Concept explainers
For the interval 103 ≤ N ≤ 106 cycles, develop an expression for the axial fatigue strength
Want to see the full answer?
Check out a sample textbook solutionChapter 6 Solutions
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
- Solve the preceding problem if the thickness of the steel plate is. t = 12 mm. the gage readings are x = 530 × 10-6 (elongation) and y = -210 -× l0-6 (shortening), the modulus is E = 200 GPa, and Poisson’s ratio is v = 0.30.arrow_forwardThe 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?arrow_forwardThree round, copper alloy bars having the same length L but different shapes are shown, in the figure. The first bar has a diameter d over its entire length, the second has a diameter d over one-fifth of its length, and the third has a diameter d over one-fifteenth of its length. Elsewhere, the second and third bars have a diameter Id. All three bars are subjected to the same axial load P. Use the following numerical data: P = 1400 kN, L = 5m,d= 80 mm, E= 110 GPa. and v = 0.33. (a) Find the change in length of each bar. (b) Find the change in volume of each bar.arrow_forward
- 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?arrow_forwardA high-strength steel bar used in a large crane has a diameter d = 2.00 in. (sec figure). The steel has a modulus of elasticity E = 29 × 10 psi and Poisson’s ratio is v = 0.29. Because of clearance requirements, the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pmaxthat is permitted?arrow_forwardRepeat Problem 2.3-18, but assume that the bar is made of copper alloy. Calculate the displacements SBand Scif P = 50 kips, L = 5 ft = 3/5 in., b1= 2.75 in., b2= 3 in., and E = 16,000 ksi.arrow_forward
- Find the equivalent stresses at Points 1 and 2 of the element with given geometry and loading conditions according to the Maximum shear stress hypothesis and the Maximum strain energy hypothesis. Shaft diameter: 20 mm, Shaft Length 120 mm, F1 = 750N, F2 = 3000N, Mb = 2400 N.mm. Steel if St37 and Safety coefficient is 2 If taken, will this stick work safely under these operating conditions? (Yield of given steel Strength 225 Mpa, Tensile strength 370 Mpa)arrow_forwardRequired information Given a bar made of brittle material, properties of Sut= 30 kpsi and Suc = 90 kpsi and with stresses of Ox= -15 kpsi, Oy= 10 kpsi, and Txy=-15 kpsi. For the given state of plane stress, determine the factor of safety using the Coulomb-Mohr theory. The factor of safety isarrow_forwardA bar of steel has the minimum properties Se = 276 MPa, Sy = 413 MPa, and Sut = 551 MPa. The bar is subjected to a steady torsional stress of 103 MPa and an alternating bending stress of 172 MPa. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use: (a) Modified Goodman criterion. (b) Gerber criterion. (c) ASME-elliptic criterion.arrow_forward
- Q3/ A 6.4mm diameter cylindrical rod fabricated from a 2014 T6 aluminum alloy is subjected to reversed tension-compression load cycling along its axis. If the maximum tensile and compressive loads are (+5340 N and -5340N respectively), determine its fatigue life. Assume the stress plotted is stress amplitude. 80 500 H70 60 400 50 9 1045 steel 300 40 2014-T6 aluminum alloy 200 20 100 Red brass H10 10 10 105 10 107 10 10 1010 Cycles to failure, N Stress amplitude, S (MPa) Stress amplitude (10 psi) 8 8 8 &arrow_forwardThe fluctuating stresses listed in the table are found at a critical location of a component made of steel with Se = 40 ksi, Sy = 80 ksi, Sut = 110 ksi and f= 0.87. These stresses are applied on the part within 10 s. What is the accumulative damage of this part? What is the life of the part in hours if this stress pattern continues to repeat for the remained of the part's life? Use Goodman criterion and Miner's rule in your solution. Loading order |Omin o max |Number of cycles -20 30 -10 50 1 3. -30 30 1arrow_forwardQ2/ A steel bolt 0.005 m² in cross-section is subjected to a Static mean load of load of 200KN. What Value of Completely reversed direct fatigue load will Produce failure in 10² Cycles? Use the GERBER relationsip and assume that the Yield sureg. Strength of the steel is 400 MN/M² and the Stress required to produce. failure at 107 Cycles under Zero mean stress Condition is 300 MN/m².arrow_forward
- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning