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The solution for the deflection y in Eq. (15.17) is given by y=P
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4e n
sin (n x/L) [EI(n /L)2 P]
(30.20)
The maximum deflection ymax of a simply supported column will usually (except for cases with a pronounced and asymmetrical initial deformation or antisymmetrical load eccentricity) occur at the column midpoint. A good approximation (probably within 10 percent) of ymax in the above-the-knee region that may be used in deflectionlimited column design is given by the coefficient of the first term in Eq. (30.20): ymax = P[c(1) (4e/ )] EI( /L)2 P (30.21)
The maximum bending moment will also usually occur at the column midpoint and at incipient yielding is closely approximated by Mmax = P e Ymid + P 4e c(1) [EI( /L)2 P] (30.22)
The immediately preceding analysis deals with the bending moment about the z axis (normal to the paper). Clearly, a similar analysis can be made with regard to bending about the y axis (Fig. 30.1). Unlike the analysis of the perfect column, where it is merely a matter of finding the buckling load about the weaker axis, in the
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INSTABILITIES IN BEAMS AND COLUMNS 30.10
LOAD CAPABILITY CONSIDERATIONS
present approach the effects about the two axes interact in a manner familiar from analysis of an eccentrically loaded short strut. We now use the familiar expression for combining direct axial stresses and bending stresses about two perpendicular axes. Since there is no ambiguity, we will suppress the negative sign associated with compressive stress: = P M(z)c(y) M(y)c(z) + + I(z) I(y) A (30.23)
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)
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)
We will express the eccentricity of the load as a fraction of the cross-section radius. Thus, e = R (4)
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