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CIE525: Assignment 3

Moment-curvature Relationships

Manish Kumar

1.1 Problem Statement

Consider the beam section below that is not drawn to scale. The section through the beam is

shown below the elevation. Assume Grade 60 rebar and '

cf = 4 ksi.

Part 1: Develop moment-curvature relationships for the following cases using hand calculations

(a) flexure producing tension at the top of the beam, no strain-hardening in the longitudinal

rebar, #4 ties at 10 inches on center; and (b) per part (a) but with #4 ties at 3 inches on center.

List all assumptions. Plot the relationships in Excel or equivalent. Comment on the results.

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Part 2: Develop moment-curvature relationships using Xtract for the two cases of Part 1; and (c)

including strain hardening in the longitudinal rebar. Plot the relationships in Excel or equivalent.

Show all of the relationships on one plot. Comment on the results.

1.2 Solution: Part1: Part a

For the part 1, a distance of 10 inches between ties is too much to realize any significant

confining effects and hence beam is treated as unconfined when obtained moment-curvature

relationships.

Three transition points on moment-curvature curves are considered are points are interpolated

between them to obtain the full curve: 1) cracking, 2) yielding and 3) ultimate. Each state is

discussed here. Geometrical parameters of given beam-section is summarized in Table 1.

Table 1: Geometrical parameters for given beam-section

Parameter Description Value

Part 1 Part 2

cl Clear cover of concrete 1.615” 1.615”

ld Diameter of longitudinal bars 1.27” 1.27”

hd Diameter of transverse bars 0. 5” 0. 5”

s Vertical spacing between hoops 10” 3” 's Clear vertical spacing between hoops 9.5” 2.5”

b Width of the section 20 20

d

Depth of the section 24 24

cb Horizontal spacing between centerlines of perimeter hoop 16.27” 16.27”

cd Vertical spacing between centerlines of perimeter hoop 20.27” 20.27”

'w Clear distance between longitudinal bars Varies Varies

Cracking

At first onset of cracking, stress in the extreme tension fiber reaches modulus of rupture of

concrete. Critical moment is then calculated using expression:

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r g

cr

t

f IM

y (1)

where rf is the modulus of rupture of concrete, gI is the moment of inertia of gross section

ignoring the contribution from reinforcements, and ty is the distance of extreme tension fiber

from neutral axis of the section. Ignoring the contribution of reinforcements to moment of inertia

and neutral axis of the section has negligible effects on moment calculations in elastic range of

behavior. Gross moment of inertia is given as:

3

12g

bdI (2)

And neglecting contribution of reinforcements, neutral axis would at the centroid of the section

and given as / 2ty d . As per ACI (2011), modulus of rupture of concrete is given by

expression:

'7.5r cf f (3)

Substituting these values back in Equation (1) gives us the cracking moment of the section. The

corresponding curvature is obtained using elastic theory:

crcr

c g

M

E I (4)

This gives the cracking point on the curve ( , )cr crM

Yielding

Following the cracking of concrete section in tension, crack propagates through the cross-section

on further application of moment and tensile force is taken by the tension reinforcements. The

moment-curvature behavior is still linear, however, only up to the point when tension

reinforcement yields. At yielding, strain in the tension reinforcement is ( / )y s sf E and neutral

axis shifts towards compression area.

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The neutral axis at yielding is given as distance kd from extreme compression fiber, where the

ratio k is calculated using expression:

2 2 '

( ') 2( ' ) ( ')d

k n n nd

(5)

Where ( / )sA bd and ''( / )sA bd are the tension and compression steel ratios, )( /s cn E E

is the modular ratio, and d and 'd are the distance of compression and tension steel from

extreme compression fiber.

Taking moment about compressive force due to concrete, yield moment is given by:

' ' '

3 3y s y s y

kd kdM A f d A f d

(6)

Since stress in the tension steel is yf , using similar triangles, stress in compression steel is

calculated as:

' 's y

d df f

d kd

(7)

Once stress in compression steel is obtained, yield moment is obtained substituting it back to

Equation (6). Curvature is then obtained as:

y

yd kd

(8)

This gives us the yielding point on the curve ( , )y yM

Ultimate

After yielding of tension steel, its stress remains constant but strain keeps increasing until

compressive strain in extreme fiber of concrete reaches the strain value of cu at maximum stress

in concrete '

cf . In order to address the nonlinearity in concrete at high strains, whitney-block is

used to approximate the parabolic stress distribution in concrete to an equivalent rectangular

stress-block representation.

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The calculation of ultimate state requires iteration. For hand calculations, let us assume that

strain in compression steel '

s exceeds the yield strain y . This assumption will be checked later.

Equilibrium of tension and compressive force ( )C T gives the depth of neutral axis c as:

' '

'

10.85

y y s s

c

A f A fc

f b

(9)

Ultimate moment is then obtained by taking moment about tension steel as:

' ' '1 11' (0.85 ) '

2 2u c s c s s

c cM C d C d d f cb d A f d d

(10)

Ultimate curvature is then calculated as:

uu

c

(11)

where u is ultimate strain in concrete at maximum stress, which is 0.003 as per ACI (2011).

This is first trial value of u . Assumption of yielding in compression is now checked by

ensuring:

' 's cu y

c d

c

(12)

If the above condition is satisfied then assumption made is true and obtained value of ( , )u uM

defines the ultimate state on the moment-curvature curve. If the condition is not satisfied further

iteration is required with new trial strain value as

'

2

s y .

A Matlab program was written to calculate the moment-curvature values for three states as per

principles explained in above sections. Values obtained have been shown in Table 2.

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Table 2: Moment-curvature values for given beam section

Moment (kip-in) Curvature 4( 10 ) Curvature-ductility

Cracking 911 0.11 0.076

Yielding 5753 1.43 1

Ultimate 5890 8.07 5.6

1.3 Solution: Part 1: Part b

As distance between ties at center is 3 inch in this case, confinement of concrete is considered

here while calculation moment-curvature values. Mander’s stress-strain model for confinement

of concrete is used here. Mander’s confinement model was derived primarily for columns under

uniaxial compression and suggested values of confinement effective co-efficient based on

experiments on columns might not be applicable for beams under pure flexure. Confinement

effective constants are calculated here from first principles suggested in Mander et al. (1988). In

case of beams under flexure, only the area above neutral axis experience compression and there

is no effect of confinement in tension. Accordingly, when effective area is calculated, the

ineffectively confined area with tension reinforcement is neglected here.

7 1.27

0.02716.27 20.27

cc

(13)

2

' 2 2 2

1

2 6 2 17.27 668n

i

i

w in

(14)

Note that only parabolic ineffectively areas between compression reinforcements and vertical

reinforcements have been considered in the above equation.

Confinement effective constant is calculated as:

668 2.5 2.51 1 1

6 16.27 20.27 2 16.27 2 20.270.59

(1 0.027)ek

(15)

Transverse reinforcement ratios are calculated as:

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22 0.54 0.0066

3 20.27

sxx

c

nA

sd

(16)

22 0.54 0.0082

3 16.27

sy

y

c

nA

sd

(17)

Effective lateral pressures in both directions are obtained as:

' 0.59 0.0066 60 0.23lx e x yhf k f ksi (18)

' 0.59 0.0082 60 0.29ly e y yhf k f ksi (19)

Using '

'

0.230.06

4

lx

c

f

f and

'

'

0.290.073

4

ly

c

f

f and using chart given in Mander et al. (1988),

we obtain:

'

'1.4cc

c

fK

f (20)

So the strength of confined core is given as: ' ' 1.4 4 5.6cc cf Kf ksi .

Using 0.0148s x y , the ultimate compressive strain in concrete can be calculated as

per Mander’s equation:

'

1.4 1.4 0.0148 60 0.10.004 0.004 0.026

5.6

s yh sm

cu

cc

f

f

(21)

The strain, cc , at compressive strength of confined concrete is calculated as:

'

'0.002 1 5 1 0.006cc

cc

c

f

f

(22)

So, 4cu

cc

.

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In order to determine the equivalent stress block parameters, Figure 1 is referred from Paulay and

Priestley (1992).

Figure 1: Concrete compressive stress block parameters for rectangular sections with

rectangular hoops (from Paulay and Priestley (1992) as reported in Whittaker (2012))

Values of stress block parameters are obtained as:

0.85 1and

So the average strength to use for equivalent rectangular stress block is

' 0.85 5.6 4.76ccf ksi

Similar procedure as used in part 1 of the problem can be used here, except for ultimate moment-

curvature calculations. For ultimate moment-curvature calculations, 0.85 is replaced by

1 calculated above and instead of ' 4cf ksi ,

'

ccf value of 5.6 ksi is used in the calculations.

Other assumption is that at large curvatures, the unconfined cover concrete has spalled and

effective width and depth of the beam is reduced to: 16.73 , 19.61b in d in .

Using the same Matlab code provided in Appendix A, moment-curvature values are obtained and

presented in Table 3.

Table 3: Moment-curvature values for given beam section

Moment (kip-in) Curvature 4( 10 ) Curvature-ductility

Cracking 911 0.11 0.076

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Yielding 5753 1.43 1

Ultimate 5380 96.23 67

1.4 Solution: Part 2

XTRACT was used and moment curvature graphs were obtained for cases described in Part 1

and with and without strain hardening of reinforcements. Obtained plots are presented in Figure

2.

Figure 2: Moment-curvature plots for given beam section

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1.5 Solution: Problem 2

XTRACT was used to obtain moment-curvature plots for column section shown in Figure 3, for

different level of axial loads applied.

Figure 3: Column section used for the analysis

Plots obtained from XTRACT for different amount of axial loads are shown in Figure 4. Plots

show that as axial load increases, strength of column section increases but its maximum

curvature or curvature-ductility decreases. Failure modes that limit the maximum curvature for

different axial load cases are summarized in Table 4.

Table 4: Failure modes of column section for different axial loads

Axial Load Failure Mode

0 Failure of longitudinal bars '0.1 c gf A Failure of longitudinal bars

'0.2 c gf A Failure of confined concrete

'0.4 c gf A Failure of confined concrete

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Figure 4: Moment-curvature plots obtained from XTRACT for different level of axial loads

(' ' '0 0 , 1 0.1 , 2 0.2 , 4 0.4c g c g c gMC kips MC f A MC f A MC f A )

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References

ACI (2011). "Building code requirements for structural concrete and commentary." Report ACI

318-11, American Concrete Institute, USA.

Mander, J., Priestley, M. J. N., and Park, R. (1988). "Theoretical stress‐strain model for confined

concrete." Journal of structural engineering, 114, 1804.

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Appendix A

%Program written for Moment-curvature analysis of reinforced concrete beam %Reinforced Concrete Design CIE525 %Written by Manish Kumar %Date 09-24-2012

%Define parameters fc=4000; e_cu=0.003; fy=60000; Ey=29000000; ey=fy/Ey; Ec=57000*sqrt(fc); n=Ey/Ec; b=20; d_gross=24; d=21.25; dc=2.75; As=5.08; r=As/(b*d); Asc=3.81; rc=Asc/(b*d);

%obtain value of beta 1 if fc<=4000 b1=0.85; elseif 4000<fc<=8000 b1=0.85-0.05*(fc-4000)/1000; else b1=0.65; end

%Initialize moment curvature matrix M=[]; phi=[]; %Cracking Ig=(b*(d_gross^3))/12; yt=d_gross/2; fr=7.5*sqrt(fc); Mcr=fr*Ig/yt; phi_cr=Mcr/(Ec*Ig); M=[M;Mcr]; phi=[phi; phi_cr];

%Yielding k=sqrt(((r+rc)*n)^2+2*(r+rc*dc/d)*n)-(r+rc)*n; fsc=((k*d-dc)/(d-k*d))*fy; My=As*fy*d*(1-k/3)+Asc*fsc*(k*d/3-dc); phi_y=ey/(d-k*d); M=[M;My]; phi=[phi;phi_y];

%Ultimate ct=0.5*d; c=0; while abs(c/ct-1)>0.0002; e_sc=((ct-dc)/ct)*e_cu; fsc=Ey*e_sc; Cs=Asc*fsc; Cc=0.85*fc*b*b1*ct; T=As*fy; c=(T-Cs)/(0.85*fc*b*b1); ct=(c+ct)/2; end

Mu=0.85*fc*b1*c*b*(d-b1*c/2)+Asc*fsc*(d-dc); phi_u=e_cu/c; M=[M;Mu]; phi=[phi;phi_u];

%graph plot(phi,0.001*M);