CHAPTER OBJECTIVES

1. CHAPTER OBJECTIVES • Discuss the behavior of columns. • Discuss the buckling of columns. • Determine the axial load needed to buckle an ideal column. • Analyze the buckling with bending of a column. • Discuss inelastic buckling of a column. • Discuss methods used to design concentric and eccentric columns.

2. CHAPTER OUTLINE • Critical Load • Ideal Column with Pin Supports • Columns Having Various Types of Supports • *The Secant Formula • *Inelastic Buckling • *Design of Columns for Concentric Loading • *Design of Columns for Eccentric Loading

3. 13.1 CRITICAL LOAD • Long slender members subjected to axial compressive force are called columns. • The lateral deflection that occurs is called buckling. • The maximum axial load a column can support when it is on the verge of buckling is called the critical load, Pcr.

4. 13.1 CRITICAL LOAD • Spring develops restoring force F = k, while applied load P develops two horizontal components, Px = P tan , which tends to push the pin further out of equilibrium. • Since  is small,  = (L/2) and tan  ≈ . • Thus, restoring spring force becomes F = kL/2, and disturbing force is 2Px = 2P.

5. 13.1 CRITICAL LOAD • For kL/2 > 2P, • For kL/2 < 2P, • For kL/2 = 2P,

6. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • An ideal column is perfectly straight before loading, made of homogeneous material, and upon which the load is applied through the centroid of the x-section. • We also assume that the material behaves in a linear-elastic manner and the column buckles or bends in a single plane.

7. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • In order to determine the critical load and buckled shape of column, we apply Eqn 12-10, • Recall that this eqn assume the slope of the elastic curve is small and deflections occur only in bending. We assume that the material behaves in a linear-elastic manner and the column buckles or bends in a single plane.

8. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • Summing moments, M = P, Eqn 13-1 becomes • General solution is • Since  = 0 at x = 0, then C2 = 0.Since  = 0 at x = L, then

9. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • Disregarding trivial soln for C1 = 0, we get • Which is satisfied if • or

10. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • Smallest value of P is obtained for n = 1, so critical load for column is • This load is also referred to as the Euler load. The corresponding buckled shape is defined by • C1 represents maximum deflection, max, which occurs at midpoint of the column.

11. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • A column will buckle about the principal axis of the x-section having the least moment of inertia (weakest axis). • For example, the meter stick shown will buckle about the a-a axis and not the b-b axis. • Thus, circular tubes made excellent columns, and square tube or those shapes having Ix≈ Iy are selected for columns.

12. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • Buckling eqn for a pin-supported long slender column, Pcr = critical or maximum axial load on column just before it begins to buckle. This load must not cause the stress in column to exceed proportional limit. E = modulus of elasticity of material I = Least modulus of inertia for column’s x-sectional area. L = unsupported length of pinned-end columns.

13. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • Expressing I = Ar2 where A is x-sectional area of column and r is the radius of gyration of x-sectional area. cr = critical stress, an average stress in column just before the column buckles. This stress is an elastic stress and therefore cr Y E = modulus of elasticity of material L = unsupported length of pinned-end columns. r = smallest radius of gyration of column, determined from r = √(I/A), where I is least moment of inertia of column’s x-sectional area A.

14. 13.2 IDEAL COLUMN WITH PIN SUPPORTS • The geometric ratio L/r in Eqn 13-6 is known as the slenderness ratio. • It is a measure of the column’s flexibility and will be used to classify columns as long, intermediate or short.

15. 13.2 IDEAL COLUMN WITH PIN SUPPORTS IMPORTANT • Columns are long slender members that are subjected to axial loads. • Critical load is the maximum axial load that a column can support when it is on the verge of buckling. • This loading represents a case of neutral equilibrium.

16. 13.2 IDEAL COLUMN WITH PIN SUPPORTS IMPORTANT • An ideal column is initially perfectly straight, made of homogeneous material, and the load is applied through the centroid of the x-section. • A pin-connected column will buckle about the principal axis of the x-section having the least moment of intertia. • The slenderness ratio L/r, where r is the smallest radius of gyration of x-section. Buckling will occur about the axis where this ratio gives the greatest value.

17. EXAMPLE 13.1 A 7.2-m long A-36 steel tube having the x-section shown is to be used a pin-ended column. Determine the maximum allowable axial load the column can support so that it does not buckle.

18. EXAMPLE 13.1 (SOLN) Use Eqn 13-5 to obtain critical load with Est = 200 GPa.

19. EXAMPLE 13.1 (SOLN) This force creates an average compressive stress in the column of Since cr < Y = 250 MPa, application of Euler’s eqn is appropriate.

20. EXAMPLE 13.2 The A-36 steel W20046 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields.

21. EXAMPLE 13.2 (SOLN) From table in Appendix B, column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5106 mm4,and Iy = 15.3106 mm4. By inspection, buckling will occur about the y-y axis. Applying Eqn 13-5, we have

22. EXAMPLE 13.2 (SOLN) When fully loaded, average compressive stress in column is Since this stress exceeds yield stress (250 N/mm2), the load P is determined from simple compression:

23. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS • From free-body diagram, M = P(  ). • Differential eqn for the deflection curve is • Solving by using boundary conditions and integration, we get

24. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS • Thus, smallest critical load occurs when n = 1, so that • By comparing with Eqn 13-5, a column fixed-supported at its base will carry only one-fourth the critical load applied to a pin-supported column.

25. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • If a column is not supported by pinned-ends, then Euler’s formula can also be used to determine the critical load. • “L” must then represent the distance between the zero-moment points. • This distance is called the columns’ effective length, Le.

26. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length

27. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • Many design codes provide column formulae that use a dimensionless coefficient K, known as thee effective-length factor. • Thus, Euler’s formula can be expressed as

28. 13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • Here (KL/r) is the column’s effective-slenderness ratio.

29. EXAMPLE 13.3 A W15024 steel column is 8 m long and is fixed at its ends as shown. Its load-carrying capacity is increased by bracing it about the y-y axis using struts that are assumed to be pin-connected to its mid-height. Determine the load it can support sp that the column does not buckle nor material exceed the yield stress. Take Est = 200 GPa and Y = 410 MPa.

30. EXAMPLE 13.3 (SOLN) Buckling behavior is different about the x and yaxes due to bracing. Buckled shape for each case is shown. The effective length for buckling about the x-x axis is (KL)x = 0.5(8 m) = 4 m. For buckling about the y-yaxis, (KL)y = 0.7(8 m/2) = 2.8 m. We get Ix = 13.4106 mm4 and Iy = 1.83106 mm4 from Appendix B.

31. EXAMPLE 13.3 (SOLN) Applying Eqn 13-11, By comparison, buckling will occur about the y-y axis.

32. EXAMPLE 13.3 (SOLN) Area of x-section is 3060 mm2, so average compressive stress in column will be Since cr < Y = 410 MPa, buckling will occur before the material yields.

33. EXAMPLE 13.3 (SOLN) NOTE: From Eqn 13-11, we see that buckling always occur about the column axis having the largest slenderness ratio. Thus using data for the radius of gyration from table in Appendix B, Hence, y-y axis buckling will occur, which is the same conclusion reached by comparing Eqns 13-11 for both axes.

34. *13.4 THE SECANT FORMULA • The actual criterion for load application on a column is limited to either a specified deflection of the column or by not allowing the maximum stress in the column exceed an allowable stress. • We apply a load P to column at a short eccentric distance e from centroid of x-section. • This is equivalent to applying a load P and moment M’ = Pe.

35. *13.4 THE SECANT FORMULA • From free-body diagram, internal moment in column is • Thus, the general solution for the differential eqn of the deflection curve is • Applying boundary conditions to determine the constants, deflection curve is written as

36. *13.4 THE SECANT FORMULA Maximum deflection • Due to symmetry of loading, both maximum deflection and maximum stress occur at column’s midpoint. Therefore, when x = L/2, = max, so

37. *13.4 THE SECANT FORMULA Maximum deflection • Therefore, to find Pcr, we require

38. *13.4 THE SECANT FORMULA The secant formula • Maximum stress in column occur when maximum moment occurs at the column’s midpoint. Using Eqns 13-13 and 13-16, • Maximum stress is compressive and

39. *13.4 THE SECANT FORMULA The secant formula • Since radius of gyration r2= I/A, • max = maximum elastic stress in column, at inner concave side of midpoint (compressive). • P = vertical load applied to the column. P < Pcrunless e = 0, then P = Pcr(Eqn 13-5) • e = eccentricity of load P, measured from the neutral axis of column’s x-sectional area to line of action of P.

40. *13.4 THE SECANT FORMULA The secant formula • c = distance from neutral axis to outer fiber of column where maximum compressive stress max occurs. • A = x-sectional area of column • L = unsupported length of column in plane of bending. For non pin-supported columns, Le should be used. • E = modulus of elasticity of material. • r = radius of gyration, r = √(I/A), where I is computed about the neutral or bending axis.

41. *13.4 THE SECANT FORMULA Design • Once eccentricity ratio has been determined, column data can be substituted into Eqn 13-19. • For max = Y, corresponding load PY is determined from a trial-and-error procedure, since eqn is transcendental and cannot be solved explicitly for PY. • Note that PY will always be smaller than the critical load Pcr, since Euler’s formula assumes unrealistically that column is axially loaded without eccentricity.

42. *13.4 THE SECANT FORMULA IMPORTANT • Due to imperfections in manufacturing or application of the load, a column will never suddenly buckle, instead it begins to bend. • The load applied to a column is related to its deflections in a nonlinear manner, so the principle of superposition does not apply. • As the slenderness ratio increases, eccentrically loaded columns tend to fail at or near the Euler buckling load.

43. EXAMPLE 13.6 The W20059 A-36 steel column shown is fixed at its base and braced at the top so that it is fixed from displacement, yet free to rotate about the y-y axis. Also, it can sway to the side in the y-z plane. Determine the maximum eccentric load the column can support before it either begins to buckle or the steel yields.

44. EXAMPLE 13.6 (SOLN) From support conditions, about the y-y axis, the column behaves as if it was pinned at the top, fixed at the base and subjected to an axial load P. About the x-x axis, the column is free at the top and fixed at the base, and subjected to both axial load P and moment M = P(200 mm).

45. EXAMPLE 13.6 (SOLN) y-y axis buckling: Effective length factor is Ky = 0.7, so (KL)y = 0.7(4 m) = 2.8 m = 2800 mm. Using table in Appendix B to determine Iy for the section and applying Eqn 13-11,

46. EXAMPLE 13.6 (SOLN) x-x axis yielding: Kx = 2, so (KL)x = 2(4 m) = 8 m = 8000 mm. From table in Appendix B, A = 7580 mm2, c = 210 mm/2 = 105 mm, and rx = 89.9 mm, applying secant formula,

47. EXAMPLE 13.6 (SOLN) x-x axis yielding: Solving for Px by trial and error, noting that argument for secant is in radians, we get Since this value is less than (Pcr)y = 5136 kN, failure will occur about the x-x axis. Also,  = 419.4103 N / 7580 mm2 = 55.3 MPa < Y = 250 MPa.