chapter 5 reinforced concrete columns

Chapter 5

Reinforced Concrete Columns

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Lecture Goals

Definitions and Types of columnsColumns

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Analysis and Design of Columns

General Information

Vertical Structural members

Transmits axial compressive loads with or without moment

transmit loads from the floor & roof to the foundation

Column:

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Types of Columns

Short Columns Slender Columns

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Stocky members Material failure

Maximum load supported is controlled by section dimensions and strength of materials

Short Reinforced Concrete Columns

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Slender Reinforced Concrete Columns

Bending deformations

Secondary moments

Instability OR buckling

Analysis and Design of “Short” Columns

General Information

Column Types:

1. Tied

2. Spiral

3. Composite

4. Combination

5. Steel pipe

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-Tie spacing h (except for seismic)

-Tie supports long bars (reduce buckling)

-Ties provide negligible restraint to lateral expose of core

Tied Columns - 95% of all columns in buildings are tied

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Pitch = 25 mm. to 75 mm.

spiral restrains lateral (Poisson’s effect)

axial load delays failure (ductile)

Spiral Columns

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Column construction

Longitudinal column reinforcement spliced every floor (Most common)

Longitudinal column reinforcement spliced every other floor to reduce congestion

Tied column under construction.

Reinforcement cage for a tied column

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Behavior of Tied and Spiral Columns

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Although this column has been deflected sideways 50 cm, it is still carrying load

Tied col. destroyed in 1971 San Fernando earthquake

Behavior of Tied and Spiral Columns

In a spiral column, the lateral expansion of theconcrete inside the spiral (referred to as the core) is restrained by the spiral. This effect increases the stresses to:

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Behavior of Tied and Spiral ColumnsTriaxial stresses in core of spiral column

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When spiral col. are eccentrically loaded, the second max. load may be less than the initial max., but the deformations at failure are large, allowing load redistribution.

Because of their greater ductility, compression-controlled failures of spiral col. are assigned a strength-reduction factor, f of 0.75, rather than the value 0.65 used for tied columns.

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Elastic Behavior The change in concrete strain with respect to time will affect the concrete and steel stresses as follows:

Concrete stress

Steel stress

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Elastic Behavior

Concrete creeps and shrinks, therefore we can not calculate the stresses in the steel and concrete due to “acting” loads using an elastic analysis.

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Therefore, we are not able to calculate the real stresses in the reinforced concrete column under acting loads over time. As a result, an “allowable stress” design procedure using an elastic analysis was found to be unacceptable. Reinforced concrete columns have been designed by a “strength” method since the 1940’s.

Note: Creep and shrinkage do not affect the strength of the member.


Ultimate Behavior, and Design under Concentric Axial loads

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Behavior, Nominal Capacity and Design under Concentric Axial

loads

'0 g st y st0.85 cP f A A f A

Let

Ag = Gross Area = b×h

Ast = Area of long. steel

fc = Concrete compressive strength fy = Steel yield strength

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loads

Maximum Nominal Capacity for Design Pn (max) 2.

0maxn rPP r = Reduction factor to account for:

Small eccentricity, Non-alignment, bending,…

r = 0.80 ( tied )

r = 0.85 ( spiral ) ACI 10.3.6.3

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loads

Reinforcement Requirements (Longitudinal Steel Ast)

3.

g

stg A

A

- ACI Code 10.9.1 requires

Let

08.001.0 g

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1% ≤ ρg ≤ 8%


loads

3.

- Minimum No. of Bars ACI Code 10.9.2

- min. of 6 bars in circular columns

with min. spiral reinforcement.

- min. of 4 bars in rectangular columns

- min. of 3 bars in triangular ties

Reinforcement Requirements (Longitudinal Steel Ast)

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loads

3.

ACI Code 7.10.5.1

Reinforcement Requirements (Lateral Ties)

Φ10 mm bar if longitudinal bar Φ32 mm bar Φ 12mm bar if longitudinal bar Φ36 mm bar Φ 16mm bar if longitudinal bars are bundled

size

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loads

3. Reinforcement Requirements (Lateral Ties) Vertical spacing: (ACI 7.10.5.2)

16 db ( db for longitudinal bars ) 48 dt ( dt for tie bar ) least lateral dimension of column

s s s

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loads

3. Reinforcement Requirements (Lateral Ties)

Arrangement Vertical spacing: (ACI 7.10.5.3)

At least every other longitudinal bar shall have lateral support from the corner of a tie with an included angle 135o.

No longitudinal bar shall be more than 15 cm. clear on either side from “support” bar.

1.)

2.)

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loads

Examples of lateral ties.

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loads

ACI Code 7.10.4

Reinforcement Requirements (Spirals )

10 mm diameter size

clear spacing between spirals 75 mm. ACI 7.10.4.325mm.

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( Pitch )


loads

Reinforcement Requirements (Spiral)

sDA

c

sps

4Core of VolumeSpiral of Volume

Spiral Reinforcement Ratio, s

sp cs 2

c

( ) ( )from:

( /4) A DD s

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loadsReinforcement Requirements (Spiral)

g cs

ch y

0.45 1A fA f

ACI Eqn. 10-5

where

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loads4. Design for Concentric Axial Loads

(a) Load Combination

Gravity:

Gravity + Wind:and

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Pu = 1.4 PDL

Check for

Tension

Pu = 1.2 PDL + 1.6 PLL

Pu = 1.2 PDL + 1.0 PLL + 1.6 PWL

Pu = 0.9 PDL ± 1.3 PWL


loads4. Design for Concentric Axial Loads

(b) General Strength Requirement un PP f

f = 0.65 for tied columns

f = 0.75 for spiral columns

where,

(c) Expression for Design

08.00.01 Code ACI gg

stg

AA

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loads

or

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Tied Cols

Spiral Cols

ACI Code Equation 10-1 (spiral) ACI Code Equation 10-2 (tied)

Example 1: Design Tied Column for Concentric Axial Load

Design tied column for concentric axial load

PDL = 670 kN; PLL = 1340 kN; Pw = 220 kN

fc = 30 MPa, fy = 414 MPa

Design a square column aim for g ≤ 0.03.

Select longitudinal transverse reinforcement.

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Determine the loading

Check the compression or tension in the column

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Pu = 1.4 PDL = 1.4 (670) = 938 kN

Pu = 1.2 PDL + 1.6 PLL Pu = 1.2 (670) + 1.6 (1340) =2948 kN

Pu = 1.2 PDL + 1.0 PLL + 1.6 PWLPu = 1.2 (670) + 1.0 (1340) + 1.6 (220) = 2496 kN

Design Axial load Pu = 2948 kNPu = 0.9 PDL + 1.3 PWL = 0.9 (670) ± 1.3 (220) = 889 kN

Example 1: Design Tied Column for Concentric Axial LoadFor a square column r = 0.80 and f = 0.65 and ≤ 0.03

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Ag = 152583 mm2

Ag = h2 , h =390.6 mm

Use h = 400 mm, Ag = 160000 mm2

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Then, calculate the corresponding area of steel Ast:

Ast = 4090.68 mm2 , use 4 Φ 28 mm + 4 Φ 25 mmAst ,prov. = 4426 mm2


Use Φ 10 mm ties compute the spacing

< 150 mm. No cross-ties needed

Stirrup design

Use Φ10 mm stirrups with 400 mm spacing in the column 39

S ≤ 16 db = 16(28) = 448 mm ≤ 48 dstirrups = 48(10) = 480 mm ≤ Smaller b or h = 400 mm

Interaction Diagrams

A, I = area and moment of inertia of the cross section, respectively

Y = distance from the centroidal axis to the most highly comp. surface (surface A–A), (+) to the right

P = axial load, (+) in compression

M = moment, (+) as shown in Fig.

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Columns subjected to eccentric loads

Interaction diagram for an elastic column, |ƒcu| = |ƒtu|41

Inte

ract

ion

diag

ram

s for

ela

stic

colu

mns

, |ƒ cu

| not

equ

al to

|ƒtu|.

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Interaction Diagram for Reinforced Concrete Columns

Calculation of Pn and Mn for a given strain distribution.43

Strain distributions corresponding to

points on the interaction diagram

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Significant Points on the Column Interaction Diagram

1. Point A- Pure Axial Load2. Point B - Zero Tension, Onset of Cracking.3. Region A–C - Compression-Controlled

Failures.4. Point C - Balanced Failure, Compression-

Controlled Limit Strain.5. Point D - Tensile-Controlled Limit.6. Region C–D - Transition Region.

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Columns subjected to eccentric loads

Strength Reduction Factors

Computation Method for Interaction Diagrams

The general case involves the calculation of Pn acting at the centroid and Mn acting about the centroid of the gross cross section, for an assumed strain distribution with ecu = 0.003

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The strain distribution will be defined by setting ecu = 0.003 and assuming a value for es1.

An iterative calculation will be necessary to consider a series of cases.

The iteration can be controlled by selecting a series of values for the neutral axis depth, c.

Large values of c will give points high in the interaction diagram and low values of c will give points low in the interaction diagram.

To find points corresponding to specific values of strain in the extreme layer of tension reinforcement, the iteration can be controlled by setting es1 = Zey, where Z is an arbitrarily chosen value.

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By similar triangles

For elastic–plastic reinforcement with the stress–strain curve

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The stresses in the concrete are represented by the equivalent rectangular stress block

If a is greater than di for a particular layer of steel

If a is less than di


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=0.85 – 0.05

(negative in tension)

The nominal axial load capacity, Pn for the assumed strain distribution is the summation of the axial forces:

A positive internal moment corresponds to a compression at the top face

If the gross cross section is not symmetrical, the moments would be computed about the centroid of the gross section, and the factored moment resistance would be


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Columns in Pure Tension

Section is completely cracked

(no concrete axial capacity)

Uniform Strain y e

N

y snt tension ii 1

P f A

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Behavior under Combined Bending and Axial Loads

Interaction Diagram Between Axial Load and Moment (Failure Envelope)

Concrete crushes before steel yields

Steel yields before concrete crushes

Any combination of P and M outside the envelope will cause failure.

Note:54

EX. 2- Calculation of an Interaction Diagram

Compute four points on the interaction diagram for the column shown below:

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Use = 35 MPa, = 414 MPa

=4Φ28mm=2463 mm2, =4Φ28mm=2463 mm2

==4826 mm2

1. Compute the concentric axial-load capacity and maximum axial-load capacity.

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=0.85 +

=

= 4616426.5 + 1997964 = 6614.39 kN

3439.48 kN

2. Compute fPn and fMn for the general case.

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2. Compute fPn and fMn for the general case.

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Interaction diagram

Ex. 2Fig. 1

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2. Compute f and fMn for balanced failure.

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Interaction Diagrams for Circular Columns

The compression zone is a segment of a circle having depth a. To compute the compressive force and its moment about the centroid of the column, it is necessary to be able to compute the area and centroid of the segment

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Nondimensional interaction diagram for rect. tied col. with bars in two faces 73

0.145 x MPa

0.14

5 x

MPa

Unsymmetrical Columns Simplified Interaction Diagrams for Columns74

Choice of Column TypeIn seismic areas or in other situations where ductility is important, spiralcolumns are used more extensively

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Design for Combined Bending and Axial Load

(Short Column)Column Types

Spiral Column - more efficient for e/h < 0.1, but forming and spiral expensive

Tied Column - Bars in four faces used when e/h < 0.2 and for biaxial bending

1)

2)

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Estimating the Column SizeTied columns

Spiral Columns

Although the ACI Code does not specify a minimumcolumn size, the min dimension of a cast-in-place tied column should not be less than 200mm. and preferably not less than 250mm. The diameter of a spiral column should not be less than 300mm.

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Bar-Spacing Requirements

Arrangement of bars at lap splices in columns78

Reinforcement Splices

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Types of lap splices required if all bars are lap spliced at every floor.

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EX. 2- Design of a Tied Column for a Given Pu and Mu

1. Select trial size, and trial reinforcement ratio.

Choose a400mm square column 81

Pu =2000 kN, Mu = 160 kN.mUnsupported length of 3.2 mMaterials: fy = 414 MPa, f’c = 27.6 MPa.

𝐴𝑔(𝑡𝑟𝑖𝑎𝑙)≥𝑃𝑢

0 .40 ( 𝑓 ′𝑐❑+ 𝑓 𝑦𝜌𝑔)

≥147885 .24𝑚𝑚2

Consider a column with bars in two faces. Use a tied column with bars in two faces.

Neglect the Slenderness Effect

Summary for the trial column. A 400 mm×400mm. tied column with bars in two faces.

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= = 0.08 m = 80 mm

= 0.2

2. Compute g

3. Use interaction diagrams to determine g

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=0.75

=

=0.145 X

=

=0.145 X

From Fig. A-6a (Interaction diagram for

0.015, 0.2

84

MPa

0.14

5 x

If the value of g computed here exceeds 0.03 to 0.04, a larger section should be chosen. If is less than 0.01, either use 0.01 (the min. allowed by ACI Section 10.9.1) or recompute, using a smaller cross section.4. Select the reinforcement.

Possible combination is (from the Table )

Try a 400 mm. square column with six 25mm bars. 85

Ast = Ag

= 0.015 X 400 X 400 = 2400 mm2

Use Six bars, 25 mm, Ast = 2945 mm2, three in each side

5. Check the maximum load capacity Pu

6. Design the lap splices.

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The tensile face for Pu/bh = 1.81 and Mu/bh2 = 0.36

=

Ld =48.9 x 25 =1.222m 1.3 Ld =1.58 m

7. Select the ties.

The ties must satisfy both ACI Code Chapter 11 and ACI Code Sec. 7.10.5

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10 mm diameter stirrups

16 longitudinal bar diameters = 16 X 25= 400 mm48 ties diameters = 48 X 10 = 480 mmLeast column dimensions = 400 mm

= = =

= 1.669 MPa , Vc = 1.669 x bd= 226.98 kN

Computation of γ and arrangement of ties 88

Design of a Circular Spiral Column for a Large Axial Load and Small Moment

1. Select the material properties, trial size, and trial reinf. ratio.

2. Compute g

3. Use interaction diagrams to determine g

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Pu = 7100 kN and Mu = 200 kN.m

f’c = 28 Mpa fy = 414 Mpa, try ρg = 0.04

Try 650mm Dia.

=0.81

==0.145 X

==0.145 X

Looking at Figs. A-12b and A-12c, due to the relatively small moment, we are on the flat part of the diagrams. For both g values read g = 0.024

4. Select the reinforcement.

5. Check the maximum load capacity.

6. Select the spiral.

The minimum-size spiral is 10mm 90

Ast = Ag

= 0.024 X = 7964 mm2 , Try 10 Φ 32 mm

ΦPn = 0.85 x 0.75 [0.85 x 28(-8042)+ 8042 x 414]= 7135.16 kN, therefore, o.k.

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MPa

MPa

From the interaction diagrams, fs is compression. Therefore, use compression lap splices (ACI Code Section 12.16.1). From the equation, these must be 960 mm. long. Because the bars are enclosed in a spiral, this can be multiplied by 0.75, giving a splice length of 720 mm. (ACI 12.17.2.5)

7. Design the lap splices.

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= 650 – 2x 37.5 =575 mmAg =gross Area=

Ach =Area of concrete core= =259672 mm2

= 65 mm

Use Φ10mm at S= 65 mm

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Engineering

chapter 5 reinforced concrete columns