Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
It is desirable to produce a cylindrical steel bar (1040 steel) with min. tensile strength (TS) = 865 MPa, a ductility of at least 10 (%EL = 10) and a diam. of 6.3 mm. What we have available are bars of 8 mm which are deformed 20%. (%CW = 20.) Show the procedure needed to achieve the properties described above. Assume that the steel bar can withstand a CW = 40% before failure occurs.

It is desirable to produce a cylindrical steel bar (1040 steel) with min. tensile strength (TS) = 865 MPa, a ductility of at least 10 (%EL = 10) and a diam. of 6.3 mm. What we have available are bars of 8 mm which are deformed 20%. (%CW = 20.) Show the procedure needed to achieve the properties described above. Assume that the steel bar can withstand a CW = 40% before failure occurs.

Elements Of Electromagnetics
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
icon
Related questions
Question
It is desirable to produce a cylindrical steel bar (1040 steel) with min. tensile strength (TS) = 865 MPa, a ductility of at least 10 (%EL = 10) and a diam. of 6.3 mm. What we have available are bars of 8 mm which are deformed 20%. (%CW = 20.) Show the procedure needed to achieve the properties described above. Assume that the steel bar can withstand a CW = 40% before failure occurs.
Transcribed Image Text:It is desirable to produce a cylindrical steel bar (1040 steel) with min. tensile strength (TS) = 865 MPa, a ductility of at least 10 (%EL = 10) and a diam. of 6.3 mm. What we have available are bars of 8 mm which are deformed 20%. (%CW = 20.) Show the procedure needed to achieve the properties described above. Assume that the steel bar can withstand a CW = 40% before failure occurs.
Expert Solution
Step 1

Given Data

  • The required diameter is:Dd=6.3 mm
  • The available diamter is:Do=8 mm
  • The ductility is:Ductility=10%EL
  • The deformation is:%CW=20%

Nwo let us check how much excess CW is to be done on the avilable steel material ,

%CW=Ao-AdAo×100%CW=Do2-Dd2Do2×100%CW=82-6.3282×100%CW=37.98%

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS