Chapter 10 Columns. 10.1 Introduction Column = vertical prismatic members subjected to compressive forces Goals of this chapter: 1.Study the stability.

Chapter 10 Columns 

10.1 Introduction

Column = vertical prismatic members subjected to compressive forces

Goals of this chapter:

1. Study the stability of elastic columns

2. Determine the critical load Pcr

3. The effective length

4. Secant formula

Previous chapters: -- concerning about

(1) the strength and

(2) excessive deformation (e.g. yielding)This chapter:

-- concerning about

(1) stability of the structure (e.g. bucking)

10.2 Stability of Structures

allow

PA

cr

PLAE

22( )sin ( )crLP K (10.1)

sin (10.2)

Concerns before:

New concern: Stable?

Unstable?

22( ) ( )crLP K

sin

4 /crP K L (10.2)

The system is stable, if

The system is unstable if

Since

4 /crP K L

4 /crP K L

A new equilibrium state may be established

22( )sin ( )LP M K

4 sinPLK

The new equilibrium position is:

4sinPLK

(10.3)

or

After the load P is applied, there are three possibilities:

1. P < Pcr – equilibrium & = 0 -- stable

2. P > Pcr – equilibrium & = -- stable

3. P > Pcr – unstable – the structure

collapses, = 90o

10.3 Euler’s Formula for Pin-Ended Columns

Determination of Pcr for the configuration in Fig. 10.1 ceases to be stable

2

2

d y M Py

dx EI EI

2

2 0d y P

ydx EI

Assume it is a beam subjected to bending moment:

(10.5)

(10.4)

2 Pp

EI

22

2 0d y

p ydx

sin cosy A px B px

Defining:

The general solution to this harmonic function is:

(10.7)

(10.6)

(10.8)

B.C.s:

@ x = 0, y = 0 B = 0

@ x = L, y = 0

Eq. (10.8) reduces to

0sinA pL (10.9)

0sinA pL

2 22

2

np

L

2 Pp

EI

(10.9)

Therefore,

1. A = 0 y = 0 the column is straight!

2. sin pL = 0 pL = n p = n /L

Since (10.6)

We have2 2

2

n EIP

L

For n = 12

2cr

EIP

L

(10.10)

-- Euler’s formula (10.11)

Substituting Eq. (10.11) into Eq. (10.6),

2

2cr

EIP

L

2 Pp

EI

2 22

2 2crP EI

pEI L EI L

(10.6)

(10.11)

Therefore,

Hence pL

Equation (10.8) becomes sinx

y AL

(10.12)

This is the elastic curve after the beam is buckled.

If P < Pcr sin pL 0

0sinA pL

1. A = 0 y = 0 the column is straight!

2. sin pL = 0 pL = n 2

2cr

EIP

L

(10.9)

Hence, A = 0 and y = 0 straight configuration

Critical Stress:

Introducing

2

2( )cr

ELr

2

2cr

cr

P EIA L A

2I Ar

Where r = radius of gyration

, ,y zxx y z

I IIr r r

A A A

Where r = radius of gyration (10.13)

L/r = Slenderness ratio

10.4 Extension of Euler’s Formula to columns with

Other End Conditions

2

2cre

EIP

L

2

2( )cr

e

ELr

(10.11')

(10.13')

Case A: One Fixed End, One Free End

Le = 2L

Case B: Both Ends Fixed

At Point C

RCx = 0

Q = 0 0VQIt

Point D = inflection point M = 0 AD and DC are symmetric

Hence, Le = L/2

2

2

d y M P Vy x

dx EI EI EI

2 Pp

EI

22

2

d y Vp y x

dx EI

Case C: One Fixed End, One Pinned End

M = -Py - Vx

Since

Therefore,

The general solution:

The particular solution:

sin cosy A px B px

2

Vy x

p EI

2 Pp

EI

Vy x

P

Substituting into the particular solution, it follows

As a consequence, the complete solution is

sin cosV

y A px B px xP

(10.16)

sinV

A pL LP

sinV

y A px xP

(10.16)

B.C.s:

@ x = 0, y = 0 B = 0

@ x = L, y = 0

(10.17)

Eq. (10.16) now takes the new form

sin cosV

y A px B px xP

Taking derivative of the question,

cosdy V

Ap pxdx P

cosV

Ap pLP

sinV

A pL LP

B.C.s: @ x = L, dy/dx = = 0

(10.18)

(10.17)

10 1710 18

( . )( . ) tan pL pL (10.19)

Solving Eq. (10.19) by trial and error,

2 24 4934 4 4934 20 19064. . / . /pL p L p L

2 22

20 19.P EIp P p EI

EI L

2

2 2

20 19.cr

e

EI EIP

L L

Since

Therefore,

Case C

Solving for Le

Le = 0.699L 0.7 L

Summary

10.5* Eccentric Loading; the Scant Formula

Secant Formula:

(10.36)

2

11

2

max

sec( )e

PA ec P L

r EA r

11

2sec( )e

P LEA r

If Le/r << 1,

Eq. (10.36) reduces to

21

maxPecAr

(10.37)

10.6 Design of Columns under a Centric

Load

10.6 Design of Columns under a Centric LoadAssumptions in the preceding sections:

-- A column is straight

-- Load is applied at the center of the column

-- < y

Reality: may violate these assumptions

-- use empirical equations and rely lab data

Facts:

1. Long Columns: obey Euler’s Equation

2. Short Columns: dominated by y

3. Intermediate Columns: mixed behavior

Test Data:

Empirical Formulas:

Real Case Design using Empirical Equations:

Two Approaches:

1. Allowable Stress Design

2. Load & Resistance Factor Design

1. For L/r Cc [long columns]:

[Euler’s eq.]

Structural Steel – Allowable Stress Design

Approach I -- w/o Considering F.S.

2

2( / )cr

EL r

2

212

( / )[ ]cr Y

c

L rC

2. For L/r Cc [short & interm. columns]:

where2

2 2c

Y

EC

1. L/r Cc :

(10.43)

Approach II -- Considering F.S.

2

21 92. . ( / )cr

all

EF S L r

2. L/r Cc :

211

2/

[ ( ) ]. . .cr Y

allc

L cF S F S C (10.45)

10.7 Design of Columns under an Eccentric Load

centric bending

max

P McA I

(10.56)

(10.57)

(I) Allowable Stress Method

(II) Interaction Method

Two Approaches:

1. The section is far from the ends

2. < y

I. Allowable-Stress Method

all

P McA I

(10.58)

-- all is obtained from Section 10.6.

-- The results may be too conservative.

II. Interaction Method

1/ /

all all

P A Mc I

1/ /

( ) ( )all centric all bending

P A Mc I

(10.59)

(10.60)(Interaction Formula)

all centric -- determined using the largest Le

Case A: If P is applied in a plane of symmetry:

1max max/ //( ) ( ) ( )

x x x x

all centric all bending all bending

M z I M x IP A

(10.61)

Case B: If P is NOT Applied in a Plane of Symmetry: