CE324P Lab 8 Wood

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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

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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?

6 xviii. 𝐼(𝑤??) = (58.21)(1.4795) 3 12 = 15.7 − −→ ?(??𝑦) = (0.636)(27.9375) 3 48(0.284)(15.7) = 64.8 ?𝑖??/𝑖? b) The moisture content for our wood specimens is simply defined as xix. ??𝑖????? ??????? = ?(𝑤𝑒?)−?(??) ?(??) ∗ 100 xx. ??𝑖????? ???????(??𝑦) = 731.1 𝑔−719.7 𝑔 719.7 𝑔 ∗ 100 = 1.58% xxi. ??𝑖????? ???????(𝑊??) = 800.3 𝑔−659.6 𝑔 659.6 𝑔 ∗ 100 = 21.3% 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 extreme fiber stress is calculated as xxii. 𝜎 = ?? 𝐼 , 𝑤ℎ??? ? = ?? 4 ??? ? = ℎ 2 Where 𝜎 is our extreme fiber stress, 𝑃 is our load at the proportional limit, ? is our length, and ℎ is our height, Using the data we collected, we get: xxiii. ?(??𝑦) = 1.888∗27.9375 4 = 13.2, ?(??𝑦) = 1.425 2 = 0.7125 ??? 𝜎(??𝑦) = 13.2∗0.7125 13.2 = 0.7125 ??𝑖 xxiv. ?(??𝑦) = 0.636∗27.9375 4 = 4.44, ?(??𝑦) = 1.4795 2 = 0.74 ??? 𝜎(??𝑦) = 4.44∗0.74 15.7 = 0.209 ??𝑖 d) The modulus of rupture is calculated as xxv. 𝜎 = ?? 𝐼 , 𝑤ℎ??? ? = ?? 4 ??? ? = ℎ 2 Where 𝜎 is our modulus of rupture, 𝑃 is our maximum, ? is our length, and ℎ is our height, Using the data we collected, we get: xxvi. ?(??𝑦) = 2.002∗27.9375 4 = 13.98, ?(??𝑦) = 1.425 2 = 0.7125 ??? 𝜎(??𝑦) = 13.98∗0.7125 13.2 = 0.755 ??𝑖 xxvii. ?(𝑤??) = 1.09∗27.9375 4 = 7.61, ?(𝑤??) = 1.4795 2 = 0.74 ??? 𝜎(𝑤??) = 7.61∗0.74 15.7 = 0.366 ??𝑖

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7 e) According to Table 1, the maximum strength in terms of bending that can be handled was 3050 psi or 3.05 ksi. From part d, we can see that our dry specimen was able to handle up to 0.755 ksi or 755 psi, which was lower than the values in Table 1. O wet specimen could only handle up to 0.366 ksi or 366 psi, which was also less than the values in Table 1, showing the effects of increased moisture to maximum strength. 4. Figure 7: Stress Strain Curve for Wood in Parallel Compression for both conditions Figure 8: Stress Strain Curve for Wood in Perpendicular Compression for both conditions

8 Figure 9: Load vs. Displacement Curve for Wood in Bending for both conditions a) From what I can see, the general trend which is seen on all loading conditions is that moisture weakens wood, reducing its strength and stiffness under both compression and bending loads. The effects of moisture are more pronounced in compression perpendicular to the grain and bending compared to compression parallel to the grain. When interpreting the plots, we can see how the wet condition curves generally lie below the dry condition curves, indicating the decrease in strength and stiffness due to moisture content. b) Continuing our discussion from part a, we can see that for all the loading conditions, the dry specimen was stronger than the wet specimen as the moisture content for the dry specimen as calculated in earlier parts was lower than the wet specimen and we know that as moisture content increases, strength decreases and vice versa. The FSP is the moisture content at which the wood cell walls are saturated with water, but the cell cavities are empty. Below the FSP, the wood behaves differently than above it. The FSP is significant in understanding the transition from the hygroscopic to the saturated state. In summary, the strength of wood specimens depends on the loading condition, but the general trend is that dry specimens are stronger than wet specimens. For parallel compression, the fibers are aligned along the load direction. When wood is wet, the FSP is reached, and beyond this point, additional moisture does not significantly affect strength. However, before reaching the FSP, the presence of water weakens the wood, making the wet specimen weaker than the dry one. For perpendicular compression, when wood is dry, the fibers are more rigid and can resist compression better. In the wet state, the FSP has been passed, causing the wood to be significantly weaker due to the collapse of the weaker cell walls. For bending, before the FSP, the dry wood is stiffer and stronger, while beyond the FSP, the wet wood becomes more flexible and weaker. The wet specimen will exhibit lower modulus of elasticity and strength due to the effects of moisture.

9 5. a) Based on our modulus of rupture value, which we calculated to be around 0.755 ksi and given the parameters that our beam has to meet, I will assume a value of 6 inches for the height, so xxviii. ? = ?? 4 = (2500)(240) 4 = 150,000 xxix. ? = ℎ 2 = 6 2 = 3 xxx. 𝜎 = ?? 𝐼 − −→ 𝐼 = ?? 𝜎 = (150,000)(3) 0.755 = 596,000 xxxi. 𝐼 = ?ℎ 3 12 −→ 596,000 = ?(6) 3 12 −→ ? = 33,111 xxxii. ? = ? ∗ 𝑤−→ 33,111 = 240 ∗ 𝑤−→ 𝑤 = 137.96 Ultimately, this comes down to a 2x6 in. beam b) Based on the values from Table 1, specifically the modulus of elasticity which we will select the structural grade, Dense Select Structural with E=1,900,000 psi or 1,900 ksi xxxiii. ? = ?? 3 48?𝐼 −→ 𝐼 = ?? 3 48?? = (2500)(240) 3 48(1)(1,900,000) = 37.89 xxxiv. 𝐼 = ?ℎ 3 12 −→ 37.89 = ?(6) 3 12 −→ ? = 2.105 xxxv. ? = ? ∗ 𝑤−→ 2.105 = 240 ∗ 𝑤−→ 𝑤 = 0.00877 ∗ 1000 = 8.77 Ultimately, this comes down to a 6 x 8 in. beam

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