Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 6, Problem 18CQ
To determine
The reason for plastic strain in crystalline material at low temperatures.
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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
Draw a tensile stress-strain curve for a typical semi-crystalline polymer such as LLDPE, and define the three main regions on the curve.
The deformation per unit length is called
O(A) Tensile stress
O(B) Compressive stress
OCC) Shear stress
O(D) Strain
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_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_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
- 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_forwardAn aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.arrow_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
- Within 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_forwardA bronze rod is rigidly attached between an aluminum rod and a steel rod as shown in the figure below. Axial loads are applied at the positions indicated. a) Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in bronze of 100 MPa, or in aluminum of 90 MPa. b) Determine the deformation of the bronze rod if the value of P is 22.50 KN. The moduli of elasticity are 200 GPa for steel, 80 GPa for bronze and 70 GPa for aluminum.arrow_forwardThe stress in an elastic material is: A Inversely proportional to the force acting B Inversely proportional to the materials yield strength C Proportional to the displacement D Inversely proportional to the strainarrow_forward
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