Manufacturing Processes for Engineering Materials (6th Edition)
Manufacturing Processes for Engineering Materials (6th Edition)
6th Edition
ISBN: 9780134290553
Author: Serope Kalpakjian, Steven Schmid
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.120P
To determine

The rolling force and the power requirement for single pass.

The rolling force and the power requirement for double pass.

Expert Solution & Answer
Check Mark

Answer to Problem 6.120P

Rolling force is 57665.23N and the power required is 5150.1W for single pass.

Rolling force is 40796N and the power required is 2577.66W for double pass.

Explanation of Solution

Given:

The initial height is ho=1.5mm .

The final height is hf=0.9mm .

The radius of the rollers is R=75mm .

The coefficient of friction is μ=0.1 .

The width of the sheet is w=25mm .

Velocity of the roll surface is v=1m/s .

Formula used:

The expression for change in height for single pass is given as,

  Δh1=hohf

The expression for change in height for double pass is given as,

  Δh2=hohf2

The expression for contact length for single passis given as,

  L1=RΔh1

The expression for contact length is given as,

  L2=RΔh2

The expression for rolling force is given as,

  F1=σoL1w

The rolling force for double pass can be given as,

  F2=σoL2w

Here, σo is the roll separating forceis given as,

The expression for roll separatingstressis given as,

  σo=Kεnn+1

Here, strength coefficient is K and strength hardening exponent is n.

The expression for the strain is given as,

  ε=ln(h0hf)

The expression for the angular velocity is given as,

  ω=vR

The expression for the arm length is given as,

  a1=L12

The expression for the arm length in double pass is given as,

  a2=L22

The expression for the torque is given as,

  T1=F1×a

The expression for the torque for double pass is given as,

  T2=F2×a

The expression for the power is given as,

  P1=2×T1×ω

The expression for the power for double pass is given as,

  P2=2×T2×ω

Calculation:

The change in height can be calculated as,

  Δh1=hohfΔh1=1.5mm0.9mmΔh1=0.6mm

The contact length can be calculated as,

  L1=RΔh1L1=75mm×0.6mmL1=6.7mm

The strain can be calculated as,

  ε=ln( h 0 h f )ε=ln( 1.5 0.9)ε=0.51

The roll separating stress can be calculated as,

  σo=Kεnn+1σo=580MPa× ( 0.51 ) 0.341+0.34σo=344.27MPa

Refer to table 2.2“Typical values of strength coefficient K and strength hardening exponent n ” for 8515 brass is,

  K=580MPan=0.34

The force for single pass can be calculated as,

  F1=σoL1wF1=344.27MPa×106Pa×6.7×103m×25×103mF1=57665.23N

The angular velocity can be calculated as,

  ω=vRω=1m/s75× 103mω=13.33s-1

The arm length can be calculated as,

  a1=L12a1=6.7× 10 3m2a1=3.35×103m

The torque can be calculated as,

  T1=F1×a1T1=57665.23N×3.35×103mT1=193.178Nm

The power can be calculated as,

  P1=2×T1×ωP1=2×193.178Nm×13.331/sP1=5150.1W

The change in height for double pass can be calculated as,

  Δh2=hohf2Δh2=1.5mm0.9mm2Δh2=0.3mm

The contact length for double pass can be given as,

  L2=RΔh2L2=75mm×0.3mmL2=4.74×103m

The arm length for the double pass can be calculated as,

  a2=L22a2=4.74× 10 3m2a2=2.37×103m

The rolling force for double pass can be given as,

  F2=σoL2wF2=344.27×106Pa×4.74×103N×25×103mF2=40796N

The torque for the double pass can be given as,

  T2=F2×a2T2=40796N×2.37×103mT2=96.686N.m

The power required for double pass can be given as,

  P2=2×T2×ωP2=2×96.686Nm×13.331/sP2=2577.66W

Conclusion:

Therefore, rolling force is 57665.23N and the power required is 5150.1W for single pass.

Therefore, rolling force is 40796N and the power required is 2577.66W for double pass.

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

Manufacturing Processes for Engineering Materials (6th Edition)

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
Types of Manufacturing Process | Manufacturing Processes; Author: Magic Marks;https://www.youtube.com/watch?v=koULXptaBTs;License: Standard Youtube License