The document discusses design considerations for rectangular reinforced concrete beams with tension steel only. It covers topics such as beam proportions, deflection control, selection of reinforcing bars, concrete cover, bar spacing, effective steel depth, minimum beam width, and number of bars. Beam proportions should have a depth to width ratio of 1.5-2 for normal spans and up to 4 for longer spans. Minimum concrete cover and bar spacings are specified to protect the steel. Effective steel depth is the distance from the extreme compression fiber to the steel centroid. Design assumptions must be checked against the final design.

1. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 1
CE 370
REINFORCED CONCRETE-I
Prof. Abdelhamid Charif
Design of rectangular RC beams
with tension steel only
Design of rectangular beams
• Before the design of an actual beam is attempted,
several important topics are first discussed :
1. Beam proportions
2. Deflection control and minimum thickness
3. Beam self weight
4. Selection of bars
5. Minimum concrete cover
6. Minimum spacing of bars and steel layers
7. Minimum beam width
8. Maximum number of bars in one layer
9. Effective steel depth
2

2. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 2
Beam proportions
• Under transverse loading, the resulting bending stress and deflection
are both inversely proportional to the moment of inertia I
• Thickness is therefore more important in bending than the width
• In normal beams, thickness is greater than width
• Exceptional shallow beams (width greater than thickness) are
sometimes used for architectural reasons only
3
12
3
bhI
I
My 
 12
3
bhI 
Beam proportions
• The most practical and economical beam
section is usually obtained for normal spans
(up to 7.5 m in length), when the ratio of d to
b is in the range of 1.5 to 2.
• For longer spans, better economy is obtained
if deep, narrow sections are used. The depth
may be as large as 3 or 4 times the width
• The overall beam dimensions are selected to
whole inches (multiples of 25 mm). This is
done for simplicity in the use of forms which
are usually available in 1- or 2-inch (25 or 50
mm) increments.
• Shallow beams are used for architectural
reasons (to hide them or reduce floor height)
b
SpanLong
4to3 bbd 
sA
SpanNormal
2to5.1 bbd 
sA
b
b = Width
h = Depth (Thickness)
d = Steel depth
4

3. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 3
Deflection control and minimum thickness
• SBC/ACI codes provide minimum thickness for beams and slabs so
that deflection calculations and checks are not required.
Minimum Thickness for deflection control
L = Span length (center to center)
Simply
supported
One end
continuous
Both ends
continuous Cantilever
Solid one-
way slab
L / 20 L / 24 L / 28 L / 10
Beams
or ribs
L / 16 L / 18.5 L / 21 L / 8
5
• Thickness less than minimum value can be used, but in that case
deflections must be determined and checked (long and tedious)
• Better comply with code minimum thickness
6
One end
continuous
Both ends
continuous
Cantilever
Simply
supported
Minimum Thickness for deflection control
Simply
supported
One end
continuous
Both ends
continuous Cantilever
Solid one-
way slab
L / 20 L / 24 L / 28 L / 10
Beams
or ribs
L / 16 L / 18.5 L / 21 L / 8
Deflection control and minimum thickness

4. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 4
Initial section dimensions
• Overall Thickness h: Assume a minimum overall
thickness (or depth) h equal to or greater than the
minimum thickness specified by the code (unless
deflections are to be calculated and checked).
• Beam Width b: The beam width can be roughly
estimated equal to about one-half of the assumed
value of h.
• Beam self weight (kN/m) :
= b × h × unit weight of concrete
7
Final section dimensions
• After performing reinforcement design, the initial
section dimensions may turn out to be inadequate
• If the reinforcement is less than minimum, this
means that the section dimensions can be reduced.
• If the designed section does not satisfy tension
control condition, then either the section dimensions
must be increased or compression steel provided.
• Increase in thickness is more effective than in width.
• Ultimate moment must be corrected after any
increase in the section dimensions
8

5. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 5
Selection of steel bars
• After the required reinforcing area is calculated, select the bars
that provide (at least) the necessary area.
• It is usually convenient to use bars of one size only in a beam,
although occasionally two sizes are also used.
• Bars for compression steel and stirrups are usually of a
different size (to avoid confusion for workmen).
• We usually choose the bar diameter and determine the
required bar number (Integer equal to or greater than):
bbps
bs
b
b
b
b
s
b
ANA
AA
d
d
A
A
A
N
:isareasteelprovidedActual
baroneofAreaareasteelRequired
diameterBar
4
,
,
2


 
9
Concrete cover
• Reinforcing steel bars must be protected from the surrounding
environment, such as fire and corrosion.
• The reinforcing bar is located at a certain minimum distance
from the surface of concrete so that a protective layer of
concrete, called cover, is available.
10

6. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 6
Why is cover needed ?
a) Bonds reinforcement to concrete
b) Protects reinforcement against
corrosion
c) Protects reinforcement from fire
(over heating causes major steel
strength loss)
d) Larger cover values are used in
foundations, garages, factories,
etc. to account for abrasion,
wear and chemical attacks.
Concrete cover
11
Minimum cover dimensions (ACI / SBC)
• Concrete not exposed to earth or weather :
 Slabs, walls, joists : 20 mm
 Beams, Columns : 40 mm
• Concrete cast against / exposed to earth : 75 mm
Minimum concrete cover
12

7. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 7
Bar spacing and layer spacing
• Whenever possible steel bars
must be arranged in a single layer
• This simplifies analysis and
design calculations, and
optimizes the bending capacity
with a greater lever arm .
• It is frequent however that more
than one steel layer is required
• Spacing between bars in any
layer and spacing between
various layers must comply with
SBC / ACI provisions.
13
Bar arrangement in layers
Bars in layers must be arranged directly above
each other as shown.
14

8. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 8
Bar spacing and layer spacing
SBC / ACI specify maximum spacing of flexural reinforcing bars
in walls & slabs not to be exceeded
15
Main bars and
stirrup arrangement
Net distance between stirrup
and outer main bar:
more)andmm(12diameterbarMain
mm)12to(8diameterStirrup
5.02



b
s
bs
d
d
dd
All spacing requirements must be considered in
determining steel depth and number of steel layers
16

9. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 9
Effective steel depth – One Layer
• Effective steel depth d is
defined as the distance from
the extreme compression fiber
to the centroid of tensile steel
• Case of one steel layer :
2
cover b
s
d
dhd 
17
Effective steel depth – Many layers
• Effective steel depth d is defined as the distance from
the extreme compression fiber to the centroid of
tensile steel.
• Case of many steel layers :
syhd 
ys = Distance from bottom to tension
steel centroid
ys Computed by complying with layer
spacing and cover requirements
18

10. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 10
19
Effective steel depth - Example
• Assume two layers and h = 600 mm
• First (bottom) layer : three bars of 20 mm
• Second layer : two bars of 16 mm
• Stirrup diameter 10 mm
• Maximum aggregate size = 18 mm
mmy
mm
dd
yy
mm
d
dy
bb
b
s
1038102560
25spacingLayer
)25,1833.1(MaxspacingLayer
22
SpacingLayer
60101040
2
Cover
2
21
12
1
1





21
2211
ss
ss
s
s
AA
yAyA
y
yhd




19
2020
Effective steel depth - Example / Cont.
syhd 
mmyhd
mmmm
AA
yAyA
y
s
ss
ss
s
52773600
739.72
21
2211





2
2
2
2
2
1
21
12.402
4
16
2
48.942
4
20
3
10360
mmA
mmA
mmymmy
s
s





20

11. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 11
2121
Effective steel depth - General
Many steel layers (Layer 1 = Bottom)
yi is computed for each layer i
syhd 


si
isi
s
A
yA
y



 
i
ii
bi
ibi
s
n
yn
An
yAn
y
21
Case of same bar diameter in all layers
Each layer i has ni bars and distance yi
2
2
Cover
1,
1
1
1





ibbi
lii
b
s
dd
Syy
d
dy
2222
Minimum beam width
• Given the bar diameter, number of bars,
stirrup diameter, and aggregate size, what is
the minimum beam width to accommodate
the bars in one layer ?
• Width b must accommodate all bars while
satisfying cover and spacing conditions
sizeaggregateMaximum
),25,33.1(MaxSpacingBar


Agg
dAggS bb
 
cover26)1(
cover25.0222)1(
min 

bsbb
bssbb
ddSn-dnb
dddSn-dnb
In practice, beam width b is not less than 200 mm
22

12. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 12
2323
Minimum beam width - Example
• Assume :
• Three bars of 20 mm
• Stirrup diameter 10 mm
• Maximum aggregate size = 18 mm
• Width b must accommodate the three bars
while satisfying cover and spacing conditions
mmb
ddSn-dnb
mmdS
bsbb
bb
23040220106252203
cover26)1(
25),25,1833.1(MaxSpacingBar
min
min



mmb 230min 
23
• Given the beam width, bar diameter, bar spacing, and
stirrup diameter, what is the maximum number of
bars that can be accommodated in one layer ?
Maximum number of bars in one layer
bb
bsb
bb
bsb
bsbbb
bsbb
Sd
ddSb
n
Sd
ddSb
n
ddSSdnb
ddSn-dnb








cover26
cover26
cover26)(
cover26)1(
max
24

13. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 13
• Beam width, bar diameter, stirrup diameter,
and bar spacing are:
Maximum number of bars in one layer
Example
5
88.5
2518
0421810625350
max 




n
n
mmSmmdmmdmmb bsb 251018350 
bb
bsb
Sd
ddSb
n



cover26
25
• The designer must first assume an initial bar diameter
and an initial number of layers, and then deduce the
effective steel depth d.
• Effective steel depth is computed at the steel centroid.
• If bar diameter is not fixed, the effective steel depth for
RC beams may be approximated as follows:
Estimation of layer number
and effective steel depth
mmhd 65One steel layer :
mmhd 90Two steel layers :
26

14. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 14
• After finishing design, the effective bar number, layer number,
and steel depth must be checked with initial assumptions.
• If the effective steel depth is greater than the initial value, the
design is safer than anticipated. New design may be performed
only for economic reasons (if bar number can be reduced).
• If the effective steel depth is less than the initial value, the
design may be unsafe . The deficiency in steel depth may
however be compensated by the provided steel area which
usually is greater than the required value (when choosing bar
number). A moment check is then required and new design has
to be performed if moment check fails (fMn < Mu) .
• It is important not to overestimate the initial steel depth, not
only because of re-design risk, but also because in practice,
smaller values are delivered.
27
Checking effective steel depth
27
28
• It can be both unsafe and uneconomical to perform tensile
steel strain check at the centroid.
• Assuming steel yielding at the centroid requires yielding of all
tension layers. The least tensioned layer with minimum depth
may not yield despite yielding at the centroid.
• It is therefore unsafe to check yielding at the centroid
• Yielding must be checked at the least tensioned layer with
minimum depth dmin
• Performing tension control check at the centroid is
uneconomical. The most tensioned bottom layer with
maximum depth reaches the tension control limit of 0.005
before the centroid.
• Tension control check must be performed at the most
tensioned bottom layer.
Tensile steel strain check with many layers

15. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 15
29
0.003
d
dt
c
Depths Strains
005.0t
yε min
dmin
s
Lumping of tension steel layers at centroid






005.0
:layersMany
005.0:layerOne
min
t
y
t
ε



Required steel
strain checks :
ts
ts
ε
ε




min
min
:layersMany
:layerOne
d , εs = Depth and strain at centroid of steel layers
dt , εt = Depth and strain at bottom layer (max. depth)
dmin , εmin = Depth and strain at minimum depth layer
Assumptions of tension-controlled section (εt ≥ 0.005)
and steel stress equal to yield stress (fs = fy)
must be checked as shown next (in analysis or design).
Checks must be performed, once compression block
depth a and neutral axis depth c have been
determined using the actual steel area.
Neutral axis depth : c = a / β1
30
Tensile steel strain check with many layers

16. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 16
31
• Compute steel strain εmin at
minimum tension depth
and check that εmin ≥ εy
c
cd 
 min
min 003.0
• If εmin < εy : Assumption of
all steel yielding is violated.
Lumping cannot be used. Use
strain compatibility method.
31
Tensile steel strain check with many layers
Yield check at minimum depth layer
d
dt
Yield limit of NA at min depth
dmin
0.003
cy
t
yε
s
y
yy
y
y
d
cc
dc



10003
3
Thus
003.0003.0
:checkdepthN.A.Equivalent
min
min
min




32
• Compute steel strain εt at
bottom layer and check that
εt ≥ 0.005
c
cdt
t

 003.0
• If εt < 0.005 : Section is not
tension controlled. Rejected
by SBC code. Increase
section or re-design with
compression steel
32
Tensile steel strain check with many layers
Tension control check at bottom layer
t
t
tt
tt
d
d
cc
dc
375.0
8
3
005.0Thus
005.0003.0003.0
:checkdepthN.A.Equivalent




d
dt
T.C. limit of NA at bottom layer
dmin
0.003
ct
minε
s
dt
0.005

17. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 17
Strain checks - Summary
• In case of one tension steel layer, perform tension control check
only εt ≥ 0.005
• In case of more than one layer, both yield check at the minimum
tension depth layer (εmin ≥ εy), and tension-control check at the
bottom layer (εt ≥ 0.005) must be performed.
• It is frequent that if the second check is satisfied, the first will
also be satisfied but exceptional cases may exist (shallow beams).
If the yield check is not satisfied, layer lumping cannot be used
and other design methods must be used.
• If the minimum depth strain satisfies tension-control condition,
there is no need to perform the check at the bottom layer (as the
bottom strain is always greater ):
If : εmin ≥ 0.005 then εt ≥ 0.005 (No need to check it)
33
Design of rectangular RC sections
with tension steel only
Compared to analysis, design problems are one hand more difficult
because of the unknown steel area and depth, but easier on the
other hand because of the known steel stress (steel yielding).
Steel depth d has to be estimated according to the expected layer
number and must be checked at the end.
d
b
sA
N.A.
c ca 1
abfC c
'
85.0
'
85.0 cf
ys fAT  bdA
bd
A
s
s




85.0850
850 '
c
y
'
c
ys
ys
'
c
f
df
bf.
fA
afAabf.TC


34

18. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 18
35
7.1
4
11
85.0
:Solution
with,0
7.1
:equationQuadratic
7.1
1
7.1
1
85.02
)(
2
:DesignOptimal:DesignSafe
90.0,005.0:controltensionAssuming
2
2
2
2
2







































'
c
n
y
'
c
u
nny'
c
y
'
c
y
y
u
'
c
y
yu
'
c
y
yysun
unun
t
f
R
f
f
ρ
bd
M
RRf
f
f
f
f
ρf
bd
M
f
f
ρfbdM
f
df
dfρbd
a
dfAMM
MMMM
f


f

f

fff
ff
f
Design of rectangular RC sections
with tension steel only (Cont.)
36
7.1
4
11
85.0
:smallesttheUse
positiveareequationquadraticofsolutionsBoth
satisfiedeasilyusuallyCondition
47.1
4
47.10
7.1
4
1
7.1
4
0
7.1
:equationQuadratic
2
2
2
2
2
2
2


















'
c
n
y
'
c
u
'
c
u
n
'
c
'
c
n
y'
c
yn
y
ny'
c
y
f
R
f
f
ρ
Mbdf
bd
M
Rf
f
R
f
f
fR
f
Rf
f
f
f
f

Design of rectangular RC sections
with tension steel only (Cont.)

19. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 19
Design of rectangular RC sections
with tension steel only (Cont.)
The required steel ratio must comply with the SBC/ACI
minimum limit. it must be greater than or equal to the
minimum steel ratio :









yy
c
ff
f 4.1
,
4
Max
'
min
  bdAs  min,MaxareaSteel 
37
The final required steel area to be adopted is :
• Using this final required steel area, compute the required bar
number, layer number and deduce effective steel depth and
actual provided steel area.
• Check effective steel depth with assumed value
• If the actual steel depth is less than assumed, the design may
be unsafe and moment check is thus necessary.
• Using the actual provided steel area, compute the stress block
and neutral axis depths a and c, and then perform steel strain
checks.
• Perform moment check if necessary (if d < assumed).
• If tension control check fails, re-design.
• With many layers, if yield check fails, use compatibility
method to check moment capacity.
• If required moment check fails (fMn < Mu) re-design.
Design of rectangular RC sections
with tension steel only (Cont.)
38

20. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 20
• After performing reinforcement design, the initial
section dimensions may turn out to be inadequate
• If the reinforcement is less than minimum, this
means that the section dimensions can be reduced.
• If the designed section does not satisfy tension
control condition, then either the section dimensions
must be increased or compression steel provided.
• Increase in thickness is more effective than in width.
• Ultimate moment must be corrected after any
increase in the section dimensions
39
Observations
Design Problem 1
(unknown dimensions)
MPa.420andMPa30Use
supported.betoarekN/m25ofloadliveaand
weight)beamincluding(notloaddead15kN/maif
beamrrectangulasupportedsimplym6.5aDesign
 y
'
c ff
SOLUTION:
Minimum thickness for a simply supported beam is
L/16 . We use a higher value of L/10 = 650 mm
h = 650 mm
40

21. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 21
kN/m07.524650.0325.0weightselfBeam
kN/m24:tunit weighConcrete
mm32550
mm8556565065depthSteel:layeroneAssume
conc
3
conc






hb
h.b
hd
Minimum Thickness for deflection control
Simply
supported
One end
continuous
Both ends
continuous Cantilever
Solid one-
way slab
L / 20 L / 24 L / 28 L / 10
Beams
or ribs
L / 16 L / 18.5 L / 21 L / 8
SOLUTION 1 (continued)
41
kN.m372.85
8
6.570.598
8
kN/m598.70)25(7.1)07.515(4.17.14.1
22




lw
M
LDw
u
u
u
0.90ssume fA
0.00963
307.1
725.34
11
420
3085.0
725.3
58532590.0
1085.372
with
7.1
4
11
85.0
2
6
2


























ρ
bd
M
R
f
R
f
f
ρ
u
n
'
c
n
y
'
c
f
SOLUTION 1 (continued)
Ultimate uniform load and ultimate moment
42

22. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 22
2
,
2
2
26.18473areasteelProvided
OKislayerOnebars283Use
95.2
28
4
5.1831
barsof#bars,mm28Use
mm5.183158532500963.0
mmAA
A
A
ρbdA
bps
b
s
s







586
325
283
64
650
  





















00333.000333.0,00326.0Max
420
4.1
,
4204
30
Max
4.1
,
4
Max
'
min
yy
c
ff
f
SOLUTION 1 (continued)
OK586141040650
2
coverEffective mm
d
dhd b
s 
Questions: What if No of bars = 2.1 ? / if d less than assumed ?
43
Tension-control check
(In case of one layer no need to check steel yield)
mm14.110
85.0
62.93
mm62.93
3253085.0
42026.1847
85.0
1
,






a
c
bf
fA
a '
c
yps
586
325
283
64
650
0.90controlledtensionisSection
OK005.0013.0
013.0
14.110
14.110586
003.0
003.0







f



t
t
t
c
cd
44
Checks must be performed using the actual provided steel area

23. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 23
Moment checking
(Not necessary)
586
325
283
64
650
OKkN.m)85.372(kN.m376.5
kN.m33.4189.0capacitymomentDesign
kN.m33.418N.mm1033.418
2
62.93
58642026.1847
22
6
,























un
n
nn
n
ypsn
MM
M
MM
M
a
dfA
a
dTM
f
f
45
• Design of a 300 x 600 mm rectangular section subjected to
an ultimate bending moment Mu = 300 kN.m
• b = 300 mm h = 600 mm
• Estimate number of layers expected and steel depth
• Expecting two steel layers, the effective steel depth is
estimated as: d = h – 90 = 510 mm
• Assume stirrup diameter of 10 mm and aggregate maximum
size of 20 mm
46
Design Problem 2
(known dimensions)
85.042025 1
'
 MPafMPaf yc

24. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 24
47
0.01147
257.1
272.44
11
420
2585.0
272.4
51030090.0
100.300
with
7.1
4
11
85.0
2
6
2


























ρ
bd
M
R
f
R
f
f
ρ
u
n
'
c
n
y
'
c
f
0.90ssume fA Required steel ratio is given by :
  00333.000333.0,00298.0Max
420
4.1
,
4204
25
Max
4.1
,
4
Max
'
min




















yy
c
ff
f

SOLUTION of Problem 2
48
2
,
2
2
min
96.18846areasteelProvided
bars20-6Use
59.5
20
4
91.1754
barsof#bars,mm20Use
91.175451030001147.0
OK
mmAA
A
A
mmρbdA
ρ
bps
b
s
s









SOLUTION 2 (continued)
Check layer number and effective steel depth

25. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 25
Maximum number of bars in one layer :
443.4
7.2620
042201067.26300
7.26)20,25,2033.1(Max
),25,33.1(MaxSpacingBar
cover26
max 








nn
mmS
dAggS
Sd
ddSb
n
b
bb
bb
bsb
49
SOLUTION 2 (continued)
Using 4 bars per layer, the effective bar spacing will be
greater than 26.7 mm
For six bars, we thus need two layers: 4+2
We may use 3+3, but 4+2 is better, why ?
50
d
300
206
sy
600
SOLUTION 2 (continued)
syhd 

si
isi
s
A
yA
y



 
i
ii
bi
ibi
s
n
yn
An
yAn
y
Case of same bar diameter in all layers
Each layer i has ni bars and distance yi
Effective steel depth :
2
260101040
2
cover 121
b
l
b
s
d
Syymm
d
dy 

26. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 26
51
d
300
206
sy
600
SOLUTION 2 (continued)
mmyhd
mmmm
n
yn
An
yAn
y
s
i
ii
bi
ibi
s
52476600:deptheffectiveCentroid
766.75
24
7.1062604









Effective steel depth :
mm
d
Syy
mm
d
dy
mmS
AggS
b
l
b
s
l
l
7.106207.2660
2
2
60101040
2
cover
7.26)25,2033.1(Max
)25,33.1(MaxSpacingLayer
12
1




Effective steel depth greater than assumed (510 mm): OK
Compression block and NA depths (using actual steel area) are :
mm
a
cmm
bf
fA
a
c
yps
10.146
85.0
1856.124
1856.124
3002585.0
42096.1884
85.0 1
'
,





Strain check (at dmin ) :
OK
c
cd
y 



 0021.000713.0
1.146
1.1463.493
003.0003.0 min
min 
controlTension005.0005.0Since min OKt  
No need to determine bottom steel strain
SOLUTION 2 (continued)
52
mmyhdd
mmyhdd
t 54060600
3.4937.106600:depthsLayer
11
2min2



27. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 27
Moment checking
(Not necessary)
OKkN.m)0.300(kN.m329.1
kN.m68.3659.0capacitymomentDesign
kN.m68.365N.mm1068.365
2
1856.124
52442096.1884
22
6
,























un
n
nn
n
ypsn
MM
M
MM
M
a
dfA
a
dTM
f
f
53
524
300
206
76
600
• Design of a 375 x 900 mm rectangular section subjected
to an ultimate bending moment Mu = 1500 kN.m
• b = 375 mm h = 900 mm
• Expecting two steel layers, the effective steel depth is
estimated as: d = h – 90 = 810 mm
• Assume stirrup diameter of 10 mm and 30 mm for bar
spacing and layer spacing.
54
Design Problem 3
81.035008.009.135420 1
'
 MPafMPaf cy

28. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 28
55
0.01856
357.1
774.64
11
420
3585.0
774.6
81037590.0
100.1500
with
7.1
4
11
85.0
2
6
2


























ρ
bd
M
R
f
R
f
f
ρ
u
n
'
c
n
y
'
c
f
0.90ssume fA Required steel ratio is given by :
  00352.000333.0,00352.0Max
420
4.1
,
4204
35
Max
4.1
,
4
Max
'
min




















yy
c
ff
f

SOLUTION of Problem 3
56
2
,
2
2
min
5.615710areasteelProvided
bars28-10Use
16.9
28
4
6.5637
barsof#bars,mm28Use
6.563781037501856.0
OK
mmAA
A
A
mmρbdA
ρ
bps
b
s
s









SOLUTION 3 (continued)
Check layer number and effective steel depth

29. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 29
Maximum number of bars in one layer :
505.5
3028
0422810630375
)given(30SpacingLayerBar /
cover26
max 







nn
mmSS
Sd
ddSb
n
lb
bb
bsb
57
SOLUTION 3 (continued)
For ten bars, we thus need two layers: 5+5
58
SOLUTION 3 (continued)
syhd 


si
isi
s
A
yA
y



 
i
ii
bi
ibi
s
n
yn
An
yAn
y
Case of same bar diameter in all layers
Each layer i has ni bars and distance yi
Effective steel depth :
2
264141040
2
cover 121
b
l
b
s
d
Syymm
d
dy 
d
375
2810
sy
900

30. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 30
59
SOLUTION 3 (continued)
performedbemustcheckMoment
mm810ofvalueAssumed
80793900:deptheffectiveCentroid
93
255
1225645 21













d
mmyhd
mm
yy
n
yn
An
yAn
y
s
i
ii
bi
ibi
s
Effective steel depth :
mm
d
Syy
mm
d
dy
b
l
b
s
122283064
2
2
64141040
2
cover
12
1


d
375
2810
sy
900
Using the actual steel area provided, the compression block and
neutral axis depths are:
mm
a
cmm
bf
fA
a
c
yps
188.286
81.0
812.231
812.231
3753585.0
4205.6157
85.0 1
'
,





Strain check :
OK
OK
c
cd
t
y






005.0005.000516.0Since
0021.000516.0
188.286
188.286778
003.0003.0
min
min
min


SOLUTION 3 (continued)
60
mmyhd
mmyhd
t 83664900
778122900:depthsLayer
1
2min



31. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 31
Moment checking
requireddesign-reNoOK
kN.m)0.1500(kN.m08.5461
kN.m27.17879.0capacitymomentDesign
kN.m27.1787N.mm1027.1787
2
812.231
8074205.6157
22
6
,
























un
un
n
nn
n
ypsn
MM
MM
M
MM
M
a
dfA
a
dTM
f
f
f
61
The moment check is necessary as the
final steel depth (807) is less than the
initially assumed value (810).
375
807
2810
188.286
a
Thank You
62