1. Mukavemet II Strength of Materials II
Stabilite (Burkulma)
Stability (Buckling)
Teaching Slides
Chapter 10:

2. Stability (Buckling)

3. Chapter Outline Stability of Structures
Euler’s Formula for Pin-Ended Columns
Extension of Euler’s Formula to Columns with Other End Conditions
Eccentric Loading; the Secant Formula
Design of Columns under a Centric Load
Design of Columns under an Eccentric Load
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4. FIGURE Buckling demonstrations.
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6. In Fig. a, the ball is said to be in stable equilibrium because, if it is slightly displaced to one side and then released, it will move back toward the equilibrium position at the bottom of the “valley.’’
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7. If the ball is on a perfectly flat, level surface, as in Fig
If the ball is on a perfectly flat, level surface, as in Fig. b, it is said to be in a neutral-equilibrium configuration. If slightly displaced to either side, it has no tendency to move either away from or toward the original position, since it is in equilibrium in the displaced position as well as the original position.
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9. THE IDEAL PIN-ENDED COLUMN; EULER BUCKLING LOAD
To investigate the stability of real columns, we begin by considering the ideal pin-ended column, as illustrated in Fig.10:

10. We make the following simplifying assumptions:
The column is initially perfectly straight, and it is made of linearly elastic material.
The column is free to rotate, at its ends, about frictionless pins; that is, it is restrained like a simply supported beam. Each pin passes through the centroid of the cross section.
The column is symmetric about the xy plane, and any lateral deflection of the column takes place in the xy plane.
The column is loaded by an axial compressive force P applied by the pins.

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12. However, if a load P=Pcr is applied to the column, the straight configuration becomes a neutral-equilibrium configuration, and neighboring configurations, like the buckled shape in Fig. 10.6b, also satisfy equilibrium requirements. Therefore, to determine the value of the critical load, Pcr, and the shape of the buckled column, we will determine the value of the load P such that the (slightly) bent shape of the column in Fig. 10.6b is an equilibrium configuration.
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14. Equilibrium of the Buckled Column
First, using the FBD of Fig. 10.6c, we get
Ax=P (from ΣFx=0 ).
Ay=0 [from ΣMB=0], and By=0.
Therefore, on the FBD in Fig. 10.6d, we show only a vertical force P acting on the pin at A, and we show no horizontal force V(x) at section x.
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15. The sign convention adopted for the moment M(x) in Fig. 10
The sign convention adopted for the moment M(x) in Fig. 10.6d is a positive bending moment. From Fig. 10.6d we get
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16. Differential Equation of Equilibrium, and End Conditions:
Substituting M(x) into the moment-curvature equation, we obtain elastic curve equation as below: or or

17. The boundary conditions for the pin-ended member are
This is the differential equation that governs the deflected shape of a pin-ended column.
It is a homogenous, linear, second-order, ordinary differential equation.
The boundary conditions for the pin-ended member are
and
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18. Solution of the Differential Equation:
The term v(x) means that we cannot simply integrate twice to get the solution. When EI is constant, there is a simple solution to this equation.
It is an ordinary differential equation with constant coefficients. For the uniform column, therefore, it becomes
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19. The general solution to this homogeneous equation is
We seek a value of and constants of integration C1 and C2 such that the two boundary conditions of this homogeneous equation are satisfied.
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20. We just have the original straight configuration.
Thus,
Obviously, if the constants C1 and C2 equal to zero, the deflection v(x) is zero everywhere.
We just have the original straight configuration.
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21. For an alternative equilibrium configuration, since C1 can not be zero, sin(λ 1L) must be zero. So, the characteristic equation is satisfied with C1≠0.
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22. Combining the equations above gives the following formula for the possible buckling loads:
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23. The deflection curve that corresponds to each load Pn is obtained by combining the following Equations.
and
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24. we get
The function that represents the shape of the deflected column is called a mode shape, or buckling mode.
The constant C, which determines the direction (sign) and amplitude of the deflection, is arbitrary, but it must be small.
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25. The existence of neighboring equilibrium configurations is analogous to the fact that the ball in Fig. 10.3b can be placed at neighboring locations on the flat, horizontal surface and still be in equilibrium.
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26. The value of P at which buckling will actually occur is obviously the smallest value given by Eq (n=1).Thus, the critical load is

27. The critical load for an ideal column is known as the Euler buckling load, who was the first to establish a theory of buckling of columns.
Leonhard Euler (1707–1783) is the famous Swiss mathematician

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29. Let us examine some important implications of the Euler buckling-load formula. We can express this in terms of critical (buckling) stress.
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31. FIGURE: Critical stress versus slenderness ratio for structural steel columns.
For slenderness ratios less than 100, the critical stress is not meaningful.
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33. EXAMPLE
A 2-m-long pin-ended column of square cross section is to be made of wood. Assuming E=13 GPa, σall=12 MPa, and using a factor of safety of 2.5 in computing Euler’s critical load for buckling, determine the size of the cross section if the column is to safely support
a) a 100-kN load,
b) a 200-kN load.
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34. a) For the 100-kN Load. Using the given factor of safety, we make
in Euler’s formula and solve for I. We have
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35. We check the value of the normal stress in the column:
Recalling that, for a square of side a, we have , we write
We check the value of the normal stress in the column:
Since σ is smaller than the allowable stress, a 100x100-mm cross section is acceptable.

36. b) For the 200-kN Load.
Solving again buckling equation for I, but making now Pcr=2.5(200)=500 kN, we have
The value of the normal stress is
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37. Since this value is larger than the allowable stress, the dimension obtained is not acceptable, and we must select the cross section on the basis of its resistance to compression. We write
A 130x130-mm cross section is acceptable

38. EXTENSION OF EULER’S FORMULA TO COLUMNS WITH OTHER END CONDITIONS
Euler’s formula was derived in the preceding section for
a column that was pin-connected at both ends.
Now the critical load Pcr will be determined for columns with different end conditions.
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39. Column with One Free End
In the case of a column with one free end A supporting a load P and one fixed end B (Fig.-a), it is observed that the column will behave as the upper half of a pin-connected column (Fig.-b).
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40. The critical load for the column of Fig
The critical load for the column of Fig.-a is thus the same as for the pin-ended column of Fig.-b and can be obtained from Euler’s formula by using a column length equal to twice the actual length L of the given column.
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41. We say that the effective length Le of the column is equal to 2L and substitute Le=2L in Euler’s formula:
Le=2L
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42. The critical stress is found in a similar way from the formula
The quantity Le /r is referred to as the effective slenderness ratio of the column and, in the case considered here, is equal to 2L/r.
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43. Column with Two Fixed Ends
Consider next a column with two fixed ends A and B supporting a load P (Fig ).
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44. The symmetry of the supports and of the loading about a horizontal axis through the midpoint C requires that the shear at C and the horizontal components of the reactions at A and B be zero (Fig ).
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45. 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 (Fig ).

46. Portion 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 midpoint E of the lower half of the column must also be zero (Fig a).

47. Since the bending moment at the ends of a pin-ended column is zero, it follows that the portion DE of the column of Fig a must behave as a pin ended column (Fig b). We thus conclude that the effective length of a column with two fixed ends is Le =L/2.
Le=L/2

48. EXAMPLE
What is the maximum compressive load that can be applied to an aluminum-alloy compression member of length L=4 m if the member is loaded in a manner that permits free rotation at its ends and if a factor of safety of 1.5 against buckling failure is to be applied?
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52. Physically, the effective length of a column is the distance between points of zero moment when the column is deflected in its fundamental elastic buckling mode. Figure illustrates the effective lengths of columns with several types of end conditions.
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57. STABILITY OF STRUCTURES
Let us deal with design of a column AB of length L to support a given load P (Fig. 10.1). The column exposed to a centric axial load P will be pin-connected at both ends.

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59. However, it may happen that, as the load is applied, the column will buckle; instead of remaining straight, it will suddenly become sharply curved
(Fig. 10.2). Photo 10.1 shows a column that has been loaded so that it is no longer straight; the column has buckled. Clearly, a column that buckles under the load it is to support is not properly designed.
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60. EULER’S FORMULA FOR PIN-ENDED COLUMNS
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61. Our approach will be to determine the conditions under which the configuration of Fig is possible.
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