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

Videos

Question
Book Icon
Chapter 2, Problem 2.96P

(a)

To determine

The true strain for the 1112 steel specimen during necking.

The true strain for the annealed 1112 steel specimen during necking.

The elongation of the 1112 steel specimen during necking.

The elongation of the annealed 1112 steel specimen during necking.

(a)

Expert Solution
Check Mark

Answer to Problem 2.96P

The true strain for the 1112 steel specimen during necking is 0.08 .

The true strain for the annealed 1112 steel specimen during necking is 0.19 .

The elongation of the 1112 steel specimen during necking is 8.32% .

The elongation of the annealed 1112 steel specimen during necking is 20.92% .

Explanation of Solution

Given:

The diameter of specimen is d=20mm .

The length of the specimen is lo=25mm .

Formula used:

The expression for the true strain for 1112 steel specimen during necking is given as,

  ε=n

Here, n is the exponent for the 1112 steel specimen.

The expression for the true strain for 1112 annealed steel specimen during necking is given as,

  εo=no

Here, no is the exponent for the 1112 annealed steel specimen.

The expression for the true strain for the 1112 steel specimen is given as,

  ε=ln(llo)

Here, l is the length of the cold rolled 1112 steel after necking.

The expression for the true strain for the 1112 annealed steel specimen is given as,

  εo=ln(lelo)

Here, le is the length of the 1112 annealed steel after necking.

The expression for the elongation of the 1112 steel specimen at that instant is given as,

  e=(llolo)×100

The expression for the elongation of the 1112 annealed steel specimen at that instant is given as,

  eo=(lel0lo)×100

Calculation:

The properties of 1112 steel specimen is given as,

The constant is K=760MPa .

The exponent is n=0.08 .

The properties of 1112 annealed steel specimen is given as,

The constant is Ko=760MPa .

The exponent is no=0.19 .

The true strain when necking occurs can be calculated as,

  ε=nε=0.08

The true strain for 1112 annealed steel specimen during necking can be calculated as,

  εo=noεo=0.19

The length of the cold rolled 1112 steel after necking can be calculated as,

  ε=ln(l l o )0.08=ln(l 25mm)e0.08=ln(l 25mm)l=27.08mm

The elongation of the 1112 steel specimen at that instant can be calculated as,

  e=( l l o l o )×100e=( 27.08mm25mm 25mm)×100e=8.32%

The length of the 1112 annealed steel after necking can be calculated as,

  εo=ln( l e l o )0.19=ln( l e 25mm)le=30.23mm

The elongation of the 1112 annealed steel specimen at that instant can be calculated as,

  eo=( l e l 0 l o )×100eo=( 30.23mm25mm 25mm)×100eo=20.92%

Conclusion:

Therefore, the true strain for the 1112 steel specimen during necking is 0.08 .

Therefore, the true strain for the annealed 1112 steel specimen during necking is 0.19 .

Therefore, the elongation of the 1112 steel specimen during necking is 8.32% .

Therefore, the elongation of the annealed 1112 steel specimen during necking is 20.92% .

(b)

To determine

The ultimate strength for 1112 steel specimen.

The ultimate strength for 1112 annealed steel specimen.

(b)

Expert Solution
Check Mark

Answer to Problem 2.96P

The ultimate strength for 1112 steel specimen is 573MPa .

The ultimate strength for 1112 annealed steel specimen is 458.13MPa .

Explanation of Solution

Formula used:

The expression for the ultimate strength for 1112 steel specimen is given as,

  Ss=Kεn

The expression for the ultimate tensile strength for 1112 steel specimen is given as,

  UTSs=Ss×en

The expression for the ultimate strength for 1112 annealed steel specimen is given as,

  Sa=Koεono

The expression for the ultimate tensile strength for 1112 annealed steel specimen is given as,

  UTSa=Saen

Calculation:

The ultimate strength for 1112 steel specimen can be calculated as,

  Ss=KεnSs=760MPa(0.08)0.08Ss=621MPa

The ultimate tensile strength for 1112 steel specimen can be calculated as,

  UTSs=Ss×enUTSs=621MPa×e0.08UTSs=573MPa

The ultimate strength for 1112 annealed steel specimen can be calculated as,

  Sa=KoεonoSa=760MPa(0.19)0.19Sa=554MPa

The ultimate tensile strength for 1112 annealed steel specimen can be calculated as,

  UTSa=SaenUTSa=554MPa×e0.19UTSa=458.13MPa

Conclusion:

Therefore, the ultimate strength for 1112 steel specimen is 573MPa .

Therefore, the ultimate strength for 1112 annealed steel specimen is 458.13MPa .

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

EBK MANUFACTURING PROCESSES FOR ENGINEE

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
An Introduction to Stress and Strain; Author: The Efficient Engineer;https://www.youtube.com/watch?v=aQf6Q8t1FQE;License: Standard YouTube License, CC-BY