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1 year ago

If the Poisson’s ratio of a 5 mm X 5 mm titanium alloy pin is 0.31 and it is elastically loaded

Engineering

1 answer:

leonid [27]1 year ago

7 0

The new dimensions of the titanium alloy pin will be that the width is 0.0775 mm and the length is 4.9225m.

<h3>What is Poisson's ratio?</h3>

The Poisson's ratio is the proportion of a material's change in width per unit width to its change in length per unit length due to strain. In order for a stable, isotropic, linear elastic material to have a positive Young's modulus, shear modulus, and bulk modulus, the Poisson's ratio must be between 1.0 and +0.5. Poisson's ratio values for the majority of materials fall between 0.0 and 0.5.

The formula for the longitudinal strain is:

= Change in length / Initial length

Based on the information, the longitudinal strain will be:

= 105 - 100 / 100

= 0.05

Poisson ratio will be illustrated as the change in the width divided by the longitudinal strain. :

0.31 = ∆w/5 / 0.05

∆w = 0.0775 mm

New side length will be the difference in the changes in the dimensions:

= w - ∆w

= 5 - 0.0775

= 4.9225m

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Answer:

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Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

$\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})$ (1)

Where:

$\epsilon$ - Emissivity, dimensionless.

$A$ - Surface area of the student, measured in square feet.

$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

$T_{s}$ - Temperature of the student, measured in Rankine.

$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

Where $\Delta t$ is the heat transfer time, measured in hours.

If we know that $\dot Q = 417.492\,\frac{BTU}{h}$ and $\Delta t = \frac{1}{3}\,h$ , then the net energy transfer is:

$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

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Answer:

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