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CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 1
4. MOMENT-CURVATURE RELATIONSHIPS
4.1 Recommended Reading
1. Paulay, T. and Priestley, M. J. N., Seismic Design of Reinforced Concrete and Masonry Buildings, Wiley InterScience, Chapter 3
2. MacGregor, J., Reinforced Concrete, Mechanics and Design, Prentice Hall, Third Edition.
4.2 Redistribution of Moments in RC Systems
4.2.1 Gravity-Load-Resisting Systems
Section 8.4 of ACI 318 permits redistribution of moment in continuous reinforced concrete flexural members.
• linear elastic analysis of a nonlinear component
For a prismatic cross section with positive and negative moment capacity nM , the maximum load w by
analysis is
• Elastic analysis: max 2
12e nMw
L=
• Plastic analysis: max max2
161.33p enM
w wL
= =
l
w kips/ft
Elastic moment diagram
Plastic moment diagram
Plastic hinges?
Prismatic cross section, nM
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 2
So, the use of plastic analysis produces a larger permissible load than elastic analysis. How can this additional capacity be realized?
• adequate ductility in the plastic hinging regions
ductility is a measure of inelastic deformation capacity beyond the yield deformation
use moment-curvature analysis to determine deformation limits
degree of concrete confinement will affect the deformation limit
maximum concrete strain maxcε
4.2.2 Lateral-Force-Resisting Systems
Redistribution of lateral forces underpins the response of framing systems subjected to earthquake and blast forces because components attain their maximum strengths at different levels of deformation. See the beam-sway mechanism below that is a preferred mechanism in earthquake engineering.
• why is beam sway a preferred mechanism?
• effect of gravity load moments on component response?
So, adequate deformation capacity must be provided for all of the hinges to form as shown
• large inelastic deformation in the concrete
• large ductility achieved through the use of appropriate details, including confinement
earthquake
gravity
sum
Flexural demands on a beam
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 3
4.3 Moment-Curvature Analysis of Unconfined Sections
4.3.1 Response Calculations
For hand calculations, the moment at three levels of curvature are established
• curvature at cracking of the cross section (at crM )
• curvature at yield of the cross section (at yM )
• curvature at the ultimate concrete strain (at uM )
The procedures are illustrated below for two unconfined sections: (1) a slab with tension rebar only, and (2) a beam with tension rebar (part 1) and tension and compression rebar.
4.3.2 Moment-Curvature Analysis of a Slab
This analysis represents the simplest of all moment-curvature analyses. Some simplifying assumptions are made to calculate the moment-curvature ( M − φ) relationships, and all are basic assumptions in flexural theory, namely,
1. Sections perpendicular to the axis of bending that are plane before bending are plane after bending, or plane sections remain plane. See the figure below.
As such, curvature and strain are related as follows:
yεφ=
where y is the distance from the neutral axis.
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 4
2. The strain in the rebar is equal to the strain in the concrete at the same level in the cross section
3. The stresses in the steel and concrete can be established from the individual stress-strain relationships
The example slab is shown below. The objective is to calculate the moment-curvature relationship for the slab section. Assume a 12-in wide section of slab for the purpose of calculation, Grade 60 rebar, and cf ′ =
4 ksi. Assume 1 in of cover to the longitudinal rebar.
Three sets of calculations are made, at cracking, at yield, and at ultimate.
Cracking
Ignoring the rebar (and having to transform the section), 3 312 6 216
12 123
gbDI in×= = =
E for the concrete is 57,000 4000 3604psi ksi=
Calculate the cracking moment, (7.5 ) (216) 34.2
1000 3r g c
crt
f I fM kip in
y
′= = = −
Calculate the curvature at the cracking moment, 34.2 0.000044 4.39 53604 216
crcr
c g
ME
E Iφ = = = = −
×
So the data point for cracking ( , )cr crMφ is (0.000043, 34.2)
Yield
For this calculation, use is made of the cracked transformed moment of inertia. The limiting strain is the yield strain in the tension steel. For information on the calculation of transformed moments of inertia, see Chapter 9 of MacGregor.
6” #4 @ 6”
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 5
The assumed stress distribution in the concrete is shown above. The depth to the neutral axis is kd. The strain in the rebar is yε . For a singly reinforced section,
22 ( )k n n n= ρ + ρ −ρ
where n is the modular ratio (= s cE E ) and sA bdρ = .
For the subject cross section, 0.56 1 4.75"2
d = − − = , 22 (0.2 ) 0.0070
12 4.75in×ρ = =
×, and 29,000 8.04
3604n = = ,
and 0.28k = . (Is this value reasonable?)
Taking moments about the centroid of the concrete compression block, which is located at a distance of
3kd below the top of the slab,
2 0.28 4.75( ) ( ) (0.4 )60(4.75 ) 103.43 3y s s s y
kdM A f jd A f d in kip in×= = − = − = −∑
The corresponding curvature is
0.0021 0.00061 6.1 4( ) 3.42
yy E
d kd
εφ = = = = −
−
So the data point for yielding ( , )y yMφ is (0.00061, 103.4)
Ultimate
See the figure below for the information needed to solve for the ultimate moment and ultimate curvature.
yε
yφ
kd
snA
bkd
d
b
Transformed Areas
Strains
cf
sf
Stresses
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 6
Assuming a rectangular (Whitney-type) stress block, calculate the depth to the neutral axis c:
1
0.4 60 0.69"0.85 0.85 4 12 0.85
s y
c
A fc
f b×= = =
′ β × × ×
The ultimate moment uM is calculated in the traditional manner, namely,
1 0.85 0.69( ) 0.4 60 (4.75 ) 106.92 2u s yc
M A f d kip inβ ×= − = × × − = −
The corresponding ultimate curvature is
max 0.003 0.0043 4.3 3 90.69
cu yE
cε
φ = = = = − = φ
So the data point for ultimate ( ,u uMφ ) is (0.0043, 106.9). Note the small difference between uM (107
kip-in) and yM (104 kip-in).
0.000 0.001 0.002 0.003 0.004 0.005Curvature 1/in
0
50
100
150
Mom
ent (
kip-
in)
c
d
by> ε
uφ
Strains
yf
Stresses
maxcε0.85 cf ′
1cβ
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 7
4.3.3 Moment-Curvature Analysis of a Beam
This sample analysis for a beam presented below builds on the slab example presented above. Two cases will be considered: (a) tension rebar only, and (b) tension and compression rebar. Key information for the analysis is presented in the table below.
cf ′ 4 ksi
yf 60 ksi
rf 0.474 ksi
gI 13,210 in4
cE 3604 ksi
ρ 0.0099
′ρ 0.0066
Part 1: No compression rebar
Cracking
13,310 (0.474) 57311
573 1.19 5/3604(13,310)
cr
crcr
M kip in
ME in
EI
= = −
φ = = = −
Yielding
2
8.04
2(0.0099)(8.04) (0.0099 8.04) 0.0099 8.04 0.3270.327 20( ) 3.00 60(20 ) 3207
3 3
0.0021 1.56 4 /( ) 13.46
y s y
yy
n
kkdM A f d kip in
E ind kd
=
= + × − × =×= − = × − = −
εφ = = = −
−
3#9 (parts a and b)
2”
22”
2#9 (part b only)
2”
15”
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 8
Ultimate
1
1
max
3.0 60 4.15"0.85 0.85 4 15 0.85
0.85 4.15( ) 3.0 60(20 ) 3282 1.022 2
0.003 7.2 4 / 4.64.15
4.6
s y
c
u s y y
cu y
A fc
f b
cM A f d kip in M
E inc
φ
×= = =′ β × × ×
β ×= − = × − = − =
εφ = = = − = φ
µ =
Part 2: Including compression rebar
Cracking (as before)
13,310 (0.474) 57311
573 1.19 5/3604(13,310)
cr
crcr
M kip in
ME in
EI
= = −
φ = = = −
Yielding
2 2
8.04
( ) 2( ) ( ) 0.301
n
dk n n nd
=
′′ ′ ′= ρ + ρ + ρ + ρ − ρ + ρ =
Now the general equation for the moment yM is
( ) ( )3 3y s y s s
kd kdM A f d A f d′ ′ ′= − + −
where the stress in the compression steel is a function of the distance k. If the stress in the tension rebar is
yf , then the strain in the compression rebar can be calculated using similar triangles, namely,
( ) 17.3( )
0.301 20 0.301 203.0 60(20 ) 2.0 17.3(2.0 ) 32383 3
0.0021 1.50 4 /( ) 13.98
s y
y
yy
kd df f ksid kd
M kip in
E ind kd
′−′ = =−
× ×= × − + × − = −
εφ = = = −
−
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 9
Ultimate
The calculation of ( , )u uMφ requires some iteration to establish the location of the neutral axis. For hand
calculations, initially assume that the strain in the compression steel s′ε exceeds the yield strain, and
check this assumption later in the analysis.
1
11
3.0 60 2.0 60 1.38"0.85 0.85 4 15 0.85
(0.85 )( ) ( ) 33212
0.003 0.00221.38
s y s s
c
u c s s
cuu
A f A fc
f b
cM f cb d A f d d kip in
c
′ ′− × − ×= = =′ β × × ×
β′ ′ ′ ′= β − + − = −
εφ = = =
Checking the assumption regarding the strain in the compression steel,
max ( ) 0.0015 0.71s c yc d
c′−′ε = ε = = ε
and so the assumption is not valid and another trial is required. Following a few iterations, 2.90"c = , 27sf ksi′ = , and
max
0.85 2.90(0.85 4 0.85 2.90 15)(20 ) 2.0 27(20 2) 33312
0.003 0.00102.90
6.7
u
cu
M kip in
c
φ
×= × × × × − + × − = −
εφ = = =
µ =
Consider now the tabulated data below.
Compression Rebar
No Yes
yM 3207 3238 ← negligible increase
yφ 0.000156 0.000150 ← no increase
uM 3282 3331 ← negligible increase
uφ 0.00072 0.0010 ← 40% increase
φµ 4.6 6.7 ← 40% increase
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 10
4.4 Moment-Curvature Analysis of Confined Sections
4.4.1 Response Calculations
For hand calculations, the moment at three levels of curvature are established as before for unconfined sections
• curvature at cracking of the cross section (at crM )
• curvature at yield of the cross section (at yM )
• curvature at the ultimate concrete strain (at uM )
The procedures are illustrated below for one confined section: the beam cross section of Section 4.3.3 with tension and compression rebar. Assume that #5 perimeter hoops at 4 inches on center confine the cross section.
cf ′ 4 ksi
yf 60 ksi
rf 0.474 ksi
gI 13,210 in4
cE 3604 ksi
ρ 0.0099
′ρ 0.0066
The first step in the calculation process is to establish the properties of the confined concrete. Consider for this example moment on the cross section that produces compression in the top of the reinforced concrete beam (x-x bending). Axes x and y are as shown.
For the cross section shown, and using the terminology of Lecture 03 (from Paulay and Priestley),
13.2
20.2
2 0.3 0.01144 13.22 0.3 0.0074
4 20.2
x
y
y
x
h in
h in
′′ ≈
′′ ≈
×ρ = =××ρ = =×
3#9
2”
22”
2#9
2”
15”
#5 ties @ 4 in. on center
x
y
Note: these dimensions will violate the minimum cover requirements of ACI 318
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 11
Is it reasonable to calculate the degree of confinement based on a weighted average of xρ and yρ if the bending is around the x-x axis and prior experience would indicate that the neutral axis will be located close to the top of the beam?
• conservative to average xρ and yρ
what will be the effect on the moment-curvature relationship?
Assuming an effectiveness coefficient of 0.75,
600.75 0.0074 0.0834
600.75 0.0114 0.1284
lx
c
ly
c
ff
f
f
′= × × =
′
′= × × =
′
From the above figure, and noting that the largest effective confining strength in this example is lyf ′ , the value of K is 1.6 and the strength of the confined core is therefore
1.6 4 6.4cc cf Kf ksi′ ′= = × =
The remaining parameters needed to defined the stress-strain relationship for the confined cross section are
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 12
The stress-strain relationship for the confined and unconfined concrete in this cross section is shown below.
To calculate the ultimate moment and ultimate curvature for this section, the stress block parameters must be established. From before,
• K = 1.6, max 0.028cε = , and 0.008ccε =
• max 0.028 3.50.008
c
cc
ε= =
ε
max
sec
sec
1.4( ) 1.4(0.0188)60 0.10.004 0.004 0.0286.4
6.40.002[1 5( 1)] 0.002[1 5( 1)] 0.0084
6.4 8000.008
57,000 57,000 4000 3604
1.28
0.0
x y yh smc
cc
cccc
c
cc
cc
c c
c
c
c
f
f
ff
fE ksi
E f psi ksi
Er
E E
x
ρ + ρ ε ×ε = + = + =′
′ε = + − = + − =
′
′= = =
ε
′= = =
= =−
ε=
1.28 1.28
12508
6.4(125 )1.28 1024
1 1.28 1 0.28 (125 )
c
cc c cc r
c
f xrf
r x x
= ε
′ ε ε= = =
− + − + + ε
0.00 0.01 0.02 0.03Compressive strain (in/in)
012345678
Com
pres
sive
stre
ss (k
si)
ConfinedUnconfined
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 13
From the above relationships, 1, 0.9, 0.9β = αβ = α = . There is now sufficient information to complete the moment-curvature analysis.
Cracking (as before)
13,310 (0.474) 57311
573 1.19 5/3604(13,310)
cr
crcr
M kip in
ME in
EI
= = −
φ = = = −
Yielding (as before)
2 2
8.04
( ) 2( ) ( ) 0.301
n
dk n n nd
=
′′ ′ ′= ρ + ρ + ρ + ρ − ρ + ρ =
( ) 17.3( )
0.301 20 0.301 203.0 60(20 ) 2.0 17.3(2.0 ) 32383 3
0.0021 1.50 4 /( ) 13.98
s y
y
yy
kd df f ksid kd
M kip in
E ind kd
′−′ = =−
× ×= × − + × − = −
εφ = = = −
−
Ultimate
For the purpose of the calculation below, the effect of the compression rebar will be ignored.
• impact of this decision will be discussed later
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 14
With confinement, the maximum concrete strains will substantially exceed the spalling strain that will be assumed to be 0.004. Therefore, the calculation at ultimate conditions should assume that the cover concrete has spalled (see the orange hatched zone on the figure below).
• b = 13.2”
• d =19.1”
Consider now the tabulated data below.
Confinement
No Yes
yM 3207 3207 ← no change
yφ 0.000156 0.000150 ← no change
uM 3282 3215 ← negligible change
uφ 0.00072 0.0119 ← increase by a factor of 17
φµ 4.6 79 ← increase by a factor of 17
What is the effect of ignoring the compression rebar in the above analysis?
• effect on the depth to the neutral axis c when compression rebar is included?
• if the maximum compression strain is unchanged, and c is (increased/decreased) by the addition of the compression rebar, what is the effect on the ultimate curvature?
1
11
3.0 60 2.36"0.9 6.4 13.2 1
( )( ) 32152
0.028 0.01192.36
79
s y
cc
u cc
cuu
A fc
f b
cM f cb d kip in
c
φ
×= = =′α β × × ×
β′= α β − = −
εφ = = =
µ =
#5 ties @ 4 in. on center
2”
3#9
2#9
2”
22”
15”
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 15
What is the effect of ignoring strain hardening in the steel rebar in the above analysis?
• effect on maximum strength?
• effect on ultimate curvature?
In summary, how is the ultimate curvature and curvature ductility of a cross section increased?
Increase in uφ , φµ ?
No Yes
Increase in ρ
Increase in ′ρ
Increase in yf
Increase in cf ′
Increase in ′′ρ
Increase in axial compression
4.5 Moment Curvature Analysis of Complex Sections
Most moment-curvature analyses undertaken in the design office make use of computer software. Three examples of such software are
• BIAX: developed by Wallace at UC Berkeley in the early 1990s.
• Xtract/UCFyber: developed by Chadwell at UC Berkeley in the late 1990s—see the link to this program at the Imbsen website: http://www.imbsen.com/xtract.htm.
• SEQMC (developed by SEQAD in the late 1990s – see the link to this program at the SC Solutions website: http://www.best.com/~solvers/seqmc.pdf.)
Students may make use of any of these programs. UCFyber can be downloaded free-of-charge at the Imbsen web site and is available (as UCFyber) on the CSEE servers.
The computer codes operate in a somewhat standard manner with different post-processing features and GUIs. Below is a short presentation on how moment-curvature relationships are established for arbitrary cross sections. Some of the presentation is adapted from Priestley, Seible, and Calvi. For the presentation below, it is assumed that the stress-strain relationship has already been established for the concrete.
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 16
To date, the moment-curvature analysis has assumed that the stress-strain relationship for rebar is elastic perfectly plastic. Such an assumption simplifies hand calculations but is substantially conservative. The figure below from Priestley, Seible, and Calvi shows monotonic tensile stress-strain curves for different grades of rebar.
• nominal yield strength versus measured yield strength
• strain range for yield plateau and maximum strain for three grades of rebar
• values of smε for the three grades of rebar
Consider the stress-strain relationship below for monotonic loading of Grade 60 rebar (from Priestley, Seible, and Calvi).
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 17
For this grade of rebar, the expected yield strength ( yef ) will exceed the nominal yield strength ( yf ) by a factor of between 1.1 and 1.3. The strain shε can be taken as 0.008 and the ultimate strain in the rebar
suε can be taken as 0.12. In the strain-hardening region of the curve ( sh s suε ≤ ε ≤ ε ), the stress in the rebar can be taken as
20.12[1.5 0.5( ) ]
0.112s
s yef f− ε
= −
For analysis involving concrete strains greater than 0.003 and 0.004, the analyst must distinguish between the confined and unconfined regions of the reinforced concrete element:
• concrete contained within the hoops is considered to be confined
• concrete outside of the hoops is considered to be unconfined
The figure below, from Priestley, Seible, and Calvi, presents nomenclature for the remainder of this lecture.
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 18
The moment-curvature analysis is an iterative procedure involving considerations of axial and moment equilibrium on the cross section and a selected vales of extreme fiber strain in compression ( cε ).
Consider the circular cross section. The solution for the rectangular cross section is similar but simpler.
From axial equilibrium on the cross section
( / 2)
( ) ( ) ( ) 1( / 2)[ ( ) ( ) ( )] ( )
D nc x c x x c x cu x si s xiix D c
P b f b b f dx A f== −
= ε + − ε + ε∑∫
where
( 0.5 )cx x D c
cε
ε = − +
From moment equilibrium on the cross section
( / 2)
( ) ( ) ( ) 1( / 2)[ ( ) ( ) ( )] ( )
D nc x c x x c x cu x si s xi iix D c
M b f b b f xdx A f x== −
= ε + − ε + ε∑∫
and from before
ccε
φ =
In the above equations, ( ), ( ),c cuf fε ε and ( )sf ε are the stresses in the confined concrete, unconfined
concrete, and rebar, respectively, as a function of the strain; and siA is the area of the rebar at distance ix
from the centroidal axis. Other terms are defined in the figure above.
The solution scheme is as follows
1. Select an extreme fiber strain and an axial load P.
2. Solve for c by trial and error using the known axial load P and the specified extreme fiber strain.
3. Calculate the moment M and the curvature φ using the above equations.
4. Select a new extreme fiber strain (up to the ultimate compression strain) and repeat steps 2 and 3.
5. Select a new axial load P.
Note that if the subject section is rectangular, the above equations are simplified as follows:
• ( )xb b= and ( )c x cb b=
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 19
4.6 Cross Section Analysis with UCFyber
The results of the analysis of the confined beam cross section using UCFyber are presented on the following two pages. The moment-curvature relationships are summarized below.
0.000 0.002 0.004 0.006 0.008 0.010Curvature (1/in)
0
1000
2000
3000
4000
5000
Mom
ent (
kip-
in)
Rebar strain hardeningRebar bilinear
Note the effect of including rebar strain hardening in the analysis
• a substantial increase in maximum strength (to be considered for capacity design)
• a substantial reduction in ultimate curvature
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 20
CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker
Module 04 Page 21