Chapter 1.5, Problem 113E

Shrinkage in Concrete Beams As concrete dries, it shrinks—the higher the water content, the greater the shrinkage. If a concrete beam has a water content of w kg/m³, then it will shrink by a factor

$S=\frac{0.032w-2.5}{10,000}$

where S is the fraction of the original beam length that disappears due to shrinkage.

1. (a) A beam 12.025 m long is cast in concrete that contains 250 kg/m³ water. What is the shrinkage factor S? How long will the beam be when it has dried?
2. (b) A beam is 10.014 m long when wet. We want it to shrink to 10.009 m, so the shrinkage factor should be S = 0.00050. What water content will provide this amount of shrinkage?

To determine

To evaluate: The shrinkage factor and length of beam when it is dried.

Answer to Problem 113E

The shrinkage factor is $\underset{\_}{0.00055}$ and the length of a beam when it has dried is $\underset{\_}{12.018\text{m}}$.

Explanation of Solution

Given:

If a concrete beam has a water content of $w{\text{kg/m}}^3$, then it will shrink by a factor $S=\frac{0.032w-2.5}{10,000}$, where $S$ is the fraction of the original beam length that disappears due to shrinkage.

That is, $S=\frac{\text{Length}\text{that}\text{disappears}\text{due}\text{to}\text{shrinkage}}{\text{Original}\text{length}\text{of}\text{beam}}$.

Calculation:

Given that the water content in the concrete is $250\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $250$ for $w$ in the equation $S=\frac{0.032w-2.5}{10,000}$.

$\begin{array}{c}S=\frac{0.032\times 250-2.5}{10,000}\\ =\frac{8-2.5}{10,000}\\ =\frac{5.5}{10,000}\\ =0.00055\end{array}$

Thus, the shrinkage factor in concrete beam is $\underset{\_}{0.00055}$.

Note that, the length of the beam is 12.025 m.

Substitute $0.00055$ for $S$ and $12.025$ for original length of beam in the equation $S=\frac{\text{Length}\text{that}\text{disappears}\text{due}\text{to}\text{shrinkage}}{\text{Original}\text{length}\text{of}\text{beam}}$.

$\begin{array}{l}0.00055=\frac{\text{Length}\text{that}\text{disappears}\text{due}\text{to}\text{shrinkage}}{12.025}\\ \text{Length}\text{that}\text{disappears}\text{due}\text{to}\text{shrinkage}=0.00055\times 12.025\\ \text{Length}\text{that}\text{disappears}\text{due}\text{to}\text{shrinkage}=0.0066\end{array}$

Therefore, the length that disappears due to shrinkage is $0.0066\text{m}$.

$\text{Length of beam when it has dried}=\left[\begin{array}{l}\text{Original length of beam}-\text{Length that}\\ \text{ disappears due to shrinkage}\end{array}\right]$

Substitute $12.025$ for original length of beam and $0.0066$ for length that disappears due to shrinkage.

$\begin{array}{c}\text{Length of beam when it has dried}=12.025-0.0066\\ \text{Length of beam when it has dried}=12.018\end{array}$

Length of beam when it has dried is $12.018\text{m}$.

Thus, length of beam when it has dried is $\underset{\_}{12.018\text{m}}$.

To determine

The water content in concrete beam.

Answer to Problem 113E

The water content in concrete beam is $\underset{\_}{234.375\text{kg}/{\text{m}}^{\text{3}}}$.

Explanation of Solution

The shrinkage factor is obtained as $0.00050$.

Substitute $0.00050$ for $S$ in the equation $S=\frac{0.032w-2.5}{10,000}$.

$\begin{array}{l}0.00050=\frac{0.032\times w-2.5}{10,000}\\ 0.00050\times 10,000=0.032w-2.5\\ 0.032w-2.5=5\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}0.032w=5+2.5\\ 0.032w=7.5\\ w=\frac{7.5}{0.032}\\ w=234.375\end{array}$

Thus, the water content in concrete beam is $\underset{\_}{234.375\text{kg}/{\text{m}}^{\text{3}}}$.