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LekaFEV [45]
4 years ago
8

4. Which of the following drive roll designs is used

Engineering
1 answer:
Yuki888 [10]4 years ago
5 0
Knurled drive rolls are used with gas and self shielded flux cored and metal cored wires, which are softer due to the flux inside and the tubular design.
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How's your day been so far? <3
Rzqust [24]
My day as been going awesome
7 0
3 years ago
The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisput
evablogger [386]

Answer:

The cult of personality that surrounded Joseph Stalin in the Soviet Union led soviet citizens to believe that there was undisputed support for Stalin both among the government and the common people. In turn, this fueled self-censorship and made political change harder. This cult of personality was achieved through propaganda and censorship, as the Communist Party had control of all mass media. This desire to make himself a "god-like" figure was also an attempt to increase acceptance of communism among the people and to boost morale.

Explanation:

7 0
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

5 0
3 years ago
At the end of the vowel, the soft palate (is lowered) so that the air flows out the (nasal passage). At the same time, the _____
Marat540 [252]

Answer:

At the end of the vowel, the soft palate (is lowered) so that the air flows out the (nasal passage). At the same time, the BACK of the tongue rises to articulate with the VELUM, preventing air from escaping through the MOUTH. The lips remain OPEN. The vocal folds continue to SOUND.

Explanation:

This is the way in which sound is produced in any language through the use of our organs of speech called articulators.

NOTE:

This question should be in English Category not Engineering. I believe that was an oversight from your end.

3 0
3 years ago
Which two events in Britain indirectly influenced the American Revolution?
AysviL [449]

Answer:

Explanation:The two events that indirectly influenced the American Revolution was the Passage of the English Bill of Rights and the English Declaration of Independence.

8 0
3 years ago
Read 2 more answers
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