Chapter 12.15, Problem 64AAP

To determine

The required amounts of components such as cement in sacks, sand and gravel in pounds, and water in gallons.

Answer to Problem 64AAP

The required amounts of components to produce $100{\text{ft}}^3$ concrete are as followed.

• Cement    - 12.14 sacks
• Sand    - 2255 lb
• Gravel    - 3670 lb
• Water     - 54.2 gal

Explanation of Solution

Write the formula for saturated surface dry density.

  $\begin{array}{c}\text{SSDD}=\text{specific gravity}\times \text{density of water}\\ =\text{specific gravity}\times 62.4{\text{lb/ft}}^3\end{array}$        (I)

Calculate the saturated surface dry density of constituents using Equation (I) as shown in Table 1.

Constituents	Specific gravity	Density of water in ${\text{lb/ft}}^{\text{3}}$	SSDD in ${\text{lb/ft}}^{\text{3}}$
Cement	3.15	62.4	197
Sand	2.65	62.4	165
Gravel	2.65	62.4	165
Water	1	62.4	62.4

Table 1

Write the formula for absolute volume of constituents per sack of cement.

  $\begin{array}{c}\text{absolute volume}\\ \text{of constituent}\end{array}=\frac{\left(\begin{array}{l}\text{weight ratio of }\\ \text{the constituent}\end{array}\right)\times \left(\begin{array}{l}\text{weight of cement}\\ \text{ per sack}\end{array}\right)}{\text{SSDD}}$        (II)

Calculate the absolute volume of constituents per sack of cement using Equation (II) as shown in Table 2.

Constituent	Weight ratio of the constituents	Weight of cement per sack	Weight of the constituents	SSDD	Absolute volume of constituent in ${\text{ft}}^{\text{3}}$
Cement	1	94 $\text{lb}$	94 $\text{lb}$	197${\text{lb/ft}}^{\text{3}}$	$0.477{\text{ft}}^3$
Sand	1.9	94 $\text{lb}$	178.6 $\text{lb}$	165${\text{lb/ft}}^{\text{3}}$	$\text{1}\text{.082}{\text{ft}}^{\text{3}}$
Gravel	3.2	94 $\text{lb}$	300.8 $\text{lb}$	165${\text{lb/ft}}^{\text{3}}$	$1.823{\text{ft}}^{\text{3}}$
Water	5.5 gal per sack of cement			$7.48{\text{gal/ft}}^3$	$0.735{\text{ft}}^{\text{3}}$
Total absolute volume of constituents per sack of cement					$4.117{\text{ft}}^3$

Table 2

On dry basis:

The required amount of sacks of cement to produce $50{\text{ft}}^3$ concrete is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{ sacks of cement}\end{array}=\frac{50{\text{ft}}^3}{\left(\begin{array}{l}\text{total absolute volume of}\\ \text{constituents per sack of cement}\end{array}\right)}\\ =\frac{50{\text{ft}}^3}{4.117{\text{ft}}^3\text{/per sack of cement}}\\ =12.14\text{sacks}\end{array}$

The required amount of sand to produce $50{\text{ft}}^3$ concrete is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{ sand}\end{array}=12.14\text{sack}\times \left(\begin{array}{l}\text{weight of cement}\\ \text{ per sack}\end{array}\right)\times \left(\begin{array}{l}\text{weight ratio}\\ \text{of sand}\end{array}\right)\\ =12.14\text{sack}\times 94\text{lb/sack}\times 1.9\\ =2168\text{lb}\end{array}$

The required amount of gravel to produce $50{\text{ft}}^3$ concrete is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{gravel}\end{array}=12.14\text{sack}\times \left(\begin{array}{l}\text{weight of cement}\\ \text{ per sack}\end{array}\right)\times \left(\begin{array}{l}\text{weight ratio}\\ \text{of gravel}\end{array}\right)\\ =12.14\text{sack}\times 94\text{lb/sack}\times 3.2\\ =3652\text{lb}\end{array}$

The required amount of water to produce $50{\text{ft}}^3$ concrete is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{water}\end{array}=12.14\text{sack}\times \left(\begin{array}{l}\text{volume of water }\\ \text{per sack of cement}\end{array}\right)\times \left(\begin{array}{l}\text{weight ratio}\\ \text{of water}\end{array}\right)\\ =12.14\text{sack}\times 5.5\text{gal/sack}\times 1\\ =66.77\text{gal}\end{array}$

On wet basis:

Since, the sand is having the moisture content of $4\%$  and the gravel having $0.5\%$.

The required amount of the wet sand is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{wet sand}\end{array}=2168\text{lb}\times \left(1+\frac{4\%}{100}\right)\\ =2168\text{lb}\times \text{1}\text{.04}\\ =2255\text{lb}\end{array}$

The weight of water present in the sand is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Weight of water}\\ \text{present in sand}\end{array}=2255\text{lb}-2168\text{lb}\\ =87\text{lb}\end{array}$

The required amount of the wet gravel is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Required amount of}\\ \text{wet gravel}\end{array}=3652\text{lb}\times \left(1+\frac{0.5\%}{100}\right)\\ =3652\text{lb}\times \text{1}\text{.005}\\ =3670\text{lb}\end{array}$

The weight of water present in the gravel is expressed as follows.

  $\begin{array}{c}\begin{array}{c}\text{Weight of water}\\ \text{present in sand}\end{array}=3670\text{lb}-3652\text{lb}\\ =18\text{lb}\end{array}$

The required weight of water is expressed as follows.

$\begin{array}{c}\text{Required weight of water}=66.8\text{gal}-\left(87\text{lb}+18\text{lb}\right)\left(\frac{1}{62.4{\text{lb/ft}}^3}\times 7.48{\text{gal/ft}}^3\right)\\ =66.8\text{gal}-12.6\text{gal}\\ =54.2\text{gal}\end{array}$

Thus, the required amounts of components to produce $100{\text{ft}}^3$ concrete are as followed.

• Cement    - 12.14 sacks
• Sand    - 2255 lb
• Gravel    - 3670 lb
• Water     - 54.2 gal