mukavemet ii strength of materials ii teaching slides chapter 10: mukavemet ii strength of materials...
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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)
Stability (Buckling)
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
Stability (Buckling)
FIGURE Buckling demonstrations.
Stability (Buckling)
Stability (Buckling)
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.’’
Stability (Buckling)
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.
Stability (Buckling)
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.
Stability (Buckling)
Stability (Buckling)
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.
Stability (Buckling)
Stability (Buckling)
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
Stability (Buckling)
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
Stability (Buckling)
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
Stability (Buckling)
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
Stability (Buckling)
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.
Stability (Buckling)
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.
Stability (Buckling)
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
Stability (Buckling)
Combining the equations above gives the following formula for the possible buckling loads:
2
22
L
EInPn
EI
P2
Stability (Buckling)
The deflection curve that corresponds to each load Pn is obtained by combining the following Equations.
0sin LnxCxCxv cossin)( 21 and
Stability (Buckling)
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.
Stability (Buckling)
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
Stability (Buckling)
Stability (Buckling)
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
Stability (Buckling)
2
2
E
cr
Stability (Buckling)
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
Stability (Buckling)
Stability (Buckling)
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
Stability (Buckling)
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.
Stability (Buckling)
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
Stability (Buckling)
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.
Stability (Buckling)
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
Stability (Buckling)
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.
Stability (Buckling)
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
Stability (Buckling)
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
Stability (Buckling)
Consider next a column with two
fixed ends A and B supporting a
load P (Fig. 10.10).
Column with Two Fixed Ends
Stability (Buckling)
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
Stability (Buckling)
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
Stability (Buckling)
Stability (Buckling)
Stability (Buckling)
Stability (Buckling)
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 (Buckling)
Stability (Buckling)
Stability (Buckling)
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.
Stability (Buckling)
Stability (Buckling)
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.
Stability (Buckling)
EULER’S FORMULA FOR PIN-ENDED COLUMNS
Stability (Buckling)
Our approach will be to determine the conditions under which the configuration of Fig. 10.2 is possible.