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TiliK225 [7]
3 years ago
14

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

(a) Compute the maximum stress level. Entry field with correct answer 280 MPa (b) Compute the minimum stress level. Entry field with correct answer -100 MPa (c) Compute the stress ratio. Entry field with correct answer -0.36 (d) Compute the magnitude of the stress range.
Engineering
1 answer:
Gwar [14]3 years ago
3 0

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

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3 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
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Answer:

Q=4.98\times 10^{-3}\ m^3/s.

Explanation:

Given that

L= 50 m

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We know that

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\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

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Q=4.98\times 10^{-3}\ m^3/s

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6 0
3 years ago
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Answer:

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Use ideal gas law:

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n is number of moles,

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3 0
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