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where c(y) and c(z) = perpendicular distances from the z axis and y axis, respectively (these axes meet the x axis at the cross-section centroid at the origin), to the outermost fiber in compression; A = cross-sectional area of the column; and = total compressive stress in the fiber which is farthest removed from both the y and z axes. For an elastic design limited by yield strength, is replaced by the yield strength; M(z) in the right side of Eq. (30.23) is the magnitude of the right side of Eq. (30.22); and M(y) is an expression similar to Eq. (30.22) in which the roles of the y and z axes interchange. Usually, in elastic design, the yield strength is divided by a chosen factor of safety to get a permissible or allowable stress. In problems in which the stress increases linearly with the load, dividing the yield stress by the factor of safety is equivalent to multiplying the load by the factor of safety. However, in the problem at hand, it is clear from the preceding development that the stress is not a linear function of the axial load and that we are interested in the behavior of the column as it enters the above-the-knee region in Fig. 30.4. Here it is necessary to multiply the applied axial load by the desired factor of safety. The same procedure applies in introducing a factor of safety in the critical-load formulas previously derived. Example 1. We will examine the design of a nominally straight column supporting a nominally concentric load. In such a case, a circular column cross section is the most reasonable choice, since there is no preferred direction. For this case, Eq. (30.23) reduces to = P Mc + I A (1)

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For simplicity, we will suppose that the principal imperfection is due to the eccentric location of the load and that the column crookedness effect need not be taken into account, so that Eq. (30.22) reduces to Mmax = P e + P 4e [EI( /L)2 P] (2)

Note that for a circular cross section of radius R, the area and moment of inertia are, respectively, A = R2 and I= R4 A2 = 4 4 (3)

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We will express the eccentricity of the load as a fraction of the cross-section radius. Thus, e = R (4)

DI-4 DI-5 DI-6 DI-7 NCD-102-4

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Then we have, from Eqs. (1) through (4), allow = P P 4P 4 + 1+ A [(EA2)/(4 )]( /L)2 P A (5)

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Usually P and L are given, and E are the properties of chosen material, and is determined from the clearances, tolerances, and kinematics involved, so that Eq. (5) is reduced to a cubic in A. At the moment, however, we are interested in comparing the allowable nominal column stress P/A with the allowable stress of the material allow for columns of different lengths. Keeping in mind that the radius of gyration r of a circular cross section of geometric radius R is R/2, we will define R =r 2 allow =p P/A E =q allow Then Eq. (5) may be written as p = 1 + 4 + 16 [ 2pq(r/L)2 1] (7) (6)

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