Wood Lab Report

Posted: August 25th, 2021

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Wood Lab Report

Introduction

I come back with the results and graphs following a request from the Timber Framed Visitor Center for preliminary tests on the tree samples that were submitted. Specifically, compression and beam bending tests were conducted within our labs from which we made a number of determinations. For example, the modulus of rupture, modulus of elasticity, displacement versus load characteristics, and ultimate strength for both parallel and perpendicular to grain rations. The beam bending test was conducted on all the four wood shapes. That is, 2×6, 2×8, LVL and I-Joist. On the other hand, compression test was performed on determine the parallel the perpendicular to grain ratio. It is, therefore, my hope that from these results and graphs, determination of the suitability of the tree samples for your upcoming project will be made easier.

Procedures That Were Used

For the beam bending test, the following arrangement was applied

For the compression test, all the four wood teams were arranged in the longitudinal direction as shown below.

On each of the above arrangements, parallel to grain and perpendicular to grain ratios were determined with the intention of finding the stronger of these samples.

Results

Modulus of Rupture Values (Beam Test).

This concept is primarily based on the elastic beam theory which asserts that should a material remain nearly elastic to the point where it reaches a maximum load is reached, the strength values obtained by both flexural test and compression test will be equal to each other. The modulus of rupture is calculated by multiplying two by the average sample breadth divided by the square of the average sample depth.

Thus, for the 2 * 6 beam, the modulus of rupture = (2 * 4)/8² = 0.125 pounds per square inch

            for the 2 * 8 beam, the modulus of rupture = (2 * 3.5)/ 8² = 0.109 pounds per square inch

            for the LVL beam, the modulus of rupture = (2* 5)/ 9² = 0.123 pounds per square inch

            for the I-Joist beam, the modulus of rupture = (2 * 4.5)/ 8² = 0.141pounds per square inch

Load vs. Displacement Plots (Beam Test)

A plot of the values of load versus displacement for the beam tests resulted in the following common characteristic.

Modulus of Elasticity Values (Beam Test)

For wooden beams, the modulus of elasticity is calculated through the formula MOE = PL³/48IΔ. In each of the tests, the values of the displacement (Δ) kept varying as shown in the following calculations. For the 2 * 6 beam, the modulus of elasticity = (200 * 9³)/ 48 * 64 * 0.6 = 0.729psi

                     For the 2 * 8 beam, the modulus of elasticity = (200 * 9³)/ 48 * 64* 0.09 = 0.528 psi

            For the for the LVL beam, the modulus of elasticity = (200 * 9³)/ 48 * 81 * 0.67 = 0.651psi

            for the I-Joist beam, the modulus of elasticity = (200 * 8³)/ 48 * 64 * 0.76 = 0.911 psi

Load vs. Displacement Plots (Compression Tests)

A plot of the load versus displacement from the compression test of the beams revealed the following characteristics

Plotting the average values that I got from each of the tests I conducted against temperature, the following characteristics emerged. From the modulus of elasticity, the following curve emerged

From the values of relative modulus of rupture in the tests, the following curve emerged

From the compression test, the following curve emerged showing the the variation of compression strength values against temperature during the experiment

Comparisons

 Primarily, axial loading of the 2 x 6 member results in a deformation that is perpendicular to the load direction and is proportional to the deformation that is parallel to the load. Thus, factor μ is calculated by dividing the transverse strain by the axial strain. From the compression lab tests that I conducted, the factor, μ, that is needed to make flexural modulus of elasticity from the 2 x 6 sample equal to modulus of elasticity for the parallel to grain clear specimen can be calculated as 200/520.83 = 0.384. It is important to note that factor μ varies from one tree species to another and also that it is majorly affected by the moisture content and specific gravity acting on the tree samples.

            To calculate the percent difference between E_{perpendicular} and E_parallel for the small clear specimens in compression, I determined the difference between the average of the values of modulus of elasticity of the small clear specimen samples in both the perpendicular and parallel directions and multiplied this difference by 100. That is, {(0.128 – 0.124)/0.128} * 100 = 3.125%

            To calculate the percent difference between the ultimate stresses of the perpendicular to grain and parallel to grain small clear specimens in compression, I divided the value of average difference between the perpendicular and parallel ultimate stresses of the small clear specimens by the perpendicular to grain stress and multiplied that value by 100. That is, {(16,500kPa – 1,700kPa)/ 16,500kPa} * 100 = 89.7%.

Discussion

Item 1. The common type of failure in compression test was the occurrence of cracks which mostly occurred along the grain direction. Also, brittle failure occurred occasioned by rupturing of wood fibers as a result of excessive compression. Excessive compression failure led to plastic deformation of the wood samples especially at the loading points.

Item 2. During lateral torsional buckling, the applied loads caused bending and torsion on the wood samples I used. Mostly, it increased when the sample in use are tall, thin and W-shaped. It failed since the samples used were relatively thick.

Item 3. When both the flexural test and the parallel to grain compressive test caused normal stresses parallel to the grain of the wood flexural modulus of elasticity and compression test modulus of elasticity must be considered separately because usually when the wood samples are loaded in bending, the portion of the wood on one side of the neutral axis is stressed in tension parallel to grain while the other side is stressed in compression parallel to grain.

Item 4. Isotropic and orthotropic materials exhibit various differences between them. The main difference is that whereas isotropic materials have uniform mechanical and thermal properties in every direction, orthotropic materials do not have uniform thermal and mechanical properties in every direction. Thus, orthotropic materials have three planes of axis. That is, longitudinal, tangential, and radial. Wood is one such material.

            Item 5. Basing on the differences of my results for the parallel and the perpendicular compression tests, it is clear that wood is stronger when compressed parallel to the grain than when perpendicular to the grain. This is so because the orientation of wood fibers is such that the entire fiber lengths can resist the applied stresses.