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COLUMNS

(Buckling, Euler and Tetmajer Formulas)

11.

(tvid- 11.a , 11.b)

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11.1 What is Column?

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The lateral bending of a column (or a bar element) before the yield (or fracture in brittle materials) limit, as a result of being exposed to a compressive load in the direction of its axis, is called buckling.For this reason, when designing a column, it is extremely important to consider not only the yield/fracture limit but also the buckling limit.Our aim in this regard is to derive formulas that will enable us to theoretically calculate the buckling limits (critical buckling loads) of columns. In the last part, the subject will be reinforced with various examples.

11.2 What is Buckling?

Vertical axis bearing rod elements are called columns. Columns, which are of vital importance in the statics and durability of structures, are generally exposed to compressive loads in the direction of their axis.

A cut column ☹.

Figure 11.1

(a)

(b)

(c)

Figure 11.2

Beam

Column

Beam

Column

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If a column is thick enough, it will carry the compressive load up to its yield limit.

In this case, strength calculations are made according to the yield limit.

a-) Thick Columns

at the instant of yielding:

b-) Thin Columns Kolonlar

If a column is thin enough, it will bend laterally as a result of axial compressive load before reaching its yield limit.

In this case, strength calculations should be made according to the buckling limit.

at the instant of buckling:

Before Yielding:

Before Buckling:

Before Yielding,

11.3 Classification of Columns

We can divide columns into two types, thick and thin columns, according to their damage types (yielding or buckling).

Note: Yielding in brittle columns is very small. For this reason, the fracture instant is taken into account instead of the yield limit..

Before Buckling,

Figure 11.3

Figure 11.4

Figure 11.5a

Figure 11.5b

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11.4. Tips of Buckling

4. Depending on boundary conditions, buckling of column may occur in different modes, and final buckled shape of column may be different in each mode. In general, a column is most easily buckled in first mode. There is a different critical buckling load for each mode.

3. In buckling, compressive loads and compressive stresses are involved, but their signs are taken as + in the calculations.

5. If there is no boundary condition difference according to the planes, the column first tries to rotate around the minimum principal inertia axis, that is, it is buckled.

(This article is explained in more detail in section 11.7).

Figure 11.6

(a)

(b)

(c)

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11.5 Euler Buckling Approach with Elastic Curve Simulation

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• This approach is based on the assumptions and equations in the elastic curve of beams

subjected to bending.

• It considers buckling under axial loads as a lateral elastic curve and allows us to calculate critical

buckling loads for different boundary conditions.

If the undeformed state of the beam is taken into account

Considering the deformed state of the beam

In Euler's approach:

• The deformed state of the elastic curve of the beam is taken into account.
• It is assumed that there is a compression force P only in the beam axis direction.

* P has no effect on the bending internal moment M.

* P has an effect on M. Like this:

If we rotate the beam longitudinally, we obtain a column.Euler's approach is also valid for columns.

Now, based on the Euler approximation logic, we will obtain the Pcr loads for columns with different boundary conditions…>>

Let's consider a vertically loaded beam, as described in Chapter 9. Let an axial horizontal force P also act on this beam:

From equation

9.2.2.a, Curvature of the beam:..>>

* This acceptance has been made regarding bending (chapter 5) and deflection of beams (chapter 9)..

* Equation 9.2.2.a is valid for both cases above.

Euler

proportion limit

Figure 11.7

Figure 11.8

Figure 11.9

Figure 11.10

Figure 11.11a

Figure 11.11b

Figure 11.11c

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• These calculations can be applied up to the proportionality limit where Hooke's law is valid (for the linear part of the compression curve).

11.6 Euler Buckling Formulas for Different Situations

Let's consider a column of length L, subjected to axial force P. The section is symmetrical and the principal axes of inertia are z and y. Therefore, I_z < I_y ,the minimum principal moment of inertia is I_min = I_z =I. The column rotates around the z axis in this plane, that is, it is buckled, and a lateral displacement occurs in the y direction.

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The general solution of this constant coefficient homogeneous differential equation is :

at x=0

--> B sin(kL)=0

Here B cannot be zero (B≠0). Because, since A = 0, y = 0 in every case. This will mean that buckling will never occur, which is not possible. Then sin(kL) should be = 0.

;

In this case, kL must be one of the values ​​π, 2π,….,nπ.

If kL= nπ is taken:

(buckling mode number: n=1, 2, 3,…. )

11.6.1 Column with ball (spherical) joint at lower and upper ends and slideable upper end

From the equilibrium of the

deformed state of the lower part of cut I-I :..>>

From equ. 9.2.2.a:

We can find the constants A and B from the boundary conditions. Since there will be no lateral displacement in the supports;

---> A=0

at x=L

P is gradually increased, the k value also increases, and when it reaches P = P_cr , buckling occurs and

sin(kL)=0

and

at the instant of buckling

buckling load in n^th mode:

in first mode( n = 1)

in 2nd mode ( n = 2)

The value of n varies depending on the boundary conditions.

Figure 11.12

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If we define

11.6.2 Lower end is built-in, upper end is free column

​

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can be written.

If

The general solution of this diff. equation is:

at x=0

at x=0

P is gradually increased. During 0<P<Pcr, no y displacement, that is, buckling, occurs. When P = P_cr , y = δ for x = L at point B and it can be said that buckling occurs at this instant.

From the equilibrium of the

deformed state of the lower part of cut I-I :..>>

From equ. 9.2.2.a :

(deflection equation)

(slope equation)

At the built-in end, the slope and deflection are zero:

at the instant of buckling:

(constant coefficient homogeneous differential equation)

Figure 11.13

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-----> D=0

----> C = -δ is found.

is taken here again,

y=0

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can be writen.

where

The general solution of this diff equation is:

,

(constant coefficient homogeneous differential equation)

Example 11.1 : How much does the buckling load change if a singular bending moment Mo, in addition to force P, is applied to the column with its lower end fixed and its upper end free?

Taking the Mo moment into account, we will reach the result with the steps in article 11.1.2:

From the equilibrium of the upper part of the deformed version of cut I-I:

From equ. 9.2.2.a

Figure 11.14.a

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is found.

..>>

Solution

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at x=L

When

(For this reason, it is accepted that the buckling occurred at this instant.)

P is increased, while the k term also increases. y displacement begins to occur.

Instant of Buckling:

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Figure 11.14.b

Deflection equation:

Let's rewrite the equations we found:

Slope equation:

Now let's find the constants C and D from the boundary conditions.

At the fixed end, both deflection and slope values ​​are zero.

Deflection at free end :

Determining the instant of buckling:

and

and

11.7 Generalization of Euler's Buckling Formula:

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Critical buckling loads are calculated similarly for each of the types shown in Figure 11.15 below, and general formulas can be written as in equations 11.5 and 11.6:...>>

Bu Formüller hangi durumda geçerlidir? … >>

Table 11.1

The plane joint does not allow free rotation in the direction perpendicular to its own plane and acts as a built-in. For this reason, Le = 0.5L is taken for the vertical direction. Buckling control should also be done for the vertical direction. (The L value for type 2 in the table is for the plane joint itself.)

Ball joints and built-in connections: do not impose any different restrictions in other directions and the Le values ​​in the table do not change. Buckling occurs around the minimum principal inertia axis of the section. These situations will be explained later.

Top:

Below:

Free

Built-in

(fixed end)

Ball or Plane Joint

Ball or Plane Joint

Ball or Plane Joint

Built-in

Built-in

Built-in

Built-in Sliding

Built-in

Buckling axis: It is the axis around which the column is forced to rotate during buckling.

I in the equations is the moment of inertia about the buckling axis. In ball joint and built-in connections, the column wants to rotate around the minimum principal inertia axis and I = I_min. In a plane joint, the I value may be different from I_min.

Figure 11.15

10

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L_e in the equations is the effective length and varies depending on the column type.

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(Because Hooke’s Law was used in the derivation of equation 11.3 in the elastic curve simulation, which forms the basis of Euler's formulas.)

Euler

Compression of material test diagram

Proportion limit

𝑖 term in equation 11.8 is radius of gyration:

It is a property of the rod that depends on its geometry and is defined as follows:

In this case, 11.6 critical buckling stress (general Euler equation) can be expressed as in equation 11.10:..>>

Attention: Euler's formulas (all equations 11.2 to 11.11) are valid up to the limit of proportionality.

Up to Proportion Limit::

Figure 11.16

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and critical value of slenderness ratio:

At proportional limit

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11.9 Tetmajer Empirical Equation

When the Proportion Limit is Exceeded :

When the proportionality limit is exceeded, the Tetmajer Empirical Equation (Equation 11.12), obtained with experimental data, is used in buckling load calculations:

Table 11.2 Constants of the Tetmajer Empirical Equation

Why is it called «Tetmajer line»?

Although the Tetmajer equation is second order, as can be seen from Table 11.2, the constant c is zero for many materials.For this reason, the Tetmajer equation is generally linear and therefore it can also be called the Tetmajer line.

Figure 11.17

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Euler

Tetmajer

Compression of material test diagram

Proportion limit

Figure 11.16

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11.10. In which axis does the first buckling occur?

• So far, we have examined the buckling of the column in the x-y plane and thought of it as trying to rotate around the z axis.
• However, depending on the boundary conditions and beam cross-section, the initial buckling of the beam may occur in the vertical plane and around the y-axis or around a different axis.
• This situation must be taken into consideration and examined when determining the critical buckling load. Because the important thing is to find the P_cr load when the column first buckles.

Buckling axis

Ball Joint and Built-in connections constrain the beam in the same way in all directions, that is, the translational or rotational freedoms are the same in all directions in the connection region. The same can be said for the free end.

In this case, the first buckling occurs when the column rotates slightly around the minimum principal inertia axis of its cross-section. For these boundary conditions, I_min is always used and the effective length is always L_e = L for ball joint and L_e = 0.5L for built-in. Depending on the cross-section geometry I_min is calculated.

11.10.1 Determining the Buckling Axis in Case the Boundary Conditions Do Not Change According to the Direction (ball joint, fixed end, free end)

The determination of the axis of the initial buckling may differ for the 2 categories of boundary conditions:

Column section alternatives (I-I section)

Non-symmetrical section

Symmetrical section

Figure 11.18

Figure 11.19

(a)

(b)

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Although connections such as plane joints allow rotation in their own plane (or, in other words, around the axis perpendicular to the plane), they do not allow rotation in the perpendicular plane (around axes parallel to the plane).

Therefore, they act as built-ins in the vertical plane. Since it allows rotation in its own plane, the effective length is L_e = L and the buckling axis of symmetrical cross-section beams (the axis around which the beam wants to rotate) is z.

However, due to the fixed behavior in the vertical plane, L_e = 0.5L and the buckling axis will be y. I=I_z should be taken in its own plane, and I=I_y should be taken in the perpendicular plane. Whichever of the critical buckling loads calculated in both planes is smaller, is the critical buckling load (P_cr) value.

Or we can say that buckling occurs first in the plane with the smaller 𝝈_c𝒓 value.

Views of the same column from different planes

From equ. 5.6:

Cross-section

11.10.2) Determination of Buckling Axis in Case of Boundary Conditions Changing According to Direction (Planar Joints)

In this case, buckling will occur first in the x-y plane, that is, around the z axis. In that case:

Buckling axis: z

Buckling axis: y

Figure 11.20.a

Figure 11.20.b

Columns

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Views of the same column from different planes

11.10.3) Determining the First Buckling Axis

from the Slenderness Ratio:

Buckling will occur in the x-y plane (around z).

Figure 11.21.a

Figure 11.21.b

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If we calculate the slenderness coefficients on the y and z axes from equation 11.8:..>>

Buckling will occur in the x-z plane (around y).

Cross-section

For the column given in two different views on the side,

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11.11 Road Map to Follow in Colon Problems:

1st Step - Buckling Load Determination:

1.2- It is decided which of the Euler or Tetmajer formulas to use. There are 2 alternatives:

1st Alternative:

Comparison of slenderness ratios

,

Euler is used.

Tetmajer is used

2nd Alternative:

1.4 – If the connection is in the form of a plane joint etc. and changes the boundary conditions depending on the direction, buckling control in those directions must also be made.

2nd Step – Compressive Load Detection:

Load that will cause the compression safety to be exceeded:

P_c-max =

3rd Step - Determination of Maximum Load:

Smaller of P_cr and P_c-max values ​​is maximum load (P_max) value that column can carry.

1.1- From Figure 11.15 and Table 11.1, the type of column and its effective length (Le) are determined.

In special problems, the rotations and translations freely allowed by the connections should be analyzed well and it should be determined correctly which support type it is suitable for and which column type it is. In some cases, the type of column can be interpreted differently depending on the plane. Additionally, if there are more than 2 connections, the buckling mode must be determined correctly from figure 11.6. If the type of column cannot be determined, the buckling formula should be calculated from the Euler beam simulation as in 11.6.

The axial compression load that a column of certain dimensions and material can carry is determined by the following steps:

From equation 11.8, the current slenderness ratio is determined:

From equation 11.9 the slenderness ratio at the proportionality limit is determined:

equ: 11.10

equ: 11.12

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It has 4 stages.

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Example 11.2

Calculate the maximum load P that the column fixed at its lower end can carry without losing its functionality. The column material is brittle and the tension-compression diagram is given below.

One end is free, the other end is built-in

1st Step: Buckling load (P_cr) determination

From Alternative 2

Let's assume we can use Euler:

Tension and compression experimental diagrams

Solution:

We will follow the roadmap steps in article 11.11

1.1-)

1.2-)

This column falls into Type 1 of the types in Figure 11.15.

Effective length from Table 11.1

Should Euler or Tetmajer be used?

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Minimum moment of inertia of the section

If we pay attention to the compression region of the diagram given in the question, the stress at the proportional limit is:

compressive yield stress

(In buckling, compressive forces and stresses are considered to have positive signs..)

(From equ 11.10 buckling stress)

Elastic modulus (E) calculation:

(a)

Figure 11.22

(b)

Figure 11.23

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1.3-)

2nd Step :

Max Load Detection:

1.4-)

There is no connection (such as a plane joint) that changes the boundary conditions depending on the direction.

Buckling load:

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Compressive Load :

3rd Step :

Example 11.3

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Tetmajer Equation (if necessary) :

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Efective length

1.1) Column type (see topic 11.7)

One end is free, the other end is built-in:

Solution:

1st Step ) Buckling load (P_cr) determination

From Table 11.1

1.2) Should Euler or tetmajer be used?

Let's go with the 1st alternative:

The cross section is not symmetrical. The column wants to rotate around the axis of minimum principal inertia.

We must calculate the slenderness ratio:

Calculate the load P that the wooden column in the figure can carry with a safety coefficient of e= 3.

We follow the road map described in article 11.11.

Figure 11.24

(a)

(b)

Figure 11.25

Figure 11.26

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Then, buckling occurs first.

Slenderness ratio :

Tetmajer should be used.

From the Tetmajer equation given in the question, the stress at buckling is:

The stress in the proportionality limit can be found from the Euler Hyperbola equation:..>>

1.3)

Critical Buckling Load:

Allowable (Safe) Buckling Load:

(Maximum Force that can be applied within safety limits)

There is no need to do the other steps in the road map.

Figure 11.27

It can be taken that : Yield stress in compression ≅ Stress at the proportional limit.

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If we first examine it in terms of compressive strength;

Example11.4

Solution:

The tube has no support. In this case, which type can we put it in?

• For this loading case, both ends can rotate in all directions.
• There is a slope angle at both ends.
• The loading and boundary conditions are symmetrical and a symmetrical buckling is expected with respect to the median axis.
• All these situations are provided by the ball joint condition of type 2 from the types in Table 11.1

L_e=L

Type 2

1.1- Determination of column type and effective length:

Buckling load (P_cr) determination

We will follow the steps in the road map in article 11.11.

Figure 11.28

Figure 11.29

compressive force that will cause the material to reach its yield limit

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1.3 ) Buckling Load:

we can say that buckling will occur first.

or if we compare in terms of loads:

If we move on from the second alternative

1.2 ) Should we use Euler or Tetmajer?

We agree that we will use Euler:

Stress at the Limit of Proportion :

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EXAMPLE 11.5

The steel bar AC shown in the figure is supported by frictionless wheels at point B and subjected to load P. The wheels can rotate in the y direction perpendicular to the shape plane. There are spherical joints in A and C. Calculate the force P that can be carried safely by taking the buckling safety coefficient of the system as e = 2.5.

Buckling in the x-z plane (around the y axis):

Solution:

Support B affects the deformation in this plane (prevents displacement in the z direction.) And the beam is buckled as in the figure 11.6.b. For this reason, the buckling mode is n = 2 in this plane.

Considering only ball joints A and C,

column type is 2. (see: Figure 11.15, table 11.1)

Type 2

Mode 2

Allowable (Safe) buckling load in n=2 mode (from Equ. 11.2):

Effective Length (from table 11.1)

radius of gyration (from equ 11.9)

Slenderness ratio (from equ 11.8)

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Figure 11.30.a

Figure 11.30.b

Figure 11.31

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Type 2

Mode 1

Buckling in the x-y plane (around the z axis):

Again, if the existence of only A and C spherical (ball) joints is accepted, the column type is still 2 since the effects of these connections are the same in all planes.

If we pay attention to the shape of the B support and consider that the wheels are frictionless, we can see that the B support will not affect the deformation in this plane. (Since it allows displacement in the y direction, it can be assumed that there is no support B in this plane). For this reason, the buckling mode is n = 1 in this plane.

Allowable load that the rod can carry

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We found that

Figure 11.32

Effective Length (from table 11.1):

radius of gyration (from equ 11.9):

Slenderness ratio (from equ 11.8)

Allowable (Safe) buckling load in n=1 mode (from Equ. 11.2):

Allowable (Safe) compressive load:

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L= 1.5 m , a= 0.5 m ,

b= 60 mm, h= 80 mm, s= 8 mm, t= 6 mm

Beam BC, which is connected to a fixed plane joint at end B, is supported by a vertical column AC at end C, and a vertical force F is applied to the beam. Check the safety of the AC column in terms of compression and buckling and determine the buckling safety coefficient.(All connections show plane joint properties. Joint B is in the x-z plane, and joints A and C are in the x-y plane.)

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EXAMPLE 11.6

Solution..>>

Given:

Figure 11.33

Figure 11.34

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From static analysis, let's calculate the axial compressive force on rod AC:

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Solution: 11.6

Let's follow the roadmap in article 11.11:

1st Step – Buckling Load determination for AC rod.

These joints allow rotation around the z axis.

From table 11.1, the effective length for this type is L_e-z=L =1.5m

1.1) Which type of column does the rod fall into? What is the effective length?

Examination in x-y plane

Type 2

Examination in x-z plane

The upper and lower connection do not allow rotation around the y-axis (built-in). The upper end allows movement in the x direction, the lower and upper ends of the rod are fixed.

If we examine the figure in the question:

The upper end acts as a sliding plane joint,

the lower end acts as a fixed plane joint.

Therefore, for this plane the rod falls into Type 2. (see: topic 11.7)

Therefore, in this plane the rod falls into Type 4.

From table 11.1, the effective length for this type is L_e-y=0.5L =0.75m

Figure 11.35

Figure 11.36

Figure 11.37

Figure 11.39

Figure 11.38

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• b= 60 mm , s= 8 mm , h= 80 mm , t= 6 mm

​

​

​

the first buckling occurs around the z axis (in the x-y plane).

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1.2) Which of the Euler or Tetmajer formulas will we use in terms of buckling?

If we use alternative 1:

Slenderness ratio at the limit of proportion:

Since buckling types differ depending on the plane, we need to determine the slenderness ratios in both planes.

Verification for Tetmajer region:

is verificated

With the 2nd alternative, determining whether Euler or Tetmajer should be used could also be done as follows:

So then Euler cannot be used, we have to use Tetmajer

(From equ.11.11)

Figure 11.40

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SAFE in terms of compression

Critical buckling load:

Buckling Control:

We have previously found that the load falling on the rod is

Then there will be no

There is no buckling, but could the compression safety have been exceeded?

Current stress :

(It was given in the question)

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Safety factor for buckling:

Figure 11.41

Example 11.8

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Considering that the buckling safety coefficient in the Euler region is desired to be 2 for the T-section rod with two fixed ends shown in the figure, calculate how much the rod can be heated (how much its temperature can be increased)..

Rectangular cross-section, length L aluminum rod B is also supported as built-in. Link A of the rod allows rotation about z and y and translation in the z direction, but prevents translation in the y direction. According to this;

Example 11.9

What should be the a/b ratio so that the buckling loads are equal in both planes (x-y and x-z planes)?

Figure 11.42

(a)

(b)

Figure 11.43