EBK MANUFACTURING PROCESSES FOR ENGINEE
6th Edition
ISBN: 9780134425115
Author: Schmid
Publisher: YUZU
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Chapter 2, Problem 2.97P
To determine
The Mohr’s circle diagram for this stress state.
The stress
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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?
In the attached picture there is a sketch of a socket wrench. Assume the wrench is held at a fixed point “A”. The yield stress of the material is known to be 500 MPa. Answer the questions below
Describe the stresses at point “A” and their causes and calculate the stresses.
Determine the factor of safety against yield assuming the Tresca yield criteria.
Determine the factor of safety against yield assuming the von Mises yield criteria using both principal stresses and “Cartesian” stresses. Do your values match or not, and is this expected? Explain.
Do the calculated values make sense with the respect to the Tresca value? Explain, why or why not?
A brass specimen is subjected to a Tensile
test in a laboratory.
The following data obtained are:
Length of Specimen = 322 mm; Diameter
of Specimen = 10 mm;
Yield Stress = 1279 MPa; Ultimate Stress
= 2446 N/mm²;
Fracture Stress = 1780 MPa; % of
Elongation = 49 %
% of Reduction in area = 27 %. Determine
the elongation at 43 kN, Load at yielding,
Maximum load, Load at fracture, Final
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Modulus as 118 GPa.
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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- Uniaxial yield stress of a material is 260 MPa Determine shear yield stress using Von Mises criteria.arrow_forwardFor some metal alloy, a true stress 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 418 MPa (60630 psi) is applied if the original length is 500 mm (19.69in) ? Assume a value of 0.22 for the strain- hardening exponent, narrow_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
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