By 5d FALL 2002 - · PDF filesame load in compression that it can in tension. 2 CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 2 ENCE 355 ©Assakkaf

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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

Third EditionCHAPTER

5d

Structural Steel DesignLRFD Method

ENCE 355 - Introduction to Structural DesignDepartment of Civil and Environmental Engineering

University of Maryland, College Park

INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS

Part II – Structural Steel Design and Analysis

FALL 2002By

Dr . Ibrahim. Assakkaf

CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 1ENCE 355 ©Assakkaf

Long, Short, and Intermediate Columns

The strength of a column and the manner in which it fails are greatly dependent on its effective length.A very short stocky steel column may be loaded until it reaches it yield point, and perhaps the strain hardening range.In essence, it can support about the same load in compression that it can in tension.

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As the effective length of a column increases, its buckling stress will decrease.The steel column is said to fail elastically if the buckling stress is less than the proportional limit of steel when the effective length exceeds a certain value.



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

strength of long columns where the axial compressive buckling stress remains below the proportional limit.


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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 practical application.



Intermediate Columns– For intermediate columns some of the

fibers will reach the yield stress and some will not.

– The member will fail by both yielding and buckling, and their behavior is said to be inelastic.

– Most columns fall into this range.


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Column Formulas

The Euler formula is used by the AISC LRFD Specification for long columns with elastic buckling.Other empirical (based on testing) equations are used by the LRFD for short and intermediate columns.With these equations, a critical or buckling stress Fcr is determined for a compression element.


Column Formulas

LRFD General Design Equation for Columns

The design strength of a compression member is determined as follows:

0.85 with =≤

=

φφφ crgcnc

crgn

FAPFAP

(1)

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Column Formulas

LRFD Critical Buckling StressTwo equations are provided by the LRFD for the critical buckling stress Fcr:

( )

>

≤=

5.1for 877.0

5.1for 658.0

2

2

cyc

cy

crF

FF

c

λλ

λλ

(2)


Column FormulasLRFD Critical Buckling Stress

The limiting λc value is given by

Where Fe = Euler buckling stress =Hence,

e

yc F

F=λ

( )22

/ rKLEπ

( )

( )EF

rKL

ErKLF

rKLE

FFF yyy

e

yc πππλ ==== 2

2

2

2

/

/

(3)

(4)

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Column Formulas

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

whereFy = yield strength of material (steel)Fe = Euler critical buckling stress

EF

rKL

FF y

e

yc π

λ == (5)


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

used to compute the critical buckling stress Fcr when λc ≤ 1.5.

– For elastic flexural buckling, Eq. 2 can be used to compute the critical buckling stressFcr when λc > 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 FormulasFigure 1. LRFD Critical Buckling Stress

Shortcolumn Intermediate

column Long column

5.1=cλ

Inelastic buckling

Elastic buckling(Euler Formula)

rKL

crcFφ


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

Manual provides computed values φc Fcrvalues for steels with Fy = 36 ksi and 50 ksi for KL/r from 1 to 200 and has shown the results in Tables 3.36 and 3.50 of the LRFD Specification located in Part 16 of the Manual.

– Also, there is Table 4 of the LRFDSpecification from which the user may obtain values for steel with any Fy values.

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Column FormulasLRFD Manual Design Tables (P. 16.I-143)


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

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Column FormulasLRFD Manual Design Tables (P. 4-25)


Maximum Slenderness Ratios

Compression members preferably should be designed with

as specified in Section B7 of the LRFD Manual.

200≤r

KL (6)

Note that LRFD Tables 3.36 and 3.50 give a value of 5.33 ksi forthe design stress φc Fcr 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 RatiosMore Simplification by the LRFD Manual for Design– It is to be noted that the LRFD Manual in

its Part 4 has further simplified the calculations required by computing the column design strength φc Fcr Ag for each of the shapes normally used as columns for commonly used effective lengths or KLvalues.

– These were determined with respect to the least radius of gyration for each section.


Example Problems

Example 1a. Using the column design stress values

shown in Table 3.50, part 16 of the LRFD manual, determine the design strength, φc Pnof the Fy = 50 ksi axially loaded column shown in the figure.

b. Repeat the problem using the column tables of 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 (cont’d)

7212W ×ft 15

ncu PP φ≤

ncu PP φ≤


Example ProblemsExample 1 (cont’d)

a. The properties of the W12 × 72 are obtained from the LRFD Manual as

in 430.0 in 27.1

in 670.0 in 00.12 in 3.12

in 04.3 in 31.5 in 1.21 2

==

===

===

w

ff

yx

tktbdrrA

( ) ( )

( ) 37.4704.3

151280.0

andcontrols ,in 31.5in 04.3 Since

Text) 5.1, (Table 1 Table from 80.0

=×

=

=<==

y

yxy

rKL

rrrK

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Table 1



For KL/r = 47 and 48, Table 3-50 of the LRFD Manual, Page 16.I-145, gives respectively the following values for φc Fcr: 36.2 ksi and 35.9 ksi.Using interpolation,

ksi 09.364748

4737.472.369.352.36

9.354837.47

2.3647=⇒

−−

=−−

⇒ crccrc

crc FFF φφ

φ

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Example Problems

Example 1 (cont’d)P. 16.I-145



Therefore,

b. Entering column tables Part 4 of the LRFD Manual with Ky Ly in feet:

( ) k 5.7611.2109.36 === gcrcnc AFP φφ

( )k 761

ft 121580.0

==

==

ncu

yy

PPLKφ

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P. 4-25


Example Problems

Example 1 (cont’d)c. Checking W12× 72 for compactness:

For flange

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

( ) 49.1350

102956.056.096.8670.020.12

2

3

=×

=<==yf F

Eth

88.3550

102949.149.17.22430.076.9 3

=×

=<==yw F

Eth

OK

OK

See Table 2

See Table 3

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Table 2. Limiting Width-Thickness Ratios for Compression Elements

Example Problems


Table 3. (cont’d) Limiting Width-Thickness Ratios for Compression Elements

Stiff

ened

Ele

men

ts

Example Problems

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Example ProblemsExample 2

Determine the design strength φc Pn of the axially loaded column shown in the figure if KL = 19 ft and 50 ksi steel is used.

2021PL ×

yxx

in 12

42.7MC18×

in 50.18

C] ofback fromin 877.0,in 3.14 ,in 554

in, 0.18 ,in 6.12[44

2

=

==

==

xIIdA

yx



( )

( )( ) ( )( ) in 87.62.35

5.96.12225.0200.5 topfrom

in 2.356.1222120 2

=×+×

=

=+

=

y

A

( ) ( )[ ] ( ) ( )

( ) ( )[ ] ( )

in 64.62.35

1554

in 92.62.35

1688

in 155412205.0877.066.1223.142

in 688,125.069.61012

5.02069.625.96.1225542

43

2

423

2

===

===

=+++=

=−++−+=

AI

r

AIr

I

I

yy

xx

y

x

Controls

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For KL/r = 34 and 35, Table 3-50 of theLRFD Manual, Page 16.I-145, gives respectively the following values for φc Fcr: 39.1 ksi and 38.9 ksi

( ) 34.3464.61912

=×

==yr

KLr

KL

ksi 03.393435

3434.341.399.381.39

9.383534.34

1.3934=⇒

−−

=−−

⇒ crccrc

crc FFF φφ

φ

Therefore, the design strength = φc Pn = φc Ag Fcr

=39.03 (35.2) = 1374 k


Example ProblemsExample 3

a. Using Table 3.50 of Part 16 of the LRFD Manual, determine the design strength φc Pnof the 50 ksi axially loaded W14 × 90 shown in the figure. Because of its considerable length, this column is braced perpendicular to its weak axis at the points shown in the figure. These connections are assumed to permit rotation of the member in a plane parallel to the plane of the flanges. At the same time, however, they are assumed to prevent translation or sideway and twisting

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of the cross section about a longitudinal axis passing through the shear center of the cross section.

– Repeat part (a) using the column tables of Part 4 of the LRFD Manual.

ncPφ

ncPφ

ft 32

ft 10

ft 10

ft 12

90W14×

General support xy direction


Example ProblemsExample 3 (cont’d)– Note that the

column is braced perpendicular to its weak y axis as shown.

ft 10

ft 10

ft 12

y

yx

x

Bracing

90W14×

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a. The following properties of the W14 × 90 can be obtained from the LRFD Manual as

Determination of effective lengths:

See Table for the K values

in 70.3 in 14.6 in 5.26 2 === yx rrA

( )( )( )( )( )( ) ft 6.9128.0

ft 10100.1ft 6.25328.0

==

====

yx

yy

xx

LKLKLK

Governs for Ky Ly


Example Problems

Table 1

Example 3 (cont’d)

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Example ProblemsExample 3 (cont’d)Computations of slenderness ratios:

Design Strength:

43.3270.31012

03.5014.6

6.2512

=×

=

=×

=

y

x

rKLr

KL Governs

( ) k 9385.264.35

ksi 4.35 gives 50-3 Table ,5003.50

c

c

===∴

=≈=

gcrcn

cr

AFP

Fr

KL

φφ

φ



b. Using columns tables of Part 4 of LRFD Manual:Note: from part (a) solution, there are two different KL values:

Which value would control? This can accomplished as follows:

ft 10 andft 6.25 == yyxx LKLK

y

yy

x

xx

rLK

rLK Equivalent =

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The controlling Ky Ly for use in the tables is larger of the real Ky Ly = 10 ft, or equivalent Ky Ly:

yx

xx

x

xxyyy rr

LKrLKrLK

/ Equivalent ==

k 938

:ioninterpolatby and 42.15For

ft 1043.1566.1

6.25 Equivalent

1.66 lescolumn tab of bottom from 90for W14

c =

=

=>==

=×

n

yy

yyyy

y

x

PLK

LKLK

rr

φ


Example Problems

Example 3 (cont’d)The Interpolation Process:

• For Ky Ly = 15 ft and 16 ft, column table (P. 4-23) of Par 4 of the LRFD Manual, gives respectively the following values for φc Pn: 947 k and 925 k. Therefore, by interpolation:

k 9381516

1542.15947925947

9251642.15

94715=⇒

−−

=−−

⇒ ncnc

nc PPP φφ

φ

Documents

By 5d FALL 2002 - · PDF filesame load in compression that it can in tension. 2 CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 2 ENCE 355 ©Assakkaf