diagramas momento curvatura

CIE 525 Reinforced Concrete Structures Instructor: Andrew Whittaker

Lecture 04

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


Lecture 04

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


Lecture 04

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.


Lecture 04

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”


Lecture 04

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


Lecture 04

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


Lecture 04

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”


Lecture 04

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 = = = −

−


Lecture 04

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


Lecture 04

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


Lecture 04

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


Lecture 04

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


Lecture 04

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


Lecture 04

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”


Lecture 04

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.

• UCFyber: developed by Chadwell at UC Berkeley in the late 1990s—see the link to this program at the Zevent website: http://www.zevent.com/framep.html.

• 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 Zevent web site and is available 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.


Lecture 04

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).


Lecture 04

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.


Lecture 04

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=


Lecture 04

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


Lecture 04

diagramas momento curvatura

Documents

momentcurvature relationships

momentcurvature analyses

levels of curvature

use of plastic analysis

ym curvature

crm curvature

established curvature

concrete large ductility