DIAGRAMA DE INTERACCIÓN DE COLUMNAS

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Introduction to Concrete Column Design Flow Charts

The Column Design Section in Strunet contains two main parts: Charts to develop strength interaction diagrams for any given section, and ready -made Column Interaction Diagrams, for quick design of a given column.

Concrete column is one of the most interes ting members in concrete structural design application. A structural design of a concrete column is quite complicated procedures. Evaluation, however, of a given column section and reinforcement is straightforward process. This is due to the fact that pure axial compression is rarely the case in column analysis. Some value of moment is always there due to end restraint, or accidental eccentricity due to out of alignment. ACI established the minimum eccentricity on a concrete column, regardless of the struct ural analysis proposed for the column, which is defined as the maximum axial compression load that a column can be designed for.

Column Design Charts in Bullets: One of the demanding aspects in concrete column design is to define the controlling points on strength interaction diagram. The column strength interaction diagram is a curve plot of points; where each point has two ordinates.

The first ordinate is bending moment strength and the second is the corresponding axial force. Both ordinates are linked with eccentricity. The shape of the curve, or the strength interaction diagram, can be defined by finding the ordinates of major seven points. Each point has specific requirement, as established by the code, and thus evaluating the requirement of this poin t will result of calculating the ordinates. The points and their respective requirements are as follows: • • • • • • •

Point 1: Pure compression.

Point 2: Maximum compression load permitted by code at zero eccentricity.

Point 3: Maximum moment strength at the maximum a xial compression permitted by code.

Point 4: Compression and moment at zero strain in the tension side reinforcement.

Pont 5: Compression and moment at 50% strain in the tension side reinforcement.

Point 6: Compression and moment at balanced conditions.

Point 7: Pure tension.

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Notations for Concrete Column Design Flow Charts a ab

= =

Ag As A’s A’st b c cb

= = = = = = =

Cc Cs C’s d

= = = =

d’

=

e eb Es f’c fy fs f’s h Mb Mn Po Pb Plim

= = = = = = = = = = = = =

Pn T

= = = = = = =

β? εs ε 's εy φ

depth of equivalent rectangular stress block, in. depth of equivalent rectangular stress block at balanced condition, in. gross area of column, in 2. area of reinforcement at tension side, in 2 . area of reinforcement at compres sion side, in2 .

Total area of reinforcement in column cross section, in 2 . column width dimension, in. distance from extreme compression fiber to neutral axis, in. distance from extreme compression fiber to neutral axis at balanced condition, in. compression force in equivalent concrete block. compression force in tension-side reinforcement, if any. compression force in compression-side reinforcement. distance from extreme compression fiber to centroid of tensi onside reinforcement distance from extreme compression fiber to centroid of compression-side reinforcement eccentricity, in. eccentricity at balanced condition, in. modulus of elasticity of reinforcement, psi. specified compressive strength of concrete, psi. specified tensile strength of reinforcement, psi. stress in tension-side reinforcement at strain ε s , ksi. stress in compression-side reinforcement at strain ε 's , ksi. overall column depth, in. nominal bending moment at balanced condition. nominal bending moment at any point. nominal axial load strength at zero eccentricity. nominal axial force at balanced condition. limit of nominal axial load value at which low or h igh axial compression can be defined in accordance with ACI 9.3.2.2. nominal axial load strength at any point. tension force in tension-side reinforcement. factor as defined by ACI 10.2.7.3. strain in tension-side reinforcement at calculated stress fs strain in compression-side reinforcement at calculated stress f’s yield strain of reinforcement. strength reduction factor

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Main Points of Column Interaction Diagram

1

Axial Compression, φ Pn

3 max. axial comp.

4

ε s=0.0

2 5

ε s=0.5εy 6 balanced

comp. control tension control

consistent φ

increasing φ pure tension Bending Moment, φ M n

7

Point 1: axial compression at zero moment.

Point 2: maximum permissible axial compression at zero eccentricit.

Point 3: maximum moment strength at maximum permissible axial compression.

Point 4: axial compression and moment strength at zero strain.

Point 5: axial compression and moment strength at 50% strain.

Point 6: axial compression and moment strength at balanced conditions.

Point 7: moment strength at zero axial force.

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Point 1: Axial Compression at Zero Moment Point 2: Maximum Permissible Axial Compression at Zero Eccentricity

Finding Point 1 ACI 10.3.5.1

Ast ,Ag ,fc′,fy

Po = 0.85fc′ ( Ag − Ast ) + fy Ast Finding Point 2

Spiral?

Pn (max) = 0.80Po ACI 9.3.2.2

Pn (max) = 0.85Po ACI 9.3.2.2

φ = 0.70

φ = 0.75

φPn (max) = 0.56Po

φPn (max) = 0.64Po

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Strength Reduction Factor

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ACI 9.3.2.2

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finding φ Axial Compression and Axial Compression with Flexure

Axial Tension and Axial Tension with Flexure

φ = 0.90

PL =

0.10fc′Ag

φ

Spiral?

φ = 0.70

φ = 0.75

PL = 0.143fc′Ag

PL = 0.133fc′Ag

Pn PL > Pn High Values of Axial Compression

Low Values of Axial Compression

Spiral

φ = 0.75

φ = 0.70

• fy ≤ 60 ksi • h − d ′ − ds ≤ 0.70h • Symmetric reinf.

Spiral?

Spiral?

PL = min(0.143fc′Ag , Pb )

φ = 0.9 −

0.2Pn PL

PL = min(0.133fc′Ag , Pb )

φ = 0.9 −

0.15Pn PL

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φ = 0.9 −

0.2Pn PL

φ = 0.9 −

0.15Pn PL

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Point 3: Maximum Moment Strength at Maximum Permissible Compression

ACI 10.2.7.3

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fc′ ≤ 4000 psi  fc′ − 4000   ≥ 0.65  1000 

β1 = 0.85

β1 = 0.85 − 0.05  finding distance c

Cc = 0.85fc′ ( β1c ) b Cs′ = A's ( fy − 0.85fc′ ) d'

εs =

( c − d ) 0.003

A's

c

fs = εsEs

h

As

Cs = As ( fs − 0.85fc′ )

b

 ( c − d ) 0.003  Cs = As  Es − 0.85fc′  c  

Pn of point 2

 ( c − d ) 0.003  Pn = 0.85fc′ ( β1c ) b + A's ( fy − 0.85fc′ ) + As  Es − 0.85fc′  c   Result: c

a = β1c Compute: Cc , C's from above Eqs.

Mn = 0.5Cc ( h − a) + Cs′ ( 0.5h − d′) − Cs ( 0.5h − ds ) Strunet.com: Concrete Column Design V1.01 - Page 6

C's Cc

a

c

d

Pn = Cc + Cs′ + Cs

0.003 ε 's

Cs

εs ds strain

stress

Point 4: Axial Compression and Moment at Zero Strain

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Finding Point 4

εs = 0.0 c =d a = β1c

See Finding β 1 in point 3

Cc = 0.85fc′ab  c − d′ 

d'

ε s′ =   0.003  c  fs′ = min(εs′Es ,fy )

A's

0.003 ε 's

C's a

Cc

c=d

h As

Cs′ = As′ ( fs′ − 0.85fc′ )

0.0 b

Pn = Cc + Cs′ h a h  Mn = Cc  −  + Cs′  − d ′   2 2 2 

e=

Mn Pn

φMn & φPn

T=0.0

See finding φ

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strain

stress

Finding Point 5

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εy =

Point 5: Axial Compression and Moment Strength at 50% Strain

fy d'

Es A's

εs = 0.5ε y

0.003 ε 's c

C's

a

Cc

h

    d c =   1 + ε s  0.003  

a = β1c

As

ε s=0.5ε y b

See Finding β 1 in point 3

Cc = 0.85fc′ab  c − d′   0.003 

ε s′ =   c

fs′ = min(εs′Es ,fy ) Cs′ = As′ ( fs′ − 0.85fc′ ) T = As ( 0.5fy )

Pn = Cc + Cs′ −T h h a h   Mn = Cc  −  + Cs′  − d ′  + T  d −  2  2 2 2  

e=

T

Mn Pn

φMn & φPn Strunet.com: Concrete Column Design V1.01 - Page 8

See finding φ

strain

stress

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Finding Point 6

εs = ε y =

Point 6: Axial Compression and Moment Strength at Balanced Conditions

fy Es

 0.003 cb =   0.003 + ε y 

  0.003 

ab = β1cb

See Finding β 1 in point 3

Cc = 0.85fc′abb d'

 c − d′  εs′ =  b s  0.003  cb 

fs′ = min(εs′Es ,fy )

A's

Cc

As

T

ε y=fy /Es b

T = fy As

Pb = Cc + Cs −T h h a  h   Mb = Cc  − b  + Cs  − d ′  + T  d −  2 2   2 2  Mb Pb

φMn & φPn

C's

a

h

Cs = As′ ( fs′ − 0.85fc′ )

eb =

c

0.003 ε 's

See finding φ

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strain

stress

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Point 7: Moment Strength at Zero Axial Force

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Finding Point 7 d' A's

Cc = 0.85fc′ab

0.003 c

Cc

a

ε 's~0.0

h

T = fy As

As

T

ε s>ε y

a=

Asfy

b

0.85fc′b

c=

a

β1

See Finding β 1 in point 3

 0.003   d − 0.003 > ε y  c 

εs = 

Cc = 0.85fc′abb T = fy As h h a  Mn = Cc  −  + T  d −  2  2 2 

φ = 0.9

See finding φ

φ Mn

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strain

stress