Columns – Combined axial force and bending (by Prof. Abdelhamid Charif) Part A: Short columns Introduction: Columns are vertical members supporting axial compression forces, bending moments and shear forces. The vertical loads from the various floors are cumulated and transmitted by the columns to the foundations. Columns play a major role in structural safety. As a compression member, the failure of a column is more dangerous than that of a beam. Stability effects (buckling) must be considered for columns and compression members especially if they are slender (long). For the majority of columns, which are referred to as “short columns”, slenderness effects can be neglected. Slender columns are studied in Part B. A column is usually subjected to an axial compression force and two bending moments (biaxial bending) transmitted by beams and girders connected to it. It is also subjected to two shear forces and a torsion moment. The first part deals with the combination of an axial force with one moment only. Biaxial bending is studied later. Types of columns: Most of RC columns are either tied (more than 90 %) or spiral (5 to 10 %). Special composite columns are sometimes used.
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Columns – Combined axial force and bending (by Prof. Abdelhamid Charif)
Part A: Short columns
Introduction:
Columns are vertical members supporting axial compression forces, bending moments and shear
forces. The vertical loads from the various floors are cumulated and transmitted by the columns
to the foundations. Columns play a major role in structural safety. As a compression member,
the failure of a column is more dangerous than that of a beam.
Stability effects (buckling) must be considered for columns and compression members
especially if they are slender (long). For the majority of columns, which are referred to as “short
columns”, slenderness effects can be neglected. Slender columns are studied in Part B.
A column is usually subjected to an axial compression force and two bending moments (biaxial
bending) transmitted by beams and girders connected to it. It is also subjected to two shear
forces and a torsion moment. The first part deals with the combination of an axial force with one
moment only. Biaxial bending is studied later.
Types of columns:
Most of RC columns are either tied (more than 90 %) or spiral (5 to 10 %). Special composite
columns are sometimes used.
Tied columns
In tied columns which may be of any shape, independent ties are used. All reinforcing bars must
be enclosed by lateral ties. Tie spacing requirement is the smallest of the following three values:
),(Min,48,16Min hbddS sb
With: db = main bar diameter, ds = tie (stirrup) diameter and (b, h) = section dimensions.
The role of ties is:
1. Hold and restrain main bars from buckling
2. Hold steel cage during construction
3. May confine concrete and provide ductility
4. Serve as shear reinforcement
Tie diameter ds should be at least 10 mm if longitudinal bars have 32 mm diameter or smaller.
For higher bar diameters, the tie diameter should be at least 12 mm.
The minimum number of bars in columns and compression members is four for rectangular or
circular ties and three for triangular ties. The maximum angle in a tie is 135o
Maximum distance between untied bar and tied one is 150 mm.
First tie at a distance of half spacing above slab and above footing.
Last tie at a distance of half spacing below lowest reinforcement bar of slab.
The next figure shows some typical tied column sections.
Typical tied columns
Spiral columns
Spiral columns are usually circular. The continuous spiral plays the same role as ties and
provides a lateral confinement opposing lateral expansion and thus improving the column
ductility. The spiral pitch S ranges from 40 to 85 mm. Spiral columns are used in regions with
high seismic activity. The spiral column ductility improves the structure capacity in absorbing
seismic energy and resisting seismic forces.
The minimum spiral reinforcement is given by: y
c
c
g
sf
f
A
A '
min 145.0
Where Ag is the gross concrete section and Ac is the confined area of concrete measured to the
outside diameter of the spiral.
The figure below highlights the behavioral difference between tied and spiral columns. Tied
columns have brittle failures. Spiral columns develop large deformations prior to failure.
Spirals may be used for any section shape but they are effective for circular shapes only.
Behavior of tied and short columns
The improved behavior of spiral columns justifies the use of a higher strength reduction factor in
compression (0.70) as compared to tied columns (0.65).
Strength reduction factor for columns
In general columns are compression members but may also be subjected to axial tension
resulting from lateral loading (wind, earthquake). The column section may then vary from a
compression-controlled case to a tension-controlled one with a linear transition zone between
the two, as shown by the next figure. The strength reduction factor depends on the value of
the steel tension strain t .
Variation of strength reduction factor
Longitudinal reinforcement
The percentage of reinforcement of columns is expressed as the ratio of the total steel area with
respect to the full concrete gross section. g
stt
A
A
The ACI / SBC limits for this percentage are 1% and 8%. In practice, because of bar splicing
(usually located at the top of each floor), it is recommended not to exceed 4 % reinforcement.
Strength of columns in axial compression
Under pure axial compression (with no bending), the nominal (ultimate) column strength is
obtained from the combination of concrete strength and steel strength as follows:
stystgc AfAAfP '
0 85.0
Where stg AA is the net concrete area.
However because of possible accidental eccentricities and resulting accidental bending, SBC
and ACI codes reduce this nominal capacity as: 0(max) rPPn
where r is a reduction factor. r = 0.80 for tied columns r = 0.85 for spiral columns
The design axial compression force is therefore:
column Spiral : 85.0595.085.070.0
column Tied : 85.052.080.065.0
'
0
'
0
(max)
stystgc
stystgc
nAfAAfPx
AfAAfPxP
Column tension strength
Only steel resists tension. The nominal and design tension strengths are then:
stynt AfP stynt AfP 90.0 (Tension is negative)
Concrete shear strength for columns
The concrete shear strength is increased by the axial compression force:
dbf
A
PV w
c
g
u
c614
1
'
if uc VV 5.0 then ties must be designed for shear.
Design of concrete section
If unknown the concrete gross section may be determined using axial force only with a
reduction factor to account for bending and by considering an initial value of steel ratio from 1
to 2 %. The minimum gross section is:
Tied column: )(40.0 ')(
tyc
utrialg
ff
PA
Spiral column:
)(50.0 ')(
tyc
utrialg
ff
PA
This approximate section design must then be followed by a check taking into account the
bending moment as shown in the next example.
Example: Design of a tied column for given loading
Design the section and reinforcement of a tied column to support the loading:
kNVmkNMkNP uuu 60.1501550
Material data is: MPafc 20' MPaf y 420
a/ Select trial section and trial steel ratio:
We select a trial steel ratio of 0.02 (2 %). 02.0t
For a tied column: )(40.0 ')(
tyc
utrialg
ff
PA
(0.50 for a spiral column)
2
3
)( 7.136443)015.042020(40.0
10.1550mm
xA trialg
This gives a 370-mm square column. We must however take a greater dimension to allow for
the bending moment. We take a 400-mm square column. b = h = 400 mm
b/ Select reinforcement
20.320040040002.0 mmxxAA gtst
One bar area for 25 mm diameter is: 22
88.4904
25mm
xAb
Required number of bars 52.688.490
0.3200
b
st
A
An
We must use an even number to obtain symmetrical steel.
We take eight bars of 25 mm diameter (four bars in each layer).
Total steel area is then: 20.3927 mmAst
We use two layers only, to optimize steel resistance as the
column is subjected to one bending moment only.
c/ Check maximum compression capacity
We must have: (max)nu PP .
For a tied column: stystgcn AfAAfPxxP '
0(max) 85.080.065.080.065.0
We find kNNPn 342.22372237342(max) which is much greater thanuP . OK
d/ Design lap splices
For columns, we must use either a Class B splice if more than half the bars are spliced or a Class
A if fewer are spliced. In a column, normally all the bars would be spliced at the same location.
The splice length for Class B is ds ll 3.1 (SBC 12.15) where dl is the development length.
For 25 bars: mmxxxxx
df
fl b
c
y
d 7.140825205
1114203
5
3
'
mmxls 18317.14083.1
SBC and ACI codes (12.17.2) allow reductions of lap splice
lengths in compression members provided enough tie (or spiral)
area is available. Reduction factors are 0.83 for tied columns
and 0.75 for spiral columns. The final splice length is then:
mmmxls 52.11520183183.0
To avoid steel congestion and bar spacing problems, splicing is performed by putting the bars to
be stopped inside the new cage.
b
h
b
h
e/ Select ties and check for shear
For 25 mm bars, we can use 10-mm ties. Spacing requirement is the smallest of the following
three values:
(a) mmdb 04016 (b) mmds 04848 (c) mmhbMin 040),(
Use a spacing of 400 mm (or smaller).
Shear strength check:
Check that uc VV 5.0 db
f
A
PV w
c
g
u
c614
1
'
with mmdd
hd s
b 5.337102
254400)
2cover( and mmbw 400
We find kNNVc 25.1706.170250
kNVkNxxV uc 6075.6325.17075.05.05.0 : OK
Ties are therefore not required to play shear reinforcement role.
f/ Check section safety with regard to bending moment
Axial compression force and bending moment are related by interaction diagrams due to the fact
that they both cause normal stresses. The P-M interaction curves will be covered later.
Instead of drawing the whole interaction diagram and checking that the point (Mu , Pu) lies
inside the safe zone, that is, un MM and un PP , we use a different method.
We will start from conditions such that un PP and then check that the corresponding nominal
moment is such that un MM
With a section subjected to a bending moment and a compressive axial force, it is reasonable to
assume the following strain distribution:
The top steel strain is greater or equal to the yield strain which is 0.0021, otherwize the stress is
equal to fy. The bottom bar strain is less than yield strain.
2
21 5.1963 mmAA ss mmd 5.621 mmd 5.3372
The neutral axis depth c is unknown
Concrete compression force:
cxxcxxxabfC cc 578040085.02085.085.0 ' (1)
Compression in top steel layer (with displaced concrete):
NxffAC cyss 5.791290)2085.0420(5.1963)85.0( '
1 (2)
Strain in bottom steel layer: c
c
c
cds
5.337003.0003.0 2
2
Stress is c
c
c
cxEf sss
5.337600
5.337003.020000022
ys 1
0.003
2s
c d2
d1
The tension force is therefore c
c
c
cxfAT ss
5.3371178100
5.3376005.196322 (3)
Compression force (1) and tension force (3) are functions of the unknown neutral axis depth c.
Total nominal force )3()2()1( TCCP scn
un PP with 065 (tied column in compression control) (steel tension strain less than y ).
So NP
P un 4.2384615
65.0
10.1550 3
Thus NTCC sc 4.2384615 (4)
Therefore we have: c
cc
5.33701178105.79129057804.2384615
We multiply the terms of this equation by c:
)5.3370(1178105.79129057804.2384615 2 cccc
or 0)5.3370(1178105.7912904.23846155780 2 cccc
03976087509.4152245780 2 cc
Second degree equation with the following positive solution: mmc 646.300
This value is less than the depth of bottom layer. The depth of the compression block is
mmxca 55.255646.30085.01 . These results confirm the strain distribution assumed.
Check assumed steel strains
0003678.0646.300
646.3005.337003.0003.0 2
2
c
cds which is less than yield strain: OK
00238.0646.300
5.62646.300003.0003.0 1
1
c
dcs which is greater than yield strain: OK
Total forces
The force values are kNCc 74.1737 kNCs 29.791 kNT 42.144
These forces verify equation (4).
Check bending moment un MM
The nominal moment is
222221
hdTd
hC
ahCM scn
We find 2.03375.042.1440625.02.029.791127775.02.074.1737 nM
mkNMn .17.245 and mkNMmkNxM un .150.21.16517.24565.0 OK
This means that the point (Mu , Pu) lies inside the safe zone of the the P-M interaction curve.
Final comments
This method of checking of section safety with respect to bending moment is rather long and
must be repeated for each load combination (Pu , Mu). The use of P-M interaction diagrams is
more effective.
Axial force – bending moment interaction diagrams
Both axial force P and bending moment M cause normal stresses:
A
PP
I
MyM
The total normal stress is: I
My
A
P
Assuming compression as positive, the total stress must not exceed the material strength:
StrengthI
My
A
P which can be transformed to: 1
ultult M
M
P
P
This linear inequality results in an interaction curve relating the axial force to the bending
moment. For elastic linear and symmetric materials, with equal strength in tension and
compression (mild steel) this P-M interaction curve is of the form:
P-M interaction diagram for elastic symmetric materials
For reinforced concrete with nonlinear stress-strain curves and where the tension strength is
provided by steel only, the P-M interaction diagram is of the form shown in the next figures.
There are two curves: (1): Nominal curve Pn-Mn (2): Design curve nP -nM
A safe design is inside or on the border of the shaded design curve. The distance between the
two curves is variable depending on the strength reduction factor. The two curves are closer in
the tension-control zone ( 90.0 ). The horizontal line limit corresponds to the code maximum
design compression force (max)nP
P-M interaction diagram for RC columns (produced by RC-TOOL software)
Importance of P-M interaction diagrams:
The P-M interaction curves are very important for column analysis and design which is much
more complex than in beams. For many load combinations, beam design requires designing for
the largest moment value. For columns, it is general never obvious which load combination
controls design. In the previous figure, point 2 is unsafe although it has smaller values of both
axial force and bending moment as compared to point 1, which is safe.
Column design requires checking that load combination points lie inside (or on the border) of
the safe design curve.
Drawing the P-M interaction curve
We consider the case of a symmetrical rectangular section. The P-M curve may be
approximately drawn from few important points.
A) Simple points
These are the pure compression and pure tension points with no bending moment.
A1) Pure compression point:
Nominal axial compression force: stystgc AfAAfP '
0 85.0
SBC/ACI Maximum nominal force:
Spiral : 85.085.085.0
Tied : 85.080.080.0
'
0
'
0
(max)
stystgc
stystgc
nAfAAfP
AfAAfPP
Design force:
Spiral : 85.0595.085.070.0
Tied : 85.052.080.065.0
'
0
'
0
(max)
stystgc
stystgc
nAfAAfPx
AfAAfPxP
A2) Pure tension point:
Nominal tension strength: stynt AfP
Designl tension strength: stynt AfP 90.0 ( 90.0 )
B) General points
A point on the interaction curve is defined by its two coordinates M (horizontal) and P (vertical).
The moment is expressed about the gross section centroid.
Consider a rectangular section with dimensions (b , h)
subjected to a bending moment about X-axis as shown.
The moment is positive if causing compression in top and
tension in bottom.
The section reinforcement is expressed in terms of steel layers
Asi with distances di from the concrete top fiber.
The total steel area is: i
sist AA
The tension steel strain t corresponds to that in the bottom layer.
The section centroid is at a distance h/2 from the top.
M
b
h
X
di
Asi
Combining bending and axial compression, section failure occurs when the top fiber strain
reaches the concrete ultimate strain of 0.003
Relations between steel strain and neutral axis depth
Although compression is considered positive, for steel, it is more
convenient to consider tension as positive.
From similar triangles, we find the following two relations linking
steel strain si (at layer i with depth di) and the neutral axis depth c:
We have: si
idc
003.0003.0 Thus
si
idc
003.0
003.0
We also have: ccd i
si 003.0
Thus
c
cdisi
003.0
A point on the P-M interaction curve is usually defined by either the steel strain or by the neutral
axis depth, and if one of the two is known, the other is easily found with the preceding relations.
Steps for the general interaction point:
a) Concrete contribution:
Knowing the neutral axis depth c, concrete compression block depth is: ca 1
with: MPafc 30' : 85.01 MPafc 30' :
7
)30(05.085.0,65.0
'
1
cfMax
The nominal concrete compression force is bafP cnc
'85.0
The displaced concrete by steel layers located inside the compression block will be considered
with steel.
b) Steel layers contribution:
Contribution of each steel layer i is computed as follows (tension is positive):
Steel strain in layer i: c
cdisi
003.0
Steel stress in layer i: sissi Ef but with ysiy fff
Nominal steel force in layer i:
concrete) displaced(With : 85.0
concrete) displaced (No :
' adifffA
adiffAT
icsisi
isisi
nsi
Concrete displaced by steel layer i located in the compression block is considered by subtracting
from the steel force a concrete compression force equal to sic Af '85.0 . This is equivalent to
adding an equal tension force to the layer
Steel Young’s modulus Es is equal to 200000 MPa or 200 GPa
si
0.003
t
c di
Total forces and moments:
The total forces and moments are obtained by combining concrete and steel contributions.
The total nominal force is: i
nsincn TPP
The total nominal moment with respect to the centroid is:
i
insincn
hdT
ahPM
222
Design values are obtained by multiplying by the strength reduction factor. The latter depends
on the known value of the tension steel strain at the bottom steel layer.
i
nsincn TPP
i
insincn
hdT
ahPM
222
Particular points on the interaction curve
These particular points are:
Pure compression point (M = 0 , c = infinity, 70.0/65.0 )
Pure tension point (M = 0 , c = - infinity, 90.0 )
Balanced point (tension steel strain = Yield strain s
y
ytE
f , 70.0/65.0 )
0.005 steel strain point ( 005.0t , 90.0 )
The last two points are the limits of the transition zone between compression-controlled sections
and tension-controlled sections.
The left hand side of the curve (M < 0) is generated similarly by inverting the section upside
down.
The next figure shows the P-M interaction diagram for the revious example and the loading
point (Pu = 1550 kN, Mu = 150 kN.m). The point lies inside the safe zone confirming the
previous result.
nsiT
0.85 fc’
h/2 di
a Pnc
si
0.003
t
c di
Interaction curve example:
Tied square column 500 x 500 mm with eight 25 mm bars in
three layers (1.57 % of steel)
MPafc 25' Thus 85.01
MPaf y 420 Thus yield strain 0021.0y
Tie diameter ds = 10 mm
Determine the particular points on the interaction diagram as well as point C with c = h
There are three steel layers (top and bottom layers with three bars each, and middle layer with
two bars). Steel areas: As1 = As3 = 1472.62 mm2 As2 = 981.75 mm