Question

Consider the following particulars in respect of a concrete mix design:
$\begin{array}{ccc}& \text{Weight}& \text{Specific Gravity}\\ \text{Cement}& 400\frac{kg}{m^3}& 3.2\\ \text{Fine aggregates}& -& 2.5\\ \text{Coarse}& 1040\frac{kg}{m^3}& 2.6\\ \text{aggregates}& & \\ \text{Water}& 200\frac{kg}{m^3}& 1.0\end{array}$
What shall be the weight of the fine aggregates?

• A. $520\frac{kg}{m^3}$
• B. $570\frac{kg}{m^3}$
• C. $690\frac{kg}{m^3}$
  
• D. $1000\frac{kg}{m^3}$

Answer 1

The correct option is C $690\frac{kg}{m^3}$


Let us assume the weight of the fine aggregated by absolute volume principle

$\frac{M_C}{{\rho }_C}=\frac{M_{FA}}{{\rho }_{FA}}+\frac{M_{CA}}{{\rho }_{CA}}+V_{\omega }=1$

$\frac{400}{3200}+\frac{x}{2500}+\frac{1040}{2600}+\frac{200}{1000}=1$

$\frac{x}{2500}+\frac{29}{40}=1$

$[x=687.5\frac{kg}{m^3}$]
$Weightoffineaggregateis690\frac{kg}{m^3}$