Buckling of Slender Columns 1

Buckling of Slender Columns (10.1-10.4)

MAE 314 – Solid Mechanics

Yun Jing

Buckling of Slender Columns 2

Introduction Up to this point, we have designed structures with constraints based on material failure.

Find the state of stress and strain due to applied loads for certain simple structures. Compare the stress state to a maximum stress criterion or the strain state to a maximum strain criterion and determine at what load the structure will fail.

In this chapter, we learn how to design structures with constraints based on structural failure. No point in the material reaches the maximum stress or strain criterion. The structure at equilibrium becomes unstable. Any small change in the loading or any imperfection in the structure will not remain at equilibrium. This is generally due to compressive loading.

Buckling of Slender Columns 3

Example: Pinned ColumnColumn in Equilibrium

rollersupports

deformablecolumn

What happens as P increases?1. The point A moves downward and the beam shortens.2. At a certain value, PCR, the beam “buckles” and suddenly changes shape.3. This new shape may not support the full load.Equilibrium state for P<PCR Equilibrium state for P=PCR

Will a structure experience material failure or structural failure first? Depends on the material, geometry, etc.

In general: Long slender columns typically fail first due instability. Short wide columns typically fail first due to the material.

Buckling of Slender Columns 4

Example: Pinned Column

Why does this occur? To find out, let’s consider a simpler system. When is the deformed equilibrium position stable?

Move point C a small amount to the right. If the system moves back to the verticalposition it is stable. If the system moves away from the verticalposition it is unstable.

Spring moment: M = K (2Δθ) Draw a free body diagram of the top bar. Sum moments about point C. Assume

KM

rigid bars

torsional spring

angle of rotationof spring

K

LPM

LPMC 2sin

2sin

2

sin

L

KP

LK

PLMC

4

22

2

Buckling of Slender Columns 5

Example: Pinned Column

What happens for different values of P? For P < 4K / L, ΣMC < 0

Point C moves back to the left. Stable Equilibrium

For P = 4K / L, ΣMC = 0 Point C does not move. Neutrally Stable Equilibrium

For P > 4K / L, ΣMC > 0 Point C moves to the right. Unstable Equilibrium

We define PCR to be the critical load P at which the system moves from a stable equilibrium to an unstable equilibrium.

L

KP

LMC

4

2

L

KPCR

4

Buckling of Slender Columns 6

Euler’s Formula

This chapter uses slender pinned column as its basis because it is a commonly used member. The same concepts can be applied to other structures. Next step is to derive Euler’s formula for pin-ended columns. Solve for deflection curve of buckled beam.

EI

M

dx

yd

2

2

Moment at some distance x: PyM

EI

Py

dx

yd

2

2

02

2

yEI

P

dx

yd

Define p2 = P / EI

022

2

ypdx

yd

Solution:pxBpxAy cossin

Buckling of Slender Columns 7

Critical Load

Boundary conditions y(0) = 0: y(L) = 0:

The lowest value of P corresponds to n = 1.

We cannot solve for the amplitude (A)using a stability analysis. Other values of n correspond to higherorder modes.

00)0( BynpLpLALy 0sin0)(

EIL

nPnLp

2

222222

EIL

PCR 2

2 x

LAxy

sin)(

n=1n=2 n=3 n=4

EIL

P

EIL

P

EIL

P

EIL

P

nCR

nCR

nCR

nCR

2

2

4

2

2

3

2

2

2

2

2

1

16

9

4

Buckling of Slender Columns 8

Critical Stress

Knowing the critical load, we can now calculate the critical stress.

Evaluate at section of beam with smallest I. For a rectangular beam, I = bh3/12.

Typical plot of σCR vs. L / r for steel (r is theradius of gyration) Critical stress depends on the slendernessratio L /r .

2

2

AL

EI

A

PCRCR

2

2

2

22

2

32

/

1

121212 hL

E

L

Eh

bhL

EbhCR

buckling determinesfailure

yield stress determinesfailure

Buckling of Slender Columns 9

Rectangular Cross-Section

What happens when the column cross-section is rectangular? i.e. In which direction does it buckle?

2211 II

2211 CRCR PP

So the beam buckles about the 2-2 axis first.EI

LPCR 2

2

Buckling of Slender Columns 10

Other End Conditions

We can extend Euler’s formula to columns with other end conditions. Replace the length L with an “equivalent” or “effective” length Le. L is the actual length of the beam & Le is the length for use in PCR. EI

LP

eCR 2

2

Buckling of Slender Columns continued 11

Example Problem

Knowing that the torsional spring at B is of constant K andthat the bar AB is rigid, determine the critical load Pcr.

Buckling of Slender Columns continued 12

Example Problem

Determine the critical load of an aluminum tube that is 1.5 m long and has a 16-mm outer diameter and a 1.25-mm wall thickness. Use E = 70 GPa.1.25 mm

16 mm

Buckling of Slender Columns continued 13

Example Problem

Column AB carries a centric load P of magnitude 15 kips. Cables BC and BD are taut and prevent motion of point B in the xz-plane. Using Euler’s formula and a factor of safety of 2.2, and neglecting the tension in the cables, determine the maximum allowable length L.Use E = 29 x 106 psi.

W10 x 22