Lec.8 strength design method t beams

Analysis & Design of
Reinforced Concrete Structures (1) Lecture.8 Strength Design Method
64
Dr. Muthanna Adil Najm
Reinforced concrete floor systems normally consist of slabs and beams that are
placed monolithically. As a result, the two parts act together to resist loads. In
effect, the beams have extra widths at their tops (flanges) resulting on a T shaped
beams (interior beams) or L shaped beams (exterior beams).
Effective flange width:
The ACI code section 8.10 gives limitations for the effective flange width of T-
beam and L-beams.
T-beam: (ACI 318-05 Section 8.10.2)
Effective flange width (b) is the smaller of:
1-
4
l
b  where: l = beam span.
2-  wbb clear distance between beams.
3- fw hbb 16
L-beam: (ACI 318-05 Section 8.10.3)
1-
12
l
bb w  where: l = beam span.
2-
2
1
 wbb clear distance between beams.
3- fw hbb 6
The analysis and design of T-beams depends on whether the neutral axis falls in
the flange or in the web.
Strength Design Method
Analysis & Design of T Beams.
b b
bw bwClear distance between beams
T-sectionL-section
d
fh
bwClear distance between beams

65
Analysis procedure:
1- Check the ACI minimum reinforcement requirements using b = bw.
db
f
f
A w
y
c
s


25.0
min, But not less than db
f
A w
y
s
4.1
min, 
2- calculate the depth of concrete compression block 'a' using b = flange width.
bf
fA
a
c
ys


85.0
A- if fha  , then analyze as a rectangular section with b = flange width and lever
arm =
2
a
d 
B- if fha  , then, theoretically the steel area shall be divided into two parts:
sfA which when stressed will balance the compression force of the overhanging
parts of the flange;   fwcysf hbbffA  85.0
 
y
fwc
sf
f
hbbf
A


85.0
This couple will produce Mn1 with a lever arm of 






2
fh
d







2
1
f
ysfn
h
dfAM
The remaining steel area sfs AA  , at stress fy is balanced by compression in the
web and produce an additional moment Mn2 .
 
wc
ysfs
bf
fAA
a



85.0
  






2
2
a
dfAAM ysfsn
wb
fh
b
d
N.A
Case (a): N.A within the flange.
wb
fh
b
d
N.A
Case (b): N.A within the web.

66
The total nominal resisting moment is the sum of the two parts.
21 nnn MMM 
3- Check the ACI requirements for maximum steel ratio:
db
A
w
s
w  and
db
A
w
sf
f 
fw   maxmax,
Where max is the maximum steel ratio defined for rectangular section.













 

005.0003.0
003.085.0 1
max
y
c
f
f
  Reduction factor 9.0













 

004.0003.0
003.085.0 1
max
y
c
f
f
  Reduction factor 9.0
Or instead, the designer can check the net tensile strain at the extreme tension steel
 t .
1
a
c  and 003.0




 

c
cd
t
If 005.0t  The section is ductile and 9.0
If 005.0004.0  t  The section is in transition zone and 9.0
If 004.0t  The section is not ductile .
Analysis Examples:
Ex.1)
Determine the design strength of the T-beam shown below if MPafc 28 and
MPafy 420 .
Sol.)
Check ACI minimum steel requirements,
  2
24554915 mmAs 
b = 1500 mm
100
600 mm
250 mm
5 Ø 25

67
0164.0
600250
2455



db
A
w
s
 > 0033.0
420
4.14.1
min 
yf
 .OK
Assume that N.A falls within the flange.
mmhmm
bf
fA
a f
c
ys
1009.28
15002885.0
4202455
85.0






 The N.A. falls within the flange, analyze as a rectangular section with b =
1500mm.
34
85.0
9.28
1


a
c
005.005.0003.0
34
34600
003.0 




 





 

c
cd
t
 Section is ductile and 9.0
mkN
a
dfAM ysu .4.543
2
9.28
60042024559.0
2













Ex.2)
Determine the design strength of the T-beam shown below if MPafc 28 and
MPafy 420 .
Sol.)
Check ACI minimum steel requirements,
  2
64328048 mmAs 
0245.0
750350
6432



db
A
w
s
 > 0033.0
420
4.14.1
min 
yf
 .OK
Assume that N.A falls within the flange.
mmhmm
bf
fA
a f
c
ys
1003.151
7502885.0
4206432
85.0






 T-section
    2
7.2266
420
1003507502885.085.0
mm
f
hbbf
A
y
fwc
sf 




b = 750 mm
100
750 mm
350 mm
8 Ø 32

68
2
3.41657.22666432 mmAA sfs 
  mm
bf
fAA
a
wc
ysfs
210
3502885.0
4203.4165
85.0







0245.0
750350
6432



db
A
w
s
w and 0086.0
750350
7.2266



db
A
w
sf
f
0181.0
005.0003.0
003.085.0 1
max 












 

y
c
f
f

0267.00086.0018.0maxmax,  fw 
max,ww  
 Section is ductile and 9.0
mkN
h
dfAM
f
ysfn .4.666
2
100
7504207.2266
2
1 












  mkN
a
dfAAM ysfsn .4.1128
2
210
7504203.4165
2
2 












The total nominal resisting moment is the sum of the two parts.
mkNMMM nnn .8.17944.11284.66621 
  mkNMM nu .3.16158.17949.0 
Design Procedure:
1- Determine the effective flange width per ACI requirements ( See page 59 )
2- Calculate the factored moment Mu.
3- Assume a = hf
Calculate As:








2
a
df
M
A
y
u
s

4- Check the assumed value of 'a'
bf
fA
a
c
ys


85.0
Where b = bf
A- if fha  , then design as a rectangular section.
B- if fha  , then, design as a ' T ' beam.
5- Calculate the amount of steel required to balanced the moment of the flange.
 
y
fwc
sf
f
hbbf
A


85.0







2
1
f
ysfu
h
dfAM  and 12 uuu MMM 
6- Calculate the amount of steel required to balanced the moment of the web.

69
2
2
2
bd
M
R u
u

 









y
u
f
R 

 2
2
2
11
1
dbA wsw 2 where:
c
y
f
f


85.0

7- Calculate total amount of steel
swsfs AAA 
8- Check ACI minimum steel requirement.
db
f
f
A w
y
c
s


25.0
min, But not less than db
f
A w
y
s
4.1
min, 
9- Check maximum steel ratio or section ductility as described previously.
Design Examples:
Ex.3)
Design a T-beam for the floor system shown below. The span is simple and is 6.0
m, the service applied moments are mkNMD .110 and mkNML .135 .
MPafc 28 and MPafy 420 .
Sol.)
1- mm
l
b 1500
4
6000
4
 (Controls)
2-  wbb clear distance between beams = 3000 mm .
3-   mmhbb fw 19001001630016 
    mkNMu .3481356.11102.1 
Assume a = hf
2
6
2301
2
100
4504209.0
10348
2
mm
a
df
M
A
y
u
s 


















Check the assumed value of 'a'
450 mm
300 mm
3000 mm 3000 mm3000 mm 3000 mm
= 100 mmfh

70
mmmm
bf
fA
a
c
ys
10027
15002885.0
4202301
85.0






 Design as a rectangular section,
2
6
2109
2
27
4504209.0
10348
2
mm
a
df
M
A
y
u
s 


















Check the obtained value of 'a'
mmmm
bf
fA
a
c
ys
1008.24
15002885.0
4202109
85.0






Calculate As with the revised 'a' value,
2
6
2104
2
8.24
4504209.0
10348
2
mm
a
df
M
A
y
u
s 


















2
2109mm .OK
Check minimum reinforcement,
  2
min, 4504503000033.0
4.1
2104 mmdb
f
AA w
y
ss  .OK
Check section ductility,
mm
a
c 2.29
85.0
8.24
1


005.0058.0003.0
2.29
2.29600
003.0 




 





 

c
cd
t
 Section is ductile and assumed reduction ( 9.0 ) is Ok.
Ex.4)
Design a T-beam for the floor system shown below. The span is simple and is 5.5
m, the service applied moments are mkNMD .270 and mkNML .575 .
MPafc 21 and MPafy 420 .
Sol.)
1- mm
l
b 1375
4
5500
4
 (Controls)
75 mm
600 mm
1800 mm 1800 mm1800 mm 1800 mm375 375 375

71
2-  wbb clear distance between beams = 1800 mm .
3-   mmhbb fw 1575751637516 
    mkNMu .12445756.12702.1 
Assume a = hf

















2
75
6004209.0
101244
2
6
a
df
M
A
y
u
s

2
5850mmAs 
Check the assumed value of 'a'
mmmm
bf
fA
a
c
ys
751.100
13752185.0
4205850
85.0






 Design as a T- section,
Calculate the amount of steel required to balanced the moment of the flange.
     2
3187
420
7537513752185.085.0
mm
f
hbbf
A
y
fwc
sf 




  mkN
h
dfAM
f
ysfu .64.677
2
75
60042031879.0
2
1 












mkNMMM uuu .36.56664.677124412 
Calculate the amount of steel required to balanced the moment of the web.
 
53.23
2185.0
420
85.0



c
y
f
f

  
66.4
6003759.0
1036.566
2
6
2
2
2 


bd
M
R u
u

  0131.0
420
53.2366.42
11
53.23
12
11
1 2
2 















y
u
f
R 


  2
2 29526003750131.0 mmdbA wsw  
Calculate total amount of steel:
2
613929523187 mmAAA swsfs 
Check minimum steel requirements:
  2
min, 7506003750033.0
4.1
6139 mmdb
f
AA w
y
ss  .OK
Check maximum steel requirements:
0273.0
600375
6139



db
A
w
s
w and 0142.0
600375
3187



db
A
w
sf
f
b = 1375 mm
75
600 mm
375 mm

72
Where max is the maximum steel ratio defined for rectangular section.
0135.0
005.0003.0
003.0
420
2185.085.0
005.0003.0
003.085.0 1
max 











 













 

y
c
f
f

0277.00142.00135.0maxmax,  fw 
0277.00273.0 max,  ww 
Section is ductile and assumed reduction factor ( 9.0 ) is Ok.

Lec.8 strength design method t beams

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