Structural stability of column

8/15/2019 Structural stability of column

1/13

August 2014 11-1

P r i n t e d f r o m : h

t t p : / / w w w . m e . m t u . e d u / ~ m a v a b l e / M o M 2 n d . h t m

Stability of Columns

• Bending due to a compressive axial load is called Buckling.

• Structural members that support compressive axial loads are called Columns.

• Buckling is the study of stability of a structure’s equilibrium.

Learning objectives

• Develop an appreciation of the phenomena of buckling and the various types ofstructure instabilities.

• Understand the development and use of buckling formulas in analysis anddesign of structures.


2/13

August 2014 11-2



Buckling Phenomenon

Energy Approach

Bifurcation/Eigenvalue Problem

Stable Equilibrium

Neutral Equilibrium

Unstable Equilibrium

Potential Energy is Minimum

P P

K θθ

θ

L

Lsinθ

(c)

O O

R

StableStable

UnstableUnstable

PL K

θ ⁄ θ θsin ⁄ =


3/13

August 2014 11-3



Snap Buckling Problem

L

P

P

(45−θ)

P

45oFs

Fs

L cos(45-θ)

L s i n ( 4 5 - θ )

0 θ 45o

< <

θ 450

>

θ 0=

Fs

P

L cos(θ−45)

Fs

P

(θ−45)

s

n

−

P

K L

L---------- 45 θ – ( )cos 45cos – ( ) 45 θ – ( )tan= 0 θ 45

0< <

P

K L L---------- θ 45 – ( )cos 45cos – ( ) θ 45 – ( )tan= θ 45

0>


4/13

August 2014 11-4



Local Buckling

crinkling

Axial Loads

Compressive

Torsional Loads


5/13

August 2014 11-5



C11.1 Linear deflection springs and torsional springs are attached to

rigid bars as shown.The springs can act in tension or compression and

resist rotation in either direction. Determine Pcr , the critical load value.

Fig. C11.1

30 in

O

P

k = 8 lbs/in

30 in

k = 8 lbs/in

K = 2000 in-lbs/rads


6/13

August 2014 11-6



Euler Buckling

Boundary Value Problem

Differential Equation:

Boundary conditions:

Solution

Trivial Solution:

Non-Trivial Solution:

where:

Characteristic Equation:

Euler Buckling Load:

• Buckling occurs about an axis that has a minimum value of I.

v(x)

M z

P

N = P

A

y

x

L

PA

EI

x2

2

d

d v P v+ 0=

v 0( ) 0= v L( ) 0=

v 0=

v x( ) A λ xcos B λ xsin+=

λ P EI ------=

λ Lsin 0=

P n

n2

π2 EI

L2

------------------= n 1 2 3. ⋅ ⋅, ,=

P cr

π2 EI

L2

------------=


7/13

August 2014 11-7



Buckling Mode: v B nπ x

L---

⎝ ⎠⎛ ⎞sin=

Mode shape 2

Mode shape 3

Pcr Pcr

Pcr Pcr

L/2

Pcr

Pcr

L/2

L/3 L/3 L/3

P cr

4π2 EI

L2

----------------=

P cr

9π2 EI

L2

----------------=

I

I I

Mode shape 1

PcrPcr

Pcr

L P cr

π2 EI

L2

------------=


8/13

August 2014 11-8



Effects of End Conditions

Case

1.

Pinned at

both Ends

2.

One end fixed,

other end free

3.

One end fixed,

other end pinned

4.

Fixed at both ends.

Differen-

tial

Equation

Boundary

Conditions

Character-

istic

Equation

Critical

Load

Pcr

[

Effective

Length—

Leff

L 2L 0.7L 0.5L

y

x

L

P

A

B

x

L

P

B

A

y

x

L

P

B

Ay

x

L

P

B

A

EI x

2

2

d

d v P v+

0=

EI x

2

2

d

d v P v+

P v L( )=

EI x

2

2

d

d v P v+

R B L x – ( )=

EI x

2

2

d

d v P v+

R B L x – ( ) M B+=

v 0( ) 0=

v L( ) 0=

v 0( ) 0=

xd

dv0( ) 0=

v 0( ) 0=

xd

dv0( ) 0=

v L( ) 0=

v 0( ) 0=

xd

dv0( ) 0=

v L( ) 0=

xd

dv L( ) 0=

λ P EI ------=

λ Lsin 0= λ Lcos 0= λ Ltan λ L= 2 1 λ Lcos – ( )

λ L λ Lsin – 0=

π2 EI

L2

------------ π2 EI

4 L2

------------ π2 EI

2 L( )2--------------=

20.13 EI

L2

------------------- π2 EI

0.7 L( )2------------------=

4π2 EI

L2

--------------- π2 EI

0.5 L( )2------------------=

P cr π

2

EI L

eff

2------------=


9/13

August 2014 11-9



Axial Stress:

Radius of gyration:

Slenderness ratio: (Leff /r).

Failure Envelopes

• Short columns: Designed to prevent material elastic failure.

• Long columns: Designed to prevent buckling failure.

σcr

P cr

A--------=

r I

A---=

σcr

P cr

A--------

π2 E

Leff

r ⁄ ( )2

-----------------------= =

A A A


10/13

August 2014 11-10



C11.2 Columns made from an alloy will be used in a construction of

a frame. The cross-section of the columns is a hollow-cylinder of thick-

ness 10 mm and outer diameter of ‘d’ mm. The modulus of elasticity is

E = 200 GPa and the yield stress is σyield = 300 MPa. Table below showsa list of the lengths ‘L’ and outer diameters ‘d’. Identify the long and the

short columns. Assume the ends of the column are built in.

L

(m)

d

(mm)

1 60

2 80

3 100

4 150

5 200

6 225

7 250


11/13


12/13

August 2014 11-12



Class Problem 1

Identify the members in the structures that you would check for buckling. Circle the

correct answers.

Structure 1 AP BP Both None




110o

FA

B

P

Structure 1 110o

F

A

B

P 40o

Structure 2

75o

F

A

B

P60

o

F

AB

P

30o

Structure 3

Structure 4


13/13

August 2014 11-13



C11.4 A hoist is constructed using two wooden bars to lift a weight

of 5 kips. The modulus of elasticity for wood is E = 1,800 ksi and the

allowable normal stress 3.0 ksi. Determine the maximum value of L to

the nearest inch that can be used in constructing the hoist.

Fig. C11.4

W

B

L ft. CA

A

A

A2 in

4 in

Cross-section AA

30o

A

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Structural stability of column