CEE380

Elementary Structures II

Columns-1

columns-1.ppt 2

Focus

• Overview of column behavior: short,

long, intermediate

• Euler buckling formula for long columns

• Concept of slenderness ratio

• How to determine axial design strength

using tables from AISC

columns-1.ppt 3

Mathematical Models

• Short or stub: Failure when the

compressive strength of the material is

exceeded

• Long: Failure by buckling

• Intermediate: Combination of strength

and stiffness

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Design Strategies

• Short

– Keep compressive stress of P/A below a

specified value

• Long

– Avoid buckling behavior

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Euler Critical Buckling Load

�

PCR

=!

2EI

KL( )2

K = ratio of effective column length to actual unbraced length

L = unbraced length [in]

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Derivation of KL

Use equilibrium to

solve for the

buckling load using

this FBD of a fixed-

pinned column

columns-1.ppt 7

�

M0! = "Py + Rx" EI # # y = 0

# # y + PEI( )y = R

EI( )x

y = A sin PEI

x$

% &

'

( ) + B cos P

EIx

$

% &

'

( ) +

Rx

P

columns-1.ppt 8

Use boundary conditions to evaluate constants and reaction

at x = 0, y = 0 ! B = 0at x = L, y = 0 :

0 = Asin PEI( )L"

#$%&'+RL

P

A =(RL

P sin PEI( )L"

#$%&'

at x + L, )y = 0 and since B = 0:

0 = PEI( )Acos P

EI( )L"#$

%&'+R

P

A =(R

P PEI( )cos P

EI( )L"#$

%&'

columns-1.ppt 9

Set the two equations for “A” equal to each

other.

�

PEI

L = tan PEI

L!

" #

$

% &

PEI

L ' 0; i.e., it can't be negative

4.493 = tan(4.493)

The only solution is in the third quadrant.

( PEI

L = 4.493

P = 4.493( )2 EI

L2( )

Rewriting, factoring out ) 2 ,we obtain

PCR

=) 2EI

0.7L( )2

where K = 0.7 * effective length factor

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K values from the steel manual: This table

is in your handout for columns.

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Design

Design Equation:

Pu ! "cPn

"cPn = "cAgFcr with "c = 0.85

Slenderness Ratio:

KL

r

KL = effective length

r=minimum radius of gyration

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AISC design strategies

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Examples of using the columns tables

Using the columns tables:

Fy = 50ksi : Find !cPn for the following members.

W12 x 106, KL = 20 ' :!cPn = 858k

W14 x 109, KL = 14 ' :!cPn = 1170k

W14 x 53, KL = 21' :!cPn = ?

KL = 20 ',!cPn = 213kKL = 21',!cPn = ?KL = 22 ',!cPn = 176k

20 " 21

20 " 22=

213" ?

213"176so ? = 194.5k

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Example using Table 3-50

GIVEN:W12x87, L=30ft, Simply-Supported ("pinned") ends

FIND: !cPn

SOLUTION:K = 1.0 from Table C-C2.1

KL = 30 ft

W12x87, radius of gyration ry = 3.07 inches from W Tables; Area A = 25.6 in2

KL

r=

30(12)

3.07= 117.3

Using Table 3-50

KL = 117, !cFCR = 15.6ksiKL = 118, !cFCR = 15.3ksi

so !cFCR = 15.5ksi

and !cPn = 25.6in2 (15.5ksi) = 396.8k

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Select the lightest available W 12 sections for the following columns:

a) Pu = 500k, L = 14 ft, pinned end supports

K = 1.0 for pinned-pinned (Table C-C2.1); KL = 14, Fy = 50 ksi, W12x58 is OK, !cPn=521k

b) Pu = 360 k, L = 12 ', fixed end supports

K = 0.65 for fixed-fixed (Table C-C2.1); KL = 7.8 '; round up to 8', Fy = 50 ksi,

W12x40,!cPn = 416k

c)Pu = 700k, L = 16.5 ft, fixed at bottom, pinned on top

K = 0.80 (Table C-C2.1); KL = 13.2', Fy = 50 ksi,

KL = 13 : W12x72,!cPn = 740k

KL = 14 : W12x72,!cPn = 717ksoKL = 13.2,!cPn = 735.2k

Examples using the Columns Tables

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Design Example

Enter the column load table for W shapes at an effective length of KL=16 ft.

Select W14 x 99, good for 1040 kips > 1000 kips.

rx/ry = 1.66

Equivalent effective length for X-X axis:

31/1.66 = 18.6 ft

Since 18.6 > 16 ft, X-X axis controls.

Re-enter table for effective length of 19 ft to satisfy axial load of 1000 kips, select

W 14 x 109.

Design the lightest W shape of Fy=50 ksi steel to support a

factored concentric load [i.e. Pu] of 1000 kips.

The effective length with respect to its minor axis is 16 ft.

The effective length with respect to the major axis is 31 ft.

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Summary

• More example problems with complex end conditions provided on hand-written

sheets.

• Overview of column behavior: short, long, intermediate

• Euler buckling formula

• Concept of slenderness ratio

• How to determine axial design strength using tables from AISC