Lecture mechanics of materials chapter 11 stability of columns

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August 2012 11-1

Stability of Columns

Learning objectives

structure instabilities.

design of structures.

Buckling Phenomenon

Energy Approach

Bifurcation/Eigenvalue Problem

Stable Equilibrium

Neutral Equilibrium

Unstable Equilibrium Potential Energy is Minimum

θ L

Lsinθ

(c)

R

Stable Stable

Unstable Unstable

PL Kθ⁄ = θ⁄ sinθ

August 2012 11-3

Snap Buckling Problem

L P

P (45−θ)

P

L cos(45-θ)

0 < <θ 45o

P

L cos(θ−45)

P

(θ−45)

P

K L L

- = (cos(45 – θ) – cos45)tan(45 – θ) 0 < <θ 450

P

K L L

- = (cos(θ 45– ) – cos45)tan(θ 45– ) θ 45> 0

Local Buckling

crinkling

Axial Loads

Compressive Torsional Loads

August 2012 11-5

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

Euler Buckling

Boundary Value Problem

Boundary conditions:

Solution

Trivial Solution:

Non-Trivial Solution:

where:

Characteristic Equation:

Euler Buckling Load:

v(x)

M z

P

N = P

A

y

x

L

P

A

EI

x2

2

d

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

v = 0

v x( ) = Acosλx +Bsinλx

EI

-=

λL

sin = 0

2

π2EI

L2

L2

-=

August 2012 11-7

L

sin

=

Mode shape 2

Mode shape 3

cr

L/2

P cr

P cr

L/2

2

EI

L2

-=

2EI

L2

-=

I

Mode shape 1

P cr

P cr

2

EI

L2

-=

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

y x L P

A

B

x L

P B

x L

P B

x L

P B

A

EI

x2

2

d

d v +Pv

0

=

EI

x2

2

d

d v +Pv

Pv L( )

=

EI

x2

2

d

d v +Pv

R B(L x– )

=

EI

x2

2

d

d v +Pv

R B(L x– ) M+ B

=

v 0 ( ) = 0

v L( ) = 0

v 0 ( ) = 0

x d

dv( )0 = 0

v 0 ( ) = 0

x d

dv( )0 = 0

v L( ) = 0

v 0 ( ) = 0

x d

dv( )0 = 0

v L( ) = 0

x d

dv( )L = 0

EI

-=

λL

sin = 0 cosλL = 0 tanλL = λL 2 1 ( – cosλL)

λLsinλL

π2EI

L2

- π2EI

4L2

- π2EI

2L

( )2

L2

- π2EI

0.7L

( )2

L2

- π2EI

0.5L

( )2

-=

L eff2

-=

August 2012 11-9

Axial Stress:

Radius of gyration:

Failure Envelopes

σcr P cr

A

-=

A

-=

σcr P cr

A

- π2E

L eff ⁄ r

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 shows

a 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)

August 2012 11-11

shown Both bars have a diameter of d = 1/4 inch, modulus of elasticity

E = 30,000 ksi, and yield stress σyield = 30 ksi Bar AP and BP have lengths of LAP= 8 inches and LBP= 10 inches, respectively Determine the factor of safety for the two-bar structures

Fig C11.3

60o

F A

B

25o

P

Class Problem 1

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

110o

F A

B

P

F A

B

P 40o

Structure 2

75o

F

A

B

P60 o

F

A B

P

30o

Structure 3

Structure 4

August 2012 11-13

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

A A

A

A

2 in

4 in

Cross-section AA

A