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2 years ago

Relation between Poisson's ration, young's modulus, shear modulus for an isotropic material

Engineering

1 answer:

poizon [28]2 years ago

8 0

Answer:

E=2 G(1+2μ)

Explanation:

Isotropic material :

Those material have same property in all direction is known as isotropic material.

Homogeneous material :

Those material have same property through out the volume is known as homogeneous material.

Relationship between Poisson ratio ,young modulus and shear modulus:

E=2 G(1+2μ)

Relationship between Poisson ratio ,bulk modulus and shear modulus:

E=2 K(1-μ)

Where

E is young modulus.

G is shear modulus.

μ is Poisson ratio.

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In DC electrode positive, how much power is at the work clamp?

Korolek [52]

Answer:

1/3 power

Explanation:

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2 years ago

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

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2 years ago

Which type of work is a textile engineer most likely to do?

pickupchik [31]

Answer:

B because as a textile engineer, your job is to help design and create fabric, including the equipment and materials needed for fabrication.

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2 years ago

A spherical container made of steel has 20 ft outer diameter and wal thickness of 1/2 inch. Knowing the internal pressure is 50

anastassius [24]

Answer:

maximum normal stress = 5975 psi

maximum shear stress = 2987.50 psi

Explanation:

Given data

dia = 20 ft

wall thickness = 1/2 inch

internal pressure = 50 psi

To find out

the maximum normal stress and the maximum shearing stress

Solution

By the Mohr's circle we will find out shear stress

first we calculate inner radius

i.e. r = (diameter/2) - t

r = (20 × 12 in )/2 - ( 1/2 )

r = 120 - 0.5 = 119.5 inch

Now we find out maximum normal stress by given formula

normal stress = ( internal pressure× r ) / 2 t

normal stress = ( 50×119.5 ) / 2 × 0.5

maximum normal stress = 5975 psi

and minimum normal stress is 0, due to very small radius

and maximum shear stress will be

shear stress = ( maximum normal stress - minimum normal stress ) / 2

shear stress = ( 5975- 0 ) / 2

maximum shear stress = 2987.50 psi

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3 years ago

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Pepsi [2]

Answer:

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Explanation:

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2 years ago