CHAPTER OBJECTIVES Discuss the behavior of columns.

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.

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

The lateral deflection that occurs is called buckling.
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.

Since  is small,  = (L/2) and tan  ≈ .
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.

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

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.

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.

Summing moments, M = P, Eqn becomes General solution is Since  = 0 at x = 0, then C2 = 0. Since  = 0 at x = L, then

Disregarding trivial soln for C1 = 0, we get Which is satisfied if or

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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

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:

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

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.

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.

Effective length

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

Effective length Here (KL/r) is the column’s effective-slenderness ratio.

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.

EXAMPLE 13.3 (SOLN) Buckling behavior is different about the x and y axes 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-y axis, (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.

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

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.

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 for both axes.

This is equivalent to applying a load P and moment M’ = Pe.
*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.

From free-body diagram, internal moment in column is
*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

*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

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

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

Since radius of gyration r2 = I/A,
*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 < Pcr unless 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.

A = x-sectional area of column
*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.

*13.4 THE SECANT FORMULA Design Once eccentricity ratio has been determined, column data can be substituted into Eqn 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.

*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.

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.

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).

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,

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,

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 mm = 55.3 MPa < Y = 250 MPa.

Consider a stress-strain diagram as shown.
*13.5 INELASTIC BUCKLING Application of Euler’s equation requires that the stress in column remain BELOW the material’s yield point when column buckles. So it only applies to long slender columns. In practice, most are intermediate columns, so we can study their behavior by modifying Euler’s equation to apply for inelastic buckling. Consider a stress-strain diagram as shown.

*13.5 INELASTIC BUCKLING Proportional limit is pl, and modulus of elasticity E is slope of line AB. A plot of Euler’s hyperbola is shown having a slenderness ratio as small as (KL/r)pl, since at this pt, cr = pl. When column about to buckle, change in strain that occurs is within a small range , so E for material can be taken as the tangent modulus Et.

Et is defined as the slope of stress-strain diagram at pt D.
*13.5 INELASTIC BUCKLING Et is defined as the slope of stress-strain diagram at pt D. At time of failure, column behaves as if it were made of a material having lower stiffness than when it behaves elastically, Et < E. In general, as slenderness ratio increases, critical stress for a column continues to rise and tangent modulus for material decreases.

This is the tangent modulus or Engesser’s eqn.
*13.5 INELASTIC BUCKLING Thus, we substitute the material’s tangent modulus Et for E into Euler’s equation, This is the tangent modulus or Engesser’s eqn.

EXAMPLE 13.7 A solid rod has a diameter of 30 mm and is 600 mm long. It is made of a material that can be modeled by the stress-strain diagram shown. If it is used as a pin-supported column, determine the critical load.

EXAMPLE 13.7 (SOLN) Radius of gyration and slenderness ratio are Applying Eqn yields

EXAMPLE 13.7 (SOLN) Assume that critical stress is elastic. From diagram, Thus, Eqn (1) becomes Since cr > pl = 150 MPa, inelastic buckling occurs. From second line segment of diagram, we have

EXAMPLE 13.7 (SOLN) Applying Eqn (1) yields Since value falls between 150 MPa and 270 MPa, it is indeed the critical stress. Critical load on the rod is therefore

*13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING
To account for behavior of different-length columns, design codes specify several formulae that will best fit the data within the short, intermediate, and long column range. Steel columns Structural steel columns are designed on the basis of formulae proposed by the Structural Stability Research Council (SSRC). Factors of safety are applied to the formulae and adopted as specs for building construction by the American Institute of Steel Construction (AISC).

Steel columns For long columns, the Euler formula is used. A factor of safety F.S. = 23/12 ≈ 1.92 is applied. Thus for design, Value of slenderness ratio obtained by

Steel columns For slenderness ratio lesser than (KL/r)c, the design eqn is

Steel columns

Aluminum columns Design equations are specified by the Aluminum Association, applicable for specific range of slenderness ratios. For a common alloy (2014-T6), we have

Aluminum columns

Timber columns Timber design formulae published by the National Forest Products Association (NFPA) or American Institute of Timber Construction (AITC).

Timber columns NFPA’s formulae for short, intermediate and long columns having a rectangular x-section of dimensions b and d (smallest dimension),

Procedure for analysis Column analysis When using any formula to analyze a column, or to find its allowable load, it is necessary to calculate the slenderness ratio in order to determine which column formula applies. Once the average allowable stress has been computed, the allowable load in the column is determined from P = allowA.

Procedure for analysis Column design If a formula is used to design a column, or to determine the column’s x-sectional area for a given loading and effective length, then a trial-and-check procedure generally must be followed if the column has a composite shape, such as a wide-flange section. One way is to assume the column’s x-sectional area, A’, and calculate the corresponding stress ‘ = P/A’.

Procedure for analysis Column design Also, with A’ use an appropriate design formula to determine the allowable stress allow. From this, calculate the required column area Areq’d = P/allow. If A’ > Areq’d, the design is safe. When making comparison, it is practical to require A’ to be close to but greater than Areq’d, usually within 2-3%. A redesign is necessary if A’ > Areq’d.

Procedure for analysis Column design Whenever a trial-and-check procedure is repeated, the choice of an area is determined by the previously calculated required area. In engineering practice, this method for design is usually shortened through the use of computer software or published tables and graphs.

EXAMPLE 13.8 An A-26 steel W250149 member is used as a pin-supported column. Using AISC column design formulae, determine the largest load that it can safely support. Est = 200(103) MPa, Y = 250 MPa.

EXAMPLE 13.8 (SOLN) From Appendix B, A = mm2; rx = 117 mm; ry= 67.4 mm. Since K = 1 for both x and y axes buckling, slenderness ratio is largest if ry is used. Thus From Eqn 13-22,

EXAMPLE 13.8 (SOLN) Here 0 < KL/r < (KL/r)c, so Eqn applies Allowable load P on column is

EXAMPLE 13.10 A bar having a length of 750 mm is used to support an axial compressive load of 60 kN. It is pin-supported at its ends and made from a 2014-T6 aluminum alloy. Determine the dimensions of its x-sectional area if its width is to be twice its thickness.

EXAMPLE (SOLN) Since KL = 750 mm is the same for x-x and y-y axes buckling, largest slenderness ratio is determined using smallest radius of gyration, using Imin = Iy: Since we do not know the slenderness ratio, we apply Eqn first,

EXAMPLE (SOLN) Checking slenderness ratio, Try Eqn 13-26, which is valid for KL/r ≥ 55;

EXAMPLE (SOLN) From Eqn (1), Note: It would be satisfactory to choose the x-section with dimensions 27 mm by 54 mm.

EXAMPLE 13.11 A board having x-sectional dimensions of 150 mm by 40 mm is used to support an axial load of 20 kN. If the board is assumed to be pin-supported at its top and base, determine its greatest allowable length L as specified by the NFPA.

EXAMPLE (SOLN) By inspection, board will buckle about the y axis. In the NFPA eqns, d = 40 mm. Assuming that Eqn applies, we have

EXAMPLE (SOLN) Here Since 26 < KL/d  50, the solution is valid.

*13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING
A column may be required to support a load acting at its edge or on an angle bracket attached to its side. The bending moment M = Pe, caused by eccentric loading, must be accounted for when column is designed. Use of available column formulae Stress distribution acting over x-sectional area of column shown is determined from both axial force P and bending moment M = Pe.

Use of available column formulae Maximum compressive stress is A typical stress profile is also shown here. If we assume entire x-section is subjected to uniform stress max, then we can compare it with allow, which is determined from formulae given in chapter 13.6. If max ≤ allow, then column can carry the specified load.

Use of available column formulae Otherwise, the column’s area A is increased and a new max and allow are calculated. This method of design is rather simple to apply and works well for columns that are short or intermediate length.

Interaction formula It is sometimes desirable to see how the bending and axial loads interact when designing an eccentrically loaded column. If allowable stress for axial load is (a)allow, then required area for the column needed to support the load P is

Interaction formula Similarly, if allowable bending stress is (b)allow, then since I = Ar2, required area of column needed to resist eccentric moment is determined from flexure formula,

Interaction formula Thus, total area A for the column needed to resist both axial force and bending moment requires that

Interaction formula a = axial stress caused by force P and determined from a = P/A, where A is the x-sectional area of the column. b = bending stress caused by an eccentric load or applied moment M; b is found from b = Mc/I, where I is the moment of inertia of x-sectional area computed about the bending or neutral axis.

Interaction formula (a)allow = allowable axial stress as defined by formulae given in chapter 13.6 or by design code specs. Use the largest slenderness ratio for the column, regardless of which axis it experiences bending. (b)allow = allowable bending stress as defined by code specifications.

Interaction formula Eqn is sometimes referred to as the interaction formula. This approach requires a trial-and-check procedure. Designer needs to choose an available column and check to see if the inequality is satisfied. If not, a larger section is picked and the process repeated. American Institute of Steel Construction specifies the use of Eqn only when the axial-stress ratio a/(a)allow ≤ 0.15.

EXAMPLE 13.12 Column is made of 2014-T6 aluminum alloy and is used to support an eccentric load P. Determine the magnitude of P that can be supported if column is fixed at its base and free at its top. Use Eqn

EXAMPLE (SOLN) K = 2. Largest slenderness ratio for column is By inspection, Eqn must be used (277.1 > 55).

EXAMPLE (SOLN) Actual maximum compressive stress in the column is determined from the combination of axial load and bending. Assuming that this stress is uniform over the x-section, instead of just at the outer boundary,

EXAMPLE 13.13 The A-36 steel W15030 column is pin-connected at its ends and subjected to eccentric load P. Determine the maximum allowable value of P using the interaction method if allowable bending stress is (b)allow = 160 MPa, E = 200 GPa, and Y = 250 MPa.

EXAMPLE (SOLN) K = 1. The geometric properties for the W15030 are taken from the table in Appendix B. We consider ry as it lead to largest value of the slenderness ratio. Ix is needed since bending occurs about the x axis (c = 157 mm/2 = 78.5 mm). To determine the allowable bending compressive stress, we have

EXAMPLE (SOLN) Then KL/r < (KL/r)c and so Eqn must be used. Assuming that this stress is uniform over the x-section, instead of just at the outer boundary,

EXAMPLE (SOLN) Applying the interaction formula Eqn yields Checking application of interaction method for steel section, we require

EXAMPLE 13.14 Timber column is made from two boards nailed together so the x-section has the dimensions shown. If column is fixed at its base and free at its top, use Eqn to determine the eccentric load P that can be supported.

EXAMPLE (SOLN) K = 2. Here, we calculate KL/d to determine which eqn to use. Since allow is determined using the largest slenderness ratio, we choose d = 60 mm. This is done to make the ratio as large as possible, and thus yield the lowest possible allowable axial stress. This is done even though bending due to P is about the x axis.

EXAMPLE (SOLN) Allowable axial stress is determined using Eqn since 26 < KL/d < 50. Thus Applying Eqn with allow = max, we have

CHAPTER REVIEW Buckling is the sudden instability that occurs in columns or members that support an axial load. The maximum axial load that a member can support just before buckling occurs is called the critical load Pcr. The critical load for an ideal column is determined from the Euler eqn, Pcr = 2EI/(KL)2, where K = 1 for pin supports, K = 0.5 for fixed supports, K = 0.7 for a pin and a fixed support, and K = 2 for a fixed support and a free end.

CHAPTER REVIEW If axial loading is applied eccentrically to the column, then the secant formula must be used to determine the maximum stress in the column. When the axial load tends to cause yielding of the column, then the tangent modulus should be used with Euler’s eqn to determine the buckling load. This is referred to as Engesser’s eqn. Empirical formulae based upon experimental data have been developed for use in the design of steel, aluminum, and timber columns.

CHAPTER OBJECTIVES Discuss the behavior of columns.

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