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1
Fifth EditionReinforced Concrete Design
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
CHAPTER
9c
REINFORCED CONCRETEA Fundamental Approach - Fifth Edition
COMBINED COMPRESSION AND BENDING: COLUMNS
ENCE 454 – Design of Concrete StructuresDepartment of Civil and Environmental Engineering
University of Maryland, College Park
SPRING 2004By
Dr . Ibrahim. Assakkaf
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 1ENCE 454 ©Assakkaf
Strength of Eccentrically Loaded Columns: Axial Load and Bending
Behavior of Eccentrically Loaded Non-Slender Columns– Stress distribution and Whitney’s Block for
beams can applied in this case.– Figure 6 shows a typical rectangular column
cross-section with strain, stress, and force distribution diagrams.
– Notice the additional nominal force Pn at the limit failure state acting at an eccentricity efrom the plastic centriod of the section.
2
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 2ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)
Strength of Eccentrically Loaded Columns: Axial Load and Bending
2h
d
d ′
y
h
b
PlasticCentroid
Figure 6a. Stresses and Forces in Columns
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 3ENCE 454 ©Assakkaf
Strength of Eccentrically Loaded Columns: Axial Load and Bending
c
003.0=cε
sε
cdc
ccd
s
s
′−=′
−=
003.0
003.0
:Strains
ε
ε
cf ′85.0
cCsC
a
sT
cCsC
sT
dd ′−
nP
ysss
ysss
fEf
fEf
≤′=′
≤=
ε
ε:Stresses
sss
sss
cc
fATfAC
bafC
=
′′=
′= 85.0:forces Internal
e
e′
GeometricCentroid
N.A.
Figure 6b. Stresses and Forces in Columns
3
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 4ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– Definitions of Terms for Figure 6
Strength of Eccentrically Loaded Columns: Axial Load and Bending
steeln compressio ofcover effectivesteel tension toload ofty eccentrici
centroid geometric toload ofty eccentricicentroid geometric of distance
axis neutral todistance
=′=′===
deeyc
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 5ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– The depth of neutral primarily determines the
strength of the column.– The equilibrium equations for forces and
moments (see Figure 6) can be expressed as follows for non-slender (short) columns:
Strength of Eccentrically Loaded Columns: Axial Load and Bending
( ) sscn TCCP −+=failureat (19)
4
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 6ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– Nominal resisting moment Mn, which is equal
to Pn e, can be obtained by writing the moment equilibrium equation about the plastic centroid.
– For columns with symmetrical reinforcement, the plastic centroid is the same as the geometric centroid.
Strength of Eccentrically Loaded Columns: Axial Load and Bending
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 7ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)
– since
Strength of Eccentrically Loaded Columns: Axial Load and Bending
( ) ( )ydTdyCayCePM sscnn −+′−+
−==
2(20)
sss
sss
cc
fATfAC
bafC
=
′′=
′= 85.0(21)
5
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 8ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– Eqs. 19 and 20 can be rewritten as
– where for rectangular section = h/2
Strength of Eccentrically Loaded Columns: Axial Load and Bending
(22)( ) ( )ydfAdyfAaybafePM
fAfAbafP
sssscnn
sssscn
−+′−′′+
−′==
+′′+′=
285.0
85.0
(23)
y
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 9ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– In Eqs. 22, the depth of the neutral axis c is
assumed to be less than the effective depth dof the section, and the steel at the tension face is actual tension.
– Such a condition changes if the eccentricity eof the axial force Pn is very small.
Strength of Eccentrically Loaded Columns: Axial Load and Bending
6
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 10ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– For such small eccentricities, where the total
cross-section is in compression, contribution of the tension steel should be added to the contribution of concrete and compression steel.
– The term Asfs in Eqs. 22 and 23 in such a case would have a reverse sign since all the steel is in compression.
Strength of Eccentrically Loaded Columns: Axial Load and Bending
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 11ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– It is also assumed that ba – As ≈ ba; that is, the
volume of concrete displaced by compression steel is negligible.
– Symmetrical reinforcement is usually used such that = As in order to prevent the possible interchange of the compression reinforcement with the tension reinforcement during bar cage placement.
Strength of Eccentrically Loaded Columns: Axial Load and Bending
sA′
7
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 12ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– If the compression steel is assumed to have
yielded and As = , Eqs 22 and 23 can be rewritten as.
Strength of Eccentrically Loaded Columns: Axial Load and Bending
sA′
( ) ( )
( )ddfAadbafePM
ydfAdyfAaybafePM
bafP
yscnn
ysyscnn
cn
′−′+
−′=′=
−+′−′+
−′==
′=
285.0
285.0
85.0 (24)
(25)
(26)or
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 13ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– Additionally, Eqs. 24 and 26 can be combined
to obtain a single equation for Pn as
– Also, from Eq. 22
Strength of Eccentrically Loaded Columns: Axial Load and Bending
( )ddfAadPePM ysnnn ′−′+
−=′=
2 (27)
bfPfAfA
cac
nssys
′+′′−
==85.0
β1 (28)
8
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 14ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– When the magnitude of or fs is less than fy,
the actual stresses can be calculated using the following equations:
Strength of Eccentrically Loaded Columns: Axial Load and Bending
(29a)( )
yssss
yssss
fcdEEf
fc
dcEEf
≤
′−=′=′
≤′−
=′=′
1003.0
or
003.0
ε
ε
(29b)
sf ′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 15ENCE 454 ©Assakkaf
Behavior of Eccentrically Loaded Non-Slender Columns (cont’d)– For fs:
Strength of Eccentrically Loaded Columns: Axial Load and Bending
(30a)( )
yt
ssss
yt
ssss
fcdEEf
fc
cdEEf
≤
−==
≤−
==
1003.0
or
003.0
ε
ε
(30b)
9
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 16ENCE 454 ©Assakkaf
Eqs. 22 and 23 determine the nominal axial load Pn that can be safely applied at an eccentricity e for any eccentrically loaded column. Te following unknowns can be identified:
1. Depth of the equivalent stress block, a.2. Stress in compression steel, 3. Stress in tension steel, fs
4. .Pn for the given e, or vice versa
Trial and Adjustment Procedure for Analysis (Design) of Columns
sf ′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 17ENCE 454 ©Assakkaf
The stresses and fs can be expressed in terms of the neutral axis c as in Eqs. 29 and 30 and thus in terms of a.The two remaining unknowns, a and Pn, can be solved using Eqs. 22 and 23.However, combining Eqs. 22 and 23 to 28 leads to a cubical equation in terms of the neutral axis depth c. Also, check must be done if and fs are less than fy.
Trial and Adjustment Procedure for Analysis (Design) of Columns
sf ′
sf ′
10
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 18ENCE 454 ©Assakkaf
The Suggested Procedure:1. For a given section geometry and
eccentricity e, assume a value for the distance c down to the neutral axis. This value is a measure of the compression block depth a since a = β1c.
2. Using the assumed value of c, calculate the axial load Pn using Eq. 22 and a = β1c.
3. Calculate and fs, respectively, using Eqs. 29 and 10.
Trial and Adjustment Procedure for Analysis (Design) of Columns
sf ′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 19ENCE 454 ©Assakkaf
The Suggested Procedure:4. Calculate the eccentricity corresponding to
the calculated load Pn in Step 3 using Eq. 23. The calculated eccentricity should match the given eccentricity e. If not, repeat the steps until a convergence is accomplished.
5. If the calculated eccentricity is larger than the given eccentricity, this indicates that the assumed c and corresponding a are less than the actual depth.
Trial and Adjustment Procedure for Analysis (Design) of Columns
11
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 20ENCE 454 ©Assakkaf
The Suggested Procedure:6. In such a case, try another cycle, assuming
a larger value of c.
Trial and Adjustment Procedure for Analysis (Design) of Columns
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 21ENCE 454 ©Assakkaf
Strain Limits Methods
Strain Limits Zones– The strain limits for compression-controlled
sections can be represented by the following strain distributions across the depth of the cross sections with εt = 0.002 for Grade 60 Steel, or generally εt = fy/Es.
– Figure 8 illustrates the behavior limits presented in Figure 7.
12
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 22ENCE 454 ©Assakkaf
Strain Limits MethodsStrain Limits Zones
003.0ε =c 003.0ε =c 003.0ε =c
c c csε′ sε′ sε′
002.0ε ≤t 005.0ε002.0 ≤≤ t 005.0ε ≥t
Strain conditionfor compression-
controlled sections(Gr 60 steel)
Strain conditionfor intermediate
behavior
Strain conditionfor tension-
Controlled sections
Figure 7. Stresses and Forces in Columns
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 23ENCE 454 ©Assakkaf
Strain Limits MethodsStrain Limits Zones
Figure 8. Strain Limit Zones and variation of Strength Reduction Factor φ
13
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 24ENCE 454 ©Assakkaf
Strain Limits MethodsStress Limits
1) Tension-controlled limit case (εt > 0.005)
375.0005.0003.0
003.0εε
ε=
+=
+=
tc
c
tdc
tdca 1β 375.0==
′−=
′−=′
ts d
dcd 67.21003.01003.0ε
yt
sss fddEf ≤
′−=′=′ 67.21000,87ε
(31)
(32)
(34)
(33)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 25ENCE 454 ©Assakkaf
Strain Limits MethodsStress Limits
1) Compression-controlled limit case (εt = 0.002)
60.0002.0003.0
003.0εε
ε=
+=
+=
tc
c
tdc
tdca 1β 60.0==
′−=′
ts d
d60.0
1003.0ε
yt
sss fddEf ≤
′−=′=′ 67.11000,87ε
(35)
(36)
(38)
(37)
14
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 26ENCE 454 ©Assakkaf
Strain Limits Methods
Stress Limits– Transition zone for limit strain with
intermediate behavior• This characterizes compression members in which
the tensile reinforcement As has yielded but the compressive reinforcement has a stress level less than fy depending on the geometry of the section.
• Intermediate φ values change linearly with εt from φ= 0.90 when εt > 0.005 to φ = 0.65 for tied columns, or φ = 0.70 for spiral columns when εt ≤ 0.002.
sA′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 27ENCE 454 ©Assakkaf
Strain Limits MethodsStress Limits– Transition zone for limit strain with
intermediate behavior (cont’d)• It should be noted that for nonprestressed flexural
members and for nonprestressed members with axial load less than 0.10 Ag, the net tensile strain εt should not be less than 0.004. Hence, in the transition zone of Fig. 8, the minimum strain value in flexural members for determining the φ value is 0.004.
• This limit is necessitated, as a φ value can otherwise become so low that additional reinforcement would be needed to give the required nominal moment strength.
cf ′
15
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 28ENCE 454 ©Assakkaf
Summary: Modes of Failure in Columns– Based on the magnitude of strain in the
tension face reinforcement (Figure 6), the section is subjected to one of the following:1. Tension-controlled state, by initial yielding of the
reinforcement at the tension side, and strain εtgreater than 0.005.
2. Transition state, denoted by initial yielding of the reinforcement at the tension side, but strain with strain εt value smaller than 0.005 but greater than 0.002.
3. Compression-controlled case by initial crushing of the concrete at the compression face.
Strain Limits Methods
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 29ENCE 454 ©Assakkaf
Strain Limits Methods
Summary: Modes of Failure in Columns (cont’d)– Accordingly, in analysis and design, the
following eccentricity limits correspond to the strain limits presented:
et ≥ limit e0.005 (c = 0.375 dt): tension-controlled
ett ≤ limit e0.005-0.002 (c = 0.375 dt – 0.60 dt): intermediate transition
ec ≤ limit e0.002 (c ≥ 0.60 dt): compression-controlled
16
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 30ENCE 454 ©Assakkaf
General Load-Moment Relationship
Axial Load-Moment Combination– Assume that Pu is applied to a cross
section at an eccentricity e from the centroid, as shown in Figs. 9a and 9b.
– Add equal and opposite forces Pu at the centroid of the cross section, as shown in Fig. 9c.
– The original eccentric force Pu may now be combined with the upward force Pu to form a couple Pue, that is a pure moment.
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 31ENCE 454 ©Assakkaf
General Load-Moment RelationshipAxial Load-Moment Combination
e
Pu
e euP
uPuP ePu
= =
Figure 9
(a)
(b) (c) (d)
17
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 32ENCE 454 ©Assakkaf
General Load-Moment Relationship
Axial Load-Moment Combination– This will leave remaining one force, Pu
acting downward at the centroid of the cross section.
– It can be therefore be seen that if a force Pu is applied with an eccentricity e, the situation that results is identical to the case where an axial load of Pu at the centroidand a moment of Pue are simultaneously applied as shown in Fig. 9d.
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 33ENCE 454 ©Assakkaf
General Load-Moment Relationship
Axial Load-Moment Combination– If Mu is defined as the factored moment to
be applied on a compression member along with a factored axial load of Pu at the centroid, the relationship between the two can expressed as
u
u
PMe = (39)
18
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 34ENCE 454 ©Assakkaf
General Load-Moment RelationshipEccentric Axial Loading in A Plane of Symmetry
C
y
fxC
y
fxC
y
fx= +
( ) ( )bendingcentric xxx fff +=
Figure 10
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 35ENCE 454 ©Assakkaf
General Load-Moment Relationship
PP A
CB
ED
Eccentric Axial Loading in A Plane of Symmetry
The stress due to eccentric loading on a beam cross section is given by
IMy
AP
x ±=σ (40)
19
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 36ENCE 454 ©Assakkaf
General Load-Moment RelationshipEquivalent Force System for Eccentric Loading
P = 4.8 kN
120 mm 80 m
m
35 mm
yP =4.8 kN
( )( ) mkN 12035608.4
mkN 192408.4⋅=−=
⋅==
z
x
MM
Mx = 192 kN·m
Mz = 120 kN·my
x
z
Figure 11
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 37ENCE 454 ©Assakkaf
General Load-Moment Relationship
Analysis of Short Columns: Large Eccentricity– The first step in the investigation of short
columns carrying loads at eccentricity is to determine the strength of given column cross section that carries load at various eccentricities.
– For this, the design axial load strength φPnis found, where Pn is defined as the nominal axial load strength at a given eccentricity.
20
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 38ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7Find the design axial load strength φPn for the tied column for the following conditions: (a) small eccentricity, (b) pure moment, (c) e = 5 in., and (d) the balanced condition. The column cross section is shown. Assume a short column. Bending about the Y-Y axis. Use = 4000 psi and fy = 60,000 psi.
cf ′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 39ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
Y
Y
XX 41 ′′
41 ′′
02 ′′bars 9#6−
21
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 40ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)(a) Small Eccentricity:
( )[ ]( ) ( )( ) ( )( )[ ]kips 723
6606280485.070.080.0
85.080.0 (max)
=+−=
+−′=
=
stystgc
nn
AfAAf
PP
φ
φφ
( )bars) 9#6 of (areain 6
in 28020142
2
−=
==
st
g
A
A
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 41ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)(b) Pure Moment:
The analysis of the pure moment condition is similar to the analysis of the case where the eccentricity e is infinite as shown in Figure 12.The design moment φMn will be found since Pu and φPn will both be zero.Assume that As is at yield, and then with reference to Figure 13, then
22
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 42ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
e = ∞
PuFigure 12
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 43ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)C1 = concrete compressive forceC2 = steel compressive forceT = steel tensile force
Since
cc
cc ss 3003.0003.03
−=′⇒=
−′
εε (41)
sss Ef ε ′=′ (42)
23
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 44ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
nMφ71 ′′
3 ′′sA′
sAc85.0
2C1C
3 ′′ 003.0cf ′85.0
T
2Z1Z
yεmoreor
c
sε ′
Figure 13
(a) (b) (c) (d)Strain Stress and Force
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 45ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)Substituting Es = 29 × 106 psi and given by Eq. 41 into Eq. 42, gives
For equilibrium in Figure 13d,
Substituting into above equation, yields
( )c
cc
cfs3873003.01029 6 −
=−
×=′
sε ′
(43)
TCC =+ 21 (44)
24
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 46ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
– The above equation can be solved for c to give
( )( )
( )( )( ) ( ) ( )( ) ( )6033485.0338785.0485.0
85.085.085.0
=−−
+
=′′−′′+
ccc
AfAfAfbcf syscssc(45a)
(45b)
on)(compressi ksi 90.1462.3
362.387
thus,andin. 62.3
=−
=′
=
sf
c
(46)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 47ENCE 454 ©Assakkaf
General Load-Moment Relationship
#3 #4 $5 #6 #7 #8 #9 #10 #111 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.562 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.123 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.684 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.245 0.55 1.00 1.55 2.20 3.00 3.95 5.00 6.35 7.806 0.66 1.20 1.86 2.64 3.60 4.74 6.00 7.62 9.367 0.77 1.40 2.17 3.08 4.20 5.53 7.00 8.89 10.928 0.88 1.60 2.48 3.52 4.80 6.32 8.00 10.16 12.489 0.99 1.80 2.79 3.96 5.40 7.11 9.00 11.43 14.04
10 1.10 2.00 3.10 4.40 6.00 7.90 10.00 12.70 15.60
Number of bars
Bar numberTable 5. Areas of Multiple of Reinforcing Bars (in2)
25
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 48ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)Therefore, the forces will be
– The internal Moments are
( ) ( )( )( )( ) kips 5.1461462.385.0485.085.085.01 ==′= bcfC c
( ) ( )( ) kips 5.343485.039.1485.02 =−=′′−′′= scss AfAfC
( ) kips-ft 8.1882
62.385.01712
5.146111 =
−== ZCM n
( ) kips-ft 3.4012
145.34222 === ZCM n
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 49ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
nMφ71 ′′
3 ′′sA′
sAc85.0
2C1C
3 ′′ 003.0cf ′85.0
T
2Z1Z
yεmoreor
c
sε ′
Figure 13
(a) (b) (c) (d)Strain Stress and Force
26
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 50ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)Therefore,
andkips-ft 2293.408.18821 =+=+= nnn MMM
( ) kips-ft 14922965.0 ==nMφ
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 51ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)(c) The eccentricity e = 5 in:
The situation of e = 5 in. is shown in Figure 14Note that in Part (a), all steel was in compression and in Part (b), the steel on the side of the column away from the load was in tension. Therefore, there is some value of the eccentricity at which steel will change from tension to compression. Since this is not known, the strain in Figure 15 is assumed.
27
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 52ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
PuFigure 14
5 ′′=e Y
Y
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 53ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
nP71 ′′=d
sA′
sAc85.0
2C1C
3 ′′ 003.0cf ′85.0
T
2Z1Z
sε
c
sε ′
Figure 15
(a) (b) (c) (d)
Assumed Strain Stress and Force
5 ′′21 ′′
Comp./tens.?
28
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 54ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)The assumptions at ultimate load are1. Maximum concrete strain = 0.0032. > εy, therefore, =fy
3. εs is tensile4. εs < εy and thus fs < fy
These assumptions will be verified later.The unknown quantities are Pu and c.The forces will be evaluated as follows:
sε ′ sf ′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 55ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)( )( )( )
( ) ( )( )
ccc
TCCP
cc
cc
Ac
cdAEAfT
AfAfCccabfC
n
ssssss
scsy
c
−−+=
−+=
=
−=
−
=
−
===
=−=
′′−′=−=′=
∑
172618.16940.46
:7c Fig.in 0 moments From
1726131787
87
kips 8.1693485.0360
85.046.401485.0485.085.0
21
2
1
ε
003.0
sε
c
sε ′
d
ccd
ccdEf
ccd
ccd
sss
s
s
−=×
−
==
−=
−=
871029003.0
and ,003.0
003.0
3ε
ε
ε
(47)
29
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 56ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)From ∑ moments = 0, taking moments about T in Figure 15d:
Eqs. 47 and 48 can be solved simultaneously for c to give
( ) ( )
( )
+
−=
+
−=
14`8.169285.01746.40
121
142
12 21
cc
CadCPn (48)
kips 733in. 86.14
==
nPc
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 57ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)Now, the assumptions can be checked:
Therefore, = fy, and based on the location of the neutral axis:
( )
( ) OK 0024.000207.01029
000,60
0024.0003.086.14
386.14
6 =′<=×
==
=
−
=′
ss
yy
s
Ef
εε
ε
sf ′
OK ksi 60ksi 53.1286.14
86.141787 <=
−
=sf
30
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 58ENCE 454 ©Assakkaf
Example 7 (cont’d)– The design moment for an eccentricity of 5
in. can be computed as follows:
– Therefore, the given column has a design load-moment combination strength of 477 kips axial load and 199 ft-kips moment.
( )( ) kips-ft 199
125477
kips 47773365.0
===
===
ePM
PP
nn
nu
φφ
φ
General Load-Moment Relationship
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 59ENCE 454 ©Assakkaf
Example 7 (cont’d)(d) The Balanced Condition Case:
The balanced condition is defined when the concrete reaches a strain of 0.003 at the same time that the tension steel reaches its yield strain, as shown in Fig. 16c.The value of cb can be calculated from
( ) in. 06.10176087
8787
87=
+=
+= d
fc
yb
General Load-Moment Relationship
31
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 60ENCE 454 ©Assakkaf
Example 7 (cont’d)
bP
71 ′′=d
sA′
sAc85.0
2C1C
3 ′′ 003.0cf ′85.0
Tyε
bc
sε ′
Figure 16
(a) (b) (c) (d)Strain Stress and Force
be
0.002
General Load-Moment Relationship
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 61ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)
Therefore, = fy = 60 ksiThe forces can computed as follows:
( ) 002.00021.0003.006.10
306.10=>=
−=′ ys εε
sf ′
( )( )( )( )( ) ( )( )( )
kips 397180170407kips 180360
kips 1703485.0360Ckips 4071406.1085.0485.0
21
2
1
=−+=−+===
=−===
TCCPT
C
b
32
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 62ENCE 454 ©Assakkaf
General Load-Moment RelationshipExample 7 (cont’d)– The value of eb may be calculated by
summing moments about T as follows:
– From which, eb = 12.0 in. Therefore, at the balanced condition:
( ) ( )
( ) ( ) ( )141702
206.1085.0174077397
142
85.07 21
+
−=+
+
−=+
b
bbe
e
CcdCeP
( )( ) kips-ft 258
1212258
kips 25839765.0
===
==
bbn
b
ePM
P
φφ
φ
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 63ENCE 454 ©Assakkaf
General Load-Moment RelationshipExample 7 (cont’d)– The results of the four parts can be tabulated
(see Table 6) and plotted as shown in Fig. 17.– This plot is called an “interaction diagram”.– In the plot, any point on the solid line
represents an allowable combination of load and moment.
– Any point within the solid line represents a load-moment combination that is also allowable, but for which this column is overdesigned.
33
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 64ENCE 454 ©Assakkaf
General Load-Moment RelationshipExample 2 (cont’d)
27827812 in
2145135 in.
1600Infinite
0 (small)723Small
Moment strength(φPne, ft- kips)
Axial load strength(φPn,kips)
e
Table 6
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 65ENCE 454 ©Assakkaf
General Load-Moment Relationship
0
100
200
300
400
500
600
700
800
0 50 100 150 200 250 300Mo me nt
kips)-(ft Moment ePnφ
(kips)nPφ
e=
0
e = 5 in.
e = eb =12 in.
Compression Failure
Balanced condition
Tensile Failure
Member ofStrength Bending
Figure 17. Column Interaction Diagram
e = ∝
34
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 66ENCE 454 ©Assakkaf
General Load-Moment Relationship
Example 7 (cont’d)– Any point outside the solid line represents an
unaccepted load-moment combination or a load-moment combination for which this column is underdesigned.
– Radial lines from the origin represent various eccentricities (slope = φPn/φPn or 1/e).
– Any eccentricity less than eb will result in compression controlling the column, and any eccentricity greater than eb will result in tension controlling the column.
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 67ENCE 454 ©Assakkaf
Design of Short Columns: Large Eccentricity
The design of a column cross section using the previous calculation approach would be a trial-and-error method and would become exceedingly tedious.Therefore, design and analysis aids have been developed that shorten the process to a great extent.A chart approach has been developed in ACI Publication SP-17 (97), ACI Design Handbook.
35
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 68ENCE 454 ©Assakkaf
Design of Short Columns: Large Eccentricity
The charts take on the general form of Figure 17 but are set up to be more general so that they will remain applicable if various code criteria undergo changes.These charts can be used for both analysis and design of columns.There are also computer programs available to aid in the design process.
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 69ENCE 454 ©Assakkaf
General Case of Columns Reinforced on All Faces: Exact Solution
In cases where columns are reinforced with bars on all faces and those where the reinforcement in the parallel faces is nosymmetrical, solutions have to be based on using first principles of Eqs. 22 and 23.These equations have to be adjusted for this purpose and strain compatibility checks for strain in each reinforcing bar layer have to be performed at all load levels.
36
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 70ENCE 454 ©Assakkaf
Figure 18 shows the case of a column reinforced on all four faces.Assume that
Gsc = center of gravity of steel compressing force.Gst = center of gravity of steel tensile force.Fsc = resultant steel compressive force = ∑ fsc
– Fst = resultant steel tensile force = ∑As fst
General Case of Columns Reinforced on All Faces: Exact Solution
sA′
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 71ENCE 454 ©Assakkaf
General Case of Columns Reinforced on All Faces: Exact Solution
Figure 18. Column Reinforced with Steel on all Faces: (a) Cross-Section; (b) Strain; (c) Forces
37
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 72ENCE 454 ©Assakkaf
Equilibrium of the internal and external forces and moments requires that
General Case of Columns Reinforced on All Faces: Exact Solution
srsccn FFcbfP −+′= 1β85.0
ststscsccn yFyFchcbfeP ++
−′= 11 β
21
2β85.0
(49)
(50)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 73ENCE 454 ©Assakkaf
The strain values in each bar layer are determined by the linear strain distribution in Figure 18 to ensure strain compatibility.The stress in each reinforcing bar is obtained using the expression
General Case of Columns Reinforced on All Faces: Exact Solution
cS
cSEEf ii
cssissi 000,87=== εε (51)
ysi ff ≤
38
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 74ENCE 454 ©Assakkaf
Circular ColumnsThe angle θ subtended by the compressive block chord shown in Figure 19b isCase 1:
Case 2:
−
=
<≤
−
2/2/cosθ
90 θ ,2
1
o
hah
ha
(52a)
−
=
−
=
>>
−−
2/2/cos and
2/2/cosθ
90 θ ,2
11
o
hhaφ
hah
ha
(52b)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 75ENCE 454 ©Assakkaf
Figure 19. Circular Columns: (a) Strain, Stress and CompressionBlock Segment; (b) Compression Segment Chord x-x Geometry
39
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 76ENCE 454 ©Assakkaf
The area of the compressive segment of the circular column in Figure 19b is
The moment of area of the compressive segment about the center of the column is
−
=4
θcosθsinθ2 radc hA (53a)
Circular Columns
=
12θsin3
3hyAc (53b)
= distance of the centroid of compressive block to section centroidy
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 77ENCE 454 ©Assakkaf
Circular Columns
( )
( )
N.A. below zonension within tebarsin stress where
1000,87
zone ecompressiv within barsin stress where
1000,87
/2 γwhere
θcos2γ
2 bar
=
≤
−=
=′
≤
−=′
′−=
−=
si
yi
si
si
yi
si
ii
f
fcdf
f
fcdf
hdh
hhd (54a)
(54b)
(54c)
40
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 78ENCE 454 ©Assakkaf
Circular Columns
Expressions for nominal axial force Pn and nominal bending moment Mn for circular columns
∑
∑
′−+′=
+′=
isisiccn
sisiccn
dhAfyAfM
AfAfP
285.0
85.0
Note: moment is taken about the circular column center.
(55a)
(55c)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 79ENCE 454 ©Assakkaf
Whitney’s Approximate SolutionRectangular Concrete Columns
– Assumptions:1. Reinforcement is symmetrically placed in single
layers parallel to axis of bending in rectangular sections.
2. Compression steel has yielded.3. Concrete displaced by the compression steel is
negligible compared to the total concrete area in compression.
4. The depth of the stress block is assumed to be 0.54d.
5. The interaction curve in the compression zone is a straight line.
41
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 80ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Rectangular Concrete Columns (cont’d)– If compression controls, the equation for
rectangular sections can be written as
18.135.0 2 +
′+
+′−
′=
dhe
fbh
dde
fAP cys
n (56)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 81ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Circular Concrete Columns– Transform the circular column to an
idealized equivalent rectangular column as shown in Figure 20.
– For compression failure, the equivalent rectangular column would have1. the thickness in the direction of bending equal to
0.8h, where h is the outside diameter of the circular column (Figure 20b).
2. the width of the idealized rectangular column to be obtained from the same gross area Ag of the circular column such that b = Ag/0.8h.
42
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 82ENCE 454 ©Assakkaf
Circular Concrete Columns (cont’d)
Figure 20. Equivalent column section(a) given circular section (Ast, totalreinforcement area; (b) equivalentrectangular section (compression failure);(c) equivalent column (tension failure)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 83ENCE 454 ©Assakkaf
Whitney’s Approximate SolutionCircular Concrete Columns (cont’d)
3. the total area of reinforcement Ast to be equally divided in two parallel layers and placed at a distance of 2Ds/3 in the direction of bending, where Ds is the diameter of the cage measured center to center of the outer vertical bars.
– For tension failure, use the actual column for evaluating Cc, but place 40% of the total steel Ast in parallel at a distance 0.75Ds as shown In Figure 20.
– Once the dimensions are established, the analysis can be similar to rectangular column.
43
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 84ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Circular Concrete Columns (cont’d)– For Tension Failure:
– For Compression Failure:
−−+
−′= 38.085.0
5.238.085.085.0
22
he
hmD
hehfP sg
cn
ρ
( ) 18.167.08.06.90.13
2 ++
′+
+=
s
cg
s
ystn
DhhefA
De
fAP
(57)
(58)
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 85ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Circular Concrete Columns (cont’d)– Definitions of Variables for Eqs. 57 and 58
h = diameter of sectionDs = diameter of the reinforcement cage center to center of the outer
vertical barse = eccentricity to plastic centroid of section
c
y
g
stg
ff
m
AA
′=
==
85.0
area concrete grossarea steel gross ρ
44
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 86ENCE 454 ©Assakkaf
Whitney’s Approximate SolutionExample 8
Obtain an equivalent rectangular cross section for the circular column shown in figure 20a. Assume that Ds = 15 in.
( )( )
( )
2
in. 103152
32
in. 63.19208.0
1420
8.0section rect. ofwidth
in. 16200.8 section rect. of thickness2
stss
s
g
AAA
Ddd
hA
=′=
===′−
===
=×=
π
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 87ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Example 9A concrete circular column 20 in. in diameter is reinforced with six No. 8 equally spaced bars as show in the figure. Using Whitney’s approximate approach, compute the nominal axial load Pn for (a) eccentricity e = 16.0 in. and (b) eccentricity e = 5.0 in.
6 #8
psi 000,60 psi 4000
:Given
==′
y
c
ff
2.5 in.
20 in.
45
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 88ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Example 9 (cont’d)a) e = 16.0 in. with axial load 6 in. outside the
circular section. It can be assumed that the section is tension-controlled.
( )
( )015.0
42074.4
4
74.4No.8 6 of areain. 155.2220
22g =====
=−=
ππρ
hAAA
D
gg
st
s
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 89ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
#3 #4 $5 #6 #7 #8 #9 #10 #111 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.562 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.123 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.684 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.245 0.55 1.00 1.55 2.20 3.00 3.95 5.00 6.35 7.806 0.66 1.20 1.86 2.64 3.60 4.74 6.00 7.62 9.367 0.77 1.40 2.17 3.08 4.20 5.53 7.00 8.89 10.928 0.88 1.60 2.48 3.52 4.80 6.32 8.00 10.16 12.489 0.99 1.80 2.79 3.96 5.40 7.11 9.00 11.43 14.04
10 1.10 2.00 3.10 4.40 6.00 7.90 10.00 12.70 15.60
Number of bars
Bar numberTable 5. Areas of Multiple of Reinforcing Bars (in2)
46
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 90ENCE 454 ©Assakkaf
Whitney’s Approximate Solution
Example 9 (cont’d)Using Eq. 57, yields
( ) 65.17400085.0000,60
85.0==
′=
c
y
ff
m
( )( ) ( ) ( )( )( )
( )
kips 79.151
38.020
1685.0205.2
1565.17015.038.020
1685.020485.0
38.085.05.2
38.085.085.0
22
n
22
=
−−+
−=
−−+
−′=
P
he
hmD
hehfP sg
cn
ρ
CHAPTER 9c. COMBINED COMPRESSION AND BENDING: COLUMNS Slide No. 91ENCE 454 ©Assakkaf
Whitney’s Approximate SolutionExample 9 (cont’d)
b) e = 5.0 in. with axial load inside the circular section. It can be assumed that the section is compression-controlled.
( )
gives 58, Eq. Using
in 2.314420 ,in 74.4 area steel total 2
22 ===
πgst AA
( )( )
( )( )
( )( )( )
kips 58.62618.1
1567.0208.05206.9
42.314
115
536074.4
18.167.08.06.90.13
2
2
=+
×+×
++
=
++
′+
+=
n
s
cg
s
ystn
P
DhhefA
De
fAP