Lec12 Continuous Beams and One Way Slabs(2) Columns (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

26-Apr-13
CE370 : Prof. Abdlehamid Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. A. Charif
Columns
1
Columns - Combined axial force and bending
Short columns
• Columns = vertical (or near vertical) members
supporting axial compression forces, bending moments
and shear forces
• Vertical loads from various floors are cumulated and
transmitted by columns to foundations
• Columns play a major role in structural safety.
• As compression members, failure of columns is more
dangerous than in beams.
2

26-Apr-13
• Stability effects (buckling) must be considered for columns and
compression members especially if they are slender (long)
• For the majority of columns, referred to as “short columns”,
slenderness effects can be neglected. Slender columns are
studied in CE 470.
• 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.
• Column reinforcement is distributed along the border in both
directions in order to resist biaxial bending.
• This course deals with the combination of an axial force with
one moment only. Biaxial bending is studied in CE 470.
3
Short columns
• RC columns are either tied (more
than 90 %) or spiral (5 to 10 %).
• Special composite columns are
sometimes used.
4
Types of columns

26-Apr-13
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:
With: db = main bar diameter,
ds = tie (stirrup) diameter
(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
 ),(Min,48,16Min hbddS sb
5
• Tie diameter ds should be at least 10 mm if longitudinal bars
have 32 mm diameter or smaller.
• For higher bar diameters, ds 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
• The maximum distance between an untied bar and a 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.
6
Tied columns

26-Apr-13
Typical tied column sections
7
Spiral columns
Spiral columns are usually circular
Continuous spiral plays same role as ties and
provides lateral confinement opposing lateral
expansion and thus improving column ductility
Spiral pitch S ranges from 40 to 85 mm
Spiral columns are used in regions with
high seismic activity. Spiral column ductility
improves the structure capacity in absorbing
seismic energy and resisting seismic forces.
Spirals may be used for any section shape but
are effective for circular shapes only.
8

26-Apr-13
• Figure highlights behavioral difference between tied and spiral
columns. Tied columns have brittle failures. Spiral columns
develop large deformations prior to failure
• Improved behavior of spiral columns justifies use of higher
strength reduction factor in compression (0.70) as compared to
tied columns (0.65)
9
Tied and spiral columns
Strength reduction factor for columns
• Columns are compression members but may be subjected to
axial tension resulting from lateral loading (wind, earthquake).
Column section may vary from compression-controlled case to
tension-controlled one, with linear transition zone in between
• Strength reduction factor depends on steel tension strain
10

26-Apr-13
Longitudinal reinforcement in columns
• The percentage of reinforcement of columns is
expressed as the ratio of the total steel area with
respect to the full concrete gross section
• The ACI / SBC limits (minimum and maximum) for
this percentage are 1% and 8%.
• In practice, because of bar splicing (usually located at
each floor slab), it is recommended not to exceed 4 %
for longitudinal reinforcement
• For spiral columns:
g
st
t
A
A

11
spiralinsidecoreofArea:
145.0
'
c
y
c
c
g
s
A
f
f
A
A







Strength of columns in axial compression
• Under pure axial compression (with no bending), the nominal
column strength is obtained from the combination of concrete
strength and steel strength as follows:
Where is the net concrete area.
• However because of possible accidental eccentricities and
resulting accidental bending, SBC and ACI codes reduce this
nominal capacity as: where r is a reduction factor.
r = 0.80 for tied columns r = 0.85 for spiral columns
• The maximum design axial compression force is therefore:
  stystgc AfAAfP  '
0 85.0
 stg AA 
0(max) rPPn 
  
  
(max)
'
00
'
00
(max)
columnSpiral:85.0595.0595.085.070.0
columnTied:85.052.052.080.065.0
nu
stystgc
stystgc
n
PP
AfAAfPP
AfAAfPP
P










12

26-Apr-13
Column tension strength
• Only steel resists tension. The nominal and design tension
strengths are then:
(Tension is negative)
Concrete shear strength for columns
• The concrete shear strength is increased by the axial
compression force:
• if then ties must be designed for shear.
styntstynt AfPAfP 90.0 
db
f
A
P
V w
c
g
u
c
614
1
'









uc VV 5.0
13
• 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:
• Spiral column:
• This approximate section design must then be followed by
a check taking into account the bending moment as will be
shown later in an example.
)(40.0 ')(
tyc
u
trialg
ff
P
A


)(50.0 ')(
tyc
u
trialg
ff
P
A


14

26-Apr-13
Axial force - bending moment interaction curves
• Both axial force P and bending moment M cause normal stress
• Assuming compression as positive, the total stress must not
exceed the material strength:
I
My
A
P
I
My
A
P
MP   :stressTotal
1
:becomesEquation
:alonemomentBending
1
:aloneforceAxial





ultult
ultult
ult
ult
ult
ult
ult
ult
ult
ult
ult
ult
ult
ult
M
M
P
P
M
M
P
P
MI
y
I
yM
PAA
P
I
My
A
P






(a)
(a)
15
This linear inequality results in an
interaction curve relating the axial force
P to the bending moment M.
For elastic linear, symmetric materials,
(equal strength in tension and
compression), this P-M interaction
curve has the form shown.
A safe combination (M , P) must lie
inside or on the border of the
shaded area.
1
ultult M
M
P
P
16

26-Apr-13
17
Elastic linear, symmetric
materials, (equal strength in
tension and compression)
Elastic linear materials, with
no tension strength
A combination (M , P) is safe if it lies inside or on the
border of the shaded area.
• For reinforced concrete with nonlinear stress-strain curves and
where tension strength is provided by steel only, the P-M
interaction curves have the form shown in the figures.
18
P-M interaction curves for RC columns

26-Apr-13
• There are two curves:
• (1): Nominal curve Pn - Mn
• (2): Design curve
• 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
• The horizontal line limit corresponds to
the code maximum design compression
force
nn MP  
90.0
(max)nP
19
P-M interaction curves for RC columns
• 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.
• For columns, it is in general never obvious
which combination controls design.
• In the 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 all load
combination points lie inside (or on the
border) of the safe design curve.
20
Importance of P-M interaction curves

26-Apr-13
Importance of P-M interaction curves
• For a given constant axial
force P, critical design is
governed by the largest
value of the moment M.
Smaller moments are
always safe.
• However for a given
constant moment M, critical
design is not obvious.
Smaller or greater axial
forces may be unsafe.
21
Problem 1: Design of a tied column for given loading
• Design the section and reinforcement of a tied column to support
the loading:
• Material data is:
• a/ Select trial section and trial steel ratio:
• We select a trial steel ratio of 0.015 (1.5 %).
• For a tied column:
• This gives a 352-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
kNVmkNMkNP uuu 60.1501550 
MPafMPaf yc 42025'

2
3
)(')( 0.123802
)015.042025(40.0
10.1550
)(40.0
mmA
ff
P
A trialg
tyc
u
trialg 





22

26-Apr-13
b/ Select reinforcement
• One bar area for 22 mm diameter is:
• Required number of bars is 6.3 but we must use an even
number to obtain symmetrical steel.
• Use six bars of 22 mm diameter (three bars in each layer).
• Total steel area is then:
• This represents a ratio of 1.43 %.
2
2400400400015.0 mmAA gtst  
2
2
13.380
4
22
mmAb 



2
8.2280 mmAst 
b
h
23
c/ Check maximum compression capacity
• We must have:
• For a tied column:
(max)nu PP 
   
OK0.1550
924.22402240924
85.052.052.080.065.0
(max)
(max)
'
00(max)



kNPP
kNNP
AfAAfPPP
un
n
stystgcn



24

26-Apr-13
d/ Design lap splices
• Depending on the axial force, splices in columns can be tension
splices or compression splices. The latter are of course shorter.
• If the axial force in the column can have positive and negative
signs, tension splice must therefore be used for the column.
• In this example the axial force is compressive.
25
mml
lmml
MPafifdf
MPafifdf
l
mmdf
f
ll
sc
dcsc
yby
yby
sc
by
c
dcsc
656
65622420071.0
420)2413.0(
420071.0
46222420043.0,
25
24.0
Max043.0,
24.0
Max
'

























SBC / ACI allow reductions of lap splice lengths in compression
members provided enough tie (spiral) area is available. Reduction
factors are 0.83 / 0.75 for tied / spiral columns. We will not perform
this reduction here. We use a splice length of 700 mm.
26
b
h
To avoid steel congestion and bar
spacing problems, splicing is
performed by putting new bars
inside the cage.
Lap splices

26-Apr-13
e/ Select ties and check for shear
• For 22 mm bars, we can use 10-mm ties. Spacing requirement is
the smallest of the following three values:
• Use a spacing of 350 mm (or smaller).
• Shear strength check:
• Check that
• Ties are therefore not required to play role of shear
reinforcement.
• If so, design stirrups as in beams
mmhbmmdmmd sb 040),(Min:(c)04848:(b)35216:(a) 
OK607.712.19175.05.05.0
2.191191192400
33910
2
22
40400)
2
cover(
614
1with5.0
'












kNVkNV
kNNVmmb
mmd
d
hd
db
f
A
P
VVV
uc
cw
s
b
w
c
g
u
cuc


27
f/ Check section safety with regard to combined
bending moment and axial compression
• Axial compression force and bending moment are related by
interaction diagrams. They both cause normal stresses.
• Construction of 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,
i.e.
we use a different method.
• We will start from conditions such that :
and then check that the corresponding nominal moment is
such that :
un PP 
un MM 
unun PPMM   and
28

26-Apr-13
• Section subjected to bending moment and compressive axial
force, it is reasonable to assume strain distribution as:
• Top steel strain greater or equal to yield strain which is 0.0021,
the stress is equal to fy. Bottom bar strain less than yield strain.
• Neutral axis depth c is the problem unknown.
mmdhd
mmd
d
d
mmAA
s
b
ss
33961400
6110
2
22
40
2
Cover
4.1140:depthsandareasLayer
12
1
2
21



(2)
(1)
NffAC
ccabfC
cyss
cc
5.454734)2585.0420(4.1140)85.0(
:concretedisplacedforaccountingsteelin topforcenCompressio
722540085.02585.085.0
:forcencompressioConcrete
'
1
'


29
Section safety with regard to combined
0.003
c
d2
d1 ys  1
ys  2
• Strain, stress and tension force
in bottom steel layer :
(3)
c
c
c
c
fAT
c
c
c
c
Ef
c
c
c
cd
ss
sss
s












339
684240
339
6004.1140
339
600
339
003.0200000
339
003.0003.0
22
22
2
2


30
0.003
c
d2
d1 ys  1
ys  2

26-Apr-13
31
(3)
(2)
(1)
c
c
c
c
fAT
NffAC
ccabfC
ss
cyss
cc






339
684240
339
6004.1140
:steelbottominforceTension
5.454734)2585.0420(4.1140)85.0(
:concretedisplacedforaccountingsteelin topforcenCompressio
722540085.02585.085.0
22
'
1
'
Compression force (1) and tension force (3) are functions of the
unknown neutral axis depth c.
(3)(2)(1)  TCCP scn
:positive)ison(compressiforcenominalTotal
• These results seem to confirm the strain distribution assumed.
32
mmcammc
cc
cc
c
c
c
NTCC
P
P
PP
TCCP
sc
u
n
un
scn
28.242069.28585.004.285:solutionPositive
02319573609.12456407225
:ofin termsequationquadraticaobtainWebytermsallMultiply
339
0684245.45473472254.2384615
4.2384615
65.0
10.1550
control)-ncompressioincolumnTied(65.0with
:valuesforceultimateanddesignEquating
1
2
3











(4)
(3)(2)(1)

26-Apr-13
(4)equationmequilibriusatisfyvaluesforceThese
53.12973.45441.2059
OK0005679.0
04.285
04.285339
003.0003.0
OK00236.0
04.285
6104.285
003.0003.0
2
2
1
1
kNTkNCkNC
c
cd
c
dc
sc
ys
ys


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




2222
:check thatmustWe
centroidsectionaboutcomputedMoment
21
h
dTd
h
C
ah
CM
MM
scn
un
33
Check assumed steel strains
0.85 fc’
h/2
d2
a Cc
Cs
T
34
• Final comments: This method of checking section safety with
respect to bending moment M is long and must be repeated for
each combination (Pu , Mu).
• The use of P-M interaction diagrams is more effective.
 
 
 
OK.150.25.158
465.24365.0.465.243
2.0339.053.129
061.02.073.454
121214.02.041.2059
2222
:check thatmustWe
21

























mkNMmkNM
MmkNM
M
h
dTd
h
C
ah
CM
MM
un
nn
n
scn
un



0.85 fc’
h/2
d2
a Cc
Cs
T

26-Apr-13
• The figure shows the P-M
interaction diagram for the
example and the loading
point.
Pu = 1550 kN
Mu = 150 kN.m
• The point lies inside the
safe zone meaning that the
combination is safe.
• Interaction curve
generated by RC-TOOL
software.
35
• We consider a symmetrical rectangular section. The P-M curve
may be approximately drawn from few important points.
• A ) Simple points : Pure compression and pure tension points
with no moment
 
  
  
  
  
styntstynt
stystgc
stystgc
n
stystgc
stystgc
n
stystgc
AfPAfP
AfAAfP
AfAAfP
P
AfAAfP
AfAAfP
P
AfAAfP
90.0:Design:Nominal
:)90.0(pointtensionPureA2)
Spiral:85.0595.085.070.0
Tied:85.052.080.065.0
:strengthdesignSBC/ACI
Spiral:85.085.085.0
Tied:85.080.080.0
:ty)eccentriciaccidentalforaccountto(reducedstrengthnominalSBC/ACI
85.0:strengthncompressioNominal
:pointncompressioPureA1)
'
0
'
0
(max)
'
0
'
0
(max)
'
0




















36
Drawing P-M interaction curve

26-Apr-13
• A point on the interaction curve is defined by its two coordinates
M (horizontal) and P (vertical).
• Moment M is expressed about gross section centroid.
• We consider a rectangular section (b , h)
subjected to moment about X-axis as shown.
• Section centroid is at a distance h/2 from the top.
• Moment is positive if causing compression in top
• Reinforcement is expressed in terms of steel layers Asi with
distances di from concrete top fiber.
• The total steel area is:
• Tension steel strain corresponds to that in the bottom layer.
• Combining bending and axial compression, failure occurs when
top fiber strain reaches concrete ultimate strain of 0.003
M
b
h
X
di
Asi

i
sist AA
37
B) General points on the interaction curve P-M
• A column is subjected to an axial force (which is
generally a compression force, but may
sometimes be a tensile force) and a bending
moment which causes (if positive) compression
in the top part and tension in the bottom.
• The sign convention is vey important and must
be respected to avoid confusion.
• We assume compression as positive as concrete
tension strength is neglected.
• The derived equations for the steel layers must be
used algebraically as these can be subjected to
either compression or tension forces.
38
Relations between steel strain and
neutral axis depth
0.003
c di
si
st

26-Apr-13
• Compression is positive.
• Any point on the P-M interaction curve is characterized by its
unique strain distribution across the section. This can be defined by
either steel strain in any layer, or by the neutral axis depth. If one of
these two is known, the other is deduced from similar triangles:
39
Relations between steel strain and
neutral axis depth
depthaxisneutralgiventhefromlayeranyin
strainsteelthedeterminetousedisrelationThis
003.0
003.0
layeranyinstrainsteelgivenafromdepth
axisneutralthedeterminetousedisrelationThis
003.0
003.0
003.0003.0
i
c
dc
cdc
i
dc
dc
i
si
i
si
si
i
si
i










0.003
c di
si
st
(a) Concrete contribution
bafC
MPaf
f
MPaf
cca
cc
c
c
c
'
1
1
85.0:forcencompressioconcreteNominal
30for65.0,
7
05.045.7
Max
30for85.0
known)depth(N.A.
:isblockncompressioofDepth










 





40
Steps for the general interaction point
Concrete displaced by steel layers located inside the compression
block will be considered with steel (in order to account for both
force and moment corrections).
0.003
c di
si
st
a Cc
Fsi
0.85 fc’
h/2 di

26-Apr-13
(b) Steel layers contribution
• Contribution of each steel layer i is computed as follows
(compression is positive) :
 








block)insidelayerbyconcrete(displacedif85.0
block)ncompressiooutsidelayer(steelif
:Force
If:withbut:Stress
003.0:Strain
'
adffA
adfA
F
fffffEf
c
dc
icsisi
isisi
si
ysiysiysiysissi
i
si


41
Concrete displaced by steel layer i located inside the compression
block (di ≤ a) is considered by subtracting from the steel force a
concrete compression force equal to 0.85 fc’Asi
Steel Young’s modulus Es is equal to 200000 MPa (200 GPa)
Reminder : Be careful with algebraic evaluation of steel
contribution (could be compression or tension)
Total forces and moments (algebraic values)
• Total forces and moments are obtained by combining concrete
and steel contributions :
















i
isicn
i
sicn
h
dF
ah
CM
FCP
222
:momentnominalTotal
:forcenominalTotal
























  i
isicn
i
sicn
h
dF
ah
CMFCP
222

42
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.
0.003
c di
si
st
a Cc
Fsi
0.85 fc’
h/2 di

26-Apr-13
• Particular points on the interaction curve
These particular points are:
1. Pure compression point (M = 0 , c = infinity, )
2. Pure tension point (M = 0 , c = - infinity, )
3. Balanced point (tension steel strain = Yield strain)
4. 0.005 tension steel strain point ( )
• Last two points are the limits of the transition zone between
compression-controlled and tension-controlled sections.
• Left hand side of curve (M < 0) is generated with section
upside down.
70.0/65.0
90.0
90.0005.0  st
70.0/65.0 
s
y
yst
E
f
43
Problem 2: Interaction curve points
Tied square column 500 x 500 mm with eight 25-mm
bars in three layers (1.57 % of steel)
Tie diameter ds = 10 mm
0021.0420
85.025 1
'


yy
c
MPaf
MPaf


mmdhd
d
hd
mm
h
dmmd
d
d
s
b
s
b
5.4375.62500
2
cover
250
2
5.6210
2
25
40
2
cover
13
21




















44
Determine all particular points as well as point with c = h.
Draw interaction curve and check safety for load combinations:
(a) Pu = 2000 kN, Mu = 200 kN.m
(b) Pu = 3000 kN, Mu = 230 kN.m
Top / bottom layers with 3 bars each, middle layer with 2 bars
Steel areas: As1 = As3 = 1472.62 mm2 As2 = 981.75 mm2
Total steel area Ast = 3926.99 mm2
Steel depths:

26-Apr-13
• 1/ Pure compression point ( = 0.65):
• Nominal axial compression force:
• SBC/ACI Maximum nominal force:
• Design force:
• 2/ Pure tension point ( = 0.90):
• Nominal tension strength:
• Design tension strength:

  stystgc AfAAfP  '
0 85.0
  kNNP 387.6878687838799.392642099.39265005002585.00 
kNPPn 710.550280.0 0(max) 
kNPn 761.35767096.550265.0(max) 

kNNAfP stynt 3358.16498.164933599.3926420 
kNAfP stynt 402.148490.0 
45
• 3/ Balanced point B ( = 0.65):
• This point is defined by
• Neutral axis depth:
• Neutral axis depth c = 257.3529 mm
• Depth of compression block:
• Concrete compression force:
• Steel layer 1: Strain (comp.)
• Stress:
• Force: With displaced concrete
0021.03  yst 
0.003
0.0021
0021.0003.0
003.0
5.437
003.0
003.0
3
3




s
dc

mmca 750.2183529.25785.01  
kNNbafC cc 21875.232475.232421850075.2182585.085.0 '

MPaff ysys 42011  
mmad 75.2185.621 
    kNNffAF csss 2072.587225.5872072585.042062.147285.0 '
111 
00227.03
353.257
5.623529.257
003.0003.0 1
1 




c
dc
s
46

26-Apr-13
• Steel layer 2: Strain: (comp.)
• Stress:
• Force: No displaced concrete
• Steel layer 3: Strain: (tension)
• Stress: fs3 = - fy = - 420 MPa
• Force: No displaced concrete
MPaEf sss 14.170000857.020000022  
mmad 75.2182502 
kNNfAF sss 8272.16195.1682714.1775.981222 
0021.03  ys 
kNNfAF sss 5004.6184.61850042062.1472333 
mmad 75.2185.4373 
0000857.0
353.257
250353.257
003.0003.0 2
2 




c
dc
s
47
• Total forces and moments:
• Total nominal force:
• Using (kN) for force and (m) for distance, total nominal moment:
 
i
sicn kNFCP 75275.23095004.6188272.162072.58721875.2324
 












i
isicn
h
dF
ah
CM
222
mkNMkNP nn .39.35934.150165.0  
mkN
Mn
.9134.552
2
5.0
4375.05004.618
2
5.0
25.08272.16
2
5.0
0625.02072.587
2
21875.0
2
5.0
21875.2324


























48

26-Apr-13
• 4/ Point D with 0.005 tensile steel strain ( = 0.90):
• This point is defined by :
• Following the same steps, we find:
• Neutral axis depth c = 164.0625 mm
• Depth of compression block: a = 139.4531 mm
• Concrete compression force: Cc = 1481.6895 kN
• Steel layer 1: Strain (compression)
• Stress fs1 = 371.4286 MPa
• Force with displaced concrete (since d1 < a):
Fs1 = As1( fs1 - 0.85 x f’
c) = 515.6799 kN
• Steel layer 2: Strain (tension)
• Stress fs2 = - 314.2857 MPa
• Force Fs2 = As2 x fs2 No displaced concrete (since d2 > a):
Fs2 = - 308.550 kN
005.03  sst  0.003
0.005
00186.01 s
00157.02 s
49
• Steel layer 3: Strain (tension)
• Stress fs3 = - 420.0 MPa
Fs3 = - 618.5004 kN
• Total nominal and design forces and moments:
• Pn = 1070.3190 kN Mn = 479.7681 kN.m
•  Pn = 963.2871 kN  Mn = 431.7913 kN.m
005.03 s
50

26-Apr-13
• 5/ Full compression Point C with c = h ( = 0.65) :
• This point is defined by the neutral axis depth c = h = 500 mm
It corresponds to the onset of tension in the
section, that is the smallest neutral axis depth
with no tension in the section.
• Depth of compression block: a = 425.0 mm
• Concrete compression force: Cc = 4515.625 kN
• Steel layer 1: Strain (compression)
• Stress fs1 = fy = 420.0 MPa
• Fs1 = As1(fs1 - 0.85 x f’
c) = 587.2072 kN
0.003
c = h
00263.01 s
51
• Steel layer 2: Strain = 0.00150 (compression)
• Fs2 = As2( fs2 - 0.85 x f’
c) = 273.6628 kN
• Steel layer 3: Strain = 0.00038 compression)
• Fs3 = 110.4465 kN
• Total nominal and design forces and moments:
• Pn = 5486.9415 kN Mn = 258.7286 kN.m
•  Pn = 3566.5120 kN  Mn = 168.1736 kN.m
52

26-Apr-13
• All these points match the curve delivered by RC-TOOL software
• The next figure represents an approximate drawing from the few
particular points.
• RC-TOOL curve is generated with hundreds of points and gives
details for any desired point.
Point Pn (kN) Mn (kN.m) Pn (kN) Mn (kN.m)
Pure compression
 = 0.65 6878.387 0 3576.761 0
Point C (c = h)
 = 0.65 5486.9415 258.7286 3566.5120 168.1736
Point B (balanced)
 = 0.65 2309.75275 552.9134 1501.340 359.394
Point D (0.005 strain)
 = 0.90 1070.3190 479.7681 963.2871 431.7913
Beam bending
 = 0.90 0 329.1185 0 296.2067
Pure tension
 = 0.90 -1649.336 0 -1484.402 0
53
Table shows all results including beam bending point (P = 0)
Approximate drawing of interaction curve
54
• The load combination point
(Pu = 2000 kN, Mu = 200 kN.m)
is inside the safe zone. This
combination is safe.
• The second combination
(Pu = 3000 kN, Mu = 230 kN.m)
is unsafe as the corresponding
point is outside the safe zone,
but close to the border.
• It is desirable for this
combination to use a more
accurate interaction curve.

26-Apr-13
P-M interaction curve generated by RC-TOOL software
55
• Both previous load
combinations are safe
• RC-TOOL software
performs safety check
for any combination.
• Beam bending point
• Beam bending point is located on the horizontal axis (P = 0).
Both the neutral axis depth and steel strain are unknown. They
must be determined using equilibrium equations. For sections
with more than two layers, this may require several iterations.
RC-TOOL performs all these iterations.
• Beam analysis and design is therefore only a particular case
with P = 0.
56

26-Apr-13
Comparing balanced point B and 0.005 steel strain point D
• Balanced point B and 0.005 strain point D are limits of
transition zone with variable strength reduction factor. Above
point B is the compression-control zone and below point D is
the tension-control zone. The axial force (nominal and design)
in point B is greater than that of point D.
• However for moments, we have:
• For nominal moments MnB > MnD
• For design moments MnB < MnD
• Transition zone generates in some cases design curves with
non-convex parts. The strain compatibility technique used in
RC-TOOL software tracks all the points whatever the non-
convexity. Some other programs determine interaction curve
using axial force looping and may not track correctly the
transition zone (between points B and D).
57
Problem 3
Tied column of section (300 x 500 mm)
with six 20-mm bars in two layers
Determine, for the P-Mx interaction curve,
the following nominal and design points:
a) Point with tension steel strain = 0.006
b) Point with neutral axis depth c = 300 mm
c) Point with neutral axis depth c = 200 mm
Clear cover = 40 mm Tie diameter = 10 mm
For Mx : Two steel layers with three bars each.
Note: For moment My , there are three layers.
As1 = As2 = 942. 48 mm2 Ast = 1884.96 mm2
d1 = Cover +db/2 + ds = 60 mm
d2 = h - d1 = 440 mm
0021.0
420
85.0
25
1
'




y
y
c
MPaf
MPaf


58
300 mm
500mm
Mx

26-Apr-13
59
a) Point with tension steel strain = 0.006 (at bottom layer 2)
mmca
mmdc
s
st
667.124667.14685.0
667.146
006.0003.0
003.0
440
003.0
003.0
:depthblockncompressioanddepthaxisNeutral
006.0
1
2
2
2










kNNbafC cc 75.794794750300667.1242585.085.0
'

    kNNffAF
mmad
MPaEf
c
dc
csss
sssys
s
124.3143141242585.055.35448.94285.0
blockncompressioinsideLayer667.12460:Force
55.354:Stress
on)(Compressi0017727.0
667.146
60667.146
003.0003.0:Strain
:1layerSteel
'
111
1
111
1
1










0.003
0.006
60
kNNfAF
mmad
MPaff
c
dc
sss
ysys
s
s
842.39539584242048.942
blockncompressiooutsideLayer667.124440:Force
0.420:Stress
(Tension)006.0
667.146
440667.146
003.0003.0:Strain
90.0006.0:2layerSteel
222
2
22
2
2
2












mkNMkNP
mkNM
M
h
dF
ah
CM
kNP
FCP
nn
n
n
i
isicn
n
i
sicn
.637.255and730.64190.0
.042.284
2
5.0
44.0842.395
2
5.0
06.0124.314
2
124667.0
2
5.0
75.794
222
:momentnominalTotal
032.713842.395124.31475.794
:forcenominalTotal








































26-Apr-13
61
b) Point with neutral axis depth c = 300 mm
0.003
2s
1s300 mm
kNNC
bafC
mmca
c
cc
625.16251625625
3002552585.085.0
0.2550.30085.0
:depthblocknCompressio
'
1


 
mmad
MPaff
c
dc
csss
ysys
s
814.3753758142585.00.42048.94285.0
0.420:Stress
on)(Compressi0024.0
0.300
600.300
003.0003.0:Strain
:1layerSteel
'
111
1
11
1
1










62
kNNfAF
mmad
MPaEf
c
dc
sss
sssys
s
894.26326389428048.942
0.2800014.0200000:Stress
(Tension)0014.0
0.300
4400.300
003.0003.0:Strain
:2layerSteel
222
2
222
2
2










mkNMkNP
mkNM
M
h
dF
ah
CM
kNP
FCP
nn
yst
n
n
i
isicn
n
i
sicn
.444.208and404.1129
65.0:controlnCompressio
0014.0:2layerbottominstrainTensile
.684.320
2
5.0
44.0894.263
2
5.0
06.0814.375
2
255.0
2
5.0
625.1625
222
:momentnominalTotal
545.1737894.263814.375625.1625
:forcenominalTotal
2












































26-Apr-13
63
c) Point with neutral axis depth c = 200 mm
0.003
2s
1s200 mm
kNNC
bafC
mmca
c
cc
750.10831083750
3001702585.085.0
0.1700.20085.0
'
1


 
mmad
MPaff
c
dc
csss
ysys
s
814.3753758142585.00.42048.94285.0
0.420:Stress
on)(Compressi0021.0
0.200
600.200
003.0003.0:Strain
:1layerSteel
'
111
1
11
1
1










64
kNNfAF
mmad
MPaff
c
dc
sss
ysys
s
842.39539584242048.942
0.420:Stress
(Tension)0036.0
0.200
4400.200
003.0003.0:Strain
:2layerSteel
222
2
22
2
2










mkNMkNP
mkNM
M
h
dF
ah
CM
kNP
FCP
nn
y
yt
tyst
n
n
i
isicn
n
i
sicn
.610.253and959.828
7793.0
0021.0005.0
0021.00036.0
25.065.0
005.0
25.065.0:zoneTransition
005.00036.0:2layerbottominstrainTensile
.433.325
2
5.0
44.0842.395
2
5.0
06.0814.375
2
170.0
2
5.0
750.1083
222
:momentnominalTotal
722.1063842.395814.375750.1083
:forcenominalTotal
2




















































26-Apr-13
65
P-M interaction curve
generated by
RC-TOOL software
66
• Circular columns are frequent in
buildings and structures, for
architectural or safety (spiral) reasons.
• They can be tied or spiral.
• For uniaxial bending (section with
vertical symmetry), steel bars must be
arranged as shown.
• With an even bar number, arrangement
(b) gives a smaller bending capacity
and must be used for safety reasons.
• An odd bar number causes bending
unsymmetry. Positive and negative
moments are slightly different in both
bar arrangements.
Circular columns

26-Apr-13
67
b
s
b
N
drr
d
d
d


2
:angleIncrement
:radiusringSteel
2
CoverdepthRadial
0



Circular columns
• Nb steel bars are arranged in a ring with a regular increment angle.
• Depending on the arrangement, bars in the same level are lumped in
a single layer and the corresponding depth, computed from the top
concrete fiber, is determined using geometric and trigonometric
relations. In general a layer contains one or two bars.
68
   
  






 


 
i
isicn rdFMM
ryArA
r
ar
r
ar
:center)circle(aboutmomentnominalTotal
unitsradianUsesin
3
2
cossin
coscos
:angleusingdeterminedaremomentandforceConcrete
32
1



Circular columns
The main difficulty
with circular
columns is the shape
of the concrete
compression block.

26-Apr-13
69
Problem 4
• Spiral circular column with 600 mm
diameter and six 28-mm bars (1.3 %).
• Spiral diameter = 10 mm, cover = 40 mm.
• Determine P-M interaction points :
 Pure compression / tension points
 Balanced point
 0.005 tension steel strain point
 Point with c = 200 mm
0021.0420
85.025 1
'


yy
c
MPaf
MPaf


22
2
321 5.36945.1231
4
28
2 mmAmmAAA stsss  
• Using bar arrangement (b), there are three steel layers with two
bars each.
70
mmdrd
mmrd
mmrrd
N
mmdrr
mmd
d
d
b
s
b
382.504618.956002:layerBottom
300:layerMiddle
618.95
6
cos236300
2
cos:layerTop
60
36
22
:angleIncrement
23664300:radiusringSteel
64101440
2
CoverdepthRadial
13
2
01
0
0





















solution 4

26-Apr-13
71
a) Pure compression point
 
 
kNP
PPrPP
kNPP
r
rPP
kNNP
P
AfAAfP
n
n
n
n
stystgc
479.4451
595.085.070.0:strengthncompressioDesign
256.635985.0
85.0:factorReductioncolumnSpiral
:forcencompressionominalmaximumSBC/ACI
478.74817481478
5.36944205.36943002585.0
85.0:forcencompressioNominal
0.70:control-ncompressioIncolumnSpiral
(max)
000(max)
0(max)
0(max)
0
2
0
'
0













b) Pure tension point
kNP
kNNAfP
nt
stynt
521.1396690.155190.0:Designl
690.155115516905.3694420:Nominal



72
0.003
0.0021
c) Balanced point (yield strain at bottom steel layer)
mmca
mmdc
s
yst
191.252695.29685.0
695.296
0021.0003.0
003.0
382.504
003.0
003.0
:depthsblockncompressioandN.A.0021.0
1
3
3
3










 
 
 
  mkNmmNM
rfyAfM
kNNC
AfC
mmA
rA
rad
r
ar
c
ccc
c
cc
.02.368.1002.368sin300
3
2
2585.0
sin
3
2
85.085.0:momentConcrete
170.23972397170
1128082585.085.0:forcencompressioConcrete
112808cossin41075.1300
cossin:areablocknCompressio
41075.183.80
300
191.252300
coscos
:angleusingdeterminedmomentandforceConcrete
63
3''
'
22
2
011











 





 
 







26-Apr-13
73
mmad
MPaEf
c
dc
csss
sssys
s
601.4744746012585.0634.4065.123185.0
blockncompressioinsideLayer695.296618.95:Force
634.406:Stress
695.296
618.95695.296
003.0003.0:Strain
:1layerSteel
'
111
1
111
1
1










kNNfAF
mmad
MPaEf
c
dc
sss
sssys
s
231.88231684.65.1231
blockncompressiooutsideLayer695.2960.300:Force
684.6:Stress
(Tension)00003342.0
695.296
0.300695.296
003.0003.0:Strain
:2layerSteel
222
2
222
2
2










74
kNNfAF
mmad
MPaff
sss
ysys
ys
230.5175172304205.1231
0.420:Stress
column)(spiral70.00021.0:3layerSteel
333
3
22
3






 
     
mkNMkNP
mkNM
M
rdFMM
kNP
FCP
nn
n
n
i
isicn
n
i
sicn
.51.399and42.164270.0
.73.570
3.0504382.0230.5173.03.0231.83.0095618.0601.47402.368
:momentnominalTotal
31.2346230.517231.8601.474170.2397
:forcenominalTotal










26-Apr-13
75
0.003
0.005
d) Point with 0.005 tensile steel strain
mmca
mmdc
s
st
772.160143.18985.0
143.189
005.0003.0
003.0
382.504
003.0
003.0
:depthsblockncompressioandN.A.005.0
1
3
3
3










 
 
 
  mkNmmNM
rfyAfM
kNNC
AfC
mmA
rA
rad
r
ar
c
ccc
c
cc
.84.265.1084.265sin300
3
2
2585.0
sin
3
2
95.12941294953
60939cossin0882.1300
0882.1348.62
300
772.160300
coscos
63
3''
'
22
2
011











 





 
 






76
mmad
MPaEf
c
dc
csss
sssys
s
19.3393391922585.068.2965.123185.0
68.296:Stress
143.189
618.95143.189
003.0003.0:Strain
:1layerSteel
'
111
1
111
1
1










kNNfAF
mmad
MPaEf
c
dc
sss
sssys
s
07.43343306966.3515.1231
66.351:Stress
(Tension)0017583.0
143.189
0.300143.189
003.0003.0:Strain
:2layerSteel
222
2
222
2
2











26-Apr-13
77
kNNfAF
mmad
MPaff
sss
ysys
ys
23.5175172304205.1231
0.420:Stress
333
3
22
3






 
     
mkNMkNP
mkNM
M
rdFMM
kNP
FCP
nn
n
n
i
isicn
n
i
sicn
.79.396and46.61590.0
.88.440
3.0504382.023.5173.03.007.4333.0095618.019.33984.265
:momentnominalTotal
84.68323.51707.43319.33995.1294
:forcenominalTotal









78
e) Point with neutral axis depth c = 200 mm
mmca 0.1700.20085.0
1  
0.003
2s
1s200 mm
3s
 
 
 
  mkNmmNM
rfyAfM
kNNC
AfC
mmA
rA
rad
r
ar
c
ccc
c
cc
.99.279.1099.279sin300
3
2
2585.0
sin
3
2
09.14001400088
65886cossin1226.1300
1226.1321.64
300
0.170300
coscos
63
3''
'
22
2
011











 





 
 







26-Apr-13
79
mmad
MPaEf
c
dc
csss
sssys
s
46.3593594632585.014.3135.123185.0
14.313:Stress
0.200
618.950.200
003.0003.0:Strain
:1layerSteel
'
111
1
111
1
1










kNNfAF
mmad
MPaEf
c
dc
sss
sssys
s
45.3693694500.3005.1231
0.300:Stress
(Tension)00150.0
0.200
0.3000.200
003.0003.0:Strain
:2layerSteel
222
2
222
2
2










80
kNNfAF
mmad
MPaff
c
dc
sss
ysys
s
23.5175172304205.1231
0.420:Stress
(Tension)0045725.0
0.200
832.5040.200
333
3
33
3
3










 
     
mkNMkNP
mkNM
M
rdFMM
kNP
FCP
nn
y
yt
st
n
n
i
isicn
n
i
sicn
.48.399and40.759
870.0
0021.0005.0
0021.00045725.0
20.070.0
005.0
20.070.0
zoneTransition005.00045725.00021.0
:factorreductionStrength
.17.459
3.0504382.023.5173.03.045.3693.0095618.046.35999.279
:momentnominalTotal
87.87223.51745.36946.35909.1400
:forcenominalTotal
3y






















26-Apr-13
81
These results match those produced by RC-TOOL software
Flowchart for P-M interaction curve
• Loop for neutral axis values from maximum compression to
maximum tension
 For each value of c :
Compute concrete contribution
Compute all steel layers contribution
Compute and save nominal force and moment
Compute and save corresponding strength reduction factor
• Repeat all the previous steps for negative moments (section turned
upside down)
• Draw nominal interaction curve and design curve by joining the
saved points
• Add the horizontal line limit corresponding to ACI / SBC design
compression strength Pn(max) 82

26-Apr-13
83
P-M interaction curve for a rectangular section
Two equal steel layers One bottom steel layer
Typical P-M interaction curves
Design of RC columns
• Design of columns is more complex than that of
beams as the required steel reinforcement cannot be
directly derived from closed form equations.
• Trial and error strategy is required for column design.
• An initial reinforcement must be assumed and then
checked using the interaction diagrams.
• If unsafe, the steel reinforcement must be increased.
• If safe but with a point very far from the border, then
the design is not economical and may be revised.
• Use of software is recommended for column design.
84

26-Apr-13
• Design of columns using P-M interaction diagrams
• Safe design requires that all load combination points
(Pu , Mu) lie inside or on border of the design curve.
• This requires many cycles of trial and error with
successive updating of reinforcement.
• Only appropriate software such as RC-TOOL can
perform these complex operations.
• An optimal design will correspond to a loading point
lying on the border of the design curve.
• RC-TOOL offers many design options including
standard design with one or two layers, two equal
layers or many layers.
• Stress and strain distributions are also delivered
85
86
Design of a square column with three steel layers

26-Apr-13
87
Design of a square column subjected to a
negative moment and a tensile axial force
RC-TOOL design
• The next two figures show design results for a
column with 500 x 500 mm section subjected to an
ultimate compression force of 2000 kN and an
ultimate moment of 350 kN.m, using two steel layers
at 60 and 440 mm depths.
• The first design is performed with symmetrical
reinforcement and delivers a steel ratio of 1.356 %.
The second design is standard (no symmetry
imposed) and gives a smaller steel ratio of 0.774 %.
In both cases, the loading point lies on the border of
the design curve.
88

26-Apr-13
RC design with two equal steel
layers.
Symmetric interaction curve.
Because of the compression
force, the top steel layer is more
stressed than the bottom one.
A standard design (no steel
symmetry imposed) should
deliver more top reinforcement.
89
Standard RC design with one
or two steel layers
Un-symmetric interaction curve.
Bottom steel is hardly required.
Optimal design delivers a very
small amount of bottom tension
steel just in order to obtain a
tension controlled section.
90

Lec12 Continuous Beams and One Way Slabs(2) Columns (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

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