CHAPTER 4
COLUMNS: COMBINED AXIAL LOAD & BENDING
Abrham E. Sophonyas Sophony as A. A.
1
4.0 Introduction A column is a vertical structural member supporting
axial compressive loads, with or without moments. The cross-sectional dimensions 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 the basis basis of geomet geometry; ry; recta rectangul ngular ar,, squar square, e, circul circular ar,, L-shaped, T-shaped, etc. depending on the structural or architectural requirements b) On the the basi basiss of of comp composi ositio tion; n; comp composi osite te colum columns, ns, infilled columns, etc.
2
c) On the basis basis of of lat later eral al reinf reinfor orcem cement ent;; tied tied column columns, s, spiral columns. d) On the the basis basis of mann manner er by by which which later lateral al stabi stabilit lity y is provided to the structure as a whole; who le; braced columns, un-braced columns. e) On the the basis basis of sens sensiti itivit vity y to to secon second d order order eff effect due to lateral displacements; sway columns, non-sway columns. f) On the the basi basiss of of degr degree ee of slend slender erne ness; ss; short short colu column mn,, slender column.
g) On the basis basis of of loadi loading ng:: axial axially ly loade loaded d colu column mn,, columns under uni-axial bending, columns under biaxial bending. 3
4.0 Introduction
4
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. horizontal. These may be vertical, inclined, or horizontal. A column is a special case of a compression member
that 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
4.1 Tied and Spiral Spiral Columns Most of the columns in buildings in nonseismic regions
are tied columns.
Occasionally, when high strength and/or ductility are ductility are Occasionally, required, the bars are placed in a circle, and the ties are replaced by a bar bent into a helix or a spiral spiral,, with a pitch from 35 to 85 mm. column. (Fig. 11-3) Such a column, is called a spiral column. restrain the the lateral expansion of the The spiral acts to restrain column core under axial loads causing crushing and, in doing so, delays the failure of the core, making the column more ductile ductile.. 6
4.1 Tied and Spiral Spiral Columns
7
4.1 Tied and Spiral Spiral Columns 4.1.1 Behavior of Tied and Spiral Columns Fig. 11-4a shows a portion of the core of a spiral column
enclosed 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% 70% of of the cylinder strength. l ateral al expansion of the concrete In spiral column, the later inside the spiral (the core) is restrained by the spiral. tension (see (see fig 11.14). These stresses the spiral in tension 8
4.1 Tied and Spiral Spiral Columns f 1 f sp f 2 f sp
s
f 2
f sp
Dc f 1
f sp
f 1 f 2 f 2
f 2
f 2
Fig 11-4 Triaxial stresses in core of spiral column
f 2
f 2 f 1 9
4.1 Tied and Spiral Spiral Columns 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 increases the the strength of concrete: f 1 = f c’+4.1f 2. tied column column in a non-seismic region, the ties are In a tied spaced roughly the width of the column apart and thus provide relatively little lateral restraint to the core. effect on the strength of Hence, normal ties have little effect
the core in a tied column.
10
4.1 Tied and Spiral Spiral Columns They do, however, act to reduce the unsupported length
of the longitudinal bars, thus reducing the danger of buckling of those bars as the bar stresses approach yield. load-deflection diagrams diagrams for a tied Fig 11-5 presents load-deflection column and a spiral column subjected to axial loads. initial parts parts of these diagrams are similar. The initial load is reached, vertical cracks and As the maximum load is crushing develop in the concret concrete e shell s hell outside the ties or spiral, and this concrete spalls off .
11
4.1 Tied and Spiral Spiral Columns column, the capacity of the When this occurs in a tied column, core that remains is less less than than the load load on on the column. concrete e core is crushed crushed,, and the reinforcements The concret buckle outward b/n ties. manner. 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 strength of the cores has been enhanced by the triaxial stresses. 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. 12
4.1 Tied and Spiral Spiral Columns
Fig 11-5
13
4.1 Tied and Spiral Spiral Columns Such a failure is much more ductile and gives warning
of impending failure, along with possible load redistribution redistribution 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 deforma deformations tions.. The tied column has failed completely, while the spiral column,, although badly damaged, is still supporting column supporting a a load. inadequate to The very minimal ties in Fig 11-6 were inadequate confine the core concrete. Had the column been detailed according to ACI Section 21.4, the column would have performed much better. 14
4.1 Tied and Spiral Spiral Columns
15
4.1 Tied and Spiral Spiral Columns
16
Sway Frames vs. Nonsway Frames A nonsway (braced) frame is one in which the lateral
stability 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. consideration. A sway (unbraced) frame is one that depends on
moments in the columns to resist lateral loads and lateral lateral deflections. The applied lateral-load lateral-load moment, Vl, and the moment due to the vertical loads, ΣP∆ shall be equilibrated equilibrated by the sum of the moments at the top and bottom of all the columns as shown in the figure below. below. 17
Sway Frames vs. Nonsway Frames
18
Sway Frames vs. Nonsway Frames
Fig. Non-sway Frame / Braced columns
Fig. Sway Frame/ Un-braced columns
19
Sway Frames vs. Nonsway Frames For design purpose, a given story in a frame can
be considered “non“non-sway” if horizontal displacements displacements do not significantly reduce the vertical load carrying capacity of the structure. In other words, a frame can be “non“non -sway sway ” if the P-∆ P-∆ moments due to lateral deflections are small compared with the first order moments due to lateral loads. In sway frames, it is not possible to consider columns independently independently as all columns in that frame deflect laterally by the same amount. 20
Slender Columns vs. Short Columns categorized in to two as short Columns are broadly categorized and slender slender columns. 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 effects and compression members can be designed based on forces determined from firstorder analyses. analyses.
21
Slender Columns vs. Short Columns When the unsupported length of the column is long,
lateral 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 from lateral 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%.
22
Slender Columns vs. Short Columns effects cannot be neglected, the When slenderness effects design of compression members, restraining restraining beams and other supporting members shall be based on the factored factored forces and moments from a second-order analysis.. analysis
Fig. Forces in slender column 23
4.2 Strength Strength of Axially Loaded Columns When a symmetrical column is subjected to a
strains , develop concentric axial load, P, longitudinal strains uniformly across the section as shown in Fig 11-8a. Because the steel & concrete are bonded together, the
strains strains in the concrete & steel are equal. strain, it is possible to compute the For any given strain, stresses in the concrete & steel using the stress-strain curves for curves for the two materials. materials. Failure occurs occ urs when Po reaches a Failure maximum: Po = f cd(Ag – – A As,tot ) + f ydAs,tot 24
4.3 Interaction Diagrams structures 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 from the column resisting a portion of the unbalanced moments at 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. 25
4.3 Interaction Diagrams
Fig. 11-9 Load and moment on column.
26
4.3 Interaction Diagrams The load P and moment M are calculated w.r.t. the
geometric centroidal axis because the moments and forces obtained from structur structural al analysis are referred to this axis. For an idealized homogeneous and elastic column with a
compressive strength, f cu, equal to its tensile strength, strength, f tu, failure failure would occur in compression when the maximum stresses reached f cu, as given by:
= f
cu
Dividing both sides by f cu gives:
= 1 f f cu
cu
27
4.3 Interaction Diagrams The maximum axial load the column can support occurs
when M = 0 and is P max = 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:
=1 M P max
max
This equation is known as an interaction equation ,
because it shows s hows the interaction interaction of, 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 resistance of the section. s ection. 28
4.3 Interaction Diagrams A point inside the diagram, such as E , represents a combination of P and M that will not cause failure
Combinations Combinations of P and M falling on the line or outside the line, such as point F , will equal or exceed the resistance of the section and hence will cause failure
Fig. 11-10 Interaction diagram for an elastic column, |f cu| = |f tu|.
29
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns concrete is not elastic el astic & has a tensile Since reinforced concrete strength strength that is lower than its compressive strength, strength, the general shape of the diagram resembles fig. 11.13
4.4.1 Strain Compatibility Solution Interaction diagrams for columns are generally
computed by assuming a series of strain distributions at the ULS, each corresponding to a particular point on the interaction diagram, 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. 30
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns 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 the
compression stress block and the stress in each layer of reinforcement. layers, Compute the forces in the concrete and the steel layers,
by multiplying the stresses by the areas on which they act. 31
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns Finally, compute the axial force P n by summing the
individual 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. represent one point poi nt on the These values of Pn and Mn represent interaction diagram. interaction diagram can be Other points on the interaction generated generated by selecting other values for the depth, c, to the neutral axis from the extreme compression fiber.
32
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns B =3.5o/oo
10o/oo (3/7)h
A 10o/oo
=2.0o/oo
33
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns Uniform compression
Onset of cracking
e c n a t s i s e r d a o l l a i x A
Balanced failure
Moment resistance
Fig 11-13 Strain distributions corresponding to points on the interaction diagram 34
In the actual design, interaction charts prepared for for
uniaxial bending can be used. The procedure involves: involves: cross section, d’ and evaluate evaluate d’/h to choose Assume a cross appropriate chart Compute: Normal force ratio: Moment ratios: ratio), if Enter the chart and pick ω (the mechanical steel ratio), the coordinate coordinate (ν, μ) lies within the families of curves. If the coordinate coordinate (ν, μ) lies outside the chart, the cross section 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 requirement
= / ℎ = / ℎ
, = /
35
36
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns Draw the interaction diagram for the column cross
section. Use C-30 concrete and S-460 steel. Show a minimum of 6 points on the interaction diagram corresponding to 1. 2. 3. 4. 5. 6.
Pure axi Pur axial al co comp mprres essi sion on Bala Ba lanc nce ed fail ilur ure e Zero Ze ro te tens nsio ion n (Ons (Onset et of crac cracki king ng)) Pure fl flexure A point point b/n b/n bala balance nced d fail failur ure e and and pure pure flex flexur ure e A point point b/n b/n pure pure axial axial comp compre ressi ssion on and and zero zero tens tension ion 37
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns
38
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns •
Solution 1. Pure axial compression s2
Cs2
Cc
c s1
Cs1
= 2 /oo Cross section
Strain distribution distributio n (ULS)
Stress resultant
39
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns Cont’d Pu = Cs2 + Cs1 + Cc =
yd = f yd/Es =
s2As2 +
s1As1 + f cdbh
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/(400 500)) (400/13.6) = 1.0 u =
1 + 1 = 2.0
NB: ν = Relative design axial force (P / (f cdA ))
40
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns
41
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns 2. Balanced failure o =3.5 /o cm
X
Cs2 Cc
s2
cd
Ts1 yd
Cross section –
= 2 /oo
strain distribution distributio n (ULS)
stress resultant
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 reinf orcement has yielded
C = T = 3400400 = 1360000N
42
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns •
Cont’d
•
cm > o and NA is within the section
– 2)/3 2)/3cm c = kx(3cm – = (254.54/400)(3 3.5-2)/(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(33.5-2))
= 0.2647 cd = 0.2647400 = 105.882 mm (h/2-h’) h’) + Ts1(h/2(h/2-h’) h’) Mu = Cc(h/2- cd) + Cs2(h/243
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns
Cont’d = 1120969.7(250-105.882) + 1360000(250-100) + 1360000(250-100) = 569551912.2 Nmm Pu= Cc+Ts1+Cs2 = Cc = 1120969.7 –
•
= P/(f cdbh) = 1120969.7/(13.6 400500) = 0.412 u = Mu/(f cdbh2) = 569551912.2/(13.64005002) = 0.419
u
NB:
u =
Mu/(f cdbh2) is called relative moment
44
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns 6. A point b/n pure axial compression and zero zero tension cm=3.0
/oo 1.0 b
(3/7)h
Cc
C cb
Cs2
e a = 2 /oo
Cs1
Cross Section Strain Distribution Distributi on Stress Resultant (ULS) 45
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns Choose cm = 3 %o (strain profile 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 reinforcemen reinforcementt from: b/114.286 = 1/214.286 b = 0.533 %o s2 = 2 + 0.533 = 2.533 %o > 2 %o reinforcement has yielded 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 Di agrams for Reinforced Concrete Columns –
–
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
d = 0.467400 = 186.788 mm
47
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns –
Cont’d
–
Pu = Cc+Cs1+Cs2 = 4122455.1 N
–
–
–
Mu = 1991788.4(250-186.788)+1360000(250100)-770666.67(250-100) = 214304927.8 Nmm = Pu/(f cdbh) = 4122455.1/(13.6 400500) = 1.516 u
= Mu/(f cdbh2) = 214304927.8/(13.64005002) = 0.1576 u
48
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns –
–
4. Pure flexure Start with cm/s1 = 3.5 / / 5 / and repeat until 0.00 u oo
cm=3.5
/oo
X s1=5.0
/oo
Cross section strain distribution
oo
Cs2 Cc
cd
Ts1
stress resultant
49
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns –
Cont’d
–
Use cm/s1 = 3.5 / / 6.227 /
–
kx= 3.5/(3.5+6.227)=0.359823 x=143.9293mm
–
s2 = ((x-100)/x)3.5 = 1.06825 /
oo
oo
oo
Cs2=3400 (1.06825/1000) 200000=726410N –
– 2)/3 2)/3cm c = kx(3cm – = (143.9293/400)(3 3.5-2)/(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/(13.6 400500) = 0.00
50
4.4 Interaction Diagrams Di agrams for Reinforced Concrete Columns –
c = kx(cm(3cm - 4) + 2)/(2 cm(3cm - 2)) = (143.9293/400)(3.5(3 3.5-4)+2)/(2 3.5(33.5-2)) = 0.149674
cd = 0.149674 400 = 59.8696 mm –
Mu = Cc(h/2- cd) + Cs2(h/2(h/2-h’) h’) + Ts1(h/2(h/2-h’) h’) = 633837.1(250-59.8696) + 726410(250-100) + 1360000(250-100)
= 433473111 Nmm –
u = Mu/(f cdbh2) = 433473111 /(13.6 400 5002) = 0.319 51
4.4 Interaction Diagrams for RC Columns 0.2
0.4
0.6
0.8
52
4.5 Biaxially Loaded Columns 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 pr incipal axes of the column section. approximation which Uniaxial loading is an idealized approximation can be made when the e/D ratio ratio with respect to one of the two principal axes can be considered to be negligible.
53
4.5 Biaxially Loaded Columns 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. se ction and reinforcing pattern, pattern, one For a given cross section can draw an interaction diagram for axial load and bending about either axis. These interaction diagrams form the two edges of an
interaction surface for axial load and bending about interaction 2 axes.
54
Interaction diagram for compression plus bi-axial bending
55
4.5 Biaxially Loaded Columns As shown in the figure above, the interaction diagram
involves a three-dimensional interaction surface for axial load and bending about the two axes. The calculation of each point on such a surface involves a double iteration: iteration: – –
The strain gradient across the section is varied, and The angle of the neutral axis is varied.
different methods for the design of o f Biaxially There are different loaded columns: – – – –
Strain compatibility method The equivalent eccentricity method Load contour method Bresler reciprocal load method 56
4.5 Biaxially Loaded Columns interaction diagrams calculated and prepared as Biaxial interaction load contours contours or P-M diagrams drawn drawn on planes of constant angles relating relating the magnitudes of the biaxial moments are more suitable for design (but difficult to derive).
contours for for Interaction diagrams are prepared as load contours biaxially loaded columns with different different reinforcement reinforcement arrangement arrangement (4-corner reinforcement, reinforcement, 8-rebar arrangement, uniformly distributed reinforcement on 2-edges, uniformly distributed reinforcement reinforcement on 4edges and so on. 57
4.5 Biaxially Loaded Columns interaction diagrams for The procedure for using interaction columns under biaxial bending involves: involves: Select cross section dimensions h and b and also h’
and b’ Calculate h’/h and b’/b and select suitable chart Compute: –
–
= / ℎ Moment ratios: = / ℎ and = /
Normal force ratio:
58
4.5 Biaxially Loaded Columns Select suitable chart which satisfy and h’/h and b’/b
ratio: 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.
59
Fig. Sample Interaction diagram
60
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
In Beam-and-column type plane frames in building
structures with beams connecting each column at each story level may be classified as non-sway for a given load case, when first-order theory is used, the horizontal displacements in each story due to the design loads (both horizontal horizontal & vertical), plus the initial i nitial sway imperfection satisfy the following following criteria. criteria.
Where: δ
≤ 0.1 is the horizontal displacement at 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,
The displacement δ in the above equation shall
be determined using stiffness values for beams and columns corresponding to the ultimate limit state. All frames including sway frames shall also be
checked for adequate resistance to failure in non-sway modes.
Determination of story buckling Load N cr Unless more accurate methods are used, the buckling load
of a story may be assumed to be equal to that of the substitute substitute beam-column frame defined in Fig. and may be determined as:
=
Where: EIe is the effective stiffness of the substitute column
designed using the equivalent reinforcement area. Le is the effective effective length. It may may be determined using the stiffness properties of the gross concrete section for both beams and columns of the substitute frame (see (see Fig. 4.4-1c 4. 4-1c )
•
In lieu of a more accurate determination, the effective stiffness of a column EIe may be taken as:
Where:
= 0.2
Ec = 1100f cd Es is the modulus of elasticity of steel Ic, Is, are the moments of inertia of the concrete and reinforcement sections, respectively, of the substitute column, with respect to the centroid of the concrete section (see Fig. 4.4-1c).
•
•
or alternatively
= (1 ) ≥ 0.4
Where: Mb is the balanced moment capacity of the substitute substitute column
1/rb is the curvature at balanced load & may be taken as
1 = 5 10−
The equivalent reinforcement areas, A s, tot, in the substitute
column to be used for calculating I s and Mb may be obtained by designing the substitute column at each floor level to carry the story design axial load and amplified sway moment at the critical section.
Slenderness Ratio ratio, λ of a column is defined as the ratio Slenderness ratio, of the effective effective length l ei to the radius of gyration. g yration. NB: λ also provides a measure of the vulnerability to failure failure of the column by elastic instability (buckling)
where: Le
= is the effective buckling length
i is the minimum radius 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 sectiona l area
Limits of Slenderness The slenderness ratio of concrete columns shall not
exceed 140. Second order moment in a column can be ignored if:
25 For sway frames, the greater of: ≤ 15/ = / For non-sway frames: ≤ 50 25
Where: M1 and M2 are the first-order end moments, M 2
being always positive and greater in magnitude than M 1, and M1 being positive if member is bent in single curvature and negative if bent in double curvature
Effective Length of Columns Effective e buckling length is the length between points Effectiv of inflection of columns and it is the length which is effective 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 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 idealized boundary conditions:
The rotational restraint at a column end in a building
frame is governed by the flexural stiffnesses stiffnesses of the members framing into it, relative to the flexural stiffness of the column itself. measures of the Hence, it is possible to arrive at measures ‘degree of fixity’ 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). alignment lignment charts or approximate These involve use of a equations.
( Nomograph): for members The alignment chart (Nomograph): that are parts of a framework
Approximate equations
approximate equations can be used The following approximate provided that the values of α 1 and α2 don’t exceed 10 (see EBCS 2). Non-sway mode
In Sway mode
= 0.4 ≥ 0.7 0.8
= .+ + +. ≥ 1.15 1.1 5 .++ Or Conservatively,
= 1 0.8 ≥ 1.15
Where α1 and α2 are as defined above and α m is
defined as:
= 2
Note that: for flats slab construction, an equivalent
beam shall be taken taken as having the width and thickness of the slab forming the column strip. restrainet is quantified by the two The effect of end restrainet end restrain factors α 1 and α2
/ = /
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
Note that: if the end of the column is fixed, theoretically,
α = 0, but an α value of 1 is recommended for 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 10 is recommended for use. foregoing recommendations is The rationale behind the foregoing that no support in reality can be truly fixed or pinned.
Design of columns, EBSC-2 1995 General: mo ments may generally be The internal forces and moments determined by elastic global analysis using either first order theory or second order theory. First-order theory, theory, using the initial geometry of the First-order 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 Individual non-sway compression members shall be
considered to be isolated elements and be designed accordingly.
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 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 ec centricity allowance for imperfections. For isolated columns:
= 300 ≥ 20
e2 is the second-order eccentricity
First order equivalent eccentricity i.
ii. ii.
For fi first-order ec eccentricity e0 is equal at both ends of a column
=
For For fir first st-o -ord rder er mom momen ents ts varyi varying ng linea linearl rly y alon along g the the length, the equivalent eccentricity 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 e 01. e01 is positive if the column bents in single curvature and negative if the column bends in double curvature.
iii. iii. For diff differen erentt eccent eccentiri iries es at at the the ends, ends, (2) (2) abov above, e, the the critical end section shall be check chec ked for first order moments:
=
Second order eccentricity i.
The se second-order ec 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 at the critical section.
ii. ii.
The The curv curvat atur ure e is appr appro oxima ximatted by: by:
1 = 5 10−
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. column.
iii. iii. The The appr approp opri ria ate val value ue of k2 may be found iteratively taking an initial value corresponding to first-order actions.
Reinforcement Reinforcement Detail Size The minimum lateral dimension of a column shall be
at least 150mm Longitudinal Reinforcement The minimum number of bars to be used in columns
shall 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 A c for
ductility.
The maximum area of steel shall be 0.08 A c, including
laps. Lateral Reinforcement Minimum bar size shall be ¼ times the largest
compression bar, bar, but not less than 6mm. exceed: 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 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 which cause appreciable sidesway, calculated by a first-order first-order elastic frame analysis.
First-order vs. Second-order Frame Analyses: A first-order frame analysis is one in which the effect of
lateral deflections on bending moments, axial forces, and lateral deflections is ignored. The resulting moments and deflections are linearly
related to the loads. effects cannot be neglected, the When slenderness effects design of compression members, restraining beams and other supporting members shall be based on the factored factored forces and moments mome nts from a second-order s econd-order analysis.
First-order vs. Second-order Frame Analyses: A second-order frame analysis, is one which considers
the effects of deflections on moments, and so on. The resulting moments and deflections include the
effects effects of slenderness and hence are nonlinear with respect to the load. The stiffness, EI, used in the analysis should represent
the stage immediately prior to yielding of the flexural flexural reinforcement since the moments are directly affected by the lateral deflections.
Methods of Second Order Analysis 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∆). (ΣP∆). (ΣP∆/l) and the This shall be loaded as sway force (ΣP∆/l)
structure shall be re-analyzed until convergence is achieved. The values can be assumed to converge when the
change in deflection between two consecutive analyses is less than 2.5 percent . (ACI) 89
Fig. 4 Determination of sway forces 90
b) Direct P-∆ analysis P- ∆ analysis mathematically It describes the iterative P-∆ as 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).
0
1 ( P u ) 0 (Vl )
Thus the second-order moments are δsMs 1 ( P u ) 0 1 . 0 Where: s and Q 1 Q v l 91
c) Ampl Amplif ifie ied d Swa Sway y Mom Momen ents ts Meth Method od:: First-order sway moments are multiplied by a sway
moment magnifier, δs, to obtain the second order moments. M1 = M1ns + δsM1s M2 = M2ns + δsM2s 1 Where: s 1 1 P u (0.75 P c ) s
(According to ACI)
1 1 N S d
N
cr
(According to EBCS)
Where: Pu, NSd is the design value of the total vertical vertical load Pc, Ncr is its critical value for failure failure in a sway mode 92
