Mixture - Concrete - Solution / Explanation

Algebra 2

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Wed Apr 22, 2015 7:41 pm

"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 ?

It tells you this in the question.... Sand is 1/3 voids so must be 2/3 actual sand particles what I called "solid rock sand"....the bits that are actual particles of sand. Similarly for the gravel that question tells you it is 45% voids so must be 55% solid gravel rock particles.

For sand..... 2/3 x 0.441 = 0.294 cu yards solid sand particles and 1/3 x 0.441 = 0.147 cu yards voids in the sand.

For gravel..... 0.55 x 0.882 = 0.485 cu yards solid gravel particles and 0.45 x 0.882 = 0.0.397 cu yards voids in the gravel.

And if you add in the cement powder = 0.22 cu yards solid cement powder particles having no voids .........

The Total of all this is... (0.22 + 0.294 + 0.485) = 1 cu yard of solids particles + (0.00 + 0.147 + 0.397) = 0.544 cu yards of voids.
This gives a total volume of the materials measured out for mixing as... (1.00 + 0.544) = 1.544 cu yards of dry mix material made up of solids+voids.

There are some minor roundings errors now in the calculations because taking calcs to 3 decimal places.........
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Wed Apr 22, 2015 8:38 pm

Thanks. How did you figure out all the small details ?
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Thu Apr 23, 2015 12:26 am

"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."

Please explain how the 1:2:4 mix would be used to obtain the same answer ? Thanks.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Thu Apr 23, 2015 6:15 am

Does that mean you still do not understand what we are calculating..????......

This is the calculation where we start as we would in "real life" with measuring out loose dry cement powder. particle sand and particle gravel in the ratio of 1 : 2 : 4 as it states in the question to give a dry mix volume of 1 cu yard. ...... 1 + 2 + 4 = 7 parts alltogether and consists of particles and voids. So the fraction by volume of each component in 1 cu yard dry materials is as follows............

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 that can be made from 1 cu yard of dry mix = 0.143 + 0.190 + 0.314 = 0.647 cu yards of solid concrete. Because concrete is all solid material having no voids.

The Ratio of "cement solid", "sand solid", and "gravel solid" in the finished solid concrete then is......
Divide by 0.143 to give ratios based on 1 unit of cement solid....gives ........

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

The dry mixture ratios are the ratios of the solids+voids in the less dense dry mixture having voids, The solid concrete ratios is the ratio of the solid particles in the concrete having no voids.

From above we found that 1 cu yard of dry material contains 0.647 cu yards of solid particles and thus can make 0.647 cu yards of solid concrete

Therefore 1/0.647 = 1.546 cu yards of dry mix will contain 1 cu yard of solid particles and can make 1 cu yard od solid concrete.

The dry mix of 1.546 cu yards can then be split into cement, sand and gravel using 1 : 2 : 4 mix because it is dry materials containg particles and voids..... as calc. above to give same answer as above.

1/7 x 1.546 = 0.221 cu yards of loose cement powder

2/7 x 1.546 = 0.442 cu yards of loose sand particles

4/7 x 1.546 = 0.883 cu yards of loose gravel particles

This is the volumes of loose particles in 1.546 cu yards of dry materials mix including voids.


1/7 x 1.546 = 0.221 cu yards of solid cement powder

2/7 x 2/3 x 1.546 = 0.294 cu yards of solid sand particles

4/7 x 55/100 x 1.546 = 0.486 cu yards of solid gravel particles

This is the volumes of solid particles only in 1.546 cu yards of dry materials mix including voids.

It takes 1.546 cu yards of loose materials mix to make 1 cu yard of solid concrete.

We had already calculated this in some form or another before...........
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Thu Apr 23, 2015 9:44 am

I understand now. Thanks very much.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Sun Apr 26, 2015 12:53 pm

A follow-up question:

I just a related statement about another 1:2:4 mixture of concrete. The statement indicates for a cubic FOOT of concrete, .22 cu. ft. of cement, .44 cu. ft. of sand, and .88 cu ft. would be used. Would those be the correct answers ? The statement did provide the calculations. Just curious. Thanks.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Sun Apr 26, 2015 2:45 pm

The statement should be : I just read a statement about another 1:2:4 mixture of concrete. Also, the .88 cu. ft. should be .88 cu. ft. of gravel.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Sun Apr 26, 2015 10:56 pm

Error - The statement did not provide calculations.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Mon Apr 27, 2015 11:47 am

Please reply. Thanks.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Wed Apr 29, 2015 4:04 pm

Please respond.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Mon May 04, 2015 8:35 pm

From your statement .... 0.22 + 0.44 + 0.88 = 1.54 cuft of dry 1:2:4 mix materials makes 1 cuft of mixed and set solid concrete.
This is the same figures as we found in the earlier question....seems reasonable.
Guest
 

Re: Mixture - Concrete - Solution / Explanation

Postby Guest » Mon May 04, 2015 10:41 pm

Thanks again.
Guest
 

Previous

Return to Algebra 2



Who is online

Users browsing this forum: No registered users and 2 guests

cron