CHAPTER OBJECTIVES

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13. Buckling 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

1. Critical Load

2. Ideal Column with Pin Supports

3. Columns Having Various Types of Supports

4. *The Secant Formula

5. *Inelastic Buckling

6. *Design of Columns for Concentric Loading

7. *Design of Columns for Eccentric Loading



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.



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 = kL/2, and disturbing force is 2Px = 2P.



13.1 CRITICAL LOAD

• For kL/2 > 2P,

• For kL/2 < 2P,

• For kL/2 = 2P,

mequilibriustable4

kLP

mequilibriuunstable4

kLP

mequilibriuneutral4

kLPcr



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.

1-132

2M

dx

dEI




• 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

2-1302

2

EIP

dx

d

3-13cossin 21

x

EIP

CxEIP

C

0sin1

LEIP

C




• Disregarding trivial soln for C1 = 0, we get

• Which is satisfied if

• or

nLEIP

0sin

LEIP

,...3,2,12

22 n

L

EInP




• 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.

2

2

L

EIPcr

Lx

C sin1




• 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.

5-132

2

L

EIPcr




• 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.

6-13

2

2

rL

Ecr




• 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.

kN2.228

m2.7

mm1000/m17041

kN/m10200

2

44262

2

2

L

EIPcr



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.

MPa100N/mm2.100

mm7075

N/kN1000kN2.228

2

222

APcr

cr



EXAMPLE 13.2

The A-36 steel W20046 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 = 45.5106 mm4,and Iy = 15.3106 mm4.

By inspection, buckling will occur about the y-y axis.

Applying Eqn 13-5, we have

kN6.1887

m4

mm1000/m1mm103.15kN/m102002

444262

2

2

L

EIPcr



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:

2

2

N/mm5.320

mm5890

N/kN1000kN6.1887

A

Pcrcr

kN5.1472

mm5890N/mm250

22

P

P



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

7-132

2

EIP

EIP

dx

d

8-13cos1

x

EIP




• 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.

9-134 2

2

L

EIPcr




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

10-13KLLe

12-13

11-13

2

2

2

2

rKL

E

KL

EIP

cr

cr




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

ratio.



EXAMPLE 13.3

A W15024 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 = 13.4106 mm4 and Iy = 1.83106 mm4 from Appendix B.



EXAMPLE 13.3 (SOLN)

Applying Eqn 13-11,

By comparison, buckling will occur about the y-y axis.

kN8.460

m8.2

m1083.1kN/m10200

kN2.1653

m4

m104.13kN/m10200

2

46262

2

2

2

46262

2

2

ycr

y

yycr

xcr

x

xxcr

P

KL

EIP

P

KL

EIP



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.

22

3N/mm6.150

m3060

N108.460 A

Pcrcr



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.

3.114

mm5.24mm/m1000m8.2

4.60mm2.66mm/m1000m4

y

x

rKL

rKL



*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

• 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

13-13 ePM

14-13cossin 21 exEIP

CxEIP

C

15-131cossin2

tan

x

EIP

sEIPL

EIP

e




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

16-1312

secmax

L

EIP

e




Maximum deflection

• Therefore, to find Pcr, we require

17-13

22

2sec

2

2

L

EIP

LEIP

LEIP

cr

cr

cr




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

18-132

sec

max

LEIP

PeM

ePM

2sec; maxmax

LEIP

IPec

AP

IMc

AP




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.

19-132

sec12max

EAP

rL

r

ecAP




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.




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.




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 W20059 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,

kN5136N5136247

mm2800

mm104.20N/mm102002

46232

2

2

y

yycr

KL

EIP



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,

xx

xx

x

x

x

x

xY

PP

PP

EAP

r

KL

r

ecAP

36

2

2

10143.1sec598.2110895.1

75801032009.8928000

sec9.89

1052001

7580250

2sec1



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, = 419.4103 N / 7580 mm2 = 55.3 MPa < Y = 250 MPa.

kN4.419N419368 xP

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