The rough material for the operation is 100mm diameter mild steel, and the length of the rough material is 150mm. To machine the component, the diameter must be turned down from 100mm to 82mm for a length of 100mm. You are going to achieve this with three cuts, with a depth of cut of 3.0mm per cut, as is shown in the sketch below. Thereafter, the billet is machined for a length of 50mm to a diameter of 64mm, again in three cuts of 3.0mm in depth each. The technical manual of the cutter supplier gives the following data for the specific cutter: f₂ = 0.2 mm/tooth/rev (Recommended feed) Vc = 160 m/min (Tangential cutting speed or surface speed) Calculate the following: 23.1 The required rotational speed for each of the three cuts The material removal rate in mm³/min 23.2 23.3 The total cutting time. Assume that the tool approach and retract times are negligible in this process Rof materiaal Rough material Finale komponent Final component

The rough material for the operation is 100mm diameter mild steel, and the length of the rough material is 150mm. To machine the component, the diameter must be turned down from 100mm to 82mm for a length of 100mm. You are going to achieve this with three cuts, with a depth of cut of 3.0mm per cut, as is shown in the sketch below. Thereafter, the billet is machined for a length of 50mm to a diameter of 64mm, again in three cuts of 3.0mm in depth each. The technical manual of the cutter supplier gives the following data for the specific cutter: f₂ = 0.2 mm/tooth/rev (Recommended feed) Vc = 160 m/min (Tangential cutting speed or surface speed) Calculate the following: 23.1 The required rotational speed for each of the three cuts The material removal rate in mm³/min 23.2 23.3 The total cutting time. Assume that the tool approach and retract times are negligible in this process Rof materiaal Rough material Finale komponent Final component

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: 2 Explain the limitations of the following expressions: a) AS = CIn(T/T), (b) AG=AH-TAS, and (c)…

A: The Gibbs function, commonly known as the Gibbs free energy, is described in terms of…

Q: Find values for the constants a, b, and c that will make f(x)=cosx and g(x)=a+bx+cx² satisfy the…

Q: aluminum alloy 2014

A: Problem asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a…

Q: Bronze alloy, the following true stresses produce the corresponding plastic true strains, before to…

Q: 1900 Fo = Forsterite 1800 Liquid En = Enstatite Qz = Quartz 1700 Fo+liq 1600 -En+lig Qz+liq 1500…

A: From the olivine mineral group, forsterite is a magnesium silicate crystal with the chemical formula…

Q: A bolt on a polyethylene reactor must be able to withstand a minimum force of 220 kN, not including…

Q: Do provide explanations too. Thank you.

A: The grain boundary and bulk diffusion creep both are the mechanism that leads to the elongation of…

Q: Aluminium with a mass of 100 kg and an initial temperature of 20, 6\deg C is to be melted in an…

A: The mass of the aluminum, The initial temperature of aluminum, The melting point of aluminum, The…

Q: Dr Downey is trying his hand at fording and has made a blade of 15N20 steel that can be assumed to…

Q: Don’t use ai pls

A: Step 1:The steel with carbon content 0.78% is called as eutectoid steel. In this project, the steel…

Q: 6. A sheet of aluminum alloy is cold-rolled 30 percent to a thickness of 0.080 in. If the sheet is…

A: According to the given dataX1=0.080 inX2=0.064 inFirst cold rolled is 30%We need to findTotal cold…

Q: Question-7. Steady-state creep data taken for an alloy at a stress level of 160 MPa are given below.…

A: Given, Data for stress level of 160 MPa The stress component , n =6.8 R = 8.3145 J/mol-K To find…

Q: | 1. A stacking fault region is a planar defect TT 2. Screw dislocation can cross slip under…

Q: Choose a common metal that is solid at 0°C and liquid at 1500°C. a) Look up the following values:…

Q: An 0.29 inch diameter copper bar is to be cold worked 57%. Find the final diameter in inches.

A: Givend=0.29 inchCW=57%

Q: produce a M=1000 h of S, = 20,000 000 psi, and a min

Q: sce M13.5 Strain measurement with rosettes Ea 500 µE Eb -200 με %3D Ec 250 με 30° 30° I| ||

A: “Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

Q: Suppose that the corrected fatigue limit of a component made from steel is S=600 f MPa. If the size,…

Concept explainers

Machining Processes and Machine Tools

Machining is represented as the process in which an excessive amount of substance is eliminated from the particular element in the form of pieces. In machining, cutting tools such as rotary tools, lathe, drill press, grinder, chop saw, welders, and many more are applied to get the desired shape. As compared to other manufacturing processes, machining is precise and versatile.

Question

this is an mechanical engineering queation.

The rough material for the operation is 100mm diameter mild steel, and the length of the rough material is
150mm. To machine the component, the diameter must be turned down from 100mm to 82mm for a length
of 100mm. You are going to achieve this with three cuts, with a depth of cut of 3.0mm per cut, as is shown in
the sketch below. Thereafter, the billet is machined for a length of 50mm to a diameter of 64mm, again in
three cuts of 3.0mm in depth each. The technical manual of the cutter supplier gives the following data for
the specific cutter:
f₂ = 0.2 mm/tooth/rev (Recommended feed)
Vc = 160 m/min (Tangential cutting speed or surface speed)
Calculate the following:
23.1 The required rotational speed for each of the three cuts
The material removal rate in mm³/min
23.2
23.3 The total cutting time. Assume that the tool approach and retract times are negligible in this
process
Rof materiaal
Rough material
Finale komponent
Final component

Formules / Formulae:
100
1
8/2
150
100
64
50
Rof materiaal
Rough material
Finale komponent
Final component
MRR=TT-Davg'd-f₂ N
Snit / Cut 1
Snit / Cut 2
Snit / Cut 3
Snit / Cut 4
Snit / Cut 5
Snit / Cut 6
t=1/(f₂.N)

Branch of science that deals with the stationary and moving bodies under the influence of forces.

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

Step 2

Step 3

Step 4

Step by step

Solved in 4 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

During 650°C, the average grain diameter of a brass material was measured as a function of time, as seen below at two distinct points of Time (min) Grain Diameter (mm) 30 3.9 × 10^-2 90 6.6 × 10^-2 a. What was the grain diameter at the start? (orig diameter) b. What is the expected grain diameter after 150 minutes at 650 °C?
Answer part (iv) and (v)
Need help on this Q
In analysis of an aluminum-zinc alloy, the mass of gelatin capsule is 0.1192g, mass of alloy is 0.2634g, mass of empty Beaker is 168.2449, mass of Beaker + Water is 272.5682g, the temperature of water is 22 degrees Celsius, the vapor pressure of water from table 1 is 19.827 mm Hg, Calculate mass of alloy sample.
the behaviour of graphite materials under stress
Pick the best soultion for the multiple choices. Modulus of resilience is: Slope of elastic portion of stress – strain curve Area under the elastic portion of the stress –strain curve Energy absorbed during fracture in a tension test Energy absorbed during fracture in an impact test Slope of the plastic portion of the stress- strain curve ( ) Fatigue failure occurs under the condition of: High elastic stress High corrosivity High stress fluctuations High temperature High rate of loading ( ) Creep failure of a material occurs most rapidly when the operating temperature is: Cryogenic temperature Equal to its melting point Grater than 0.4 times its melting point in K Greater than 200 oF Close to boiling point ( ) Young’s modulus of a material is indicative of its: Tensile strength Yield Strength Ductility Stiffness Corrosion Resistance…
Discuss the relationship between the grain size and hardness
Current Attempt in Progress During the recovery of a cold-worked material, which of the following statement(s) is (are) true? Some of the internal strain energy is relieved. O All of the internal strain energy is relieved. There is some reduction in the number of dislocations. There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state. The electrical conductivity is recovered to its precold-worked state. The thermal conductivity is recovered to its precold-worked state. The metal becomes more ductile, as in its precold-worked state. O Grains with high strains are replaced with new, unstrained grains.
Neat solution
1. Suppose that you have measured the strain aging kinetics of a high-strength low-alloy (HSLA) steel and obtained the following results: Aging Temperature (°C) Aging Time (min) ΔΥ (MPa) 150 240 40 130 1800 40 Determine (using a mathematical rather than graphical approach) the time necessary to obtain a strengthening increment, AY, of 40 MPa at an ageing temperature of 120°C. State any assumptions made in order to arrive at your approximation.
Tag question Consider the graph below for 3 metals which have been cold-worked. From the graphs, which of the following statements are true? Stress Select one or more: a. C. Strain & A e. B C b. B and C have similar hardness; C has greater brittleness B is softer than C and has nearly similar ductility A is harder than C and is more ductile d. A and B have similar hardness; A has greater brittleness A is harder than C and is less ductile f. B is harder than C but has nearly similar ductility 5 up i A C
6.8 Suppose that a cast Al-1 wt% Si alloy, contains equilibrium Si particles 1 µm in diameter. This alloy is to be heat treated by first putting all of the Si particles back into solution. (a) Draw a set of schematic curves indicating what will happen during this process. Temperature °C (b) Indicate the initial conditions, the final conditions (once a new equilibrium is reached) and the boundary conditions during dis- solution. (c) Estimate how long the alloy should be held at 550°C so as to ensure that all of the Si is dissolved. Al-Si 1500 1300 1100 900 700 500 300 660.452°C 0 -(A1) 10 AI 12.6 10 20 20 30 30 Atomic Percent Silicon 40 50 Thom L 60 Them 577+1°C 40 50 60 Weight Percent Silicon 70 70 80 80 90 90 100 1414°C (Si)- 100 Si