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Concrete Mixture: A Gauss Jordan Elimination Application)

This document discusses concrete mixtures and provides a sample problem to calculate the amounts of cement, sand, gravel, and water needed to produce 50 cubic yards of concrete given weight proportions and a water-cement ratio. A concrete mixture ratio of 1 part cement to 2 parts sand to 3 parts gravel is provided. The sample problem sets up equations based on the given proportions, densities of the materials, and amount of concrete desired. It then uses Gauss-Jordan elimination to solve the system of equations for the amounts of each constituent.
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0% found this document useful (0 votes)
356 views3 pages

Concrete Mixture: A Gauss Jordan Elimination Application)

This document discusses concrete mixtures and provides a sample problem to calculate the amounts of cement, sand, gravel, and water needed to produce 50 cubic yards of concrete given weight proportions and a water-cement ratio. A concrete mixture ratio of 1 part cement to 2 parts sand to 3 parts gravel is provided. The sample problem sets up equations based on the given proportions, densities of the materials, and amount of concrete desired. It then uses Gauss-Jordan elimination to solve the system of equations for the amounts of each constituent.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Concrete Mixture

(A GAUSS JORDAN ELIMINATION APPLICATION)

Concrete - Concrete is the most commonly used man-made material on earth. It is an important construction
materials used extensively in buildings, bridges, roads, and dams. Its uses range from structural applications, to
paviours, kerbs, pipes and drains.

A concrete mixture ratio of 1 part cement, 2 parts sand, and 3 parts aggregate will produce a concrete mix of
approximately 3000 psi. Mixing water with the cement, sand, and stone will form a paste that will bind the
materials together until the mix hardens.

Sample problem:

A concrete mix was designed to have a weight proportions of [1:1.9:2.8] with a water cement ration of 7
gallons of water per sack of cement. How much of each constituent is needed to produce 50 yd^3 of
concrete?

Given:

1 sack of cement = 94 lb

Densities:

Cement = C = 195 lb/ft^3 GRAVEL = G = 165 lb/ft^3

SAND = S = 165 lb lb/ft^3 WATER = W = 62.4 lb/ft^3

Solution:
𝑪 𝑺
= ; 𝟏. 𝟗𝑪 − 𝑺 = 𝟎 (𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏)
𝟏 𝟏.𝟗
𝑪 𝑮
= ; 𝟐. 𝟖𝑪 − 𝑮 = 𝟎 (𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐)
𝟏 𝟐.𝟖
𝟏 𝒔𝒂𝒄𝒌 𝟕 𝒈𝒂𝒍𝒍𝒐𝒏𝒔
= ; 𝟓𝟖. 𝟑𝟖𝟕𝟗𝑪 − 𝟗𝟒𝑮 = 𝟎 (𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟑)
𝑪 𝑾
𝟑 𝒇𝒕 𝟑 𝒇𝒕𝟑 𝒇𝒕𝟑 𝒇𝒕𝟑 𝒇𝒕𝟑
𝟓𝟎 𝒚𝒅𝟑 ( ) = 𝑪𝒍𝒃 (𝟏𝟗𝟓𝒍𝒃) + 𝑺𝒍𝒃 (𝟏𝟔𝟓𝒍𝒃) + 𝑮𝒍𝒃 (𝟏𝟗𝟓𝒍𝒃) + 𝑾𝒍𝒃 (𝟏𝟗𝟓𝒍𝒃)
𝟏 𝒚𝒅

𝑪 𝑺 𝑮𝑪 𝑾
𝟏𝟑𝟓𝟎 = + + + (𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟒)
𝟏𝟗𝟓 𝟏𝟔𝟓 𝟏𝟔𝟓 𝟔𝟐.𝟒

Matrix:
C S G W
𝟏. 𝟗 −𝟏 𝟎 𝟎 𝟎
𝟐. 𝟖 𝟎 −𝟏 𝟎 𝟎
𝟓𝟖. 𝟑𝟖𝟕𝟗 𝟎 𝟎 −𝟗𝟒 = [ 𝟎 ]
𝟏 𝟏 𝟏 𝟏
[ 𝟏𝟗𝟓 𝟏𝟔𝟓 𝟏𝟔𝟓 𝟔𝟐.𝟒 ]
𝟏𝟑𝟓𝟎

Make the pivot in the 1st column by dividing the 1st row by 1.9:
𝟏 −𝟎. 𝟓𝟐𝟔𝟑 𝟎 𝟎 𝟎
𝟐. 𝟖 𝟎 −𝟏 𝟎 𝟎
𝟓𝟖. 𝟑𝟖𝟕𝟗 𝟎 𝟎 −𝟗𝟒 = [ 𝟎 ]
𝟏 𝟏 𝟏 𝟏
[ 𝟏𝟗𝟓 𝟏𝟔𝟓 𝟏𝟔𝟓 𝟔𝟐.𝟒 ]
𝟏𝟑𝟓𝟎
Eliminate the 1st column:

𝟏 −𝟎. 𝟓𝟐𝟔𝟑 𝟎 𝟎 𝟎
𝟎 𝟏. 𝟒𝟕𝟑𝟕 −𝟏 𝟎 𝟎
𝟎 𝟑𝟎. 𝟕𝟑𝟎𝟓 𝟎 −𝟗𝟒 = [ 𝟎 ]
𝟏 𝟏
[𝟎 𝟎. 𝟎𝟎𝟖𝟖 𝟏𝟔𝟓 𝟔𝟐.𝟒 ]
𝟏𝟑𝟓𝟎
Make the pivot in the 2nd column by dividing the 2nd row by 1.4737:

𝟏 −𝟎. 𝟓𝟐𝟔𝟑 𝟎 𝟎 𝟎
𝟎 𝟏 −𝟎. 𝟔𝟕𝟖𝟔 𝟎 𝟎
𝟎 𝟑𝟎. 𝟕𝟑𝟎𝟓 𝟎 −𝟗𝟒 = [ 𝟎 ]
𝟏 𝟏
[𝟎 𝟎. 𝟎𝟎𝟖𝟖 𝟏𝟔𝟓 𝟔𝟐.𝟒 ]
𝟏𝟑𝟓𝟎
Eliminate the 2nd column:
𝟏 𝟎 −𝟎. 𝟑𝟓𝟕𝟏 𝟎 𝟎
𝟎 𝟏 −𝟎. 𝟔𝟕𝟖𝟔 𝟎 𝟎
𝟎 𝟎 𝟐𝟎. 𝟖𝟓𝟐𝟖 −𝟗𝟒 = [ 𝟎 ]
𝟏
[𝟎 𝟎 𝟎. 𝟎𝟏𝟐𝟎 ] 𝟏𝟑𝟓𝟎 𝟔𝟐.𝟒
Make the pivot in the 3rd column by dividing the 3rd row by 20.8528:
𝟏 𝟎 −𝟎. 𝟑𝟓𝟕𝟏 𝟎 𝟎
𝟎 𝟏 −𝟎. 𝟔𝟕𝟖𝟔 𝟎 𝟎
𝟎 𝟎 𝟏 −𝟒. 𝟓𝟎𝟕𝟖 = [ 𝟎 ]
𝟏
[𝟎 𝟎 𝟎. 𝟎𝟏𝟐𝟎 ] 𝟏𝟑𝟓𝟎
𝟔𝟐.𝟒

Eliminate the 3rd column:

𝟏 𝟎 𝟎 −𝟏. 𝟔𝟎𝟗𝟗 𝟎
[𝟎 𝟏 𝟎 −𝟑. 𝟎𝟓𝟖𝟗] = [ 𝟎 ]
𝟎 𝟎 𝟏 −𝟒. 𝟓𝟎𝟕𝟖 𝟎
𝟎 𝟎 𝟎 𝟎. 𝟎𝟕𝟎𝟏 𝟏𝟑𝟓𝟎
Make the pivot in the 4th column by dividing the 4th row by 0.0701:

𝟏 𝟎 𝟎 −𝟏. 𝟔𝟎𝟗𝟗 𝟎
[𝟎 𝟏 𝟎 −𝟑. 𝟎𝟓𝟖𝟗] = [ 𝟎 ]
𝟎 𝟎 𝟏 −𝟒. 𝟓𝟎𝟕𝟖 𝟎
𝟎 𝟎 𝟎 𝟏 𝟏𝟗𝟐𝟒𝟕. 𝟐𝟎𝟒𝟓
Eliminate the 4th column:

𝟏 𝟎 𝟎 𝟎 𝟑𝟎𝟗𝟖𝟔. 𝟓𝟎𝟗𝟔
[𝟎 𝟏 𝟎 𝟎] = [𝟓𝟖𝟖𝟕𝟒. 𝟑𝟔𝟖𝟑]
𝟎 𝟎 𝟏 𝟎 𝟖𝟔𝟕𝟔𝟐. 𝟐𝟐𝟕𝟎
𝟎 𝟎 𝟎 𝟏 𝟏𝟗𝟐𝟒𝟕. 𝟐𝟎𝟒𝟓
C = 30,986.5096 lb.
S = 58,874.3683 lb.
G = 86,762.2270 lb.
W = 19,247. 2045 lb.

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