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Ket [755]
3 years ago
13

Describe how to mix and apply body filler?

Engineering
1 answer:
malfutka [58]3 years ago
8 0
This is all you need to mix body filler.
Start with a golf-ball size glob of filler.
Squeeze a ribbon of hardner across the filler.
Stir the mixture quickly, but do not "whip" it. ...
Mix until overall color is even.
Pick up an appropriate amount of filler for the task at hand.
You might be interested in
). 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
What are the mechanical properties of a geotextile that are of most importance when using it as a separator in an unpaved road s
Alekssandra [29.7K]

hey thanks for the point's i really needed them.. Cheers!

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6 0
3 years ago
The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi
vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

R=1+\frac{\rho * K_{d}  }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65

The time will take will be:

t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days

b) For toluene:

R=1+\frac{2.6*3.3  }{0.37} = 24.19

t=\frac{12*24.19}{0.18} =1612.67days

6 0
3 years ago
Read 2 more answers
An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
N76 [4]

Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

Explanation:

4 0
3 years ago
Read 2 more answers
A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t
Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

tan\phi=\frac{\bigtriangleup l}{h}

tan\phi=\frac{0.08}{1.38}

tan\phi= 0.05797

For small angle of \phi, tan\phi can take as\phi.

\phi = 0.05797 rad.

Thus, the shear strain is 0.05797 rad.

7 0
3 years ago
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