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

2

Total steel area Ast = 3926.99 mm2

Steel depths: mmdd

d sb 5.6210

2

2540

2cover1

mm

hd 250

22

mmdd

hd sb 5.4375.6250010

2

2540500

2cover3

1/ Pure compression point ( = 0.65):

Nominal axial compression force: stystgc AfAAfP '

0 85.0

kNNxxxP 387.6878687838799.392642099.39265005002585.00

SBC/ACI Maximum nominal force: kNPPn 710.550280.0 0(max)

Design force: kNxPn 761.35767096.550265.0(max)

2/ Pure tension point ( = 0.90):

Nominal tension strength: kNNxAfP stynt 3358.16498.164933599.3926420

Designl tension strength: kNAfP stynt 402.148490.0

3/ Balanced point B ( = 0.65):

This point is defined by 0021.03 yst

Neutral axis depth: 0021.0003.0

003.05.437

003.0

003.0

3

3

s

dc

Neutral axis depth c = 257.3529 mm

Depth of compression block: mmxca 750.2183529.25785.01

Concrete compression force:

kNNxxxbafP cnc 21875.232475.232421850075.2182585.085.0 '

Steel layer 1:

Strain 00227.03353.257

3529.2575.62003.0003.0 1

1

c

cds (compression)

Stress MPaff ys 4201 because strain greater than yield strain (in absolute value)

Force: mmadi 75.2185.62 So there is displaced concrete

kNNxffAT cssns 2072.587225.5872072585.042062.147285.0 '

111

Steel layer 2: Strain 0000857.0353.257

353.257250003.0003.0 2

2

c

cds

Stress MPaxEf sss 14.170000857.020000022

Force: mmad 75.2182502 So there is no displaced concrete

kNNxfAT ssns 8272.16195.1682714.1775.981222 (Compression)

Steel layer 3: Strain s3 = 0.0021 (tension) Stress fs3 = fy = 420 MPa

Force: mmad 75.2185.4373 So there is no displaced concrete

kNNxfAT ssns 5004.6184.61850042062.1472333 (Tension)

Total forces and moments:

The total nominal force is:

0.003

0.0021

i

nsincn kNTPP 75275.23095004.6188272.162072.58721875.2324

Using (kN) for forces and (m) for distances, the total nominal moment with respect to the

centroid is:

i

insincn

hdT

ahPM

222

mkN

M n

.9134.5522

5.04375.05004.618

2

5.025.08272.16

2

5.00625.02072.587

2

21875.0

2

5.021875.2324

= 0.65 Thus Pn = 1501.34 kN and Mn = 359.394 kN.m

4/ Point D with 0.005 tensile steel strain ( = 0.90):

This point is defined by 005.03 yst

Following the same steps, we find:

Neutral axis depth c = 164.0625 mm

Depth of compression block: a = 139.4531 mm

Concrete compression force: Pnc = 1481.6895 kN

Steel layer 1: Strain 00186.01 s (compression) Stress fs1 = -371.4286 MPa

Force with displaced concrete (since d1 < a): Tns1 = As1(fs1 + 0.85 x f’c) = -515.6799 kN

Steel layer 2: Strain s2 = 0.00157 (tension) Stress fs2 = 314.2857 MPa

Force Tns2 = As2 x fs2 No displaced concrete (since d2 > a): Tns2 = 308.550 kN

Steel layer 3: Strain s3 = 0.005 (tension) Stress fs3 = 420.0 MPa

Force Tns3 = As3 x fs3 No displaced concrete (since d3 > a): Tns3 = 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

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: Pnc = 4515.625 kN

Steel layer 1: Strain 00263.01 s (compression) Stress fs1 = - fy = - 420.0 MPa

Force with displaced concrete (since d1 < a): Tns1 = As1(fs1 + 0.85 x f’c) = -587.2072 kN

Steel layer 2: Strain s2 = -0.00150 (compression) Stress fs2 = -300.0 MPa

Force with displaced concrete (since d2 < a): Pns2 = As2(fs2 + 0.85 x f’c) = -273.6628 kN

Steel layer 3: Strain s3 = -0.00038 compression) Stress fs3 = -75.0 MPa

0.003

0.005

0.003

c = h

Force Tns3 = As3 x fs3 No displaced concrete (since d3 > a): Tns3 = -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

The table below shows all the point results including the beam bending point (P = 0).

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

Max tension

= 0.90

-1649.336

0

-1484.402

0

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.

Approximate drawing of P-M interaction curve

Interaction curve generated by RC-TOOL software

Beam bending point:

The 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 used 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.

Comparing balanced point B and 0.005 steel strain point D:

Balanced point B and 0.005 strain point D are the limits of the transition zone with a 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

This transition zone generates in some cases design curves with non-convexity parts. The strain

compatibility technique used in RC-TOOL software tracks all the points whatever the non-

convexity. Some other programs determine the interaction curve using axial force looping

(determine the moments for a given axial force). This technique may not track correctly the

transition zone (between points B and D).

Design of columns using P-M interaction diagrams

Safe design requires that all load combination points (Pu , Mu) lie inside or on the border of the

design curve. This requires many cycles of trial and error with successive updatings 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. The next two figures show design results for the same

column 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. Stress and strain distributions are also delivered.

RC design of a column with two equal steel layers

Standard RC design of a column with one or two steel layers