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shepuryov [24]
3 years ago
13

Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control

inputs b, c, d, and e. Each data input should be0, 1, a, or a′.(Hint: Start with a minterm expansion of F and combineminterms to eliminate a and a′ wherepossible.)(3pts)

Engineering
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

See explaination

Explanation:

please see attachment for the detailed diagram used in solving the given problem.

It is attached as an attachment.

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). 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

5 0
3 years ago
How fast is a 2012 nissan sentra<br>speed and acceleration ​
Alika [10]

Answer:

it has 15 horsepower to 300 horsepower and it weighs 2,906 to 3,131

Explanation:

its torque is 142 to 180

it has a inline 4 engine

there's a SE-R which has a turbo

4 0
3 years ago
A(n)___ branch circuit supplies two or more receptacles or outlets for lighting and appliances
Juli2301 [7.4K]

Answer: General purpose branch circuit

Explanation:

General purpose branch circuit are the type of circuits that are used mainly to supply light to two or more receptacle outlets for small appliances. This circuits are about 120v can be used either in residential, commercial and industrial buildings.

6 0
3 years ago
Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.
Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is \frac{F}{A}

also,

  • Dimensional formula of <em>Pressure is </em>M^{1}L^{-1}T^{-2}
  • Dimensional formula of <em>length is L </em>
  • Dimensional formula of <em>mass is M</em>
  • Dimensional formula  of <em>velocity is </em>L^{1} T^{-1}

So, as given W=\frac{P*L^{3} }{M*V^{2} }

Dimensional formula of W =\frac{M^{1}L^{-1}T^{-2}  L^{3}  }{M^{1} L^{2}T^{-2}   }

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

#SPJ10

6 0
2 years ago
2.
den301095 [7]
Jae pain seems the most off
4 0
3 years ago
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