Mukavemet II Strength of Materials II Teaching Slides Chapter 10: Mukavemet II Strength of Materials...

Mukavemet IIMukavemet IISStrength of trength of Materials IIMaterials II

Teaching SlidesChapter 10:

Mukavemet IIMukavemet IISStrength of trength of Materials IIMaterials II

Stabilite (Burkulma)

Stability (Buckling)


Chapter OutlineChapter 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


FIGURE Buckling demonstrations.


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.’’


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.

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:

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.


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.


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.

Equilibrium of the Buckled Column


The sign convention adopted for the

moment M(x) in Fig. 10.6d is a positive

bending moment. From Fig. 10.6d we

get

Substituting M(x) into the moment-curvature equation, we obtain elastic curve equation as below:

Differential Equation of Equilibrium, and End Conditions:

)()()(" xPvxMxEIv

vEI

P

EI

xM

dx

vd

)(2

2

or

or

0)()(" xPvxEIv


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

0)0( v 0)( Lvand


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

Solution of the Differential Equation:

EI

P2 0" 2 vv 0" vPvEI


The general solution to this homogeneous equation is

xCxCxv cossin)( 21

We seek a value of and constants of integration C1 and C2 such that the two boundary conditions of this homogeneous equation are satisfied.


Thus,

00)0( 2 Cv

0sin0)( 1 LCLv

Obviously, if the constants C1 and C2 equal to zero, the deflection v(x) is zero everywhere.We just have the original straight configuration.


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.

,....2,1,0sin nL

nL nn


Combining the equations above gives the following formula for the possible buckling loads:

2

22

L

EInPn

EI

P2


The deflection curve that corresponds to each load Pn is obtained by combining the following Equations.

0sin LnxCxCxv cossin)( 21 and


we get

L

xnCxv

sin

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.


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.

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

2

22

L

EInPn

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


Let us examine some important implications of the Euler buckling-load formula. We can express this in terms of critical (buckling) stress.

22

/ rL

E

e

cr

r

Le2

2

E

cr


2

2

E

cr


FIGURE: Critical stress versus slenderness ratio for structural steel columns.For slenderness ratios less than 100, the critical stress is not meaningful.

2

2

E

cr


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.

EXAMPLE


Using the given factor of safety, we make

a) For the 100-kN Load.

in Euler’s formula and solve for I. We have

We check the value of the normal stress in the column:

Recalling that, for a square of side a, we have , we write

Since σ is smaller than the allowable stress, a 100x100-mm cross section is acceptable.


Solving again buckling equation for I, but making now Pcr=2.5(200)=500 kN, we have

b) For the 200-kN Load.

The value of the normal stress is

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


EXTENSION OF EULER’S FORMULA TOCOLUMNS 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.


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).

Column with One Free End


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.


We say that the effective length Le of the column is equal to 2L and substitute Le=2L in Euler’s formula:

2

2

ecr L

EIP

2

2

2

2

42 L

EI

L

EIPcr

Le=2L


The critical stress is found in a similar way from the formula

22

/ rL

E

e

cr

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.

r

Le2

2

E

cr


Consider next a column with two

fixed ends A and B supporting a

load P (Fig. 10.10).

Column with Two Fixed Ends


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. 10.11).

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. 10.12).

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. 10.13a).

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.

10.13a must behave as a pin ended

column (Fig. 10.13b). We thus conclude

that the effective length of a column with

two fixed ends is Le =L/2.

2

2

ecr L

EIP

42/ 2

2

2

2

L

EI

L

EIPcr

Le=L/2


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?

EXAMPLE


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 10.13 illustrates the effective lengths of columns with several types of end conditions.

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.


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.


EULER’S FORMULA FOR PIN-ENDED COLUMNS


Our approach will be to determine the conditions under which the configuration of Fig. 10.2 is possible.

Mukavemet II Strength of Materials II Teaching Slides Chapter 10: Mukavemet II Strength of Materials...

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