This experiment explores the nature of five different types of wood. By using the angle at which the grain was cut, as well as initial measurements of mass length and thickness of each member, this experiment reveals the strength and properties of five different wood samples. Ultimately, the Baldwin
Universal Testing Machine, shown in Figure 1 below, was used to compress samples of Douglas fir (short
and long samples), Yellow Pine (samples with knots as well as samples without knots), and a dowel of
Oak. This compression revealed relationships between load and deflection which allow for the calculation of flexural modulus of elasticity, modulus of rupture and applied cross sectional area of each wood sample.
Figure 1: Baldwin Universal Testing Machine (Brenner, Course Guide CD, 2010)

This experiment followed two separate procedures, one for a long sample of wood, the other, for much shorter samples. It is first necessary to obtain a sample of each variety of wood to be tested, regardless of the method to be used. These samples must have their length, width, height, mass and
grain-cut angle measured (an example of which is pictured below in Figure 2), regardless of the size of
the sample. Shorter samples will require an aluminum block which will compress the sample, it is necessary to know the width of this block in such a case.
Figure 2 - A 25 Degree Grain-Cut Angle (Brenner, my.fit.edu/~jbrenner, 2010)
With these initial measurements made, it is necessary to move to the Baldwin Universal Testing
Machine and set up a compression test. It is important to know the distance between the rollers on the top and bottom of the sample. Once a sample is then installed into the experimental setup, it is necessary to place a dial gauge at the midpoint of the (not yet) applied load. An initial measurement should be taken at load=0lbs.
The Baldwin Universal Testing Machine should then be turned on, and dial gauge readings should be taken at regular intervals of the applied load, it is important that the dial gauge reading at the maximum load experienced is read, as this load will be known to be the failure strength of the wood.
For a less detailed result, one can record only the failure strength of each sample, as was done in this report after the first sample (Long Wood: Douglas Fir). Should a dowel need to be examined, the procedure should be altered thusly: no aluminum block or rollers should be used, the failure strength can be determined by the first deformation seen within the dowel when placed vertically within the testing machine.

2000
1500
1000
500
0
0,000
-500
0,100 0,200 0,300 0,400 0,500 0,600
Deflection (in)
Figure 3 - Graphed Data
Figure 3 (pictured above) illustrates the load vs deflection relationship of the long Douglas Fir
sample. This relation allows for the calculation of the slope ‘a’ to be used in the equation for Flexural
Modulus of Elasticity. Because the plot is largely linear, with only a slight deviation at the uppermost portion, it can be concluded that the Douglas Fir sample underwent very little to no plastic deformation.
Equations used include:
𝐹𝑚𝑜𝐸 = 𝑎 ∗ 𝐿
3
4𝑤 ∗ ℎ 3
𝑀𝑜𝑅 =
3 ∗ 𝐹 ∗ 𝐿
2𝑤 ∗ ℎ 2
𝑈𝐶𝑆 = 𝑤
𝐹
′ 𝑤
𝑈𝐶𝑆 𝑑𝑜𝑤𝑒𝑙
=
𝐹
. 25𝜋𝐷 2
Where ‘F’ is the failure strength and ‘L’ ‘h’ and ‘w’ are the dimensions of the sample.
Ultimately, the failure of each sample was marked by a loud sound, which served as a que to
record the load measurement. The results displayed in Table 1 below reflect the data gathered.

Table 1 (Brenner & Lam, my.fit.edu/~jbrenner, 2010)
FMOE (Douglas Fir)
MOR (Douglas Fir)
UCS (Yellow Pine)
UCS (Yellow pine w/knots)
UCS (Douglas Fir)
UCS (Oak)
1.95* 10
6
psi
12400 psi
580 psi
580 psi
800 psi
7440 psi
1.677 ∗ 10
6
± 68000 𝑝𝑠𝑖
18700 ± 18900 𝑝𝑠𝑖
940 ± 3 𝑝𝑠𝑖
1570 ± 5 𝑝𝑠𝑖
1040 ± 3 𝑝𝑠𝑖
10300 ± 6𝑝𝑠𝑖
The above results illustrate a significant deviation from the data that is considered to be acceptable by the literature values. These deviations can be explained by the multiple sources or significant error within the experiment.
First and foremost, it should be said that a direct comparison between the experimental and literature values displayed is not wholly appropriate. Because of the rigorous conditions under which the literature values are collected, the experimental data is to be considered inferior. Literature data is collected with wood treated for standardized water content, one of the largest sources of error within the experiment is that no such control could be placed. The extra hydrogen bonds made available by the excess water would make the wood inter-fibrous bonds significantly stronger, resulting in the
increased Ultimate Compressive Strength (UCS) in all cases in Table 1.
The experimental data differs further from the experimental data in that each literature value assumes a grain-cut angle which is perfect (180 degrees), this was not the case with any but one of the samples in the experimental data, shown in Table 4 in the Appendix. The imperfect angle of the grain would create a horizontal component to the applied load, effectively making the UCS of the wood in question greater.
Each wood sample failed in the sample fashion: delamination, with the exception of the oak dowel which experienced a fibrous rupture. These failure mechanisms are to be expected with the respective stresses placed on the wood samples. Representative pictures of such failures are included below.

Figure 4 - Representative Failure Mechanisms
In the results section, it was mentioned that Figure 3 revealed that the douglas fir experienced
little to no plastic deformation. This information indicates that douglas fir is very brittle as opposed to ductile, in that a ductile substance readily experiences plastic deformation prior to rupture.

Table 2 - 95% Confidence Intervals (Datafit)
95% Confidence Intervals
Variable Value a
95% (+/-)
5651.911 228.2708
Lower
Limit
Upper
Limit
5423.64 5880.182 b -490.221 64.93406 -555.155 -425.287
Table 3 - Spreadsheet Data
Long Wood Douglas Fir
Numbers dial gauge (in) load (lb)
0.100
0.114
0.129
0.145
0
100
200
300
0.160
0.175
0.190
0.206
0.224
0.240
0.257
0.273
400
500
600
700
800
900
1000
1100
0.290
0.308
0.326
0.345
0.364
0.384
0.406
0.434
0.467
1200
1300
1400
1500
1600
1700
1800
1900
2000

Table 5 – Sample Calculations
𝐹𝑚𝑜𝐸 = 𝑎∗𝐿 3
4𝑤∗ℎ 3
𝐹𝑚𝑜𝐸 =
(5651.911)∗(16.125)
3
4∗(1.377)∗(1.369) 3
= 1.677 ∗ 10 6 𝑝𝑠𝑖
𝐸𝑟𝑟𝑜𝑟
𝐹𝑚𝑜𝐸
𝐿 3
= (
4𝑤ℎ 3
) 𝑑 𝑎
+ (
3𝑎𝐿 2
4𝑤ℎ 3
) 𝑑
𝐿 𝑎𝐿 3
− (
4𝑤 2 ℎ 3
) 𝑑 𝑤
3𝑎𝐿 3
− (
4𝑤ℎ 4
)𝑑 ℎ
𝐸𝑟𝑟𝑜𝑟
𝐹𝑚𝑜𝐸
= (
(16.125)
3
4(1.377)(1.369) 3
) (228.27) + (
3(5651.911)(16.125)
2
4(1.377)(1.369) 3
) (. 125) −
(
(5651.911)(16.125) 3
4(1.377) 2 (1.369) 3
) (. 001) − (
3(5651.911)(16.125) 3
4(1.377)(1.369) 4
) (. 001) = 67724 𝑝𝑠𝑖
𝑀𝑜𝑅 =
3∗𝐹∗𝐿
2𝑤∗ℎ 2
𝑀𝑜𝑅 =
3∗2000∗16.125
2(1.377)∗(1.369) 2
= 18744 𝑝𝑠𝑖
𝐸𝑟𝑟𝑜𝑟
𝑀𝑜𝑅
= (3 ∗
𝐿
2∗𝑤∗ℎ 2
) 𝑑
𝐹
+ (3 ∗
𝐹
2∗𝑤∗ℎ 2
) 𝑑
𝐿
− (3 ∗ 𝐹 ∗
𝐿
2∗𝑤 2 ∗ℎ 2
) 𝑑 𝑤
− (3 ∗ 𝐹 ∗
𝐿 𝑤∗ℎ 3
) 𝑑 ℎ
𝐸𝑟𝑟𝑜𝑟
𝑀𝑜𝑅
= (3 ∗
16.125
2∗1.377∗1.369
2
) (10) + (3 ∗
16.125
2∗1.377
2 ∗1.369
2
) (. 001) − (3 ∗ 2000 ∗
2000
2∗1.377∗1.369
2
) (. 125) − (3 ∗ 2000 ∗
16.125
1.377∗1.369
3
) (. 001) = 189750 𝑝𝑠𝑖
𝑈𝐶𝑆 = 𝑤
𝐹
′ 𝑤
𝑈𝐶𝑆 =
2840
(2.006)(1.307)
= 1083 𝑝𝑠𝑖
𝐸𝑟𝑟𝑜𝑟
𝑈𝐶𝑆
= (
1 𝑤∗𝑤 ′
) 𝑑
𝐹
− ( 𝑤
𝐹
′2 ∗𝑤
) 𝑑 𝑤
′
− ( 𝑤 ′
𝐹
∗𝑤 2
) 𝑑 𝑤
𝐸𝑟𝑟𝑜𝑟
𝑈𝐶𝑆
1
= (
2.006∗1.307
) (10) − (
2840
2.006
2 ∗1.307
) (. 001) − (
2840
2.006∗1.307
2
) (. 001) = 2.45