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ss7ja [257]
2 years ago
11

A cylindrical specimen of a metal alloy 45.8 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 378 MPa ca

uses the specimen to plastically elongate to a length of 54.2 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 45.8 mm to a length of 55.8 mm

Engineering
1 answer:
Sliva [168]2 years ago
8 0

Answer:

390.242 MPa

Explanation:

Attached is the full solution.

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Conduct online research and write a short report on the origin and evolution of the meter as a measurement standard. Discuss how
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Answer:

People have come up with all sorts of inventive ways of measuring length. The most intuitive are right at our fingertips. That is, they are based upon the human body: the foot, the hand, the fingers or the length of an arm or a stride.

In ancient Mesopotamia and Egypt, one of the first standard measures of length used was the cubit. In Egypt, the royal cubit, which was used to build the most important structures, was based on the length of the pharaoh’s arm from elbow to the end of the middle finger plus the span of his hand. Because of its great importance, the royal cubit was standardized using rods made from granite. These granite cubits were further subdivided into shorter lengths reminiscent of centimeters and millimeters.

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Fragment of a Cubit Measuring Rod

Credit: Gift of Dr. and Mrs. Thomas H. Foulds, 1925

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5 0
2 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

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Explanation:

Given

A = 45

B = -60

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a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

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Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

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p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

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Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

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Answer:

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Explanation:

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