Design of Compression Members

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68402

Slide #
1

Design of Compression Members

Monther Dwaikat

Assistant Professor

Department of Building Engineering

An
-
Najah National University


68402:
Structural Design of Buildings II

61420:
Design of Steel Structures

62323:
Architectural Structures II

68402

Slide #
2


Short and long columns



Buckling load and buckling failure modes



Elastic and Inelastic buckling



Local buckling



Design of Compression Members



Effective Length for Rigid Frames



Torsional and Flexural
-
Torsional Buckling



Design of Singly Symmetric Cross Sections


Design of Compression Members

68402

Slide #
3

Axially Loaded Compression Members



Columns



Struts



Top chords of trusses



Diagonal members of trusses


Column

and
Compression member

are often used
interchangeably

68402

Slide #
4

Axially Loaded Compression Members


Commonly Used Sections:


W/H shapes


Square and Rectangular or round HSS


Tees and Double Tees


Angles and double angles


Channel sections


68402

Slide #
5

Columns


Failure modes (limit states):


Crushing

(for short column)


Flexural

or
Euler Buckling

(unstable under bending)


Local Buckling

(thin local cross section)

68402

Slide #
6

Short Columns



Compression

Members
:

Structural

elements

that

are

subjected

to

axial

compressive

forces

only

are

called

columns
.

Columns

are

subjected

to

axial

loads

thru

the

centroid
.



Stress
:

The

stress

in

the

column

cross
-
section

can

be

calculated

as




f

-

assumed

to

be

uniform

over

the

entire

cross
-
section
.




Short

columns

-

crushing

A
P

f
68402

Slide #
7

Long Columns


This

ideal

state

is

never

reached
.

The

stress
-
state

will

be

non
-
uniform

due

to
:


Accidental eccentricity of loading with respect to the centroid


Member out
-
of

straightness (crookedness), or


Residual

stresses

in

the

member

cross
-
section

due

to

fabrication

processes
.


Accidental

eccentricity

and

member

out
-
of
-
straightness

can

cause

bending

moments

in

the

member
.

However,

these

are

secondary

and

are

usually

ignored
.


Bending

moments

cannot

be

neglected

if

they

are

acting

on

the

member
.

Members

with

axial

compression

and

bending

moment

are

called

beam
-
columns
.


“Long”

columns

68402

Slide #
8

Long Columns


The

larger

the

slenderness

ratio

(
L/r
),

the

greater

the

tendency

to

buckle

under

smaller

load


Factors

affecting

tendency

to

buckle
:


end

conditions


unknown

eccentricity

(concentric

&

eccentric

loads)


imperfections

in

material


initial

crookedness


out

of

plumbness


residual

stress


buckling

can

be

on

one

or

both

axes

(major

or

minor

axis)


68402

Slide #
9

Column Buckling


Consider

a

long

slender

compression

member
.

If

an

axial

load

P

is

applied

and

increased

slowly,

it

will

ultimately

reach

a

value

P
cr

that

will

cause

buckling

of

the

column
.

P
cr

is

called

the

critical

buckling

load

of

the

column
.


P
cr
P
cr
P
P
(a)
(b)
P
cr
P
cr
P
P
P
P
(a)
(b)
Figure 1.

Buckling of axially loaded compression members

68402

Slide #
10

Buckling Load


P
M



Now assume we have a pin connected column. If we apply a similar
concept to that before here we find

P
cr

P


The internal resisting moment M in the column
is

P

P
cr

M




We can write the relationship between the
deflected shape and the Moment M

EI
P
EI
M
dx
d
cr





2
2
0
2
2


EI
P
dx
d
cr


x



x



cr
P
-

The load at which bucking starts to happen
(Critical buckling load)

2
2
L
EI
P
cr


68402

Slide #
11

Column
Buckling


What

is

buckling?


Buckling

occurs

when

a

straight

column

subjected

to

axial

compression

suddenly

undergoes

bending

as

shown

in

the

Figure

1
(b)
.

Buckling

is

identified

as

a

failure

limit
-
state

for

columns
.



The

critical

buckling

load

P
cr

for

columns

is

theoretically

given

by

Equation

(
3
.
1
)
:



[
3
.
1
]


I
-

moment of inertia about axis of buckling.


K
-

effective length factor based on end boundary conditions.




2
2
L
K
I
E
P
cr


68402

Slide #
12

Effective Length


KL
-

Distance between inflection points in column.


K
-

Effective length factor


L
-

Column unsupported length

KL=L

KL=0.5L L

KL=0.7L L

K = 1.0

K = 0.5

K = 0.7

68402

Slide #
13

Column Buckling

Boundary
conditions

Table C
-
C2.2

Approximate Values of Effective Length Factor, K

68402

Slide #
14

Ex. 3.1
-

Buckling Loads


Determine

the

buckling

strength

of

a

W

12

x

50

column
.

Its

length

is

6

m
.

For

minor

axis

buckling,

it

is

pinned

at

both

ends
.

For

major

buckling,

is

it

pinned

at

one

end

and

fixed

at

the

other

end
.




x
y
68402

Slide #
15

Ex. 3.1
-

Buckling Loads

Step

I
.

Visualize

the

problem


For

the

W
12

x

50

(or

any

wide

flange

section),

x

is

the

major

axis

and

y

is

the

minor

axis
.

Major

axis

means

axis

about

which

it

has

greater

moment

of

inertia

(I
x

>

I
y
)
.


Step

II
.

Determine

the

effective

lengths


According

to

Table

C
-
C
2
.
2
:


For pin
-
pin end conditions about the minor axis


K
y

= 1.0 (theoretical value); and K
y

= 1.0 (recommended design
value)


For pin
-
fix end conditions about the major axis


K
x

= 0.7 (theoretical value); and K
x

= 0.8 (recommended design
value).


According to the problem statement, the unsupported length for
buckling about the major (x) axis = L
x

= 6 m.

68402

Slide #
16

Ex. 3.1
-

Buckling Loads


The unsupported length for buckling about the minor (y) axis = L
y

=
6 m.


Effective length for major (x) axis buckling = K
x

L
x

= 0.8 x 6 = 4.8 m.


Effective length for minor (y) axis buckling = K
y

L
y

= 1.0 x 6 = 6 m.



Step III.
Determine the relevant section properties


Elastic modulus of elasticity = E = 200 GPa (constant for all steels)


For W12 x 50:

I
x

= 163x10
6

mm
4
.

I
y

= 23
x10
6

mm
4





68402

Slide #
17

Ex. 3.1
-

Buckling Loads


Step IV.

Calculate the buckling strength



Critical load for buckling about x
-

axis = P
cr
-
x
=




P
cr
-
x

=



=

13965

kN
.



Critical load for buckling about y
-
axis = P
cr
-
y

=





P
cr
-
y

=




= 1261 kN.




Buckling strength of the column = smaller (P
cr
-
x
, P
cr
-
y
) =
P
cr

= 1261 kN.



Minor (y) axis buckling governs.





2 6
2
200 163 10
4800

  


2
2
y
y
y
L
K
I
E



2 6
2
200 23 10
6000

  


2
2
x
x
x
L
K
I
E

68402

Slide #
18

Ex. 3.1
-

Buckling Loads


Notes:


Minor axis buckling usually governs for all doubly
symmetric cross
-
sections. However, for some cases,
major (x) axis buckling can govern.


Note that the steel yield stress was irrelevant for
calculating this buckling strength.


68402

Slide #
19


Let

us

consider

the

previous

example
.

According

to

our

calculations

P
cr

=

1261

kN
.

This

P
cr

will

cause

a

uniform

stress

f

=

P
cr
/A

in

the

cross
-
section
.


For

W
12

x

50
,

A

=

9420

mm
2
.

Therefore,

for

P
cr

=

1261

kN
;

f

=

133
.
9

MPa
.


The

calculated

value

of

f

is

within

the

elastic

range

for

a

344

MPa

yield

stress

material
.


However,

if

the

unsupported

length

was

only

3

m,

P
cr

=would

be

calculated

as

5044

kN,

and

f

=

535
.
5

MPa
.


This

value

of

f

is

ridiculous

because

the

material

will

yield

at

344

MPa

and

never

develop

f

=

535
.
5

kN
.

The

member

would

yield

before

buckling
.



Inelastic Column Buckling


68402

Slide #
20

Inelastic Column Buckling


Eq
.

(
3
.
1
)

is

valid

only

when

the

material

everywhere

in

the

cross
-
section

is

in

the

elastic

region
.

If

the

material

goes

inelastic

then

Eq
.

(
3
.
1
)

becomes

useless

and

cannot

be

used
.


What happens in the inelastic range?


Several other problems appear in the inelastic range.


The

member

out
-
of
-
straightness

has

a

significant

influence

on

the

buckling

strength

in

the

inelastic

region
.

It

must

be

accounted

for
.


The

residual

stresses

in

the

member

due

to

the

fabrication

process

causes

yielding

in

the

cross
-
section

much

before

the

uniform

stress

f

reaches

the

yield

stress

F
y
.



The

shape

of

the

cross
-
section

(W,

C,

etc
.
)

also

influences

the

buckling

strength
.



In

the

inelastic

range,

the

steel

material

can

undergo

strain

hardening
.


All

of

these

are

very

advanced

concepts

and

beyond

the

scope

of

this

course
.

68402

Slide #
21

AISC Specifications for Column Strength



The

AISC

specifications

for

column

design

are

based

on

several

years

of

research
.



These

specifications

account

for

the

elastic

and

inelastic

buckling

of

columns

including

all

issues

(member

crookedness,

residual

stresses,

accidental

eccentricity

etc
.
)

mentioned

above
.


The

specification

presented

here

will

work

for

all

doubly

symmetric

cross
-
sections
.


The design strength of columns for the flexural buckling
limit state is equal to

c
P
n


Where,



c

= 0.9


(Resistance factor for
compression members)


68402

Slide #
22

Inelastic Buckling of Columns

P
F
F


Elastic

buckling

assumes

the

material

to

follow

Hooke’s

law

and

thus

assumes

stresses

below

elastic

(proportional)

limit
.


If

the

stress

in

the

column

reaches

the

proportional

limit

then

Euler’s

assumptions

are

violated
.


L/r

Stress “F”

Euler assumptions

2
2
)
/
(
r
L
E
F
cr


Elastic Buckling
(Long Columns)

Inelastic Buckling
(Short columns)

Proportional

limit

68402

Slide #
23

AISC Specifications for Column
Strength


P
u





P
n


P
n

=

A
g

F
cr






[E
3
-
1
]











[E
3
-
4
]







F
e
-

Elastic

critical

Euler

buckling

load


A
g

-

gross member area




K
-

effective length factor


L
-

unbraced length of the member



r
-

governing radius of gyration

The 0.877 factor in Eq (E3
-
3) tries to account for initial crookedness.



2
2
r
KL
E
F
e
















y
e
y
y
e
F
y
F
cr
F
E
r
KL
F
F
E
r
KL
F
F
71
.
4
877
.
0
71
.
4
658
.
0


Inelastic [E3
-
2]

Elastic [E3
-
3]

68402

Slide #
24

AISC Specifications For Column
Strength


c
=
E
F
r
L
K
y

F
cr
/
F
y
1.0
1.5
0.39
F
cr
=
F
y









2
c
877
.
0
F
cr
=
F
y


2
c
658
.
0


c
=
E
F
r
L
K
y


c
=
E
F
r
L
K
y

F
cr
/
F
y
1.0
1.5
0.39
F
cr
=
F
y









2
c
877
.
0
F
cr
=
F
y









2
c
877
.
0
F
cr
=
F
y


2
c
658
.
0

F
cr
=
F
y


2
c
658
.
0

r
KL
y
F
E
71
.
4
y
e
F
y
F
cr
F
F







658
.
0
e
cr
F
F
877
.
0

Inelastic Buckling
(Short columns)

Elastic Buckling
(Long columns)

68402

Slide #
25

AISC Specifications For Column
Strength



For

a

given

column

section
:


Calculate

I,

A
g
,

r



Determine

effective

length

K

L

based

on

end

boundary

conditions
.



Calculate

KL/r



If

KL/r

is

greater

than

,

elastic

buckling

occurs

and

use

Equation

(E
3
.
4
)


If

KL/r

is

less

than

or

equal

to

,

inelastic

buckling

occurs

and

use

Eq
.

(E
3
.
3
)


Note

that

the

column

can

develop

its

yield

strength

F
y

as

KL/r

approaches

zero
.


y
F
E
71
.
4
y
F
E
71
.
4
68402

Slide #
26

Ex. 3.2
-

Column Strength


Calculate the design strength of W14 x 74 with length of 6
m and pinned ends. A36 steel is used.



Step I.

Calculate the effective length and slenderness ratio for the
problem


K
x

= K
y

= 1.0


L
x

= L
y

= 6 m


Major axis slenderness ratio = K
x
L
x
/r
x

= 6000/153.4 = 39.1


Minor axis slenderness ratio = K
y
L
y
/r
y

= 6000/63 = 95.2



Step II.

Calculate the buckling strength for governing slenderness
ratio


The governing slenderness ratio is the larger of (K
x
L
x
/r
x
, K
y
L
y
/r
y
)



68402

Slide #
27

Ex. 3.2
-

Column Strength


K
y
L
y
/r
y

is larger and the governing slenderness ratio;














MPa



Therefore,





MPa


Design column strength =

c
P
n

= 0.9 (A
g

F
cr
) = 0.9 (14060x154)/1000=
1948.7 kN.


Design strength of column = 1948.7 kN.

200000
/4.71 4.71 133.7
248
y y y
y
E
K L r
F
  


0.658 154
y e
F F
cr y
F F
 


2 2
2
2
3.1416 200000
217.8
95.2
/
e
y y y
E
F
K L r


  
68402

Slide #
28

Local Buckling Limit State

Figure 4.

Local buckling of columns


The AISC
specifications for
column strength
assume that column
buckling is the
governing limit state.
However, if the
column section is
made of thin (slender)
plate elements, then
failure can occur due
to
local

buckling

of
the flanges or the
webs.


68402

Slide #
29

Local Buckling Limit State


Local buckling is another limitation
that represents the instability of the
cross section itself.




If local buckling occurs, the full
strength of the cross section can
not be developed.


68402

Slide #
30


If

local

buckling

of

the

individual

plate

elements

occurs,

then

the

column

may

not

be

able

to

develop

its

buckling

strength
.



Therefore,

the

local

buckling

limit

state

must

be

prevented

from

controlling

the

column

strength
.


Local

buckling

depends

on

the

slenderness

(width
-
to
-
thickness

b/t

ratio)

of

the

plate

element

and

the

yield

stress

(F
y
)

of

the

material
.



Each

plate

element

must

be

stocky

enough,

i
.
e
.
,

have

a

b/t

ratio

that

prevents

local

buckling

from

governing

the

column

strength
.



The

AISC

specification

provides

the

slenderness

(b/t)

limits

that

the

individual

plate

elements

must

satisfy

so

that

local

buckling

does

not

control
.





Local Buckling Limit State

68402

Slide #
31

Local Buckling Limit State


Local buckling can be prevented by limiting the width to thickness ratio
known as “

” to an upper limit

r


y
f
F
E
t
b
56
.
0



y
w
F
E
t
h
49
.
1



b
h
w
t
f
t
b
f
t
68402

Slide #
32


The

AISC

specification

provides

two

slenderness

limits

(

p

and


r
)

for

the

local

buckling

of

plate

elements
.


Local Buckling Limit State

Compact
Non
-
Compact
Slender
Compact
Non
-
Compact
Slender
b
t
F
Axial shortening,

Axial Force, F
F
y
Compact
Non
-
Compact
Slender
Compact
Non
-
Compact
Slender
b
t
F
Axial shortening,

Axial Force, F
F
y
68402

Slide #
33





If

the

slenderness

ratio

(b/t)

of

the

plate

element

is

greater

than


r

then

it

is

slender
.

It

will

locally

buckle

in

the

elastic

range

before

reaching

F
y


If

the

slenderness

ratio

(b/t)

of

the

plate

element

is

less

than


r

but

greater

than


p
,

then

it

is

non
-
compact
.

It

will

locally

buckle

immediately

after

reaching

F
y


If

the

slenderness

ratio

(b/t)

of

the

plate

element

is

less

than


p
,

then

the

element

is

compact
.

It

will

locally

buckle

much

after

reaching

F
y


If

all

the

plate

elements

of

a

cross
-
section

are

compact,

then

the

section

is

compact
.


If

any

one

plate

element

is

non
-
compact,

then

the

cross
-
section

is

non
-
compact


If

any

one

plate

element

is

slender,

then

the

cross
-
section

is

slender
.



Local Buckling Limit State

68402

Slide #
34

Local Buckling Limit State


Cross

section

can

be

classified

as


compact
”,


non

compact


or


slender


sections

based

on

their

width

to

thickness

ratios


If

the

cross
-
section

does

not

satisfy

local

buckling

requirements

its

critical

buckling

stress

F
cr

shall

be

reduced


If



then

the

section

is

slender,

a

reduction

factor

for

capacity

shall

be

computed

from







It

is

not

recommended

to

use

slender

sections

for

columns
.



r



AISC Manual for Steel
Design

68402

Slide #
35


The

slenderness

limits


p

and


r

for

various

plate

elements

with

different

boundary

conditions

are

given

in

the

AISC

Manual
.


Note

that

the

slenderness

limits

(

p

and


r
)

and

the

definition

of

plate

slenderness

(b/t)

ratio

depend

upon

the

boundary

conditions

for

the

plate
.



If

the

plate

is

supported

along

two

edges

parallel

to

the

direction

of

compression

force,

then

it

is

a

stiffened

element
.

For

example,

the

webs

of

W

shapes


If

the

plate

is

supported

along

only

one

edge

parallel

to

the

direction

of

the

compression

force,

then

it

is

an

unstiffened

element
.

Ex
.
,

the

flanges

of

W

shapes
.


The

local

buckling

limit

state

can

be

prevented

from

controlling

the

column

strength

by

using

sections

that

are

compact

and

non
-
compact
.


Avoid

slender

sections


Local Buckling Limit State

68402

Slide #
36

Local Buckling Limit State

68402

Slide #
37

Ex. 3.3


Local Buckling

200000
1.49 1.49 42.3
248
r
y
E
F

    

Determine

the

local

buckling

slenderness

limits

and

evaluate

the

W
14

x

74

section

used

in

Example

3
.
2
.

Does

local

buckling

limit

the

column

strength?



Step I. Calculate the slenderness limits


See
Tables in previous slide
.


For the flanges of I
-
shape sections in pure compression




For the webs of I
-
shapes section in pure compression





Use

E = 200000 MPa

200000
0.56 0.56 15.9
248
r
y
E
F

    
68402

Slide #
38

Ex. 3.3


Local Buckling


Step

II
.

Calculate

the

slenderness

ratios

for

the

flanges

and

webs

of

W
14

x

74


For the flanges of I
-
shape member, b = b
f
/2 = flange width / 2


Therefore, b/t = b
f
/2t
f
.




For W 14 x 74, b
f
/2t
f
= 6.41

(
See Section Property Table
)



For the webs of I shaped member, b = h


h is the clear distance between flanges less the fillet / corner radius of
each flange


For W14 x 74, h/t
w

= 25.4


(
See Section Property Table
).



Step III.
Make the comparisons and comment


For the flanges, b/t <

r
. Therefore, the flange is non
-
compact


For the webs, h/t
w

<

r
. Therefore the web is non
-
compact


Therefore, the section is non
-
compact


Therefore,
local buckling will not limit the column strength
.

68402

Slide #
39

Design of Compression Members


Steps for design of compression members


Calculate the factored loads P
u


Assume (a cross section) or (
KL/r ratio between 50 to 90)


Calculate the slenderness ratio
KL/r and the ratio F
e


Calculate

c
F
cr
based on value of

F
e



Calculate the Area required A
g



Choose a cross section and get (
K
x
L/r
x

)and (
K
y
L/r
y
) (
KL/r
)

max



Recalculate

c

F
cr

and thus check


Check local buckling requirements

cr
c
u
required
F
P
A


u
cr
c
g
n
c
P
F
A
P






2
2
r
KL
E
F
e


68402

Slide #
40

Ex. 3.4


Design Strength


Determine

the

design

strength

of

an

ASTM

A
992

W
14

x

132

that

is

part

of

a

braced

frame
.

Assume

that

the

physical

length

L

=

9

m,

the

ends

are

pinned

and

the

column

is

braced

at

the

ends

only

for

the

X
-
X

axis

and

braced

at

the

ends

and

mid
-
height

for

the

Y
-
Y

axis
.




Step I.

Calculate the
effective lengths
.


From Section Property Table


For W14 x 132:

r
x

= 159.5 mm;

r
y

= 95.5 mm;

A
g

=25030 mm
2



K
x

= 1.0

and

K
y

= 1.0


L
x

= 9 m

and

L
y

= 4.5 m


K
x
L
x

= 9 m and

K
y
L
y

= 4.5 m


68402

Slide #
41

Ex. 3.4


Design Strength



344 620.5
0.658 344 272.8
cr
F
  

Step

II
.

Determine

the

governing

slenderness

ratio


K
x
L
x
/r
x

=

9000
/
159
.
5

=

56
.
4


K
y
L
y
/r
y

=

4500
/
95
.
5

=

47
.
1


The

larger

slenderness

ratio,

therefore,

buckling

about

the

major

axis

will

govern

the

column

strength
.




Step III.

Calculate the column strength



2 2
2
2
200000
620.5
56.4
e
x x x
E
F
K L r
 

  
MPa

200000
/4.71 4.71 113.6
344
x x x
y
E
K L r
F
  
MPa

68402

Slide #
42

Ex. 3.4


Design Strength

5
.
13
56
.
0



y
r
F
E





Step IV.

Check the local buckling limits


For the flanges, b
f
/2t
f

= 7.15 <



For the web, h/t
w

= 17.7<



Therefore, the section is non
-
compact.
OK
.

5
.
13
56
.
0



y
r
F
E

0.9 25030 272.8 1000 6145
n
P kN
    
68402

Slide #
43

Ex. 3.5



Column Design


A

compression

member

is

subjected

to

service

loads

of

700

kN

DL

and

2400

kN

of

LL
.

The

member

is

7
.
8

m

long

&

pinned

at

each

end
.

Use

A
992

steel

and

select

a

W

shape
.



Step I.

Calculate the factored design load P
u


P
u

= 1.2 P
D

+ 1.6 P
L

= 1.2 x 700 + 1.6 x 2400 = 4680 kN.



Step II.

Calculate F
cr

by assuming KL/r = 80





2 2
2
2
200000
308.4
80
e
E
F
KL r
 

  
MPa



344 308.4
0.658 344 215.3
cr
F
  
200000
/4.71 4.71 113.6
344
y
E
KL r
F
  
MPa

68402

Slide #
44

Ex. 3.5



Column Design


Step

III
.

Calculate

the

required

area

of

steel


A = 4680*1000/(0.9*215.3) = 24156 mm
2



Step IV.
Select a W shape from the Section Property Tables


Select W14 x 132. It has A = 25030 mm
2

OR W12 x 136 A = 25740 mm
2


Select W14 x 132 because it has lower weight.


K
y
L
y
/r
y

=7800/95.5 = 81.7


F
e

= 295.7


F
cr

= 211.4


P
n

= 4897.3 kN


OK


W14 x 145 is the lightest.




Section

is

non
-
compact

but

students

have

to

check

for

that


Note

that

column

sections

are

usually

W
12

or

W
14
.

Generally

sections

bigger

than

W
14

are

not

used

as

columns
.



68402

Slide #
45

Effective Length


Specific Values of
K

shall be known


For compression elements connected as rigid frames the effective length
is a function of the relative stiffness of the element compared to the
overall stiffness of the joint. This will be discussed later in this chapter



Values for
K

for different end conditions range from 0.5 for
theoretically fixed ends to 1.0 for pinned ends and are given by:


Table C
-
C2.2 AISC Manual

End conditions

K

Pin
-
Pin

1.0

Pin
-
Fixed

0.8

Fixed
-
Fixed

0.65

Fixed
-
Free

2.1

Recommended
design values (not
theoretical values)

68402

Slide #
46


If we assume all connections are pinned
then:
K
x

L

= 3 m and
K
y

L

= 6 m


However the rigidity of the beams affect
the rotation of the columns. Thus in rigid
frames the K factor can be determined
from the relative rigidity of the columns


Determine a G factor

3
m
3
m
'
10

L
K
x
'
20

L
K
y



g
g
g
c
c
c
L
I
E
L
I
E
G
/
/

Where “c” represents column and “g” represents girder


The G value is computed at each end of the member and K is
computed factor from the monograms in




g
g
c
c
L
I
L
I
G
/
/
AISC Manual


Figure C
-
C2.2

K Factor for Rigid Frames

3 m

6 m

68402

Slide #
47

Effective Length of Columns in Frames



So

far,

we

have

looked

at

the

buckling

strength

of

individual

columns
.

These

columns

had

various

boundary

conditions

at

the

ends,

but

they

were

not

connected

to

other

members

with

moment

(fix)

connections
.



The

effective

length

factor

K

for

the

buckling

of

an

individual

column

can

be

obtained

for

the

appropriate

end

conditions

from

Table

C
-
C
2
.
2

of

the

AISC

Manual

.


However,

when

these

individual

columns

are

part

of

a

frame,

their

ends

are

connected

to

other

members

(beams

etc
.
)
.



Their

effective

length

factor

K

will

depend

on

the

restraint

offered

by

the

other

members

connected

at

the

ends
.



Therefore,

the

effective

length

factor

K

will

depend

on

the

relative

rigidity

(stiffness)

of

the

members

connected

at

the

ends
.

68402

Slide #
48

Effective Length of Columns in Frames


The

effective

length

factor

for

columns

in

frames

must

be

calculated

as

follows
:


First,

you

have

to

determine

whether

the

column

is

part

of

a

braced

frame

or

an

unbraced

(moment

resisting)

frame
.



If

the

column

is

part

of

a

braced

frame

then

its

effective

length

factor

0

<

K



1


If

the

column

is

part

of

an

unbraced

frame

then

1

<

K






Then,

you

have

to

determine

the

relative

rigidity

factor

G

for

both

ends

of

the

column


G

is

defined

as

the

ratio

of

the

summation

of

the

rigidity

(EI/L)

of

all

columns

coming

together

at

an

end

to

the

summation

of

the

rigidity

(EI/L)

of

all

beams

coming

together

at

the

same

end
.





It

must

be

calculated

for

both

ends

of

the

column





b
b
c
c
L
I
E
L
I
E
G
c: for columns

b: for beams

68402

Slide #
49

Effective Length of Columns in Frames


Then,

you

can

determine

the

effective

length

factor

K

for

the

column

using

the

calculated

value

of

G

at

both

ends,

i
.
e
.
,

G
A

and

G
B

and

the

appropriate

alignment

chart



There

are

two

alignment

charts

provided

by

the

AISC

manual,



One

is

for

columns

in

braced

(sidesway

inhibited)

frames
.

0

<

K



1


The

second

is

for

columns

in

unbraced

(sidesway

uninhibited)

frames
.

1

<

K






The

procedure

for

calculating

G

is

the

same

for

both

cases
.


68402

Slide #
50

Effective Length

Monograph or

Jackson and Moreland

Alignment Chart

for Unbraced Frame


68402

Slide #
51

Effective Length

Monograph or

Jackson and Moreland

Alignment Chart

for braced Frame


68402

Slide #
52

Ex. 3.6


Effective Length Factor


Calculate

the

effective

length

factor

for

the

W
12

x

53

column

AB

of

the

frame

shown
.

Assume

that

the

column

is

oriented

in

such

a

way

that

major

axis

bending

occurs

in

the

plane

of

the

frame
.

Assume

that

the

columns

are

braced

at

each

story

level

for

out
-
of
-
plane

buckling
.

Assume

that

the

same

column

section

is

used

for

the

stories

above

and

below
.


10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
5.4 m

5.4 m

6 m

4.5 m

3.6 m

3.0 m

3.0 m

68402

Slide #
53

Ex. 3.6


Effective Length Factor


Step

I
.

Identify

the

frame

type

and

calculate

L
x
,

L
y
,

K
x
,

and

K
y

if

possible
.


It

is

an

unbraced

(sidesway

uninhibited)

frame
.



L
x

=

L
y

=

3
.
6

m


K
y

=

1
.
0


K
x

depends

on

boundary

conditions,

which

involve

restraints

due

to

beams

and

columns

connected

to

the

ends

of

column

AB
.


Need

to

calculate

K
x

using

alignment

charts
.



Step II.
Calculate K
x


I
xx

of W 12 x 53 = 425 in
4


I
xx

of W14x68 = 753 in
4


021
.
1
360
.
6
493
.
6
12
20
723
12
18
723
12
12
425
12
10
425
L
I
L
I
G
b
b
c
c
A












68402

Slide #
54

Ex. 3.6


Effective Length Factor





Using G
A

and G
B
: K
x

= 1.3

-

from
Alignment Chart on Page
16.1
-
242



Step III.
Design strength of the column


K
y
L
y

= 1.0 x 12 = 12 ft.


K
x

L
x

= 1.3 x 12 = 15.6 ft.


r
x

/ r
y

for W12x53 = 2.11


(KL)
eq

= 15.6 / 2.11 = 7.4 ft.


K
y
L
y

> (KL)
eq


Therefore, y
-
axis buckling governs. Therefore

c
P
n

= 547 kips


835
.
0
360
.
6
3125
.
5
12
20
723
12
18
723
12
15
425
12
12
425
L
I
L
I
G
b
b
c
c
B












68402

Slide #
55

Ex. 3.8


Column Design





Design

Column

AB

of

the

frame

shown

below

for

a

design

load

of

2300

kN
.



Assume

that

the

column

is

oriented

in

such

a

way

that

major

axis

bending

occurs

in

the

plane

of

the

frame
.



Assume

that

the

columns

are

braced

at

each

story

level

for

out
-
of
-
plane

buckling
.


Assume

that

the

same

column

section

is

used

for

the

stories

above

and

below
.



Use

A
992

steel
.





68402

Slide #
56

Ex. 3.8


Column Design



10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
5.4 m

5.4 m

6 m

4.5 m

3.6 m

3.0 m

3.0 m

68402

Slide #
57

Ex. 3.8


Column Design


Step I
-

Determine the design load and assume the steel material.


Design Load = P
u

= 2300 kN.


Steel yield stress = 344 MPa (A992 material).



Step II.
Identify the frame type and calculate L
x
, L
y
, K
x
, and K
y

if
possible.


It is an unbraced (sidesway uninhibited) frame.


L
x

= L
y

= 3.6 m


K
y

= 1.0


K
x

depends on boundary conditions, which involve restraints due
to beams and columns connected to the ends of column AB.


Need to calculate K
x

using alignment charts.


Need to select a section to calculate K
x

68402

Slide #
58

Ex. 3.8


Column Design


Step III
-

Select a column section


Assume minor axis buckling governs.


K
y

L
y

= 3.6 m


Select section W12x53


K
y
L
y
/r
y

= 57.2

F
e

= 604.4

F
cr

= 271.1



c
P
n

for y
-
axis buckling = 2455.4 kN



Step IV
-

Calculate K
x


I
xx

of W 12 x 53 = 177x10
6

mm
4




I
xx

of W14x68 = 301x10
6

mm
4





68402

Slide #
59

Ex. 3.8


Column Design










Using GA and GB
: K
x

= 1.3

-

from Alignment Chart

177 177
3 3.6
1.02
301 301
5.4 6
c
c
A
b
b
I
L
G
I
L
 

 
 
  



177 177
3.6 4.5
0.836
301 301
5.4 6
c
c
B
b
b
I
L
G
I
L
 

 
 
  



68402

Slide #
60

Ex. 3.8


Column Design


Step V
-

Check the selected section for X
-
axis buckling


K
x

L
x

= 1.3 x 3.6 = 4.68 m


K
x

L
x
/r
x

= 35.2


F
e

= 1590.4

F
cr

= 314.2



For this column,

c
P
n

for X
-
axis buckling = 2846.3



Step VI
-

Check the local buckling limits


For the flanges, b
f
/2t
f

= 8.69 <




For the web, h/t
w

= 28.1

<




Therefore, the section is non
-
compact. OK, local buckling is not a
problem

5
.
13
56
.
0



y
r
F
E

9
.
35
49
.
1



y
r
F
E