Lec06 Analysis and Design of T Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

Lec06 Analysis and Design of T Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

1. 25-Feb-13
CE370: Prof. A. Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif

Analysis and Design of T-Beams
•T-shaped beams are frequently used in structures
•There are two types of T-beams :
 Beams directly cast and delivered as isolated
T-beams (especially in bridges)
T-shaped beams resulting from interaction of slabs
with beams (building slabs)
•The obvious advantage of T-beams is to reduce useless
tensile concrete quantity
T-Beams
2

2. 25-Feb-13
CE370: Prof. A. Charif 2
Isolated T-beams

T-Beams
Slab interaction
T-beams

3
A beam is considered as such even if it has a larger base (heel) to
accommodate tension steel.
The shape and size of concrete on the tension side, assumed to be
cracked, has no effect on the theoretical resisting moments.
T-Beams
Precast T-Beams for bridges
4

3. 25-Feb-13
CE370: Prof. A. Charif 3
T-Beams
RC floors normally consist of slabs and beams that are cast
monolithically. The two act together to resist loads and because of
this interaction, the effective section of the beam is a T or L section.
T-section for
interior beams

L-section for
exterior beams
5
T-Beams
The effective flange width resulting from slab interaction
depends on the slab type (one-way or two-way slab).
The present course focuses on one-way slabs.
6

4. 25-Feb-13
CE370: Prof. A. Charif 4
Exterior beam – Minimum of:
Beam tributary width
Web width plus six times slab
thickness
Web width plus one-twelfth
of beam span
Interior Beam – Minimum of:
Beam tributary width
Web width plus 16 times slab
thickness
One-quarter of beam span
Effective flange width in one way slabs
7
Beam tributary width
Beam tributary width is the transverse width of the slab
transmitting load to the beam (supported by the beam)
It is determined using mid-lines between beam lines
For exterior beams, offsets are included

Figure shows tributary widths for beams A and B
A
B
C
L
t
L
t
8

5. 25-Feb-13
CE370: Prof. A. Charif 5
Flange Dimensions
(Isolated T-Beams)
For an isolated T-beam:
The flange thickness may
not be less than one-half
the web width
The effective flange
width may not be larger
than four times the web
width
wf
w
f
bb
b
h 4
2

9
Location of neutral axis and
Rectangular stress block depth a
•Neutral axis (N.A.) for T-beams can
fall either in the flange or in the web
•At ultimate state, it is the depth of
the rectangular stress block that
counts.
If the stress block falls in the flange, as is very frequent for
positive moments, the rectangular beam formulas apply.
In this case tensile concrete is assumed to be cracked, and its
shape has no effect (other than weight).
The section is analyzed as a rectangular one using the flange
width b = b
f
10

6. 25-Feb-13
CE370: Prof. A. Charif 6
Location of rectangular
stress block (Contd.)
If the stress block falls in the
web, the compression area
does not consist of a single
rectangle, and the rectangular
beam design procedure does
not apply.
The section is in this case
divided in two parts as will be
seen later (decomposition
method)

11
Case of a negative moment
•If the T-section is subjected to a negative moment,
then the flange will be in tension
•The section can thus be analyzed as an inverted
rectangular one using the web width b = b
w
12

7. 25-Feb-13
CE370: Prof. A. Charif 7
Minimum Steel for T-Beams
•Minimum steel quantity is imposed by codes for the
same reasons as in rectangular beams

db
ff
f
A
w
yy
c
s









4.1
,
4
Max
'
min,
Exception: For a statically determinate T-beam subjected
to a negative moment (cantilever), the minimum steel
area is:
According to SBC/ACI: db
f
f
A
w
y
c
s
2
'
min,

13
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif

Analysis of T-Beams

8. 25-Feb-13
CE370: Prof. A. Charif 8
15
Analysis of T-Beams
Yielding of Tension Steel
•With a larger compression area from the flange, a T-
beam is usually under reinforced and tension steel
should yield before failure.
•We therefore first present the method of analysis in
case of steel yielding.
•Case of steel not yielding is presented later. methodion decomposit use in web,Block :If
continue flange,in Block OK :If
85.0
85.0:mequilibriu Force
'
'
f
f
fc
ys
ysfc
ha
ha
bf
fA
afAabf



16
Location of rectangular stress block
in Analysis Case
•Assume stress block in
flange
•Express force equilibrium
•Deduce value of a and
check. 2
a
d w
b s
A f
b f
h fc
abf
'
85.0 a ys
fA

9. 25-Feb-13
CE370: Prof. A. Charif 9
Analysis steps for T-Beams nn
f
f
f
fc
ys
s
MM
c
cd
β
a
c
a
ha
b
ha
bf
fA
a
A


and , Calculate 6.
case yielding no goto yieldingnot steel If.5
003.0strain steel and depth axis neutral Calculate .4
of valuenew find tomethodion decomposit Use
webreachesblock Stress If
4 goto , width flange usingsection r rectangula a as
analysis continue flange,in block Stress If
thicknessflange depth withblock Compare .3
85.0
depth its determine and flangein block stress Assume 2.
area steel minimumCheck 1.
s
1
'
min





17
Decomposition Method (1)
•When the stress block exceeds the flange, the section is divided:
T-section = W-section + F-section
•W-section = Compression part of the web (unknown depth a)
•F-section = Overhanging parts of the flange (known depth h
f)
18 w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d
Total steel area and nominal moment are decomposed: nwnfnswsfs
MMMAAA 

10. 25-Feb-13
CE370: Prof. A. Charif 10
Decomposition Method (2)
19 sfssw
y
fwc
sfysffwcf
AAA
f
hbbf
AfAhbbfC 

 and
)(85.0
)(85.0
'
'
Force equilibrium in the flange part gives the two steel areas : w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d
Force equilibrium in the web gives the compression block depth : w
'
c
ysw
yswwwcw
bf.
fA
afATabfC
850
85.0
'
 































2

2

2

2

a
dfA
h
dfAMMM
a
dfAM
h
dfAM
ysw
f
ysfnwnfn
yswnw
f
ysfnf
Decomposition Method (3)
20
Using appropriate lever arms, the nominal moments are thus : w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d

11. 25-Feb-13
CE370: Prof. A. Charif 11
Problem 1 m 1.5 is width tributaryBeam thick.mm 100 is that slabfloor awith
integrallycast isspan m 9 with beam The MPa. 420 and MPa 30
figure. in theshown beam Tinterior ofmoment design Determine

y
'
c ff
21 600d 250 286 widthEffective 100 mmd
mmd
t
655
545
min


Solution 1 mm 1500
mm 22504/90004Span /
mm 18501001625016
1500width Tributary
Min
: widthFlange Effective









bhb
mml
fw
t  
OK A mm 5.369428
4
6 and
mm 0.50060025000333.0,00326.0Max
600250
420
4.1
,
4204
30
Max
4.1
,
4
Max
Checking
min s
22
2
min min
'
min
min




























ss
ss
w
yy
c
s
s
AA
AA
db
ff
f
A
A

22

12. 25-Feb-13
CE370: Prof. A. Charif 12 :flangein block stress Assume 0.90 controlTension 005.0 005.00313.0
72.47
72.47545
003.0003.0:checkStrain
mm 72.47
85.0
56.40
flangein isblock stressOK mm 100 a
mm 56.40
15003085.0
4205.2694
85.0
min
min
min
1
'





















t
f
fc
ys
εε
c
cd
ε
β
a
c
h
bf
fA
a
Solution 1 – Cont.
23 600d 250 286 mm1500 widthEffective 100 mmd
mmd
t
655
545
min

 layer bottomat control-nfor tensiocheck must Then we
: 005.0but If :Note
minmin
 
y mkNM
mkNmmNM
M
a
dfAM
n
n
n
ysn
.6.80955.8999.0 :momentDesign
.55.899.1055.899
2
56.40
6004205.3694
2
:moment Nominal
6

















Solution 1 – Cont.
The T-beam can therefore resist any ultimate moment
equal to or less than 809.6 kN.m
24 600d 250 286 mm1500 widthEffective 100 mmd
mmd
t
655
545
min



13. 25-Feb-13
CE370: Prof. A. Charif 13
Problem 2
Analysis of a T-section with six 20-mm
bars in two layers as shown.
Net layer spacing S
l = 30 mm
Stirrup diameter = 10 mm MPafMPaf
yc
42020
'

Steel depths :



mmddmmdd
mm
dd
AA
AdAd
d
mmdSdd
mmd
d
hdd
t
ss
ss
bl
s
b
t
490540
515
2
49050540)2030(540) (
54060600)10
2
20
40(600)
2
cover (
2min1
21
21
2211
12
1









600
75
525
300



Total steel area is : 2
2
96.1884
4
20
6 mmA
s


It is greater than the minimum steel area: 2
min
'
min
515515300
420
4.1
515300
420
4.1
,515300
4204
20
Max
4.1
,
4
Max
mm
x
A
db
f
db
f
f
A
s
w
y
w
y
c
s
















Solution 2
First assume compression block in the flange (a ≤ h
f ) and
if true analyze as a rectangular section (b
f , h).
If not, use decomposition method.

14. 25-Feb-13
CE370: Prof. A. Charif 14
Force equilibrium C = T gives the compression block depth is : mm
bf
fA
a
fc
ys
616.77
6002085.0
42096.1884
85.0
'




This value is greater than the flange thickness a > h
f
Compression block is thus in the web
Use decomposition method:
T-section = W-section + F-section







nfnwn
sfsws
MMM
AAA  
2
2
'
246.974
714.910
420
75)300600(2085.0
and
85.0
mmAAA
mmA
AAA
f
hbbf
A
sfssw
sf
sfssw
y
fwfc
sf







Solution 2 – Cont. mm
a
c
hamma
bf
fA
afAabf
f
wc
ysw
yswwc
39.94
85.0
232.80
)(232.80
3002085.0
420246.974
85.0
85.0
1
'
'







New value of compression
block depth is obtained
from the force equilibrium
in the web part : control-Tension 005.0005.0
0021.001257.0
39.94
39.94490
003.0003.0
min
min
min
OK
OK
c
cd
t
y








Steel strain check at minimum depth:
Solution 2 – Cont.
No need to calculate strain at bottom layer.

15. 25-Feb-13
CE370: Prof. A. Charif 15
Nominal moment: mkNM
mkNMMM
mkNmmNM
a
dfAM
mkNmmNM
h
dfAM
n
nfnwn
nw
yswnw
nf
f
ysfnf
.26.339958.37690.0
.958.376
.315.194.194314612
2
232.80
515420246.974
2
.364.182.182643693
2
75
515420714.910
2


































Solution 2 – Cont.
Problem 3 MPa 420 MPa 30 
y
'
c
ff
30 750d 350 328 750 widthEffective 100 mmd
mmd
t
860
640
min


Compute the design moment for the shown T-beam with all
dimensions in mm.  
OK A
mm 0.875
75035000333.0,00326.0Max
750350
420
4.1
,
4204
30
Max
4.1
,
4
Max
mm 0.643432
4
8
min s
2
min
min
'
min
22






























s
s
s
w
yy
c
s
s
A
A
A
db
ff
f
A
A


16. 25-Feb-13
CE370: Prof. A. Charif 16 2
2
'
43.400557.24280.6434
57.2428
420
)100350750(3085.0)(85.0
:With
mmAAA
mm
f
hbbf
A
sfssw
y
fwc
sf





 mm
a
c
mmhmm
bf.
fA
a
f
w
'
c
ysw
75.221
85.0
49.188
: isdepth axis Neutral
)100(49.188
3503085.0
42043.4005
850
: isdepth block n Compressio
1






Solution 3
31 nfnwnsfsws
f
fc
ys
MMMAAA
mmhmm
bf
fA
a






:ion decomposit using Analyzein web liesblock Stress
1003.141
7503085.0
4206434
85.0
:flangein block stress Assume
' mkNM
mkNMMM
mmN
a
dfAM
mmN
h
dfAM
c
cd
n
nwnfn
yswnw
f
ysfnf
y
t
.48.16352.181790.0 :moment Design
.2.1817 :moment nominal Total
.102.1103
2
49.188
75042043.4005
2

.100.714
2
100
75042057.2428
2

:are moments Nominal
layer bottomat control-nfor tensiocheck must Then we
: 005.0but If :Note
0.90 controlTension OK 005.0005.0
00566.0
75.221
75.221640
003.0 003.0:check Strain
6
6
minmin
min
min
min










































Solution 3 – Cont.
32

17. 25-Feb-13
CE370: Prof. A. Charif 17
33
Analysis of T-beams with tension steel not yielding   

























2
:check003.0
:continueif1
4
1
285.0
600
formula. rectangle Use:)( flange in thefirst it assume still weweb,
in thelikely more isblock n compressio theyielding, steel noith Although w
6000030
003.0 :ThenIf
1
1
'
a
dfAMεEf
c
cd
haca
P
dP
c
bf
A
P
ha
c
cd
A
c
cd
.EAT
c
cd
EAfAT
ssnsssyss
f
fc
s
f
sss
ssssssys




Situations where steel does not yield at failure in T-beams are very
rare. The case of one tension steel layer is treated here. With many
layers, non yielding must be solved using strain compatibility
method (as was done with rectangular beams).
34
Analysis of T-beams with tension steel not yielding
Decomposition (more likely)  
 
 
sssyss
w
fwf
wc
s
sfwfcwcwf
wcwcwfwfcf
wfsf
εEf
c
cd
ca
QP
PdQP
c
b
hbb
Q
bf
A
PPdcQPc
c
cd
AhbbfcbfCCT
cbfabfChbbfC
CCC
c
cd
ATha




























:check003.0and: Deduce
1
)(
4
1
2
)(
: issolution Positive
)(
85.0
600
with0)(
60085.085.0
85.085.085.0
600:methodion decomposit useif
1
2
11
'
2
'
1
'
1
'''

18. 25-Feb-13
CE370: Prof. A. Charif 18
35
Analysis of T-beams with tension steel not yielding
Decomposition - Continued  
sfssw
s
fwfc
sf
sswnw
f
ssfnf
nwnfn
sssyss
AAA
f
hbbf
A
a
dfAM
h
dfAM
MMM
εEf
c
cd
ca























and
)(85.0
22
:check003.0
'
1

36 600d 100 750 350 328
Problem 4 MPa 420 MPa 20 
y
'
c
ff
Compute the design moment for the shown T-beam with all
dimensions in mm. The large base (heel) allows many bars in a
single layer and does not change the T-section behavior.  
OK A
mm 0.700
60035000333.0,00266.0Max
600350
420
4.1
,
4204
20
Max
4.1
,
4
Max
mm 0.643432
4
8
min s
2
min
min
'
min
22






























s
s
s
w
yy
c
s
s
A
A
A
db
ff
f
A
A


19. 25-Feb-13
CE370: Prof. A. Charif 19 2
2
'
95.481405.16190.6434
05.1619
420
)100350750(2085.0)(85.0
:With
mmAAA
mm
f
hbbf
A
sfssw
y
fwc
sf





 mm
a
c
mmhmm
bf.
fA
a
f
w
'
c
ysw
86.399
85.0
88.339
: isdepth axis Neutral
)100(88.339
3502085.0
42095.4814
850
: isdepth block n Compressio
1






Solution 4
37 nfnwnsfsws
f
fc
ys
MMMAAA
mmhmm
bf
fA
a






:ion decomposit using Analyzein web liesblock Stress
10094.211
7502085.0
4206434
85.0
:flangein block stress and yielding steel Assume
' mmc
QP
b
hbb
Q
bf
A
P
QP
PdQP
c
c
cd
w
fwf
wc
s
y
s
20.3631
)45.13430.763(
60030.7634
1
2
)45.13430.763(
45.134
85.0350
100)350750(
30.763
85.03502085.0
6434600
)(
85.0
600
with1
)(
4
1
2
)(
:ion decomposit theuse and webin the isit hat directly t assumerather We
check and flange in theblock n compressio first the assumeagain can We
web)in the isblock stresst likely tha still(but : yieldingNot
00150.0
86.399
86.399600
003.0 003.0:strain Steel
2
11
'2
s














































Solution 4 – Cont.
38

20. 25-Feb-13
CE370: Prof. A. Charif 20 2
2
'
1
76.469524.17380.6434
24.1738
2.391
001)350750(2085.0)(85.0
2.391001956.0200000
: confirmedyieldingNo
001956.0
2.363
2.363600
003.0003.0
assumed as webin theblock nCompressio
72.30820.36385.020.363
mmAAA
mm
f
hbbf
A
MPaεEf
c
cd
mmcammc
sfssw
s
fwfc
sf
sss
ys
s


















Solution 4 – Cont.
39 mkNM
mkNMMM
mmNM
a
dfAM
mmNM
h
dfAM
n
nwnfn
nw
sswnw
nf
f
ssfnf
.19.7756.119265.0yielding) No(65.0
.6.1192
.106.818
2
72.308
6002.39176.4695
2
.100.374
2
100
6002.39124.1738
2
6
6


































Solution 4 – Cont.
40

21. 25-Feb-13
CE370: Prof. A. Charif 21
CE 370
REINFORCED CONCRETE-I
Prof. A. Charif

Design of T Beams
Design of T-Beams
The same minimum thickness for deflection control is used for
rectangular or T-beams. Design steps are :
•Determine all dimensions of T-section
•Estimate steel depth according to expected number of layers
•Determine location of compression block
•If stress block is in flange, design as a rectangular section using
the flange width
•If stress block is in web, use decomposition method for design
•Check minimum steel
•Perform strain checks using actual provided steel

•Location of compression block determined by comparing
design moment capacity of full flange with ultimate moment.
42 CE 370 : Prof. Abdelhamid Charif

22. 25-Feb-13
CE370: Prof. A. Charif 22 2
fh
d w
b s
A f
b f
h ffc
hbf
'
85.0
Moment capacity of full flange nff
f
ffcnff
M
h
dhbfM










flange full ofmomentDesign
2
85.0:flange full ofmoment Nominal
'
43 CE 370 : Prof. Abdelhamid Charif
Location of rectangular stress block
•The location of the stress block (flange or web) is determined
with the following steps:
1.Compute design moment capacity of full flange
2.Compare this moment with given ultimate moment methodion decomposit e Us
)( webinblock Stress:If
h with widtbeamr rectangula a asDesign
)( flangeinblockStress:If
2
85.0
'
funff
f
funff
f
ffcnff
haMM
b
haMM
h
dhbfM













44 CE 370 : Prof. Abdelhamid Charif

23. 25-Feb-13
CE370: Prof. A. Charif 23
Decomposition Method in Design (1)
•When the stress block exceeds the flange, the section is divided:
T-section = W-section + F-section
•W-section = Compression part of the web (unknown depth a)
•F-section = Overhanging parts of the flange (known depth h
f)
45 CE 370 : Prof. Abdelhamid Charif w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d
Decomposition Method in Design (2) nffnf
f
ysfnf
y
fwc
sfysffwc
sfswsfs
MM
h
dfAM
f
hbbf
AfAhbbf
AAAA













:Note
2
:part flange ofmoment Nominal
)(85.0
)(85.0
: givespart flange of mEquilibriu
'
'
46 CE 370 : Prof. Abdelhamid Charif w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d

24. 25-Feb-13
CE370: Prof. A. Charif 24
Decomposition Method in Design (3)  
nfuwu
w
nfu
f
ysfnf
MMM
b
MM
h
dfAM














moment ultimate reduced a tosubjectedwhen
th width section wir rectangula a as designed thereforeispart webThe
part. webby the taken is moment ultimate remaining The
2
toequalmoment ultimatean resistspart Flange
47 CE 370 : Prof. Abdelhamid Charif w
b s
A f
bb f
h a d 2
a
d w
b sw
A a w
b sf
A 2/)(
w
bb f
h 2
fh
d
Decomposition Method in Design (4)
Steel area component A
sw is the solution of quadratic equation : 22'
'
with
7.1
4
11
85.0
db
MM
db
M
R
f
R
f
dbf
A
w
nfu
w
wu
wn
c
wn
y
wc
sw













Total steel area A
sf + A
sw must then be compared to the minimum
value A
smin .

The new value of the stress block depth (to be used for strain check)
is obtained from force equilibrium in the web using the actual
provided steel area. wc
ypsw
sfpspsw
bf
fA
aAAA
'
,
,,
85.0

48 CE 370 : Prof. Abdelhamid Charif

25. 25-Feb-13
CE370: Prof. A. Charif 25
Design Problem-1 MPafMPaf
mmdb
kN.m
y
'
c
w
420 30 : Take
expected).layer (onely respective 450 and 300 asgiven are and
m 6 isspan Beam 300.0 ofmoment ultimatean
tosubjected when belowshown systemfloor for the beam T Design the
 mm 300
w
b mm 100
f
h mm 450d m 3 s
A s
A s
A m 3 m 3 m 3
49 CE 370 : Prof. Abdelhamid Charif
Solution 1 mmb
mm
mmhb
mmm
ffw 1500
15004/6000Span/4
19001001630016
30000.3
2
0.3
2
3.0
width Tributary
ofLesser : widthFlange Effective










 mmb
haMM
mkNM
mkNmmNM
h
dhbfM
f
funff
nff
nff
f
ffcnff
1500th width section wir rectangula a asDesign
)( flangein block Stress
.0.1377153090.0momentDesign
.0.1530.101530
2
100
45010015003085.0
2
85.0
:flange full ofmoment Nominal
6
'






















50 CE 370 : Prof. Abdelhamid Charif

26. 25-Feb-13
CE370: Prof. A. Charif 26
Required steel area A
s is given by : OK islayer One)5.1963(254 use We
4.1803
307.1
974.0.14
11
420
45015003085.0
0974.1
450150090.0
10300
with
7.1
4
11
85.0
2
,
2
2
6
2'
'
mmA
mmA
R
db
M
R
f
R
f
dbf
A
ps
s
n
f
u
n
c
n
y
fc
s






























Solution 1 – Cont.
51 CE 370 : Prof. Abdelhamid Charif
Designed section OK A 450
450300
420
4.1
,
4204
30
Max
4.1
,
4
Max
Checking
min s
2
min
min,
'
min,
min





















ss
s
w
yy
c
s
s
AmmA
A
db
ff
f
A
A 450 300 254 1500 width Effective 100    
0.90 controlTension OK 005.00502.0
003.0
365.25
365.25450
003.0
365.25
85.0
56.21
depth axis Neutral
) 100 ( 56.21
15003085.0
4205.1963
85.0
depth block Stress
1
'
,




















t
t
f
fc
yps
ε
c
cd
ε
mm
β
a
c
mmhmm
bf
fA
a
52 CE 370 : Prof. Abdelhamid Charif

27. 25-Feb-13
CE370: Prof. A. Charif 27
Design Problem-2 mm 375
w
b mm 75
f
h mm 700h m .81 s
A s
A s
A m .81 m .81 m .81 mmhd
MPafMPaf
mmhb
kN.m
y
'
c
w
61090 : asdepth steel theestimate welayers, 2 Expecting
420 22
ly respective 700 and 375 asgiven are and
m 5.4 isspan Beam 1250.0 ofmoment ultimatean
tosubjected when belowshown systemfloor for the beam T Design the


53 CE 370 : Prof. Abdelhamid Charif
Solution 2 mmb
mm
mmhb
mmm
ffw 1350
13504/5400Span/4
1575751637516
18008.1
2
8.1
2
1.8
width Tributary
ofLesser : widthFlange Effective










 methodion decomposit usingDesign
)( in webblock Stress).1250(
.0.6.975108490.0momentDesign
.0.1084.101084
2
75
6107513502285.0
2
85.0
:flange full ofmoment Nominal
6
'




















funff
nff
nff
f
ffcnff
hamkNMM
mkNM
mkNmmNM
h
dhbfM


54 CE 370 : Prof. Abdelhamid Charif

28. 25-Feb-13
CE370: Prof. A. Charif 28 mkNmmNM
h
dfAM
mmA
f
hbbf
AfAhbbf
A
nf
f
ysfnf
sf
y
fwfc
sfysffwfc
sf
.86.782.1086.782
2
75
6104208.3255
2
:part flange ofcapacity Nominal
8.3255
420
)753751350(2285.0
)(85.0
)(85.0
: steel required givespart flange in the mEquilibriu
6
2
'
'





















Solution 2 – Cont. mkNM
MMM
b
wu
nfuwu
w
.426.54586.7829.01250
moment ultimate reduced aunder
th width section wir rectangula a as designedpart Web


55 CE 370 : Prof. Abdelhamid Charif
Steel area component A
sw given by : 2
2
6
2'
'
7.2731
227.1
343.44
11
420
6103752285.0
343.4
61037590.0
10426.545
with
7.1
4
11
85.0
mmA
R
db
M
R
f
R
f
dbf
A
sw
wn
w
wu
wn
c
wn
y
wc
sw





























Solution 2 – Cont. 2
5.59877.27318.3255 :areasteelTotal mmAAA
swsfs

56 CE 370 : Prof. Abdelhamid Charif

29. 25-Feb-13
CE370: Prof. A. Charif 29 OK A 5.762
610375
420
4.1
,
4204
22
Max
4.1
,
4
Max
Checking
min s
2
min
min,
'
min,
min





















ss
s
w
yy
c
s
s
AmmA
A
db
ff
f
A
A
Solution 2 – Cont. )5.6157( 2810 requires This
5.5987 :areasteelrequiredTotal
2
,
2
mmA
mm
ps
 375 2810 3501 widthEffective 75 610
57 CE 370 : Prof. Abdelhamid Charif required. isdesign -re if see check tomoment Perform
mm) (610 valueAssumed0.607
2
5783028
63664700)141040(
OKbarsfor ten layers Two5
05.5
3028
0422810630375
cover26
:spacinglayer and spacingbar for mm 30 Assuming
:layer onein 28 bars ofnumber Maximum
21
11min2
1
max















mm
dd
d
mmdSdddd
mmhdd
n
n
Sd
ddSb
n
lb
t
bb
bsb
Solution 2 – Cont.
58 CE 370 : Prof. Abdelhamid Charif

30. 25-Feb-13
CE370: Prof. A. Charif 30
Solution 2 – Cont. layer bottomat control-nfor tensiocheck must Then we
: 005.0but If :Note
0.90 controlTension OK 005.000548.0
46.204
46.204578
003.0003.0
46.204
85.0
79.173
depth axis Neutral
)75(79.173
3752285.0
4207.2901
85.0
: isdepth block Stress
7.29018.32555.6157 :part websteel Actual
entreinforcem provided actual theusing performed bemust Checks
minmin
min
min
min
1
'
,
2
,,















y
f
wc
ypsw
sfpspsw
ε
c
cd
ε
mm
β
a
c
mmhmma
bf
fA
a
mmAAA
59 CE 370 : Prof. Abdelhamid Charif
Moment check

60 375 3501
f
b 75 607 2810 79.173
a
The moment check is necessary as
the final steel depth (607) is less than
the initially assumed value (610). requireddesign -re NoOK).1250(
.35.127161.141290.0 :moment Design
.61.1412 :moment nominal Total
.1086.633
2
79.173
6074207.2901
2

.1075.778
2
75
6074208.3255
2

:are moments Nominal
6
,
6
































mkNMM
mkNM
mkNMMM
mmN
a
dfAM
mmN
h
dfAM
un
n
nwnfn
ypswnw
f
ysfnf



31. 25-Feb-13
CE370: Prof. A. Charif 31
Design Problem 3
Design the shown T-section for an
ultimate bending moment of 440 kN.m MPafMPaf
yc
42025
'
 600
75
525
300
Expecting two tension steel layers, and with
25-mm layer spacing, the effective steel depth
at the centroid is estimated as :
d = h – 90 = 600 – 90 = 510 mm
The full flange moment capacity is: mkNMmkNM
mkNmmNM
h
dhbfM
unff
nff
f
ffcnff
.0.440.65.40683.4519.0
.83.451.1083.451
2
75
510756002585.0
2
85.0
6
'



















Compression block is thus in the web.
Decompose as follows: T-section = W-section + F-section    
mkNmmNM
h
dfAM
mm
f
hbbf
A
AAAMMM
nf
f
ysfnf
y
fwfc
sf
sfswsnfnwn
.914.225.10914.225
2
75
51042039.1138
2
39.1138
420
753006002585.085.0
6
2
'






















 funff
hamkNMmkNM  .0.440.65.406
Solution 3 – Cont.

32. 25-Feb-13
CE370: Prof. A. Charif 32
The web is designed as rectangular section for an ultimate moment: mkNMMM
nfuwu
.68.236914.2259.0440 
The steel area component A
sw is the solution of a quadratic
equation given by: 2
2
2
6
2'
'
9.248239.11385.1344 : is area steel Total
5.1344
257.1
3702.34
11
420
5103002585.0
3702.3
51030090.0
1068.236
7.1
4
11
85.0
mmAAA
mm
db
M
R
f
R
f
dbf
A
sfsws
w
wu
wn
c
wn
y
wc
sw























Solution 3 – Cont.
CE 370 : Prof. Abdelhamid Charif 64 mm
dd
d
mmddd
mmdd
t
5.517
2
: is cemtroidat depth steel Effective
49520255402025
540101040600
21
1min2
1




 2
min s
2
min
'
min,
min
9.2482OK A 0.510
510300
420
4.1
,
4204
25
Max
4.1
,
4
Max
Checking
mmAAmmA
db
ff
f
A
A
sss
w
yy
c
s
s





















Use eight 20-mm bars in two layers = 2513.27 mm2.
The steel depths are :
Solution 3 – Cont.
The final steel depth is just greater than the assumed value : OK
Moment check is not necessary.

33. 25-Feb-13
CE370: Prof. A. Charif 33
Compression block and neutral axis depths are computed using
the actual steel area :
Strain check: ControlTensionOK005.00109.0
0021.00109.0
565.106
565.106495
003.0003.0
min
min
min







 OK
c
cd
y mm
a
cmm
bf
fA
a
mmA
AAA
mmA
wc
ypsw
psw
sfpspsw
ps
565.106
85.0
58.90
58.90
3002585.0
42088.1374
85.0
88.137439.113827.2513
: ispart in web area steel Actual
27.2513
4
20
8: is area steel Actual
1
'
,
2
,
,,
2
2
,









Solution 3 – Cont.
Thank you
66 CE 370 : Prof. Abdelhamid Charif