Concept explainers
(a)
The comparison between the magnitude of the total surface energy to the total bond energy.
Answer to Problem 3.9P
The magnitude of total energy of sphere is
Explanation of Solution
Given:
The surface energy of
The cohesive energy of
The lattice parameter of FCC Ni is
The total radius of the sphere is
Formula used:
The total surface energy of the sphere is given by
Here,
The atomic diameter is given by
Here,
The total bond energy is given by,
Here,
The ratio of total surface energy and bond energy is given by
Calculation:
The atomic diameter of the Ni atom is calculated as
Substitute
The atomic diameter of
The total surface energy of the sphere is calculated as,
Substitute
The total bond energy is calculated as
Substitute
Substitute
Conclusion:
Therefore, the magnitude of the total energy of the sphere is
(b)
Whether the
Answer to Problem 3.9P
Yes, the
Explanation of Solution
Formula used:
The total energy of the Ni atom is given by,
Calculation:
The total energy is calculated as,
Substitute
The total energy of the sphere is negative which signifies that the
Conclusion:
Therefore, the
Want to see more full solutions like this?
Chapter 3 Solutions
Materials Science And Engineering Properties
- In an engineering application, the material is a strip of iron with a fixed crystallographic structure subject to a tensile load during operation. The part failed (yielded) during operation and needs to be replaced with a component with better properties. You are told that two other iron strips had failed at yield stresses of 110 and 120 MPa, with grain sizes of 30 microns and 25 microns respectively. The current strip has a grain size of 20 microns. The diameter of the rod is 1 mm and the load applied is 100 N. What is the yield stress of the new part C and would you recommend it for operation? Select one: Oa. 133.5 MPa, yes O b. OC. Od Oe. 120.5 MPa, no 129.5, yes 140.5, no 123.5 MPa, yesarrow_forwardFive classes of ceramic materials have been defined.arrow_forwardWhich of the following statements are true of dislocations? Select one or more: a. Dislocations can be viewed with high powered microscopy and not with the naked eye b. Dislocations can move under stress c. Dislocations can arise due to shear deformation of the lattice Od. Dislocations primarily enable high stiffness in metalsarrow_forward
- Which of the following statements are true of dislocations? Select one or more: O a. Dislocations can move under stress O b. Dislocations can arise due to shear deformation of the lattice O C. Dislocations can be viewed with high powered microscopy and not with the naked eye Od. Dislocations primarily enable high stiffness in metalsarrow_forwardCalculate the unit cell edge length for an 57 wt% Ag- 43 wt% Pd alloy. All of the palladium is in solid solution, and the crystal structure for this alloy is FCC. Room temperature densities for Ag and Pd are 10.49 g/cm3 and 12.02 g/cm3, respectively, and their respective atomic weights are 107.87 g/mol and 106.4 g/mol. Report your answer in nanometers.arrow_forwardAt a temperature of 60°F, a 0.02-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=α=12.5 x 10-6/°F] bar with a width of 2.8 in. and a thickness of 0.85 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.6 in. and a thickness of 0.85 in. The supports at A and C are rigid. Assume h1=2.8 in., h2=1.6 in., L1=26 in., L2=40 in., and Δ=Δ= 0.02 in. Determine(a) the lowest temperature at which the two bars contact each other.(b) the normal stress in the two bars at a temperature of 225°F.(c) the normal strain in the two bars at 225°F.(d) the change in width of the aluminum bar at a temperature of 225°F.arrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning