Pearson eText for Manufacturing Processes for Engineering Materials -- Instant Access (Pearson+)
6th Edition
ISBN: 9780137503520
Author: Serope Kalpakjian, Steven Schmid
Publisher: PEARSON+
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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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1.
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?
By performing torsion tests, which develop pure shear in a ductile specimen, does the maximum distortion energy theory accurate results?
Is the maximum normal stress theory useful to predict the accurate failure of brittle material?
Chapter 2 Solutions
Pearson eText for Manufacturing Processes for Engineering Materials -- Instant Access (Pearson+)
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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- Why do we define engineering and true stresses for tension/compression loading but not for shear loading?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_forwardDo not give answer in image and hand writingarrow_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_forwardIn True stress-true-strain curve in tension of solid metal cylinder 45 mm high and 8 mm in diameter, two pairs of values of stress and strain were given for the specimen metal after it had yielded (1) true stress = 217 MPa, and true strain = 0.35; and (2) true stress = 259 MPa, and true strain = 0.68. Based on these data points, determine the following: a) The average flow stress that the metal experiences if it is subjected to a stress that is equal to its strength coefficient K. b) The work done that the metal experiences if it is subjected to elongation in height of 45% c) If during the deformation the relative speed = 20 mm/s, determine the strain rate at h = 50 mm and h = 70 mm.arrow_forwardA steel rotating-beam test specimen has an ultimate strength, Sut, of 120 kpsi. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude, Orev, of 74 kpsi. Take the value of the fatigue strength fraction, f, from Fig. 6-23. The life of the specimen is cycles.arrow_forward
- Question # 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_forwardLet's assume that the stress-strain elongation curve of a high-strength steel is given by this equation. o = 1.4*10º"ɛ 3 2.9*10*ɛ ² + 2.1*10°z Calculate the yield limit value of this steel. Oa (kgf/cm³) (For Yield Strength, use a straight line from the starting point and Calculate wwwwwww ww w m by moving the line parallel to the given curve)arrow_forwardIf the ultimate strength of a steel rotating-beam test specimen was around 1600 MPa. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude of 900 MPa.arrow_forward
- Illustrate the effect caused by the longitudinal shear stress?arrow_forward140 900 1040 Steel Percent cold work = 0 %3D 120 Yield strength = 200 MPa 800 Yield strength = 29 ksi 700 100 600 Brass -80 500 400 60 Copper 300 40 200 10 20 40 30 50 60 70 Percent cold work Tensile strength (MPa) Tensile strength (ksi)arrow_forwardDon't provide handwritten solution and don't try if you don't knowarrow_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