An 8-cm-square bar of steel is found to contain a1-mm corner crack, oriented perpendicular to the length ofthe bar. If an axial stress is applied from 0 to 420 MPa at afrequency of once every 10 minutes, how long will it take forthe rod to fracture? The properties of the bar are:KIC = 90 MPa √m , σys = 1500 MPa, and the crack growthrelation is da/dN = 2 x 10-37 ΔK4, with da/dN and ΔKbeing given in units of m/cycle and Pa √m , respectively

An 8-cm-square bar of steel is found to contain a1-mm corner crack, oriented perpendicular to the length ofthe bar. If an axial stress is applied from 0 to 420 MPa at afrequency of once every 10 minutes, how long will it take forthe rod to fracture? The properties of the bar are:KIC = 90 MPa √m , σys = 1500 MPa, and the crack growthrelation is da/dN = 2 x 10-37 ΔK4, with da/dN and ΔKbeing given in units of m/cycle and Pa √m , respectively

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter2: Axially Loaded Members

Section: Chapter Questions

Problem 2.5.1P: The rails of a railroad track are welded together at their ends (to form continuous rails and thus...

See similar textbooks

Similar questions

The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate the clacking sound of the wheels) when the temperature is 60°F. What compressive stress ?? =6.5×10-6 /? is produced in the rails when they are heated by the sun to 120"F if the coefficient of thermal expansion a = the modulus of elasticity E = 30 × 106 psi?
A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by =18,0001+30000.03(=ksi) in which is nondimensional and has units of kips per square inch (ksi). (a) Construct a stress-strain diagram for the material. (bj Determine the elongation, of the wire due to the Forces P. (c) IF the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the proportional limit?
Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in ) that can be stored in each or the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Material Weight Density (lb/in3) Modulus of Elasticity (ksi) Proportional Limit (psi) Mild sleel 0.284 30,000 36,000 Tool steel 0.284 30,000 75,000 Aluminum 0.0984 10,500 60,000 Rubber (soft) 0.0405 0.300 300
The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm andagagelengthof50.8mm.IfaloadofP=60kNisapplied to the specimen, determine its new diameter and length. Take n = 0.35.
L b F What is the change in length (L) of the titanium bar (E = 16.5E6 psi) given: L = 36 in., b = 3 in., h = 0.25 in, and F = 3 kips Answer to above question is 8.72E-3 in, the shear modulus for the bar is 6.145E6 psi, poisons ratio is 0.343. My question is what is the strain (in macrostrain) in the bar in a direction perpendicular to the direction of the applied force? (Answer is 82.8, I need to understand how to reach this answer.)
The top lid of a nuclear reactor is being held by clamping screws which are made of AISI 4340 steel tempered at 260°C. They are stressed to a maximum level of 1050 MPa during a pressurization test before starting the reactor. Find the factor of safety guarding against yielding and the factor of safety guarding against crack propagation if the initial cracks in the material have B=1.3 and a=1.2 mm
A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the modulus of elasticity. Stress (ksi) 240 200 160- 120- 80 40- 0+ 0 0 Upper scale O 79100 ksi E O 94000 ksi O 86700 ksi O 63200 ksi O 51900 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)
A brittle material made of ASTM grade 30 cast iron has ultimate strength in tension of 31 kpsi and ultimate strength in compression of 109 kpsi and undergoes the following plane of stress state: o, = 20 kpsi and Tyy = 10 kpsi (CW). a) using the 3D Mohr's circle, find all three principal normal stresses. b) using the Brittle-Coulomb-Mohr theory and Modified-Mohr theory, find the factors of safety for the given plane of stress state.
A bar of 4340 steel has an Sy = 50 ksi. Which of these statements is true: If Sy = 50 ksi, than the normal stress in the bar is always 50 ksi. If a normal stress of 40 ksi is applied to the bar, it will undergo plastic deformation. The Sy = 50 ksi indicates bar is behaving in a brittle manner. O Any normal stress above 50 ksi will cause plastic deformation in the bar.
A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the yield strength (0.20% offset). Stress (ksi) 180 150- 120- 90 60 30- 0 0 0 Upper scale O 77 ksi O 56 ksi O 103 ksi O 86 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.00 0.004 0.006 0.008 0.010 0.014 Strain (in./in.)
A cast iron cylinder has outside and inside diameter 200mm and 125mmrespectively. If the ultimate strength of cast iron is 150MPa calculate according to each of the following theories of failure the internal pressure that will cause fracture.(a) Maximum principal stress theory. (b) Maximum principal strain theory. (c) Maximum strain energy theory. Assume no longitudinal stress in the cylinder. Which of the results obtained you consider should be applied to this case. Take:µ = 0.25, σU = 150MPa. Ans: (a) 65.73MPa. (b) 59.24MPa, (c) 55.33MPa.
Q2 A tensile test specimen having a diameter of (10 mm) und a page length of (S0 mm) was tested to fracture The stieisitrain curve from the tensiotn test is shown in Figure (9) The lower plat is the expanded tegion OAT and associatod with the strain valuns tiven on the lower scale Determine, 480 Lpper veile AB 241 Loser scale + The ultimaie stress, be The propotional limit, 120 e The fracture strees 4 The modulus of elasticity, OAK 0.04 0.0s 0.00 0.12 0.16 1.20 0K2 DANN 0.006 0.008 0.010 e T'he offiet yield stuen at 0.2%, Strain tmm/um) G The asial force ucting on the specireni whert t i uxtended by Figuie (3) a sans