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azamat
1 year ago
15

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

Learn more about Poisson on:

brainly.com/question/7879375

#SPJ1

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2 years ago
A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a
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2 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
3 years ago
A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying
Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius,r_{2} = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, r_{1} = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

 τmax  = \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0
3 years ago
Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o
lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω  cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

7 0
3 years ago
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