Chapter 7, Problem 7.1QP

The design engineer specifies a concrete strength of 5500 psi. Determine the required average compressive strength for:

1. a. a new plant where s is unknown
2. b. a plant where s = 500 psi for 22 test results
3. c. a plant with extensive history of producing concrete with s = 400 psi
4. d. a plant with extensive history of producing concrete with s = 600 psi

To determine

The required average compressive strength for a new plant when standard deviation (s) is unknown.

Answer to Problem 7.1QP

The required average compressive strength for a new plant when s is unknown is $\underset{\_}{6,900\text{psi}}$.

Explanation of Solution

Given information:

The concrete strength ${f^{\prime }}_c$ is 5,500 psi.

Calculation:

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+s$

Here, s is the standard deviation.

Take standard deviation as 1,400 psi for compressive strength greather then 5,000 psi.

Substitute 5,500 psi  for ${f^{\prime }}_c$ and 1,400 psi for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+1,400\\ =6,900\text{psi}\end{array}$

Therefore, required average compressive strength for a new plant when standard deviation (s) is unknown is $\underset{\_}{6,900\text{psi}}$.

To determine

The required average compressive strength for a new plant when standard deviation (s) is 500 psi.

Answer to Problem 7.1QP

The required average compressive strength for a new plant when standard deviation (s) is 500 psi $\underset{\_}{6,240\text{psi}}$.

Explanation of Solution

Given information:

The concrete strength ${f^{\prime }}_c$ is 5,500 psi.

Calculation:

Determine the modification factor for 22 number of test $F_{22}$ using the interpolation.

$F_{22}=\left(F_{25}-F_{20}\right)\left(\frac{T_{22}-T_{20}}{T_{25}-T_{20}}\right)+F_{20}$

Here, $F_{25}$ is the modification factor for the 25 number of tests, $F_{20}$ is the modification factor for the 20 number of tests, $T_{22}$, $T_{20}$, and $T_{25}$ are the number of tests.

Substitute 1.03 for $F_{25}$, 1.08 for $F_{20}$, 22 for $T_{22}$, 20 for $T_{20}$, and 25 for $T_{25}$.

$\begin{array}{c}{F}_{22}=\left(1.03-1.08\right)\frac{\left(22-20\right)}{25-20}+1.08\\ =1.06\end{array}$

Determine the modified standard deviation $s^{\prime }$.

$s^{\prime }=s\left(F_{22}\right)$

Substitute 500 psi for s and 1.06 for $F_{22}$.

$\begin{array}{c}{s}^{\prime }=500\times 1.06\\ =530\text{psi}\end{array}$

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+1.34s$

Here, s is the standard deviation.

Substitute 5,500 psi for ${f^{\prime }}_c$ and 530 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+1.34\times 530\\ =6,210\text{psi}\end{array}$

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+2.33s-500$

Substitute 5,500 psi for ${f^{\prime }}_c$ and 530 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+2.33\times 530-500\\ =6,235\text{psi}\end{array}$

Take the maximum value from the calculated ${f^{\prime }}_{cr}$ values.

Use ${f^{\prime }}_{cr}$ value as 6,240 psi.

Therefore, the required average compressive strength for a new plant when standard deviation (s) is 500 psi is $\underset{\_}{6,240\text{psi}}$.

To determine

The required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 400 psi.

Answer to Problem 7.1QP

The required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 400 psi is $\underset{\_}{6,040\text{psi}}$.

Explanation of Solution

Given information:

The concrete strength ${f^{\prime }}_c$ is 5,500 psi.

Calculation:

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+1.34s$

Substitute 5,500 psi for ${f^{\prime }}_c$ and 400 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+1.34\times 400\\ =6,036\text{psi}\end{array}$

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+2.33s-500$

Substitute 5,500 psi for ${f^{\prime }}_c$ and 400 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+2.33\times 400-500\\ =5,932\text{psi}\end{array}$

Take the maximum value from the calculated ${f^{\prime }}_{cr}$ values.

Use ${f^{\prime }}_{cr}$ value as 6,040 psi.

Therefore, the required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 400 psi is $\underset{\_}{6,040\text{psi}}$.

To determine

The required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 600 psi.

Answer to Problem 7.1QP

The required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 600 psi is $\underset{\_}{6,400\text{psi}}$.

Explanation of Solution

Given information:

The concrete strength ${f^{\prime }}_c$ is 5,500 psi.

Calculation:

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+1.34s$

Substitute 5,500 psi for ${f^{\prime }}_c$ and 600 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+1.34\times 600\\ =6,304\text{psi}\end{array}$

Determine the required average compressive strength for a new plant ${f^{\prime }}_{cr}$.

${f^{\prime }}_{cr}={f^{\prime }}_c+2.33s-500$

Substitute 5,500 psi for ${f^{\prime }}_c$ and 600 for s.

$\begin{array}{c}{f^{\prime }}_{cr}=5,500+2.33\times 600-500\\ =6,398\text{psi}\end{array}$

Take the maximum value from the calculated ${f^{\prime }}_{cr}$ values.

Use ${f^{\prime }}_{cr}$ value as 6,400 psi.

Therefore, the required average compressive strength for a plant with extensive history of producing concrete with standard deviation (s) of 600 psi is $\underset{\_}{6,400\text{psi}}$.