1

Members with Axial Load and Bending

INTRODUCTION

Concrete compression members are defined as members that carry predominantly axial loads

(although bending moment about one or both axes may exist). Columns are most frequently

thought of as the compression members that Ahold up@ floors in a building. Compression

members may be part of concrete trusses, arch ribs, or shells. ____________________________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

Concrete columns are reinforced with two types of steel. The first is longitudinal steel,

which helps in resisting the compression forces and also counteracts the tensile forces that result

from bending moments. There is also lateral reinforcement consisting of ties or spirals. There

are three general types of reinforced concrete columns used in buildings and bridges and they are

shown in Figure 1.

Longitudinal Bars Lateral Ties

Spiral Pitch

Spiral

WF Section

(A) Tied Column

(B) Spiral Column ( C) Composite Column

Figure 1: Types of reinforced concrete columns.

2

Columns of types (A) and (B) are the most common. Type (C) columns are frequently used in

composite building frames where a steel erection skeleton is encased in concrete to enhance

stiffness of the frame when subjected to lateral loads.

Columns may be categorized as short or slender. The strength of a short column is _______

______________________________________________________________. A slender column

has its strength _________________________________________________________________

______________________________________________________________________________

_____________________________________________________________________________.

It is the philosophy of the ACI code to determine the capacity of a column as if it where short

and then modify the bending moment and axial load capacities to account for slenderness effects.

As a result, thorough discussion of the uniaxial, biaxial and pure compression strengths of short

concrete columns will be undertaken followed by discussions pertaining to slenderness effects.

TIE EFFECT ON COLUMN BEHAVIOR

The (3) types of concrete columns shown in Figure 3.1 all behave in significantly different ways

at their ultimate load. The composite column will be neglected in subsequent discussions.

As axial load is applied to a reinforced concrete column, compatibility of strain requires that the

stress in the longitudinal steel is given by,

where ss c

c

Ef nf nE

= =

in the elastic range. n is the modular ratio, sf is the stress in the steel, and cf is the stress in the

concrete. The axial load in the column can then be written as,

( ) ( ) ( )

( )1

c g st c st c g st s st

c g st

P f A A nf A f A A f A

f A n A

= − + = − +

= + −

where gA is the gross area of the concrete cross section and stA is the area of the longitudinal

steel.

The ultimate strength of the reinforced concrete column can be computed recognizing the

nonlinear material behavior at ultimate load where,

3

0.85 and c c s yf f f f′= =

and therefore,

( )0.85 0.85c c y st c g st y stP f A f A f A A f A′ ′= + = − +

At ultimate load, the longitudinal steel carries a significantly larger portion of the applied load

than it does in the elastic range. This is the major motivation for the presence of lateral ties and

spiral reinforcement in concrete columns.

The effects of ties on the behavior of a reinforced concrete column can be seen through

examination of column behavior up to failure. The longitudinal steel and the concrete share the

applied axial loading. The concrete column shortens as it is loaded (as does the steel). In

addition, the concrete begins to expand laterally as a result of the Poisson effect. As the nominal

axial capacity is reached, the concrete cover on the steel begins to spall. Not soon after, the tied

column forms a conical failure surface and the longitudinal reinforcing steel buckles between the

ties. The spirally reinforced column has lateral reinforcement for the longitudinal steel at much

closer spacing than the tied column. The spiral therefore tends to confine the concrete core after

shell spalling until it yields or fractures. Ductility in the column is also increased with spirals. A

schematic load-deformation response of the two column types is shown in Figure 2.

Tied Column Spiral Column

P

P

Axial Deformation

Spiral shellspalls

Tied column

n

Figure 2: Behavior of Reinforced Concrete Columns At Ultimate Load.

The AACI Spiral@ offsets the strength lost through spalling of the concrete shell by ensuring that

4

the confinement of the concrete core is sufficient to allow a small increase in load after large

deformation prior to complete failure. Therefore, the AACI Spiral@ column has high ductility and

toughness when compared to the tied column.

The nominal strength of a reinforced concrete column can be written to account for the

presence of tie or spiral reinforcement in the form of the general strength equation below:

(1)

where sk is a constant that varies from 1.5 to 2.5 (avg. 1.95) to account for spiral reinforcement,

syf is the yield strength of the spiral reinforcement and suv is the volume of spiral steel per unit

column length. The net area of concrete, cA , is based on the gross area of concrete for a tied

column and core area of concrete for spiral columns. If one considers a tied column only the

first two terms in the above equation are considered. All three terms are used in the case of a

spirally reinforced column.

ACI CODE PROVISIONS FOR REINFORCEMENT

The provisions set forth by the ACI 318-95 code for concrete column reinforcement are

described in this section. The provisions for lateral and longitudinal reinforcement are in

various locations of the code, but for the most part sections 7.8, 7.10 and 10.9 include the

coverage for this reinforcement. As discussed previously, lateral reinforcement in a concrete

column has significant impact on the ultimate strength of the member. The lateral reinforcement

requirements of the code are grouped within section 7.10. The requirements for spiral ties will

be the first to be discussed.

Section 7.10.4 of the code covers the requirements for spiral reinforcement. Standard spiral

sizes are 3/8", 1/2" and 5/8" diameter. The pitch of spiral reinforcement is typically 2 to 3

inches. The code provisions are enumerated as follows:

a.) The minimum size of spiral reinforcement is diameter.

b.) The clear spacing between spirals shall be . If the

5

spacing of spirals becomes too close together, concrete aggregate will not be able to flow

around the reinforcement, creating voids and insufficient bond.

c.) The anchorage of spiral steel shall be provided by of

spiral bar or wire at each end of the spiral unit.

d.) Splices in spiral reinforcement shall be lap type of or welded.

e.) Spirals shall extend from top of footing or slab in any story to level of lowest horizontal

reinforcement in the members framing into the column.

f.) If beams or brackets are not located on all sides of the column where spiral reinforcement

is to be terminated, ties shall extend above termination of spiral to bottom of slab or drop

panel.

g.) In the case of column capitals, spirals shall extend to a level at which the diameter or

width of capital is two times that of the column.

h.) Spirals shall be held firmly in place and true to line during construction.

ACI Code section 7.10.5 covers the requirements for tie reinforcement. These code

provisions are listed below:

a.) All non-prestressed reinforcement (longitudinal) shall be enclosed by lateral ties at least

in diameter for bars #10 and smaller and at least in for #11, #14,

#18 and bundled bars.

b.) Vertical tie spacing shall not exceed:

1. longitudinal bar diameters (largest bar if multiple sizes are used,

2. tie bar or wire diameters,

3. of the compression member.

c.) Ties shall be located vertically not more than one-half a tie spacing above the top of

footing or slab in any story and shall be spaced to not more than one-half a tie spacing

below the lowest horizontal reinforcement in a slab or drop panel above.

d.) Where beams or brackets frame into all four sides of a column, the ties may be

terminated no more than below the lowest reinforcement in the shallowest

of such beams or brackets.

e.) Ties shall be arranged such that every corner bar and alternate longitudinal bar shall have

lateral support provided by the corner of a tie with an included angle of no greater than

6

135 deg and no bar shall be farther than 6" clear on each side along the tie from such a

laterally supported bar. Figure 3 clarifies the longitudinal bar spacing with respect to

those bars that must be tied.

< 6" (typ.) > 6" (typ.)

< 135 deg.

( a ) ( b ) ( c )

( d ) ( e ) ( f )

( g )

Figure 3: Valid Tie Arrangements for Rectangular and Square Columns.

In addition to the code requirements for tie and stirrup steel, there are also requirements

pertaining to the longitudinal bars in a reinforced concrete column. Code section 10.9 sets

forth limits for the longitudinal reinforcement in columns and these limits are listed below:

a.) The area of steel used as longitudinal reinforcement shall be .

Practically speaking, the upper limit on reinforcement in a column should be around

4-5% of the gross concrete area. The reason for this is that congestion at beam to

column joints will become excessive if greater than 5% steel is used.

b.) The minimum number of longitudinal bars in compression members shall be 4 for bars

within rectangular ties or circular ties, 3 for bars within triangular ties, and 6 for bars

enclosed by spirals.

Additional requirements within ACI 318-99 for spiral reinforcement can be understood

through examining the confinement stress and the added strength of the column that results. The

7

confinement stress the hoop provides to the concrete can be computed using hoop stress

principals.

f ’

A fsp y

2

A fsp y

d

s

c

Spiral

Distribution ofConfinementStress

Figure 4: ACI Analogy to Analytically Determine Confinement Stress and the

Volumetric Ratio of Spiral Reinforcement.

The spiral pitch is defined as s and the confinement stress is labeled, 2f ′ . Equilibrium in the

lateral direction is written as,

( ) ( )22 0sp y cA f f sd′− =

which results in the confinement stress being written as,

2

2 y sp

c

f Af

sd′ = (2)

The ratio of the volume of spiral steel to the volume of core concrete is written as,

2

422

cs sp

c

d Ad s

ρ ππ =

from which the spiral area can be computed,

4

s csp

d sA ρ= (3)

Plugging equation (3) into equation (2) results in,

2 2s yf

fρ

′ = (4)

The compressive strength of the spirally confined core concrete has been found by experiment to

be closely represented by,

8

(5)

where 0.85 cf ′ is the compressive strength of unconfined concrete and 2f ′ is the confinement

stress provided by the spiral reinforcement.

The strength provided by the spiral can be computed realizing that the increase in

compressive strength as a result of spiral confinement is 24.0 f ′ as computed using (5).

Therefore, multiplying by the area of the concrete core gives,

strength from spiral 4.0 22s y

c s c y

fA A f

ρρ

= =

The strength contribution of the concrete shell is given by,

( )strength from shell 0.85 c g cf A A′= −

where cA is the area of the concrete core and gA is the gross area of the concrete section.

The basis behind spiral design by ACI 318-99 is that the strength gained from the spiral

should be at least equal to the strength lost when the concrete shell spalls. Therefore,

(6) Equation (6) can be simplified to,

0.425 1g cs

c y

A fA f

ρ ′

= −

(7)

which forms the basis of ACI Code section 10.9.3. This section states that the volume ratio of

spiral reinforcement to total concrete core volume shall not be less than,

0.45 1g cs

c y

A fA f

ρ ′

= −

(8)

Equation (7) is rounded up to an even value to form equation (8) which is ACI equation 10-6.

The last section of the ACI Code pertaining to column reinforcement is section 7.8, which

covers special reinforcement details for columns. The provisions pertaining to offset bars are

listed below and are pictured schematically in Figure 5.

a.) The slope of the inclined portion of offset bars with the axis of the column shall not

9

exceed .

b.) Bar portions above and below the offset shall be .

c.) Horizontal support at offset bends shall be provided by lateral ties, spirals or parts of

floor construction.

.

d.) Where a column face is offset greater than 3", longitudinal bars shall not be offset bent

and separate dowels shall be provided

LAP

min(3", s /2)1

2

s

s /22

6" (max.)

s

1

(a ) Offset less than 3 inches

( b ) Offset greater than or equal to 3 inches

l

LAPss /2

sdc

2

CompressionDowels

2

1

max. slope = 1:6

parallel to col. above

min(3", s /2)1

Figure 5: Special Considerations for Column Dimension Offsets.

SHORT COMPRESSION MEMBERS

A short column has its capacity determined by the material strength of the components used in

the cross-section (steel and concrete). There is no detrimental behavior (added moments for

10

example) that results from the interaction of bending and axial thrust such as P-∆ or P-δ effects.

The goal of this section is to present means of analysis and design for short columns subjected to

pure compression, compression with uniaxial (single axis) bending, and compression with

biaxial bending.

Concentrically Loaded Short Columns All column members whether structural steel, wood or concrete have bending moments applied

to them as a result of out-of-plumb, imbalance of loading, or slightly offset beam to column

connections. As a result, a true concentrically loaded column subjected to zero bending moment

never occurs in practice. The effect of this is accounted for in the ACI Code when defining the

strength of a short concrete column subjected to pure axial load.

Consider the column section subjected to axial load and bending moment at its end in Figure

6.

P

Me= PM

P

( a ) ( b ) Figure 6: Equivalent Column Eccentricity: (a) Axial Loading and Bending

Moment; (b) Equivalent System with Eccentricity.

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

The ACI Code utilizes the equivalent eccentricity when defining the strength of short

concentrically loaded concrete columns. Columns with relatively small eccentricity compared to

the concrete cross section are characterized as having compression over the entire cross section.

If this column is overloaded, it will fail under a combination of crushing of the concrete and

yielding of the steel in compression on the more heavily compressed side. Columns with large

eccentricity have tension over portions of the cross-section and therefore, at ultimate load, the

11

cross section may fail as a result of concrete crushing, tensile steel yielding or a combination of

both phenomena.

Design of concrete columns is based on the factored overload stage. At this stage, the

factored loads must not exceed the nominal strengths multiplied by reduction factors,

and n u n uP P M Mφ φ≥ ≥

In the cases of uniaxial and biaxial bending of columns, the factored loads act in combination

with moments and therefore, the adequacy of the concrete cross section under combined loads

cannot be determined through the simple comparison shown above. The member must be

designed for the maximum bending moment that can accompany a given design axial load.

More elaborate techniques are therefore used.

The capacity of a concentrically loaded concrete column is derived from loads carried by both

longitudinal steel and concrete. Using an equation previously derived, this capacity can be

written as,

( )0.85o c g st st yP f A A A f′= − + (9)

Failure for the concentrically loaded column occurs with uniform compression over the cross-

section. As the result, the strains in the concrete and steel are uniform over the cross-section.

As discussed previously, columns are never really free from bending moment. Therefore, the

ACI Code limits the concentrically loaded strength of the column through an assumed minimum

eccentricity. Furthermore, it was previously shown that the type of lateral reinforcement also

affects the ultimate strength of the concentrically loaded column. ACI Code section 10.3.5 gives

the design axial load strength of tied and spiral reinforced concrete columns. The maximum

design strength allowed for a reinforced concrete column with lateral reinforcement consisting of

spirals is computed using,

(10)

which is ACI code equation (10-1). The maximum allowed design strength for a tied column is

determined by,

12

( )(max) 0.80n oP Pφ φ= (11)

It should be noted that the two factors, 0.85 for spiral columns and 0.80 for tied columns account

for minimum eccentricity ratios, e h , of 0.05 and 0.10, respectively. The φ factors for the

concentrically loaded member also take the type of lateral reinforcement into consideration.

Recalling previous discussion, the spirally reinforced concrete column has a slightly higher axial

load carrying capacity and greater ductility than the tied column. Therefore, the φ factors are

assigned as follows:

a.) ________ for the spirally reinforced column,

b.) ________ for the tied column.

Strain Compatibility Analysis Consider the reinforced concrete column section with axial load acting at an equivalent

eccentricity e shown in Figure 7.

hy

d’

h2

d

b

A’

A

0.85 f’

,’

,

c a

, = 0.003

C

C

T

e e’

P

N.A.P.N.AP.C.

s

s

s

s

s

c

s

nc

c

( b ) Strains( a ) Cross Section ( c ) Stresses and Internal Forces

Figure 7: Fundamental Concepts for Strain Compatibility Analysis.

The assumption that the strains in the concrete and steel are compatible as the member deforms

requires that the strain in the concrete is identical to that in the adjacent reinforcing steel. It

should be noted that the stresses are NOT THE SAME. Stresses in the materials are a

function of the modular ratio, n. The strains, stresses and internal forces in the column are also

shown in the figure. Several terms in Figure 7 need definition: PNA is the plastic neutral axis of

the section, NA is the neutral axis of the section corresponding to the geometric centroid

13

(obtained through moment-area principles), PC is the plastic centroid of the section and sA′ is the

area of the steel subjected to compressive stress.

Equilibrium for the short column depicted in Figure 7 allows the nominal axial force at failure

to be written as follows:

(12)

The nominal bending moment, which is rewritten for analytic purposes as nP e , is obtained

through writing the moment equilibrium equation about the plastic centroid of the cross section,

( ) ( ) ( )2n n c s saM P e C y C y d T d y′= = − + − + − (13)

It should be noted that the plastic centroid and the geometric centroid are the same for columns

with symmetric reinforcement layout. It is assumed that at ultimate load for the column

(combined uniaxial bending and axial load) the following hold true,

0.85c c

s s s

s s s

C f baC A fT A f

′=′ ′=

=

Substituting the above relations into equations (12) and (13) gives the axial load and bending

moment at failure of the column,

0.85n c s s s sP f ba A f A f′ ′ ′= + − (14)

and

( ) ( ) ( )0.85 2n c s s s saM f ba y A f y d A f d y′ ′ ′ ′= − + − + − (15)

The depth to the neutral axis, c , is assumed to be less than the effective depth, d, of the

cross-section and the tension steel is assumed to be in tension. If the eccentricity of the load is

very small, the total cross-section is in compression (including the Atension@ zone steel). In this

case, the contribution of the tension steel should be added to the compressive strength of the

column. That is, the minus sign in equation (14) should be changed to a plus sign. The signs in

equation (15) should also be watched to ensure there is tensile steel. In addition, it is also

assumed that the volume of concrete compression area displaced by reinforcing steel is

negligible. The error in this last assumption is not significant for small bar sizes, but could be

14

for large bar sizes. It should be noted that the axial load strength of the concrete column

computed using equation (14) cannot exceed that of the tied column given by (11).

The magnitude of the load eccentricity dictates whether or not the compression and/or tension

steel reach their ultimate stress, yf . If failure of the cross section occurs with yielding in the

tension area steel, the stress in the steel should be replaced with the yield stress. If the stresses in

the tension and compression area steel are less than yield, similar triangles can be used to

compute the necessary stresses,

( )0.003s s s s y

c df E E f

cε

′−′ ′= = ≤ (16)

( )0.003s s s s y

d cf E E f

cε

−= = ≤ (17)

Interaction Diagrams for Concrete Columns

As one might surmise, determining the state of stress in the reinforcing steel at failure can be

difficult and tedious when analyzing a short steel column. Furthermore, using strain

compatibility analysis becomes too tedious for design applications (of course, with computer

software nothing is too tedious). An alternate approach to design of concrete columns is to

construct interaction diagrams defining combinations of bending moment and axial load at

failure for a column for a full range of eccentricities. A column with zero eccentricity is

subjected to pure compression and a column with infinite eccentricity is subjected to pure

moment.

For any loading combination defining the failure of a concrete column, there is a unique axial

load and bending moment (eccentricity) that defines the state of failure. For a given column

cross-section and reinforcing steel layout, these combinations define a yield surface for the

reinforced concrete compression member. Such a surface is called an interaction diagram and a

schematic example is shown in Figure 8.

15

Figure 8: Interaction Diagram for Nominal Column Strength and Uni-Axial Bending.

The radial lines in the figure illustrate a constant eccentricity of axial load.

_______________

______________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

For a load path defined by constant eccentricity; as the axial load (and corresponding bending

moment) increases, the load path will eventually reach the yield surface and failure of the

column will occur.

There are two distinct conditions of failure in a concrete as indicated in the interaction

diagram:

a.) Tension failure of the steel through yielding (corresponding to large eccentricities),

b.) Compression failure of the concrete (corresponding to small eccentricities).

A third condition ______________________________________________________________

CompressioControls

TensioControls

P

P

P

Radial lines show

Pe =

M

Me ∞

e,

o

n

n

n Mo

n(max)

n

16

____________________________________________________________________________

The balanced condition is attained when the strain in the concrete reaches its ultimate value

( 0.003c in inε = ) at the same time the strain in the reinforcing steel reaches its yield value yε .

The eccentricity in the member corresponds to the balanced eccentricity and the axial load in the

member (on the failure) corresponds to the balanced axial load.

The limit state for members with large eccentricities is defined by yielding of the tension

steel. If the eccentricity is larger than the balanced eccentricity or the axial load (on the yield

surface) is less than the balanced axial load, failure will occur through yielding of the steel. The

stress in the compression steel may/may not be at yield, and the actual stress should be compute

using strain compatibility (similar triangles).

If the eccentricity is less than the balanced eccentricity and the stress in the tensile

reinforcement is less than the yield stress, concrete crushing will be the controlling failure

mechanism. The stress in the compression steel will most probably be at the yield stress and

therefore, compatibility of strain should be used to check the state of strain in the compression

steel.

17

EXAMPLE 1: CONSTRUCTING THE INTERACTION DIAGRAM Consider a concrete column having the geometry listed below and cross-section shown below.

Assume that 6,000cf psi′ = and 60,000yf psi= .

3 “

11 “14 “

12 “

(4)#11

0.003 in./in. 0.85 f’

A’ f

C

A f

( a ) Cross Section ( b ) Strains ( c ) Stresses

,’

c

ss

,sss

s

c

Solution:

1. The first point on the interaction surface to be created is that corresponding to the

concentric axial loading.

The strain and stress distribution for this case is shown in the figure above. The nominal

axial loading corresponding to pure axial compression is,

2. Another key point on the interaction surface is the balanced point defining the

balanced condition at failure.

Using strain compatibility we have,

( )87,000 11.0 6.5187,000 60,000bc in = ⋅ = +

6.51 3.00.003 0.0016 0.002076.51s yin inε ε− ′ = ⋅ = < =

18

( )29,000 0.0016 46,400sf psi′ = ⋅ =

The depth of the stress block can be computed as follows,

( )

( )

1

1

0.05 6,000 4,0000.85 0.75

1,000

0.75 6.51 4.88b ba c in

β

β

⋅ − = − =

= = =

The nominal axial capacity is then computed using equation (14),

( )( )( ) ( )( )

( )( )0.85 6,000 12.0 4.88 2 1.56 46, 400

2 1.56 60,000 256, 224nbP

lbs

= +

− =

The nominal moment capacity assumed to act in conjunction with the nominal axial

loading at the balanced condition is computed using equation (15),

14.0 72 2hy in= = =

The eccentricity corresponding to the balanced condition is therefore,

2,689,743 10.5256, 224b

lb ine inlb⋅

= =

3.) The third point on the interaction diagram is that corresponding to the pure

bending condition (zero axial loading).

The contribution of the compression steel to the moment capacity is neglected and

therefore, oM can be computed as follows,

( )( )( )( )

2 1.56 60,0003.06

0.85 0.85 6,000 12s y

c

A fa in

f b= = =

′

The nominal bending moment is computed in the usual manner,

( )( ) 3.062 1.56 60,000 11.0

2 21,772,784

no s yaM A f d

lb in

= − = −

= ⋅

The remaining points on the interaction surface correspond to strain distributions that are

not as well defined as the three previously computed points. Two additional points will

19

be computed: one in the compression controls region and one in the tension controls

region.

4.) Compute a point in the compression controls region of the diagram.

The schematic strain diagram and internal forces are shown below.

3 “

11 “14 “

12 “

(4)#11

0.003 in./in. 0.85 f’

A’ f = C

C

A f = T

( a ) Cross Section ( b ) Strains ( c ) Stresses

,’c

ss

, < ,ys s

s

c

ss

s

When compression failure occurs, the depth of the neutral axis is greater than that at the

balanced condition,

14.0 72 2hy in= = =

Any arbitrary value for the depth of the neutral axis can be chosen. In the present case,

10 inches is chosen for c. The maximum compressive strain in the concrete remains

fixed at the ultimate strain ! It should be noted that ANY neutral axis depth could be

chosen as long as it is GREATER than the depth to the neutral axis at the balanced

condition. Using the assumed depth, 10c in= , the strain in the compression steel can be

computed as,

The strain and stress in the tension steel can be computed in a similar manner to that done

previously,

( )

11 100.003 0.000310

0.0003 29,000,000 8700.0

s y

sf psi

ε ε− = ⋅ = < = =

The depth of the compression block is computed as,

20

( )1 0.75 10 7.5a c inβ= = =

The forces in the compression steel, tension steel, and concrete are therefore,

Using the above values, the nominal axial capacity is computed as the algebraic

summation,

459,000 187, 200 27,144 619,056nP lb= + − =

The nominal moment capacity and eccentricity corresponding to the nominal axial

capacity shown above is computed using equation (15),

( ) ( )7.5459,000 7.0 187,000 7.0 3.0 27,144 11.0 7.0 2,349,126

2

3.79

n

n

n

M lb in

Me inP

= − + − + − = ⋅

= =

5.) Compute a point in the tension controls region.

3 “

11 “14 “

12 “

(4)#11

0.003 in./in. 0.85 f’

A’ f = C

C

A f = T

( a ) Cross Section ( b ) Strains ( c ) Stresses

,’ c

s

s

, > ,yy s

s

c

ss

s

The assumed strain distribution and internal forces are shown above. In this case, the

depth to the neutral axis will be less than that for the balanced condition. The strain

diagram will give indication as to why this must be the case. A value for the depth to the

neutral axis is assumed to be 4.2 inches. The depth of the compression block, the strain

in the compression steel, and the stress in the compression steel, are computed as,

21

The forces in the compression steel, tension steel, and the concrete are computed as,

( )( )( )( )( )

0.85 6,000 12 3.15 192,780

3.12 24,857 77,554

3.12 60,000 187, 200

c

s

s

C lb

C lb

T lb

= =

= =

= =

The nominal axial loading and bending moment capacity are computed as,

The eccentricity corresponding at this condition is,

Five points on the interaction diagram have now been computed. These points can be used to

construct the nominal strength interaction diagram for the beam-column cross-section. It should

be noted that no ACI requirements have been satisfied to this point. The interaction diagram

computed is based purely on theoretical cross-section strength. The interaction diagram that we

computed is given below,

22

P (kips)

M (kip -in)

1,231 k

(2,689.7 k-in, 256.2 k)

(2,349.1 k-in, 619.0 k)

(2,103.9 k-in, 83.1 k)1.772.8 k-in

200

400

600

800

1000

1200

1400

500 1000 1500 2000 2500 3000

Computed InteractionDiagram

Actual InteractionDiagram

Nominal Strength Interaction Diagram

23

CIRCULAR COLUMNS AND OTHER ACI REQUIREMENTS

The interaction diagram just computed is based purely on cross-section strength and does not include reduction (workmanship) factors. The ACI Code allows the reduction factor to change (in a beneficial manner) depending on the magnitude of axial load in the column. Furthermore, circular columns provide unique challenges in design and analysis. Each of these special conditions will be addressed. The interaction diagram will then be modified to include reduction factors and other ACI Code requirements. Circular Column Analysis A circular column poses unique challenges in the analysis of a column. Difficulty arises in hand computations due to the following:

1.) ______________________________________________________________________, 2.) _______________________________________________________________________

______________________________________________________________________. It obviously would be advantageous to create an equivalent rectangular (square) column to simplify the calculations without sacrificing too much accuracy in the analytical results. An empirical method is available for the creation of an equivalent rectangular column for

circular column analysis. This equivalent section can be created using the following steps;

1.) Thickness in the direction of bending set equal to 0.8h ;

2.) Width is obtained from the same gross area, gA , of the circular column with

0.8gb A h= ;.

3.) Area of reinforcement is replaced with (2) parallel layers placed at 2 3sD in the

direction of bending. sD is the diameter of the longitudinal steel placement

configuration.

4.) Special consideration for tension failure: Use the actual column (and strain

compatibility) for evaluating the compressive force, cC , but place 40 percent of the

total cross-section steel in a parallel layer at a distance 0.75 sD .

The usual rectangular analysis techniques can be used if the above steps are taken in the

development of the equivalent rectangular column. The method provides satisfactory results for

most cases.

24

EXAMPLE 2: COMPUTATION OF EQUIVALENT RECTANGULAR COLUMN Determine the equivalent rectangular column dimensions and steel layout for the circular column

shown below.

Assume that there are (8) -#8 bars in the column with

#4 spiral ties and 1 1/2" clear cover.

Solution:

Following the procedure previously established, the area of the cross section and the diameter of

the reinforcement cage are,

( )

( ) ( )

220314.2 . .

420 2 1.5 2 0.5 1.0 15.0

s

s

A sq in

D in

π= =

= − − − =

Thus, the dimensions of the equivalent rectangular section are:

( )0.8 20 16

314.2 19.616

h in

b in

= =

= =

The area and distance between the equivalent parallel reinforcement for the compression controls

section is,

( )2 15 10

3

2st

s s

d d in

AA A

′− = =

′= =

The distance between and the area of the parallel reinforcement for the tension controls case

are,

A = total area of steel

D

20 “

st

s

25

( )0.75 15 11.250.4s s st

d d inA A A

′− = =

′= =

The equivalent cross sections for the compression controls and tension controls cases are shown

in the figure below. Incidentally, the balanced condition falls under the compression

controls condition for computation of the strength of the equivalent section.

st

A = A / 2

sA’ = A / 2

19.63 “

sts

sts

A’ = 0.4 A

A = 0.4 A

s st

( a ) compression failure ( b ) tension failure

It should be noted that geometric relations are available for determining the area of the circle

segment depicted in figure (b). The steel manual (LRFD) or (ASD) has such relations as part of

section 6 in the design manual.

26

ACI REDUCTION FACTOR IN TENSION CONTROLS REGION

The ACI Code has made the evaluation of the design interaction diagram a little more

complicated in that the reduction factors change in the region of the diagram where the axial load

is small. One can think of the beam-column with small axial loading as a beam. Essentially, the

N factor is increased from that required for the axially loaded column (0.70 or 0.75 depending on

the ties) to that of pure bending, 0.90.

ACI Code section 9.3.2.2 covers this aspect of reduction factor increase. The increase in N

can occur provided that the effective depth between tension and compression reinforcement in

the column is not less than 70% of the total height and the yield strength of the reinforcement

does not exceed 60,000 psi. Two simple formulas can be used to compute the reduction factor.

These equations depend on the lateral reinforcement type: for tied columns,

0.90 0.20 0.700.10

ntied

c g

Pf A

φφ

= − ≥ ′ (18)

and the following for spiral columns,

0.90 0.15 0.750.10

nspiral

c g

Pf A

φφ

= − ≥ ′ (19)

The following criteria must be used when increasing φ ,

a.) If 0.10nb c gP f Aφ ′< , use 0.10 0.70c g nbf A P′ = in equation (18) and use 0.10 0.75c g nbf A P′ =

in equation (19).

b.) No increase in φ can be taken above the balanced point.

27

EXAMPLE 3: COMPUTATION OF DESIGN INTERACTION DIAGRAM Each point on the interaction diagram previously created as part of Example 1 will be addressed

and the correct reduction factors applied to develop points on the strength interaction surface for

the uniaxially loaded column.

Solution:

We will use the five points on the cross-section strength interaction diagram computed

previously and adjust to account for the appropriate strength reduction factor(s).

1.) The first point on the interaction diagram considered was the case of pure axial

loading. The nominal capacity and design capacity for this point are computed

below.

( )( )

(max)

(max)

1, 231, 2000.8 1, 231, 200 984,960

0.7 984,960 689, 472

n

n

n

P lbP lb

P lbφ

=

= =

= =

2.) The second point on the diagram corresponded to the balanced condition. In this

case, all reduction factors are those for the tied column and therefore,

( )

( )

256, 2240.7 256, 244 179,3572,689,7430.7 2,689,743 1,882,820

nb

nb

nb

nb

P lbP lb

M lb inM lb in

φ

φ

=

= =

= ⋅

= = ⋅

3.) The third point on the diagram corresponded to the case of pure bending.

( )

1,772,7840.9 1,772,784 1,595,506

no

no

M lb inM lb inφ

= ⋅

= = ⋅

4.) The fourth point corresponded to a case where the failure was controlled by

compression (above the balanced point). The design axial loads and moments are

therefore,

( )

( )

619,0560.7 619,056 433,3392,349,1260.7 2,349,126 1,644,388

n

n

n

n

P lbP lb

M lb inM lb in

φ

φ

=

= =

= ⋅

= = ⋅

5.) The last point was for a case where tension controlled the failure. This is the place

28

where the reduction factor can be adjusted.

83,1342,103,884

n

n

P lbM lb in

== ⋅

We now need to determine if the reduction factor can increase. Assuming that

0.70φ = , we can compute the following,

( )

( )( )( )0.70 83,134

0.10 0.1 12 14 6,000 100,800 179,357

0.10 0.70

n

c g nb

n c g

P lb

f A lb P lb

P f A

φ

φ

φ φ

=

′ = = < =

′< ⇒ =

It is also obvious that

0.10

nb n

c g

P Pf A

φ φ>′>

As a result, equation (18) can be used without modification. Several trials may be

required for the results to converge to the correct reduction factor. The first trial

yields,

( )

( )

0.20 58,1940.90 0.785

100,8000.785 583,134 65, 222nP lb

φ

φ

= − =

= =

The second trial is,

( )0.20 65, 2220.90 0.77

100,800φ = − =

which is sufficiently close to the first trial. Therefore, the design axial loading and

bending moment can be computed as,

( )

( )

83,1340.77 83,134 64,0132,103,8840.77 2,103,884 1,619,991

n

n

n

n

P lbP lb

M lb inM lb in

φ

φ

=

= =

= ⋅

= = ⋅

The design interaction diagram can now be generated. One should take note of the extra work

involved to increase the reduction factor from 0.70 to 0.77. As a designer, one should always

keep in mind whether or not effort is really providing added value to a design. However, if an

29

evaluation (or analysis) is being done, the extra 0.07 may be the difference between “calling” a

column adequate and telling a client to “reinforce or replace”. One should certainly appreciate

the added value in this instance. The strength and design interaction diagram for this column

cross-section is shown below.

P (kips)

M (kip -in)

200

400

600

800

1000

1200

1400

500 1000 1500 2000 2500 3000

984.96 k

689.5 k

P - M diagramu

n n

u

P - M diagram0.7(1231)=862

30

HOMEWORK ASSIGNMENT NUMBER 1

31

DESIGN AIDS FOR COLUMN STRENGTH

As illustrated through the previous examples, strain compatibility would be an extremely

cumbersome procedure to use for design of columns subjected to axial load and uniaxial

bending. (Biaxial bending using this procedure would be even more unthinkable in a design

office setting.) One does not know the shape of the interaction surface until the column is

chosen and one cannot choose the column size and reinforcement until the interaction surface is

known. This trial and error procedure makes the strain compatibility method undesirable for

design.

Design aids and/or computer programs are usually employed to facilitate design of concrete

columns. The two most popular are SP-17 by the American Concrete Institute and the CRSI

Handbook by the Concrete and Reinforcing Steel Institute. Figure 9 illustrates an example of a

chart included in the SP-17 Handbook. The chart is an interaction diagram for a given column

size and reinforcement configuration. The charts are generalized for many different cross

sectional sizes through non-dimensional ordinates.

Figure 9: Design chart for column analysis and design.

32

The x-axis of the interaction diagram is the non-dimensional bending moment written using the

moment or axial load acting at an eccentricity,

or n n

c g c g

P e Mf A h f A h

⋅′ ′

The y-axis is the non-dimensional axial load,

n

c g

Pf A′

The diagrams are denoted WX-Y.Z indicating X ksi concrete, Y ksi grade steel, Z ratio of inside

dimension of reinforcement layout (center to center) to overall height of cross-section. W

indicates: R - rectangular tied column, L - rectangular column with reinforcement placed in (2)

layers parallel to the axis of bending, C- spirally reinforced circular column and S - rectangular

column with spiral reinforcement layout.

Reinforcement ratios of 0.01 to 0.08 are given. Recall, that using ratios larger than

approximately 4 percent can result in excessive congestion of reinforcement in the concrete

columns and joints. Practical construction issues should always be considered in design of the

concrete column.

The alignment charts as included in SP-17, allow the direct design of eccentrically loaded

columns throughout a wide range of practical geometry and reinforcement ratios. Use of the

charts can take place in two typical manners: (assuming factored axial load and moment or

eccentricity is given).

1.) Design total section: cross-section and steel.

a.) Select trial cross-section dimensions: b, h.

b.) Calculate the ratio ( based on cover, estimate bar sizes and tie requirements.

c.) Calculate Pn / f ’c Ag and Mn / f ’c Agh.

d.) Plot the point corresponding to the values computed in part c.) on the interaction

diagram corresponding to the desired concrete strength, steel strength,

reinforcement layout and ( value computed in part b.) Read the required

reinforcement ratio. (Make sure the point plotted is within the interaction surface

corresponding to your reinforcement ratio.)

e.) Calculate the total steel are Ast=Dgbh and detail accordingly.

33

2.) Section sizes and steel based on assumed reinforcement ratio.

a.) Select a steel reinforcement ratio, Dg.

b.) Choose a trial value of h and calculate the eccentricity rato (e/h) and (.

c.) Using the reinforcement ratio, eccentricity ratio and the correct chart read off the

corresponding value of Pn / f ’c Ag and compute the gross area of the cross

section.

d.) Calculate the corresponding width, b.

e.) Revise the trial cross section as required to obtain a well proportioned cross

section.

f.) Calculate the area of steel based on the cross-section dimensions and the assumed

reinforcement ratio.

The use of the SP-17 design aids for column design for eccentrically loaded concrete columns

will be illustrated through several examples.

34

EXAMPLE 4 - USE OF SP-17 TO DESIGN FOR UNIAXIAL BENDING AND COMPRESSION

A column is to be designed to carry a factored axial load of 518 kips and a factored moment

equal to 530 kip-ft. Use the assumed material strengths of 60,000 psi for reinforcement and

4,000 psi for concrete. A steel ratio of 0.03 is determined to be economical from cost studies

during preliminary design. Bending is assumed to occur about the major (strong) axis. Also,

assume that the reinforcement is located parallel to the major axis of bending.

Determine the required overall column dimensions and the total area of steel. Assume that

the reinforcement will be arranged in two layers adjacent to the outer faces of the column and

parallel to the strong axis of bending.

Solution:

Select a trial dimension perpendicular to the axis of bending. In this case choose h=24". Using a

concrete cover of 2.4 in. to the bar centers,

Therefore, using the concrete strength, steel strength, ( value and the reinforcement arrangement

given, the correct chart for this design problem is L4-60.9.

The eccentricity ratio for the load can now be computed as,

Using L4-60.9, the eccentricity ratio and reinforcement ratio are used to locate a point on the

interaction diagram (refer to the next sheet). The design charts have been created in such a

manner, that the ordinate and absissa can easily be used as “ray” lines or slopes.

35

The point on the chart corresponding to the intersection of the ray line e/h with the interaction

diagram for Dg = 0.03 yields,

The section size required is . The total area of steel required is computed as,

Thus, a placed in two(2)

layers of four (4) bars parallel to the major axis as indicated in L4-60.9.

36

EXAMPLE 5 – USE OF SP-17 FOR UNIAXIAL BENDING AND COMPRESSION DESIGN

An exterior column that is part of a two story structure is to be designed for a service axial

DL=150 kips, maximum service axial LL=347 kips, service dead load moment of 100 kip-ft and

a service live load moment of 192 kip-ft. The minimum live load compatible with the full

service live load moment is 106 kips. This last live load is obtained when no live loading is

placed on the roof, but a full live load is placed on the second floor. Architectural considerations

require that a rectangular column be used, with dimensions b=18" and h=24". Use 4,000 psi

concrete and 60,000 psi material strengths. Also, assume reinforcement to be distributed around

perimeter and that bending occurs about the major axis.

Determine:

1.) Compute the required column reinforcement for the condition of full live loading,

2.) Check to ensure that the column is adequate for the condition of no live load applied to

the roof framing.

Solution:

Part (a):

The factored axial load and bending moments for the full load condition are computed below,

( ) ( )( ) ( )

1.4 150 1.7 347 800

1.4 100 1.7 192 466.4 5600u

u

P kips

M k in k in

= + =

= + = ⋅ = ⋅

Quantities necessary for design aid use are calculated below:

Assuming 2.0" cover to the center of the longitudinal steel,

( )24 2 2.40.8

24γ

−= =

37

The material strengths, reinforcement layout and ( computed above require that chart R4-60.8

be used. Plotting the point on the interaction diagram corresponding to the factored axial and

bending stresses above on the design chart (below) results in Dg = 0.018.

Thus, the area of steel required in the 18"x 24" cross-section is,

Use a 18"x 24" column with (17)-#8 bars spread out around the perimeter, five (4) each side as

shown below (this is more than is needed, but it will prevent field errors in placement).

38

Part (b):

Consideration of the absent roof live load is prudent as a result of the interaction diagram tending

to "become smaller" as the axial load in the member decreases. In a certain sense, the axial load

in the member is helpful in relieving tension stresses caused by bending moment. As the axial

load decreases, the tensile stresses will increase possibly deeming the cross-section inadequate.

The minimum axial load and corresponding bending moment are,

( ) ( )1.4 150 1.7 106 390.2

466.4 5600u

u

P kipsM k ft k in

= + =

= ⋅ = ⋅

Values to be used in the design chart are,

( )( )

( )( )( )

390 0.320.7 4 432

5600 0.190.7 4 432 24

u

c g

u

c g

Pf A

Mf A h

φ

φ

= =′

= =′

Plotting the two values on the design chart indicates that the required Dg is approximately 0.023,

which greater than that previously required. We chose 12-#8 bars, which provides 9.48 square

inches. This translates to 0.022gρ = . We’ll consider this adequate.

NOTE: The lateral tie layout size and spacing should be computed in addition to the

longitudinal reinforcement.

39

DESIGN AND ANALYSIS FOR BIAXIAL LOADING - SHORT COLUMNS

In addition to uniaxial bending, columns can have bending moments applied about both principal

axes. This case of applied loads is commonly termed biaxial bending. Biaxial bending is a

result of many different framing and loading situations, including,

1.) corner columns in tier buildings.

2.) unbalanced loading on the beams or slab system framing into the concrete column.

Using the basic principles of equilibrium and strain compatibility is difficult when a single axis

of bending is considered. If bending about two axes occurs, the resulting neutral axis is no

longer parallel to one of the sides of the member. It is inclined. As one might expect, utilizing

strain compatibility analysis in a situation such as this can be prohibitively difficult for practical

analysis and design.

It should be noted that circular columns (provided the reinforcement is spread equally around

the perimeter) posses polar symmetry so that a resultant moment can be created,

2 2( )u res ux uyM M M= +

and design or analysis can proceed with the circular column with factored axial load, Pu , and

uniaxial bending Mu(res) defined above.

Failure Surfaces

The concept of the failure surface is used to facilitate design and analysis of reinforced concrete

columns subjected to biaxial bending and compression. The interaction diagram used to analyze

and design columns subjected to uniaxial bending is a portion of a three dimensional failure

surface. This three-dimensional failure surface defining biaxial bending strength can be

visualized by revolving (spinning) the uni-axial interaction diagram 90 degrees about the axial

load axis.

Notation used to describe the failure surfaces needs to be developed. A rectangular

reinforced concrete column is subjected to bending moments about both of its principal axes.

The bending moments are assumed to be rewritten in terms of eccentricities about both the x-

and y-axes as shown in Figure 10.

40

x-axis

y-axise

e

( Reinforcing bars notshown for clarity )

P

M = P ex

y

nx

ny

n

M = P en

x

y

n

Figure 10: Notation for Biaxial Bending of Compression Members.

Three types of failure surfaces have been defined in the literature. The most basic surfaces are

defined on a three axis coordinate system consisting of the nominal axial load or inverse of

nominal axial load and eccentricities about the x- and y-axes. Figure 11(a) illustrates the first

failure surface and Figure 11(b) illustrates the second failure surface, which is commonly called

the reciprocal failure surface.

Failure SurfaceS ( P , e ,e )1 n x y

P

e

ey

x

Failure SurfaceS ( 1 / P , e , e )

1 / P

e

e

n

y

x

n yx2

(a) Failure Surface, 1S (b) Failure Surface, 2S

Figure 11: Failure Surface 1S and Reciprocal Failure Surface 2S

The third failure surface is a three dimensional representation of the interaction diagram used

previously. In this case, the nominal axial load and bending moments about both axes are used.

41

The eccentricities present in Figures 11(a) and 11(b) are replaced by biaxial bending moments.

The three dimensional respresentation is slightly more intuitive as a result of our previous

discussions of the case of uniaxial bending. Figure 12 illustrates the more intuitive interaction

surface.

P

M

M

Failure SurfaceS ( P , M , M )

P - M InteractionCurvesn

y

x

n3 nx ny

n

Figure 12: Failure Surface 3S Commonly Used in Beam-Column Design.

The failure surfaces illustrated have been used in the creation of design and analysis

procedures. Three such procedures are:

1. The Bresler reciprocal load method,

2. The Bresler load contour method,

3. The PCA load contour method.

All methods are quite useful, however, we will concentrate on the load contour methods in this

course because their implementation is more closely related to the case of uniaxial bending.

Bresler Load Contour Method The use of failure surface S3 affords a much more “feeling” oriented approach to the design and

analysis of biaxially loaded columns. The failure surface looks more like the interaction diagram

discussed previously for uniaxially loaded columns and therefore, the load contour method is

consistent in its look and feel with the methods used for uni-axial bending.

The load contour method approximates the failure surface using a family of curves

42

corresponding to constant values of nP . Figure 13 illustrates the concept of load contours on

failure surface S3.

P

M M

M M

P , Mnoy

nx

nyn

n n

M

y

Plane at constant Pn

Load contour

Mnox

nP

M

x

Figure 13: Load Contours for Constant Axial Loading on Failure Surface 3S .

Each layer (contour) represents all combinations of biaxial bending moment that will cause

failure at the axial load corresponding to the contour level. Therefore, the contour is nothing

more than a slice through the interaction surface. An approximate expression can be written for

each load “plane” as the interaction equation below,

(20)

nxM and nyM are the nominal biaxial moment strengths in the direction of the x- and y-axis,

respectively. These moments are the vector equivalent of the nominal uniaxial moment strength.

Mnox and Mnoy are the nominal uniaxial moment strengths with bending considered in the

direction of the x- and y-axis, respectively. The values of the exponents in (20) depend on the

amount, distribution and location of longitudinal reinforcement, the dimensions of the column,

and the strength and elastic properties of the steel and concrete. It is reasonably accurate to

43

assume that the exponents are the same variable and therefore,

1.0nynx

nox noy

MMM M

αα + =

(21)

The result of this is shown in Figure 14 as a range of contours,

Figure 14: Biaxial Moment Interaction Curves Defined by Equation (21).

A simplification to equation (21) that results in consistently conservative results is to set α

equal to unity,

1.0nynx

nox noy

MMM M

+ = (22)

Equation (22) becomes overly conservative when axial loads are large and there are low

percentages of reinforcement in the column. The linear simplification given by equation (22)

should only be used in situations where,

0.1n c gP f A′<

This limit on axial loading ensures that one is within the upper-reaches of the failure surface.

Upon examination of Figure 13, one should recognize that the upper portions of the failure

surface are not as “buldgy” as the lower regions.

PCA Load Contour Method The third method for analysis and design of reinforced concrete columns subjected to biaxial

44

bending is an extension of the Bresler load contour method. This method, termed the PCA load

contour method (after a technique developed by the Portland Cement Association) is based upon

the Bresler interaction equation for a given axial load shown here as equation(21). A typical

load contour computed using equation (21) is shown in Figure 15.

Figure 15: Load Contour of Failure Surface 3S Along Plane of Constant nP

The PCA method is defined in such a manner that at point B the biaxial moment strengths nxM

and nyM are in the same ratio as the uniaxial moment strengths; noxM and noyM . This can be

written as,

nx nox

ny noy

M MM M

=

The load contour can be non-dimensionalized as shown in Figure 16.

Figure 16: Non-Dimensional Load Contour

The load resistence of the concrete column can be plotted in a dimensionless form, with the

following parameters,

45

nyn nx

o nox noy

MP MP M M

The second two terms above are designated as relative moments. The failure surface that can be

generated using the above parameters has the advantage of being symmetric about the vertical

plane that splits the two relative moment coordinate planes. This new failure surface 4S is shown

in Figure 17.

Figure 17: Non-Dimensional Failure Surface 4S Used in the PCA Load Contour Method

This procedure yields results sufficiently accurate for design for square and rectangular sections,

and for equal and unequal distribution of reinforcement along the faces of the member.

The relationship that relates the non-dimensional coordinate β to the exponent in equation

(21) is given below,

Equation (21) may then be rewritten in a non-dimensional form as,

46

log0.5log0.5loglog

1.0nynx

nox noy

MMM M

ββ + =

(23)

A plot of the curves implied by equation (23) is shown in Figure 18 (also in SP-17 Handbook).

Figure 18: Non-Dimensional Biaxial Moment Strength Relationship.

Design charts giving values for β for varying reinforcement layouts, material strengths and

axial load ratios are available (Figure 19).

Figure 19: Typical Biaxial Bending Design Chart (taken from SP-17).

47

A method suitable for relatively rapid design of reinforced concrete sections can now be

developed. The design procedure can be expedited if the curves for $ are approximated using

two (2) straight lines as shown in Figure 20.

Figure 20: Bilinear Approximation of Non-Dimensional Load Contour.

Upon examination of Figure 20, it can be seen that the equation for the upper line is,

1 1.0 for ny ny noynx

noy nox nx nox

M M MMM M M M

ββ

−+ ⋅ = >

which can be rewritten as,

(24)

For rectangular sections with longitudinal reinforcement distributed equally on all faces,

equation (24) can be approximately rewritten as,

1 noyny nx noy

nox

Mb bM M Mh h M

ββ

− + ≈ ≈

(25)

The equation describing the lower line of the figure is,

1 1.0 for ny ny noynx

noy nox nx nox

M M MMM M M M

ββ

−⋅ + = <

or,

(26) Once again, for rectangular sections with equally distributed reinforcement on all faces,

48

1nx ny nox

bM M Mh

ββ

− + ≈

(27)

Use of design equations (25) and (27) requires that the b/h or h/b be chosen and the value of

β be assumed. β can be assumed to be 0.65 for an initial guess in biaxial bending analysis

assuming a lightly loaded column is to be checked.

Manual Design Procedure for Biaxial Bending The designer should develop a mechanical procedure for design of columns subjected to biaxial

bending. A procedure for manual design is included below for simplicity:

1. Choose the value of β at 0.65 or use the design charts for β to make an estimate.

2. If ny nxM M is greater than b h , use equation (25) to calculate an approximate

equivalent uniaxial moment strength noyM . If ny nxM M is less than b h , use equation

(27) to calculate an approximate equivalent uniaxial moment strength noxM .

3. Design the section using any one of the methods for uniaxial bending previously

described to provide axial load strength nP and an equivalent uniaxial bending moment

strength noxM or noyM .

4. Verify the section obtained using either load contour approach.

Design Examples for Biaxial Bending of Concrete Columns Use of the load contour design and analysis procedures must be illustrated through several

examples. The following publication; Notes on ACI 318-99 - Building Code Requirements for

Reinforced Concrete with Design Applications, Edited by S.K. Ghosh and Basile G. Rabbat,

published by the Portland Cement Association is an excellent source for design examples for

biaxial columns. Therefore, two examples taken from this text will be used to illustrate the

design and analysis procedures for biaxially loaded columns discussed in this chapter.

49

EXAMPLE 6 - DESIGN OF RECTANGULAR COLUMN FOR BIAXIAL BENDING

50

51

52

53

EXAMPLE 7 - DESIGN OF CIRCULAR COLUMN FOR BIAXIAL BENDING

54

55

HOMEWORK NUMBER 2

56

This Page Intentionally Left Blank

57

SLENDERNESS EFFECTS Up until this point, we have been speaking about short columns in which material failure of the

member has been the governing limit state. There are situations, however, where the member

can fail as a combination of overall member stability and material failure. The limit state at a

“buckling failure” of the member is the motivation for ACI Code provisions concerning

slenderness effects.

A typical eccentrically loaded column is shown in Figure 21. This moment can also result

from concentrated bending moments. Of course, this bending moment could then be written in

terms of the applied axial load at an eccentricity.

Figure 21: Eccentrically Loaded Slender Column.

The initial moment in the member (called first order moment) can be written as,

(28)

As the axial load, P , is applied, the bending moments induced in the member cause it to

displace laterally an additional amount, δ . As a result of this lateral deformation, the original

P P

P

P

e M = P ( e

58

moment in the member given by equation (28) is increased. Using the free body diagram of the

member taken up to the point of maximum lateral deformation, the maximum bending moment at

the member mid-height can be written as,

( )cM Pe P P eδ δ= + = + (29)

The final bending moment at the member mid-height is then a sum of the first order bending

moment, Pe , and what is often called the second order moment, Pδ . Therefore, the moment

that the column is designed for must be that expressed in equation (29).

The effect that the second order moments have on the interaction diagram for the reinforced

concrete column can be seen in Figure 22.

B

C

A A’

Pe P*

O

P, axial load

M, bending moment

Short columninteraction diagram

Material failure

Short column

MMMP 4

Stability failure

Figure 22: Slenderness Effects and the Interaction Diagram.

Various load-moment curves are plotted on the short column interaction diagram shown in

Figure 22. Three separate columns are depicted in the figure, given by the load moment curves;

O-A=, O-B and O-C.

59

1. Line O-A=: (relatively short column: O-A is approximated line). This column is termed a

short column and failure is through essentially material failure and secondary effects are

minimal.

2. Line O-B: (column of moderate length). This column fails when the load-moment curve

intersects the interaction diagram at point B and material failure occurs. This type of

failure is expected in most practical building columns

3. Line O-C: (long or very slender column). Failure of this type of column occurs without

full cross-section strength being obtained as a result of very significant second-order

effects.

Effective Length for Columns

In general, reinforced concrete columns reside within two types of frameworks. Each of these

frame types has significant bearing on the behavior of the members within. Frames can be

classified as;

1. Braced Frames where the ends of the columns are ________ allowed to translate with

respect to one another.

2. Unbraced (Sway) Frames where the ends of the columns _________ allowed to translate

with respect to one another.

Second order effects are commonly quantified based on the critical load for the column

(sometimes called the Euler buckling load),

( )

2

2cru

EIPKLπ

= (30)

Various terms in the above equation will be defined for review purposes;

E = the modulus of elasticity of the member,

I = the moment of inertia of the member (minimum value for minimum critical load),

K = effective length of the column depending on the bracing at the member ends (should

be consistent with I),

Lu = unbraced length of the column (should be consistent with I).

The determination of the effective length factor and the moment of inertia for a reinforced

60

concrete column becomes the challenge when considering slenderness effects and the critical

buckling load.

Reinforced concrete columns have end conditions that are similar to those in structural steel

columns. One might argue, however, that all concrete columns have nearly “fixed” end

conditions (or at least fully restrained – no pins). Consider the four idealized column end

conditions shown in Figure 23.

K = 1.0 K = 0.50 K = 1.0 K = 2.0

* )L

KL

L

KL

L

KL

u

uu u

u

u

u*

)

L = KLu

Figure 23: Idealized End Conditions and Effective Lengths for Braced and Unbraced

Concrete Columns

The pin-pin column shown in Figure 23 has the end conditions for which the Euler buckling load

or critical load equation was derived. The effective length factor is introduced to modify the

critical load equation so that differing end conditions on the column can be introduced. The

effective length, KLu, is the length of a fictitious pin-pin column that has the same critical load as

the column under consideration.

One should note that for members within an actual building frame, the restraints at the ends

of the columns are not fully fixed nor truly pinned. Therefore, an alternate means for computing

the effective length factors must be employed. For braced columns, the effective length factor

can be obtained using an alignment chart such as that shown in Figure 24(a).

61

(a) Braced Column (b) Unbraced Column

Figure 24: Jackson-Moreland Alignment Charts

The following definition is in order,

which is the relative stiffness of the compression members to flexural members framing in a

plane at one end of the column. Lc is the length of the compression member measured from

centerline of joint to centerline of joint. L is the length of the flexural member measured

centerline of support to centerline of support. The summation for the columns framing into the

joint should include the column under consideration. The alignment chart can be used to

compute effective length factors for the unbraced column (Figure 24(b)).

It should be noted that properly designed and detailed foundation systems never really

provide zero rotational restraint. Therefore, it is common practice to allow a value of ψ = 10.0 at

the Ahinged@ end of the compression member for use in the alignment chart for unbraced

columns. The case of precast concrete columns should be carefully scrutinized.

62

Moment Magnification As discussed previously, the interaction of gravity load and displacement results in additional

deformation and increased bending moments called second-order effects. The current state of

design office practice requires that first-order analysis be used as a basis for approximating

second order effects (Computer technology advances are changing this, and soon second-order

analysis programs for individual member and frame analysis will be widely available.) The ACI

Code does allow second order analysis programs to be used to compute bending moments in

columns either individually or as part of structural frames. This section will discuss the typical

means and theory for magnifying first order moments for braced and unbraced columns (frames).

Moment Magnification for Braced Columns The column shown in Figure 25 is considered braced. It is subjected to end moments and axial

compression as shown.

o

P

M

M

P

* * M + P ( * + * )P ( * + * )s oo

o

Mo so s o

(a) Loaded column (b) First ordercolumn moment

(c) Second ordermoment

(d) Total moment

Figure 25: First-Order and Second-Order Moments in Braced Column.

The deflection under the action of bending moment Mo is considered to be the first order

deflection, δo. This is the quantity usually obtained via standard computer programs and analysis

63

techniques. When the axial load, P, is applied, the deflection increases by the amount δs. The

total deflection is termed the second order deflection and it is given by, δ=δ o+δs. The primary

moment in the column is Mo and the secondary moments are called Pδ moments. The total

moment indicated in Figure 3.36(d) is the magnitude the column should be designed for. The

total moment can be obtained using the first order moment and application of a magnification

factor.

A magnification factor can be derived using conjugate beam principles. Assume that the

deflected shape of the column and moment diagram for the secondary moment is a half-sine

wave. The area under one-half the second order moment diagram is given by,

( )02

2sP LAreaEI

δ δπ

= ⋅ + ⋅ ⋅ (31)

and the centroid of the area is located at L/π from the support. The second order deflection is

therefore,

( )02

2s sP L LEI

δ δ δπ π

= ⋅ + ⋅ ⋅ ⋅

( )2

02s sPL

EIδ δ δ

π= ⋅ + (32)

One should notice that equation (32) contains the Euler buckling load (excluding the effective

length factor) as shown in equation (30). Thus,

Grouping terms in the above equation and solving for the second order deflection gives,

1

Es o

E

P PP P

δ δ= ⋅−

The final deflection δ is a sum of first and second order deflections,

11 1

Eo s o o o

E E

P PP P P P

δ δ δ δ δ δ= + = + ⋅ = ⋅− −

(33)

Examination of equation (33) reveals that as the axial load approaches the Euler buckling load,

the deformation at the column mid-height approaches infinity.

64

We can assume that the bending moments vary in the same way as the deflections.

Therefore, the maximum bending moment at the column mid-height is given by,

1

oo o

E

PM M P MP Pδδ= + = +

− (34)

The first order bending moment for the column shown in Figure 25(a) is given by,

2

8o

oM L

EIδ =

The axial load may be written algebraically as,

2

2E

P EIPP L

π= ⋅

Substituting the above two equations into equation (34) gives,

(35)

where Cm is equation (35) is a factor that depends on the end moments that the column is

subjected to. AF is the amplification factor for second order moments and Mnt is the moment in

the column with no lateral translation of the ends. It should be noted that PE in equation (35) is

the result of the pin-pin column being used in the derivation.

A value for Cm can be easily derived for the condition of unequal end moments at the ends of

the column. It is assumed that no transverse loads exist between the ends of the column. The

expression for Cm is the same as that used in the AISC Specifications for steel design. This

expression is given below,

1

2

0.6 0.4mMCM

= + (36)

The ratio 1 2M M is ________ if the member is bent in single curvature and ________ if the

member is bent in reverse curvature.

65

Equations (35) and (36) are a fast and relatively accurate means for computing the bending

moment of a braced compression member with end moments. As stated previously, the

derivation of equation (35) assumes that no transverse loads exist on the column. If the

compression member under consideration has transverse loading, it is suggested that Cm factors

corresponding to the loading conditions considered be obtained from the AISC Specifications or

a text on structural stability such as; Structural Stability - Theory and Implementation by Wai-

Fah Chen and Eric M. Lui.

Moment Magnification - Unbraced Columns Similar magnification of moments can occur for the case of unbraced columns. Consider the

unbraced frame shown in Figure 26.

P P

H

)1 2

L

L b

Figure 26: Unbraced Frame Used for Moment Magnification Derivation.

The horizontal deflection of the frame (story) is given by ∆. Each column has an axial load

applied at the joint. It should be noted that this axial load could be a result of many stories above

this one in question. There are two approaches to the computation of a magnification factor for

first order moments in sway frames. The first is utilizes the story stiffness concept and the

second uses the multiple column buckling concept.

The first method to derive magnification factors is the story stiffness approach. The basic

66

assumptions made in the derivation of the magnification term using this procedure are;

1. Each story can behave independently of other stories,

2. Additional column moments from the P ⋅∆ effect are caused by a force P L∑ ∆ .

The additional lateral force that is assumed to exist comes from equilibrium of the column(s) in

the deformed configuration of the frame as illustrated in Figure 27.

P P

)

1 2

L

)

P )

Li3P )

P )

L1

2P )

L

2P )

L

L1

Figure 27: Equivalent Lateral Force(s) for Second-Order Analysis.

The story stiffness (in sway mode) can be defined as,

horizontal forcelateral displacementFS =

It is assumed that we can set the first order stiffness equal to the final stiffness,

o

H H P L∑ ∑ +∑ ∆=

∆ ∆ (37)

Using a little algebra, equation (37) can be solved for the displacement that includes second

order effects, ∆, as,

1

1o

oPHL

∆ = ⋅∆∑ ∆−∑

(38)

Assuming the moments in the member will be proportional to the lateral deformations of the

member. The final moments can be written as,

67

1

1lt

oM MP

HL

= ⋅∑ ∆−∑

(39)

where ltM is the moment as a result of lateral translation of the frame.

The second method to derive second order moment magnification terms utilizes the story

buckling concept. Equation (33) can be used (with a minor substitution) if it is assumed that

story instability occurs with all columns in that story become unstable simultaneously (in a sway

buckling mode). Using this assumption and replacing EP P with EP P∑ ∑ , equation (33)

becomes,

1

1o

E

PP

∆ = ⋅∆∑

−∑

(40)

where: P∑ is the sum of all gravity loading applied to the columns in the story; and EP∑ is

sum of all Euler critical buckling loads for the columns in the story. As before, with rigid joints

and elastic behavior, the moments that include second order effects are written as,

(41)

where all terms have been defined to this point.

General ACI Provisions for Moment Magnification The approximate procedure allowed by the Code uses magnified moments from an "ordinary

first- order" analysis. The ACI Commentary defines first-order frame analysis as an elastic

analysis, which does not include the added internal forces resulting from deflection. This is

(was) the usual analysis standard. Computer programs are now beginning to utilize second-order

solution algorithms. As a result, the magnification procedure will have a limited life as software

becomes more advanced. One should be very careful using second-order analysis via

commercial software.

When structural steel members are used in the moment magnifier equations, the modulus of

elasticity and section properties are well defined. However, when reinforced concrete is used,

68

these quantities are not so easily determined. First of all, cracking in the columns and beams

affect the cross sectional properties significantly. Thus, all the moment magnification terms are

affected. Furthermore, the duration of loads applied to the members and corresponding creep

can also affect the slenderness and stability of a compression member.

ACI Code section 10.11.1 outlines simplified values of section properties and modulus of

elasticity for use in the expressions contained in the moment magnification procedure. The Code

allows the following properties for members within a structure;

a.) The modulus of elasticity for the members in the structure can be computed using,

1.5c c cE w f ′= (42)

and for concrete strength: 5,000 12,000cpsi f psi′≤ ≤ ,

( )1.5

640,000 1.0 10145

cc c

wE f ′= + × ⋅

(43)

b.) The moments of inertia can be taken as the following:

Beams: ________ gI

Columns: ________ gI

Walls Cracked: ________ gI

Uncracked: ________ gI

Flat plates and flat slab: ________ gI

c.) Area: ________ gA

The moments of inertia are to be divided by ( )1 dβ+ when,

a.) sustained lateral loads act,

b.) stability checks are made with dβ as a creep factor.

The value of dβ is defined as,

maximum factored sustained loadtotal factored axial loaddβ = (44)

Another section property needed when computing critical loads using the moment

69

magnification procedure is the radius of gyration. The radius of gyration for the cross section,

according to 10.11.2 may be taken as,

a.) 0.3 h for rectangular and members, where h is the overall height in the direction

stability is being considered,

b.) 0.25 D for circular members where D is the overall diameter,

c.) the computed radius of gyration for the gross concrete cross section for other shapes.

ACI section 10.11.3 defines the unsupported length of the compression member. The

unsupported length lu is taken as the clear distance between floor slabs, beams other members

capable of providing lateral support in the direction being considered for stability/slenderness. If

column capitals or haunches exist, the unsupported length must be defined using the lower

extremity of the haunch or capital.

Section 10.11.4 defines the sway and non-sway column. The definition of sway and non-

sway extends to both stories and frames. A column and story within a structure can be

considered as non-sway if the following are satisfied.

a.) The increase in column end moments due to second order effects does not exceed five

(5) percent of the first-order end moments.

b.) The stability index given by,

u o

u c

PQV l

∑ ∆=

is less than 0.05. uP∑ and uV are the total vertical load acting on the story and the story

shear, respectively. o∆ is the first order relative story displacement and cl is the length

of the compression members within the story measured from center to center of framing

beams/slabs. One should note the stability index term within equation (39).

It is sometimes more useful to think of the amplification factor rather than the stability index.

Recalling the amplification for first-order (sway) effects,

1 1 1.0531 1 0.05FA

Q= = =

− −

it can be seen that when the stability index is at its limit for braced behavior, the second order

effects are 5.3% of the first order (relatively small). Thus, the stability index check is saying

70

that when second order effects are small, the frame can be considered as braced.

There is a limit as to how slender a compression member can be when using the moment

magnifier approach. The common measure is the slenderness ratio, uKL r . ACI section 10.11.5

dictates that if the slenderness ratio for a compression member exceeds 100, a second order

analysis considering material non-linearity, cracking, member curvature, lateral loads, duration

of loads, shrinkage, creep, foundation interaction and factored loads shall be used in lieu of the

moment magnifier approach. In other words, if uKL r is greater than 100, the analysis required

may be too complicated for design. If possible, a larger section should be chosen.

Biaxial bending is a distinct possibility when a slender column is considered. As directed by

ACI 10.11.6, the moment about each principal axis shall be magnified separately based on the

conditions of restraint corresponding to that axis.

As was seen throughout the derivation of expressions used for moment magnification in non-

sway and sway frames, the critical loads for compression members must be evaluated. The ACI

Code section 10.12.1 requires that the calculation of K be based on values of E and I used in

section 10.11.1. In lieu of using the alignment chart for evaluation of the effective length factor

for a braced column, the ACI Commentary allows the following two expressions to be used as an

upper-bound estimate,

( )( )min

0.70 0.05 1.0

0.85 0.05 1.0A Bk

k

ψ ψ

ψ

= + + ≤

= + ≤ (45)

The smaller of the two expressions should be used.

The ACI commentary also allows upper bound solutions using equations for un-braced

columns. The expressions are broken down into two categories; columns with both ends

restrained, and columns with restraint at one end, and hinged at the other.

a.) Compression member restrained at both ends;

[ ],m A Bavgψ ψ ψ=

71

202 1

202 0.9 1

mm m

m m

k

k

ψψ ψ

ψ ψ

−< = +

≥ = + (46)

b.) Compression member hinged at one end;

2.0 3.0 resk ψ= +

where resψ is the value at the restrained end.

ACI Moment Magnification for Non-Sway Frames

The ACI provisions for moment magnification in non-sway frames are contained in section

10.12 of the Code. The basic overall procedure and philosophy of the code is to determine

whether or not slenderness should be considered, and then magnify the moments in the

compression member in the appropriate manner. In a non-sway frame, slenderness effects and

hence magnification of moments can be neglected if,

1

2

34 12ukl Mr M≤ − (47)

where 1 2 0.50M M > − . The term 1 2M M is positive if the compression member is bent in

single curvature. ukl is effective length of the section and the radius of gyration, r, is evaluated

as outlined in 10.11.2.

The magnified moment in the compression member is written as an amplification term

multiplied by the larger of the two end moments that result in the member with no lateral

translation of the frame. This can be simply written as,

2c nsM Mδ= (48)

where cM is the magnified non-sway moment, 2M is the larger of the two end moments and

nsδ is the non-sway amplification factor given by,

1.01

0.75

mns

u

c

CP

P

δ = ≥−

(49)

where,

72

( )

2

2cu

EIPklπ

= (50)

The similarity in the above expressions to those derived is apparent. The quantity, EI, shall be

taken as,

0.2

1c g s se

d

E I E IEI

β+

=+

(51)

where seI is the moment of inertia of the reinforcement about the member cross section

centroidal axis. The member stiffness can also be taken as,

0.41

c g

d

E IEI

β=

+ (52)

The factor, mC , which modifies the amplification factor according to the moments applied at the

member ends is taken as,

1

2

0.6 0.4 0.4mMCM

= + ≥ (53)

where 1 2M M is positive if the member is bent in single curvature. mC can be taken as 1.0 if

transverse loads exist between the member ends. 2M is the larger of the two end moments.

As one might expect there is also a minimum factored moment the slender non-sway member

must be designed for according to the Code. This minimum moment is computed as,

( )2,min 0.6 0.03uM P h= + (54)

about each axis separately. The member height, h, is to be taken in inches. For members in

which 2,minM exceeds 2M , the value of mC in equation shall be taken equal to 1.0 or based on

the ratio of computed end moments.

ACI Moment Magnification for Sway Frames A similar magnification procedure to that done for non-sway frames is performed for the case of

sway frames. The ACI 318-95 procedure has been modified over that contained in previous

editions. The revised procedure consists of the following steps:

1. Compute the magnified sway moments, s sMδ . This calculation can be done using

73

second-order elastic frame analysis, an approximate moment magnification technique

included in the Code, or sway magnifiers from previous editions of the Code. sM is the

first order moment as a result of frame sway.

2. The magnified sway moments, s sMδ , are added to the unmagnified non-sway moment

nsM at each end of the column. The non-sway moments may be computed using a first

order elastic analysis.

3. If the column is slender and the loads on it are high, it is checked to see whether the

moments at points between the ends are higher than those at the ends of the compression

member.

The above procedure is imbedded in the ACI Code provisions contained in section 10.13 for

sway columns.

The effective length factor for the sway column shall be determined using the relationships

found in section 10.11.1. A main difference in the computation or attainment of K from the

alignment charts in the case of the sway column versus that of the non-sway column is in the

definition of βd.

1. Except as provided below in (b),

max. factored sustained shear in storytotal factored shear in storydβ =

the maximum sustained story shear could result from lateral earth pressure acting on a

portion of a frame. It will be zero for most building / bridge structures.

2. For stability checks of sway frames carried out in accordance with ACI 10.13.6,

max. factored sustained axial loadtotal factored axial loaddβ =

In all cases, the effective length factor shall be taken as greater than or equal to 1.0.

Slenderness may be neglected for a member allowed to sway when the slenderness ratio

adheres to the following;

22uklr<

74

The bending moments acting at each end of a beam-column need to be magnified for the

design. Therefore, the moments at the ends of plane frame compression members shall be taken

as,

1 1, 1,

2 2, 2,

ns s s

ns s s

M M MM M M

δδ

= +

= + (55)

The magnified moments at the ends of the member can be computed in two ways based on the

story stiffness or story buckling concept. Using the story stiffness method,

1

ss s s

MM MQ

δ = ≥−

(56)

where Q is the stability index previously defined. If the amplification factor for sway moments,

sδ exceeds 1.50, a second order analysis or the story buckling amplification factor shall be used.

The magnified moment computed based on the story buckling concept is given by,

1

0.75

ss s s

u

c

MM MPP

δ = ≥∑

−∑

(57)

where uP∑ is the sum of all factored vertical loads applied to the story under consideration and

cP∑ is the sum of all critical loads for the compression members contained in the story under

consideration. cP is computed using the effective length factors and methods already

established.

For slender columns with high axial loads, the point of maximum moment within the

member may not be at the ends of the member as is assumed thus far. The maximum moment

may actually reside within the ends of the member and this situation has not been addressed in

the code provisions to this point. Therefore, if an individual compression member has,

35u

u c g

lr P f A>

′ (58)

it shall be designed for the factored axial load, Pu, and the moment Mc calculated using the non-

sway magnified moment given by equation (48) in which M1 and M2 are computed using the

magnified moments for the sway column given by equations (55).

The strength and stability of the structure under gravity loads alone must also be maintained.

75

Therefore,

1. When the magnified moments are computed using a second order elastic analysis, the

ratio of second-order lateral deflections to first order lateral deflections for 1.4 DL and

1.7 LL plus lateral load applied to the structure shall not exceed 2.5.

2. When the magnified moments are computed according to the story shear method given

by equation (56), the value of the stability index computed using 3Pu for 1.4 DL plus 1.7

LL shall not exceed 0.60.

3. When the magnified moments are computed using the story buckling concept given by

equation (57), the amplification factor corresponding to factored dead and live loads shall

be positive and not exceed 2.5.

In all three of the cases above, βd shall be taken as the ratio of the maximum factored sustained

axial load to the total factored axial load for the story under consideration.

One should always keep in mind that magnifying the moments acting on a column to account

for second order effects implies that the beams framing into the column must have their moments

magnified also.

Design Examples - Slender Columns As one might expect, the design of a braced or unbraced slender concrete column is a rather

cumbersome task. The main issue is the fact that the effects of slenderness are not known until

the cross-section is chosen and the cross-section will dictate the effects of slenderness. As a

result, a semi iterative procedure is required for the design of slender columns.

A schematic procedure for the design of slender sway and non-sway columns can be

described as follows:

1. Select a trial column section to carry the factored axial load, Pu, and the moment, Mu=M2

from the elastic first order frame analysis assuming short column behavior.

2. Determine if the frame (and column) is braced or unbraced.

3. Determine the unsupported length and effective length factor for the column.

4. Check for consideration of slenderness effects (depends on braced or unbraced nature of

the column).

76

5. If slenderness is not a problem, design as a short column and stop this procedure.

6. If moments in the frame are small, check to determine if minimum moments and/or

eccentricities control.

7. Depending on the bracing characteristics of the column, compute the amplified moments.

8. Check the adequacy of the column to resist the factored axial load and the amplified

factored moment. Use the SP-17 charts in the usual manner. Revise the column

reinforcement and cross-section as necessary.

9. If the column cross-section is close to its capacity (slight overstress) refine the

calculations for effective length and amplified moments further and adjust column

accordingly.

The above procedure will be used in the design of braced and unbraced reinforced concrete

columns. It should be noted that the moments given in the examples are intended to alleviate the

time consuming attainment of bending moments through frame analysis. One should keep in

mind that these moments must be determined through analysis and this is often a major step in

concrete column design.

77

Example 8 - Design of a Slender Column (Braced Case)

78

79

80

81

82

Example 9 - Design of a Slender Column (Unbraced Case)

83

84

85

86

87

88

COMPUTER ANALYSIS FOR CONCRETE COLUMNS The final stage in the saga of concrete columns is design and analysis via computer. As we have

seen, the strain compatibility method can be prohibitively cumbersome for column analysis and

even more so for design. The design handbooks are nice to use, however, in many cases the

situations that you run into in practice do not always fit the assumptions used to develop the

charts and tables. One could interpolate or make some conservative assumptions in order to fit

the situation at hand to the tables/charts, but the use of computer programs is probably the best

method for analysis and design. Furthermore, it is probably the most accurate method of all

(assuming no bugs in the program exist).

The concrete column analysis and design program called PCACOL will be introduced. The

Portland Cement Association in Skokie, Illinois publishes this program. We have obtained

authorization to distribute copies of the programs to the students for educational use.

Unfortunately, the program only designs and analyzes according to ACI 38-89. For

demonstration purposes, this is okay. THE SOFTWARE IS NOT TO BE USED FOR

PROFESSIONAL (CONSULTING) PURPOSES.

The program is a one-stop shop for complete column analysis and design. The programs

coverage includes irregular column analysis, biaxial bending, uniaxial bending and slenderness

effects and many other facets of reinforced concrete column analysis.

Use of the program will be illustrated through in-class demonstrations on the computer. Two

example problems will be illustrated. The data used in the examples is listed below. And the in-

class demonstration will provide the know-how to use the program for solving the examples.

89

Example 10 - Design of a Slender Unbraced Column A portion of a ten story office building is shown below. Design the first interior first story

columns.

Columns are not braced against sidesway and the column under consideration bends in double

curvature. First floor clear height is as shown above. Clear heights for the floors above and

below (basement level) is 11=-0@. Building layout is 7 by 3 bays with 26 in. square beams

framing into the column in both directions. Material properties for the structure are: 5,000 psi

concrete and 60,000 psi steel.

Service loads on the interior column(s) are given below (moments about the x-axis - major axis):

PD = 363 kips PL = 128 kips PW = 0 kips

MD = 16.5 ft-kips ML = 13.5 ft-kips MW = 50 ft-kips at top of column

MD = 42 ft-kips ML = 34 ft-kips MW = 50 ft-kips at bottom of column

Assume the following values for moment magnification and effective length computation:

3Pu = 13,422 kips 3Pc = 38,472 kips Pu = 544 kips Pc = 1878

kb = 0.78 ks = 1.30

Use the computer program PCACOL to design the slender, uni-axially bent, unbraced column.

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Example 11 - Design for Biaxial Loads Determine the size and reinforcement of a rectangular column with the following factored loads:

Pu = 1700 kips, Mux = 2900 ft-kips and Muy = 1200 ft-kips. Assume that the reinforcing bars

are equally distributed on all column sides. Use the block stress profile, 4,000 psi concrete and

Gr. 60 steel. Assume that the column is non-slender. Architectural considerations have limited

the sides of the column to a minimum of 30 inches and a maximum of 48 inches. Use PCACOL

for the design.

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HOMEWORK ASSIGNMENT #3

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