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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
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 axialcompressive buckling stress remains
below the proportional limit.
Long, Short, and IntermediateColumns
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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 practicalapplication.
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 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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 6ENCE 355 Assakkaf
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.
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 7ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 8ENCE 355 Assakkaf
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)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 9ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 10ENCE 355 Assakkaf
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)
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 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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 12ENCE 355 Assakkaf
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
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 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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 14ENCE 355 Assakkaf
Column Formulas
LRFD 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 Formulas
LRFD Manual Design Tables (P. 4-25)
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 17ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 18ENCE 355 Assakkaf
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.
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 19ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 20ENCE 355 Assakkaf
Example Problems
Example 1 (contd)
7212W ft15
ncu P P
ncu P P
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 21ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 24ENCE 355 Assakkaf
Example Problems
Example 1 (contd)P. 16. I-145
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 25ENCE 355 Assakkaf
Example ProblemsExample 1 (contd)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 26ENCE 355 Assakkaf
Example ProblemsExample 1 (contd)
P. 4-25
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 27ENCE 355 Assakkaf
Example ProblemsExample 1 (contd)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 28ENCE 355 Assakkaf
Table 2. Limiting Width-Thickness Ratios forCompression Elements
Example Problems
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 29ENCE 355 Assakkaf
Table 3. (contd) Limiting Width-ThicknessRatios for Compression Elements
Example Problems
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 30ENCE 355 Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 31ENCE 355 Assakkaf
Example ProblemsExample 2 (contd)
( )
( )( ) ( )( )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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 32ENCE 355 Assakkaf
Example ProblemsExample 2 (contd)
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
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 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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 34ENCE 355 Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 35ENCE 355 Assakkaf
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 36ENCE 355 Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 37ENCE 355 Assakkaf
Example Problems
Table 1
Example 3 (contd)
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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 38ENCE 355 Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 39ENCE 355 Assakkaf
Example ProblemsExample 3 (contd)
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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CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 40ENCE 355 Assakkaf
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
CHAPTER 5d. INTRODUCTION TO AXIALLY LOADED COMPRESSION MEMBERS Slide No. 41ENCE 355 Assakkaf
Example ProblemsExample 3 (contd)
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