Mixture - Concrete - Solution / Explanation

[Guest]

Here is another concrete related problem:

A 1:2:4 mixture of concrete contains 1 part of cement, 2 parts of sand, and 4 parts of gravel/rocks. Determine amount of cubic yards of each to make 1 cubic yard of concrete.

33 1/3 % of voids (air space) is allowed in the sand and 45 % of voids are allowed in the gravel/rocks.

Let x = cubic yards of cement.

Let y = cubic yards of sand.

Let z = cubic yards of gravel/rocks.

The z cu. yds. of gravel/rocks count as only .55 z cu. yd. of the finished concrete, and the y cu. yd. of sand count as only 2/3 y of the solid, finished concrete.

From equations:

1) x + 2/3 y + .55 z = 1 cu. yd.

2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1

3x + 12x + 6.6x = 3 (multiplying by 3)

21.6x = 3

21.6x/21.6 = 3/21.6

x = .138888 cu. yd. of cement.

y = .276666 cu. yd. of sand.

z = .55552 cu. yd. of gravel/rocks.

I don't think those are correct. Where did I make the error(s) ? Also, I don't understand the equations.

[Guest]

Consider 1 cu yard of Dry Mixure + Just enough water to fill the voids makes 1 cu. yard of concrete.......

Dry normal mixture........ 7 parts alltogether...1/7 is cement....2/7 is sand........4/7 is gravel.......

But sand has 33.33% voids ...... and gravel has 45% voids..........not told anything about the cement so assume no measureable voids.

2/7 x 1/3 = 2/21 cu. yards of void space in sand

4/7 x 45/100 = 180/700 cu. yards of void space in gravel

2/21 + 18/70 is total void space in mix that is left for water

(20 + 54) /210 = 74/210 = 0.352 cu yards of water to fill voids and complete the 1 cu. yard of concrete.

One of the important things about concrete mix is the cement to water ratio.......
The cement to water ratio is (1/7) / 0.352 = 0.406 by volume ......which is typical for concrete mix.....

[Guest]

Dry normal mixture........ 7 parts alltogether...1/7 is cement....2/7 is sand........4/7 is gravel....... I understand.
But sand has 33.33% voids ...... and gravel has 45% voids..........not told anything about the cement so assume no measureable voids. The problem did not state.

Please explain about the equations and where I made the error(s). Thanks.

[Guest]

You did not explain why you did each step or what you were trying to calculate at each step.
You need to work through the problem logically if you expect anyone else to follow it.
My first observation is you have too many unknowns and too many equations
How was the amount of cement for instance unknown when the question told you it was 1 part of cement etc........
I also assumed that this info.... "33 1/3 % of voids (air space) is allowed in the sand and 45 % of voids are allowed in the gravel/rocks".... was given as part of the question and that you did not dream it up.

My logic was to start with 1 cubic yard of dry materials and make use of the information given in the question regarding the percentage for voids in the materials and fill these voids with water. If we had a 1 cubic yard container full of the dry materials it would still be able to take water and not overflow until the voids are filled. We were not told anything else in the question about some of the material dissolving etc or anything else, so I just considered filling the voids around the particles of sand and gravel with water and still be left with 1 cu. yard of concrete mixed with water still having a total volume of 1 cu. yard.
It is like a saucepan full of potatoes.....It may be full of potatoes but is will still hold maybe another 2 or 3 pints of water within the voids around the potatoes.....This is what we do to get them cooked.

[Guest]

A 1:2:4 mixture of concrete contains 1 part of cement, 2 parts of sand, and 4 parts of gravel/rocks. Determine amount of cubic yards of each to make 1 cubic yard of concrete.

The following was provided with problem:

33 1/3 % of voids (air space) is allowed in the sand and 45 % of voids are allowed in the gravel/rocks.

Let x = cubic yards of cement.

Let y = cubic yards of sand.

Let z = cubic yards of gravel/rocks.

The z cu. yds. of gravel/rocks count as only .55 z cu. yd. of the finished concrete, and the y cu. yd. of sand count as only 2/3 y of the solid, finished concrete.

From equations:

1) x + 2/3 y + .55 z = 1 cu. yd.

2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1


This was my attempt at the equation:

3x + 12x + 6.6x = 3 (multiplying by 3)

21.6x = 3

21.6x/21.6 = 3/21.6

x = .138888 cu. yd. of cement.

y = .276666 cu. yd. of sand.

z = .55552 cu. yd. of gravel/rocks.

[Guest]

What is the point of posting this same thing again without explanation..........

1) x + 2/3 y + .55 z = 1 cu. yd............I don't think this is correct???????????

For 1 cu. yard of dry mixture..........
Let............
x = cu.yds cement in 1 cu.yd of mix
y = cu.yds sand in 1 cu.yd of mix
z = cu.yds in 1 cu.yd of mix
Then...............
x + 2/3y + 0.55z + voids = 1 cu.yd
...........................

[Guest]

1) x + 2/3 y + .55 z = 1 cu. yd............I don't think this is correct??????????? It was provided with problem. I didn't understand the equations.

[Guest]

That is why you need to provide all the facts in the question properly and correctly. There is no point in guessing.
What does a cubic yard of concrete mean. What does a cu. yard of dry materials mean. What does a cu.yard of dry material mean with "voids". How do you remove the voids and is that solid concrete or solid dry materials. There is no point in guessing what these mean it needs to be provided as facts in the question.

[Guest]

That was the complete problem.

[Guest]

An omission:

No voids in the cement. Sorry for the error.

Please reply. Thanks.

[Guest]

I believe I have figured out the solution:


1) x + 2/3 y + .55 z = 1 cu. yd.

2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1

3x + 4x + 6.6x = 3 (multiplying by 3).

13.6x = 3 (combining like terms).

13.6x/13.6 = 3/13.6 (dividing by 13.6 to get x alone).

x = .22058 = .22 cu. ft. of cement.

y = .44116 = .44 cu. ft. of sand.

z = .88232 = .88 cu. ft. of gravel.

Correct ??

[Guest]

Numerically your solution is calculated OK
But you do not explain your reasoning or what you are calculating.....
The question states that sand and gravel has voids in it because it is loose particles of various sizes.
Your Eqn (1) adds the solid material and ignores the voids and you say it is equal to 1 cu yard.?????
Sand and gravel is supplied as loose particles and so cannot be put back together as a solid.
The question states that the mix is ratio of 1 to 2 to 4 parts of the loose materials not 1 to 2 to 4 of solid rock
How would you measure it to get it mixed in the required ratios.

(1/7) + (2/3 x 2/7) + (.55 x 4/7) + voids = 1 cu yard


The equation (1) below is the one I am querying..????

1) x + 2/3 y + .55 z = 1 cu. yd. .....This is solid Rock with no voids ...not sand and gravel particles.


2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1

3x + 4x + 6.6x = 3 (multiplying by 3).

13.6x = 3 (combining like terms).

13.6x/13.6 = 3/13.6 (dividing by 13.6 to get x alone).

x = .22058 = .22 cu. ft. of cement......I assume you mean cu yards???????

y = .44116 = .44 cu. ft. of sand.......I assume you mean cu yards????????????

z = .88232 = .88 cu. ft. of gravel.......I assume you mean cu yards?????



Do you not agree with my earlier posted solution where I considered 1 cu yard of dry material and fill the voids with water to make 1 cu metre of concrete.????

[Guest]

"x = .22058 = .22 cu. ft. of cement......I assume you mean cu yards??????? Yes.

y = .44116 = .44 cu. ft. of sand.......I assume you mean cu yards???????????? Yes.

z = .88232 = .88 cu. ft. of gravel.......I assume you mean cu yards?????" Yes.



"Do you not agree with my earlier posted solution where I considered 1 cu yard of dry material and fill the voids with water to make 1 cu metre of concrete.???? " Yes.

I relooked at the problem and had omitted - The voids in the gravel are filled in the finished concrete. The equations, according to the problem, are due to no voids in the cement.

[Guest]

Please respond. Thanks.

[Guest]

"Do you not agree with my earlier posted solution where I considered 1 cu yard of dry material and fill the voids with water to make 1 cu yard of concrete.???? " Yes..........OK.....so it contains 1/7 cu yard of cement, 2/7 cu yard of sand, 4/7 cu yards of gravel as measured in dry materials.

I relooked at the problem and had omitted - The voids in the gravel are filled in the finished concrete. The equations, according to the problem, are due to no voids in the cement. ??????????????? What does that mean????

The question says there are voids in the sand and in the gravel. OK I agree....The voids are there because of the particles. It is not possible to get all the small particles to fit into all the voids like a jigsaw and say there are no voids. The voids will be there either filled with air or filled with water or some mixture.

That is why in my solution I started with 1 cu yard of dry mix 1:2:4 (the ratios quoted are for the ratio of dry materials with voids in the dry mix)... So in effect I started with 1/7 cu yard of cement, 2/7 cu yards of sand and 4/7 cu yards of gravel....that gave me 1 cu yard of dry mix but it had voids that were filled with air. I then worked out the volume of the voids and when voids are filled with water the total volume of the wet mix is still 1 cu yard. And the amounts of the cement, sand and gravel will still be the same as when dry but if now mixed the particles will have re-distributed and the volume of the loose particles will probably be different and more compacted.
That is why things like this are worked out on a weight or mass basis and results used as test data for conversion from dry volume to wet mixed volume.

The query I have with your Eqn(1)....... 1) x + 2/3 y + .55 z = 1 cu. yd. ....is if... x, y and z are I assume the amount of solid cement, solid sand and solid gravel (actually rock) and you said that the equation equalled 1 cu yard....Of what...solid rock....How can you measure it....it is all in the form of particles and contains voids.....Is Eqn1 not equal to the solid particle material that would be in 1 cu Yard of mix and the rest of the 1 cu yard would be voids...So how do you quantify this in your equation. Also if you are using solid volumes how can you say y = 2x, or z = 4x.

The volume of sand is twice the volume of cement when in particle form with voids ie. in its natural state, and similarily for the gravel the volume of gravel contains voids when it is 4 times the volume of cement used.

So except you can explain each step, then I think your calculations are incorrect, or you are trying to calculate something else.

What I think you or the question may be wanting to find is..... "What is the volume of dry mix materials with voids that corresponds to 1 cu yard of solid materials without voidsas solud concrete.?"

and that is a different question...........???????????...........

[Guest]

"So except you can explain each step, then I think your calculations are incorrect, or you are trying to calculate something else." I was trying to solve this problem.


Here is a quote from the problem:

"There being no voids in the cement, we have the equations:

1) x + 2/3y + .55z = 1 (cu. yd. of concrete).

2) y = 2x and z = 4x from the ratios of the mixture.

Substituting the values from y and z from (2) in (1), we have x + 4x/3 + 2.2x = 1."

Here is the question:

"How many cubic yards each of cement, sand, and gravel, mixed in the ratios 1-2-4 (i.e., 1 part cement to 2 parts sand to 4 parts gravel) are required to make one cubic yard of solid concrete, allowing 33 1/3 % voids in the sand and 45 % voids in the gravel ?

[Guest]

Here is a quote from the problem:..........I have added my comments...........

"There being no voids in the cement, we have the equations:

1) x + 2/3y + .55z = 1 (cu. yd. of concrete)......And this is the bit I don't agree with.....
It says....(cement powder solids only) + (sand solids only) + (gravel solids only) = 1 cu yards of solids only concrete.
This is OK from a solids only point of view and dealing with the finished concrete.
But your next 2 equations for y and z indicate the ratios of the dry material that contain voids, so their relationships will not be the same when in solid concrete so you cannot substitute them into the solids only equation.

2) y = 2x and z = 4x from the ratios of the mixture...?????...Yes the ratios of the mixture containing voids

Substituting the values from y and z from (2) in (1), we have x + 4x/3 + 2.2x = 1."..........I don't think the reasoning behind tis is correct?????

Here is the question:

"How many cubic yards each of cement, sand, and gravel, mixed in the ratios 1-2-4 (i.e., 1 part cement to 2 parts sand to 4 parts gravel) are required to make one cubic yard of solid concrete, allowing 33 1/3 % voids in the sand and 45 % voids in the gravel ?

I will interperate the above question to what I think it means........

""How many cubic yards each of cement powder, loose sand, and loose gravel, mixed in the ratios 1-2-4 (i.e., 1 part cement powder to 2 parts loose sand to 4 parts loose gravel) are required to make one cubic yard of solid concrete, allowing 33 1/3 % voids in the sand and 45 % voids in the gravel ?""

You cannot "avoid" the voids in the loose material...this is what it is like when you buy it. The mixture ratios are the ratios of the loose materials containing voids as measured out for mixing. The mixing and setting process then allows the small particles to fill the larger voids and become a more solid structure.

[Guest]

From your earlier post:


"For 1 cu. yard of dry mixture..........
Let............
x = cu.yds cement in 1 cu.yd of mix
y = cu.yds sand in 1 cu.yd of mix
z = cu.yds in 1 cu.yd of mix
Then...............
x + 2/3y + 0.55z + voids = 1 cu.yd"
.........................

Would this be the equation to start:

x + 2/3y + .055z + 33.33 + 0.45 = 1 cu. yd.

I don't know how to proceed.

[Guest]

Would this be the equation to start:

x + 2/3y + .055z + 33.33 + 0.45 = 1 cu. yd.
The 33.33 and the .45 are not in cu yards units, they are the percentage of the sand and gravel volume that is voids.
If you said .....x + 2/3y + .055z + 0.333y + 0.45z = 1 cu. yd......that would be correct but its the same as saying x + y + x = 1 cu yard. It does not help solving the problem....I think.???.
I will give another solution based on my first and considering the discussions so far..............

So if we continue on with my first solution......................

Consider 1 cu yard of Dry Mixure but instead of just enough water to fill the voids and make 1 cu. yard of concrete.......We will add what ever water is required to mix it and give the required "slump" test. Then when the concrete is mixed and set in position we will assume there are enough small particles of cement and sand etc. to fill all the voids and become completely solid.............

Dry normal mixture with voids ratios........ 7 parts alltogether...1/7 is cement....2/7 is sand........4/7 is gravel.......

But sand has 33.33% voids ...... and gravel has 45% voids..........not told anything about the cement so assume no measureable voids.

2/7 x 1/3 = 2/21 = 0.0952 cu. yards of void space in sand

4/7 x 45/100 = 180/700 = 0.257 cu. yards of void space in gravel

2/21 + 18/70 is total void space in mix that is left for water etc and smaller particles to mix.

(20 + 54) /210 = 74/210 = 0.352 cu yards of voids to fill by the smaller particles so a reduction in overall volume.

Volume of the hardened and set solid concrete will be (1.00 - 0.352) = 0.648 cu yards of solid concrete from 1 cu yard of dry materials.

Therefore ( 1/0.648) = 1.543 cu yards of dry material with voids needed to make 1 cu yard of solid concrete.

These dry materials will be made up of................

1/7 x 1.543 = 0.22 cu yards of cement powder

2/7 x 1.543 = 0.441 cu yards of sand includes voids

4/7 x 1.543 = 0.882 cu yards of gravel includes voids

In terms of "Solid rock sand" + voids = 2/3 x 0.441 = 0.294 cu yards solids + 0.147 cu yards voids

In terms of "Solid rock gravel" + voids = 0.55 x 0.882 = 0.485 cu yards solids + 0.397 cu yards voids

................................

....OR below do same calculations but using information regarding the volume of solids in the dry mix..............

1/7 = 0.143 cu yards of solid cement powder

2/7 x 2/3 = 4/21 = 0.190 cu yards of solid sand particles

4/7 x 55/100 = 0.314 cu yards of solid gravel particles

Volume of solid concrete from 1 cu yard of dry mix = 0.143 + 0.190 + 0.314 = 0.647 cu yards of solid concrete

The Ratio of "cement solid", "sand solid", and "gravel solid" in the finished solid concrete then is......
Divide by 0.143 to give....and ........

1 to 1.329 to 2.196 OR 1 : 1.329 : 2.196 by volume of solid concrete.....not the same as the 1 : 2 : 4 from the dry mixture

Therefore 1/0.647 = 1.546 cu yards of dry mix needed for 1 cu yard of solid concrete

The dry mix can then be split into cement, sand and gravel using 1 : 2 : 4 mix as calc. above to give same answer as above.

..........................

Regards your calculations.....From the above solution we have found that............

1) x + 2/3 y + .55 z = 0.648 cu. yd.......the volume of solids assuming x, y, z represent the dry volume with voids

You ignored the voids in equation 1 and also in y =2x and z = 4x

Numerically both these errors must have cancelled the error and you ended up with the "correct figures" only you said it was just for solids, but it is actually for dry mix containing voids.

[Guest]

"In terms of "Solid rock sand" + voids = 2/3 x 0.441 = 0.294 cu yards solids + 0.147 cu yards voids

In terms of "Solid rock gravel" + voids = 0.55 x 0.882 = 0.485 cu yards solids + 0.397 cu yards voids"

How did you calculate the .0.147 and 0.397 ?