ELASTIC BUCKLING. without experiencing a sudden change in its configuration.

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1 EASTIC BUCKING So far we have discussed: (1) the strength of the structure, i.e., its ability to support a specified load without experiencing excessive stress; () the ability of the structure to support a specified load without undergoing unacceptable deformations. we will study the stability of the structure, i.e., its ability to support a given load without experiencing a sudden change in its configuration. Consider a long straight slender member subjected to axial force. Such a member is called a COUMN. Assume both ends are pinned. Top end is restricted to move side ways. As the load is increased, theoretically there should not be any transverse deflection perpendicular to the member axis. Due to a small disturbance in the transverse direction or due to some imperfection in the column the member will bend as shown in figure. On the release of load, if the column return to its straight vertical position, then it is in stable condition under the applied load. As the load is increased further, for a particular load, transverse deflection increases without any increase of the axial load. (Unstable condition). i The load at which unstable condition is attained is called critical load. W h ll h if i ilib i i di b d h ill i i i l ilib i i i We shall note that if its equilibrium is disturbed, the system will return to its original equilibrium position as long as does not exceed a certain value cr, called the critical load. However, if > cr, the system will move away from its original position. In the first case, the system is said to be stable, and in the second case, it is said to be unstable.

2 EUER'S FORMUA FOR IN ENDED COUMNS The free body diagrams of the column and a section of the column at a distance x is shown above d M dx EI EI d v M v Setting dv dx v 0 EI dv 0 v dx EI we can write the above equation as The above is the second order differential equation, the general solution for the above equation is given by v Asin xbcosx Boundary conditions are at x =0, v=0 and at x=, v=0 Therefore B= 0; we obtain Asin x 0 A = 0 is trivial solution; therefore sin = 0 That leads to n EI The smallest of the values of defined by above equation is that corresponding to n = 1. We thus have cr EI

3 The expression obtained is known as Euler's formula, after the Swiss mathematician eonhard Euler ( ). The Deflection equation is given by v Asini x which is the equation of the elastic curve after the column has buckled (see figure). We note that the value of the maximum deflection, v max = A, is indeterminate. This is due to the fact that the differential equation is a linearized approximation of the actual governing differential equation for the elastic curve. If < cr, the condition sin = 0 cannot be satisfied, and the solution does not exist. We must then have A = 0, and the only possible configuration for the column is a straight one. Thus, for < cr the straight configuration of is stable. In the case of a column with a circular or square cross section, the moment of inertia I of the cross section is the same about any centroidal axis, and the column is as likely to buckle in one plane as another, except for the restraints which h may be imposed dby the end connections. For other shapes of cross section, the critical load should be computed by making I = I min If buckling occurs, it will take place in a plane perpendicular to the corresponding principal axis of inertia. The value of the stress corresponding to the critical load is called the critical stress and is denoted by cr. Setting I = Ar, where A is the cross-sectional area and r its radius of gyration, we have cr cr EAr E A A / r The quantity /r is called the slenderness ratio of the column. It is clear, that the minimum value of the radius of gyration r should be used in computing the slenderness ratio and the critical stress in a column.

4 the critical stress is proportional to the modulus of elasticity of the material, and inversely proportional to the square of the slenderness ratio of the column. The plot of cr versus /r is shown in Figure for structural steel, assuming E = 00 Ga and S y = 50 Ma. We should keep in mind that no factor of safety has been used in plotting cr. We also note that, if the value obtained for cr is larger than the yield strength S y y, this value is of no interest to us, since the column will yield in compression and cease to be elastic before it has a chance to buckle. Our analysis of the behavior of a column has been based so far on the assumption of a perfectly aligned centric load. In practice, this is seldom the case. Taking into account the effect of the eccentricity of the loading. lead to a smoother transition from the buckling failure of long, slender columns to the compression failure of short, stubby columns. It will also provide us with a more realistic view of the relation between the slenderness ratio of a column and the load which causes it to fail. EXTENSION OF EUER'S FORMUA TO COUMNS WITH OTHER END CONDITIONS El Euler's formula was for a column which hwas pin-connected at both ends. In the case of a column with one free end A supporting a load and the other end fixed we observe that the column will behave as the upper half of a pin-connected column. The critical load for the column of is thus the same as for the pin-ended column of Figure (b) and may be obtained from Euler's formula by using a column length equal to twice the actual length of the given column. We say that the effective length e of the column of Figure (a) is equal to and substitute e = in Euler's formula: Fixed Free column

5 Fixed Fixed Column The symmetry of the supports and of the loading about a horizontal axis through the midpoint i C requires thatt the shear at C and the horizontal components of the reactions at A and B be zero. It follows that the restraints imposed upon the upper half AC of the column by the support at A and by the lower half CB are identical. ortion AC must thus be symmetric about its midpoint D, and this point must be a point of inflection, where the bending moment is zero. A similar reasoning shows that the bending moment at the mid-point E of the lower half of the column must also be zero. Since the bending moment at the ends of a pin-ended column is zero, it follows that the portion DE of the column of must behave as a pin-ended column (b). We thus conclude that the effective length of a column with two fixed ends is e = /. Fixed inned Column In the case of a column with one fixed end B and one pin-connected end A supporting a load we must write and solve differential equation of the elastic curve to determine the effective length of the column. From the free-body diagram of the entire column we first note that a transverse force V is exerted at end A, in addition to the axial load, and that V is statically indeterminate. Considering now the free-body diagram of a portion AQ of the column, we find that the bending moment at Q is M v Vx

6 dv M v Vx dx EI EI EI Set then EI dv Vx v dx EI The solution for the above differential equation consist of general solution plus particular solution articular solution is given by V V v x x EI V v Asin xbcosx x The constants A, B and V can be solved using Boundary conditions dv at x=0, v=0; at x=, v=0 and 0 dx Using these boundary conditions, we get B=0; V A sin = (a) V A cos = (b) From (a) and (b), we obtain tan Solving by trial and error the above transcendal equation, we get EI Since, we obtain cr EI Using effective length, we can write EI 0.19EI e e

7 ECCENTRIC OADING THE SECANT FORMUA The load applied to a column is never perfectly centric. e the eccentricity of the load, i.e., the distance between the line of action of and the axis of the column. Replace the given eccentric load by a centric force and a couple M A = Fe Now, no matter how small the load and the eccentricity e, the couple M A will cause some bending of the column.

8 As the eccentric load is in-creased, both the couple M A and the axial force increase, and both cause the column to bend further. Viewed in this way,the problem of buckling is not a question of dt determining ii how long the column can remain straight and stable under an increasing load, but rather how much the, column can be permitted to bend under the increasing load, if the allowable stress is not to be exceeded and if the deflection v max is not to become excessive. Secant Formula Drawing the free body diagram of a portion AQ of the column choosing the coordinate axes as shown, we find that t the bending moment at Q is M v M A v e dv M v e dx EI EI EI Set EI dv then v e dx

9 The solution for the above differential equation consist of general solution plus particular solution articular solution is given by v = e Therefore complete solution is given by v A sin x B cos x e The constants A, B can be obtained using Boundary conditions dv at x=0, v=0; at x=, v=0 and 0 dx Using these boundary conditions, we get B = e; A sin = e(1 cos ) (a) Using the trigonometric relations sin = sin( /)cos( /); and cos( )=1- sin( /) A = e tan Substituting A and B we get v etan sin xcosx1 The maximum deflection is at x=/ vmax etan sin( / ) cos( / ) 1 = esec 1 = e sec 1 EI From the above expression v max becomes infinite when EI

10 While the deflection does not actually become infinite, it nevertheless becomes unacceptably large, and should not be allowed to reach the critical value cr EI Which is the same critical load under centric load Solving for EI from the above equation and substituting in the equation for v max we get vmax e sec 1 cr Maximum bending moment will be at mid span of the column Mmax ( e v ) max e sec cr Maximumcompressive stress is given by Mc max A I ec ec = 1 sec 1 sec A r cr A r EI

11 We note that, since max does not vary linearly with the load, the principle of superposition does not apply to the determination of the stress due to the simultaneous application i of several loads; the resultant load must first be computed, and corresponding compressive stress must be computed. For the same reason, any given factor of safety should be applied to the load, and not to the stress. Making I = A r and solving for the ratio /A in front of the bracket, we write A max ec 1 e 1 sec r EA where the effective length is used to make the formula applicable to various end conditions. This formula is referred to as the secant formula; it defines the force per unit area, /A, which causes a specified maximum stress ( max in a column of given effective slenderness ratio, e /r, for a given value of the ratio ec/r,wheree is the eccentricity of the applied load. We note that, since /A appears in both members, it is necessary to solve a transcendental equation by trial and error to obtain the value of /A corresponding to a given column and loading condition.

12 The above equation was used to draw the curve shown in Figure below for a steel column, assuming the values of E and S y shown in the figure. These curves make it possible to determine the load per unit area /A, which causes the column to yield for given values of the ratios e lr and ec/r We note that, for small values of e /r, the secant is almost equal to 1 in the equation for /A), and IA may be assumed equal to A max ec 1 r a value which could be obtained by neglecting the effect of the lateral deflection of the column. On the other hand, for large values of e /r, the " curves corresponding to the various values e of the ratio ec/r get very close to Euler's curve, and thus that the effect of the eccentricity of the loading on the value of /A becomes negligible. The secant formula is chiefly useful for intermediate values of e lr. However, to use it effectively, we should know the value of the eccentricity e of the loading, and this quantity, unfortunately, is seldom known with any, degree of accuracy.

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