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8/19/2019 CH 4 - AE Design of Columns
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CHAPTER 4
COLUMNS:
COMBINED AXIAL LOAD & BENDING
Abrham E.Sophonyas A.
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4.0 Introduction
A column is a vertical structural member supporting
axial compressive loads, with or without moments. The cross-sectional dimensions of a column are
generally considerably less than its height.
Columns support vertical loads from floors and roof and
transmit these loads to the foundations.
Columns may be classified based on the following
criteria:
a) On the basis of geometry; rectangular, square, circular,L-shaped, T-shaped, etc. depending on the structural
or architectural requirements
b) On the basis of composition; composite columns, in-
filled columns, etc. 2
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c) On the basis of lateral reinforcement; tied columns,
spiral columns.
d) On the basis of manner by which lateral stability isprovided to the structure as a whole; braced columns,
un-braced columns.
e) On the basis of sensitivity to second order effect due
to lateral displacements; sway columns, non-swaycolumns.
f) On the basis of degree of slenderness; short column,
slender column.g) On the basis of loading: axially loaded column,
columns under uni-axial bending, columns under
biaxial bending.
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4.0 Introduction
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4.0 Introduction The more general terms compression members and
members subjected to combined axial loads & bending
are used to refer to columns, walls, and members in
concrete trusses and frames.
These may be vertical, inclined, or horizontal.
A column is a special case of a compression memberthat is vertical.
Although the theory developed in this chapter applies to
columns in seismic regions, such columns require
special detailing to resist the shear forces and repeated
cyclic loading from the EQ.
In seismic regions the ties are heavier & more closely
spaced. 5
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4.1 Tied and Spiral Columns
Most of the columns in buildings in nonseismic regions
are tied columns.
Occasionally, when high strength and/or ductility are
required, the bars are placed in a circle, and the ties
are replaced by a bar bent into a helix or a spiral, with apitch from 35 to 85 mm.
Such a column, is called a spiral column. (Fig. 11-3)
The spiral acts to restrain the lateral expansion of the
column core under axial loads causing crushing and, indoing so, delays the failure of the core, making the
column more ductile.
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4.1 Tied and Spiral Columns
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4.1.1 Behavior of Tied and Spiral Columns
Fig. 11-4a shows a portion of the core of a spiral columnenclosed by one and a half turns of a spiral.
Under a compressive load, the concrete in this column
shortens longitudinally under the stress f 1 and so, to
satisfy the Poisson’s ratio, it expands laterally.
This lateral expansion is especially pronounced at
stresses in excess of 70% of the cylinder strength.
In spiral column, the lateral expansion of the concreteinside the spiral (the core) is restrained by the spiral.
These stresses the spiral in tension (see fig 11.14).
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4.1 Tied and Spiral Columns
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Fig 11-4 Triaxial stresses
in core of spiral column
f 1
f 2
f 2
f 2
f 1
f 2
f 1
f 2
f 2
f spf sp
s
f sp
f sp
Dc
f 1
4.1 Tied and Spiral Columns
f 2
f 2
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For equilibrium the concrete is subjected to lateral
compressive stresses, f 2.An element taken out of the core (see fig) is subjected
to triaxial compression which increases the strength of
concrete: f 1 = f c’+4.1f 2.
In a tied column in a non-seismic region, the ties are
spaced roughly the width of the column apart and thus
provide relatively little lateral restraint to the core.
Hence, normal ties have little effect on the strength ofthe core in a tied column.
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4.1 Tied and Spiral Columns
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They do, however, act to reduce the unsupported length
of the longitudinal bars, thus reducing the danger ofbuckling of those bars as the bar stresses approach
yield.
Fig 11-5 presents load-deflection diagrams for a tied
column and a spiral column subjected to axial loads.
The initial parts of these diagrams are similar.
As the maximum load is reached, vertical cracks and
crushing develop in the concrete shell outside the tiesor spiral, and this concrete spalls off .
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4.1 Tied and Spiral Columns
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When this occurs in a tied column, the capacity of the
core that remains is less than the load on the column. The concrete core is crushed, and the reinforcements
buckle outward b/n ties.
This occurs suddenly, w/o warning, in a brittle manner.
When the shell spalls off a spiral column, the column
does not fail immediately because the strength of the
cores has been enhanced by the triaxial stresses.
As a result, the column can undergo large deformations,eventually reaching a 2nd maximum load, when the
spirals yield and the column finally collapses.
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4.1 Tied and Spiral Columns
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Fig 11-5
4.1 Tied and Spiral Columns
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Such a failure is much more ductile and gives warning
of impending failure, along with possible loadredistribution to other members
Fig 11-6 and 11-7 show tied and spiral columns,respectively, after an EQ. Both columns are in the same
building and have undergone the same deformations. The tied column has failed completely, while the spiral
column, although badly damaged, is still supporting aload.
The very minimal ties in Fig 11-6 were inadequate toconfine the core concrete.
Had the column been detailed according to ACI Section21.4, the column would have performed much better.
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4.1 Tied and Spiral Columns
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4.1 Tied and Spiral Columns
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4.1 Tied and Spiral Columns
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Sway Frames vs. Nonsway Frames
A nonsway (braced) frame is one in which the lateralstability of the structure as a whole is provided by walls,
bracings, or buttresses, rigid enough to resist all lateral
forces in the direction under consideration.
A sway (unbraced) frame is one that depends on
moments in the columns to resist lateral loads and
lateral deflections. The applied lateral-load moment, Vl,
and the moment due to the vertical loads, ΣP∆ shall be
equilibrated by the sum of the moments at the top and
bottom of all the columns as shown in the figure below.
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Sway Frames vs. Nonsway Frames
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Fig. Non-sway Frame / Braced
columnsFig. Sway Frame/ Un-braced
columns
Sway Frames vs. Nonsway Frames
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For design purpose, a given story in a frame canbe considered “non-sway” if horizontaldisplacements do not significantly reduce thevertical load carrying capacity of the structure.
In other words, a frame can be “non-sway” if theP-∆ moments due to lateral deflections are smallcompared with the first order moments due tolateral loads.
In sway frames, it is not possible to considercolumns independently as all columns in thatframe deflect laterally by the same amount.
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Sway Frames vs. Nonsway Frames
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Columns are broadly categorized in to two as short
and slender columns.
Short columns are columns for which the strength is
governed by the strength of the materials and the
geometry of the cross section. In short columns, Second-order effects are negligible.
In these cases, it is not necessary to consider
slenderness effects and compression members can
be designed based on forces determined from first-
order analyses.
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Slender Columns vs. Short Columns
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When the unsupported length of the column is long,lateral deflections shall be so high that the moments
shall increase and weaken the column.
Such a column, whose axial load carrying capacity is
significantly reduced by moments resulting fromlateral deflections of the column, is referred to as a
slender column or sometimes as a long column.
“Significant reduction”, according to ACI, has been
taken to be any value greater than 5%.
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Slender Columns vs. Short Columns
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When slenderness effects cannot be neglected, the
design of compression members, restraining beamsand other supporting members shall be based on the
factored forces and moments from a second-order
analysis.
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Slender Columns vs. Short Columns
Fig. Forces in slender column
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4.2 Strength of Axially Loaded Columns
When a symmetrical column is subjected to a
concentric axial load, P, longitudinal strains , developuniformly across the section as shown in Fig 11-8a.
Because the steel & concrete are bonded together, the
strains in the concrete & steel are equal.
For any given strain, it is possible to compute the
stresses in the concrete & steel using the stress-strain
curves for the two materials.
24
Failure occurs when Po reaches amaximum:
Po = f cd(Ag – As,tot ) + f ydAs,tot
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4.3 Interaction Diagrams
Almost all compression members in concrete structures
are subjected to moments in addition to axial loads.
These may be due to misalignment of the load on the
column, as shown in Fig 11-9b, or may result from the
column resisting a portion of the unbalanced momentsat the ends of the beams supported by the columns (Fig
11-9c).
The distance e is referred to as the eccentricity of load.
These two cases are the same, because the eccentric
load can be replaced by an axial load P plus a moment
M=Pe about the centroid.
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4.3 Interaction Diagrams
Fig. 11-9 Load and moment
on column.
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The load P and moment M are calculated w.r.t. the
geometric centroidal axis because the moments andforces obtained from structural analysis are referred to
this axis.
For an idealized homogeneous and elastic column with a
compressive strength, f cu, equal to its tensile strength,
f tu, failure would occur in compression when the
maximum stresses reached f cu, as given by:
Dividing both sides by f cu gives:
27
4.3 Interaction Diagrams
f cu
f cu = 1
= f cu
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The maximum axial load the column can support occurs
when M = 0 and is Pmax = f cuA. Similarly, the maximum moment that can be supported
occurs when P = 0 and M is Mmax = f cuI/y.
Substituting Pmax and Mmax gives:
This equation is known as an interaction equation,
because it shows the interaction of, or relationship
between, P and M at failure. Points on the lines plotted in this figure represent
combinations of P and M corresponding to the
resistance of the section.28
Pmax Mmax = 1
4.3 Interaction Diagrams
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Fig. 11-10 Interaction diagram for an
elastic column, |f cu| = |f tu|. 29
4.3 Interaction Diagrams
A point inside the diagram, such as
E , represents a combination of Pand M that will not cause failure
Combinations of P and M falling onthe line or outside the line, such as
point F , will equal or exceed the
resistance of the section and hence
will cause failure
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4.4 Interaction Diagrams for Reinforced
Concrete Columns Since reinforced concrete is not elastic & has a tensile
strength that is lower than its compressive strength,
the general shape of the diagram resembles fig. 11.13
4.4.1 Strain Compatibility Solution
Interaction diagrams for columns are generallycomputed by assuming a series of strain distributions
at the ULS, each corresponding to a particular point on
the interaction diagram, and computing the
corresponding values of P and M.
Once enough such points have been computed, the
results are summarized in an interaction diagram.
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Assume strain distribution and select the location of the
neutral axis.
Compute the strain in each level of reinforcement from
the strain distribution.
Using this information, compute the size of thecompression stress block and the stress in each layer of
reinforcement.
Compute the forces in the concrete and the steel layers,by multiplying the stresses by the areas on which they
act.
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4.4 Interaction Diagrams for Reinforced
Concrete Columns
i i f i f d
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Finally, compute the axial force Pn by summing theindividual forces in the concrete and steel, and the
moment Mn by summing the moments of these forces
about the geometric centroid of the cross section.
These values of Pn and Mn represent one point on the
interaction diagram.
Other points on the interaction diagram can be
generated by selecting other values for the depth, c, tothe neutral axis from the extreme compression fiber.
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4.4 Interaction Diagrams for Reinforced
Concrete Columns
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33
10o/oo
(3/7)h
=2.0o/oo
A
B
10o/oo
=3.5o/oo
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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34
Fig 11-13 Strain distributions
corresponding to points on
the interaction diagram
Moment resistance
A x i a l l o a d
r e s i s t a n c e
Onset of cracking
Balanced failure
Uniform compression
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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In the actual design, interaction charts prepared foruniaxial bending can be used. The procedure involves:
Assume a cross section, d’ and evaluate d’/h to chooseappropriate chart
Compute:
Normal force ratio: = /ℎ Moment ratios:
= /ℎ
Enter the chart and pick ω (the mechanical steel ratio), ifthe coordinate (ν, μ) lies within the families of curves. Ifthe coordinate (ν, μ) lies outside the chart, the crosssection is small and a new trail need to be made.
Compute , = / Check Atot satisfies the maximum and minimum provisions Determine the distribution of bars in accordance with the
charts requirement
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f f d
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Draw the interaction diagram for the column crosssection. Use C-30 concrete and S-460 steel.
Show a minimum of 6 points on the interaction
diagram corresponding to1. Pure axial compression
2. Balanced failure
3. Zero tension (Onset of cracking)
4. Pure flexure5. A point b/n balanced failure and pure flexure
6. A point b/n pure axial compression and zero tension
37
4.4 Interaction Diagrams for Reinforced
Concrete Columns
f f d
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4.4 Interaction Diagrams for Reinforced
Concrete Columns
i i f i f d
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• Solution
1. Pure axial compression
39
s2
s1
Cs2
Cs1
Ccc
= 2/oo
Cross section Strain distribution Stress resultant
(ULS)
4.4 Interaction Diagrams for Reinforced
Concrete Columns
4 4 I i Di f R i f d
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Cont’d
Pu = Cs2 + Cs1 + Cc = s2As2 + s1As1 + f cdbh
yd = f yd/Es = 400/200000 = 0.002
reinforcement has yielded
Pu = f ydAs,tot/2 + f ydAs,tot/2 + f cdbh = f ydAs,tot + f cdbh
u = Pu / f cdbh = (f ydAs,tot )/ (f cdbh)+ (f cdbh)/(f cdbh)
= f yd/f cd + 1 = + 1 ; where is called the
mechanical reinforcement ratio and equal to= (6800/(400500))(400/13.6) = 1.0
u = 1 + 1 = 2.0
NB: νu= Relative design axial force (Pu/ (f cdAc)) 40
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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4.4 Interaction Diagrams for Reinforced
Concrete Columns
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2. Balanced failure
– x/3.5 = d/(3.5+2)
x = (400/5.5)3.5 = 254.5454mm s2/(254.54-100)=3.5/254.54
s2 = 2.125 %0 > 2 %0 reinforcement has yielded
Cs2 = Ts1 = 3400400 = 1360000N 42
s2
ydTs1
cm=3.5 /oo
X Cs2Cc
= 2/oo
Cross section strain distribution stress resultant
(ULS)
cd
4.4 Interaction Diagrams for Reinforced
Concrete Columns
4 4 I t ti Di f R i f d
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•Cont’d
• cm > o and NA is within the section
c = kx(3cm – 2)/3cm
= (254.54/400)(33.5-2)/(33.5) = 0.5151 Cc = cf cdbd = 0.5113.6400400 = 1120967.7 N
• c = kx(cm(3cm-4)+2)/(2cm(3cm-2))
= (254.54/400)(3.5(3
3.5-4)+2)/(2
3.5(3
3.5-2))
= 0.2647 cd = 0.2647400 = 105.882 mm
Mu = Cc(h/2- cd) + Cs2(h/2-h’) + Ts1(h/2-h’)
43
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– Cont’d
= 1120969.7(250-105.882) + 1360000(250-100) +1360000(250-100)
= 569551912.2 Nmm
• Pu= Cc+Ts1+Cs2 = Cc = 1120969.7
u = P/(f cdbh)
= 1120969.7/(13.6400500)
= 0.412u = Mu/(f cdbh2) = 569551912.2/(13.64005002)
= 0.419
NB: u = Mu/(f cdbh
2
) is called relative moment 44
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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6. A point b/n pure axial compression and zero
tension
45
Cs2
Cs1
Cc
= 2/oo
Cross Section Strain Distribution Stress Resultant
(ULS)
cm=3.0 /oo 1.0
C
e
b
(3/7)h
cba
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– Choose cm = 3 %o (strain profile passes also through C)
– Strain in the bottom concrete fiber cb from:
a = ((4/7)/(3/7))1 = 4/3 = 1.33
cb = 2 -1.33 = 0.667 %o
(entire cross section under compression as assumed) – Determine strain in reinforcement from:
– b/114.286 = 1/214.286 b = 0.533 %o
s2 = 2 + 0.533 = 2.533 %o
> 2 %o
reinforcement hasyielded and
– from e/185.714 = 1.33/285.714 e = 0.867 %o
s1 = 2-0.867 = 1.133 %o < 2 %o reinforcement has
not yielded 46
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– Cont’d
– cm > o and NA outside of the section
c = (1/189)(125+64cm-16cm2)
= (1/189)(125+64(3)-16(3)2) = 0.915344
Cc = cf cdbd = 0.91534413.6400400 = 1991788.4 N;
Cs2 = (As,tot/2)f yd = 3400400 = 1360000 N;
Cs1 = (As,tot/2)s1 = 3400(1.133/1000)200000
= 770666.67 N – c = 0.5-(40/7)(cm-2)
2/(125+64cm-16cm2)
= 0.5-(40/7(3-2)2/(125+643-1632) = 0.467
cd = 0.467400 = 186.788 mm 47
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– Cont’d
– Pu = Cc+Cs1+Cs2 = 4122455.1 N
– Mu = 1991788.4(250-186.788)+1360000(250-
100)-770666.67(250-100) = 214304927.8 Nmm
– u = Pu/(f cdbh) = 4122455.1/(13.6400500) =
1.516 – u = Mu/(f cdbh
2) = 214304927.8/(13.64005002)
= 0.1576
48
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– 4. Pure flexure
– Start with cm/s1 = 3.5 /oo / 5 /oo and repeat until
u 0.00
49
cm=3.5 /oo
X
Cs2Cc
Cross section strain distribution stress resultant
cd
s1=5.0 /oo
4.4 Interaction Diagrams for Reinforced
Concrete Columns
Ts1
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–Cont’d
– Use cm/s1 = 3.5 /oo / 6.227 /oo
– kx= 3.5/(3.5+6.227)=0.359823 x=143.9293mm
–
s2 = ((x-100)/x)3.5 = 1.06825 /oo
Cs2=3400(1.06825/1000)200000=726410N
– c = kx(3cm – 2)/3cm
= (143.9293/400)(33.5-2)/(33.5) = 0.291285
Cc = cf cdbd = 0.29128513.6400400 = 633837.1N
– Pu=Cc+Ts1+Cs2 = 633837.1-1360000+726410 = 247.1N
u = Pu/(f cdbh) = 247.1/(13.6400500) = 0.0050
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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– c = kx(cm(3cm - 4) + 2)/(2cm(3cm - 2))
= (143.9293/400)(3.5(33.5-4)+2)/(23.5(33.5-2))
= 0.149674
cd = 0.149674400 = 59.8696 mm – Mu = Cc(h/2- cd) + Cs2(h/2-h’) + Ts1(h/2-h’)
= 633837.1(250-59.8696) + 726410(250-100) +
1360000(250-100)
= 433473111 Nmm
– u = Mu/(f cdbh2)
= 433473111 /(13.6 400 5002) = 0.31951
4.4 Interaction Diagrams for Reinforced
Concrete Columns
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0.2 0.4 0.6 0.8
4.4 Interaction Diagrams for RC Columns
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All columns are (in a strict sense) to be treated as
being subject to axial compression combined with
biaxial bending, as the design must account for
possible eccentricities in loading (emin at least) with
respect to both major and minor principal axes of thecolumn section.
Uniaxial loading is an idealized approximation which
can be made when the e/D ratio with respect to one
of the two principal axes can be considered to benegligible.
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4.5 Biaxially Loaded Columns
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Also, if the e/D ratios are negligible with respect to
both principal axes, conditions of axial loading may
be assumed, as a further approximation.
For a given cross section and reinforcing pattern, one
can draw an interaction diagram for axial load andbending about either axis.
These interaction diagrams form the two edges of an
interaction surface for axial load and bending about
2 axes.
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4.5 Biaxially Loaded Columns
I i di f i l
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Interaction diagram for compression plus
bi-axial bending
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As shown in the figure above, the interaction diagram
involves a three-dimensional interaction surface foraxial load and bending about the two axes.
The calculation of each point on such a surface involvesa double iteration:
–
The strain gradient across the section is varied, and – The angle of the neutral axis is varied.
There are different methods for the design of Biaxiallyloaded columns:
– Strain compatibility method
– The equivalent eccentricity method
– Load contour method
– Bresler reciprocal load method
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4.5 Biaxially Loaded Columns
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Biaxial interaction diagrams calculated and prepared as
load contours or P-M diagrams drawn on planes of
constant angles relating the magnitudes of the biaxial
moments are more suitable for design (but difficult to
derive). Interaction diagrams are prepared as load contours for
biaxially loaded columns with different reinforcement
arrangement (4-corner reinforcement, 8-rebar
arrangement, uniformly distributed reinforcement on2-edges, uniformly distributed reinforcement on 4-
edges and so on.
57
4.5 Biaxially Loaded Columns
ll d d l
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The procedure for using interaction diagrams for
columns under biaxial bending involves:
Select cross section dimensions h and b and also h’
and b’
Calculate h’/h and b’/b and select suitable chart
Compute:
– Normal force ratio: = /ℎ – Moment ratios: = / ℎ and = /
58
4.5 Biaxially Loaded Columns
ll d d l
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Select suitable chart which satisfy and h’/h and b’/bratio:
Enter the chart to obtain ω
Compute
, = /
Check Atot satisfies the maximum and minimum
provisions
Determine the distribution of bars in accordance
with the chart’s requirement.
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4.5 Biaxially Loaded Columns
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Fig. SampleInteraction
diagram
Anal sis of col mns according to EBCS 2
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Analysis of columns according to EBCS 2
(short and slender)
Classification of Frames
A frame may be classified as non-sway for a given
load case if the critical load ratio for that load case
satisfies the criterion: ≤ 0.1
Where: Nsd is the design value of the total vertical load
Ncr is its critical value for failure in a sway mode
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In Beam-and-column type plane frames in buildingstructures with beams connecting each column at eachstory level may be classified as non-sway for a given
load case, when first-order theory is used, thehorizontal displacements in each story due to thedesign loads (both horizontal & vertical), plus the initialsway imperfection satisfy the following criteria.
≤ 0.1 Where: δ is the horizontal displacement at the top of the
story, relative to the bottom of the story
L is the story height
H is the total horizontal reaction at the bottom of the story
N is the total vertical reaction at the bottom of the story,
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The displacement δ in the above equation shall
be determined using stiffness values for beams
and columns corresponding to the ultimatelimit state.
All frames including sway frames shall also be
checked for adequate resistance to failure innon-sway modes.
D t i ti f t b kli L d N
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Determination of story buckling Load Ncr
Unless more accurate methods are used, the buckling load
of a story may be assumed to be equal to that of thesubstitute beam-column frame defined in Fig. and may bedetermined as:
=
Where: EIe is the effective stiffness of the substitute columndesigned using the equivalent reinforcement area.
Le is the effective length. It may be determined using the
stiffness properties of the gross concrete section for bothbeams and columns of the substitute frame (see Fig. 4.4-1c )
I li f t d t i ti th ff ti
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• In lieu of a more accurate determination, the effectivestiffness of a column EIe may be taken as:
= 0.2
Where: Ec = 1100f cd
Es is the modulus of elasticity of steel
Ic, Is, are the moments of inertia of the concrete andreinforcement sections, respectively, of the substitute
column, with respect to the centroid of the concretesection (see Fig. 4.4-1c).
• or alternatively
= (1 ) ≥ 0.4 • Where: Mb is the balanced moment capacity of the
substitute column
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1/rb is the curvature at balanced load & may be taken as
1 = 5
10−
The equivalent reinforcement areas, As, tot, in the substitute
column to be used for calculating Is and Mb may be obtained
by designing the substitute column at each floor level to
carry the story design axial load and amplified sway momentat the critical section.
Sl d R ti
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Slenderness Ratio
Slenderness ratio, λ of a column is defined as the ratio
of the effective length lei to the radius of gyration.NB: λ also provides a measure of the vulnerability to
failure of the column by elastic instability (buckling)
=
where: Le is the effective buckling length
i is the minimum radius of gyration.
The radius of gyration is equal to
= Where: I is the second moment of area of the section,
A is cross sectional area
Limits of Slenderness
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Limits of Slenderness The slenderness ratio of concrete columns shall not
exceed 140.
Second order moment in a column can be ignored if:
For sway frames, the greater of: ≤ 25
15/
= / For non-sway frames: ≤ 50 25 Where: M
1
and M2
are the first-order end moments, M2
being always positive and greater in magnitude than M1,
and M1 being positive if member is bent in single curvature
and negative if bent in double curvature
Effective Length of Columns
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Effective Length of Columns
Effective buckling length is the length between points
of inflection of columns and it is the length which iseffective against buckling.
The greater the effective length, the more likely the
column is to be buckle.
Le = kL
where k is the effective length factor (i.e., the ratio of
effective length to the unsupported length whose value
depends on the degrees of rotational and translation
restraints at the column ends.
For idealized boundary conditions:
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y
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The rotational restraint at a column end in a building
frame is governed by the flexural stiffnesses of the
members framing into it, relative to the flexuralstiffness of the column itself.
Hence, it is possible to arrive at measures of the
‘degree of fixity’ at column ends, and thereby arrive at
a more realistic estimate of the effective length ratio of
a column than the estimates (for idealized boundary
conditions).
These involve use of alignment charts or approximateequations.
The alignment chart (Nomograph): for members
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The alignment chart (Nomograph): for members
that are parts of a framework
Approximate equations
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Approximate equations
The following approximate equations can be used
provided that the values of α1 and α2 don’t exceed 10(see EBCS 2).
Non-sway mode
= 0.4 0.8 ≥ 0.7 In Sway mode
=
.+
+
+.
.++ ≥ 1.15
Or Conservatively,
1 0 8 ≥ 1 15
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Where α1 and α2 are as defined above and αm is
defined as:
= 2 Note that: for flats slab construction, an equivalent
beam shall be taken as having the width and
thickness of the slab forming the column strip.
The effect of end restrainet is quantified by the two
end restrain factors α1 and α2
= / /
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Where Ecm is modulus of elasticity of concrete
Lcol is column height
Lb is span of the beam
Icol, Ib are moment of inertia of the column and beam
respectively
β is factor taking in to account the condition of
restraint of the beam at the opposite end
= 1.0 opposite end elastically or rigidly restrained
= 0.5 opposite end free to rotate
= 0 for cantilever beam
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Note that: if the end of the column is fixed, theoretically,α = 0, but an α value of 1 is recommended for use.
On the other hand, if the end of the member is pinned,the theoretical value of α is infinity, but an α value of 10is recommended for use.
The rationale behind the foregoing recommendations isthat no support in reality can be truly fixed or pinned.
Design of columns, EBSC-2 1995
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General:
The internal forces and moments may generally be
determined by elastic global analysis using either first
order theory or second order theory.
First-order theory, using the initial geometry of the
structure, may be used in the following cases – Non-sway frames
– Braced frames
– Design methods which make indirect allowances for
second-order effects.
Second-order theory, taking into account the influence
of the deformation of the structure, may be used in all
cases.
Design of Non-sway Frames
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Design of Non sway Frames
Individual non-sway compression members shall be
considered to be isolated elements and be designedaccordingly.
Design of Isolated Columns
For buildings, a design method may be used which
assumes the compression members to be isolated.
The additional eccentricity induced in the column by
its deflection is then calculated as a function of
slenderness ratio and curvature at the critical section
Total Eccentricity
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Total Eccentricity
The total eccentricity to be used for the design of
columns of constant cross-section at the critical
section is given by:
= Where: ee is equivalent constant first-order
eccentricity of the design axial load
ea is the additional eccentricity allowance
for imperfections.
For isolated columns: = 300 ≥ 20
e2 is the second-order eccentricity
First order equivalent eccentricity
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q y
i. For first-order eccentricity e0 is equal at both ends of
a column = ii. For first-order moments varying linearly along the
length, the equivalent eccentricity is the higher of the
following two values: = 0.6 0.4
= 0.4 Where: e01 and e02 are the first-order eccentricities at the
ends, e02 being positive and greater in magnitude than e01.
e01 is positive if the column bents in single curvature and
negative if the column bends in double curvature.
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iii. For different eccentiries at the ends, (2) above, the
critical end section shall be checked for first order
moments:
=
Second order eccentricity
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Second order eccentricity
i. The second-order eccentricity e2 of an isolated
column may be obtained as
=
10 (1 ) Where: Le is the effective buckling length of the column
k1= λ/20 - 0.75 for 15≤ λ ≤ 35
k1= 1.0 for λ >35
l/r is the curvature at the critical section.
ii. The curvature is approximated by:1 =
5 10
−
Wh d i h ff i l di i i h l f
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Where: d is the effective column dimension in the plane of
buckling
k2 =Md /Mb Md is the design moment at the critical section including
second-order effects
Mb is the balanced moment capacity of the column.
iii. The appropriate value of k2 may be found iteratively
taking an initial value corresponding to first-order
actions.
Reinforcement Detail
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Reinforcement Detail
Size
The minimum lateral dimension of a column shall beat least 150mm
Longitudinal Reinforcement
The minimum number of bars to be used in columnsshall be:
Four in rectangular columns (One at each corner)
Six in circular columns
The minimum bar diameter shall be 12 mm.
The minimum area of steel shall be 0.008 Ac for
ductility.
Th i f t l h ll b 0 08 A i l di
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The maximum area of steel shall be 0.08 Ac, including
laps.
Lateral Reinforcement Minimum bar size shall be ¼ times the largest
compression bar, but not less than 6mm.
Center to center spacing shall not exceed:
12 times minimum diameter of longitudinal bars
Least dimension of column
300 mm.
Spiral or circular ties may be used for longitudinal
bars located around the perimeter of a circle. The
pitch of spirals shall not exceed 100 mm.
Design of Sway Frames
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Design of Sway Frames
Nonsway moments, Mns, are the factored end
moments on a column due to loads that cause
no appreciable sidesway as computed by a
first-order elastic frame analysis. They result
from gravity loads.
Sway moments, Ms, are the factored end
moments on a column due to loads whichcause appreciable sidesway, calculated by a
first-order elastic frame analysis.
First-order vs. Second-order Frame
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First order vs. Second order Frame
Analyses:
A first-order frame analysis is one in which the effect oflateral deflections on bending moments, axial forces,
and lateral deflections is ignored.
The resulting moments and deflections are linearlyrelated to the loads.
When slenderness effects cannot be neglected, the
design of compression members, restraining beams and
other supporting members shall be based on thefactored forces and moments from a second-order
analysis.
First order vs Second order Frame
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A second-order frame analysis, is one which considersthe effects of deflections on moments, and so on.
The resulting moments and deflections include the
effects of slenderness and hence are nonlinear withrespect to the load.
The stiffness, EI, used in the analysis should represent
the stage immediately prior to yielding of the flexural
reinforcement since the moments are directly affectedby the lateral deflections.
First-order vs. Second-order Frame
Analyses:
Methods of Second Order Analysis
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y
a) Iterative P-∆ Analysis:
When a frame is displaced by an amount ∆ due to
lateral loads, additional moments shall be induced
due to the vertical loads (ΣP∆).
This shall be loaded as sway force (ΣP∆/l) and thestructure shall be re-analyzed until convergence is
achieved.
The values can be assumed to converge when the
change in deflection between two consecutiveanalyses is less than 2.5 percent . (ACI)
89
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Fig. 4 Determination of sway forces
90
b) Di t P ∆ l i
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b) Direct P-∆ analysis
It describes the iterative P-∆ analysis mathematicallyas an infinite series.
An estimate of final deflections is obtained directly
from the first-order deflections (from the sum of the
terms in the series).
Thus the second-order moments are δsMs Where: and
)()(1 0
0
Vl P u
0.11
1
Q s
vl
P Q u 0
)(
91
c) Amplified Sway Moments Method:
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) p y
First-order sway moments are multiplied by a sway
moment magnifier, δs, to obtain the second ordermoments.
M1 = M1ns + δsM1s
M2 = M
2ns + δ
sM
2s
Where: (According to ACI)
(According to EBCS) Where: Pu, NSd is the design value of the total vertical load
Pc, Ncr is its critical value for failure in a sway mode
1)75.0(1
1
cu s
P P
cr Sd
s
N N 1
1
Recommended