Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Question
Chapter 6, Problem 6ETSQ
To determine
The reason for the difficulty in dislocation motion in covalently bonded materials such as diamond.
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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 metals
Which 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 metals
Narrow 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. P2.121
Chapter 6 Solutions
Materials Science And Engineering Properties
Ch. 6 - Prob. 1CQCh. 6 - Prob. 2CQCh. 6 - Prob. 3CQCh. 6 - Prob. 4CQCh. 6 - Prob. 5CQCh. 6 - Prob. 6CQCh. 6 - Prob. 7CQCh. 6 - Prob. 8CQCh. 6 - Prob. 9CQCh. 6 - Prob. 10CQ
Ch. 6 - Prob. 11CQCh. 6 - Prob. 12CQCh. 6 - Prob. 13CQCh. 6 - Prob. 14CQCh. 6 - Prob. 15CQCh. 6 - Prob. 16CQCh. 6 - Prob. 17CQCh. 6 - Prob. 18CQCh. 6 - Prob. 19CQCh. 6 - Prob. 20CQCh. 6 - Prob. 21CQCh. 6 - Prob. 22CQCh. 6 - Prob. 23CQCh. 6 - Prob. 24CQCh. 6 - Prob. 25CQCh. 6 - Prob. 26CQCh. 6 - Prob. 27CQCh. 6 - Prob. 28CQCh. 6 - Prob. 29CQCh. 6 - Prob. 30CQCh. 6 - Prob. 31CQCh. 6 - Prob. 32CQCh. 6 - Prob. 33CQCh. 6 - Prob. 34CQCh. 6 - Prob. 35CQCh. 6 - Prob. 36CQCh. 6 - Prob. 37CQCh. 6 - Prob. 38CQCh. 6 - Prob. 1ETSQCh. 6 - Prob. 2ETSQCh. 6 - Prob. 3ETSQCh. 6 - Prob. 4ETSQCh. 6 - Prob. 5ETSQCh. 6 - Prob. 6ETSQCh. 6 - Prob. 7ETSQCh. 6 - Prob. 8ETSQCh. 6 - Prob. 9ETSQCh. 6 - At the ultimate tensile strength. (a) The true...Ch. 6 - Prob. 11ETSQCh. 6 - Prob. 12ETSQCh. 6 - Prob. 13ETSQCh. 6 - Prob. 14ETSQCh. 6 - Prob. 15ETSQCh. 6 - Prob. 16ETSQCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Compare the engineering and true secant elastic...Ch. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - An iron specimen is plastically deformed in shear...Ch. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10PCh. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Estimate the elastic and plastic strain at the...Ch. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.1DPCh. 6 - Prob. 6.2DP
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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_forwardDraw a tensile stress-strain curve for a typical semi-crystalline polymer such as LLDPE, and define the three main regions on the curve.arrow_forwardFCC single crystal is alligned such that [001] is parallel to the stress axis, σ=100MPa.arrow_forward
- The constant a for a steel material having an ultimate strength of 106.5 ksi isarrow_forwardGiven that the yield limit of the ideal plastic material is the interface as shown in the figure d₁ b₂ (a) find the plastic bending moment of (b) aarrow_forward5) A single zinc crystal is loaded in tension with the normal to its slip plane at 60° to the tensile axis and the slip direction at 40° to the tensile axis. a) Calculate the resolved shear stress when a tensile stress of 0.69 MPa is applied. b) What tensile stress is necessary to reach the critical resolved shear stress of 0.94 MPa?arrow_forward
- At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 14.4 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 9.6 × 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and Care rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. 105.3°F 75.3°F O 147.3°F 86.6°F 113.4°F B ↑ 2 in. ↓ (2) 44 in. 0.04-in. gaparrow_forwardAt a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 12.7 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 8.6 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. 90.2°F O 69.9°F 139.2°F 103.5°F O 111.0°F B ↑ 2 in. ↓ 44 in. -0.04-in. gaparrow_forwardPermanent deformations in structures refer to: A. linear elastic deformations B. deflection diagram C. linear inelastic deformation D. plastic deformationarrow_forward
- At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 13.4 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 10.1 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and Care rigid. Determine the lowest temperature at which the two bars contact each other. (1) ↑ 3 in. 32 in. O 75.9°F O 146.5°F O 105.8°F O 122.3°F O 111.3°F 2 in. (2) 44 in. -0.04-in. gaparrow_forwardAt a temperature of 60°F, a 0.04-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 2.5 in. and a thickness of 0.75 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 1.7 in. and a thickness of 0.75 in. The supports at A and C are rigid. Assume h1=2.5 in., h2=1.7 in., L1=31 in., L2=46 in., and Δ=Δ= 0.04 in. (A) Determine the lowest temperature, Tcontact, at which the two bars contact each other. (B) Find a geometry-of-deformation relationship for the case in which the gap is closed. Express this relationship by entering the sum δ1+δ2, where δ1 is the axial deflection of Bar (1), and δ2 is the axial deflection of Bar (2). δ1+δ2= _____in. (C) Find the force in the Bar (1), F1, and the force in Bar (2), F2, at a temperature of 225oF. By convention, a tension force is positive and a compression force is negative. IN KIPS (D) Find σ1 and σ2,…arrow_forward
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