Feb-16
‫بسم هللا الرحمن الرحيم‬
CIV 223
Reinforced Concrete Structures - I
Spring 2015
Dr. Alaa Helba
Office hours:
Sunday 11 - 1:00 pm
Review (lecture 1)
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Lec. # 2 - CIV 223 - Dr. A. Helba sp.2016
Feb-16
Review (lecture 1)
lecture 1: Loads and Loads transfer in skeleton reinforced concrete structures
lecture 1: Materials and their main properties used reinforced concrete structures
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Lecture # 2
R.C. Fundamentals
Analysis of RC Sections under Flexure (B.M)
(Behavior of RC beams in different stages of loading)
 Stage I : Uncracked Section (up to M = Mcr)
 Stage II : Cracked Section in elastic stage (Mw)
 Stage III: Cracked Section in ultimate stage (Mu)
Basic units / Dimensions
For units: Use SI units (N, kN, mm and m)
For a rectangular R.C. section let:
b – width of rectangular R.C. section
t – total height of R.C. section
d – effective depth (distance to steel) h
As – Area of main steel
(Ex.:for 3f16 As=3*p(16)2/4=603mm2)
c - cover [assume c=50 mm (arranging steel in a row)
c= 70 (if arranging As in 2rows) – c=100 (if 3rows)]
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Behavior of RC beams in different stages
of loading (from beginning up to failure)
d=270 mm
b= 150 mm
2F12 mm
As=226mm2
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Flexural Behavior or RC Beams
3 main stages of the beam sections are observed:
Stage I: Beam Before cracking (uncracked
section)
Stage II: beam at Service load (after cracking
limit& before yield) cracked section in elastic
range.
Stage III: Beam at ultimate stage up to failure
(cracked section in ultimate stage)
Stage I: Beam Before cracking (uncracked section)
compression
tension
compression
tension
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Stage II: cracked section in elastic range.
compression
tension
compression
tension
Stage III: cracked section in ultimate stage
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Flexural strain & Stress in RC Sections
M
compression stress
t
Neutral Axis (N.A.)
As
b
tension stress
strain
Stress
Behavior of a Rectangular singly-reinforced sec. sub. to B.M
(elastic range)
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Properties of materials
Es - Modulus of Elasticity of steel = 200 kN/mm2
Ec - Modulus of Elasticity of concrete = 4400 fcu
n – modular ratio:
n
Es
Ec
Code suggests: n =10 in deflection & Mcr calculations
n = 15 in working stage analysis Mwork)
fcu – characteristic strength of conc. (20/25/30 N/mm2)
fy – yield stress of steel (240 / 360 / 400 N/mm2)
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• Assumptions:
Flexure behavior:
– Plane sections remain
plane (linear strain distn
up to failure)
– Hooke’s Law applies (in
working stage f = E * e )
– After cracking limit
Concrete tensile
strength is neglected
– Concrete and steel are
totally bonded (at steel
level es = ec)
Flexural Behavior
The moment-
Stage III
curvature diagram
shows the different
Stage II
stages of the beam
up to failure.
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Stage I
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Stage 1: Uncracked section
Analysis of Uncracked Section
Steps:
for a rectangular section (b x t) reinforced with
steel reinforcement of As:
t
1- determine the position of the neutral axis,N.A
– calculate shift e)
N.A
As
2- calculate the moment of inertia Ix-x (IN.A)
of the section (concrete and steel) (no cracks).
b
3- calculate the cracking moment Mcr. (M= f * I / y)
Step 1: N.A. of the Section
to assign the Centroid of the total area
Let 𝐴𝑐 = 𝐴𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 𝑏𝑡
If Asteel is transformed to an
equivalent concrete)
Atransformed steel = n As
(n = 10 before cracking)
𝐴𝑡𝑜𝑡𝑎𝑙 = 𝑏𝑡 + 10 As
Moments of Areas (@ C.L. of concrete sec tion)
t

bt (0)  nAs   c   (bt  nAs )(e)
2

e  eccentricity ( shift from centre)
t
e  ....... mm , ytension  yt   e
2
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Step 2:
Calculation of Inertia of the virtual section IN.A
t/2
e  eccentricity ( shift from centre)
e
t/2
I N.A
bt 3

 bt (e) 2  nAs ( yt  c) 2
12
For T - section :
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Step 3:
Calculate the cracking moment Mcr
M cr 
f ctr I N . A
yt
Calculation of Mcracking
Procedure:
Transform the area of steel to
equivalent concrete, nAs
using the modular ratio:
n
Es
 10
Ec
1- Calculate the location of Moments of Areas (@ C.L. of con. sec .
t

the N.A.
bt (0)  nAs   c   (bt  nAs )(e)
2


2- Calculate the total moment
of Inertia @ N.A.
e  eccentricity ( shift from centre)
3- Calculate the cracking
bt 3
I tr 
 bt (e) 2  nAs ( yt  c) 2
12
moment Mcr based on
fbottom = fctr = 0.6√fcu.
f I
M cr 
12
ctr
tr
yt
Lec. # 2 - CIV 223 - Dr. A. Helba sp.2016
Feb-16
Example # 1
Calculation of cracking moment Mcr
for a singly reinforced rectangular section
16 2
for 5 f 16 As  5(p
)  1005 mm 2
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Moments of Areas @ middle line (@ C.L. of concrete sec tion)
t

bt (0)  nAs   c   (bt  nAs )(e)
2

10(1005)400  50   [(250)(800)  10(1005)]e
e  16.7 mm
t
 e  400  16.7  383.3 mm
2
bt 3
I
 bt (e) 2  nAs ( yt  c) 2
12
250(800) 3

 250(800)(16.7) 2  10(1005)(383.3  50) 2
12
 1.19  1010 mm 4
yt 
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Cracking Moment
M cr 
f ctr I
yt
f ctr  0.6
f cu
 0.6 25  3
yt  383.3
N / mm 2
mm
I  1.19  1010 mm 4
Cracking Moment M cr
f ctr I 3(1.19  1010 )


yt
383.3
 9.31 10 7 N . mm
 93.1
kN . m
Example -2
Calculation of cracking moment Mcr – T - section
Be  800 mm
120 mm
t  700 mm
510 mm
70 mm
6 F 18
b  200 mm
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Example -2
Calculation of cracking moment Mcr
Be  800 mm
120 mm
N.A
t  700 mm
e
h/2
510 mm
6 F 18
70 mm
b  200 mm
182
)  1524 mm 2
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Moments of Areas @ middle line (@ C.L. of rect. sec tion)
for 6 F 18 As  6(p
t t
t

bt (0)  ( B  b)t s (  s )  nAs   c   [bt  ( B  b)t s  nAs ](e)
2 2
2

800  200(120)(350  60)  10(1524)(350  50)
 [(200)(700)  600(120)  10(1524)]e
e  71.77
mm
t
 e  350  71.8  421.8 mm
2
bt 3
t t
I
 bt (e) 2  ( B  b)t s3 / 12  ( B  b)t s (  s  e) 2
12
2 2
2
 nAs ( yt  c)
yt 
200(700) 3
 200(700)(71.8) 2  600(120) 3 / 12 
12
600(120)(350  60  71.8) 2  10(1524)(421.8  50) 2  8.55  109 mm 4

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Cracking Moment
M cr 
f ctr I
yt
f ctr  0.6
f cu
 0.6 30  3 .29
yt  421.8
mm
I  8.55  109
mm 4
Cracking Moment M cr
N / mm 2
f ctr I 3.29(8.55  109


yt
421.8
M cr  66.7  10 6 N .mm  66.7 kN .m
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