I'm not understanding how you are determining that? It looks to me like the tensile fiber stress at the elastic limit is 9000psi and the compressive stress at the elastic limit is 4620psi. This is for the air dried shagbark hickory. Am I misreading that?
No, you are not misreading the data supplied in the publication. The first scan shows the formula and raw data used for calculating the fiber stress at the elastic limit in the static bend test. As far as I can tell, the formula presumes the specimen to be homogeneous and having the neutral axis at the center of the cross section. I believe this is a common practice, as there are few uses for wood in tension near the elastic limit and the difficulty of conducting tension tests along with corresponding compression tests on the same sample, in order to actually determine the location of the NA in each specimen tested in bending
I cannot think of any reason why MOE data obtained from a dedicated compression test should differ from data obtained from a bend test unless there were actual differences between tension MOE and compression MOE.
For the air dried shagbark hickory tested above, the MOE at the elastic limit in bending is 1980 (x1000) vs. 2280 from the dedicated compression test., or 300 lower.
Since the bend test acts equally on the back and the belly of the specimen, and the MOE at the elastic limit represents both tension and compression stiffnesses, I have assumed it must be an average of the two. In order for the the bend test average to be 300 lower, the stiffness in tension must be 600 lower, or 1680. This makes tension MOE 76% of compression MOE. or compression MOE 36% higher than tension, depending on your point of reference. My statement of " the back is 36% less stiff.", in the post above is incorrect math. (it implies a difference of 820).
The last two scans are the data and formulas for the compression test
