Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Question
Chapter 6, Problem 12ETSQ
To determine
The ratio of tensile fracture strength of ceramic to compressive fracture strength.
Expert Solution & Answer
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Students have asked these similar questions
For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results at a stress of 510 MPa when the maximum (or critical) internal crack length is 6 mm. For the same alloy, will fracture occur at a stress level of 380 MPa when the maximum internal crack is 9.0 mm? Why or why not?
Select the most appropriate answer based on your calculation.
Select one:
a. It will not fracture
b. Not enough information
c. It will fracture
A component made of Aluminium 6061-T651 has an edge crack
with length equal to 15 mm, as shown
The yield
stress of the material is oy = 280 MPa and its fracture toughness
(Kic) ranges from 40 MN/m3/2 to 60 MN/m³/2
Consider now the same edge crack in a semi-infinite plate.
Determine the critical stress to avoid fracture.
P
I
I
I
I
←
15 mm
P↓
100 mm
30 mm
2- What is the largest size (mm) internal through crack that a thick plate of aluminium alloy
7075-T651 can support at an applied stress of (a) three-quarters of the yield strength and (b)
one-half of the yield strength? Assume Y = 1.
for 7075-T651, KỊC = 24.2 MPa ym and oYS = 495 MPa.
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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Similar questions
- 2. Please estimate the number of cycles to failure of a steel specimen under tensile fatigue loading with the following parameters. The R ratio is 3, mean stress 200 MPa, yield strength 450 MPa, ultimate tensile strength 560 MPa, Young’s modulus 200 GPa, KIC = 140 MPa . Assume the initial crack length is 0.1 mm.arrow_forwardA steel 0.6 inch×1.2 inch steel 90 m long is subjected to a 45 KN tensile load along its lenght.If poison's ratio is 0.3 Find: A. The deformation along its lenght. B. The deformation along its thickness. C. The defirmation along uts width. D. The lateral strain.arrow_forwardA ceramic part is used under a complete reverse cyclic stress with a stress amplitude (S) of 250 MPa. The yield strength and fracture toughness of materials is 550 MPa and 12.5 MPa*sqrt(m), respectively. Y is 1.4. What is the critical surface crack length?arrow_forward
- A high-strength steel has a yield strength of 1380 MPa and fracture toughness of 91 MPavm. A surface crack of 2.5 mm is found at the surface. At what max. applied stress level will catastrophic failure occur? (Y = 1.00 for internal crack, and Y=1.12 for surface crack)arrow_forwardProblem 4: A tensile test is carried out on a bar of a mild steel of diameter 2 cm. the bar yields under a load of 150 kN and breaks finally at a load of 70 kN. Estimate; 1-the tensile stress at the yield point 2-the ultimate tensile stress 3-the average stress at the breaking point, if the diameter of the fractured neck is 1 сm.arrow_forwardEngineering strain of a mild steel sample is recorded as 0.100%. The true strain is 1. 2. 3. 4. 0.010% 0.055% 0.099% 0.101%arrow_forward
- 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. P2.121arrow_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_forward= Syc = 100 kpsi and a true strain at fracture = 0.55. Estimate the factor of safety for the following principal stress states: A hot-rolled steel has a yield strength of Syf of ɛf = | (b) ox = 60 kpsi, ơy |(c) ơx |(d) ox = -40 kpsi, ơy |(a) 0x = 70 kpsi, ơy = 70 kpsi, txy 40 kpsi, txy 40 kpsi, txy = = -60 kpsi, Txy = 0 kpsi -15 kpsi 45 kpsi 0 kpsi, ơy = %3D 15 kpsi | (e) 01 = 30 kpsi, o2 = 30 kpsi, 03 = 30 kpsiarrow_forward
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