Introduction to Axially Loaded Compression Members

8/12/2019 Introduction to Axially Loaded Compression Members

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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 2ENCE 355 Assakkaf

As the effective length of a columnincreases, its buckling stress willdecrease.The steel column is said to failelastically if the buckling stress is lessthan the proportional limit of steel whenthe effective length exceeds a certainvalue.

Long, Short, and Intermediate

Columns


Long Columns Long columns usually fails elastically. The Euler formula predicts very well the

strength of long columns where the axialcompressive buckling stress remains

below the proportional limit.

Long, Short, and IntermediateColumns


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Short Columns The failure stress equals to the yield stress

for short columns. For a column to fall into this class, it would

have to be so short as to have no practicalapplication.

Long, Short, and Intermediate

Columns


Intermediate Columns For intermediate columns some of the

fibers will reach the yield stress and somewill not.

The member will fail by both yielding and

buckling, and their behavior is said to beinelastic. Most columns fall into this range.

Long, Short, and IntermediateColumns


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Column Formulas

The Euler formula is used by the AISCLRFD Specification for long columnswith elastic buckling.Other empirical (based on testing)equations are used by the LRFD forshort and intermediate columns.With these equations, a critical orbuckling stress F cr is determined for acompression element.


Column Formulas

LRFD General Design Equation forColumns

The design strength of a compressionmember is determined as follows:

0.85 with ==

cr g cnccr g n

F A P

F A P (1)


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Column Formulas

LRFD Critical Buckling StressTwo equations are provided by the LRFDfor the critical buckling stress F cr :

( )

>

=

5.1for877.0

5.1for658.0

2

2

c yc

c y

cr F

F

F

c

(2)


Column FormulasLRFD Critical Buckling Stress

The limiting c value is given by

Where F e = Euler buckling stress =

Hence,

e

yc F

F =

( )22

/ r KL

E

( )

( ) E

F

r KL

E

r KL F

r KL E

F

F

F y y y

e

yc

==== 22

2

2

/

/

(3)

(4)


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Column Formulas

LRFD Critical Buckling StressSo the limiting c value to be used in Eq. 2is given by

where

F y = yield strength of material (steel) F e = Euler critical buckling stress

E

F

r KL

F

F y

e

yc

== (5)


Column FormulasLRFD Critical Buckling Stress For inelastic flexural buckling, Eq. 2 can be

used to compute the critical buckling stress F cr when c 1.5.

For elastic flexural buckling, Eq. 2 can beused to compute the critical buckling stress

F cr

when c

> 1.5. Eq. 2 include the estimated effects of

residual stresses and initial out-of-straightness of the members.

Eq. 2 is presented graphically in Fig. 1.


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Column Formulas

Figure 1. LRFD Critical Buckling StressShortcolumn

Intermediatecolumn Long column

5.1=c

Inelastic buckling

Elastic buckling(Euler Formula)

r KL

cr c F


Column FormulasLRFD Critical Buckling Stress To facilitate the design process, the LRFD

Manual provides computed values c F cr values for steels with F y = 36 ksi and 50 ksifor KL/r from 1 to 200 and has shown theresults in Tables 3.36 and 3.50 of theLRFD Specification located in Part 16 ofthe Manual.

Also, there is Table 4 of the LRFDSpecification from which the user mayobtain values for steel with any F y values.


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Column Formulas

LRFD Manual Design Tables (P. 16. I-143)


Column FormulasLRFD Manual Design Tables (P. 16. I-145)


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Column Formulas

LRFD Manual Design Tables (P. 4-25)


Maximum Slenderness RatiosCompression members preferablyshould be designed with

as specified in Section B7 of the LRFD

Manual.

200r

KL (6)

Note that LRFD Tables 3.36 and 3.50 give a value of 5.33 ksi for

the design stress c F cr when KL/r = 200. If KL/r > 200, it isthen necessary to substitute into the column formulas to get thethe stress.


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Maximum Slenderness Ratios

More Simplification by the LRFDManual for Design It is to be noted that the LRFD Manual in

its Part 4 has further simplified thecalculations required by computing thecolumn design strength c F cr A g for each ofthe shapes normally used as columns forcommonly used effective lengths or KLvalues.

These were determined with respect to theleast radius of gyration for each section.


Example ProblemsExample 1

a. Using the column design stress valuesshown in Table 3.50, part 16 of the LRFDmanual, determine the design strength, c P nof the F y = 50 ksi axially loaded columnshown in the figure.

b. Repeat the problem using the column tablesof part 4 of the Manual.c. Check local buckling for the section selected

using the appropriate values from Table 5.2.


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Example Problems

Example 1 (contd)

7212W ft15

ncu P P

ncu P P


Example ProblemsExample 1 (contd)

a. The properties of the W12 72 areobtained from the LRFD Manual as

in430.0 in27.1

in670.0 in00.12 in3.12

in04.3 in31.5 in1.21 2

========

w

f f

y x

t k

t bd

r r A

( ) ( )

( )37.47

04.3151280.0

and

controls ,in31.5in04.3SinceText)5.1,(Table1Tablefrom80.0

==

=


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Example Problems

Example 1 (contd)P. 16. I-145



Therefore,

b. Entering column tables Part 4 of theLRFD Manual with K y L y in feet:

( ) k 5.7611.2109.36 === g cr cnc A F P

( )k 761

ft121580.0

====

ncu

y y

P P

L K


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P. 4-25



c. Checking W12 72 for compactness:For flange

For web, noting h = d 2k = 12.3 2 (1.27) =9.76 in

( )49.13

501029

56.056.096.8670.020.12

2

3

==


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Table 2. Limiting Width-Thickness Ratios forCompression Elements

Example Problems


Table 3. (contd) Limiting Width-ThicknessRatios for Compression Elements

Example Problems


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Example Problems

Example 2Determine the design strength c P n of theaxially loaded column shown in the figure if

KL = 19 ft and 50 ksi steel is used.20

21

PL

y

x x

in12

42.7MC18

in50.18

C]of backfromin877.0

,in3.14,in554

in,0.18,in6.12[44

2

=====

x

I I

d A

y x



( )

( )( ) ( )( )in87.6

2.355.96.12225.0200.5

topfrom

in2.356.12221

20 2

=+=

=+

=

y

A

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( )

in64.62.35

1554

in92.62.35

1688

in155412

205.0877.066.1223.142

in688,125.069.61012

5.02069.625.96.1225542

43

2

423

2

===

===

=+++=

=+++=

A

I r

A I

r

I

I

y y

x x

y

x

Controls


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For KL/r = 34 and 35, Table 3-50 of theLRFD Manual, Page 16.I-145, givesrespectively the following values for c F cr :39.1 ksi and 38.9 ksi

( )34.34

64.61912 ===

yr KL

r KL

ksi03.393435

3434.341.399.381.39

9.3835

34.34

1.3934=

=

cr ccr c

cr c F F

F

Therefore, the design strength = c P n = c A g F cr

=39.03 (35.2) = 1374 k


Example ProblemsExample 3

a. Using Table 3.50 of Part 16 of the LRFDManual, determine the design strength c P nof the 50 ksi axially loaded W14 90 shownin the figure. Because of its considerablelength, this column is braced perpendicularto its weak axis at the points shown in thefigure. These connections are assumed topermit rotation of the member in a planeparallel to the plane of the flanges. At thesame time, however, they are assumed toprevent translation or sideway and twisting


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Example Problems

Example 3 (contd)of the cross sectionabout a longitudinalaxis passing throughthe shear center ofthe cross section.

Repeat part (a) usingthe column tables ofPart 4 of the LRFDManual.

nc P

nc P

ft32

ft10

ft10

ft12

90W14

General support xy direction


Example ProblemsExample 3(contd) Note that the

column is bracedperpendicular toits weak y axis asshown.

ft10

ft10

ft12

y

yx

x

Bracing

90W14


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Example Problems

Example 3 (contd)a. The following properties of the W14 90can be obtained from the LRFD Manual as

Determination of effective lengths:

See Table for the K values

in70.3 in14.6 in5.26 2 === y x r r A

( )( )( )( )( )( ) ft6.9128.0

ft10100.1

ft6.25328.0

======

y x

y y

x x

L K

L K

L K

Governs for K y L y


Example Problems

Table 1

Example 3 (contd)


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Example Problems

Example 3 (contd)Computations of slenderness ratios:

Design Strength:

43.3270.31012

03.5014.6

6.2512

==

==

y

x

r KL

r KL

Governs

( ) k 9385.264.35ksi4.35gives50-3Table,5003.50

c

c

=====

g cr cn

cr

A F P F r

KL



b. Using columns tables of Part 4 of LRFDManual:Note: from part (a) solution, there are twodifferent KL values:

Which value would control? This canaccomplished as follows:

ft10 andft6.25 == y y x x L K L K

y

y y

x

x x

r

L K

r L K

Equivalent=


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Example Problems

Example 3 (contd)

The controlling K y L y for use in the tables is largerof the real K y L y = 10 ft, or equivalent K y L y:

y x

x x

x

x x y y y r r

L K r L K

r L K /

Equivalent ==

k 938

:ioninterpolat byand42.15For

ft1043.1566.1

6.25 Equivalent

1.66lescolumn tabof bottomfrom90for W14

c ==

=>==

=

n

y y

y y y y

y

x

P

L K

L K L K

r r



The Interpolation Process: For K y L y = 15 ft and 16 ft, column table (P. 4-

23) of Par 4 of the LRFD Manual , givesrespectively the following values for c P n: 947 kand 925 k. Therefore, by interpolation:

k 9381516

1542.15947925947

92516

42.1594715 =

=

ncnc

nc P P

P

Introduction to Axially Loaded Compression Members

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