chapter 5 reinforced concrete columns
TRANSCRIPT
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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Analysis and Design of “Short” Columns
-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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Analysis and Design of “Short” Columns
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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Analysis and Design of “Short” Columns
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Analysis and Design of “Short” Columns
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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Analysis and Design of “Short” Columns
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.
Analysis and Design of “Short” Columns
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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Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
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%
Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
loads
Examples of lateral ties.
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Behavior, Nominal Capacity and Design under Concentric Axial
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 )
Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
loadsReinforcement Requirements (Spiral)
g cs
ch y
0.45 1A fA f
ACI Eqn. 10-5
where
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Behavior, Nominal Capacity and Design under Concentric Axial
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
Behavior, Nominal Capacity and Design under Concentric Axial
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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Behavior, Nominal Capacity and Design under Concentric Axial
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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Example 1: Design Tied Column for Concentric Axial Load
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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Example 1: Design Tied Column for Concentric Axial Load
Then, calculate the corresponding area of steel Ast:
Ast = 4090.68 mm2 , use 4 Φ 28 mm + 4 Φ 25 mmAst ,prov. = 4426 mm2
Example 1: Design Tied Column for Concentric Axial Load
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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Computation Method for Interaction Diagrams
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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Computation Method for Interaction Diagrams
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
Computation Method for Interaction Diagrams
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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
Computation Method for Interaction Diagrams
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Computation Method for Interaction Diagrams
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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
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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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