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?
Uniaxial yield stress of a material is 260 MPa
Determine shear yield stress using Von Mises
criteria.
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 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 Length & Final diameter. Take Young's Modulus as 118 GPa.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 tensile test it can be shown that the percent cold work (%CW) is related to strain according to the following relationship: %CW=(εε+1) × 100 Using this equation, compute the percent cold work experienced by naval brass (for which the stress-strain behavior is shown in the Animated Figure 6.12) when a stress of 415 MPa (60190 psi) is applied.arrow_forward
- For 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_forward8. A copper wire of diameter 0.80mm fails at an Engineering stress = 248.2MPA. Its ductility is measured as 75% reduction in area. Determine the True stress and True strain at failure.arrow_forwardI need the answer as soon as possiblearrow_forward
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