Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Question
Chapter 9, Problem 10CQ
To determine
The reason for plastic strain in crystalline material at temperature below one half of the melting temperature.
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Chapter 9 Solutions
Materials Science And Engineering Properties
Ch. 9 - Prob. 1CQCh. 9 - Prob. 2CQCh. 9 - Prob. 3CQCh. 9 - Prob. 4CQCh. 9 - Prob. 5CQCh. 9 - Prob. 6CQCh. 9 - Prob. 7CQCh. 9 - Prob. 8CQCh. 9 - Prob. 9CQCh. 9 - Prob. 10CQ
Ch. 9 - Prob. 11CQCh. 9 - Prob. 12CQCh. 9 - Prob. 13CQCh. 9 - At temperatures above the equi-cohesive...Ch. 9 - Prob. 15CQCh. 9 - Prob. 16CQCh. 9 - Prob. 17CQCh. 9 - Prob. 18CQCh. 9 - Prob. 19CQCh. 9 - Prob. 20CQCh. 9 - Prob. 21CQCh. 9 - Prob. 22CQCh. 9 - Prob. 23CQCh. 9 - Prob. 24CQCh. 9 - Prob. 25CQCh. 9 - Prob. 26CQCh. 9 - Prob. 27CQCh. 9 - Prob. 28CQCh. 9 - Prob. 29CQCh. 9 - Prob. 30CQCh. 9 - Prob. 31CQCh. 9 - Prob. 32CQCh. 9 - Prob. 33CQCh. 9 - Prob. 34CQCh. 9 - Prob. 35CQCh. 9 - Prob. 1ETSQCh. 9 - Prob. 2ETSQCh. 9 - Prob. 3ETSQCh. 9 - Prob. 4ETSQCh. 9 - Prob. 5ETSQCh. 9 - Prob. 6ETSQCh. 9 - Prob. 7ETSQCh. 9 - Prob. 8ETSQCh. 9 - Prob. 9ETSQCh. 9 - Prob. 10ETSQCh. 9 - Prob. 11ETSQCh. 9 - Prob. 12ETSQCh. 9 - Prob. 9.1PCh. 9 - Prob. 9.2PCh. 9 - Prob. 9.3PCh. 9 - Prob. 9.4PCh. 9 - Prob. 9.5PCh. 9 - Prob. 9.6PCh. 9 - Prob. 9.7PCh. 9 - Prob. 9.8PCh. 9 - Prob. 9.9PCh. 9 - Prob. 9.10PCh. 9 - For silver at a tensile stress of 7 MPa and a...Ch. 9 - For germanium at a tensile stress of 410 MPa and a...Ch. 9 - Prob. 9.13PCh. 9 - Prob. 9.14PCh. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22P
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- An iron specimen is plastically deformed in shear by 1%, and it has u dislocation density of 1 10 14 m/ m 3 Assume that the dislocation density did not change in the 1% strain of thisspecimen, the Burger's vector (b) is a 2 [1 1 1] the slip plane is (110). the shear stress isapplied to the (110) plane, and the lattice parameter of the BCC iron is 0.286 nm. Calculate the magnitude of the Burger's vector for these dislocations in iron. Calculate the average distance moved by the mobile dislocations as a result of the 1% shear strain.arrow_forwardQus :arrow_forwardConsidering a finite cylinder of single crystal aluminum with a diameter (D) that has an axial unit screw dislocation at the center, what is the stress field around this dislocation in terms of D (diameter of cylinder) and r (radius which is the distance from dislocation center)arrow_forward
- The intensity of stress which causes unit strain is called O(A) Unit mass O(B) Modulus of rigidity OC) Bulk modulus O(D) Modulus of Elasticityarrow_forwardNarrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at T₁ = 70°F, all stresses are zero. Knowing that the temperature will be slowly raised to T₂ and then reduced to T₁, determine (a) the highest temperature T₂ that does not result in residual stresses, (b) the temperature T₂ that will result in a residual stress in the aluminum equal to 58 ksi. Assume aa = 12.8 x 10-6/°F for the aluminum and a = 6.5 × 10-6/°F for the steel. Further assume that the aluminum is elastoplastic with E = 10.9 × 106 psi and ay = 58 ksi. (Hint: Neglect the small stresses in the plate.) Fig. P2.121arrow_forwardA copper rod is deformed using a uniaxial tensile force of 16000 N. Deformation continues until sufficient strain hardening has occurred such that the applied force is too small to allow further deformation. After deformation, the rod has a diameter of 0.01 m and a length of 1.5 m. Assume that copper follows the strain hardening lawwith K of 310 MPa and n=0.54 Please calculate the true strain after the deformation ?arrow_forward
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- Given your understanding of what initiates and controls failure in materials, which of the following will increase the failure strength or lifetime of a test piece or component and why? a. Decreasing the difference between the maximum and minimum stress values, as this effects the stress concentration factor b. Decreasing the temperature below the brittle-ductile transition temperature, to make it harder C. Polishing to reduce surface defects Od. Increasing its volume, to give a larger cross sectional area Oe. Increasing the grain size so there are less grain boundaries to initiate failurearrow_forward1. The most important mechanical properties of brittle materials is Tensile strength compressive strength O rigidity hardness Creeparrow_forwardA cylindrical specimen of cold-worked steel has a Brinell hardness of 240. If the specimen remained cylindrical during deformation and its original radius was 11.8 mm, determine its radius after deformation. For steel, the dependence of tensile strength on percent cold work is shown in Animated Figure 7.19b. i mmarrow_forward
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