A tank with some water in it begins to drain. The function v ( t ) = 46 − 3.5 t determines the volume of the water in the tank (
in gallons) given a number of minutes t since the water began draining. What is the vertical intercept of f ? What does your answer to part (a) represent? Select all that apply. The weight of the tank when it is empty. How many minutes it takes for all of the water to drain from the tank. The number of gallons of water in the tank when it starts draining. List all horizontal intercepts (or "roots") of f . What does your answer to part (b) represent? Select all that apply. The weight of the tank when it is empty. The number of gallons of water in the tank when it starts draining. How many minutes it takes for all of the water to drain from the tank.
1 answer:
Answer with Explanation:
Part a)
The volume of water in the tank as a function of time is plotted in the below attached figure.
The vertical intercept of the graph is 46.
Part b)
The vertical intercept represents the volume of water that is initially present in the tank before draining begins.
Part c)
To find the time required to completely drain the tank we calculate the volume of the water in the tank to zero.
Part d)
The horizontal intercept represents the time it takes to empty the tank which as calculated above is 13.143 minutes.
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Answer:
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:
... Answer