By 5d FALL 2002 - · PDF filesame load in compression that it can in tension. 2 CHAPTER 5d....
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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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 2ENCE 355 ©Assakkaf
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, Short, and Intermediate Columns
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 3ENCE 355 ©Assakkaf
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.
Long, Short, and Intermediate Columns
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 4ENCE 355 ©Assakkaf
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.
Long, Short, and Intermediate Columns
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 5ENCE 355 ©Assakkaf
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.
Long, Short, and Intermediate Columns
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 6ENCE 355 ©Assakkaf
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.
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 7ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 8ENCE 355 ©Assakkaf
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)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 9ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 10ENCE 355 ©Assakkaf
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)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 11ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 12ENCE 355 ©Assakkaf
Column FormulasFigure 1. LRFD Critical Buckling Stress
Shortcolumn Intermediate
column Long column
5.1=cλ
Inelastic buckling
Elastic buckling(Euler Formula)
rKL
crcFφ
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 13ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 14ENCE 355 ©Assakkaf
Column FormulasLRFD Manual Design Tables (P. 16.I-143)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 15ENCE 355 ©Assakkaf
Column FormulasLRFD Manual Design Tables (P. 16.I-145)
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 16ENCE 355 ©Assakkaf
Column FormulasLRFD Manual Design Tables (P. 4-25)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 17ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 18ENCE 355 ©Assakkaf
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.
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 19ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 20ENCE 355 ©Assakkaf
Example Problems
Example 1 (cont’d)
7212W ×ft 15
ncu PP φ≤
ncu PP φ≤
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 21ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 22ENCE 355 ©Assakkaf
Example ProblemsExample 1 (cont’d)
Table 1
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 23ENCE 355 ©Assakkaf
Example ProblemsExample 1 (cont’d)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 24ENCE 355 ©Assakkaf
Example Problems
Example 1 (cont’d)P. 16.I-145
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 25ENCE 355 ©Assakkaf
Example ProblemsExample 1 (cont’d)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 26ENCE 355 ©Assakkaf
Example ProblemsExample 1 (cont’d)
P. 4-25
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 27ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 28ENCE 355 ©Assakkaf
Table 2. Limiting Width-Thickness Ratios for Compression Elements
Example Problems
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 29ENCE 355 ©Assakkaf
Table 3. (cont’d) Limiting Width-Thickness Ratios for Compression Elements
Stiff
ened
Ele
men
ts
Example Problems
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 30ENCE 355 ©Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 31ENCE 355 ©Assakkaf
Example ProblemsExample 2 (cont’d)
( )
( )( ) ( )( ) 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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 32ENCE 355 ©Assakkaf
Example ProblemsExample 2 (cont’d)
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 33ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 34ENCE 355 ©Assakkaf
Example ProblemsExample 3 (cont’d)
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 35ENCE 355 ©Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 36ENCE 355 ©Assakkaf
Example ProblemsExample 3 (cont’d)
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 37ENCE 355 ©Assakkaf
Example Problems
Table 1
Example 3 (cont’d)
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 38ENCE 355 ©Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 39ENCE 355 ©Assakkaf
Example ProblemsExample 3 (cont’d)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 40ENCE 355 ©Assakkaf
Example ProblemsExample 3 (cont’d)
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 41ENCE 355 ©Assakkaf
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