NYIT Instructors: Alfred Sanabria and Rodrigo...

NYIT

Instructors: Alfred Sanabria and Rodrigo Suarez

Massive stone columns, used

from Stonehenge to Ancient

Greece were stabilized by

their own work

With steel and concrete

technology columns have

become increasingly slender

Slenderness is a

property that relates a

column length with its

cross-sectional

dimensions.

Slenderness is critical

in the strength of

columns due to

buckling

Buckling is the sudden uncontrolled lateral displacement of a column, at which point no additional load can be supported.

A sudden failure mode for members under high compressive stresses that occurs at a load that is lower than the maximum capacity of the material yield strength.

http://www.youtube.com/watch?v=TUE7DKN

BIrU

Short columns fail by crushing due to

material failure (yield), long, slender columns

fail by buckling

Since short columns fail by crushing (material

failure) the formula for the compressive

stress is simply:

allowableactual

aA

Pσσ ≤=



stress is simply:

allowableactual

aA

Pσσ ≤=

Axial Load on Column

Allowable Axial Stress

Column Cross-

Sectional Area

Actual Axial Stress



stress is simply:

allowable

actualrequired

PA

σ≥

Axial Load on Column

Allowable Axial Stress Column Cross-

Sectional Area

Buckling happens because loads aren’t perfectly axial and there are material irregularities that develop some bending in the member.

Euler’s equation:

2

min

2

L

EIPcritical

π=

Buckling happens because loads aren’t perfectly axial and there are material irregularities that develop some bending in the member.

Euler’s equation:

2

min

2

L

EIPcritical

π=

Effective Length of the column

Axial Load at

which the

column buckles

The radius of gyration of a column cross

section is a measure of the distribution of

area (mass) around the centroid of the cross

section

2ArI

A

Ir

=

=

Radius of gyration Moment of Inertia (smallest)

Cross-Sectional Area




section

2

2

2

22

2

min

2

)/(

)(

rL

E

AL

ArE

LA

EI

A

Pcriticalcritical

πππσ ====

2ArI =




section

2

2

)/( rL

Ecritical

πσ =

Slenderness ratio

For columns, the radius of gyration and moment of inertia that matter are the smaller ones.

The higher the slenderness ratio (long, thin column), the lower the load at which the column will buckle

The most efficient columns have rx=ry W sections used as columns are usually “boxy”

and 10” to 14” deep

Determine the critical buckling stress for a 30’

long w12x65 steel column. Assume simple pin

connections at the top and bottom.

Fy=36ksi (A36 steel)

E=29,000 ksi

rx=5.28”

ry=3.02”





E=29,000 ksi

rx=5.28”

ry=3.02”

2

2

)( rL

Ecrit

πσ =

2.119

2.11902.3

)1230(

19.6828.5

)1230(

Use

ininx

rL

ininx

rL

y

x

==

==





E=29,000 ksi

rx=5.28”

ry=3.02”

ksi

ksi

rL

Ecrit

1.20

)2.119(

)000,29(

)(

2

2

2

2

=

=

=

π

πσ

In the simple analysis, columns were assumed

to be pinned at either end, so the column

would buckle in a smooth curve.

The length that is free to buckle is greatly

influenced by the end supports.

A K factor is used to increase or reduce the

effective length of the column to take the end

supports into account.

K= 1.0 0.5 0.7 2.0

2

min

2

)(KL

EIPcritical

π=

K= 1.0 0.5 0.7 2.0

K=1

K=0.7 K=2

K=0.5

Just as rigid end connections reduce the buckling length of a column, lateral bracing can also increase the column capacity by reducing the buckling length.

Lateral bracing is usually provided in the weak buckling direction

Lb

Lb

2

2

)(KL

EIPcritical

π=

2

2

)(KL

EIPcritical

π=

Slenderness ratios

must be calculated

for both axes to

determine which

direction governs

2

2

)(KL

EIPcritical

π=

Determine the critical buckling load for a 4x8 S4S Douglas Fir column that is 18’ long and braced at midheight against the weak direction of buckling. E=1,300 ksi


6


2

2

)(KL

EIPcritical

π=


2

2

)(KL

EIPcritical

π=

Strong Axis Weak Axis

4

3

3

72

)6)(4)(12/1(

121

in

bhI

=

=

=

4

3

3

32

)4)(6)(12/1(

121

in

bhI

=

=

=


2

2

)(KL

EIPcritical

π=


in

ftinxLb

216

)/12('18

=

=

in

ftinxLb

108

)/12('9

=

=

4

3

3

72

)6)(4)(12/1(

121

in

bhI

=

=

=

4

3

3

32

)4)(6)(12/1(

121

in

bhI

=

=

=


2

2

)(KL

EIPcritical

π=


432inI =

472inI =

inLb 216= inLb 108=

k

in

inksi

KL

EIPcritical

8.19

)]216)(1[(

)72)(300,1(

)(

2

42

2

2

=

=

=

π

π

k

in

inksi

KL

EIPcritical

2.35

)]108)(1[(

)32)(300,1(

)(

2

42

2

2

=

=

=

π

π

What are the two conditions that need to be investigated?


Weak Axis


Weak Axis

Strong Axis

In reality columns do not transition abruptly from short (crushing) to long (buckling.

There is an intermediate column range in which columns fail by a combination of buckling and crushing.

The analysis and design of steel columns takes this into account

AISC only recognizes two types of columns

for design, short/intermediate and long

The value for Kl/r where columns transition

from short/intermediate to long is known a Cc

and defined as:

y

cF

EC 71.4=

y

cF

EC 71.4=

)50(43.113 ksiCc =

)36(78.133 ksiCc =

Cc is the theoretical

value between

inelastic and elastic

behavior.

)50(43.113 ksir

KL≤

)50(43.113 ksir

KLCc ==

)50(43.113 ksir

KL>

y

F

F

cr FF e

y

= 658.0

2

2

==

r

KL

EFecritical

πσ

ecr FF 877.0=

)(67.1 ASDc =Ω

Design K-Values from AISC

Table 4-1

Table C-36

Table C-50

Find the maximum Axial Load on your column-P(actual) Find the KLx and KLy Use Table 4-1 to find a trial section using the KLy (Tables are only

for KLy) Find KLx (equivalent) by using rx/ry from table Use LARGEST KL/r for the rest of the design Using the largest KL/r, enter Table 9.1 or 9.2 to obtain a respective

Fa Calculate the P(allowable)=(Fa)x(A) of the trial section Check to see if P(allowable)>P(actual) If the P(allowable) is too low, pick a larger section and repeat If P(allowable) is larger then P(actual), check

efficiency…P(actual)/P(allowable) should be around 0.8 or 0.9 If your column seems overdesigned, pick a smaller section and

start again Repeat until you find an ADEQUATE and EFFICIENT section.

Find the maximum Axial Load on your column-P(actual) Find the KLx and KLy Use Table 4-1 to find a trial section using the KLy (Tables are only

for KLy) Find KLx (equivalent) by using rx/ry from table Use LARGEST KL/r for the rest of the design Figure out which of the two equations you need to use based on

Cc Calculate F(critical) using the appropriate equation Calculate the P(allowable)=(Fcr/Ωc)x(A) of the trial section Check to see if P(allowable)>P(actual) If the P(allowable) is too low, pick a larger section and repeat If P(allowable) is larger then P(actual), check

efficiency…P(actual)/P(allowable) should be around 0.8 or 0.9 If your column seems overdesigned, pick a smaller section and

start again Repeat until you find an ADEQUATE and EFFICIENT section.

NYIT Instructors: Alfred Sanabria and Rodrigo...

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