CHOU Soklin, Ph.D.
7 3 2018
Reinforced Concrete Design
Lecturer:
CHOU Soklin, Ph.D.
Depar arttment nt::
Civ ivil il Engin inee eerrin ing g
Academic Year:
2017/2018
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5. AXIALLY AXIALLY LOADED COLUMNS 5.1 Introduction •
Col Columns umns are are membe emberrs used used pri primaril arily y to suppo upport rt axia axiall compressive loads.
•
Perfect vertical alignment of columns in a multistory building is not possible, causing loads to be eccentric relative to the center of columns.
•
The eccentric loads will will cause moments in columns. Therefore, a column subjected to pure axial loads does not exist in concrete buildings.
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CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY AXIALLY LOADED COLUMNS 5.1 Introduction •
However, it can be assumed that axially loaded columns are those with relatively small eccentricity, e eccentricity, e,, of about 0.1 h 0.1 h or or less, where h where h is is the total depth of the column and e and e is is the eccentric distance from the center of the column. CL
P e h
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5. AXIALLY AXIALLY LOADED COLUMNS 5.2 Types of Columns
Col Colum umns ns may may be clas classi sifi fied ed based based on the the foll follow owin ing g dif differe ferent nt categories:
1. Based on loading, columns may be classified as follows: a. Axially a. Axially loaded columns, columns, where loads are assumed acting at the center of the column section. b. Eccentrically b. Eccentrically loaded columns, columns, where loads are acting at a distance e from the center of the column column section. section. The distance distance e e could could be along the x or y axis, causing moments either about the x or y axis. 4
CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY AXIALLY LOADED COLUMNS 5.2 Types of Columns
c. Biaxially c. Biaxially loaded columns, columns, where the load is applied at any point on the column section, causing moments about both the x and y axes simultaneously. simultaneously. 2. Based on length, columns may be classified as follows: a. Short a. Short columns, columns, where the column’s failure is due to the crushing of concrete or the yielding of the steel bars under the full load capacity of the column. b. Long b. Long columns, columns, where buckling effect and slenderness ratio must be taken into consideration in the design. 5
5. AXIALLY AXIALLY LOADED COLUMNS 5.2 Types of Columns 3. Based on column ties, columns may be classified as follows: a.
Tied column columnss containing steel ties to confine the main longitudinal bars in the columns. Ties are normally spaced uniformly along the height of the column.
b.
Spir Spiral al col colum umns ns containing containing spirals spirals (spring-t (spring-type ype reinforc reinforcemen ement) t) to hold the main longitudinal reinforcement and to help increase the column ductility before failure.
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CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY AXIALLY LOADED COLUMNS 5.2 Types of Columns
4. Based on the shape of the cross section: Square Square column column,, rectan rectangul gular ar column column,, round round column column,, L-shape L-shaped d column, octagonal column, or any desired shape with an adequate side width or dimensions. 5. Based on frame bracing: In braced frames, columns resist mainly gravity loads (non-sway column), and shear walls resist lateral loads and wind loads. In unbraced frames, columns resist both gravity and lateral loads (sway column), which reduces the load capacity of the columns. 7
5. AXIALLY AXIALLY LOADED COLUMNS 5.2 Types of Columns
6. Based on materials: Column Columnss may be reinfo reinforce rced, d, prestr prestress essed, ed, compos composite ite (conta (containi ining ng rolled steel sections such as I-sections), or a combination of rolled steel sections and reinforcing bars.
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CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY AXIALLY LOADED COLUMNS 5.3 Behavior of of Axially Loaded Loaded Columns
When the load on the column is increased to reach its design strength, the concrete will reach the maximum strength and the steel will reach its yield strength, f strength, f y. The nominal load capacity of the column can be written as follows:
Po 0.85 f c' ( Ag Ast ) Ast f y where A g :grosssection of concrete concretearea area A st :longitudinalsteelarea 9
5. AXIALLY AXIALLY LOADED COLUMNS 5.3 Behavior of of Axially Loaded Loaded Columns •
Fo For
a
tied
column,
the
conc concre rete te fail failss by crus crushi hing ng and and shea sheari ring ng outw outwar ard, d, the the longitudinal steel bars fail by buckling outward between ties, and the column failure occurs suddenly. •
A spiral spiral column undergoes a marked yielding, yielding, followed by considerable considerable deformation before complete failure. A hoop tension develops in the spiral, and for a closely spaced spiral the steel may yield . A sudden failure is not expected.
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CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY LOADED COLUMNS 5.4 ACI Code Limitations
1. For axially as well as eccentrically loaded column, the strength reduction factored:
0.65 for tied columns 0.75 for spirally reinforced columns 2. Longitudinal steel percentage: 1% s
A st bh
8%
3. At least 4 bars for rectangular column and 6 bars for circular column. Spacing between the longitudinal bars must be less than 150 mm and the clear cover is at least 40 mm.
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5. AXIALLY LOADED COLUMNS 5.4 ACI Code Limitations
4. Minimum ratio of spiral reinforcement:
A g
f ' 1 c Ach f yt
s 0.45
A g : gross area of section Ach :area of core of spirally reinforced column measured to the outside diameter of spiral f yt : yield strength of spiral reinforcement 12
CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY LOADED COLUMNS 5.4 ACI Code Limitations
5. Minimum diameter of spiral bar is 10 mm and their clear spacing should be between 25 mm and 75 mm. 6. Minimum diameter of ties is 10 mm to enclose longitudinal bars of smaller than No. 32 bar. Minimum diameter of ties is 13 mm to enclose longitudinal bars of larger than No. 32 bar.
7. Center to center spacing of ties: s st min 48 Dstirrup ,16 Dbar , b Where D stirrup : diameter of stirrup Dbar : diameter of longitudinal bar b : smaller size of column
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5. AXIALLY LOADED COLUMNS 5.5 Design Equations
Because a perfect axially loaded column does not exist, some eccentricity occurs on the column section, thus reducing its load capacity, P o , should be multiplied by a factor equal to 0.8 for tied column and 0.85 for spirally reinforced column. Thus, the axial load strength of columns: For tied column: Pu Pn (0.80) 0.85 fc' ( Ag Ast ) Ast f y For spiral column: Pu Pn (0.85) 0.85 fc' ( Ag Ast ) Ast f y 14
CHOU Soklin, Ph.D.
7 3 2018
5. AXIALLY LOADED COLUMNS 5.5 Design Equations
These equations can be written as: Pn K 0.85 fc' Ag Ast ( f y 0.85 f c' ) tocalculateaxialloadstrengthof column Pn KAg 0.85 fc' g ( f y 0.85 f c' ) to design secion of column where
0.80:for tied column
K
0.85:for spiral column 0.65 for tied columns
0.75 for spirally reinforced columns 15
5. AXIALLY LOADED COLUMNS 5.5 Design Equations
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.1 Introduction •
Vertical members such as column of a building frame are subjected to combined axial loads and bending moments. These forces develop due to external loads, such as dead, live, and wind loads.
•
From Figure below, columns AB and CD are subjected to an axial compressive force and a bending moment.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.1 Introduction •
The ratio of the moment to the axial force is usually defined as the eccentricity e, where e = M n /P n. The eccentricity e represents the distance from the plastic centroid of the section to the point of application of the load
e
P n
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.1 Introduction Ex. 6.1. Determine the plastic centroid of the section shown below. Given:
f ′c = 28 MPa and f y = 420 MPa. 65mm
A
350mm
270mm A s 2
A s1
3No.32
65mm
2No.32
x 400mm
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.2 Design Assumptions for Columns The design limitations for columns, according to the ACI Code are as follows: 1. Strains in concrete and steel are proportional to the distance from the neutral axis. 2. Equilibrium of forces and strain compatibility must be satisfied. 3. The maximum usable compressive strain in concrete is 0.003. 4. Strength of concrete in tension can be neglected. 5. The stress in the steel is f s s Es f y 6. The concrete stress block may be taken as a rectangular shape with concrete stress of 0.85 f ′ c
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram When a normal force is applied on a short reinforced concrete column, the following cases may arise, according to the location of the normal force with respect to the plastic centroid: 1. Axial Compression ( P0). A large axial load is acting at the plastic centroid; e = 0 and M n = 0. Failure of the column occurs by crushing of the concrete and yielding of steel bars. 2. Maximum Nominal Axial Load P n,max . A normal force acting on the section with minimum eccentricity, emin. P n,max = 0.80 P 0 for tied columns and 0.85 P 0 for spirally reinforced columns. In this case, failure occurs by crushing of the concrete and the yielding of steel bars.
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram
3. Compression Failure . A large axial load acting at a small eccentricity. The range of this case varies from a maximum value of P n = P n,max to a minimum value of P n=P b (balanced load). Failure occurs by crushing of the concrete on the compression side with a strain of εcu = 0.003 and f s < f y. In this case P n > P b and e < eb . 4. Balanced Condition ( P b). A balanced condition is reached when the strain of concrete reaches 0.003 and the strain in the tensile reinforcement reaches ε y = f y/ E s simultaneously. Failure of concrete occurs at the same time as the steel yields.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram
5. Tension Failure . This is the case of a small axial load with large eccentricity, that is, a large moment. Before failure, tension occurs in a large portion of the section, causing the tension steel bars to yield before actual crushing of the concrete. At failure, the strain in the tension steel is greater than the yield strain, ε s > ε y, whereas the strain in the concrete reaches 0.003. The range of this case extends from the balanced to the case of pure flexure. When tension controls, P n < P b and e > eb .
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram
6. Pure Flexure . The section in this case is subjected to a bending moment, M n, whereas the axial load is P n =0. Failure occurs as in a beam subjected to bending moment only. Thisimagecannotcurrently bedi splayed.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.3. Load-Moment Interaction Diagram
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.4. Strength Reduction Factor
The strength reduction factor, Ø, to be used for columns may vary according to the following cases:
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.4. Strength Reduction Factor
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.5. Balanced Condition: Rectangular Sections
A balanced condition occurs in a column section when a load is
•
applied on the section and produces, at nominal strength, a strain of 0.003 in the compressive fibers of concrete and a strain ε s = ε y = f y /E s in the tension steel bars simultaneously. When the applied eccentric load is greater than P b, compression
•
controls; if it is smaller than P b, tension controls in the section.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.5. Balanced Condition: Rectangular Sections
The analysis of a balanced column section can be explained in steps:
1. Calculate the distance from the extreme compressive fibers to d cb 0.003d 0.003 cb the neutral axis cb: y s cb 0.003 y 30
CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.5. Balanced Condition: Rectangular Sections
The analysis of a balanced column section can be explained in steps: 2. Calculate the balanced compressive force P b: From equilibrium, Pb Cc Cs T 0 where Cc 0.85 f c'abb ; Cs As' f s' 0.85 f c' ; T As f y f s' s' Es
c d ' c
0.003E s
Pb 0.85 f c'abb As' f s' 0.85 f c' As f y
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.5. Balanced Condition: Rectangular Sections
The analysis of a balanced column section can be explained in steps: 3. Calculate the eccentricity,eb , from the plastic centroid: Taking moment about the plastic centroid:
The balanced eccentricity is eb
M b P b
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.5. Balanced Condition: Rectangular Sections Ex. 6.2. Determine the balanced compressive force P b; then determine
eb and M b for the section shown in Figure below. Given: f ′c = 28 MPa and f y = 420 MPa.
65mm 210mm 485mm 550mm d " 210mm
4No.29bars
275mm
4No.29bars
65mm 350mm
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure •
When a column is subjected to an eccentric force with large eccentricity e, tension failure is expected. The column section fails due to the yielding of steel and crushing of concrete when the strain in the steel exceeds ε y.
•
In this case the nominal strength, P n, will be less than P b or the eccentricity, e = M n /P n, is greater than the balanced eccentricity, eb.
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure •
Because it is difficult in some cases to predict if tension or compression controls, it can be assumed a tension failure will occur
when e > d . This assumption should be checked later.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure
8. Check whether the tension steel is yielding: s y
10. Compute the strength of column: Pn Pu and M n M u 38
CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.6. Strength of Columns for Tension Failure Ex. 6.3. Determine the nominal compressive strength, P n, for the section
given in Figure below, if e = 500 mm. P n
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure •
If the compressive applied force, P n, exceeds the balanced force, P b, or the eccentricity, e = M n /P n is less than eb, compression failure is expected.
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure •
Because it is difficult to predict compression or tension failure whenever a section is given, compression failure can be assumed when e < 2d /3, which should be checked later. 1. Calculate the distance to the neutral axis for a balanced section: cb
0.003d 0.003 y
2. Calculate P n from equilibrium equation: Pn Cc Cs T where Cc 0.85 fc' ab; C s As' ( f s' 0.85 f c' ); T As f s Since, thecompressionsteelis yielding, s' y f s' f y
Pn 0.85 fc' ab As' ( f y 0.85 f c' ) As f s
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure
3. Calculate P n by taking moments about the tension steel, A s
Pn e ' Cc d Pn
a
h
Cs ( d d ') where e ' e d " and As As' d " d 2 2
1
' 0.85 f c ab d
e'
As' ( f y 0.85 f c' )( d d ') 2
a
d c d a / 1 0.003 0.003 From the strain diagram, s c a / 1 1d a 0.003E s f s s Es a
0.85 fc'b 3 ' 2 2 a 0.85 fc b(e ' d ) a A s' ( f y 0.85 f c' )(e ' d d ') 0.003E s Ase ' a 0.003 E s Ase ' 1d 0
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure
Aa3 Ba 2 Ca D 0 Where
A 0.425 fc'b B 0.85 fc'b(e ' d ) C A s' ( f y 0.85 f c' )(e ' d d ') 0.003 Es As e ' D 0.003E s As e ' 1d
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure
4. Check that the tension steel is not yield: s
d c c
0.003 y
5. Check whether the compression steel is yielding: s'
c d ' c
0.003 y
If s' y , repeat steps 2 through 4 6. Check the conditions for compression controlled section: e eb and Pn Pb 45
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure
7. Compute the strength of the column: Pn and M n
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.7. Strength of Columns for Compression Failure Ex. 6.4. Determine the nominal compressive strength, P n, for the section
given in Figure below, if e = 250 mm. P n
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6. COLUMNS UNDER AXIAL LOAD AND BENDING Ex. 6.5. Draw the interaction diagram for the column section below: P n
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.8. Rectangular Columns With Side Bars •
In some column sections, the steel reinforcement bars are distributed around the four sides of the column section.
•
The side bars are those placed on the sides along the depth of the section in addition to the tension and compression steel, A s and A′ s, and can be denoted by A ss. ' s
h Sixside bars (three on each)
Twoside bars (one on each)
s
b
b
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.8. Rectangular Columns With Side Bars •
In this case, the same procedure explained earlier can be applied, taking into consideration the strain variation along the depth of the section and the relative force in each side bar either in the compression or tension zone of the section. These are added to those of C c, C s and T to determine P n in which cu s1
' c
C s1 C c C s 2
s 2
c
s 3
h
s 4
b
0.85
s
C s3 T 1 T 2
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.8. Rectangular Columns With Side Bars Pn Cc Cs T
P n
Pn 0.85 f c' ab Cs T d '
Takingmoment about PC
M n Pn e 0.85 f c' ab d
d " 2
d "
C s (d d i' d ") T s (d
d i'
PC
h d
a
e
d ") As f y d "
A s'
e'
A s b
Where C s As' ( f s' 0.85 f c' ) and Ts As f s
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.8. Rectangular Columns With Side Bars Ex. 6.6. Determine the balanced compressive force P b; then determine
eb and M b for the section shown in Figure below. Given: f ′c = 28 MPa P b
and f y = 420 MPa.
65mm 110mm 110mm 570mm 110mm 110mm 65mm 570mm
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.9. Load Capacity of Circular Columns •
The values of load capacity P b and the nominal moment M n for circular sections can be determined using the equations of equilibrium, as was done in the case of rectangular sections.
•
The main problem is to find the depth of the compressive block a and the stresses in the reinforcing bars.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.9. Load Capacity of Circular Columns
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.9. Load Capacity of Circular Columns
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.9. Load Capacity of Circular Columns Ex. 6.7. Determine the balanced load Pb and balanced moment Mb of
the circular column as shown in Figure below. The circular column with diameter of 410 mm is reinforced by 8 No. 32 bares. Use f c’ = 28 MPa and f y = 420 Mpa.
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.10. Design of Columns for Tension Failure •
Tension failure occurs when P n < P b or the eccentricity e > eb. In the design of columns, P u and M u are given, and it is required to determine the column size and its reinforcement.
•
It may be assumed (as a guide) that tension controls when: M u P 530mm for h 610mm u M u 610mm for h 610mm P u
•
In this case, a section may be assumed, and then A s and A′ s are determined. The ACI charts may be used to determine ρ g for a given section with A s = A′ s .
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.10. Design of Columns for Tension Failure
1. Assume the dimension of the section b and h, then calculate d, d’ and e = M u /P u such that e ≥ 530 mm for h < 610 mm or e ≥ 610 mm for h ≥ 610 mm or e > d 2. Assume A s = A’ s , P u = ØP n and f’ s= f y
3. Check reinforcement ratio: 1% g
A s As ' bh
8% 58
CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.10. Design of Columns for Tension Failure 4. Check the strength of column: ØP n ≥ P u , ØM n ≥ M u
5. Calculate the stirrup requirement
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.10. Design of Columns for Tension Failure Ex.6.8. Determine the necessary reinforcement for a 400×560-mm.
rectangular tied column to support a factored load P u = 1000 kN and a factored moment M u = 550×103 kN.m. Use f ′ c = 28 MPa and f y = 420 MPa.
560mm
400mm 60
CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.11. Design of Columns for Compression Failure
Opposite to the conditions of the tension failure: 1. Assume the dimension of the section b and h, then calculate d, d’ and e = M u /P u . 2. Assume A s = A’ s , Ø = 0.65 and f s = f y
e 0.5 ' Pu d d ' bhf c ' A s 3he f y 1.18 d 2 3. Check reinforcement ratio: 1% g
A s As ' bh
8%
4. Check the strength of column: ØP n ≥ P u , ØM n ≥ M u
5. Calculate the stirrup requirement 61
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.11. Design of Columns for Compression Failure Ex.6.9. Determine the necessary reinforcement for a 400×560-mm.
rectangular tied column to support a factored load P u = 1000 kN and a factored moment M u = 300 kN.m. Use f ′ c = 28 MPa and f y = 410 MPa.
560mm
400mm 62
CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12. Biaxial Bending
If the load P n is acting anywhere such that its distance from the x-axis is e y and its distance from the y-axis is e x, then column will be subjected to a combination of an axial load P n, a moment about the xaxis = M nx= P ne y and a moment about the y-axis = M ny = P ne x. The column section in this case is said to be subjected to biaxial bending.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12. Biaxial Bending •
The neutral axis is at an angle with respect to both axes
•
It’s lengthy to determine the location of the neutral axis, strains, concrete compression area, and internal forces and their point of application.
•
The strength of columns under axial load and biaxial bending is estimated by several methods.
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CHOU Soklin, Ph.D.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.1. Square and Rectangular Columns under Biaxial Bending
Bresler Reciprocal Method
•
Square or rectangular columns with unequal bending moments about their major axes will require a different amount of reinforcement in each direction. An approximate method of analysis of such sections was developed
•
by Boris Bresler and is called the Bresler reciprocal method.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.1. Square and Rectangular Columns under Biaxial Bending
Bresler Reciprocal Method
The load capacity of the column under biaxial bending can be determined by using the following expression:
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.1. Square and Rectangular Columns under Biaxial Bending
Bresler Reciprocal Method
For P n < 0.10 P n0, the axial force may be neglected and the section can be
designed as a member subjected to pure biaxial bending:
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.1. Square and Rectangular Columns under Biaxial Bending
Bresler Load Contour Method
The failure surface shown in the Figure of Biaxial Interaction Surface is cut at a constant value of P n, giving the related values of M nx and M ny . 1.5
M For rectangular column: nx M ox
1.75
For square column:
M nx M ox
1.5
M ny M oy M ny M oy
1
1.75
1
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.1. Square and Rectangular Columns under Biaxial Bending Ex. 6.10. The section of a short tied column is 410×610 mm and is
reinforced with eight No. 32 bars. Determine the design load on the section ØP n if it acts at e x = 400 mm and e y = 300 mm. Use f ′c = 28 MPa, f y = 420 MPa, and use the Bresler reciprocal equation. A
y
65mm 240mm
B P n
8No.32bars
x
610mm 240mm 65mm
C
D y 410mm
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.2. Circular Columns under Biaxial Bending
Circular columns with reinforcement distributed uniformly about the perimeter of the section have almost the same moment capacity in all directions. If a circular column is subjected to biaxial bending about the x and y axes, the equivalent uniaxial M u moment can be calculated using the following equations:
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.12.2. Circular Columns under Biaxial Bending Ex. 6.11. Determine the load capacity P n of a 510 mm-diameter column
reinforced with 10 No. 10 bars when e x = 100 mm and e y = 150 mm. Use f ′ c = 28 MPa and f y = 420 MPa.
510mm
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.13. Parme Load Contour Method
A point B on the load contour (of a horizontal plane at a constant P n shown in Figure below) is defined such that the biaxial moment capacities M nx and M ny are in the same ratio as the uniaxial moment capacities M 0x and M 0y; that is,
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.13. Parme Load Contour Method
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.13. Parme Load Contour Method Ex. 6.12. The section of a short tied column is 410×610 mm and is
reinforced with eight No. 32 bars. Determine the design load on the section ØP n if it acts at e x = 400 mm and e y = 300 mm. Use f ′c = 28 MPa, f y = 420 MPa, and use the Parme Contour Approach (PCA). A
y
65mm 240mm
B P n
8No.32bars
x
610mm 240mm 65mm
C
D y 410mm
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.14. Equation of Failure Surface
The biaxial bending strength of an axially loaded column can be represented by a three-dimensional interaction curve.
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6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.14. Equation of Failure Surface
A general equation for the analysis and design of reinforced concrete short and tied rectangular columns is supposed to represent the failure surface and interaction diagrams of columns subjected to combined biaxial bending and axial load.
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CHOU Soklin, Ph.D.
7 3 2018
6. COLUMNS UNDER AXIAL LOAD AND BENDING 6.14. Equation of Failure Surface Ex. 6.13. The section of a short tied column is 410×610 mm and is
reinforced with eight No. 32 bars. Determine the design load on the section ØP n if it acts at e x = 400 mm and e y = 300 mm. Use f ′ c = 28 MPa, f y = 420 MPa, and use the Equation of Failure Surface. A
y
65mm 240mm
B P n
8No.32bars
x
610mm 240mm 65mm
C
D y 410mm
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7. SLENDER COLUMNS 7.1 Introduction •
In the analysis and design of short columns discussed in the previous two chapters, it was assumed that buckling, elastic shortening, and secondary moment due to lateral deflection
had minimal effect on the ultimate strength of the column; thus, these factors were not included in the design procedure. •
However, when the column is long, these factors must be considered. The extra length will cause a reduction in the column strength that varies with the column effective height, width of the section, the slenderness ratio, and the column end conditions.
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7 3 2018
7. SLENDER COLUMNS 7.1 Introduction •
A column with a high slenderness ratio will have a considerable reduction in strength, whereas a low slenderness ratio means that the column is relatively short and the reduction in strength may not be significant.
•
The slenderness ratio is the ratio of the column height, l , to the radius of gyration, r , where r
I / A.
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7. SLENDER COLUMNS 7.1 Introduction
The moment of inertia of the section can be calculated as follow: For rectanglular section:
I x y 2 dA
y dxdy 2
b
b
1
h
2 2h 2b y 2 dxdy x 2 b y 3 2 3 h2 2 2 1 h3 h3 bh 3 b 3 8 8 12 2 I y x dA x 2 dxdy h
b
1 2 2h 2b x 2 dxdy y 2 h x3 2 3 b2 2 2 1 b3 b3 hb 3 h 3 8 8 12 h
b
h
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CHOU Soklin, Ph.D.
7 3 2018
7. SLENDER COLUMNS 7.1 Introduction
The moment of inertia of the section can be calculated as follow: For circular section:
I 0 y 2 dA
r sin , dA rdrd
We have:
I 0
0
R
0
R
R 4 4
rd
r
r 3 sin 2 drd
1 r 4 4 0
y
dA d dr
2
sin(2 ) 2 4 0 D 4 64
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7. SLENDER COLUMNS 7.1 Introduction
The radius of gyration of the section can be calculated as follow: For rectanglular section: bh3 r x
I x A
2 12 h 0.288h bh 12
hb3 r y
I y A
2 12 b 0.288b bh 12
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CHOU Soklin, Ph.D.
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7. SLENDER COLUMNS 7.1 Introduction
The radius of gyration of the section can be calculated as follow: For rectanglular section: D 4 r x ry
I 0 A
64 D 2
D2 16
0.25 D
4
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7. SLENDER COLUMNS 7.1 Introduction In general, columns may be considered as follows:
1. Long with a relatively high slenderness ratio, where lateral bracing or shear walls are required. (Nonsway column ) 2. Long with a medium slenderness ratio that causes a reduction in the column strength. Lateral bracing may not be required, but strength reduction must be considered. (Sway column) 3. Short where the slenderness ratio is relatively small, causing a slight reduction in strength. This reduction may be neglected, as discussed in previous chapters. (Short column) 84
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7 3 2018
7. SLENDER COLUMNS 7.1 Introduction •
High lateral deflections may cause secondary moments (P-Δ effect).
•
If the secondary moment, M ′ , is added to the applied moment on the column, M a, the final moment is M = M a + M ′.
•
An approximate method for estimating the final moment M is to multiply the applied moment M a by a factor called the magnifying moment factor δ, in which δ ≥ 1 and M = δ M a . 85
7. SLENDER COLUMNS 7.2 Effective Length Factor, K •
The slenderness ratio l/r can be calculated accurately when the effective length of the column ( Kl u /r ) is used.
•
The effective length factor, K, represents the ratio of the distance between points of zero moment ( Inflection Point, IP ) in the column and the unsupported height of the column in one direction.
•
Example: For a column with hinge at both ends, K = l u /l u = 1. For a column is fixed at both ends, K = 0.5l u /l u = 0.5.
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CHOU Soklin, Ph.D.
7 3 2018
7. SLENDER COLUMNS 7.2 Effective Length Factor, K
87
7. SLENDER COLUMNS 7.2 Effective Length Factor, K
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CHOU Soklin, Ph.D.
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7. SLENDER COLUMNS 7.2 Effective Length Factor, K •
However, in the concrete frame, neither a perfect frictionless hinge nor perfectly fixed ends can exist. Thus, the alignment chart is necessary to estimate the value K.
•
It is necessary first to calculate the end restraint factors ψA and ψB at the top and bottom of the column, respectively, where:
•
Then plot ψA and ψB on the alignment chart and connect the two points to intersect the middle line, which indicates the K value. 89
7. SLENDER COLUMNS 7.2 Effective Length Factor, K •
l c is length center to center of joints in a frame and l is the span length of flexure, center to center of joints.
•
The restraint factor at one end shall include all columns and beams meeting at the joint.
•
For a hinged end, ψ is infinite and may be assumed to be 10.0.
•
For a fixed end, ψ is zero and may be assumed to be 1.0.
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7. SLENDER COLUMNS 7.2 Effective Length Factor, K
91
7. SLENDER COLUMNS 7.2 Effective Length Factor, K
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CHOU Soklin, Ph.D.
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7. SLENDER COLUMNS 7.2 Effective Length Factor, K
Sway frames
Nonsway frames
93
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames •
The first step in determining the design moments in along column is to determine whether the frame is braced or unbraced against sidesway.
•
If lateral bracing elements, such as shear walls and shear trusses, are provided or the columns have substantial lateral stiffness, then the lateral deflections produced are relatively small and their effect on the column strength is substantially low.
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7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames •
It can be assumed (ACI Code, Section 6.6.4.4.1) that a story within a structure is nonsway if Q
P u 0
0.05
Vus l c
Where
P u : The story total factored vertical load V us : The horizontal story shear in the story
0 : The first-order relative lateral deflection between the top and bottom of the story due to V us
l c : The length of the column, measured from center to center of the joints in the frame 95
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
Moments in Nonsway Frames
The slenderness effect must be considered when: kl u r
34 12
1
M2
where 34 12
M 1 M 2
40
where M 1 and M 2 are the factored end moments of the column and M 2 is greater than M 1. The ratio M 1 /M 2 is considered positive if the member is bent in single curvature and negative for double curvature. If e emin 15 0.03h M 2 M 2,min Pu 15 0.03h 96
CHOU Soklin, Ph.D.
7 3 2018
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
Moments in Nonsway Frames
97
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
Moments in Nonsway Frames
The procedure for determining the magnification factor δns in nonsway frames can be summarized as follows: 1. Determine if the frame is braced against sidesway and find the unsupported length, l u, and the effective length factor, K ( K may be assumed to be 1) 2. Calculate the member stiffness, EI
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7 3 2018
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
EI
Moments in Nonsway Frames
0.2 Ec I g Es I se 1 dns
Where: Ec 4700 f c' (MPa) E s 2 105 MPa I g = gross moment of inertia of the section about the axis considered I se = moment of inertia of the reinforcing steel dns
1.2 P D 1.2 P D 1.6P L 99
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
Moments in Nonsway Frames
3. Determine the Euler buckling load, P c: P c
2 EI ( kl u ) 2
4. Calculate the value of the factor C m to be used in the equation of the moment-magnifier factor. where M 1 /M 2 is positive if the column is bent in single curvature and negative if the member is bent in double curvature. Cm =1 or
2
M 1
if M 2 M 2,min Pu 15 0.03h 100
CHOU Soklin, Ph.D.
7 3 2018
7. SLENDER COLUMNS 7.3 Columns in Nonsway Frames Magnified
Moments in Nonsway Frames
5. Calculate the moment magnifier factor δns
6. Design the compression member using the axial factored load, P u and a magnified moment, M c: Where
2
c
ns M 2
M 2,min Pu 15 0.03h
In nonsway frames, the lateral deflection is expected to be less than or equal to H /1500,where H is the total height of the frame. 101
7. SLENDER COLUMNS 7.4 Columns in Sway Frames •
A sway (unbraced) frame is one that depends on moments in the columns to resist lateral loads and lateral deflections.
•
The story shears from the columns above and below a given floor gives rise to a sway force acting on that floor. M top Pe Vl P M t ,ns s M t , s bottom
Pe Vl P M b,ns s M b, s
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7. SLENDER COLUMNS 7.4 Columns in Sway Frames
103
7. SLENDER COLUMNS 7.4 Columns in Sway Frames •
It can be assumed (ACI Code, Section 6.6.4.4.1) that a story within a structure is sway if Q
P u 0 Vus l c
Where
0.05
P u : The story total factored vertical load V us : The horizontal story shear in the story
0 : The first-order relative lateral deflection between the top and bottom of the story due to V us
l c : The length of the column, measured from center to center of the joints in the frame 104
CHOU Soklin, Ph.D.
7 3 2018
7. SLENDER COLUMNS 7.4 Columns in Sway Frames Magnified
Moments in Sway Frames
The slenderness effect must be considered when: kl u 22 r The procedure for determining the magnification factor δ s in sway frames can be summarized as follows: 1. Determine if the frame is unbraced against sidesway and find the unsupported length, l u, and the effective length factor, K taken from the alignment chart. 2. Calculate the member stiffness, EI
0.2 Ec I g Es I se 1 ds 105
7. SLENDER COLUMNS 7.4 Columns in Sway Frames Magnified
Moments in Sway Frames
Where: Ec 4700 f c' (MPa) E s 2 105 MPa I g = gross moment of inertia of the section about the axis considered I se = moment of inertia of the reinforcing steel ds
V sustained V factored
3. Determine the Euler buckling load: P c
2 EI ( kl u ) 2 106
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7. SLENDER COLUMNS 7.4 Columns in Sway Frames Magnified
Moments in Sway Frames
4. Calculate the moment magnifier factor δ s a. Magnifier method:
b. Approximate second-order analysis: 107
7. SLENDER COLUMNS 7.4 Columns in Sway Frames Magnified
Moments in Sway Frames
5. Calculate the magnified end moments M 1 and M 2 at the ends of an individual column, as follows:
where M 1ns and M 2ns: the moments obtained from the no-sway condition M 1s and M 2s: the moments obtained from the sway condition. The design magnified moment: M c = max{ M 1, M 2}. 108
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7. SLENDER COLUMNS 7.5 Section Design •
A simple approximate formula for determining the initial size of the column A g where the total steel ratio ρ g is assume to 1%:
A g (trial) •
P u 0.4 f c' g f y
The compression and tension steel areas can be estimated as follow: For tension-controlled section, A s As'
Pu (e 0.5h 0.2d ) f y (d d ')
e 0.5 Pu d d ' bhf c' ' For compression-controlled section, A s As 3he f y 1.18 d 2 109
7. SLENDER COLUMNS Ex. 7.1. The column section shown in Figure below carries an axial load P D
=605 kN and a moment M D =157 kN⋅m due to dead load and an axial load P L =489 kN and a moment M L = 126 kN ⋅m due to live load. The column is part of a frame that is braced against sidesway and bent in single curvature about its major axis. The unsupported length of the column is l u =5.8 m, and the moments at both ends of the column are equal. Check the adequacy of the column using f ′ c = 28 MPa and f y = 420 MPa. 65mm 4No. 29 bars 560mm 4No. 29 bars 65mm 360mm
110