EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Question
Chapter 2, Problem 2.71P
To determine
The two different and specific examples where the maximum shear stress and distortion energy theory give the same answer.
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Check out a sample textbook solutionStudents have asked these similar questions
What is the Distortion energy failure theory? Under what loading types and materials would you use it?
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following principal stresses:
(a) determine the minimum factor of safety based on yielding according to the maximum-
shear-stress theory.
(b) determine the minimum factor of safety based on yielding according to the
distortion-energy theory.
Chapter 2 Solutions
EBK MANUFACTURING PROCESSES FOR ENGINEE
Ch. 2 - Prob. 2.1QCh. 2 - Prob. 2.2QCh. 2 - Prob. 2.3QCh. 2 - Prob. 2.4QCh. 2 - Prob. 2.5QCh. 2 - Prob. 2.6QCh. 2 - Prob. 2.7QCh. 2 - Prob. 2.8QCh. 2 - Prob. 2.9QCh. 2 - Prob. 2.10Q
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
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- A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states:arrow_forwardBy performing torsion tests, which develop pure shear in a ductile specimen, does the maximum distortion energy theory accurate results?arrow_forward1. A part made of Aluminum 6061-T6 has a yield strength = 400 MPa. For each stress state below, draw all 3 Mohr's circles, find the principal stresses, and calculate the safety factor against yield using both the distortion-energy (von Mises) and maximum shear stress (Tresca) criterions. (If relevant) A clearly labeled diagram (or diagrams) clearly pertaining to your analysis with a coordinate system and relevant labels. Final answer with appropriate units and significant figures. You can use the fprintf() command in MATLAB to format numerical results A 2-3 sentence reflection on your answer. Does it make sense? Why or why not? What are some implications?arrow_forward
- Is the maximum normal stress theory useful to predict the accurate failure of brittle material?arrow_forwardWhy do we define engineering and true stresses for tension/compression loading but not for shear loading?arrow_forwardWhen can the shear stress be determined from the torsion formula?arrow_forward
- Which predicts the lower yield strength for most combinations of applied stress––the Tresca or the von Mises yield criterion? Under what stress conditions are the predictions equal?arrow_forward1. A ductile hot-rolled steel bar has a yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories, determine the factors of safety for the following plane stress state: 75 MPa 50 MPa 50 MPa 2. Consider a bar of AISI 1015 cold-drawn steel. Using the distortion-energy and maximum-shear-stress theories to determine the factors of safety for a stress state with the following plane principal stresses: 0A = 30 kpsi, OB = 15 kpsi.arrow_forwardFor a point on a steel specimen, the principal stresses are calculated as 01 = 200 MPa and 2 = -40 MPa. Calculate the required yield strength of the material according to the Von Mises yield criterion if a safety factor of 2.1 is used. Give your answer in MPa to 3 significant figures.arrow_forward
- "The maximum principal stress yield criterion is an appropriate choice for ductile materials but the maximum principal strain criterion is preferable". Is this true or false?arrow_forwardDefine about the Selection of Shear Strength Parameters ?arrow_forwardQuestion # 2 A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories, analyze the following plane stress states/principal stresses. (d) ơ, = -12 kpsi, o, = 15 kpsi, 7y = -9 kpsi (e) 0z = -24 kpsi, ơ, = -24 kpsi, 7y = –15 kpsi = 30 kpsi TA = 30 kpsi, OB = 30 kpsi, oB = -30 kpsi (g)arrow_forward
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Understanding Failure Theories (Tresca, von Mises etc...); Author: The Efficient Engineer;https://www.youtube.com/watch?v=xkbQnBAOFEg;License: Standard youtube license