EBK MANUFACTURING PROCESSES FOR ENGINEE
EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.75P

(a)

To determine

The force vs. reduction in height curve in open die forging of cylinder for no friction between the flat dies and the specimen.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial thickness of the specimen is ho=10mm .

The initial radius of the specimen is do=25mm .

The friction coefficient is μ=0 .

Formula used:

The expression for the flow stress is given as,

  σf=Kεn ……(1)

Here, σf is the flow stress, K is the strength coefficient, ε is the true strain, n is the strain hardening coefficient.

The expression for the true strain is given as,

  ε=ln(hohf)

Here, hf is the final thickness.

The expression for the final radius by equating the volume is given as,

  df2=do2hohf

The expression for the forging force is given as,

  F=paπdf24

Here, pa is the average pressure.

The expression for the average pressure is given as,

  pa=σf(1+μdf3hf)

The expression for final height for 10% reduction in height in given as,

  ( h o h f h o )×100%=10%hf=0.9ho

The expression for final height for 20% reduction in height in given as,

  ( h o h f h o )×100%=20%hf=0.8ho

The expression forfinal height for 30% reduction in height in given as,

  ( h o h f h o )×100%=30%hf=0.7ho

The expression for final height for 40% reduction in height in given as,

  ( h o h f h o )×100%=40%hf=0.6ho

The expression for final height for 50% reduction in height in given as,

  ( h o h f h o )×100%=50%hf=0.5ho

Calculation:

For 10% reduction,

The final height can be calculated as,

  hf=0.9hohf=0.9×10mmhf=9mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm9mmdf=26.35mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 9mm)ε=0.1053

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.10536)0.54σf=93.44MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=93.44MPa×(1+ 0×26.35mm 3×9mm)pa=93.44MPa

The forging force can be calculated as,

  F=paπdf24F=93.44MPa×3.14×( 26.35mm)42F=50928N( 1MN 10 6 N)F=0.050928MN

For 20% reduction,

The final height can be calculated as,

  hf=0.8hohf=0.8×10mmhf=8mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm8mmdf=27.95mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 8mm)ε=0.223

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.223)0.54σf=140.135MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=140.135MPa×(1+ 0×27.95mm 3×8mm)pa=140.135MPa

The forging force can be calculated as,

  F=paπdf24F=140.135MPa×3.14×( 27.95mm)42F=85936N( 1MN 10 6 N)F=0.085MN

For 30% reduction,

The final height can be calculated as,

  hf=0.7hohf=0.7×10mmhf=7mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm7mmdf=29.88mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 7mm)ε=0.3566

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.3566)0.54σf=180.525MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=180.525MPa×(1+ 0×29.88mm 3×7mm)pa=180.525MPa

The forging force can be calculated as,

  F=paπdf24F=180.525MPa×3.14×( 29.88mm)42F=126522N( 1MN 10 6 N)F=0.1265MN

For 40% reduction,

The final height can be calculated as,

  hf=0.6hohf=0.6×10mmhf=6mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm6mmdf=32.275mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 6mm)ε=0.5108

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.5108)0.54σf=219.17MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=219.17MPa×(1+ 0×32.275mm 3×6mm)pa=219.17MPa

The forging force can be calculated as,

  F=paπdf24F=219.17MPa×3.14×( 32.275mm)42F=179218N( 1MN 10 6 N)F=0.1792MN

For 50% reduction,

The final height can be calculated as,

  hf=0.5hohf=0.5×10mmhf=5mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm5mmdf=35.35mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 5mm)ε=0.69

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.69)0.54σf=258.437MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=258.437MPa×(1+ 0×35.35mm 3×5mm)pa=258.437MPa

The forging force can be calculated as,

  F=paπdf24F=258.437MPa×3.14×( 35.35mm)42F=253514N( 1MN 10 6 N)F=0.2535MN

For μ=0 ,

    Reduction (in %
    )
    Forging force (in MN
    )
    100.0509
    200.085
    300.1265
    400.1792
    500.2535

The plot between forging force and reduction in height is shown in figure (1) below,

  EBK MANUFACTURING PROCESSES FOR ENGINEE, Chapter 6, Problem 6.75P , additional homework tip  1

Figure (1)

(b)

To determine

The force vs. reduction in height curve in open die forging of cylinder for μ=0.1 between the flat dies and the specimen.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial thickness of the specimen is ho=10mm .

The initial radius of the specimen is do=25mm .

The friction coefficient is μ=0.1 .

Formula used:

The expression for the flow stress is given as,

  σf=Kεn …… (1)

Here, σf is the flow stress, K is the strength coefficient, ε is the true strain, n is the strain hardening coefficient.

The expression for the true strain is given as,

  ε=ln(hohf)

Here, hf is the final thickness.

The expression for the final radius by equating the volume is given as,

  df2=do2hohf

The expression for the forging force is given as,

  F=paπdf24

Here, pa is the average pressure.

The expression for the average pressure is given as,

  pa=σf(1+μdf3hf)

The expression for final height for 10% reduction in height in given as,

  ( h o h f h o )×100%=10%hf=0.9ho

The expression for final height for 20% reduction in height in given as,

  ( h o h f h o )×100%=20%hf=0.8ho

The expression forfinal height for 30% reduction in height in given as,

  ( h o h f h o )×100%=30%hf=0.7ho

The expression for final height for 40% reduction in height in given as,

  ( h o h f h o )×100%=40%hf=0.6ho

The expression for final height for 50% reduction in height in given as,

  ( h o h f h o )×100%=50%hf=0.5ho

Calculation:

For 10% reduction,

The final height can be calculated as,

  hf=0.9hohf=0.9×10mmhf=9mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm9mmdf=26.35mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 9mm)ε=0.1053

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.10536)0.54σf=93.44MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=93.44MPa×(1+ 0.1×26.35mm 3×9mm)pa=102.56MPa

The forging force can be calculated as,

  F=paπdf24F=102.56MPa×3.14×( 26.35mm)42F=58898N( 1MN 10 6 N)F=0.05889MN

For 20% reduction,

The final height can be calculated as,

  hf=0.8hohf=0.8×10mmhf=8mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm8mmdf=27.95mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 8mm)ε=0.223

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.223)0.54σf=140.135MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=140.135MPa×(1+ 0.1×27.95mm 3×8mm)pa=156.454MPa

The forging force can be calculated as,

  F=paπdf24F=156.454MPa×3.14×( 27.95mm)42F=95943.4N( 1MN 10 6 N)F=0.0959MN

For 30% reduction,

The final height can be calculated as,

  hf=0.7hohf=0.7×10mmhf=7mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm7mmdf=29.88mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 7mm)ε=0.3566

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.3566)0.54σf=180.525MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=180.525MPa×(1+ 0.1×29.88mm 3×7mm)pa=206.211MPa

The forging force can be calculated as,

  F=paπdf24F=206.211MPa×3.14×( 29.88mm)42F=144524N( 1MN 10 6 N)F=0.1445MN

For 40% reduction,

The final height can be calculated as,

  hf=0.6hohf=0.6×10mmhf=6mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm6mmdf=32.275mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 6mm)ε=0.5108

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.5108)0.54σf=219.17MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=219.17MPa×(1+ 0.1×32.275mm 3×6mm)pa=258.47MPa

The forging force can be calculated as,

  F=paπdf24F=258.47MPa×3.14×( 32.275mm)42F=211354N( 1MN 10 6 N)F=0.2113MN

For 50% reduction,

The final height can be calculated as,

  hf=0.5hohf=0.5×10mmhf=5mm

The final radius can be calculated as,

  df2=do2hohfdf2= ( 25mm )2×10mm5mmdf=35.35mm

The true strain can be calculated as,

  ε=ln( h o h f )ε=ln( 10mm 5mm)ε=0.69

The flow stress can be calculated as,

  σf=Kεnσf=315MPa×(0.69)0.54σf=258.437MPa

Refer to table 2.2 “Typical values of strength coefficient K and strength hardening exponent n ” for annealed copper is,

  K=580MPan=0.34

The average pressure can be calculated as,

  pa=σf(1+ μ d f 3 h f )pa=258.437MPa×(1+ 0.1×35.35mm 3×5mm)pa=319.34MPa

The forging force can be calculated as,

  F=paπdf24F=319.34MPa×3.14×( 35.35mm)42F=313256N( 1MN 10 6 N)F=0.3132MN

For μ=0.1

    Reduction (in %
    )
    Forging force (in MN
    )
    100.05889
    200.0959
    300.1445
    400.2113
    500.3132

The plot between forging force and reduction in height is shown in figure (2) below,

  EBK MANUFACTURING PROCESSES FOR ENGINEE, Chapter 6, Problem 6.75P , additional homework tip  2

Figure (2)

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Continuity equation A y x dx D T معادلة الاستمرارية Ly X Q/Prove that ди хе + ♥+ ㅇ? he me ze ོ༞“༠ ?
Q Derive (continuity equation)? I want to derive clear mathematics.
motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

Chapter 6 Solutions

EBK MANUFACTURING PROCESSES FOR ENGINEE

Ch. 6 - Prob. 6.11QCh. 6 - Prob. 6.12QCh. 6 - Prob. 6.13QCh. 6 - Prob. 6.14QCh. 6 - Prob. 6.15QCh. 6 - Prob. 6.16QCh. 6 - Prob. 6.17QCh. 6 - Prob. 6.18QCh. 6 - Prob. 6.19QCh. 6 - Prob. 6.20QCh. 6 - Prob. 6.21QCh. 6 - Prob. 6.22QCh. 6 - Prob. 6.23QCh. 6 - Prob. 6.24QCh. 6 - Prob. 6.25QCh. 6 - Prob. 6.26QCh. 6 - Prob. 6.27QCh. 6 - Prob. 6.28QCh. 6 - Prob. 6.29QCh. 6 - Prob. 6.30QCh. 6 - Prob. 6.31QCh. 6 - Prob. 6.32QCh. 6 - Prob. 6.33QCh. 6 - Prob. 6.34QCh. 6 - Prob. 6.35QCh. 6 - Prob. 6.36QCh. 6 - Prob. 6.37QCh. 6 - Prob. 6.38QCh. 6 - Prob. 6.39QCh. 6 - Prob. 6.40QCh. 6 - Prob. 6.41QCh. 6 - Prob. 6.42QCh. 6 - Prob. 6.43QCh. 6 - Prob. 6.44QCh. 6 - Prob. 6.45QCh. 6 - Prob. 6.46QCh. 6 - Prob. 6.47QCh. 6 - Prob. 6.48QCh. 6 - Prob. 6.49QCh. 6 - Prob. 6.50QCh. 6 - Prob. 6.51QCh. 6 - Prob. 6.52QCh. 6 - Prob. 6.53QCh. 6 - Prob. 6.54QCh. 6 - Prob. 6.55QCh. 6 - Prob. 6.56QCh. 6 - Prob. 6.57QCh. 6 - Prob. 6.58QCh. 6 - Prob. 6.59QCh. 6 - Prob. 6.60QCh. 6 - Prob. 6.61QCh. 6 - Prob. 6.62QCh. 6 - Prob. 6.63QCh. 6 - Prob. 6.64QCh. 6 - Prob. 6.65QCh. 6 - Prob. 6.66QCh. 6 - Prob. 6.67QCh. 6 - Prob. 6.68QCh. 6 - Prob. 6.69QCh. 6 - Prob. 6.70QCh. 6 - Prob. 6.71QCh. 6 - Prob. 6.72QCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. 6.83PCh. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - Prob. 6.86PCh. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - Prob. 6.93PCh. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - Prob. 6.102PCh. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Prob. 6.107PCh. 6 - Prob. 6.108PCh. 6 - Prob. 6.109PCh. 6 - Prob. 6.110PCh. 6 - Prob. 6.111PCh. 6 - Prob. 6.112PCh. 6 - Prob. 6.113PCh. 6 - Prob. 6.114PCh. 6 - Prob. 6.115PCh. 6 - Prob. 6.116PCh. 6 - Prob. 6.117PCh. 6 - Prob. 6.118PCh. 6 - Prob. 6.119PCh. 6 - Prob. 6.120PCh. 6 - Prob. 6.121PCh. 6 - Prob. 6.122PCh. 6 - Prob. 6.123PCh. 6 - Prob. 6.124PCh. 6 - Prob. 6.125PCh. 6 - Prob. 6.126PCh. 6 - Prob. 6.127PCh. 6 - Prob. 6.128PCh. 6 - Prob. 6.129PCh. 6 - Prob. 6.130PCh. 6 - Prob. 6.131PCh. 6 - Prob. 6.132PCh. 6 - Prob. 6.133PCh. 6 - Prob. 6.134PCh. 6 - Prob. 6.135PCh. 6 - Prob. 6.136PCh. 6 - Prob. 6.137PCh. 6 - Prob. 6.138PCh. 6 - Prob. 6.139PCh. 6 - Prob. 6.140PCh. 6 - Prob. 6.142DCh. 6 - Prob. 6.143DCh. 6 - Prob. 6.144DCh. 6 - Prob. 6.145DCh. 6 - Prob. 6.146DCh. 6 - Prob. 6.147DCh. 6 - Prob. 6.149D
Knowledge Booster
Background pattern image
Mechanical Engineering
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
An Introduction to Stress and Strain; Author: The Efficient Engineer;https://www.youtube.com/watch?v=aQf6Q8t1FQE;License: Standard YouTube License, CC-BY