The elastic portion of the stress-strain diagram for an aluminum alloy is shown in the figure below (Figure 1). The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.1 mm. Take v=0.35. Figure σ (MPa) 490 0.007 € (mm/mm) 1 of 1 Part A If a load of P = 25 kN is applied to the specimen, determine its new diameter. Express your answer to four significant figures and include appropriate units. μÅ d= Submit Part B Value L = Submit Request Answer Determine its length. Express your answer to four significant figures and include appropriate units. Value Provide Feedback Units Request Answer ? Units ?

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A: Step 1: Step 2: Step 3: solution is given above Step 4:

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A: Given that  Mdulus of elasticity,E=200 GPa=200×109Pa Horizontal force,H=150 kN=150×103N Length,L=3.5…

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Q: If the applied load is 49 kN, determine the new diameter of the specimen.

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A: l=12 mm sheering stress=180mPa The Stress of an aluminum plate: 70mPA

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A: Answer: Normal stress,  σ =2.73×108 N/m2 Shear stress,  τ =1.51×108 N/m2

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Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).