Problem 10.89

Given:

A = (125 mm)² = 1.5625 x 10⁴ mm²

adjusted allowable stress (compression) parallel to the grain = 9.2 MPa

E = 12 GPa

effective length = 3.6 m

e = 50 mm

To use the allowable-stress method, these computations must be made:

Note that "c" in equation 10.51 is not the "c" used in Mc/I; its value is 0.8 for sawn lumber columns (page 641). The slenderness ratio L/d = 3.6 m/0.125 m = 28.8.

I won't repeat equation 10.51 again; substituting into it yields a column stability factor C_p = 0.41347. Then

The moment of inertia of the cross section is d⁴/12 = (0.125)⁴/12 = 2.035x10^-5 m⁴. Algebraically rearranging the stress equation...