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Deffense [45]
3 years ago
9

To become familiar with the general equations of plane strain used for determining in-plane principal strain, maximum in-plane s

hear strain, and average normal strain. The state of strain at a point has components of ϵx=380.0×(10−6), ϵy=−250.0×(10−6), and γxy=180.0×(10−6).a. Equivalent in-plane strains on the oriented element Determine the equivalent in-plane strains on an element rotated counterclockwise at an angle of θ = 60.0 ∘ . Find ϵx',ϵy', γxy' Express your answers, separated by commas, to three significant figures.b. In-plane principal strains on the oriented element Determine the in-plane principal strains on the oriented element. Find ϵ1, ϵ2. Express your answers, separated by a comma, to three significant figures.c. Maximum in-plane shear strain and average normal strain on the oriented element Determine the maximum in-plane shear strain and the average normal strain on the oriented element. Express your answers, separated by a comma, to three significant figures.

Engineering
1 answer:
lukranit [14]3 years ago
3 0

Answer:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

Explanation:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

The explanation is shown in the attachment. I hope i have been able to help.

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Answer:

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8 0
2 years ago
Free ideas free points. You will be reported for answering "no" or I don't know
KengaRu [80]

Answer:

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5 0
2 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
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Answer:

N = 38546.82 rpm

Explanation:

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A_{1}= \frac{\pi }{4}\times 150^{2}

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The centrifugal force acting on the flywheel is fiven by

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Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

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                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

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∴ R_{2} - R_{1} = 50 mm

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