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C-1
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 strutsor 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 bucklingat
loads for less then yielding forces.
P
P
P
P
(a) (b)
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C-2
For slender columns, Euler (1759) predicted the critical buckling load (Pcr)
also known as Euler Buckling Load as:
)1(2
2
CL
EIPcr
where: E = Young Modulus of Elasticity.
I = Minor moment of Inertia.
L = Unbraced length of column.
Derivation of Euler Buckling Load:
0"
2
2
yEI
Py
EI
M
dx
yd
cr
Solution of this differential equation:
y = A cos (cx) + B sin (cx)
where:
, A and B are constants.
EI
Pc
cr
x
L
y
x
y
Pcr Pcr
Pcr
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C-3
From boundary conditions:
y = 0 @ x = 0, and
y = 0 @ x = L, we get (A = 0) and (B sin cL = 0)
if B 0, then cL = n where n = 0, 1, 2, 3
cL =
)(
/
2
2
2
2
22
2
2
C
rL
EA
L
ArE
L
EIP
LEI
P
cr
cr
2
2
g
cr
cr
rL
E
A
PF
---- Euler Buckling Critical Load
where: r = minor radius of gyration
The critical buckling load
is a function of the sectionproperties (A, L, r) and the
modulus of elasticity for
material, and is not a
function of the strength or
grade of the material.
Note:
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C-4
Example C-1
Find the critical buckling load for W 12 x 50, supported in a pinned-pinnedcondition, and has an over-all length of 20 feet?
Solut ion:
22
r
Lcr
EF
rmin= ry= 1.96 inch (properties of section).
ksiFcr 19
290002
96.11220
2
Note:
The steel grade is not a factor affecting buckling,
also note Fcr
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Eccentric Loading; The Secant Formula
r
L
EA
P
r
ec
A
P eY
2
1sec1
2max
La carga no esta aplicada en el centroideEl miembro no es perfectamente recto
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C-5
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:
)3(2
2
CL
IEP tcr
where:
Et= Tangent Modulus of Elasticity
Et< 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.
Columnas largas: fallan por
inestabilidad elstica
Columnas intermedias: falla por
inestabilidad inelstica ( esfuerzo de
compresin es mayor que el esfuerzode fluencia)
Columnas cortas: No presentan
inestabilidad, la columna falla por
agotamiento de la resistencia del
material
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C-6
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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C-7
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:
4CE
For,
EF
2
rKl
t2
cr2
rKL
2
cr
where:K = Effective length factor.
(Kl) = Effective length.
(Kl/r) = Effective slenderness ratio.
see commentary
(C C2.2) (page 16.1-240)
The Euler buckling formula (C-1) is based on:
1Perfectly straight column. (no crookedness).
2Load is concentric (no eccentricity).3Column is pinned on both ends.
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C-8
AISC (Chapter E) of LRFD code stipulates:
Pu(factored load) cPn
where:
Pu= Sum of factored loads on column.
c = Resistance factor for compression = 0.90
Pn= Nominal compressive strength = FcrAg
Fcr= Critical buckling Stress. (E3 of LFRD)
a) for
3.4-E
EF
3.3-E0.877FF
0.44FFor4.71
r
Kl
3.2-EF0.658F
0.44FFor4.71r
Kl
2
rK L
2
e
ecr
yeF
E
ycr
yeF
E
y
eF
yF
y
b) for
where:
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C-9
The above two equations of the LRFD
code can be illustrated as below:where:
E
F
r
Kl
y
c
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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C-10
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?
Example C-2
Solution:
ksi30.56(96.77)
x2900
r
Kl
EFe
(0k)20096.772.48
240
r
Kl
2
2
2
2
max.
ksi21.99360.611
36x(0.658)F0.658F 1.178yF
F
cre
y
Kl =1 x 20 x 12 = 240 in
rmin= ry= 2.48
0.44 Fy= 0.44 x 36 =15.84 ksi
Fe 0.44 FyEqu. E-3.2
(controls)
cPn= 0.9 x Fcrx Ag= 0.9 x (21.99) x 21.8
= 433.44 kips (Answer)
Also from (table 4-22) LFRD Page 4-320
cFcr= 19.75 ksi (by interpolation)
cPn= cFcrAg= 430.55 kips
(much faster)
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C-11
For must profiles used as column, the
buckling of thin elements in the sectionmay 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 tothickness 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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C-12
Depending on their ( b/t ) ratios (referred to as ) ,
sections are classified as:a) Compact sections are those with flanges fully welded
(connected) to their web and their:
p (AISC B4)
b) Non compact Sections:pr (B4)
c) Slender Section:
> 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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C-13
Example C-3
Determine the design
compressive strength
(cPn) for W 12 x 65
column shown below,
(Fy= 50 ksi)?
From properties:
Ag =19.1 in2rx= 5.28 in
ry= 3.02 inksi40.22550x0.8045
F0.658F
3.2)(EEqu.ksi)22(F0.44F
ksi96.2(54.55)x29000
r
KlEF
31.793.02
12x8x1
r
LK
54.555.28
12x24x1
r
LK
y
96.2
50
cr
ye
2
2
2
2
e
y
yy
x
xx
Solut ion:
cPn= 0.9 x FcrAg= 0.9 x 40.225 x 19.1 = 691.5 kips
A) By direct LRFD
(controls)
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C-14
B) From Table (4.22) LRFD
Evaluate = = 54.55
Enter table 4.22 (page 4318 LRFD)
cFc= 36.235 ksi (by interpolation)
Pn= Fcx Ag = 692.0 kips
C) From (Table 4.1 LRFD)
maxr
Kl
ft13.7
1.75
1x24
r
r
LK(KL)
y
x
xxy
Enter table (4.1 ) page 4.17 LFRD with (KL)y= 13.7
Pn= 691.3 kips (by interpolation).
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C-15
Example C-4
Solut ion:
Find the maximum load capacity(Pn) of the W 14 x 53 (A-36)
column shown in figure ?
PP
25 ft.
15 ft.
10 ft.
A
C
A
B
C
x xx
x
x-axis Lx= 25 ft, kx= 0.8, rx= 5.89 in.
y-axisSection (AB) Ly= 15 ft, ky= 0.8, ry= 1.92 in.
Section (BC) Ly= 10 ft., ky= 1.0, ry= 1.92 in.
751.92
12150.8
r
Kl
415.98
12250.8
r
Kl
max
y
x
Enter table (4-22) , Fc= 24.1 ksi
Column capacity Pn= FcrAg= 24.1 x 15.6 = 376 kips
(controls)
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C-16
Design with Columns Load Table (4) LFRD:-
A) 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 unsymmetricalshapes the structural tees and the single and double-angles
require special consideration and are covered later.
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C-17
EXAMPLE C - 5
A compression member is subjected to service loads of 165 kips dead loadand 535 kips live load. The member is 26 feet long and pinned in each end.
Use (A572Gr 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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C-18
EXAMPLE C - 6Select the lightest W-shape that can resists a factored compressive load Puof
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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C-19
Example C-7
Select the lightest W10 section made of
A 572-Gr50 steel to resist a factored load
of (600 kips) ?
Solut ion:
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 > 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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C-20
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. Ifthe strength is too small (unsafe) or too large (uneconomical), another trial must bemade. A systematic approach to making the trial selection is as follows.
1) 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 Fcris the yield stress Fy.
2) From the requirement that cPnPu, let
cAgFcrPuand
3) Select a shape that satisfies this area requirement.
4) Compute Fcrand cPnfor the trial shape.
5) 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 Fcrfound for the current trial shape as a value for Step 1
6) Check local stability (check width-thickness ratios). Revise if necessary.
crc
u
F
P
gA
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C-21
3.2.EEqu.LRFD(15.84)F0.44F
ksi22.9111.8
x29000
r
Kl
EF
ye
2
2
2
2
e
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.
Solut ion:Try Fcr= 24 ksi (two-thirds of Fy):
Required2848
2490
1054in
F
PA
crc
ug .
)(.
Try W18 x 192:Ag= 56.4 in
2> 48.8in2
(OK)200111.82.79
26(12)
r
KL
min
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C-22
(OK)200109.52.85
26(12)
r
KL
in62.83.in68.8 A
:234xW18Try
in62.830.9(18.64)
1054
F
PARequired
:192)xW18theforcomputedjustvalue(theksi18.64FTry
(N.G.)
105494318.64x56.4x0.9FA0.9P
ksi18.64
36x0.532x360.658F0.658F
min
2g
2
crc
ug
cr
kkipscrgnc
yF
F
cr
2
22.9
36
e
y
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C-23
234xW18aUseAnswer
(OK)42.236
25313.8th
(OK)15.836
952.8
2t
b
:cheackedbemustratios
thicknes-widththesotables,loadcolumntheinnotisshapeThis
(OK)1054118519.15x68.8x0.9FA0.9P
ksi19.1536x0.53236x0.658F0.658F
3.2)-(Equ.ELFRDUse0.44FF
23.87ksi
109.5
29000EF
w
f
f
kkipscrgnc
y
F
F
cr
ye
2
2
2
r
Kl
2
e
23.87
36
e
y
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C-24
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:
No lateral movement is allowed
(0.5 < K < 1.0) (sideway prevented)
Unb raced Frames:
Lateral movement possible
(1.0 < K < 20.0) (sideway allowed)
a) Diagonal
bracing
b) Shear Walls
(masonry,
reinforcement concrete
or steel plate)
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C-25
gg
ccB
gg
ccA
/LI
/LIG
/LI
/LIG
* For fixed footing G = 1.0
* For pinned support G = 10.0
where
A is top of column
where
B is bottom of column
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C-26
In the rigid frame shown below, Determine Kxfor columns(AB) & (BC). Knowing that all columns webs are in the plane.
Column (AB):
Joint (A):
0.94
169.21586
1830/181350/201070/12833/12
/LI
/LIG
gg
ccA
Example C 9:-
Solut ion:
A
B
C
W24 x 55
W24 x 55
W24 x 68
W24 x 68W12x120
W12x120
W12x96
12'
15'
12'
18'20'20'
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C-27
For joint B,:-
0.95169.2160.5
169.21070/151070/12
/LI/LIG
gg
cc
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 girdersthat 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 beapproximated 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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C-28
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.
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.
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 beinvestigated. 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.
1.
2.
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C-29
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 symmetrysuch 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.
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When length exceeds requirements for
a single section, built-up compression
section are used as shown below:
The code provides details for built-upsection under LRFD EG.
C-30
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Calculate the capacity of the built-up column shown below.
Lx= Ly= 25 ft, Kx= 1.6, Ky= 1.0 Fy= 42 ksi ?
Solut ion:-
82.903.62
12251.0
r
lK
(controls)108.844.41
12251.6rlK
in3.6222.70
298.0
A
Ir
in4.4122.70
440.5
A
Ir
in22.7047.352A
in298)8(0.953)7.35(57.252I
in440.50.25)(58x1102I
y
yy
x
xx
yy
y
xxx
2
g
43
21
1212
yy
42
21
xx
C-31
Example C 10 :-
From table 4.22 page 4.321
cFcr = 18.3 ksi
Design Nominal Strength = cFcr Ag=18.3 x 22.70
= 415.4 kips.
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