lec 8 ce 806 behavior of short columns

CE 806Reinforced Concrete Members

NUST Institute of Civil Engineering

Columns

Dr. Wasim Khaliq

Reinforced Concrete Columns

Reinforced Concrete Columns

Behavior of Columns

Confinement in Concrete Columns

Concrete columns under monotonically increasing concentric compression show extremely brittle behavior unless confined with transverse reinforcement commonly in the form of closely spaced steel spirals or hoops. Hence, confinement has a significant influence on strength and ductility in the columns. At low levels of stress in the concrete, the transverse reinforcement is hardly stressed; hence the concrete is unconfined.

The concrete becomes confined when at stresses approaching the uniaxial strength, the transverse strains become very high because of progressive internal cracking and the concrete bears out against the transverse reinforcement, which then applies a confining reaction to the concrete. Thus the transverse reinforcement provides passive confinement.  Tests by many investigators have shown that confinement by transverse reinforcement can considerably improve the stress-strain characteristics of concrete at high strains. Richart et al. found, for example, that the equation as follows, for the strength of concrete confined by fluid pressure, applies approximately to concrete confined by circular spirals.

  f'cc = f'c + 4.1 fl

where f'cc = axial compressive strength of confined specimen f'c = uniaxial compressive strength of unconfined specimen fl = lateral confining pressure

The strength and ductility of concrete, therefore, are greatly increased under conditions of triaxial compression.

Figure (a): shows stress-strain curves obtained from three sets of concrete cylinders confined by circular spirals. Each set was for a different unconfined strength of concrete. The increase in strength and ductility with content of confining steel is very significant. Tests have demonstrated that circular spirals confine concrete much more effectively than rectangular or square hoops.

(Contd.)

Figure (a)

Figure (b)

In Figure (b) we have load-strain curves from concrete prisms tested by which contained various amounts of square ties. The effect of the different transverse steel contents on the ductility is quite appreciable, but the effect on strength is much smaller.

The reason for the considerable difference between the confinement by circular steel spirals and confinement by rectangular or square steel hoops is illustrated in Figure (c).

Circular spirals, because of their shape, are in axial hoop tension and provide a continuous confining pressure around the circumference, which at large transverse strains approximates fluid confinement.

However, square hoops can apply only confining reactions near the corners of the hoops because the pressure of the concrete against the sides of the hoops tends to bend the sides outwards, as in Figure (c).

Therefore a considerable portion of the concrete cross section may be unconfined. Because of internal arching between the corners, the concrete is confined effectively only in the corners and the central region of the section.

(Contd.)

Figure (c)

A smaller spacing leads to more effective confinement, as illustrated in Figure (d). The conerete is confined by arching of the concrete between the transverse bars and if the spacing is large it is evident that a large volume of the concrete cannot be confined and may spall away.

Nevertheless, square confining steel does produce a significant increase in ductility, and some enhancement of strength has been observed by many investigators.

A larger bar diameter leads to more effective confinement. Transverse bars of small diameter will act merely as ties between the corners because the flexural stiffness of the hoop bar is small.

With a larger transverse bar, the area of concrete effectively confined will be larger because of the greater flexural stiffness of the hoop side.

In the case of a circular spiral this variable has no significance: given its shape, the spiral will be in axial tension and will apply a uniform radial pressure to the concrete.

Figure (d)

Confinement in concrete columns is, therefore, greatly improved if

The reinforcement is placed at a relatively close spacing and is well anchored by hooks etc.

Additional supplementary overlapping hoops or cross ties with several legs crossing the section are included.

The longitudinal bars are well distributed within the section.

The ratio of volume of transverse reinforcement to volume of concrete core or the yield strength of the transverse reinforcement is increased.

Spirals or circular hoops are used instead of rectangular and supplementary cross ties.

Columns

Based on strength of materials and geometry columns may divided into 2 Categories

– Short Columns – most columns in ordinary practice– Slender Columns – use of high strength materials and improved

dimensioning

Strength of columns is evaluated on the basis of the following principles• A linear strain distribution exists across the thickness of the column.• There is no slippage between the concrete and steel (that is the strain in in

steel and in the adjoining concrete is the same).• The maximum allowable concrete strain at failure for the purpose of

strength calculations is 0.003 in/in.• The tensile resistance of the concrete is negligible and is disregarded in the

computations.

ACI Code Provisions

ACI Code Provisions

ACI Code Provisions

Variation in strength reduction factor ϕ

Column Strength and ACI Requirements

ACI Requirements

ACI Requirements

Required Strength

Capacity of Columns in Compression

Example 1

A non slender tied columns is subjected to axial load only. It has the geometry shown in figure and is reinforced with three No. 9 bars on each of the two faces. Calculate the nominal and design axial load strength Pn(max) for the column. fy = 60,000 psi and f′c = 4000 psi

  12″

20″

3 #9 bars

3 #9 bars

Example 2

A non slender spiral columns is subjected to axial load only. It has the geometry shown in figure and is reinforced with 6 No. 8 bars. Calculate the maximum and design axial load strength Pn(max) for the column. fy = 60,000 psi and f′c = 4000 psi

20″6 #8 bars 

Axial Load and Bending

Axial Load and Bending

Strain Compatibility Analysis and Interaction Diagram

P – M Interaction Diagram

Point A - Pure Axial LoadUniform axial compression without moment, this is the largest axial load the column can support

Point B - Zero Tension, Onset of crackingTensile stresses in concrete are overcome and section gets cracked

Region A-C - Compression Controlled FailureCrushing of compression face before extreme layer of tensile reinforcement yields

Point C - Balanced Failure, Compression controlled Limit StrainStrain in Concrete = εc → εu 0.003

Strain in Steel → εt= εy

Point D - Tensile Controlled LimitStrain in Concrete = εc → εu 0.003

Strain in Steel → εt= 0.005Region C-D – Transition RegionFrom brittle failure at C to ductile failure at D

P – M Interaction Diagram

Stresses and Strains Compatibility in Eccentrically Loaded Columns

Strain Compatibility Analysis and Interaction Diagram

Capacity of Eccentricity Loaded Columns

Strain Compatibility Analysis and Interaction Diagram

Example 3

When large bending moments are present, it is most economical to place most steel along the outer faces parallel to the axis of bending, unlike axial compression where most steel is distributed equally along all faces.

Example

Distributed Reinforcement

ϒ = 0.6 – 0.9

ρg = As/Ag= 0.01 – 0.08

εt = 0.002 - 0.005

fs/fy

e/h

𝑲 𝒏=𝑷𝒏

( 𝒇 ′𝒄 𝑨𝒈)

𝑹𝒏=𝑷𝒏𝒆

( 𝒇 ′𝒄 𝑨𝒈𝒉)

Graphs A-5 to A-16

Design Aids

Example 5

Pu = 1.2x222+1.6x297=742 KMu = 1.2x136+1.6x194=474 K-ftColumn size = 20x25Cover = 2.5 in

ϒ = (25-5)/25=0.80ρg = 0.023As = 0.023 x 500 = 11.5 ~ 12 in2

Use 12 # 9 bars evenly placed on all faces of column

Pu = 1.2x222+1.6x166=532 KMu = 1.2x136+1.6x194=474 K-ftColumn size = 20x25Cover = 2.5 in

ϒ = (25-5)/25=0.80ρg = 0.017As = 0.017 x 500 = 8.5 ~ 9 in2

Use 12 # 8 bars (9.48 in2) evenly placed on all faces of column

Example 5

Assume h = 25 inCover = 2.5 inϒ = (25-5)/25=0.80e = Mu/Pu = 492x12/481 = 12.3e/h = 12.3/25 = 0.49Use graph A-11Kn = 0.51

Use 15 x 25 in column with Ast = 0.003x15x25=11.25 in2

Use 8 # 11 bars (12.48 in2) evenly placed on two faces of the column

Circular Columns

Calculations are carried out similar to rectangular columns, except that for circular columns the concrete compression zone subject to the equivalent rectangular stress distribution has the shape of a segment of a circle.

Shape of the compression zone and the strain variation in the different groups of bars make longhand calculations awkward, no new principles are involved and computer solutions are easily developed.

Design or analysis of spirally reinforced columns is usually carried out by means of design aids, such as Graphs

Biaxial Bending

Column Section with Biaxial Bending – Strain Compatibility

Analysis and Design Biaxial Bending in Columns

1. Reciprocal Load Method

2. Load Contour Method

3. Equivalent Eccentricity Method

4. Strain Compatibility Method

Bresler Reciprocal Load Method

Bresler, B. “Design Criteria for Reinforced Concrete Columns Under Axial Load and Biaxial Loading.” Journal of the American Concrete Institute, Vol. 57, No. 5, November 1960, pp.481-490.

Bresler based his analysis on an assumption of a number of possible “Failure Surface” in three dimensions.

Failure Surface 1 - Failure point defined as a function of axial load and eccentricities.Failure Surface 2 - Similar basis with 1 failure point defined as function of 1/pn, ex, ey

Bresler Reciprocal Load Method

Bresler reasoned:1. The failure surface is too complicated to exactly define.2. An acceptable approximation could be defined by a plane which passes through threepoints which could be found by conventional (uniaxial bending) analysis.

Bresler Reciprocal Load Method

Bresler Reciprocal Load Method

Bresler Reciprocal Load Method

Bresler’s reciprocal load equation derived from the geometry of the approximate plane:

This procedure is acceptably accurate for design purposes provided Pn≥0.1P0.

If Pn<0.1P0, it would be more accurate to neglect the axial force entirely and to calculate the section for biaxial bending only.ACI strength reduction factors do not change the development in any fundamental way as long as the ϕ factor is constant for all columns.

Bresler Load Contour Method

• The load contour method is based on representing the failure surface of by a family of curves corresponding to constant values of Pn

• The general form of these curves can be approximated by a non dimensional interaction equation

Bresler Load Contour Method

α → Section, Rft, fy, f’c

α falls in the range from 1.15 to 1.55 for square and rectangular columns• In practice, the values of Pu, Mux, and

Muy are known from the analysis of the structure. For a trial column section, the values of Mnx0 and Mny0 corresponding to the load Pu/ϕ can easily be found by the usual methods for uniaxial bending.

• Then replacing Mnx with Mux/ ϕ and Mny with Muy/ ϕ and using α1 = α2 = α in Load Contour Eq, or alternatively by plotting (Mnx/ ϕ)/Mnx0 and (Mny/ ϕ)/Mny0 in figure for α, it can be confirmed that a particular combination of factored moments falls within the load contour (safe design)

Parme Load Contour MethodImproved Bresler Approach

β = 0.55-0.70 [Normal Range] = 0.65 for design

Equivalent Eccentricity Method

Equivalent Eccentricity Method

Example (Bi-Axial Bending)

Example (Contd.)

Example (Contd.)

Example (Contd.)

Example (Contd.)

Bi-axial Bending Design Constants

Bi-axial Bending Design Constants

Bi-axial Bending Design Constants

Bi-axial Bending Design Constants

Equivalent Eccentricity Method (Example)