STEM Problem Detail

Problem Statement

Prompt

Give step-wise answer for Example 2.4

Solution Rendering...

## Design of Concrete Mix

Total aggregate content (saturated and surface dry basis) is given by:

$$
\gamma = \gamma_0 - \gamma_c - \gamma_w
$$

where,
\[
\begin{aligned}
\gamma_0 &= \text{wet density of concrete (kg/m}^3\text{)} \\
\gamma_c &= \text{cement content (kg/m}^3\text{)} \\
\gamma_w &= \text{free-water content (kg/m}^3\text{)}
\end{aligned}
\]

**Step 5:**

Determine the proportion of fine aggregate depending upon water--cement ratio,
maximum size of aggregate, grading zone of fine aggregate and workability level
from Fig. 2.3. Also determine the coarse aggregate content.

Fine aggregate content (kg/m$^3$):
\[
= \text{total aggregate content} \times \text{proportion of fine aggregate}
\]

Coarse aggregate content (kg/m$^3$):
\[
= \text{total aggregate content} - \text{fine aggregate content}
\]

The coarse aggregate may be subdivided among 10 mm, 20 mm and 40 mm sizes.
If 10 mm and 20 mm size aggregate are used, they may be taken in the ratio
$1:1$ or $1:2$. If 10 mm, 20 mm and 40 mm size aggregates are to be used,
they may be taken in a ratio of $1:2:3$.

## British Mix Design Method

...

**Step 2:**

Determine the free-water content (kg/m$^3$) from Table 2.12 depending upon the
maximum size and type of aggregate to give a concrete of the specified slump or
Vee-Bee time.

...

### Table 2.12: Approximate water content (kg/m$^3$)

| Maximum size of aggregate (mm) | Type of aggregate | 0--10 mm | 10--30 mm | 30--60 mm |
|---|---|---|---|---|
| 10 | Uncrushed | 150 | 180 | 205 |
| 10 | Crushed | 180 | 205 | 230 |
| 20 | Uncrushed | 135 | 160 | 180 |
| 20 | Crushed | 170 | 190 | 210 |
| 40 | Uncrushed | 115 | 140 | 160 |
| 40 | Crushed | 155 | 175 | 190 |

## Numerical Example

The target mean strength is:

$$
\sigma_t = \sigma_{ck} + k\sigma = 15 + 1.64 \times 3.5 = 21~\text{MPa}
$$

The standard deviation is taken as $3.5$ for a good quality control.

Water--cement ratio based on strength is $0.56$ from Fig. 2.1 for 43 grade
cement and based on durability is $0.60$ from Table 2.5.
Adopt a water--cement ratio equal to $0.60$.

Table 2.5 gives minimum cement content as $240~\text{kg/m}^3$.
The amount of water per m$^3$ of concrete:

$$
= 240 \times 0.60 = 144~\text{kg}
$$

**Step 3:**
Let us adopt maximum size of coarse aggregate as $20~\text{mm}$.

**Step 4:**
Let us adopt fineness modulus of the mixed aggregate to be $4.7$.

**Step 5:**
The proportion of fine to mix aggregate is given by:

$$
P_m = \frac{6.5 - 4.7}{6.5 - 2.0} \times 100 = 40\%
$$

Therefore, use $40~\text{kg}$ fine aggregate with $60~\text{kg}$ coarse aggregate
in a total mixture of $100~\text{kg}$.

Volume of fine aggregate:

$$
= \frac{40}{1700}~\text{m}^3
$$

Volume of coarse aggregate:

$$
= \frac{60}{1600}~\text{m}^3
$$

Ratio of fine to coarse aggregate by volume:

$$
= 1:1.60
$$

**Step 6:**

Apparent volume of cement:

$$
= \frac{240}{1500} = 0.160~\text{m}^3
$$

If there are $60\%$ voids in cement, its solid volume:

$$
= 0.160 \times 0.40 = 0.064~\text{m}^3
$$

Volume of water:

$$
= \frac{144}{1000} = 0.144~\text{m}^3
$$

Total solid volume of aggregate in $1~\text{m}^3$ of concrete:

$$
= 1 - 0.064 - 0.144 = 0.792~\text{m}^3
$$

Solid volume of fine aggregate:

$$
= 0.792 \times \frac{1}{1 + 1.60} = 0.305~\text{m}^3
$$

Since there are $40\%$ voids in fine aggregate, its apparent volume:

$$
= \frac{0.305}{0.6} = 0.50~\text{m}^3
$$

Solid volume of coarse aggregate:

$$
= 0.792 \times \frac{1.60}{2.60} = 0.487~\text{m}^3
$$

Since there are $35\%$ voids in coarse aggregate, its apparent volume:

$$
= \frac{0.487}{0.65} = 0.75~\text{m}^3
$$

### Mix Proportions

Mix proportions by volume:
\[
\text{Cement : Fine aggregate : Coarse aggregate}
= 0.160 : 0.50 : 0.75
= 1 : 3.125 : 4.68
\]

Mix proportions by weight:

$$
1 \times 1500 : 3.125 \times 1700 : 4.68 \times 1600
= 1 : 3.54 : 5
$$

Water--cement ratio:

$$
= 0.60 \quad \text{by weight}
$$

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Problem #62