Questions: Problem 3-1: A concrete cylinder having a diameter of 6.00 in. and gage length of 12 in. is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress-strain diagram using scales of 1 in. =0.05 ksi and 1 in. =0.2(10^-3) in./in. . From the diagram, determine approximately the modulus of elasticity. Load (kip) Contraction (in.) 0 0 5.0 0.0006 9.5 0.0012 16.5 0.0020 20.5 0.0026 25.5 0.0034 30.0 0.0040 34.5 0.0045 38.5 0.0050 46.5 0.0062 50.0 0.0070 53.0 0.0075

Problem 3-1: A concrete cylinder having a diameter of 6.00 in. and gage length of 12 in. is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress-strain diagram using scales of 1 in. =0.05 ksi and 1 in. =0.2(10^-3) in./in. . From the diagram, determine approximately the modulus of elasticity. Load (kip) Contraction (in.) 0 0 5.0 0.0006 9.5 0.0012 16.5 0.0020 20.5 0.0026 25.5 0.0034 30.0 0.0040 34.5 0.0045 38.5 0.0050 46.5 0.0062 50.0 0.0070 53.0 0.0075
Transcript text: Problem 3-1: 3-1. A concrete cylinder having a diameter of 6.00 in . and gage length of 12 in . is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress-strain diagram using scales of $1 \mathrm{in} .=0.05 \mathrm{ksi}$ and $1 \mathrm{in} .=0.2\left(10^{-3}\right) \mathrm{in} /$ in. . From the diagram, determine approximately the modulus of elasticity. \begin{tabular}{|l|l|} \hline Load $($ kip) & Contraction (in.) \\ \hline 0 & 0 \\ \hline 5.0 & 0.0006 \\ \hline 9.5 & 0.0012 \\ \hline 16.5 & 0.0020 \\ \hline 20.5 & 0.0026 \\ \hline 25.5 & 0.0034 \\ \hline 30.0 & 0.0040 \\ \hline 34.5 & 0.0045 \\ \hline 38.5 & 0.0050 \\ \hline 46.5 & 0.0062 \\ \hline 50.0 & 0.0070 \\ \hline 53.0 & 0.0075 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Calculate Stress and Strain

The stress, \(\sigma\), is calculated using the formula: \[ \sigma = \frac{\text{Load (kip)}}{\text{Area (in}^2\text{)}} \] The area, \(A\), of the cylinder is: \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{6.00}{2}\right)^2 = 28.2743 \, \text{in}^2 \]

The strain, \(\epsilon\), is calculated using the formula: \[ \epsilon = \frac{\text{Contraction (in.)}}{\text{Gage Length (in.)}} = \frac{\text{Contraction (in.)}}{12} \]

Step 2: Calculate Stress and Strain for Each Data Point

| Load (kip) | Contraction (in.) | Stress (ksi) | Strain (\(10^{-3}\)) | |------------|-------------------|--------------|----------------------| | 0 | 0 | 0 | 0 | | 5.0 | 0.0006 | 0.1769 | 0.05 | | 9.5 | 0.0012 | 0.3361 | 0.10 | | 16.5 | 0.0020 | 0.5838 | 0.17 | | 20.5 | 0.0026 | 0.7249 | 0.22 | | 25.5 | 0.0034 | 0.9012 | 0.28 | | 30.0 | 0.0040 | 1.0610 | 0.33 | | 34.5 | 0.0045 | 1.2202 | 0.38 | | 38.5 | 0.0050 | 1.3613 | 0.42 | | 46.5 | 0.0062 | 1.6445 | 0.52 | | 50.0 | 0.0070 | 1.7680 | 0.58 | | 53.0 | 0.0075 | 1.8755 | 0.63 |

Step 3: Determine Modulus of Elasticity

The modulus of elasticity, \(E\), is the slope of the initial linear portion of the stress-strain curve. Using the first few points:

\[ E = \frac{\Delta \sigma}{\Delta \epsilon} = \frac{0.3361 - 0.1769}{0.10 - 0.05} = 3.184 \, \text{ksi} \]

Final Answer

The modulus of elasticity is approximately \(3.184 \, \text{ksi}\).

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