C-1

1
Compression Members
Chap. (6)
Columns
Are compression members which are subjected to
concentric axial compressive forces. These are to
be found in trusses and as a lateral bracing
members in frame building. Short columns are
sometimes referred to as to as struts or
stanchions.
Beam-Columns
Are members subjected to combined axial
compressive and bending stresses These are found
in single storey of multi-storey framed
structures. These are treated independently in
this course (chap. 12 in your text book).
Columns Theory
Stocky columns (short) fail by yielding of the
material at the cross section, but most columns
fail by buckling at loads for less then
yielding forces.
(a) (b)
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2
Critical Buckling Load of Columns
For slender columns, Euler (1759) predicted the
critical buckling load (Pcr) also known as
Euler Buckling Load as
Pcr
where E Young Modulus of Elasticity.
I Minor moment of Inertia. L
Unbraced length of column.
Derivation of Euler Buckling Load
where
Solution of this differential equation y A
cos (cx) B sin (cx)
, A and B are constants.
C-2
3
Euler Buckling Formula
From boundary conditions
y 0 _at_ x 0, and y 0 _at_ x L, we get (A 0)
and (B sin cL 0) if B ? 0, then cL n?
where n 0, 1, 2, 3 ? cL ?
---- Euler Buckling Critical Load
where r minor radius of gyration
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Euler Critical Buckling Load
Example C-1
Find the critical buckling load for W 12 x 50,
supported in a pinned-pinned condition, and has
an over-all length of 20 feet?
Solution
rmin ry 1.96 inch (properties of section).
Pcr Fcr A 19.1 x 14.7 280.8 kips
Note The steel grade is not a factor affecting
buckling, also note Fcr ltlt Fy.
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Stress Strain Relations in Compression
For short (stocky) columns Equation (C-2) gives
high values for (Fcr), sometimes greater then
proportional limit, Engessor (1889) proposed to
use (Et) instead of (E) in Euler formula
where
Et Tangent Modulus of Elasticity Et lt E When
(Fcr) exceeds (Fpl), this is called Inelastic
Buckling, constantly variable (Et) need to be
used to predict (Fcr) in the inelastic
zone. Shanley (1947), resolved this inconsistency.
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Column Strength Curve
Depending on (L/r) value the column
buckling strength was presented as shown by
Shanley.
Residual Stresses-
Due to uneven cooling of hot-rolled
sections, residual stresses develop as seen here.
The presence of residual stresses in almost all
hot-rolled sections further complicates the issue
of elastic buckling and leads towards inelastic
buckling.
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Effective Slenderness Ratio
The Euler buckling formula (C-1) is based on
1 Perfectly straight column. (no crookedness).
2 Load is concentric (no eccentricity).
3 Column is pinned on both ends.
where
K Effective length factor. (Kl) Effective
length. (Kl/r) Effective slenderness ratio.
The Previous conditions are very difficult to
achieve in a realistic building condition,
especially the free rotation of pinned ends. Thus
an effective slenderness factor is introduced
to account for various end conditions
Thus
see commentary (C C2.2) (page 16.1-240)
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AISC Column Design Requirements
AISC (Chapter E) of LRFD code stipulates Pu
(factored load) ? ?c Pn where Pu Sum of
factored loads on column. ?c Resistance
factor for compression 0.90 Pn Nominal
compressive strength Fcr Ag Fcr Critical
buckling Stress. (E3 of LFRD)
a) for
b) for
where
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AISC Column Design Criteria
The above two equations of the LRFD code can be
illustrated as below where

• The code further stipulates
• that an upper value for column
• should not exceed (200).
• ? For higher slenderness ratio,
• Equation (E-3.3) controls and
• (Fy) has no effect on (Fcr).

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Capacity of Compression Members
Example C-2
Determine the design compressive strength
(?cPn) of W 14x74 with an untraced length of (20
ft), both ends are pinned, (A-36) steel is used?
Solution
Kl 1 x 20 x 12 240 in rmin ry 2.48
?c Pn 0.9 x Fcr x Ag 0.9 x (21.99) x 21.8
433.44 kips (Answer)
? Also from (table 4-22) LFRD Page 4-320 ?c
Fcr 19.75 ksi (by interpolation) ?c Pn
?c Fcr Ag 430.55 kips (much faster)

• 0.44 Fy 0.44 x 36 15.84 ksi
• Fe 0.44 Fy ? Equ. E-3.2
• (controls)

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Local Buckling Vs. Total Buckling
For must profiles used as column, the buckling
of thin elements in the section may proceed the
ever-all bucking of the member as a whole, this
is called local bucking. To prevent local bucking
from accruing prior to total buckling. AISC
provides upper limits on width to thickness
ratios (known as b/t ratio) as shown here.
See AISC (B4) (Page 16.1-14) See also Part 1
on properties of various sections.
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Compact, Non compact Slender Sections

• Depending on their ( b/t ) ratios (referred to as
  ?) ,
• sections are classified as
• Compact sections are those with flanges fully
  welded
• (connected) to their web and their
• ? ?? ?p (AISC B4)
• b) Non compact Sections
• ?p ? ? ? ?r (B4)
• c) Slender Section
• ? gt ?r (B4)
• Certain strength reduction factors (Q) are
  introduced for slender
• members. (AISC E7). This part is not required as
  most section
• selected are compact.

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Capacity of Compact Sections as Column.
Example C-3
Determine the design compressive strength (?c Pn)
for W 12 x 65 column shown below, (Fy 50 ksi)?
(controls)
Solution
A) By direct LRFD
From properties Ag 19.1 in2 rx 5.28 in ry
3.02 in
?c Pn 0.9 x Fcr Ag 0.9 x 40.225 x 19.1
691.5 kips
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B) From Table (4.22) LRFD Evaluate
54.55 Enter table 4.22 (page 4
318 LRFD) ?cFc 36.235 ksi (by
interpolation) Pn ?Fc x Ag 692.0 kips C)
From (Table 4.1 LRFD)
Enter table (4.1 ) page 4.17 LFRD with (KL)y
13.7 Pn 691.3 kips (by interpolation).
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Variable Slenderness Ratios
Example C-4
Find the maximum load capacity (?Pn) of the W 14
x 53 (A-36) column shown in figure ?
Solution
x-axis
Lx 25 ft, kx 0.8, rx 5.89 in.
y-axis
Section (AB) Ly 15 ft, ky 0.8, ry 1.92
in. Section (BC) Ly 10 ft., ky 1.0, ry
1.92 in.
Enter table (4-22) , ?Fc 24.1 ksi Column
capacity ?Pn ?Fcr Ag 24.1 x 15.6 376 kips
(controls)
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LRFD DESIGN OF COLUMNS
Design with Columns Load Table (4) LFRD-

• Design with Column Load Table (4) LFRD
• The selection of an economical rolled shape to
  resist a given compressive load is simple with
  the aid of the column load tables. Enter the
  table with the effective length and move
  horizontally until you find the desired design
  strength (or something slightly larger). In some
  cases, Usually the category of shape (W, WT,
  etc.) will have been decided upon in advance.
  Often the overall nominal dimensions will also be
  known because of architectural or other
  requirements. As pointed out earlier, all
  tabulated values correspond to a slenderness
  ratio of 200 or less. The tabulated unsymmetrical
  shapes the structural tees and the single and
  double-angles require special consideration and
  are covered later.

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LRFD DESIGN OF COLUMNS
EXAMPLE C - 5
A compression member is subjected to service
loads of 165 kips dead load and 535 kips live
load. The member is 26 feet long and pinned in
each end. Use (A572 Gr 50) steel and select a
W14 shape.
SOLUTION
Calculate the factored load
Pu 1.2D 1.6L 1.2(165) 1.6(535) 1054
kips ? Required design strength ?cPn 1054
kips From the column load table for KL 26 ft, a
W14 ? 145 has design strength of 1230 kips.
ANSWER
Use a W14 ? 145, But practically W14 ? 132 is OK.
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LRFD DESIGN OF COLUMNS
EXAMPLE C - 6
Select the lightest W-shape that can resists a
factored compressive load Pu of 190 kips. The
effective length is 24 feet. Use ASTM A572 Grade
50 steel.
SOLUTION
The appropriate strategy here is to fined the
lightest shape for each nominal size and then
choose the lightest overall. The choices are as
follows.
W4, W5 and W6 None of the tabulated shape will
work. W8 W 8 ? 58, ?cPn 205 kips W10 W10 ?
49, ?cPn 254 kips W12 W12 ? 53, ?cPn 261
kips W14 W14 ? 61, ?cPn 293 kips
Note that the load capacity is not proportional
to the weight (or cross-sectional area). Although
the W8 ? 58 has the smallest design strength of
the four choices, it is the second heaviest.
ANSWER
Use a W10 ? 49.
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LRFD DESIGN OF COLUMNS
Example C-7
Select the lightest W 10 section made of A
572-Gr50 steel to resist a factored load of (600
kips) ?
Solution
Assume weak axis (y-y) controls buckling Enter
design tables of AISC (Section 4) with KyLy 9
ft. Select W 10 x 54 (capacity 625 k gt 600 k
OK)
Check strong axis buckling strength Enter
table for W10 x 54 with (KL)eq. 10.53
ft. Capacity 595.8 kips (by interpolation) N.G.
Select W10 x 60 capacity 698 kips for KyLy 9
ft. capacity 666 kips for
(KL)eq. 10.5 ft.
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LRFD DESIGN OF COLUMNS
B) Design for sections not from Column Load
Tables
For shapes not in the column load tables, a
trial-and-error approach must be used. The
general procedure is to assume a shape and then
compute its design strength. If the strength is
too small (unsafe) or too large (uneconomical),
another trial must be made. A systematic approach
to making the trial selection is as follows.

• Assume a value for the critical buckling stress
  Fcr. Examination of AISC Equations
• E3-2 and E3-3 shows that the theoretically
  maximum value of Fcr is the yield stress Fy.
• 2) From the requirement that ?cPn ? Pu, let
• ?cAgFcr ? Pu and
• Select a shape that satisfies this area
  requirement.
• Compute Fcr and ?cPn for the trial shape.
• Revise if necessary. If the design strength is
  very close to the required value,
• the next tabulated size can be tried. Otherwise,
  repeat the entire procedure,
• using the value of Fcr found for the current
  trial shape as a value for Step 1
• 6) Check local stability (check width-thickness
  ratios). Revise if necessary.

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LRFD DESIGN OF COLUMNS
Example C-8
Select a W18 shape of A36 steel that can resist a
factored load of 1054 kips. The effective length
KL is 26 feet.
Solution
Try Fcr 24 ksi (two-thirds of Fy)
Required
Try W18 x 192
Ag 56.4 in2 gt 48.8in2
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Example on Design of Columns Contd.
C-22
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Example on Design of Columns Contd.
C-23
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Effective Length Factor - Real Conditions
The effective length factor (K) was introduced in
page (C-7) for six ideal conditions, these are
not encountered in practical field conditions.
LRFD commentary provides both real conditions and
standard ideal conditions (C-C2.2) (page 16.1-239
to 242)
Braced Frames
Unbraced Frames
No lateral movement is allowed (0.5 lt K lt 1.0)
(sideway prevented)
Lateral movement possible (1.0 lt K lt 20.0)
(sideway allowed)

• Diagonal
• bracing

b) Shear Walls (masonry,
reinforcement concrete or steel plate)
C-24
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Alignment Charts (LRFD P-16.1 - 241)
where A is top of column
where B is bottom of column
For fixed footing G 1.0 For pinned support
G 10.0
C-25
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Use of Alignment Charts
Example C 9-
In the rigid frame shown below, Determine Kx for
columns (AB) (BC). Knowing that all columns
webs are in the plane.
Solution
Column (AB) Joint (A)
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For joint B,-
From the alignment chart for sideways
uninhibited, with GA 0.94 and GB 0.95, Kx
1.3 for column AB.
Column (BC)
For joint B, as before, G 0.95 For joint C, at
a pin connection the situation is analogous to
that of a very stiff column attached to
infinitely flexible girders that is, girders of
zero stiffness. The ratio of column stiffness to
girder stiffness would therefore be infinite for
a perfectly frictionless hinge. This end
condition is only be approximated in practice, so
the discussion accompanying the alignment chart
recommends that G be taken as 10.0. From the
alignment chart with GA 0.95 and GB 10.0, Kx
1.85 for column BC.
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Flexural, Torsional Flexural - Torsional
Buckling
When an axially loaded compression member becomes
unstable overall (that is, not locally unstable),
it can buckle in one of three ways, as shown in
figure.
1.
Flexural buckling. We have considered this type
of buckling up to now. It is a deflection caused
by bending, or flexure, about this axis
corresponding to the largest slenderness ratio
(Figure a). This is usually the minor principal
axis the one with the smallest radius of
gyration. Compression members with any type of
cross-sectional configuration can fail in this
way.
2.
Torsional buckling. This type of failure is
caused by twisting about the longitudinal axis of
the member. It can occur only with doubly
symmetrical cross sections with very slender
cross-sectional elements (Figure b). Standard
hot-rolled shapes are not susceptible to
torsional buckling, but members built up from
thin plate elements may be and should be
investigated. The cruciform shape shown is
particularly vulnerable to this type of buckling.
This shape can be fabricated from plates as shown
in the figure, or built up from four angles
placed back to back.
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Flexural-torsional buckling. This type of
failure is caused by a combination of flexural
buckling and torsional buckling. The member bends
and twists simultaneously (Figure c). This type
of failure can occur only with unsymmetrical
cross sections, both those with one axis of
symmetry such as channels, structural tees,
double-angle shapes and equal-leg single angles
and those with no axis of symmetry, such as
unequal-leg single angles.
3.
The AISC Specification requires an analysis of
torsional or flexural-torsional buckling when
appropriate. Section E3 of the Specification
covers double-angle and tee-shaped members, and
Appendix E3 provides a more general approach that
can be used for any unsymmetrical shape.
C-29
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Built-Up Column Sections
When length exceeds requirements for a single
section, built-up compression section are used as
shown below
The code provides details for built-up section
under LRFD EG.
C-30
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Capacity of Built-up Members
Example C 10 -
Calculate the capacity of the built-up column
shown below. Lx Ly 25 ft, Kx 1.6, Ky 1.0
Fy 42 ksi ?
Solution-
From table 4.22 page 4.321 ?cFcr 18.3
ksi Design Nominal Strength ?cFcr Ag
18.3 x 22.70
415.4 kips.
C-31