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MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

Dr. Jason E. Charalambides

Copyright J. CharalambidesCopyright J. Charalambides

A Few Basics

! The material:There is no actual material that is homogenous and is called concrete.It is a composite of aggregates cementitious substances andreinforcement. Some parts are very brittle, some parts are very elastic. It is a product of composite action and chemical curing and it is veryspecial!

! Curing:" Concrete’s drying is not a hardening method similar to that of clay

drying. There is a chemical reaction that takes place betweencement and water and that process needs a certain amount of timeto optimize the hardening.

" It is critical for the first 7 days to have proper temperature andmoisture conditions, and up to 2 weeks it is very important.

" 30% of concrete’s strength, or even more, can be lost if concretedries fast


pq^qf`p=lc=_b^j=^`qflk






Flexure Theory for ReinforcedConcrete! Applying flexural stress to a sample concrete beam:


ciburRb=qeblRv=clRRbfkclR`ba=`lk`Rbqb! Fundamentals:! In order to have

equilibrium…" M=T * jd" M=C * jd

! For a conventional elasticbeam we have σ=My/I,

or

σ=M⋅cI


ciburRb=qeblRv=clRRbfkclR`ba=`lk`Rbqb! The Transformed

stress (Whitney) block:

" Concrete does nottake tension. Thecompressive stressdistribution is seenin the purple color.

" The idealizedcompressive stressdistribution isindicated by thegreen blocksuperimposed overthe actual stressblock.


ciburRb=qeblRv=clRRbfkclR`ba=`lk`Rbqb! The Transformed

stress (Whitney) block:" According to ACI

code (318 par22.2.2.1), themaximum usablestrain (ε) in concreteshould not exceed0.003 although theactual limit strain atthe point of failurecan vary between0.0025 and 0.01! …and the ACI

code is a user’smanual to live by!


ciburRb=qeblRv=clRRbfkclR`ba=`lk`Rbqb! =The Transformed

stress (Whitney)block:" The compression

to be calculated isthe volume of thatblock, i.e.:

" It should be notedthat the resultantforce of theWhitney block islocated at the mid-height of a, theproduct of β1*c.

C=0.85( f ' c)⋅a⋅bw

ACI Code22.2.4.1


ciburRb=qeblRv=clRRbfkclR`ba=`lk`Rbqb! =The Transformed stress (Whitney) block:


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! The “beta 1” β1 coefficient is variable:

" For f’c<4ksi β1=0.85" For f’c>8ksi β1=0.65" For f’c values between 4ksi and 8ksi

! The “rho” ρ ratio of tensile steel reinforcement is equivalent tothe division of the cross sectional area of steel by the beamscross sectional area:

! “ρ” actual should be larger than “ρ” minimum:

β1=0.85−0.5⋅f ' c−4000 lbf

inch2

1000 lbfinch2

ρmin=max (200f y,3 √ f ' c

f y) 9.6.1.2

ρ=Asbw⋅d

ACI-318-14Table 22.2.4.3


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! Addressing the “epsilon t” (strain in tension)parameter:

" Where “c” is the distance from the Neutral axis to theextreme fiber and

" “d” is the effective depth of the beam (from extreme fiber tocentroid of rebars that is)

" “εy” is the strain of yield. We want the yield strain to beless than the actual strain experienced in tension! Thatis ...we want the steel to yield and deform (in plasticdeformation, not elastic)

ε t=0.003⋅(d−c)c


fk=`i^pp=bu^jmibp

! Example 1" Given f`c=3ksi, fy=50ksi,

b=10”, d=17”, As=3#7,solve for C, c, and verifythat the yield strain (εy) isless than the strain appliedin tension (εt)

ε t=0.003⋅(d−c)c

cd−c

= 0.003ε t

ε t=0.0093 ε y=f y

29,000ksiε y=0.0017

a=3.529incha= Cbw⋅0.85⋅f ' c

c=4.152inchc= aβ1

As=1.8inch2 T=As⋅f y=C C=90kip β1=0.85


fk=`i^pp=bu^jmibp

! Example 2" Given f`c=6ksi, fy=75ksi,

b=10”, d=18”, As=3#10,solve for C, c, and verifythat εy<εt

ε t=0.003⋅(d−c)c

cd−c

= 0.003ε t

ε t=0.0042 ε y=f y

29,000ksiε y=0.0026

a=5.603incha= Cbw⋅0.85⋅ f ' c

c=7.471inchc= aβ1

As=3.81inch2 T=As⋅f y=C C=285.75kip β1=0.75

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