A circular rod of hot-rolled 4340 steel (4340-HR steel – 4mm diameter, 1.35m long)is subjected to a tensile load (applied normal to the circular cross-section). Themodulus of elasticity of steel is 207 GPa. What is the maximum load that can beapplied without causing permanent deformation?UltimateYieldStrength, S.Strength, S,ksiMPaksiMPaMaterial6.41110124163003-A6061-A6061-T6Aluminum188.55alloys4531040276Naval brass-A5437617117CopperalloysCartridge brass-CR765246343466455422901020-HR1045-HR4340-HRSteel alloys926386041415110411329108760035241Stainlesssteels303-A316-A440C-A845794229011075970483TitaniumCommercial8055170482Table 5.3alloy*The numerical values given are representative, and values for specificmaterials could vary with composition and processing. A = annealed,HR = hot-rolled, CR = cold-rolled, and T = tempered.Ultimate and Yield%3DStrengths of SelectedMetals*

A circular rod of hot-rolled 4340 steel (4340-HR steel - 4mm diameter, 1.35m long) is subjected to a tensile load (applied normal to the circular cross-section). The modulus of elasticity of steel is 207 GPa. What is the maximum load that can be applied without causing permanent deformation? Ultimate Yield Strength, S. Strength, S, ksi MPa ksi MPa Material 41 110 124 310 6. 8. 16 3003-A 6061-A 6061-T6 Aluminum 55 alloys 18 45 40 276 117 434 17 54 76 376 524 Naval brass-A Copper alloys 63 Cartridge brass-CR 290 1020-HR 1045-HR 4340-HR 66 92 455 638 42 60 Steel alloys 414 910 151 1041 132 241 290 600 35 Stainless steels 303-A 316-A 440C-A 87 84 110 579 42 759 70 483 Titanium Commercial 80 551 70 482 Table 5.3 alloy *The numerical values given are representative, and values for specific materials could vary with composition and processing. A = annealed, HR = hot-rolled, CR = cold-rolled, and T = tempered. Ultimate and Yield Strengths of Selected Matalat

A circular rod of hot-rolled 4340 steel (4340-HR steel - 4mm diameter, 1.35m long) is subjected to a tensile load (applied normal to the circular cross-section). The modulus of elasticity of steel is 207 GPa. What is the maximum load that can be applied without causing permanent deformation? Ultimate Yield Strength, S. Strength, S, ksi MPa ksi MPa Material 41 110 124 310 6. 8. 16 3003-A 6061-A 6061-T6 Aluminum 55 alloys 18 45 40 276 117 434 17 54 76 376 524 Naval brass-A Copper alloys 63 Cartridge brass-CR 290 1020-HR 1045-HR 4340-HR 66 92 455 638 42 60 Steel alloys 414 910 151 1041 132 241 290 600 35 Stainless steels 303-A 316-A 440C-A 87 84 110 579 42 759 70 483 Titanium Commercial 80 551 70 482 Table 5.3 alloy *The numerical values given are representative, and values for specific materials could vary with composition and processing. A = annealed, HR = hot-rolled, CR = cold-rolled, and T = tempered. Ultimate and Yield Strengths of Selected Matalat

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.7.1P: A high-strength steel bar used in a large crane has a diameter d = 2.00 in. (sec figure). The steel...

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A 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?
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?
An aluminum bar has length L = 6 ft and diameter d = 1.375 in. The stress-strain curse for the aluminum is shown in Fig. 1.34. The initial straight, line part of the curve has a slope (modulus of elasticity) of 10.6 × 106 psi. The bar is loaded by tensile forces P = 44.6 k and then unloaded. (a) That is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional limit? hint: Use the concepts illustrated in Figs. l.39b and 1.40.
A prismatic bar of length L = 1.8 m and cross-sectional area A = 480 mm" is loaded by forces P{= 30 kN and A = 60 kN (see figure) The bar is constructed of magnesium alloy having a stress-strain curve described by the Ram berg-Osgood equation: 45.000 618 UW id which u has units of mega pascals (MPa). (a) Calculate the displacement 8t- of the end of the bar when the load P:acts alone. (b) Calculate the displacement when the load P, acts alone. (c) Calculate the displacement when both loads act simultaneously.
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A bar is subjected to a combination of a steady load of 60kN and a load fluctuating between - 10kN and 90 kN. The corrected endurance limit of the bar is 150 MPa, the yield strength of the material is 480 MPa and the ultimate strength of the material is 600MPa. The bar cross-section is square with side a. If the factor of safety is 2, the value of a (in mm), according to the modified Goodman‟s criterion, is ____