Column Design(4.11-13)

MAE 316 Strength of Mechanical ComponentsY. Zhu

Column Design1

Introduction

Column Design2

Long columns fail structurally (buckling failure) Short columns fail materially (yielding failure) We will examine buckling failure first, and then transition

to yielding failure. P

P

Long

2

2

LEIPcr

P

P

ShortAP crcr

Buckling of Columns (6.2) For a simply supported beam (pin-pin)

Column Design3

y, v

x

F.B.D y, v

v(x)

0

)(

)()(

2

2

2

2

PvdxvdEI

PvxMdxvdEI

xPvxM

The solution to the above O.D.E is:

xEIPBx

EIPAxv cossin)(

Buckling of Columns (6.2)

Apply boundary conditions: v(0)=0 & v(L)=0

Either A = 0 (trivial solution, no displacement) or

The critical load is then

The lowest critical load (called the Euler buckling load) occurs when n=1.

Column Design4

xEIPBx

EIPAxv cossin)(

0sin&0 LEIPAB

nLEIPL

EIP 0sin where n = 1, 2,

22)(LEInPcr

2

2

LEIPcr

Buckling of Columns (6.2) The corresponding deflection at the critical load is

The beam deflects in a half sine wave.

The amplitude A of the buckled beam cannotbe determined by this analysis. It is finite, however. Non-linear analysis is required to proceed further.

Column Design5

LxAxv sin)(

Pcr

A

Buckling of Columns (6.2) Define the radius of gyration as

Pcr can be written as

If L/rg is large, the beam is long and slender. If L/rg is small, the beam is short and stumpy.

To account for different boundary conditions (other than pin-pin), use effective length, Le.

Column Design6

AIrg 2

2

2

)/( gcr

rLE

AP

crge

cr

rLE

AP 2

2

)/(

Buckling of Columns (6.2) Effective lengths for various boundary conditions

Design values (AISI)

Column Design7

(a) fixed-free, (b) pin-pin, (c) fixed-pin, (d) fixed-fixed, (e) fixed-non-rotating

(a) Le = 2.1L, (b) Le = L, (c) Le = 0.8L, (d) Le = 0.65L, (e) Le = 1.2L

Critical Stress (6.3) Plot cr=Pcr/A vs. L/rg

Long column: buckling occurs elastically before the yield stress is reached. Short column: material failure occurs inelastically beyond the yield stress. A Johnson Curve can be used to determine the stress for an intermediate

column

Column Design8

?

No failure

Failure

Sy

Johnson Curve

Note:Sy = yield strength = ySu = ultimate strength = uSp = proportional limit

Critical Stress (6.3) Johnson equation

As Le/rg 0, Pcr/A Sy To find the transition point (Le/rg)c between the Johnson and

Euler regions

Column Design9

2

2

2

4

g

eyy

cr

rL

ES

SAP

ycg

e

g

eyy

SE

rL

rL

ES

S

rgLe

E

JohnsonEuler

2

2

2

2

2

2

2

4

Design of Columns (6.6) Procedure

Calculate (Le/rg)c and Le/rg If Le/rg (Le/rg)c use Eulers equation

If Le/rg (Le/rg)c use Johnsons equation

Notice we havent included factor of safety yet

Column Design10

crge

cr

rLE

AP 2

2

)/(

2

2

2

4

g

eyy

cr

rL

ES

SAP

ExampleFind the required outer diameter, with F.S.=3, for the column shownbelow. Assume P = 15 kN, L = 50 mm, t = 5 mm, Sy = 372 MPa, E =207,000 MPa, and the material is 1018 cold-drawn steel.

Column Design11

t

do

ExampleFind the maximum allowable force, Fmax, that can be applied withoutcausing pipe to buckle. Assume L = 12 ft, b = 5 ft, do = 2 in, t = 0.5 in,and the material is low carbon steel.

Column Design12

t

do

F

L

b