EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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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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What is the Distortion energy failure theory? Under what loading types and materials would you use it?
(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.
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:
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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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
- 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_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_forwardBy performing torsion tests, which develop pure shear in a ductile specimen, does the maximum distortion energy theory accurate results?arrow_forward
- Is the maximum normal stress theory useful to predict the accurate failure of brittle material?arrow_forwardWhen can the shear stress be determined from the torsion formula?arrow_forwardWhy do we define engineering and true stresses for tension/compression loading but not for shear loading?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
- Define about the Selection of Shear Strength Parameters ?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_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