Question

The cement content in a mix design is 378 kg/m3, water content 170 kg, sand is 30% of the total aggregate, entrapped air is 1%, specific gravity of cement, coarse aggregate and fine aggregate are respectively 3.15, 2,60 and 2.60. the fine aggregate will be approximately:  
1. 510 kg
2. 550 kg
3. 600 kg
4. 450 kg
5.

Answer 1

Correct Answer - Option 2 : 550 kg

Concept:

\(V = \left[ {W + \frac{C}{{{S_c}}} + \frac{1}{P} \times \frac{{{f_a}}}{{{S_{fa}}}}} \right] \times \frac{1}{{1000}}\)

\(\frac{{{C_a}}}{{{S_{ca}}}} = \frac{{1 - P}}{P} \times \frac{{{f_a}}}{{{S_{fa}}}}\)

where

V = absolute volume of fresh concrete, which is equal to gross volume (m3) minus the volume of entrapped air,

W = Mass of water (kg) per m3 of concrete

C = Mass of cement (kg) per m3 of concrete

Sc = Specific gravity of cement

P = Ratio of Fine Aggregate to total aggregate by absolute volume

fa, Ca = Total masses of Fine Aggregate and Coarse Aggregate (kg) per m3 of concrete respectively

Sfa, Sca = Specific gravities of saturated, surface dry fine aggregate and coarse aggregate respectively

Given:

W = 170 kg

C = 378 kg

Sc = 3.15

P = 0.30

Sfa = Sca = 2.60

V = 1 – 0.01 =0.99

Calculations:

\(0.99 = \left[ {170 + \frac{{378}}{{3.15}} + \frac{1}{{0.30}} \times \frac{{{f_a}}}{{2.60}}} \right] \times \frac{1}{{1000}}\)

\(0.99 \times 1000 = \left[ {170 + \frac{{378}}{{3.15}} + \frac{1}{{0.30}} \times \frac{{{f_a}}}{{2.60}}} \right]\)

\(990 - 170 - 120 = \left[ {\frac{1}{{0.30}} \times \frac{{{f_a}}}{{2.60}}} \right]\)

\({f_a} = 0.30 \times 2.60 \times 700 = 546\;kg\)