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

Videos

Question
Book Icon
Chapter 2, Problem 2.82P
To determine

The expression for force versus piston travel d up to d = 10 mm.

Expert Solution & Answer
Check Mark

Answer to Problem 2.82P

The expression for Force F versus piston travel d is

  F=8835.732d+294522ln(1+d25)0.4

And the value of force F at d =10 mm is 127996.71 N

Explanation of Solution

Given data:

  Diamter of cylindrical slug, D1=25mmHeight of cylindrical slug, d=25mm,Diamterof cavity in a rigid die, D2=25mm,Pressureacting of matrix is, pm=150ΔVVomMPa,Stress of the slug material, σ=600ε0.4.

A cylindrical is surrounded by a compressible matrix. Cylindrical slug is compressed vertically by a force F which is to be determined. When the force F is acted then there will be the generation of stress inside the cylindrical slug and pressure is also generated inside the compressible matrix. The force acted will be equal to the force generated inside the cylindrical slug plus the force generated inside the matrix.

Calculation:

Drawing the free body diagram of the given system,

  Manufacturing Processes for Engineering Materials (6th Edition), Chapter 2, Problem 2.82P

From the free body diagram, it can be written as,

  F=Fmatrix+Fcylinder...........(1)

Area of the matrix and cylinder can be given as −

  Area of the matrix,Am=π4(D12D22)Am=π4(502252)Am=1472.622mm2

  Area of the cylinder,Ac=π4D22Ac=π4×252Ac=490.87mm2

  Change in the volume for matrix can be given as,ΔV=Am[25(25d)]ΔV=Am×dΔV=1472.622×d

Calculating the force exerted by the matrix:-

  Fmatrix=pm×AreaFmatrix=150×ΔVVom×AreaVolume of the matrix will be, Vom=Am×25Fmatrix=150×ΔVAm×25×AmFmatrix=150×ΔV25Fmatrix=150×1472.622×d25Fmatrix=8835.732d...............(2)

The strain developed inside the cylinder for deformation of height d is −

  Strain,e=d25

True strain inside the cylinder would be −

  True strain, ε=ln(1+e)ε=ln(1+d25)

Writing true stress-true strain relation,

  σ=600ε0.4σ=600ln(1+d25)0.4

The force will be given as,

  Fc=σAcFc=600ln(1+d25)0.4×490.87Fc=294522ln(1+d25)0.4...................(3)

From equation (1), (2) and (3),

  F=Fmatrix+FcylinderBut,Fmatrix=8835.732dFc=294522ln(1+d25)0.4Therefore,F=8835.732d+294522ln(1+d25)0.4

The above is the required expression for force versus d .

Putting d = 10 mm we get,

  F=8835.732d+294522ln(1+d25)0.4F=8835.732×10+294522ln(1+1025)0.4F=88357.32+39639.39F=127996.71N

Hence, the expression for Force F versus piston travel d is

  F=8835.732d+294522ln(1+d25)0.4

And the value of force F at d =10 mm is 127996.71 N

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
2. Express the following complex numbers in rectangular form. (a) z₁ = 2еjл/6 (b) Z2=-3e-jπ/4 (c) Z3 = √√√3e-j³/4 (d) z4 = − j³
A prismatic beam is built into a structure. You can consider the boundary conditions at A and B to be fixed supports. The beam was originally designed to withstand a triangular distributed load, however, the loading condition has been revised and can be approximated by a cosine function as shown in the figure below. You have been tasked with analysing the structure. As the beam is prismatic, you can assume that the bending rigidity (El) is constant. wwo cos 2L x A B Figure 3: Built in beam with a varying distributed load In order to do this, you will: a. Solve the reaction forces and moments at point A and B. Hint: you may find it convenient to use the principal of superposition. (2%) b. Plot the shear force and bending moment diagrams and identify the maximum shear force and bending moment. (2%) c. Develop an expression for the vertical deflection. Clearly state your expression in terms of x. (1%)
Question 1: Beam Analysis Two beams (ABC and CD) are connected using a pin immediately to the left of Point C. The pin acts as a moment release, i.e. no moments are transferred through this pinned connection. Shear forces can be transferred through the pinned connection. Beam ABC has a pinned support at point A and a roller support at Point C. Beam CD has a roller support at Point D. A concentrated load, P, is applied to the mid span of beam CD, and acts at an angle as shown below. Two concentrated moments, MB and Mc act in the directions shown at Point B and Point C respectively. The magnitude of these moments is PL. Moment Release A B с ° MB = PL Mc= = PL -L/2- -L/2- → P D Figure 1: Two beam arrangement for question 1. To analyse this structure, you will: a) Construct the free body diagrams for the structure shown above. When constructing your FBD's you must make section cuts at point B and C. You can represent the structure as three separate beams. Following this, construct the…

Chapter 2 Solutions

Manufacturing Processes for Engineering Materials (6th Edition)

Ch. 2 - Prob. 2.11QCh. 2 - Prob. 2.12QCh. 2 - Prob. 2.13QCh. 2 - Prob. 2.14QCh. 2 - Prob. 2.15QCh. 2 - Prob. 2.16QCh. 2 - Prob. 2.17QCh. 2 - Prob. 2.18QCh. 2 - Prob. 2.19QCh. 2 - Prob. 2.20QCh. 2 - Prob. 2.21QCh. 2 - Prob. 2.22QCh. 2 - Prob. 2.23QCh. 2 - Prob. 2.24QCh. 2 - Prob. 2.25QCh. 2 - Prob. 2.26QCh. 2 - Prob. 2.27QCh. 2 - Prob. 2.28QCh. 2 - Prob. 2.29QCh. 2 - Prob. 2.30QCh. 2 - Prob. 2.31QCh. 2 - Prob. 2.32QCh. 2 - Prob. 2.33QCh. 2 - Prob. 2.34QCh. 2 - Prob. 2.35QCh. 2 - Prob. 2.36QCh. 2 - Prob. 2.37QCh. 2 - Prob. 2.38QCh. 2 - Prob. 2.39QCh. 2 - Prob. 2.40QCh. 2 - Prob. 2.41QCh. 2 - Prob. 2.42QCh. 2 - Prob. 2.43QCh. 2 - Prob. 2.44QCh. 2 - Prob. 2.45QCh. 2 - Prob. 2.46QCh. 2 - Prob. 2.47QCh. 2 - Prob. 2.48QCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.61PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71PCh. 2 - Prob. 2.72PCh. 2 - Prob. 2.73PCh. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. 2.76PCh. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. 2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. 2.82PCh. 2 - Prob. 2.83PCh. 2 - Prob. 2.84PCh. 2 - Prob. 2.85PCh. 2 - Prob. 2.86PCh. 2 - Prob. 2.87PCh. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - Prob. 2.90PCh. 2 - Prob. 2.91PCh. 2 - Prob. 2.92PCh. 2 - Prob. 2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - Prob. 2.99PCh. 2 - Prob. 2.100PCh. 2 - Prob. 2.101P
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
Understanding Failure Theories (Tresca, von Mises etc...); Author: The Efficient Engineer;https://www.youtube.com/watch?v=xkbQnBAOFEg;License: Standard youtube license