Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 6, Problem 20CQ
To determine
The reason for plastic strain in Glass.
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The deformation per unit length is called
O(A) Tensile stress
O(B) Compressive stress
OCC) Shear stress
O(D) Strain
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
Qus :
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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- Estimate the elastic and plastic strain at the ultimate tensile strength in the low-carbon steel specimen in Figure 6.16.arrow_forward6)arrow_forwardIn 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_forward
- The modulus of elasticity are slightly higher for ceramic materials, Polymers have modulus values that are smaller than both metals and ceramics.arrow_forwardAn 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_forwardA single crystal of BCC iron is subjected to tensile stress of 100 MPa along the [001] direction. Which of the two slip systems ((211) –[ 111 ] or (321)–[ 111 ]) is going to yield first?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_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_forwardQ7> Ductile-to-brittle transition temperature (DBTT) is a very important parameter in the design of metallic materials for engineering applications. It has been well known that most of BCC and HCP metals show the DBT phenomenon; however, there is no DBTT in FCC metals. (a) Explain the reason in terms of deformation and fracture. You must compare the BCC and FCC. (b) The ductile fracture surface consists of many dimples. Explain their formation mechanism from the concept of point defects. (c) There are two types in the brittle fracture. Explain and Compare them.arrow_forward
- Which one is a linear defect in the crystalline materials? (A) external surfaces B) vacancies (c) dislocations (D) grain boundariesarrow_forwardThe lowest stress at which permanent .deformation can be measuredarrow_forwardWithin elastic limit, stress is O(A) Inversely proportional to strain O(B) Directly proportional to strain OCC) Square root of strain O(D) Equal to strainarrow_forward
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