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4 years ago

A primary incentive for entrepreneurs is

1 answer:

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Profit is an important incentive that leads entrepreneurs to accept the risks of business failure. Entrepreneurs are individuals what are willing to take risks in order to develop new products and start new businesses. They recognize opportunities, enjoy working for themselves, and accept challenges.

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Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo

noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

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3 years ago

). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially

Delicious77 [7]

Answer:

$\frac{e'_z}{e_z} = 0.87142$

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

εx = - [ σx - ν( σy + σz ) ] / E

εy = - [ σy - ν( σx + σz ) ] / E

εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

εx = [ ν*σz ] / E

εy = [ ν*σz ] / E

εz = - [ σz ] / E .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E

εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note: σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2

σx' ( 1 - v ) = [ ν*σz ] / 2

σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

εz' = - [ σz - ν( σy' + σx' ) ] / E

εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

$\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142$ ... Answer