Consider a cylindrical specimen of a steel alloy (Please see the figure) 0.33 in. in diameter and 3.15 inches long that is pulled in tension. given: -Elongation is 0.018 in -Tensile stress = 200 ksi -Tensile strain = 6.83*10^-3 -Ultimate stress = 281.82 ksi -Ultimate strain = 0.05 solve: Fracture point (stress and strain values)

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A: Using Poisson ratio formula and Young's modulus formula

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A: Given data:A=150 mm2 Load,P=5000 Nα=11.7×10-6/°CE=200 GPa or 200×103 N/mm2

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A: The strain can be calculated using Hookes law.

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A: a}

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Q: A cylindrical tensile specimen has an undeformed diameter of 8.6 mm. It is subjected to a tensile…

A: Hello dear this is the solution of your problem The answer is found to be Change in diameter =0.37mm…

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A: The graph between strss-strain is shown in the graph below.

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Material Properties

Construction in civil engineering demands high strength, durability, and longevity. These targets can be achieved by better design and analysis, along with the correct choice of materials. Different materials have different properties and hence, they respond and performs differently under the various condition of external loadings. It is therefore a critical task of an engineer to select the right material based on the property that the material is supposed to function in real-time. When speaking about the property, there are different categories of material properties like chemical property, physical property, mechanical property, electrical property, magnetic property, thermal property, etc. These material properties form the basis of material selection in various applications.