Problem 10.90

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 interaction method, most computations are identical to those in Problem 10.89 and will not be repeated here:

c = 0.125/2 = 0.0625 m

I = 2.035 x 10^-5 m⁴

Equation 10.60 must be used:

The problem provided the allowable stress in bending - 12.8 MPa. Substituting all of these values into the above equation and solving for P, I get that P must be less than 34.7 kN.