Ex. Create an estimated S-N diagram for an aluminum bar and define its equation what is the corrected fatigue strength at 7 N = 2 * 10 cycles ? Out = 310 MPa, the forged bar is 3.81 cm in round. The load is fully reversed torsion.
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- 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?Find: approximate yield strength (Sy) of 6061 T6 aluminum sheet LT at room temperature? approximate modulus of elasticity of 250 maraging steel?A circular bar - shown below, made of AISI 1040 OQT 1300 steel (SU = 600 MPa), is loaded with an axial tensile loading P. Calculate the modified endurance strength (SE) and the modified fatigue strength (SF) for the bar if D = 36 mm, d = 30 mm, and r = 2.4 mm. Use the number of cycles to fracture to be 5 × 105. Assume that the surface finish factor, kf, is 0.85, the size factor, kS, is 0.85, and the notch sensitivity factor, qn, is 0.75. Use the following equations if needed,
- What is the approximate yield strength of 250 maraging steel at room temperature? (No need to do a 0.2% offset, just estimate from the graph)Q6:1f the rod fracture at 35% elongation find the percentage difference between engineering and true stress at fracture point?A cylindrical steel bar 8 mm in diameter is loaded 1000 cycles per day with a load of 15560 N. How long until fatigue failure takes place? Note: stress force/area 500 450 400 350 Steel Allby 300 250 200 150 Brass Alloy 100 50 1.00E-07 Cycles to Failure, N 1.00E+04 1.00E+05 1.00E-06 1.00E+08 100E+09 1.00E+10 , The bar should not fatigue. ) The bar will fail in one cycle. ) 100 days ) 1000 days Stress Amplitude, MPa
- Please Follow this Properly, I Like your answer Structure, Accuracy, Concept, Please No submit Handwriting.a) A bar of normalized AISI 4130 is subjected to a steady torsional stress of 103 MPa, an alternating torsional stress of 69 MPa, and alternating bending stress of 83 MPa. It has an endurance limit of 276 MPa. Using Gerber and ASME-elliptic criteria, find the safety factors guarding against a fatigue failure and the expected life of the part. Analyse your results. b) A bar of 25 mm diameter cold-drawn SAE 1045 steel is loaded in rotating bending. Determine its endurance limit or strength.A 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.
- please help solve and explain and include FBD pleaseFigure 1 below shows (1) an unnotched bar and (2) a notched bar of the same minimum cross-section. Both bars are machines from AISI 1050 normalized steel. For each bar, estimate (a) the value of static tensile load P causing a fracture and (b) the value of a fluctuating axial load from 0 to P that would be just on the verge of producing eventual fatigue fracture (in about 1 to 5 million cycles). Reflect on the effect of the notch. r = 2.5 mm 35 mm 30 mm 30 mm 30 mm 30 mm r = 2.5 mm (1) (2) Figure 12. A steel of rotating-beam test specimen has an ultimate strength of 1100Mpa. Estimate the fatigue strength of the specimen corresponding to a life of 150k cycles of stress reversal.