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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:
3
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);
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