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Rudiy27
3 years ago
14

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using

the Maximum Shear Stress Theory the material will __________
a. fail / not fail
b. fail /fail
c. not fail/fail
d. not fail/not fail
Engineering
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
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Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

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Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

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3 years ago
Removing Shingles from a roof is called a
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D remodeling roofing
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When Ontario
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Answer:

Explanation:

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3 years ago
A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12
nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × 10^{6} \frac{lbm}{in^{2} }

Beam deflection is = 0.15 in

Explanation:

Given data

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We know that deflection of the beam in this case is given as

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This is the value of Modulus of elasticity.

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Answer:

solution in the picture attached

Explanation:

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