CE324P Lab 8 Wood

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Dec 6, 2023

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1 Name: Nicholas Garcia Date: October 30, 2023 Assignment CE 324P Laboratory 8: Wood Lab Session: Monday 9AM, Unique Number 16640 Teaching Assistant: Farzana Rahman Lab Partners: Vinay Shah, Austin Simon, Mariana Ponce Ramirez, William Westfield, Casey Colburn 1. To calculate stress and strain for both dry and wet samples for the compression parallel to the grain, we have to measure the initial length, initial diameter, diameter under the applied load, and the applied load itself. To calculate stress, the applied load is divided by the cross-sectional area of the specimen. The equations for engineering stress and strain are as follows: i. Engineering Stress: σ= ? 𝐴 𝑜 Engineering Strain: ε= ∆? ? 𝑜 Where P is the applied load, 𝐴 ? is the initial cross-sectional area, ∆? is the displacement, ? ? is initial length. Given our length, width, and force, we were able to calculate our parameters as follows: ii. σ= 0.0014 𝑘𝑖?? 5.766 𝑖? 2 = 0.000243 ksi ε= 9.48∗10 −5 𝑖? 4 𝑖? = 2.36999 ∗ 10 −5 Upon doing this, we acquired the following plots. Figure 1: Stress Strain Curve for Dry Wood in Compression Parallel

2 Figure 2: Stress Strain Curve for Wet Wood in Compression Parallel a) For wood under parallel compression, our chord modulus can be determined by taking data points around 25% and 75% of the maximum stress and finding the slope. iii. ?ℎ??? ??????? = 𝜎 75% −𝜎 25% ? 75% −? 25% In our case for both dry and wet wood, our chord modulus is calculated to be iv. ?ℎ??? ???????(??𝑦) = 2.21 𝑘?𝑖−0.73 𝑘?𝑖 0.017−0.011 = 246.67 ??𝑖 v. ?ℎ??? ???????(𝑊??) = 1.33 𝑘?𝑖−0.45 𝑘?𝑖 0.014−0.008 = 146.67 ??𝑖 b) The moisture content for our wood specimens is simply defined as vi. ??𝑖????? ??????? = ?(𝑤𝑒?)−?(??) ?(??) ∗ 100 vii. ??𝑖????? ???????(??𝑦) = 77.2 𝑔−75 𝑔 75 𝑔 ∗ 100 = 2.93% viii. ??𝑖????? ???????(𝑊??) = 87.7 𝑔−75.2 𝑔 75.2 𝑔 ∗ 100 = 16.62% Based on these moisture contents, the dry specimen would be classified under oven-dry conditions and the wet specimen would be classified under air-dry conditions as the moisture content is not high enough to be considered green conditions. c) The proportional limit is found by finding the maximum of the elastic region. Where it is exactly is pretty subjective, but I found the proportional limit for the dry and wet specimens to be 2.5 ksi and 1.5 ksi respectively. d) The compressive strength of the wood samples is equal to the maximum stress that it was able to withstand. The compressive strength of the dry and wet specimens are 2.95 ksi and 1.78 ksi respectively.

3 e) According to Table 1, the maximum strength in terms of compression parallel to the grain that can be handled was 2250 psi or 2.25 ksi. From part d, we can see that our dry specimen was able to handle up to 2.95 ksi or 2,950 psi, which was higher than the values in Table 1. However, our wet specimen could only handle up to 1.78 ksi or 1,780 psi, which was less than the values in Table 1, showing the effects of increased moisture to maximum strength. 2. Similarly to question 1, to calculate stress and strain for both dry and wet samples for the compression parallel to the grain, we have to measure the initial length, initial diameter, diameter under the applied load, and the applied load itself. To calculate stress, the applied load is divided by the cross-sectional area of the specimen. The equations for engineering stress and strain are as follows: ix. Engineering Stress: σ= ? 𝐴 𝑜 Engineering Strain: ε= ∆? ? 𝑜 Where P is the applied load, 𝐴 ? is the initial cross-sectional area, ∆? is the displacement, ? ? is initial length. Given our length, width, and force, we were able to calculate our parameters as follows: x. σ= 0.02 𝑘𝑖?? 11.85 𝑖? 2 = 0.001838 ksi ε= 2.52∗10 −5 𝑖? 6.125 𝑖? = 4.11652 ∗ 10 −6 Upon doing this, we acquired the following plots. Figure 3: Stress Strain Curve for Dry Wood in Compression Perpendicular

4 Figure 4: Stress Strain Curve for Wet Wood in Compression Perpendicular a) For wood under perpendicular compression, the chord modulus is more subjective and can vary. I determine my range based on a region where it is linear. For dry, it would be from (0.2-0.35 ksi) and for wet, it would be from (0.1-0.15 ksi). So, the chord modulus would be: xi. ?ℎ??? ???????(??𝑦) = 0.348 𝑘?𝑖−0.198 𝑘?𝑖 0.0099−0.0076 = 65.2 ??𝑖 xii. ?ℎ??? ???????(𝑊??) = 0.15 𝑘?𝑖−0.10 𝑘?𝑖 0.011−0.008 = 16.67 ??𝑖 b) The moisture content for our wood specimens is simply defined as xiii. ??𝑖????? ??????? = ?(𝑤𝑒?)−?(??) ?(??) ∗ 100 xiv. ??𝑖????? ???????(??𝑦) = 162.1 𝑔−156.7 𝑔 156.7 𝑔 ∗ 100 = 3.45% xv. ??𝑖????? ???????(𝑊??) = 176.8 𝑔−139.7 𝑔 139.7 𝑔 ∗ 100 = 26.56% Based on these moisture contents, the dry specimen would be classified under oven-dry conditions and the wet specimen would be classified under air-dry conditions as the moisture content is not high enough to be considered green conditions. c) The proportional limit is found by finding the maximum of the elastic region. Where it is exactly is pretty subjective, but I found the proportional limit for the dry and wet specimens to be 0.35 ksi and 0.15 ksi respectively. d) The compressive strength of the wood samples is equal to the maximum stress that it was able to withstand. The compressive strength of the dry and wet specimens are 0.73 ksi and 0.38 ksi respectively.

5 e) According to Table 1, the maximum strength in terms of compression parallel to the grain that can be handled was 660 psi or 0.66 ksi. From part d, we can see that our dry specimen was able to handle up to 0.73 ksi or 730 psi, which was higher than the values in Table 1. However, our wet specimen could only handle up to 0.38 ksi or 380 psi, which was less than the values in Table 1, showing the effects of increased moisture to maximum strength. 3. Figure 5: Load vs. Displacement for Dry Wood in Bending Figure 6: Load vs. Displacement for Wet Wood in Bending a) To determine our chord modulus, we determine our modulus of elasticity, which is: xvi. ? = ?? 3 48?𝐼 , 𝑤ℎ??? 𝐼 = ?ℎ 3 12 Where ? is our modulus of elasticity, 𝑃 is our load, ? is our length, 𝛿 is our displacement, and 𝐼 is our moment of inertia. Using the data we collected, we get: xvii. 𝐼(??𝑦) = (54.76)(1.425) 3 12 = 13.2 − −→ ?(??𝑦) = (1.888)(27.9375) 3 48(0.57)(13.2) = 114 ?𝑖??/𝑖?

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