How To Calculate The Quantity Of Cement To Be Used In Stabilization Process.

Stabilization refers to changing of some engineering properties of material ( for this case is sub base material ) enough to allow road construction to take place.

The main objective of stabilization  of sub base layer  material is to strengthen the engineering properties of the sub base material such as shrinkage, compaction and bearing capacity.

Understanding the quantity of cement to be used in stabilization is a key economic factor to be considered during stabilization process.

THE PURPOSES  OF CEMENT STABILIZATION.

To increase the density of sub base material and thus lead to fewer voids. The presence of excess voids has negative impacts on the sub base material as they space for moisture to penetrate and make the material less stable by allowing it to shift under changing pressure, temperature and moisture content .

To increase the engineering properties of sub base material such as shrinkage, bearing capacity and compaction.

To improve stiffness and tensile strength of the sub base material.

FACTORS AFFECTING THE QUANTITY OF CEMENT.

Thickness of sub base.

Carriageway width.

Cement ratio.

Weight of the sub base material.

Length of the road.

CALCULATIONS ON THE QUANTITY OF CEMENT TO BE USED IN STABILIZATION.

For the purpose of understanding , consider the following assumptions :

Length of the road X is 50M

Thickness of sub base layer is 0.15M

Carriageway width is 7.5M

Cement ratio is 2.5%

Weight of sub base material is 1.60tones/M³

NB: Cement ratio is obtained in the laboratory by doing a Unconfined Compressive Strength of stabilized cement which is vital in showing the percentage of cement which will stabilize the sub base material. Also the weight of the sub base material is obtained from the laboratory.

Consider the following initials which will be used :

Volume of sub base material in 1M^{2 =} V

Weight of sub base material per 1M²=W

Weight of cement per 1M²⁼WC

Thickness of sub base layer=T

Cement ratio=C

The procedures for calculating are as follows;

Step 1: To calculate the area in which one bag of cement (50Kg) will cover.

V=1M *1M * 0.15M

V=0.15M³

WC=T*W

WC=0.15M*1.60tones/M³

WC=0.24tones/M²

W=WC*C

W=O.24tones/M²   *  2.5%

W=0.006tones/M²=6Kg/M²

Thus;

1M² will require 6 Kg of cement

How many M² will require 50Kg of cement ?

Area= ( 50Kg *1M²)/6Kg

Area=8.33M², Therefore 1 bag of cement (50Kg) will require to stabilize an area of 8.33M²

Step2: Marking the road with chalk/lime.

After knowing the area to be covered with one bag of cement rectangular boxes are to be drawn as shown below :

Area = length * width

8.33M² =L*(7.5M/2)

8.33M² =3.75*L

L = 2.20M

The rectangular boxes are drawn from the centre of the road to both sides of the road. The width  of the box is 3.75M and the length is 2.20M. each box will require one bag of cement.

How many boxes are to be marked and how many bags of cement are required for road X ?

NB: Number of cement bags = number of rectangular boxes to be drawn

Number of cement bags = total road area (A_T) /area which require one bag of cement (A₁)

A_T = 50M * 7.5M

A_T   = 375M²

A₁ = 8.33M²

Number of cement bags = 375M²/8.33M² = 45.01 ̴ 45

(The answer  should be approximated to the nearest whole number )

The quantity of cement = 45 * 50 kg

The quantity of cement is 2250 kg

The quantity of cement to be used in stabilization is 2. 25 tones.