D.= 0.250" D, = 0.125" %3D L- 6.250" If Cutting speed v=211 FPM (feet per minute) and f= 0.0018…

A: Given:Do=0.250 inDi=0.125 inv=211 feet/minf=0.0018 in/rev

Q: In a slab milling operation , L_w = 200 mm table feed = 18 mm/min, diameter of cutter = 34 mm depth…

A: Given Data: The length of the slab is, Lw=200 mm. The feed rate of the table is, f=18 mm/min. The…

Q: cutting at V = 30 m/min, f 0.3 mm of cut are increased together by 20

Q: In 3-axis milling, the table feedrate is 756.00 mm/min. Using a 10mm endmill to cut the material…

Q: If a fairing operation is executed at a rotational speed of 40.0 rev / min, the feed is 0.015 in /…

A: As the outer and inner diameter is not given so I am assuming it as D0 and Direspectivelya)…

Q: Q4: Answer the following questions: 1. Increasing cutting speed is undesirable at any time to raise…

Q: 3. A turning operation is carried out on aluminum (100 BHN), the cutting conditions are: v = 5.6…

Q: Q4. A hole with 40 mm diameter an 50mm depth is to be drilled in mild steel component. The cutting…

Q: The face milling cutter of D = 200 mm used to cut a block which is 200 mm = H, 1=480 mm, if f- 0.7…

A: Machine time is the ratio of length divided by cutting speed. Here, Length = length of block +…

Q: 1. An engine lathe is used to turn a cylindrical work part, cutting speed 2.2 m/s, feed 0.25 mm/rev,…

Q: In a turning operation, the workpiece diameter is Dm-40.00 mm and the diameter after the operation…

A: GIVEN DATA - D=40mm D2=24mm V=132m/min f=0.03mm/rev WE HAVE TO CALCULATE MRR

Q: In a turning operation on carbon steel with hardness- 175 HB. the cutting speed - 225 m/min, feed-…

A: GivenV=225m/minf=0.25mm/revd=7.5mmη=90% Cutting energy for Brinell hardess 175Hb=1.6J/mm3U=1.6J/mm3

Q: Q.4 (i) Find the time required for taking a complete cut on a plate 600 × 900 mm², if the .cutting…

A: As per the bartleby policy we can solve the only first question from the multiple questions. Q.4 (i)…

Q: A 12" x 4" block of soft aluminum alloy (u, = 2.5 hp.min/in³) is face milled at a depth of cut of…

A: Given Data: Depth of cut DC=18inch Length of block l=12 in Breadth of block b=4 in ut of aluminium…

Q: Since the cutting plane angle is 9.5°, the cutting depth is 0.4 mm, the chip thickness is 2.9 mm,…

A: 1.We know that rack angle=Cutting plane angle=9.50

Q: The workpiece of size o 50 mm x 90 mm length is to be reduced to o 42 mm x 85 mm. If the depth of…

A: We have provided data from the question, Initial diameter= 50 mm; Initial length= 90 mm Final…

Q: “Milling is an intermitted cutting operation”. Explain it

A: Intermittent cutting is that cutting when the cutting is interrupted or we can say at some time tool…

Q: H.W. The face milling cutter of D= 200 mm used to cut a block which is 200 mm = H, 1 =480 mm, if f=…

Q: 3. A slot of 20 mm width is to be produced using horizontal milling machine. The ks 5,000 N/mm?,…

Q: In 3-axis milling, the table feedrate is 703.00 mm/min. Using a 10mm endmill to cut the material…

Q: A surface of 43 mm x 246 mm will be machined by using a shaper and 7 mm deep stock will be removed…

A: Given,

Q: HW3: A block of steel has a length I. = 600 mm and a width W = 450 mm. The thickness is Machined 10…

Q: The following data is given for slab milling of a 300 mm long 50 mm wide mild steel block: Cutter of…

Q: If shear strength is 6000 kg/cm2, width of cut is 10mm, cutting speed is 30 m/min, coefficient of…

A: Given:shear strength,ζ=6000 Kg/cm2 =6000*9.8110-4ζ=588.6 MPab=10 mmVelocity of cutting,V=30…

Q: During orthogonal cutting of mild steel with a 10° rake angle tool, the chip thickness ratio was…

A: Given, Orthogonal rake angle, α=10° Chip thickness ratio, r=0.4

Q: An orthogonal cutting operation is being carried out under the following conditions: t0-D0,38 mm,…

A: Given data:-uncut chip thickness, t=0.38mmchip thickness, tc=0.65mmwidth of cut, b=2.5mmcutting…

Q: (j) In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 250 m/min,…

A: Given data : stainless steel with hardness = 200 HB cutting speed (v) = 250 m/min feed (f) = 0.3…

Q: (e) Briefly describe types of chips that occur in metal cutting. (f) For orthogonal cutting, the…

A: Types of chip 1) continuous chip - the condition for the formation of continuous chip without built…

Q: In a turning operation on cast iron, the nose radius on the tool = 1.5 mm, feed = 0.22 mm/rev, and…

A: Given data: Nose radius of the tool = 1.5 mm Feed = 0.22 mm/rev Speed = 1.8 m/s Need to determine…

Q: In a surface broaching process, a broach having 8 mm pitch and cut per teeth 0.01 mm is used for…

Q: 5. AT-slot is to be cut in a C.I. slab 260 mm long as shown make the T-slot if the cutting speed is…

A: Given Cutting speed, v = 25 m/min Diameter, d = 20 mm Number of teeth, z = 6 Find…

Q: 1. “Milling is an intermitted cutting operation”. Explain it.

A: Intermittent cutting means the cutting is not continuous. That means at some time the tool is in…

Q: 2:(b) “Milling is an intermitted cutting operation”. Explain it.

A: Intermittent cutting means the cutting is not continuous. In other words at some time cutting tool…

Q: b) A mild steel having length of 350mm and diameter of 12.5 cm, is being reduced to diameter 9.5 cm,…

Q: late the machining time to drill the flange as shown below. Take cutting speed 20 m/min, feed for 20…

A: Soln:- depth = 20 mm Number of Revolution required = depth/feed = 20/ 0.2 = 100 rev. V=…

Q: Q3 A) A shaft with 150mm in diameter and 1 m length is to be turned in an engine lathe. Cutting…

Q: D. = 0.250" D, = 0.125" L- 6.250" If Cutting speed v=211 FPM (feet per minute) and f= 0.0018 in/rev;…

Q: 1. A turning operation is performed on a cylindrical work part whose diameter 150 mm and length 375…

Q: Plate 200 mm x 90 mm is to be maintained on a shaper under following condition, v= 13.5 m/ min Feed…

A: Given: L= 200 mm B= 90 mm v= 13.5 m/min feed, f=0.57 mm/stroke approach and overtravel along length=…

Q: In a turning operation, the workpiece diameter is Dm=44.00 mm and the diameter after the operation…

A: Given:Dm=44 mmDc=22 mmV=15 m/minf=0.03 mm/rev

Q: D. 0.250" D, = 0.125" L-6.250" If Cutting speed v=211 FPM (feet per minute) and f= 0.0018 in/rev;…

A: Cutting speed v=211 ft/minor v=211×12 =2532 in/minf=.0018 in/revD0=.250''Df=.125''L=6.250''

Concept explainers

Forming and Shaping

Metal forming is a category of metalworking process through which the material can be given the desired shape and size by subjecting it to plastic deformation, where the component undergoes bulk deformation process upon application of loads. Unlike the shearing and machining process, the material does not undergo up to a failure point as indicated in the stress-strain curve. Hence, the stresses induced during the forming process are greater than yield strength but less than the failure strength of the material. The process does not lead to the loss of material, so it can be said to be an economical process. During forming, the material is subjected to tensile forces, compressive forces, bending, and shearing loading conditions. The material should possess the property of ductility to undergo the forming process.

Question

D. = 0.250"
D, = 0.125"
L= 6.250"
If Cutting speed v=211 FPM (feet per minute) and f= 0.0018 in/rev; Find
(a) N=?
(b) f=?
(c) Tm=?

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

The following data is given for slab milling of a 300 mm long 50 mm wide mild steel block: Cutter of diameter= 60 mm; Numbers of teeth =12; Cutter speed = 120 rev/min; Depth of cut = 3.2 mm; Feed is 0.25 mm/tooth.Determine (i) Table feed in mm/min, (ii) MRR (iii) Power (iv) Torque and (v) Total machining time. Assumesuitable approach and over-run and specific energy of mild steel as 5 Ws/mm3.
For turning operation, using the table below, find an approximate value for the required speed of rotation of the chuck, in rpm, if the cutting speed for the material 5417 mm/s and the radius of the wp is 0.01 m. Show details of your answer. n | we| o
In the machining operation with orthogonal cutting, the cutting plane angle is 9.5°, the cutting depth 0.4 mm, chip thickness 2.9 mm, cutting width 3.1 mm, shear strength 185 MPa in the process; a) Find the rake angle (α). b) Find the shear deformation (γ). c) Find the shear force. (Note: The shear force is 1.4 times the thrust.)
The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is 200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min) is ___. A. 7.539 B. 7.90 C. 8.25 D. 9.00
A cylindrical workpiece has a 14.0 in diameter and has 30 in long. It is to be turned in an engine lathe at the following cutting conditions: cutting speed = 400 ft/min, feed = 0.015 in/rev, and depth of cut = 0.150 in. Determine the time to complete one turning pass if the approach and overtravel allowances are each 0.25 in (total = 0.50 in). Note: 1 ft = 12 in
in shaft turning Steel raw material, diameter 40 mm., length 200 mm. to obtain a lathe workpiece diameter 35 mm., length 150 mm, as shown in the figure, with a cutting speed of 25 m/min and a rough-cutting feed rate (f) 0.2 mm/rev. The feed rate for finishing (f) 0.1 mm per revolution determines a depth for rough turning of not more than 1 mm and a depth for finishing of not more than 0.5 mm. Find 1) the speed of rotation used per minute. 2) Details of turning Such as how many rounds of rough turning, how many rounds of fine turning? 3) From the available data Find the approximate turning time in minutes.
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=6°, Fc= 515 N, and Ft=210 N. Calculate the percentage of the total energy that is dissipated in the shear plane.
Machining time required reduce a mild steel rod of diameter 60mm, initial length (L) 125mm to final length (L1) 120mm to 58 mm by turning a carbide insert. If the approach length and over run length are 5 mm and 1 mm respectively, take Cutting speed as 20 m/min and feed is =0.2 mm/rev, and the depth of cut is 0.5mm
2 1.57 Show that, for the same shear angle, there are two rake angles that give the same cutting ratio.
If a fairing operation is executed at a rotational speed of 40.0 rev / min, the feed is 0.015 in / rev and the depth of cut is 0.180 in, it determines: a) cutting time to complete the fairing operation; and b) the cutting speed; and c) The speed of metal removal at the beginning and end of the cut d) The power dissipated.
Calculate the time required for completing a 5mm deep finishing cut on a 150mm wide, 600mmlong face of a 25mm thick steel block using a face milling cutter of 150mm diameter with 6teeth. The cutting condition are Vc =1.5m/sec and fz = 0.1mm.
Manufacturing Process In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 200m/min, feed = 0.25 mm/rev, and depth of cut =7.5 mm. Determine how much power will the lathe machine draw in performing this operation if its mechanical efficiency = 90 %. Use specific energy value is: U = 2.8 J/mm