CE2155 - Stability of Compression Members

Stability of Compression Members

CE2155 Structural Mechanics and Materials

byAssoc Professor T. H. Wee

Department of Civil EngineeringEmail: [email protected]

CE2155 – Structural Mechanics and Materials

• Long slender members subjected to an axial compressive force are called columns and the lateral deflection that occurs is called buckling.

• The maximum axial load a column can support when on the verge of buckling is called the critical load, Pcr.

Critical Load


Consider a two-bar mechanism consisting of rigid bars that are pin connected at their ends, and a spring of stiffness, k.

Critical Load

• When a small vertical force P is applied at the top, the equilibrium can be upset by displacing the pin by a small amount Δ.

• When the bars are displaced, the spring will produce a restoring force

F = k Δ


The applied load P develops two horizontal components Px = P tan θ which tend to push the pin A further out of equilibrium.

• For equilibrium, 2 P tan θ < k Δ

• Since θ is small, Δ ≈ θ (L / 2), and tan θ ≈ θ, hence

2 P θ < k θ (L / 2)which gives

4kLP <

This is a condition for stable equilibrium since the force developed by the spring would be adequate to restore the bars back to their vertical position.

Critical Load


• On the other hand, if 2 P θ > k θ (L / 2), or

then the mechanism would be in unstable equilibrium.

• In other words, if this load P is applied, and a slight displacement occurs at A, the mechanism will tend to move out of equilibrium and not be restored to its original position.

• The intermediate value of P, defined by requiring 2 P θ = k θ (L / 2) is the critical load. Here

• This loading represents a case of the mechanism being inneutral equilibrium.

4kLP >

4kLPcr =

Critical Load


• Since Pcr is independent of the displacement θ of the bar, any slight disturbance given to the mechanism will not cause it to move further out of equilibrium nor will it restore to its original position. Instead the bar will remain in the deflectedposition

• The transition point where the load is equal to the critical load P = Pcr is called the bifurcation point. Physically Pcr represents the load for which the mechanism is on the verge of buckling.

• Like the two-bar mechanism, the critical buckling loads on columns supported in various ways can also be obtained.

Critical Load


• In the derivation of the critical load, the equilibrium of the column was upset by displacing the pin by a small amount Δ.

• In reality, the eccentricity in loading, homogeneity in the material and regularity in the cross-section of a column is never perfect.

• This imperfection inherently results in the column being upset by a lateral displacement, v.

Critical Load


Consider a perfectly straight column made of homogeneous material with applied load P through the centroid of the cross section.

Column with Pin Supports

When the critical load Pcr is reached, the column is on the verge of becoming unstable.

Whether or not the column will remain stable or become unstable when subjected to an axial load will depend on its ability to restore itself, which is based on its resistance to bending (similar to the restoring action of the spring).


• When the internal moment M deforms the element of a beam, the angle between the cross sections becomes dθ, as shown in the figure.

• The arc dx represents a portion of the elastic curve that intersects the neutral axis for each cross section.

• The radius of curvature for this arc is defined as the distance ρ, which is measured from the centre of curvature O’ to dx.

• Any arc on this element other than dx is subjected to a normal strain.

Relationship between Internal Moment and Deflected Shape


The strain in arc ds, located at a position y from the neutral axis, is

However

and so

or

( )ds

sdds ′−=ε

( ) θ−ρ=′θρ==dysd

ddxds and

( )[ ]θρ

θ−ρ−θρ=ε

ddyd

yε

=ρ1



If the material is homogeneous and behaves in a linear-elastic manner, then Hook’s law applies

Also, since the flexural formula applies

Combining the above equations and thereafter substituting into

we have yε

=ρ1

Eσ

=ε

IMy

=σ

EIM

=ρ1



The elastic curve for a beam can be expressed mathematically as v = f(x). To obtain this equation, we must first represent the curvature (1/ρ) in terms of v and x. In most calculus books it is shown that this relationship is

since

23

2

2

2

1

1

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+

=ρ

dxdv

dxvd

EIM

=ρ1



Therefore we can write

Consequently, the slope of the elastic curve, which is determined from dv/dx will be very small, and its square will be negligible compared with unity. Therefore the above equation simplifies to

23

2

2

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+

=

dxdv

dxvd

EIM

or 2

2

dxvd

EIM

= Mdx

vdEI =2

2



To determine the critical load of the column, the equation whichrelates the internal moment to its deflected shape is considered:

When the column is in its deflected position, the internal bending moment can be determined using the method of sections as

Equating moments, therefore

Mdx

vdEI =2

2

PvM −=

Pvdx

vdEI −=2

2



and the general solution is

The two constants of integration are determined from the boundary conditions at the ends of the column.

02

2

=⎟⎠⎞

⎜⎝⎛+ v

EIP

dxvd

Pvdx

vdEI −=2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛= x

EIPcosCx

EIPsinCv 21



Since v = 0 at x = 0, then C2 = 0. And since v = 0 at x = L, then

C1 = 0 is trivial since it requires the column to always remain straight even when column becomes unstable. Therefore,

01 =⎟⎟⎠

⎞⎜⎜⎝

⎛L

EIPsinC

0=⎟⎟⎠

⎞⎜⎜⎝

⎛L

EIPsin



is satisfied if

or

0=⎟⎟⎠

⎞⎜⎜⎝

⎛L

EIPsin

πnLEIP

=

,.....,,nL

EInP 3212

22

==π



The smallest value of P is obtained when n = 1, so the critical load for the column is therefore:

Note that n represents the number of waves in the deflected shape of the column. For example, if n = 2, then two waves will appear in the buckled shape and the column will support a critical load that is 4Pcrjust prior to buckling.

,.....,,nL

EInP 3212

22

==π

2

2

LEIPcr

π=(sometimes

referredto as Euler formula and Euler load.)



• It is important to realise that column will buckle about the weakest axis (least moment of inertia).

• In addition, the critical stress, which is the average stress in the column just before the column buckles should be less than the yield stress, otherwise yielding precedes. The critical stress σcr is given by

where r is the smallest radius of gyration of the column determined from

and I is the least moment of inertia of the column’s cross-sectional area A.

( )22

r/LE

APcr

crπσ ==

A/Ir =



The A-36 steel W200 X 46 member 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. σY = 250 MPa.Solution:From Appendix B,

2N/mm 5.3205890

10006.1887=

×==

APcr

crσ

46462 mm 103.15,mm 105.45,mm 5890 ×=×== yx IIA

( )( )( ) kN 6.18874

1000/1103.15102002

4462

2

2

=××

==ππ

LEIPcr

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

When fully loaded, the average compressive stress in the column is

Since this stress exceeds the yield stress,

(Ans) kN 5.14725890

250 =⇒= PP

Largest axial load = 1472.5 kN and failure will take place by yielding.

Example Problem:


• Euler formula was developed for the case of a column having ends that are pinned.

• L in the equation represents the unsupported distance between the points of zero moment.

• If the column is supported in other ways, then Euler’s formula can be used to determine the critical load provided “L”represents the distance between zero-moment points. The distance is called the column’s effective length, Le and can be obtained from

where K is a dimensionless coefficient called effective-length factor. Therefore

Columns with Various Types of Supports

LKLe =

( )22

KLEIPcr

π=

( )22

r/KLE

APcr

crπσ ==


Effective Length of Columns with Various Types of Supports


The aluminium column is fixed at its bottom and is braced at itstop by cables so as to prevent movement at the top along the x axis. If it is assumed to be fixed at its base, determine the largest allowable load P that can be applied.

Use a factor of safety for buckling of FS = 3.0.

Take Eal = 70GPa, σY = 215MPa, A = 7.5(10-3)m2, Ix = 61.3(10-6)m4, Iy = 23.2(10-6)m4.

Example Problem:


Solution

( )( )

( )( )

MN 31.1 , kN 424 2

2

2

2

====y

ycrx

xcr KLEIP

KLEIP ππ

( ) ( ) m 1052 ==xKL

kN 14103

424===

.FSPP cr

allow

For x–x axis buckling, K = 2,

For y–y axis buckling, K = 0.7, ( ) ( ) m 5.357.0 ==yKL

The critical loads for each case are

The allowable load and critical stress are

( ) MPa 215 MPa 5561057

424 3 <=== − ..A

Pcrcrσ

Allowable load = 141 kN and failure will occur about the x-x axis by buckling.

Example Problem:


Example Problem:


Example Problem (Solution)




( )22

r/KLE

APcr

crπσ ==



Column with Eccentric Loading

• Unlike Euler formula, Secant formula takes into consideration theeccentric loading in a column.

• Consider a load P is applied to a column at a short eccentric distance, e from the centroid of the cross-section. This loading on the column is statically equivalent to the axial load P and bending moment M’ = Pe.


From the free body diagram of the arbitrary section, the internal moment in the column is

The differential equation for the deflection curve is therefore

or

and has a general solution as

( )vePM +−=

( )vePdx

vdEI +−=2

2

eEIPv

EIP

dxvd

−=+2

2

exEIPcosCx

EIPsinCv −+= 21



Applying boundary condition;x = 0, v = 0, so C2 = ex = L, v = 0, which gives

Hence, the deflection curve can be written as

exEIPcosCx

EIPsinCv −+= 21

( )[ ]( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

−=

21

1L

EIPtane

LEI/PsinLEI/PcoseC

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+⎟

⎟⎠

⎞⎜⎜⎝

⎛= 1

2x

EIPcosx

EIPsinL

EIPtanev



Maximum DeflectionDue to symmetry of loading, both the maximum deflection and maximum stress occur as the column’s midpoint. Therefore, when x = L / 2, v = vmax, so substituting x= L / 2 into

the vmax can be found as

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟⎠

⎞⎜⎜⎝

⎛= 1

2L

EIPsecevmax

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+⎟

⎟⎠

⎞⎜⎜⎝

⎛= 1

2x

EIPcosx

EIPsinL

EIPtanev



The Secant Formula• The maximum stress in the column

can be determined by realizing that it is caused by both the axial load and the moment.

• Maximum moment occurs at the column’s midpoint and it has a magnitude of

Substituting vmax;

( )maxvePM +=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2L

EIPsecPeM



The Secant FormulaThe maximum stress in the column is compressive and it has a value of

Since the radius of gyration is defined as r2 = I / A, the secant formula is derived as

IMy

AP

max +=σ

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

2L

EIPsec

IPey

AP

maxσ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

EAP

rLsec

rey

AP

max 21 2σ



The Secant Formula

σmax = maximum elastic stress in the columnP = vertical load applied to the columne = eccentricity of the load Py = distance from the neutral axisA = cross sectional area of the columnL = unsupported length of the column in the plane of

bending. For supports other than pins, the effective length (Le) should be used.

E = modulus of elasticity for the materialr = radius of gyration

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

EAP

rLsec

rey

AP

max 21 2σ



The W200 X 59 A-36 steel column 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.

Given in steel table for W200 X 59 A-36 column:

Ix = 61.2 x 106 mm4

Iy = 20.4 x 106 mm4A = 7580 mm2

rx = 89.9 mmry = 51.9 mmdepth, d = 210E = 200 x 103 MPaσY = 250 MPa

Example Problem:


( )( )

( )( )( )

kN 5136400070

10420102002

632

2

2

=×

××==

..

KLEI

Py

yycr

ππ

For y–y axis buckling, it is subjected to an axial load P.

Ky = 0.7

Solution:From the support conditions it is seen that about the y-y axis the column behaves as if it were pinned at its top and fixed at the bottom. About the x-x axis the column is free at the top and fixed at the bottom, and it is subjected to both axial load P and moment M = P(200mm).

Example Problem:


( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

EAP

rKLsec

rey

AP x

x

x

x

xY 2

1 2σ

For x–x axis yielding, it is subjected to an axial load P and moment M. Kx = 2

Solution:

( )[ ]

MPa 250 MPa 3557580

10419.4

(Ans) kN 4419N 419368error, and by trial Pfor Solving

10143159821108951

3

x

36

=<=×

==

==

×+=× −

Ycr

x

xx

.AP

.P

P.sec.P.

σσ

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

×××××

+=7580102109892

400029891052001

7580250 32

xx P.

sec.

P

Failure will occur about the x-x axis by buckling.

Example Problem:


R. C. Hibbeler“Mechanics of Materials" (7th SI edition) Prentice Hall 2008


CE2155 - Stability of Compression Members

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