spring 2016 dr. alaa helba - delta...
TRANSCRIPT
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 1
CIV 223
Reinforced Concrete Structures - I
Spring 2016
Office hours: Sunday 11 - 1:00 pm
بسم هللا الرحمن الرحيم
Dr. Alaa Helba
For units: Use SI units (N, kN, mm, m)
For a rectangular R.C. Beam section let: b – width of rectangular R.C. section (Beam sec.)
t – total height of R.C. section
d – effective depth (distance to c.g of steel)
As – Area of steel
(Ex.: for 4f18 As=4*p(18)2/4 = 1018 mm2)
c - cover [50 mm/(1 row)–70/(2rows)–100/(3rows)]
Es - Modulus of Elasticity of steel = 200 kN/mm2
Ec - Modulus of Elasticity of concrete = 4400 fcu
n – modular ratio (Code suggests: n =10 in deflection & Mcr calculations
n = 15 in working stage analysis Mworking)
fcu – characteristic strength of conc. (20 or 25 or 30 N/mm2)
fy - yield stress of steel (240 or 360 or 400 N/mm2)
s
c
E
En
Basic units / Dimensions / Properties Review (lecture 2)
t
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 2
Calculation of Mcracking
Procedure: Transform the area of steel to
equivalent concrete, nAs
using the modular ratio:
1- Calculate the location of
the N.A. (e = ,yt = t/2 - e)
2- Calculate the total moment
of Inertia @ N.A.
3- Calculate the cracking
moment Mcr based on
fbottom = fctr = 0.6 fcu t
tctrcr
y
IfM
223
)()(12
)(
cynAebtbt
I
centrefromshifttyeccentricie
tst
10c
s
E
En
))((2
)0(
.sec...(@
enAbtct
nAbt
conofLCAreasofMoments
ss
Review (lecture 2)
Example # 1 Calculate the cracking moment Mcr
+ve and Mcr -ve for the
singly reinforced rectangular section shown with dimensions
b x t = 250 x 800 mm, if As = 5 f 16 mm. materials used are
concrete fcu=25 N/mm2 and st. grade 360/520.
Solution
a- case of Mcr +ve
1- For - section : Calculation of Mcr
+ve and Mcr-ve for beam sections
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 3
410
223
223
22
1019.1
)503.383)(1005(10)7.16)(800(25012
)800(250
)()(12
3.3837.164002
7.16
)]1005(10)800)(250[(50400)1005(10
))((2
)0(
)sec..(@@
1005)4
16(5165
mm
cynAebtbt
I
mmet
y
mme
e
enAbtct
nAbt
tionconcreteofLClinemiddleAreasofMoments
mmAfor
ts
t
ss
s
pf
mkN
mmN
y
IfMMomentCracking
mmI
mmy
mmN
ff
y
IfM
MomentCracking
t
ctrcr
t
cuctr
t
ctrcr
.1.93
.1031.9
3.383
)1019.1(3
1019.1
3.383
/3256.0
6.0
7
10
410
2
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 4
Example # 1 - b
- case of Mcr -ve
for the rectangular sections of the same dimensions with the
same AS (in tension side) the positive cracking moment Mcr +ve = the negative cracking moment Mcr
-ve
2- For T - section : Calculation of Mcr
+ve and Mcr-ve for beam sections
for the T - sections of the same dimensions with the same AS
(in tension side) the positive cracking moment Mcr +ve ≠ the
negative cracking moment Mcr -ve
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 5
186mm70
mm510
mm120
mmb 200
mmB 800
Calculate the cracking moment Mcr +ve and Mcr
-ve for the
singly reinforced T - section shown with dimensions b x t =
200 x 700 mm and B x ts = 800 x 120 mm, if As = 5 f 16
mm. materials used are concrete fcu=30 N/mm2 and st. grade
400/600.
Example # 2
Solution
a- case of Mcr +ve
Example # 2 - b
- case of Mcr -ve
186
mm580
mm120
mmb 200
mmB 800
- Solve it yourself.
- (without calculations), For T – sections of same dimensions
and steel area, Whichever is bigger Mcr+ve or Mcr
-ve , and why?
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 6
11
Lecture # 3
R.C. Fundamentals
Analysis of RC Sections under Flexure
Stage I: Uncracked Section
Stage II: Cracked Section in the working stage
Stage III: Cracked Section at ultimate stage
Analysis of cracked Section:
1- Determine the position of
the neutral axis, N.A , (z)
2- calculate the inertia of
the cracked section (Icr).
3-Calculate the working moment
Mw using the maximum allowable stresses
fc(all) in conc. and fs(all) for steel as follows:
Stage 2: Cracked section in Working stage
a rectangular section reinforced with steel As.
Steps
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 7
Calculation of working moment Mw
1- for a singly reinforced section
n As
(n=15)
fc
fs/n
Calculation of working moment Mw for a cracked Doubly Reinforced section
1- Determine the position of N.A) (z): use (n = 15)
Taking moments of areas @ the N.A.
2- Calculate the inertia:
0dnAznAz2
b
0zdnA)2
z(bz
ss
2
s
2s
3
cr zdnA3
bzI
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 8
z
IfM
crallc
w
)(
1
3- Calculate the working moment Mw
Mw is the lesser of Mw1 , Mw2
)(
)(
2zd
In
f
Mcr
alls
w
Calculation of working moment Mw
2- for a doubly reinforced section
n As’
n As
(n=15)
fc
fs/n
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 9
Calculation of working moment Mw for a cracked Doubly Reinforced section
1- Determine the position of N.A) (z):
Taking moments of areas @ the N.A.
2- Calculate the inertia:
0)()(2
2
'''2
''
dAdAnzAAnzb
zdnAdznAz
bz
ssss
ss
2''23
3dznAzdnA
bzI sscr
z
IfM
crallc
w
)(
1
3- Calculate the working moment Mw
Mw is the lesser of Mw1 , Mw2
)(
)(
2zd
In
f
Mcr
alls
w
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 10
Working stresses used in Calculation of working
moment (stage II - cracked section)
Allowable stresses in concrete and steel according to ECP
Code 203 – Version 2007 From Table 5 – 1:
- Concrete in Case of - sections > 200 mm under flexure:
30 25 20 fcu
10.5 9.5 8 fc all - flexure
400 360 240 fy
220 200 140 fs all
- Steel reinforcement
Table 5-1
Working
stresses in
concrete
and steel
according
to ECP 203
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 11
Mw is the least of Mw1 , Mw2
z
IfM
cravc
w
)(
1
M+vew
)(avcf
)(allcf
For T - section use :
fc(av) = 2/3 fc(all)
For -section use:
fc = fc(all) = allowable compression
(bending) stress in concrete
working stage (from table)
█-section
working stresses in conc. in Case of
T- sections under flexure:
Calculation of working moment M+vew
for T-sec. 1- Determine the position of N.A) (z):
Taking moments of areas @ the N.A.
02
2
2
dnAznAzB
zdnAz
Bz
ss
s
mmz
assume z < ts
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 12
in case of z < ts
the assumption is O.k.
Complete the solution, calculate Icr then Mw is the
lesser of Mw1 & Mw2
in case of z > ts then the assumption is wrong
Calculate another z Taking moments of areas @ the
new N.A. then Complete the solution, calculate Icr
then Mw is the lesser of Mw1 & Mw2
- Calculation of working moment M+vew for T-sec.
Case of z < ts
2- Calculate the inertia:
23
3zdnA
BzI scr
z
If
Mcrallc
w
)(
13
2
Mw is the lesser of Mw1 , Mw2
)(
)(
2zd
In
f
Mcr
alls
w
3- Calculate the working moment Mw
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 13
- Calculation of working moment M+vew for T-sec.
Case of z > ts
1- Taking moments of areas @ the N.A.
0]2
)([])([2
)2
()(2
2
ssss
ss
s
tbBdnAztbBnAz
b
zdnAt
ztbBz
bz
mmz
2- Calculate the inertia:
2233
)2
()(12
)(3
zdnAt
ztbBt
bBbz
I ss
ss
cr
z
If
Mcrallc
w
)(
13
2
)(
)(
2zd
In
f
Mcr
alls
w
3- Calculate the working moment Mw
Mw is the lesser of
Mw1 , Mw2
3- Calculation of working moment M-vew for T-sec.
Solve as rectangular
section as before
M-vew
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 14
Example - 3
‘ 2 For the section shown
Calculate Mw
Solution
‘ 2
‘ 2
1- Determine the position of N.A) (z): use (n = 15)
Taking moments of areas @ the N.A.
2- Calculate the inertia:
0)()(2
2
'''2
''
dAdAnzAAnzb
zdnAdznAz
bz
ssss
ss
As = 5 * 201 = 1005 mm2 ,
A’s = 3 * 113 = 339 mm2
mmz
zz
zz
234
0924843.161
0)]50(339)750(1005[15)3391005(15125
2
2
2''23
3dznAzdnA
bzI sscr
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 15
‘ 2
‘ 2 3- Calculate the working moment Mw:
2''23
3dznAzdnA
bzI sscr
49
223
1025.5
50234)339(15234750)1005(153
)234(250
mm
mkNmmN
z
IfM
crallc
w
.1.213.101.213
234
)1025.5(5.9
6
9)(
1
Mw is the lesser of Mw1 , Mw2
Mw = 135.7 kN.m
mkN
zd
In
f
Mcr
alls
w
.7.135
10)234750(
)1025.5(15
200
)( 6
9)(
2
Example - 4
Calculation of working moment Mw – T - section
186mm70
mm510
mm120
mmb 200
mmh 700
mmB 800For the section shown
Calculate Mw
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 16
Solution
1- Determine the position of N.A) (z):
Taking moments of areas @ the N.A.
02
2
2
dnAznAzB
zdnAz
Bz
ss
s
186
mmB 800
mmb 200
As = 6 * 254 = 1524 mm2
mmz
zz
zz
3.163
05.3600415.57
0)630)(1524(15)1524(15400
2
2
mmts 120
the assumption is wrong
z > ts
Taking moments of areas @ the new N.A.
> ts = 120 mm
mmz
zz
zz
7.167
01872186.948
0]2
120)200800()630)(1524(15[
)]120)(200800[()1524(15[100
2
2
2
186
mmB 800
mmb 200
mmb 200
mmts 120
0]2
)([])([2
)2
()(2
2
ssss
ss
s
tbBdnAztbBnAz
b
zdnAt
ztbBz
bz
z > ts
Taking moments of areas @ the new N.A.
Feb-16
Lec. # 3 - CIV 223 - Dr. A. Helba spring
2016 17
z > ts
49
22
33
101.6
)7.167630)(1524(15)607.167(120)200800(
12
)120)(200800(
3
)7.167(200
mm
Icr
186
mm800
mm200
mmb 200
mmts 120
2- Calculate the inertia Icr :
22
33
)2
()(
12)(
3
zdnAt
ztbB
tbB
bzI
ss
s
scr
z
IfM
cravc
w
)(
1
3- Calculate the working moment Mw :
mkN
z
IfM
cravc
w
.4.230
107.167
)101.6)(5.9(3
2
6
9
)(
1
Mw is the lesser of Mw1 , Mw2
Mw = 193.5 kN.m
mkN
zd
In
f
Mcr
alls
w
.5.193
10)7.167630(
)101.6(15
220
)( 6
9)(
2