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CHAPTER NINE COLUMNS 1

9 CHAPTER 9: COLUMNS

9.1 First-Order versus Second-Order Analysis

A first-order analysis is based on the initial geometry of the structure, assuming elastic behavior. On the other hand, second-order analysis considers the influence of lateral drift, cracking, member curvature, shrinkage and creep on the forces in the structure.

9.2 Sway and Nonsway Frames

9.2.1 Nonsway Frames

Structural frames whose joints are restrained against lateral displacement by attachment to rigid elements or bracing are called “nonsway frames”, shown in Figure 9.1.

According to ACI Code 10.10.5.1 a column in a structure is nonsway if the increase in column end moments due to second-order effects does not exceed 5 percent of the first-order end moments. Moreover, ACI Code 10.10.5.2 assumes a story within a structure is nonsway if:

cus

ulV

PQ ∑= o∆ (9.1)

is less than or equal to 0.05, where Q is the stability index which is the ratio of secondary

moment due to lateral displacement and primary moment, ∑ uP is the total factored

vertical load in the story, usV is the factored horizontal story shear, cl is length of column

measured center-to-center of the joints in the frame, and o∆ is the first-order relative

deflection between the top and bottom of that story due to usV .


Figure 9.1: Nonsway frame

9.2.2 Sway Frames

Structural frames, not attached to an effective bracing element, but depend on the bending stiffness of the columns and girders to provide resistance to lateral displacement are called “sway frames”, shown in Figure 9.2.

Figure 9.2: Sway frame

9.3 Effective Length Factor “k” of Columns of Rigid Frames

The effective length of column is the length of a column hinged at both ends and having the same buckling load as the column being considered. Thus, the effective length factor k, is the ratio of the effective length to the original length of column.


The ACI Code R10.10.6.3 recognizes the Jackson and Moreland Alignment Charts, shown in Figure 9.3, to estimate the effective length factor k for a column of constant cross section in a multibay frame. The effective length factor k is a function of the relative stiffness at each end of the column. In these charts, k is determined as the intersection of a line joining the values of ψ at the two ends of the column. The relative stiffness of the beams and

columns at each end of the column ψ is given by Eq. (9.2)

∑∑=

bbb

ccc

lIElIE

//

ψ (9.2)

where,

cl = length of column center-to-center of the joints

bl = length of beam center-to-center of the joints

cE = modulus of elasticity of column concrete

bE = modulus of elasticity of beam concrete

cI = moment of inertia of column cross section about an axis perpendicular to the plane of

buckling being considered.

bI = moment of inertia of beam cross section about an axis perpendicular to the plane of

buckling being considered.


(a) (b)

Figure 9.3: Alignment chart; (a) nonsway frames; (b) sway frames

∑ indicates a summation of all member stiffnesses connected to the joint and lying in the

plane in which buckling of the column is being considered.

Consider the two-story frame shown in Figure 9.4. To determine the effective length factor k for column EF,

Figure 9.4: Two-story frame


)/()/()/()/(

21

21

LIELIEHIEHIE

EHbBEb

EFcDEcE +

+=ψ

and

)/()/()/(

21

2

LIELIEHIE

FIbCFb

EFcF +

=ψ

ACI Code 10.10.6.3 specifies that for columns in nonsway frames, the effective length factor k should be taken as 1.0, unless analysis shows that a lower value is justified.

The ACI Code 10.10.4.1 specifies that the influence of cracking along the length of the member, presence of axial loads, and effects of duration of loads be taken into consideration when calculating k by using reduced values of moment of inertia as follows:

Beams -------------------------------- gI35.0

Columns------------------------------ gI70.0

Uncracked Walls -------------------- gI70.0

Cracked Walls ----------------------- gI35.0

where gI is the gross moment of inertia.

As an alternate to using alignment charts to determine k, the following simplified equations are used for computing the effective length factors for nonsway and sway frame members.

For columns in nonsway frames, the smaller of Eq. (9.3) and Eq. (9.4) is used

( ) 0.105.07.0 ≤++= BAk ψψ (9.3)

0.105.085.0 min ≤+= ψk (9.4)

where Aψ and Bψ are the values of ψ at the two ends of the column, and minψ is the

smaller of the two values.

For columns in sway frames restrained at both ends, k is taken as

for 2<mψ

mmk ψ

ψ+

−= 1

2020 (9.5)

for 2≥mψ


mk ψ+= 19.0 (9.6)

where mψ is the average of ψ values at the two ends of the column.

For columns in sway frames hinged at one end, k is taken as

ψ3.00.2 +=k (9.7)

where ψ is the values at the restrained end of the column.

9.4 The ACI Procedure for Classifying Short and Slender Columns According to ACI Code 10.10.1, columns can be classified as short when their effective slenderness ratios satisfy the following criteria:

For nonsway frames

( ) 0.40/1234 21 ≤−≤ MMrlk u (9.8)

or

For sway frames 22/ ≤rlk u (9.9)

Furthermore, compression members are considered braced against sidesway when bracing elements have a total stiffness, resisting lateral movement of that story, of at least 12 times the gross stiffness of the columns within the story.

where k = effective length factor

ul = unsupported length of member, defined in ACI Code 10.10.1.1 as clear distance between floor slabs, beams, or other members capable of providing lateral support, as shown in Figure 9.5.

Figure 9.5: Unsupported length of member


r = radius of gyration associated with axis about which bending occurs. For rectangular cross sections r = 0.30 h, and for circular sections, r = 0.25 h as specified by ACI Code

10.10.1.2.

h = column dimension in the direction of bending.

1M = smaller factored end moment on column, positive if member is bent single curvature,

negative if bent in double curvature.

2M = larger factored end moment on column, always positive.

Chart 9.1 summarizes the process of column design as per the ACI Code.

Chart 9.1: Column Ddesign

Example (9.1): The frame shown in Figure 9.6 consists of members with rectangular cross sections, made of the same strength concrete. Considering buckling in the plane of the figure, categorize column bc as long or short if the frame is:

Column Design

Sway frame Non-sway frame

22≤r

klu

10022 ≤<r

klu

100>r

klu

)M/M(1234r

kl21

u −≤

−>≥

2

1uMM

1234r

kl100

100>r

klu

Neglect slenderness

(short)

Moment magnification

method (slender)

Exact P-∆ analysis (slender)


a. Nonsway

b. Sway.

Figure 9.6: Frame and loads on column bc

Solution: a. Nonsway:

For a column to be short,

( ) 0.40/1234 21 ≤−≤ MMrlk u

cmlu 3403030400 =−−=

k is conservatively taken as 1.0.

( )( ) 38.32353.0

3401/ ==rlk

( ) ( ) 38.3240 astaken 1.4240/271234/1234 21 >=−−=− MM

i.e., column is classified as being short.

b. Sway:

The column is classified as being short when 22/ ≤rlk u

[ ( )( ) ( )][ ( )( ) ( )] [ ( )( ) ( )]

406.075012/603035.090012/603035.0

40012/35307.033

3

=+

=cψ


[ ( )( ) ( ) ] [ ( )( ) ( )][ ( )( ) ( )] [ ( )( ) ( )]

945.075012/603035.090012/603035.045012/40307.040012/35307.0

33

33

=++

=bψ

Using the appropriate alignment chart, k = 1.14, and ( )( ) 2291.36353.034014.1

>==rlk u

i.e., column is classified as being slender, or long.

ψ could have been evaluated using Eq. (9.5)

675.02

945.0406.0=

+=mψ

25.1675.0120

675.020=+

−=k

9.5 Short Columns Subjected to Axial force and Bending

Generally, columns are subjected to axial forces in addition to some bending moments. These moments are generally due to:

§ Unsymmetrically placed floors, as shown in Figure 9.7.

Figure 9.7 Unsymmetrically placed floors

§ Lateral loading such as wind or earthquake loads, shown in Figure 9.8.


Figure 9.8: Rigid Frame

§ Loads from eccentric loading such as crane loads acting on corbels.

§ End restraints resulting from monolithic action between floor beams and columns.

§ Accidental eccentricity resulting from column misalignment or other execution deficiencies.

9.5.1 Interaction Diagram

Unlike pure axial or bending loading, where unique strength exists for a particular section, combined axial load and bending result in an infinite number of strength combinations evaluated by using equilibrium equations and compatibility of strains. When these strength combinations are plotted, a strength curve called “Interaction Diagram” is produced. To plot an interaction diagram for a particular cross section, at least five strength combinations are required, as shown in Figure 9.9. All strength combinations located in the area under the curve are possible safe strength capacities while combinations located outside the curve are failure cases.

§ Point “A”:

This point on the curve represents pure axial compression capacity of the column cross section, where the eccentricity e is equal to zero. The nominal axial capacity nAP is given

by Eq. (9.10)

( ) ysgsggcnA fAAAfP +−′= 85.0 (9.10)

where sgA is total column reinforcement.


Figure 9.9: Strength interaction diagram

§ Point “C”:

This point on the curve represents pure flexural capacity of the column cross section analyzed as doubly reinforced section, where the eccentricity e is equal to infinity. The nominal flexural capacity nCM is given by Eq. (9.11)

( ) ( )ddCadCM scnC ′−+−= 2/ (9.11)

§ Point “B”:

This point is characterized by its maximum bending strength and represents a balanced failure of the column section. Crushing of the concrete occurs simultaneously with yielding of the reinforcement, or

003.0== cuc εε , and yt εε = .

§ Point “D”:


Points along the curve between points A and B are characterized by compression failure of the section. Failure is initiated by crushing of the concrete before the initiation of yielding of the reinforcement, or

003.0== cuc εε , and yt εε < .

The eccentricity e is smaller than the eccentricity at balanced failure be , where the

eccentricity increases when moving from point A towards point B along the interaction curve.

§ Point “E”:

Points along the curve between points B and C are characterized by tension failure of the section. Failure is initiated by yielding of the reinforcement, or

003.0== cuc εε , and 005.0t ≥ε .

The eccentricity e is larger than the eccentricity at balanced failure be , where the

eccentricity increases when moving from point B towards point C along the interaction curve.

§ Point “F”:

This point on the curve represents pure axial tension capacity of the column cross section where the eccentricity e is equal to zero. The nominal axial capacity nFP is given by Eq.

(9.12)

ysgnF fAP = (9.12)

where sgA is total column reinforcement.

9.5.2 Modes of Failure

Three modes of failure are possible for columns subjected to axial force plus bending.

9.5.2.1 Balanced Failure

Figure 9.10 shows a rectangular section subjected to an axial load with eccentricity adjusted so as to produce a balanced failure using the principle of static equivalence.


Figure 9.10: Balanced failure

The distance to the neutral axis from the extreme compression fiber is given as

df

xy

b

+=

61206120 (9.13)

The compressive force resisted by concrete is given by

bxfC bccb 185.0 β′=

The tensile force resisted by reinforcement in the tension side is

ysb fAT =

The compressive force resisted by reinforcement on the compression side of the cross section is

( )cyssb ffAC ′−′= 85.0


In the last of the previous equations, it is assumed that compression reinforcement does

yield, or ys εε ≥′ . This can be easily checked from similar triangles, or

′−=′

b

bs x

dx003.0ε

If yielding of the reinforcement does not occur, where ys εε <′ , the compressive force in

compression reinforcement is given as ( )csssb ffAC ′−′= 85.0

where the stress in the compression reinforcement is evaluated using Hook’s law, or

sss Ef ε ′=

From equilibrium of forces in the vertical direction

bsbcbnb TCCP −+= (9.14)

Substituting ,, sbcb CC and bT in Eq. (9.14), one gets

( ) yscysbcnb fAffAbxfP −′−′+′= 85.085.0 1β (9.15)

From equilibrium of moments, by taking the moments about the centroid of the cross section,

( ) ( ) dTdddCdadCMeP bsbcbnbbb ′′+′′−′−+′′−−== 2/ (9.16)

and nb

nbb P

Me =

The balanced strain condition represents the dividing point between compression-controlled sections and the transition zone of the strength interaction diagram.

9.5.2.2 Tension Failure

Figure 9.11shows a rectangular section subjected to an axial load with eccentricity chosen larger than be .


Figure 9.11: Tension failure

The compressive force resisted by concrete is given by bxfC cc 185.0 β′=

where dx

005.0003.0003.0

=+

and d375.0x =

The tensile force resisted by reinforcement in the tension side is ys fAT =

The compressive force resisted by reinforcement on the compression side of the cross section is ( )cyss ffAC ′−′= 85.0 when ys εε ≥′ .

If yielding of the reinforcement does not occur, where ys εε <′ , the compressive force in

compression reinforcement is given as ( )csss ffAC ′−′= 85.0

where the stress in the compression reinforcement is evaluated using Hook’s law, or

sss Ef ε ′=


TCCP scn −+= (9.17)


Substituting ,, sc CC and T in Eq. (9.17), one gets

( ) yscyscn fAffAbxfP −′−′+′= 85.085.0 1β (9.18)

From equilibrium of moments,

( ) ( ) ( )'ddC2/adC''deP scn −+−=+ is used to evaluate e

The nominal flexural strength ePM nn = .

9.5.2.3 Compression Failure

When the nominal compression strength nP exceeds the balanced nomial compression

strength nbP , or when the eccentricity e is less than be or when tε at the extreme layer

of steel at the face opposite the maximum compression force is less than yε , the section is

compression controlled.

Figure 9.12 shows a rectangular section subjected to an axial load with eccentricity chosen smaller than be .

Figure 9.12: Compression failure

The compressive force resisted by concrete is given by


bafC cc ′= 85.0

Since the reinforcement does not yield, the tensile force resisted by reinforcement in the tension side is

( )

−

=

−

===a

adAx

xdEAEAfAT scusssssss16120 β

εε

The compressive force resisted by reinforcement on the compression side of the cross section is

( )cyss ffAC ′−′= 85.0 for ys εε ≥′ .


TCCP scn −+= (9.19)


( ) ( )

−−′−′+′=

aad

AffAbafP scyscn1612085.085.0

β (9.20)

From equilibrium of moments, by taking the moments about the axial load ( ) ( ) ( ) 02/ =+′′+′′−′−−+′′−−− edTddedCdeadC sc

( ) ( ) ( ) 02/ =+′′−′′+′++−+′′+++− edTddedCdeadC sc (9.21)


( ) ( )( )

( ) ( ) 06120

85.02/85.0

=′′+

−

−

′′+′++−′−′+′′+++−′

dea

adA

ddedffAdeadbaf

s

cysc

β (9.22)

Eq. (9.22) which is a cubic equation in terms of a , can be written in the following form:

023 =+++ DaCaBaA (9.23)

where,

bfA c′= 425.0

( )ddebfB c −′′+′= 85.0

( )( ) ( )( )edAdddeffAC scys +′′+−′′+′+′−′= 612085.0


( ) ( )dedAD s +′′−= 16120 β

Eq. (9.23) can be solved using Newton-Raphson iteration method, or any available mathematical software. When this iteration method is used,

( ) DaCaBaAaf +++= 23 (9.24)

and the first derivative of this function is given by

( ) CaBaAaf ++=′ 23 2 (9.25)

Assume a trial value for a , named oa

The first iteration value 1a is given as

( )( )o

oo af

afaa

′−=1 (9.26)

A second iteration is evaluated using 1a evaluated from Eq. (9.26). Repeat for more

iterations until you reach a converged value for a . Then, a is substituted in Eq. (9.20) to get nP . The nominal flexural strength is evaluated by multiplying nP by e .

Example (9.6): For the column cross section shown in Figure 9.13, plot a nominal strength interaction diagram, using five strength combinations at least.

Use 2/250 cmkgfc =′ , 2/4200 cmkgf y = , and ( ) 26 /1004.2 cmkgE s = .

F

Figure 9.13: Column cross section

Solution:


d = 55 –6.25 = 48.75 cm

d ′ = 6.25 cm

d ′′ = (55.0 – 6.25 –6.25)/2 = 21.25 cm

( )002.0

1004.24200

6 ==yε

§ Point “A”:

This is a case of pure axial compression load, the nominal axial load is

( ) ( ) ( )[ ] tonsPn 65.565420027.3927.39355525085.01000

1=+−×=

§ Point “B”:

The strength combination at this point corresponds to a balanced failure.

( ) cmxb 91.2875.4842006120

6120=

+

=

( ) cmab 573.2491.2885.0 ==

( )( )( ) tonsCcb 76.1821000/35573.2425085.0 ==

tonsTb 49.821000420064.19 =

=

ys εε >=

−

=′ 0023.091.28

25.691.28003.0

( ) tonsC sb 31.7825085.042001000

64.19=×−=

From equilibrium of forces in the vertical direction,

tonsPnb 58.17849.8231.7876.182 =−+=


( ) ( )

( ) mt

Mnb

.97.6125.21100

49.82

25.2125.675.48100

31.7825.212/57.2475.48100

76.182

=+

−−+−−=


The eccentricity causing balanced failure meb 347.058.17897.61

==

§ Point “E”:

005.0003.0003.0

dx

+= and

( ) cm281.1875.48375.0d375.0x ===

ys 00197.0281.18

25.6281.18003.0 εε ≅=

−

=

0254.066.0

25.666.0003.0 −=

−

=′sε

( )( )( )( ) tons57.1151000/35281.1885.025085.0Cc ==

tonsT 49.821000420064.19 =

=

( ) tonsC s 31.7825085.042001000

64.19=×−= , assuming that ys εε ≥′


TCCP scn −+=

tons39.11349.8231.7857.115Pn =−+=


( ) ( ) ( )'ddC2/adC''deP scn −+−=+

( ) ( ) ( )25.675.4831.782

281.1885.075.4857.11525.21e39.113 −+

−=+

and cm15.51e=

m.t97.56100

15.5139.111Mn =

=

§ Point “D”:

The eccentricity is set at 0.25 m (smaller than 0.347 m) in order to locate a compression failure strength combination.


( )( ) cmkgA /75.371835250425.0 ==

( )( )( ) kgB 75.1859375.4825.21253525085.0 −=−+=

( )( )( )( ) kg.cm

C 5852781.372525.21612064.19

75.4825.2125.62525085.0420064.19=++

−++×−=

( )( )( )( ) ( ) 28 .103035529.275.482525.2185.0612064.19 cmkgD −=+−=

( ) ( )823 103035529.237.585278175.1859375.3718 −+−= aaaaf

( ) 37.58527815.3718725.11156 2 +−=′ aaaf

Let oa = 25 cm as a first trial

( ) 75.3755138025 −=f

( ) 12.1189575025 =′f

The first iteration value 1a is given as

( )( ) cm

ff

a 15.2812.1189575075.3755138025

2525

251 =+=′

−=

( ) 5.261941615.28 =f

( ) 1364641715.28 =′f

( )( ) cm

ff

a 96.2713646417

5.261941615.2815.2815.28

15.282 =−=′

−=

( ) 75.3723796.27 =f

( ) 1353454796.27 =′f

( )( ) cm

ff

a 96.2713534547

75.3723796.2796.2796.27

96.273 =−=′

−=

O.K0024.085.0/96.27

25.685.0/96.27003.0 ys εε >=

−

=′

( )( )( ) ( )

( )tons

Pn

32.22896.27

96.2775.4885.01000612064.19

25085.042001000

64.191000/3596.2725085.0

=

−

−

×−+=


( ) mtM n .08.5725.032.228 ==

§ Point “F”:

This is a case of pure axial tension load, the nominal axial load is

( ) tonsfAP ysyn 93.1641000/420027.39 === .

§ Point “C”:

This is a case of pure bending moment where the nominal flexural strength is calculated as follows:

( )( )( )( ) tonsxxCc 32.61000/3585.025085.0 ==

tonsT 49.821000420064.19 =

=

( ) tonsC s 31.7825085.042001000

64.19=×−= , assuming that ys εε ≥′


049.8231.7832.6 =−+x , and x = 0.66 cm

0254.066.0

25.666.0003.0 −=

−

=′sε , which means that the compression reinforcement is

stressed in tension, a contradiction to the equilibrium equation.

tonsxx

xCs

−=

×−

−

=23.751023.11625085.025.66120

100064.19

From equilibrium of forces,

49.8223.751023.11632.6 =−+x

x

023.751533.3332.6 2 =−+ xx

( ) ( )( )( )

64.12829.141533.33

32.6223.75132.64533.33533.33 2

±−=

+±−=

x

x

x = 8.568 cm, and the other root is negative (rejected)


mt

M n

.472.36100

25.675.48568.8

23.751023.116

2568.885.075.48

100568.832.6

=

−

−+

−

=

Figure 9.14 shows the resulting strength interaction diagram for the given cross section.

Figure 9.14: Strength interaction diagram

9.5.3 Design Interaction Diagrams

The design interaction diagram for a tied or spirally-reinforced column is evaluated by carrying out three modifications on the nominal strength interaction curve, shown in Figure 9.15 as follows:

a. All nominal strength combinations on the curve are multiplied by the strength reduction factor Φ . This factor is equal to 0.65 for tied columns and 0.75 for spirally reinforced columns.

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