帮我解一下这个材料力学题目A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2
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帮我解一下这个材料力学题目
A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load = 168,000 N is reached at a gage length = 64.2 mm.
a) Determine yield strength Y,
(b) Determine modulus of elasticity E
(c) Determine tensile strength TS.
还是英文,不给点悬赏分啊.
a)Sy=98000/(200*10^-6)=490Mpa
b)E=FL/(elongation)*A=106.5Gpa
c)assuming volume does not change,implies 50*200=64.2*A,A=155.76mm^2,
Sut=F/A=168000/(155.76*10^-6)=1.08Gpa
帮我解一下这个材料力学题目A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2
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