This document provides information on reinforced concrete design including: - Concrete and steel properties such as modulus of elasticity and grades/strengths of reinforcing bars. - Minimum concrete cover requirements for reinforcement. - Load factors and combinations for ultimate strength design. - Flexural design procedures for reinforced concrete beams including assumptions, stress/strain diagrams, and analysis for cases where steel yields or does not yield. - Requirements for reinforcement spacing, minimum member thicknesses, and ductility.

1. REINFORCED CONCRETE DESIGN

2. CONCRETE PROPERTIES
For concrete weighing from 1,500 to 2500 kg/m3
'
043
.
0
5
.
1
fc
w
E c
c 
fc’ = 28 day compressive strength in MPa
Wc = unit weight of concrete in kg/m3
Modulus of Elasticity
For Normal weight concrete
'
4700 fc
Ec 
fc’ = 28 day compressive strength in MPa
Wc = unit weight of concrete in kg/m3

3. REINFORCING BARS
GRADES AND STRENGTH OF REINFORCING BARS
(Non Prestressed)
ASTM
SPECS
GRADE Min.Yeild
Strength
(ksi)
Min.Yeild
Strength
(MPa)
Min
Tensile
Strength
(ksi)
Min
Tensile
Strength
(MPa)
A615 40
60
40
60
276
414
70
90
483
620
A616 50
60
50
60
345
414
80
90
552
620
A617 40
60
40
60
276
414
70
90
483
620
A618 60 60 414 80 552
Modulus of Elasticity of steel shall be taken as 200,000 MPa

4. CONCRETE PROTECTION FOR REINFORCEMENT
( NON PRESTRESSED)
CAST IN PLACE CONCRETE
Minimum
Cover,mm
a) Concrete cast against and permanently exposed to earth 75
b) Concrete exposed to earth or weather:
20 mm bars through 36 mm bars
16 mm bar, W31 or D31 wire, and smaller
50
40
c) Concrete not exposed to earth or weather or in contact with ground:
Slabs, walls, joist
32 mm bar and smaller
Beams,columns
Primary reinforcement ,ties,stirrups,spirals
Shells,folded plates members:
20 mm bars and larger
16 mm bars,W31or D31 wire,or smaller
20
40
20
15

5. PRECAST CONCRETE
( manufactured under plant Conditions)
Minimum
Cover,mm
a) Concrete exposed to earth or weather:
Wall Panels
32 mm bar and smaller
Other members
20 mm bars through 32 mm bars
16 mm bars,W31or D31 wire,or smaller
20
40
30
b) Concrete not exposed to earth or weather or
in contact with ground:
Slabs, walls, joist
32 mm bar and smaller
Beams,columns
Primary reinforcement
ties,stirrups,spirals
Shells,folded plates members:
20 mm bars and larger
16 mm bars,W31or D31 wire,or smaller
15
10
15
10

6. ULTIMATE STRENGTH DESIGN(USD)
Required Strength ( Load Factors)
1.Required strength U to resist dead load D and live load L
shall at least be equal to : U =1.2D +1.6L
2. If resistance to structural effects of specified wind load
W, are included in the design,the following combinations of D,L and W
shall be investigated to determine the greatest required strength U
a) U =0.75(1.4D + 1.7L + 1.7W)
Where load combination shall include full value and zero
value of L to determine the most severe condition, and
b) U = 0.9D +1.3L
For any combination of D, L and W required strength U
shall not be less than c) U =1.4D +1.7L

7. 3. If resistance to structural effects of specified earthquake load E, are
included in the design,the following combinations of D,L and E shall
be investigated to determine the greatest required strength U
a) U =1.32D + 1.1(f1)L + 1.1E
Where load combination shall include full value and zero value of
L to determine the most severe condition, and
b) U = 0.99D +1.1E
For any combination of D, L and E required strength U shall not be less
than
c) U =1.4D +1.7L
4. If resistance to earth pressure H, are included in the design,the
following combinations of D,L and H shall be investigated to
determine the greatest required strength U
a) U =1.4D + 1.7L + 1.7H
Where D or L reduces the effect of H
b) U = 0.9D +1.7H
For any combination of D, L and E required strength U shall not be less
than
c) U =1.4D +1.7L

8. 5. If resistance to loadings due to weight and pressure offluids
with well defined densities and controllable height F are
included in the design, such loading shall have a factor of 1.4
and be added to all loading combinations that include live load.
6. If resistance to impact effects are taken into account in
design,such effect shall be included with live load L.
7. Where structural effects T of differential settlement, creep,
shrinkage,expansion of creep compensating concrete or
temperature change maybe significant in design, required
strength U shall be a least equal to
a) U =0.75(1.4D + 1.7L + 1.4T)
but required strength U shall not be less than
b) U =1.4D +1.7L

9. STRENGTH REDUCTION FACTOR Ф
STRENGTH REDUCTION FACTOR Ф shall be as follows:
1. Flexure, without axial load 0.90
2. Shear and Torsion 0.85
3. Bearing on concrete except on
Post tension anchorage zone 0.70
4. Post tension anchorage zone 0.85
5. Axial tension and axial tension with flexure 0.90
6. Axial load and axial load with flexure
Both axial load and moment shall be multiplied by Ф
7. Axial compression and axial compression with flexure
a) Members with spiral reinforcement 0.75
b) Other reinforced members 0.70
except that for low values of axial compression Ф shall be
permitted to increase in accordance with the following:
For members in which fy does not exceed 415 MPa with symmetric
reinforcement,and with ( h –d’-d)/h not less than 0.7, Ф shall be permitted to
increase linearly to 0.9, as ФPn decreases from 0.10fc’Ag to zero.
For other reinforced members, Ф shall be permitted to increase linearly to 0.9, as
ФPn decreases from 0.10fc’Ag or ФPb whichever is smaller to zero.

10. DESIGN AND ANALYSIS FOR FLEXURE (BEAMS)
• Basic Assumptions
• 1. Strain in concrete and the reinforcement shall be assumed
directly proportional to the distance from the neutral axis ,except, for
deep flexural members with overall depth-to-clear span ratio greater
than 2/5 for continuous spans and 4/5 for simple span a non linear
distribution of strain shall be considered.
• 2. Maximum usable strain at extreme concrete compression fiber
shall be 0.003
• 3. Stress in reinforcement below specified yield strength fy for grade of
reinforcement used shall be taken Es times steel strain. For strain
greater than corresponding to fy, stress in the reinforcement shall be
considered independent of strain and equal to fy.
• 4. Tensile strength of concrete shall be neglected in axial and flexural
calculations.

11.  5. Relationship between concrete compressive stress distribution and
concrete strain shall be assumed to be
rectangular.trapezoidal,parabolic or any other assumed shape that
result in prediction of strength in substantial agreement with results of
comprehensive tests.
 6. Requirements of 5 may be considered satisfied by an equivalent
rectangular stress distribution defined by the following:
Concrete stress distribution of 0.85fc’ shall be assumed uniformly
distributed over an equivalent compression zone bounded by the
edges of the cross section and a straight line located parallel to the
neutral axis at a distance “a” from the fiber of maximum compressive
strain.
Distance c from fiber of maximum strain to the neutral axis shall be
measured in a direction perpendicular to the neutral axis.

12. Compression Zone
(stress in concrete) (maximum usable strain of concrete)
0.85fc’ 0.003
εs (strain of steel)
c
7. Factor β1 shall be taken as follows:
β1 = 0.85 if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not be less
than 0.65
a
NA
c
a 1



13. SYMBOLS AND NOTATIONS
 a = depth of equivalent rectangular stress block, mm
 c = distance from extreme compression face, mm
 As = area of non prestressed tension reinforcement, mm2
 As’ = area of non prestressed compression reinforcement, mm2
 b = width of compression face of the member, mm
 bw = width of the web, mm
 d = distance from extreme compression face to center of tension
reinforcement, mm
 d’ = distance from extreme compression face to center of
compression reinforcement, mm
 fc’ = specified compressive strength of concrete, MPa
 fy = specified yield strength of non prestressed reinforcement,MPa
 fs = calculated tensile stress in reinforcement at service loads,MPa
 fs’ = calculated compressive stress in reinforcement at service loads,MPa

14. • Mu = factored moment at section; ultimate moment capacity,design
strength
• Mn = nominal moment capacity
• Ф = strength reduction factor
• pb = reinforcement ratio producing balance strain condition
• p = ratio of non prestressed tension reinforcement = As/bd
• p’ = ratio of non prestressed compression reinforcement = As’/bd
• pmin = minimum required ratio of non prestressed tension
reinforcement
• pmin = 1.4/fy
• Ec = modulus of elasticity of concrete,MPa
• Es = modulus of elasticity of reinforcement,MPa
• bf = flange width of T –beams
• t = flange thickness of T- beams

15.  Balanced strain condition
This exist at a cross section when tension reinforcement reaches
the strain corresponding to its yield strength fy just as concrete
compression reaches its assumed ultimate strain of 0.003.
Compression
Zone
d
ab
0.85fc’
C = 0.85fc’abb
0.003
T = Asbfy
s
y
s
E
f


b
cb
Stress Diagram Strain Diagram
Asb = balance steel area
∑ Fx = 0
C = T
0.85fc’abb = Asbfy
From the Stress Diagram
EQ.1

16. bd
A
df
a
fc sb
y
b

'
85
.
0
divide both sides of EQ.1 by bdfy
Let
bd
A
p sb
b  then
y
b
b
df
a
fc
p
'
85
.
0

s
b
d
c



003
.
0
003
.
0
200000
y
s
f


1

b
b
a
c 
fy
d
ab


600
600 1

From the strain diagram
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



EQ. 2
EQ. 3
Substituting EQ 3 in EQ 2

17. MAXIMUM PERMISSIBLE TENSILE STEEL RATIO
PMAX = 0.75PB
This limitation is to ensure that the steel reinforcement
will yield first to ensure ductile failure.
Minimum permissible tensile steel ratio
pmin = 1.4/fy
The provision for minimum amount of reinforcement applies to
beams which for architectural and other reasons are much larger in
cross section as required by strength consideration. With very small
amount of tensile reinforcement, the computed moment strength as
a reinforced concrete member is smaller than that of the
corresponding plain concrete section computed from its modulus of
rupture. Failure in this case is quite sudden.

18. Overreinforced beam
A design in which the steel reinforcement is more than that required for
balanced strain condition. If the beam is overeinforced, the steel will not
yield before failure. As the load is increased, deflections are not
noticeable although the compression concrete is highly stressed, and
failure occurs suddenly without warning to the user of the structure.
Underreinforced beam
A design in which the steel reinforcement is lesser than that required for
balanced strain condition. If the ultimate load is approached , the steel will
begin to yield although the compression concrete is understressed. As the
load is increased, the steel will continue to elongate, resulting into
appreciable deflections and large visible cracks in the tensile concrete.
Failure under this condition is ductile and will give warning to the user of
the structure to decrease the load or apply remedial measure.

19. SPACING LIMITS OF REINFORCEMENT
Beams
 The minimum clear spacing between parallel bars in a layer should
be db( bar diameter) but not less 25 mm.
 Where parallel reinforcement is placed in two or more layers,bars in
the upper layer should be directly placed above bars in the bottom
layer with clear distance between layers not less than 25 mm.
Columns
 In spirally reinforced or tied reinforced compression members, clear
distance between longitudinal reinforcement shall not be less than
1.5db nor 40 mm.
Walls and Slabs
 In walls and slabs other than concrete joist construction, primary
reinforcement shall be spaced not farther than three times the slab
or wall thickness nor 450 mm.

20. MINIMUM THICKNESS OF NON-PRESTRESSED BEAMS AND
ONE WAY SLABS UNLESS DEFLECTIONS ARE COMPUTED
Member Simply Supported One end continuous Both ends continuous Cantilever
Solid One-Way
Slab
L/20 L/24 L/28 L/10
Beams or ribbed
one way slab L/16
L/18.5 L/21 L/8
Span Length L in millimeters
Values given shall be used directly for members with normal density concrete
(Wc = 2300 kg/m3) and Grade 60 (415 MPa) reinforcement. For other conditions,
the values shall be modified as follows:
For structural lightweight concrete having unit weights of 1500 -2000 kg/m3 the
values shall be multiplied by ( 1.65 – 0.0005 Wc) but not less than 1.09, where Wc
is the unit mass in kg/m3.
For fy other 415 MPa, the values shall be multiplied by
( 0.4 +fy/700)

21. FLEXURAL ANALYSIS : BEAMS REINFORCED FOR TENSION
 Case I : Steel yields at failure (pmax ≥ p , fs ≥ fy)
As
Compression
zone a C = 0.85fc’ab
T = Asfy
(d – a/2) Mu
0.85fc’
b
d
Stress Diagram
b
fc
f
A
a
y
s
'
85
.
0

Depth of concrete stress block Ultimate moment capacity
Mu = Ф 0.85fc’ab(d – a/2)
Mu = Ф Asfy (d – a/2)

22. GENERAL PROCEDURE FOR ANALYSIS : CASE I
GIVEN:b,d,AS fc’,fy
REQUIRED : MU
1. Check for ductility requirements
p = As/bd
β1 = 0.85 if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not
be less than 0.65
pmax = 0.75pb pmin =1.4/fy
pmin ≤ p≤ pmax
2. Solve for the depth of the concrete stress block
3. Check for minimum depth if necessary
4. Determine MU
Mu = Ф 0.85fc’ab(d – a/2) or Mu = Ф Asfy (d – a/2)
Units:
If As is in mm2, fc’ and fy in MPa, a,b and d in mm then Mu is in
N.mm. Dividing this by 106 changes N.mm to kN.m
5. Solve for any other requirement if there are any.
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



b
fc
f
A
a
y
s
'
85
.
0


23. PROBLEMS
 CE Board May 1981
 A rectangular beam with b =250 mm and d =460 mm is reinforced for
tension only with 3 – 25 mm bars. The beam is simply supported on
a span of 6 m and carries a uniform dead load of 12 kN/m. Calculate
the uniform live load the beam can carry. Concrete weighs 23 kN/m3
and steel covering is 60 mm. fc’ = 20.7 MPa, fy =276 MPa. Also
check for minimum depth requirement.
 Solution
005
.
0
276
4
.
1
4
.
1
013
.
0
)
460
(
250
62
.
1472
62
.
1472
4
)
25
(
3
min
2
2








fy
p
bd
A
p
mm
A
s
s

  03711
.
0
276
)
276
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
600
600
'
85
.
0 1





y
y
b
f
f
fc
p


24. p
p
p b 


 0278
.
0
)
03711
.
0
(
75
.
0
75
.
0
max
Steel yields at failure
mm
b
fc
f
A
a
y
s
4
.
92
250
)
7
.
20
(
85
.
0
)
276
(
62
.
1472
'
85
.
0



m
kN
a
d
f
A
M y
s
u .
37
.
151
)
10
(
)
2
4
.
92
460
(
276
)
62
.
1472
(
9
.
0
)
2
( 6





m
kN
W
W
L
W
M
u
u
u
u
/
64
.
33
8
)
6
(
37
.
151
8
2
2


 L
D
u W
W
W 7
.
1
4
.
1 


25. 460
60
250
Weight of the beam
WB=bDWc
m
kN
WB /
3
23
)
52
.
0
(
25
.
0 

Total dead load
m
kN
WD /
15
3
12 


m
kN
W
W
L
L
/
42
.
7
7
.
1
)
15
(
4
.
1
64
.
33



Minimum required depth
mm
mm
f
L
d
y
520
85
.
297
)
700
276
4
.
0
(
16
6000
)
700
4
.
0
(
16
min 






26. CE Board May 1985
A 350 mm x 500 mm rectangular beam is reinforced for tension only with 5 of
28 mm diameter bars. The beam has an effective depth of 446 mm.
fc’ =34.5MPa, fy = 414 MPa. Determine the following Ultimate moment
capacity in kN.m

27. 350
446
As
2
2
76
.
3078
4
)
28
(
5
mm
As 


00338
.
0
414
4
.
1
0197
.
0
)
446
(
350
76
.
3078
min 




p
bd
A
p s
814
.
0
)
30
5
.
34
(
008
.
0
85
.
0
)
30
'
(
008
.
0
85
.
0
1
1








 fc
  03412
.
0
414
)
414
600
(
)
600
(
814
.
0
)
5
.
34
(
85
.
0
600
600
'
85
.
0 1





y
y
b
f
f
fc
p

p
p
p b 


 0256
.
0
)
03412
.
0
(
75
.
0
75
.
0
max
Steel yields at failure

28. mm
b
fc
f
A
a
y
s
18
.
124
350
)
5
.
34
(
85
.
0
)
414
(
76
.
3078
'
85
.
0



m
kN
a
d
f
A
M y
s
u .
4
.
440
)
10
(
)
2
18
.
124
446
(
414
)
76
.
3078
(
9
.
0
)
2
( 6






29. Case II : Steel does not yield at failure ( pmax < p , fs < fy )
General Procedure for Analysis : Case II
Given: b,d,As,fc’,fy
Required : Mu
1. Check for ductility requirements
p = As/bd
β1 = 0.85 if fc’≤ 30 MPa
β1 = 0.85 - 0.008( fc’- 30) if fc’ > 30 MPa but β1 shall not
be less than 0.65
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1




30. pmax = 0.75pb
p > pmax
2. Using the stress and strain diagram solve for fs and a.
Stress Diagram Strain Diagram
b
0.85fc’
C=0.85fc’ab
T =Asfs εs
d
c
d-c
d-a/2
0.003
As
Mu NA
From the stress Diagram :
∑F =0 C = T 0.85fc’ab = Asfs EQ 1

31. From the strain diagram
:
c
c
d
s 

003
.
0

EQ 2
1
200000


a
c
fs
s


substitute in EQ 2, combine with EQ1 to solve for fs and a
3. Determine Mu
Mu = Ф 0.85fc’ab(d – a/2)
Mu = Ф Asfs (d – a/2)
Problem:
A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm,
fc’ =27.6 MPa,fy =414 MPa. Calculate the ultimate moment capacity.

32. 300
500
As
2
2
49
.
4825
4
)
32
(
6
mm
As 


00338
.
0
414
4
.
1
032
.
0
)
500
(
300
49
.
4825
min 




p
bd
A
p s
  02850
.
0
414
)
414
600
(
)
600
(
85
.
0
)
6
.
27
(
85
.
0
600
600
'
85
.
0 1





y
y
b
f
f
fc
p

p
p
p b 


 0214
.
0
)
02850
.
0
(
75
.
0
75
.
0
max
Tension Steel
does not yield at failure
∑F =0 C = T 0.85fc’ab = Asfs
1
.
46
.
1
49
.
4825
300
)
6
.
27
(
85
.
0
EQ
a
f
f
a
s
s




33. c
c
d
s 

003
.
0

1
200000


a
c
fs
s


2
.
)
425
(
600
85
.
0
85
.
0
)
85
.
0
(
500
600
85
.
0
85
.
0
500
600
500
)
003
.
0
(
200000
1
1
EQ
a
a
f
a
a
f
a
a
f
a
a
f
s
s
s
s












34. mm
a
a
a
a
a
a
a
a
EQ
EQ
22
.
260
2
)
53
.
174657
(
4
)
76
.
410
(
96
.
410
0
53
.
174657
96
.
410
600
255000
46
.
1
)
425
(
600
46
.
1
2
.
1
.
2
2
2













MPa
MPa
fs 414
94
.
379
22
.
260
)
22
.
260
425
(
600




m
kN
M
a
d
f
A
M
u
s
s
u
.
34
.
610
)
10
(
)
2
22
.
260
500
(
94
.
379
)
49
.
4825
(
9
.
0
)
2
( 6






35. Plate #1 : Beams Reinforced for tension
1. A simply supported beam 6 m long is 350 mm wide has an effective
depth of 500 mm. It supports a uniform dead load of 12 kN/m and
a concentrated live load applied at the midspan. If it is reinforced
with 6 of 22 mm diameter bars, fc’ = 20.7 MPa, fy = 414 MPa, concrete
weighs 23 kN/m3, determine the maximum value of this concentrated load .
Use concrete cover of 70 mm.
2. A rectangular beam reinforced for tension has b = 300 mm, d = 480 mm
The beam is reinforced with 7 of 25 mm bars with fc’ = 21 MPa, fy =415 MPa.
If the beam is a cantilever beam 3 m long and supports a uniform dead load
of 15 kN/m( including its own weight) applied along its entire length, calculate
the maximum value of the concentrated live load that can be applied at the
free end.
3. A reinforced concrete beam rectangular beam 300 mm wide has an effective
depth of 450 mm and is reinforced for tension only. Assuming fc’ = 27 MPa.
fy = 350 MPa, determine the required steel area that would produce balance
strain condition.

36. 2
2
8
.
2280
4
)
22
(
6 mm
As 

 02137
.
0
414
)
414
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0



b
p
013
.
0
)
500
(
350
8
.
2280



bd
A
p s
p
p
p b 

 016
.
0
75
.
0
max
Tension steel
Yields at failure
p
f
p
y



 00338
.
0
414
4
.
1
4
.
1
min
28
.
153
350
)
7
.
20
(
85
.
0
)
414
(
8
.
2280
'
85
.
0



b
fc
f
A
a
y
s
m
kN
a
d
f
A
M y
s
u .
78
.
359
)
10
(
)
2
28
.
153
500
(
414
)
8
.
2280
(
9
.
0
)
2
( 6





Solution to #1
Weight of the beam
m
kN
WB /
59
.
4
)
57
.
0
)(
35
.
0
(
23 

Total Dead
m
kN
WD /
59
.
16
59
.
4
12 



37. 4
7
.
1
8
4
.
1
7
.
1
4
.
1
2
L
P
L
W
M
M
M L
D
L
D
u 



4
)
6
(
7
.
1
8
)
6
)(
59
.
16
(
4
.
1
78
.
359
2
L
P

 kN
PL 1
.
100

Solution to #2
2
2
1
.
3436
4
)
25
(
7 mm
As 

 02161
.
0
415
)
415
600
(
)
600
(
85
.
0
)
21
(
85
.
0



b
p
024
.
0
)
480
(
300
1
.
3436



bd
A
p s
p
p
p b 

 0162
.
0
75
.
0
max
Tension steel does
Not Yield at failure
0.85fc’ab = Asfs 0.85(21)300a=3436.1fs fs =1.56a EQ.1

38. c
c
d
s 

003
.
0

EQ 2
1
200000


a
c
fs
s


1
1
003
.
0
)
200000
(


a
a
d
fs


600
)
( 1
a
a
d
fs



600
]
)
480
)(
85
.
0
[(
a
a
fs


a
a
fs
600
244800 

EQ. 1 = EQ.2
a
a
a
600
244800
56
.
1

 a
a 600
244800
56
.
1 2


a
a 6
.
384
923
,
156
2

 0
923
,
156
6
.
384
2


 a
a

39. 0
923
,
156
6
.
384
2


 a
a
mm
a 04
.
248
2
)
156923
(
4
)
6
.
384
(
6
.
384 2





y
s f
MPa
f 


 94
.
386
04
.
248
)
04
.
248
(
600
244800
m
kN
a
d
f
A
M s
s
u .
97
.
425
)
10
(
)
2
04
.
248
480
(
94
.
386
)
1
.
3436
(
9
.
0
)
2
( 6





)
2
(
a
d
f
A
M s
s
u 
 

40. PL
15 kN/m
3 m
L
P
L
W
M
M
M L
D
L
D
u 7
.
1
2
4
.
1
7
.
1
4
.
1
2




3
7
.
1
2
)
3
(
15
4
.
1
97
.
425
2
L
P


kN
PL
65


41. FLEXURAL ANALYSIS: BEAMS REINFORCED FOR TENSION &
COMPRESSION ( DOUBLY REINFORCED BEAMS )
As’
As
As’
As2
b STRESS DIAGRAMS STRAIN DIAGRAM
0.85fc’ 0.003
d’ a
d-a/2 Mu1
Mu2
Mu
C2=As’fs’ c εs’
c-d’
d-c
d-d’
C1=0.85fc’ab
T1=As1fy T2=As2fy
As1
εs
d
Criteria for adding compression reinforcement : p > 0.75pb
d’
Compression is resisted by concrete Compression is resisted by As’

42. Compression reinforcement is provided to ensure ductile failure ( tension steel must
yield) thus the stress in tension steel must always be equal to fy. On the other hand
the stress in compression steel may be equal to or less than fy. This stress must
always be checked.
Maximum permissible tensile steel area – NSCP states that for members with
compression reinforcement, the portion of pb equalized by compression reinforcement
need not be multiplied by the 0.75 factor thus
fy
f
A
bd
p
A s
s
b
s
,
'
max 75
.
0 

Stress in compression steel
From the strain diagram
c
d
c
s '
003
.
0
'



200000
'
' s
s
f


1

a
c 
a
d
a
fs
)
(
600 '
1
' 


also and
then

43. Other double reinforced beam formulas ( derived from stress diagrams)
Mu = Mu1 +Mu2
As = As1 + As2
C1 = T1
0.85fc’ab =As1fy
C2 = T2
As2 fy = As’fs’ if fs’ = fy As2 =As’
Mu1 =ФO.85fc’ab(d-a/2) Mu1 = ФAs1fy(d-a/2)
Mu2 =ФAs2fy(d-d’) Mu2 =ФAs’fs’(d-d’)
If fs’=fy
Mu2 =ФAs’fy(d-d’)

44. FLEXURAL ANALYSIS :DOUBLY REINFORCED BEAMS
CASE 1 :COMPRESSION AND TENSION STEEL YIELDS AT FAILURE
GIVEN :B,D,D’,AS,AS’,FC’,FY
REQ’D :MU
1. Assume that compression steel yields at failure
fs’ = fy As’ = As2 As1 = As - As’
2. Solve for a
3. Solve for fs’
b
fc
f
A
a y
s
'
85
.
0
1

a
d
a
fs
)
(
600 '
1
' 


4. If fs’ ≥ fy
Mu1 = ФAs1fy(d-a/2)
Mu2 =ФAs’fy (d-d’)
Mu = Mu1 +Mu2
5. Check for yielding of tension steel
'
max 75
.
0 s
b
s A
bd
p
A 
 ≥ As

45. Problem#1:
Determine the permissible ultimate moment capacity of the beam
shown in figure. fc’= 20.7 MPa, fy = 345 MPa.
2 of 28 mm
4 of 36 mm
600 mm
60 mm
350 mm
Case 2 : Compression steel does not yield at failure
tension steel yields at failure
Given :b,d,d’,As,As’,fc’,fy
Req’d :Mu

46. 1. Assume that compression steel yields at failure
fs’ = fy As’ = As2 As1 = As - As2
2. Solve for a
b
fc
f
A
a y
s
'
85
.
0
1

a
d
a
fs
)
(
600 '
1
' 


3. Solve for fs’
4. If fs’< fy
From the stress diagrams
∑F = 0 C1 + C2 = T1 + T2
0.85fc’ab + As’fs’ = As1fy + As2fy As1fy + As2fy = Asfy
0.85fc’ab + As’fs’ = Asfy EQ.A
a
d
a
fs
)
(
600 '
1
' 

 EQ.B
5.Using EQ. A and EQ. B solve for a and fs’

47. 6. Solve for Mu
Mu1 = Ф0.85fc’ab(d-a/2)
Mu2 =ФAs’fs’(d-d’)
Mu = Mu1 +Mu2
7. Check for yielding of tension steel
'
max 75
.
0 s
b
s A
bd
p
A 
 ≥ As
As’ = 775mm2
As =3625 mm2
350 mm
63 mm
600 mm
Problem#2:
Determine the permissible ultimate moment capacity of
the beam shown in figure. fc’= 27.5 MPa, fy = 345 MPa.

48. Problem#1:
Determine the permissible ultimate moment capacity of the beam
shown in figure. fc’= 20.7 MPa, fy = 345 MPa.
2 of 28 mm
4 of 36 mm
600 mm
60 mm
350 mm
Solution to Problem #1
2
2
5
.
4071
4
)
36
(
4 mm
As 


2
1 2840
5
.
1231
5
.
4071
' mm
As
As
As 




2
2
5
.
1231
4
)
28
(
2
' mm
As 


Assume that compression steel yields at failure

49. mm
b
fc
f
A
a
y
s
1
.
159
350
)
7
.
20
(
85
.
0
)
345
(
2840
'
85
.
0
1



y
s f
MPa
a
d
a
f 




 67
.
407
1
.
159
)])
60
(
85
.
0
[
1
.
159
(
600
)
(
600 '
1
' 
Compression steel yields at failure
m
kN
a
d
f
As
Mu y .
94
.
458
10
)
2
1
.
159
600
)(
345
(
2840
9
.
0
)
2
( 6
1
1 



 
m
kN
d
d
f
As
Mu y .
48
.
206
10
)
60
600
)(
345
(
5
.
1231
9
.
0
)
'
(
' 6
2 



 
m
kN
Mu
Mu
Mu .
42
.
665
48
.
206
94
.
458
2
1 





50. Check for yielding of tension steel yields
  0275
.
0
345
)
345
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
f
f
600
600
'
fc
85
.
0
p
y
y
1
b 





failure
at
yields
steel
nsion
te
mm
5
.
4071
mm
7
.
5562
5
.
1231
)
600
(
350
)
0275
.
0
(
75
.
0
A
A
bd
p
75
.
0
A
fy
fs'
fy
f
A
bd
p
75
.
0
A
2
2
max
s
'
s
b
max
s
,
s
'
s
b
max
s










51. mm
b
fc
f
A
a
y
s
18
.
120
350
)
5
.
27
(
85
.
0
)
345
(
2850
'
85
.
0
1



y
s f
MPa
a
d
a
f 




 65
.
332
18
.
120
)])
63
(
85
.
0
[
18
.
120
(
600
)
(
600 '
1
' 
Compression steel does not yield at failure
Solution to #2
2
1 2850
775
3625
' mm
As
As
As 




∑Fx =0 0.85fc’ab + As’fs’ = Asfy
0.85(27.5)350a + 775fs’ = 3625(345)
10.56a +fs’ = 1613.7
fs’ =1613.7 – 10.56 a EQ.1

52. a
a
a
d
a
fs
)])
63
(
85
.
0
[
(
600
)
(
600 '
1
' 




2
.
)
55
.
53
(
600
'
EQ
a
a
fs 


2
.
1
. EQ
EQ 
a
a
a
)
55
.
53
(
600
56
.
10
7
.
1613



32130
600
56
.
10
7
.
1613 2


 a
a
a
0
32130
7
.
1013
56
.
10 2


 a
a
0
6
.
3042
96
2


 a
a
mm
a 12
.
121
2
)
6
.
3042
(
4
)
96
(
96 2




12
.
121
)
55
.
53
12
.
121
(
600
' 

s
f
y
s f
MPa
f 
 7
.
334
'
)
2
(
1
1
a
d
f
As
Mu y 
 
)
'
(
'
'
2 d
d
fs
As
Mu 
 
6
1
10
)
2
12
.
121
600
)(
345
)
2850
(
9
.
0 

Mu
m
kN
Mu .
36
.
477
1 
6
2
10
)
63
600
(
7
.
334
)
775
(
9
.
0 

Mu
m
kN
Mu .
36
.
125
2 
m
kN
Mu
Mu
Mu .
72
.
602
2
1 


2
'
max 6
.
6500
'
75
.
0 mm
f
fs
A
bd
p
A
y
s
b
s 


0365
.
0
345
)
345
600
(
)
600
(
7
.
334
)
5
.
27
(
85
.
0



b
p
yields
TS
3625 2
max mm
As 

53. bf
t
bw
d
Cf =0.85fc’(bf-bw) t
a
bw
Tf =Asffy Tw =Aswfy
Cw=0.85fc’abw c
0.85fc’ 0.85fc’ 0.003
d-t/2 d-a/2
Muf Muw
Mu
As Asf Asw
(bf –bw)
εs
Compression is resisted by the Compression is resisted by the
overhanging flange Web
Where
bf = flange width bw = width of the web t = thickness of the slab
Stress Diagrams Strain
Diagram
T – BEAMS
Reinforced concrete floor systems, roof ,decks etc. are almost always
monolithic. Forms are built for beam soffits and sides and for the underside of
the slabs, and the entire construction is poured at once. Beam reinforcement and
stirrups extend up into the slab thus part of the slab will act with the upper part of
the beam to resist longitudinal compression. The resulting shape of the beam is
in the form of a T rather than rectangular. Figure below shows the stress and
strain diagrams of a T- beam.

54. Code requirements of T- beams
1. In T-beam construction, the flange and the web shall be built
integrally or effectively bonded together
2. The width of the flange effective as a T- beam shall not exceed ¼ of
the span , and the effective overhanging flange on each side of the
web shall not exceed :
a) eight times the slab thickness
b) ½ the clear spacing to the next web
3. For beams with slab on one side only, the effective overhanging
flange shall not exceed :
a) 1/12 the span length of the beam
b) 6 times the slab thickness
c) ½ the clear distance to the next web
w
f
b
w
p
p
p
p 

 )
(
75
.
0
max
For yielding of tension steel
Where
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



d
b
A
p
w
s
w

d
b
A
p
w
sf
f


55. bw’ S3 bw S2 bw S1
bf’ bf
t
For interior beam:
bf is the smallest of
1. bf = L/4
2. bf = bw + 16t
3. bf = S1 /2 + S2 /2 + bw
For End beams:
bf’ is the smallest of
4. bf’ = L/12 + bw’
5. bf’ = bw’ + 6t
6. bf’ = S3/2 + bw’
For Symmetrical interior beam (S1 =S2 = S3)
bf is the smallest of
7. bf = L/4
8. bf = bw + 16t
9. bf = center to center spacing of beams

56. T- BEAMS FORMULAS
 From the stress diagrams sw
sf
s A
A
A 

uw
uf
u M
M
M 

y
w
f
sf
w
f
y
sf
f
f
f
t
b
b
fc
A
t
b
b
fc
f
A
T
C
)
(
'
85
.
0
)
(
'
85
.
0





w
sw
sw
w
w
w
b
fc
fy
A
a
fy
A
ab
fc
T
C
'
85
.
0
'
85
.
0



)
2
(
)
2
(
)
(
'
85
.
0
t
d
fy
A
M
t
d
t
b
b
fc
M
sf
uf
w
f
uf







)
2
(
)
2
(
'
85
.
0
a
d
f
A
M
a
d
ab
fc
M
y
sw
uw
w
uw







57. bf
z
z
t
As As
NA
NA
The compression block of a T- beam can fall either within the flange
only or partly in the web. If it falls within the flange, the rectangular
beam formulas apply, if it falls partly on the web the beam should be
considered as a T- beam .
Criterion for selection of analysis:
0.85fc’bfz = Asfy
f
y
s
b
fc
f
A
z
'
85
.
0

if z < t wide rectangular beam
if z > t T-beam

58. FLEXURAL ANALYSIS OF T- BEAMS ( Z > T)
 Given: bw,t,As,d,center to center spacing of beams
(assuming symmetrical interior beam), L, fc’,fy
Required: MU
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Determine if it is to be analyzed as T- beam
f
y
s
b
fc
f
A
z
'
85
.
0


59. z > t
3. Solve for Asf
y
w
f
sf
f
t
b
b
fc
A
)
(
'
85
.
0 

d
b
A
p
d
b
A
p
w
sf
f
w
s
w


y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



w
f
b
w p
p
p
p 

 )
(
75
.
0
max
4. Check for yielding of tension steel
5. Solve for Muf
)
2
(
)
2
(
)
(
'
85
.
0
t
d
fy
A
M
t
d
t
b
b
fc
M
sf
uf
w
f
uf








60. 6. Solve for Asw sf
s
sw A
A
A 

w
y
s
b
fc
f
A
a
'
85
.
0

)
2
(
)
2
(
'
85
.
0
a
d
f
A
M
a
d
ab
fc
M
y
sw
uw
w
uw






uw
uf
u M
M
M 

7. Solve for a
8. Solve for Muw
9. Solve for Mu

61. Problem :
A reinforced concrete T- beam spaced at 2.0 m on centers has a
span of 3.0 m with a slab thickness of 100 mm. The effective depth
is 750 mm and the width of the web is 350 mm. The beam is
reinforced with steel of area 5200 mm2. If fc’ = 20.7 MPa and
fy= 345 MPa, calculate the ultimate moment capacity.
Solution
Determine the effective flange width bf
bf = L/4 = 3000/4 = 750 mm
bf = bw + 16t = 350+16(100) = 1950 mm
bf = center to center spacing of beams = 2000 mm
Use bf = 750 mm
Determine if it is to be analyzed as T- beam
mm
mm
b
fc
f
A
z
f
y
s
100
94
.
135
)
750
)(
7
.
20
(
85
.
0
)
345
(
5200
'
85
.
0




Analyze as T - beam

62. 0078
.
0
)
750
(
350
2040
0198
.
0
)
750
(
350
5200






d
b
A
p
d
b
A
p
w
sf
f
w
s
w
0275
.
0
345
)
345
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

)
(
75
.
0
max f
b
w p
p
p 

2
2040
345
100
)
350
750
)(
7
.
20
(
85
.
0
)
(
'
85
.
0
mm
f
t
b
b
fc
A
y
w
f
sf 




w
w p
p 


 02649
.
0
)
0078
.
0
0275
.
0
(
75
.
0
max
tension steel yields at failure

63. 6
10
)
2
100
750
(
345
)
2040
(
9
.
0
)
2
(




t
d
fy
A
M sf
uf 
m
kN
M
M
M uw
uf
u .
42
.
1092
03
.
649
39
.
443 




2
3160
2040
5200 mm
A
A
A sf
s
sw 




mm
mm
b
fc
f
A
a
w
y
sw
100
03
.
177
350
)
7
.
20
(
85
.
0
)
345
(
3160
'
85
.
0




m
kN
a
d
f
A
M y
sw
uw .
03
.
649
10
)
2
03
.
177
750
(
345
)
3160
(
9
.
0
)
2
( 6





m
kN
Muf .
39
.
443


64. Flexural analysis of T- beams ( z < t)
Given: bw,t,As,d,center to center spacing of beams(assuming
symmetrical interior beam), L, fc’,fy
Required: MU
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
f
y
s
b
fc
f
A
z
'
85
.
0

2. Determine if it is to be analyzed as T- beam
z < t
z=a

65. )
2
(
z
d
f
A
M y
s
u 

3. Solve for MU
4. Check for yielding of tension steel
bd
A
p s

y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



p
p
p b 
 75
.
0
max
Problem :
A reinforced concrete T- beam has an effective flange width of
1500 mm , slab thickness of 100 mm, effective depth of 600 mm
and the width of the web is 250 mm. The beam is reinforced with
steel of area 4500 mm2. If fc’ = 20.7 MPa and fy= 345 MPa,
calculate the ultimate moment capacity.

66. mm
mm
b
fc
f
A
z
f
y
s
100
8
.
58
)
1500
)(
7
.
20
(
85
.
0
)
345
(
4500
'
85
.
0




Analyze as wide rectangular beam
solution
m
kN
z
d
f
A
M y
s
u .
3
.
797
10
)
2
8
.
58
600
(
345
)
4500
(
9
.
0
)
2
( 6





005
.
0
)
600
(
1500
4500



bd
A
p s
0275
.
0
345
)
345
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

p
p
p b 

 0206
.
0
75
.
0
max
tension steel yields at failure

67. Plate #2 Doubly Reinforced beams
1. A rectangular concrete beam has a width of 300 mm and an effective
depth to bottom bars of 450 mm. The beam is reinforced with six
32 mm bars and two 28 mm top bars located 65 mm from the top
of the beam. If fc’=34.5 MPa, fy =345 MPa, calculate the ultimate
moment capacity of the beam.
2. Calculate the ultimate moment capacity of the beam shown in figure.
fc’=34.5 MPa, fy = 415 MPa.
As’ = 1850mm2
As =4820 mm2
80 mm
700 mm
400 mm

68. Non- Rectangular Beams
To deal with these beams, the code requirements and principles
of rectangular beams are applied. Stress and strain diagrams plays
an important role in establishing the formulas that will analyze these
beams.
375 mm
75 mm
375 mm
3 of 16 mm
Problem
Compute the ultimate moment capacity
Of the beam shown in figure. Assume
fc’=21 MPa, fy = 345 MPa

69. 375 mm
75 mm
375 mm
3 of 16 mm
Ac
0.003
s

c
375 - c
x
a
C=0.85fc’Ac
T=Asfy
375 - 2a/3
Solution
2
2
2
.
603
4
)
16
(
3
mm
As 


Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy 0.85(21)Ac =603.2(345) Ac =11,658.48 mm2
Stress diagram Strain Diagram

70. ax
Ac
2
1

450
375

a
x
1
.
833
.
0 EQ
a
x 

)
833
.
0
(
2
1
48
.
11658 a
a
 mm
a 3
.
167

mm
a
c 82
.
196
85
.
0
3
.
167
1




From the strain diagram
c
c
s 

375
003
.
0

200000
s
s
f


82
.
196
82
.
196
375
003
.
0
)
200000
(


s
f
y
s f
MPa
f 
 18
.
543 tension steel yields at failure
By similar triangles
m
kN
a
fy
A
M s
u .
34
.
49
10
)
3
3
.
167
]
2
[
375
(
345
)
2
.
603
(
9
.
0
)
3
2
375
( 6






71. 400 mm
75 mm
400 mm
3 of 16 mm
Problem
Compute the ultimate moment capacity of the beam shown in
figure. Assume fc’=20.7 MPa, fy = 345 MPa

72. 400 mm
75 mm
400 mm
3 of 16 mm
Ac
0.003
s

c
400 - c
x
a
C=0.85fc’Ac
T=Asfy
400 - 2a/3
Solution
2
2
2
.
603
4
)
16
(
3
mm
As 


Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy 0.85(20.7)Ac =603.2(345) Ac =11,829.8 mm2
Stress diagram Strain Diagram

73. ax
Ac
2
1

475
400

a
x
1
.
842
.
0 EQ
a
x 

)
842
.
0
(
2
1
8
.
11829 a
a
 mm
a 63
.
167

mm
a
c 21
.
197
85
.
0
63
.
167
1




From the strain diagram
c
c
s 

375
003
.
0

200000
s
s
f


21
.
197
21
.
197
400
003
.
0
)
200000
(


s
f
y
s f
MPa
f 
 98
.
616 tension steel yields at failure
By similar triangles
m
kN
a
fy
A
M sf
u .
98
.
53
10
)
3
63
.
167
]
2
[
400
(
345
)
2
.
603
(
9
.
0
)
3
2
400
( 6






74. BEAM DEFLECTIONS
'
7
.
0 fc
fr 
Unless stiffness values are obtained by a more comprehensive
analysis,immediate deflection shall be computed with the
modulus of elasticity of concrete and with an effective moment
of inertia as follows, but not greater than Ig .
cr
a
cr
g
a
cr
e I
M
M
I
M
M
I

























3
3
1
Where
t
g
r
cr
y
I
f
M 
fr = modulus of rupture of concrete
fc’ in MPa

75. Ma = maximum moment in member at stage deflection is computed.
Ig = moment of inertia of gross concrete section about centroidal
axis, neglecting reinforcement.
Icr = moment of inertia of cracked section transformed to concrete
yt = distance from centroidal axis of gross cross section , neglecting
reinforcement, to extreme fiber in tension
b b
x
d-x
nAs
As
d
TRANSFORMED SECTION
n = modular ratio
c
s
E
E
n 
To Locate nuetral axis:
Moment of area of Moment of area of
concrete about NA = steel about NA
)
(
2
2
x
d
nA
x
b s 

Nuetral axis

76. To determine Icr
2
3
)
(
3
x
d
nA
bx
I s
cr 


yt = distance from centroidal axis of gross section
neglecting reinforcement, to extreme fiber in tension.
Unless stiffness values are obtained by a more comprehensive analysis,
additional long term deflection resulting from creep and shrinkage of
flexural members shall be determined by multiplying the immediate
deflection caused by the sustained load considered, by the factor
'
50
1 p




where p’ shall be the value of reinforcement ratio for non prestress
compression reinforcement at midspan for simple and continuous spans,
and at support for cantilever. It is permitted to assume the time-dependent
factor ξ for sustained load to be equal to

77. 5 years or more ………………………….. 2
12 months ………………………….. 1.4
6 months ………………………….. 1.2
3 months ………………………….. 1.0
Problem:
A concrete beam 6 m long is 300 mm wide and 600 mm deep and
carries a dead load of 9 kN/m and live load of 12 kN/m. The beam
is reinforced for tension only with four 25 mm bars with an effective
depth to tension bars of 530 mm. fc’ = 20.7 MPa , fy = 345 MPa ,
fr = 2.832 MPa, Ec = 20,000 MPa, Es = 200,000 MPa.
Covering of bars is 70 mm.
a) Calculate the maximum instantaneous deflection due to dead load
and live load.
b) Calculate the deflection due to the same loads after five years
assuming that 30% of the live load is sustained.

78. Gross moment of inertia
4
6
3
3
10
5400
12
)
600
(
300
12
mm
x
bh
Ig 


mm
N
x
y
I
f
M
t
g
r
cr .
10
97
.
50
300
10
)
5400
(
832
.
2 6
6



m
kN
Mcr .
97
.
50

b=300 b
x
530-x
nAs
As
d=530
10
20000
200000



c
s
E
E
n
)
(
2
2
x
d
nA
x
b s 

2
2
2
.
1963
4
)
25
(
4
mm
As 


Transformed section

79. )
(
2
2
x
d
nA
x
b s 
 )
530
)(
5
.
1963
(
10
2
300
2
x
x


x
x 9
.
130
69377
2

 0
69377
9
.
130
2


 x
x
mm
x 96
.
205
2
)
69377
(
4
)
9
.
130
(
9
.
130 2





2
3
2
3
)
96
.
205
530
)(
5
.
1963
(
10
3
)
96
.
205
(
300
)
(
3





 x
d
nA
bx
I s
cr
4
6
10
38
.
935
,
2 mm
x
Icr 
a) m
kN
L
W
M T
a .
5
.
94
8
)
6
)(
9
12
(
8
2
2





80. 4
6
10
1
.
322
,
3 mm
x
Ie 
Effective moment of inertia
cr
a
cr
g
a
cr
e I
M
M
I
M
M
I

























3
3
1
a) Instantaneous deflection
mm
I
E
L
W
e
c
T
44
.
3
)
10
(
1
.
3322
)
20000
(
384
)
6000
)(
21
(
5
384
5
6
4
4





81. a) Long term deflection
Since only 30% l of the live load is sustained
m
kN
WT /
7
.
14
)
9
(
3
.
0
12 


mm
I
E
L
W
e
c
T
41
.
2
)
10
(
1
.
3322
)
20000
(
384
)
6000
)(
7
.
14
(
5
384
5
' 6
4
4




2
)
0
(
50
1
2
50
1 '





p


Long term deflection
mm
L 26
.
8
)
41
.
2
(
2
44
.
3
' 



 



82. Compression
Zone
d
a
0.85fc’
C = 0.85fc’ab
0.003
T = Asfy
s
y
s
E
f


b
c
Stress Diagram Strain Diagram
Design for Flexure : Beams Reinforced for tension
Derivation of designing formulas
1
.
)
2
( EQ
a
d
f
A
M y
s
u 

  2
.
'
85
.
0
EQ
b
fc
f
A
a
y
s


d-a/2 Mu

83. )
]
'
85
.
0
[
2
(
b
fc
f
A
d
f
A
M
y
s
y
s
u 
 
EQ.2 in EQ.1
)
]
'
85
.
0
[
2
(
d
b
fc
d
f
A
d
bd
bd
f
A
M
y
s
y
s
u 
 
)
]
'
85
.
0
[(
2
( d
fc
f
bd
A
d
bdf
bd
A
M
y
s
y
s
u 
 
bd
A
p s

'
85
.
0 fc
f
m
y

Let and

84. )
2
( d
pm
d
pbdf
M y
u 
 
)
2
1
(
2 pm
pf
bd
M y
u 
 
)
2
1
(
pm
pf
R y
u 

Let
u
u R
bd
M 2

 For proportioning of section
y
y
u f
m
p
pf
R
2
1
2


)
2
1
(
pm
pf
R y
u 

Coeffecient of resistance

85. mfy
f
m
p
pf
R y
y
u
2
}
2
1
{
2


2
2
2
p
m
p
R
mf
u
y


0
2
2
2


 u
y
R
mf
m
p
p
2
2
4
)
2
(
2 2
y
u
mf
R
m
m
p




86. 2
2
4
)
2
(
2
2
2
y
u
f
m
mR
m
m
p



2
]
2
1
[
)
2
(
2 2
y
u
f
mR
m
m
p



2
2
1
2
2
y
u
f
mR
m
m
p



2
]
2
1
1
[
2
y
u
f
mR
m
p



)
2
1
1
(
1
y
u
f
mR
m
p 

 Actual tensile steel ratio

87. NSCP COEFFECIENTS FOR CONTINUOUS BEAMS AND SLABS
NSCP states that in lieu of frame analysis, the following approximate moments
and shear are permitted for design of continuous beams and one way slabs
provided that :
There are two or more spans
Spans are approximately equal,with the large of two adjacent spans not greater
than the shorter by more than 20 %
Loads are uniformly distributed
Unit live load does not exceed three times the unit dead load
The members are prismatic
Positive moment
End spans
Discontinuous end unrestrained WULn2/11
Discontinuous end integral with support WULn2/14
Interior spans WULn2/16

88. Negative moment at exterior face of first interior support
Two spans WULn2/9
More than two spans WULn2/10
Negative moment at other faces of interior supports WULn2/11
Negative moment at face of all supports for
Slabs with span not exceeding 3 m; and beams
where ratio of column stiffness to beam stiffness
exceeds eight at each end of the span WULn2/12
Negative moment at interior face of exterior supports
for members built integrally with supports
Where support is a spandrel beam WULn2/24
Where support is a column WULn2/16
Shear in end members at face of first interior support 1.15WULn/2
Shear at face of all other supports WULn/2
Where Ln = the clear span for positive moment or shear and average of
adjacent clear spans for negative moments.

89. L1 L2
16
2
1
wL

2
1
wL
11
2
2
wL
9
2
n
wL

2
2
1 L
L
Ln


14
2
1
wL
2
15
.
1 2
wL
2
15
.
1 n
wL
Shear
Moment
Shear and moment for continuous beams or slab with two spans
discontinuous edge integral with support,discontinous end unrestrained
column column

90. L1 L2
16
2
1
wL

2
1
wL
16
2
2
wL
10
2
n
wL

2
2
1 L
L
Ln


14
2
1
wL
2
15
.
1 n
wL
2
15
.
1 n
wL
Shear
Moment
Shear and moment for continuous beams or slab with more than
two spans and discontinuous end integral with support
column column column
Spandrel beam
L3
2
3
wL
14
2
3
wL
10
2
n
wL

2
3
2 L
L
Ln


24
2
3
wL


91. L1 L2
2
1
wL
16
2
2
wL
10
2
n
wL

2
2
1 L
L
Ln


11
2
1
wL
2
15
.
1 n
wL
2
15
.
1 n
wL
Shear
Moment
Shear and moment for continuous beams or slab with more than
two spans and discontinuous end unrestrained
column column
L3
2
3
wL
11
2
3
wL
10
2
n
wL

2
3
2 L
L
Ln



92. Case 1 : Design for balanced strain condition with given dimensions
Given : b,d, fc’ and fy
Required : Steel area that would produce balance strain condition
General Procedure:
1. Solve for pb
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



2. Solve for Asb Asb = pbbd
Problem:
A reinforced concrete rectangular beam 300 mm wide has an effective
depth of 460 mm and is reinforced for tension only. If fc’=20 MPa,
fy = 300 MPa, determine the balance steel area in mm2.
032
.
0
300
)
300
600
(
)
600
(
85
.
0
)
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

Asb = pbbd=0.032(300)460 = 4416 mm2
Solution

93. )
2
1
(
pm
pf
R y
u 

u
u
R
M
bd


2
'
85
.
0 fc
f
m
y

Case 2 : Design of cross section and reinforcement from given loads
and type of beam dimensions
Given : fc’ , fy, Loads, type of beam, Wc
Required : cross section dimension ,size and number of steel bars
General Procedure:
1.Assume the weight of the beam ( DL) as 20 to 25% of (DL + LL). add
this to the given dead load.
2.By any available method, determine designing moment Mu.
3.Assume a value of steel ratio p ( 0.3pb to 0.6pb but not less than pmin).
This will provide enough allowance for the rounding off of values of
number of bars to be used.
4.Solve for the following design constants :
5. Solve for bd2

94. 6. Try a ratio of d/b ( from d= b to d = 2b ) , then solve for b
and d. Alternatively b may be assumed until a reasonable value of d is
attained. Provide concrete cover to get total thickness. Check for
minimum depth if required.
7. Compute actual weight ( W = bDWc) and compare with assume
Weight.
8. Solve for actual p
9.Solve for As: As = pbd
10.Solve for the number of bars by dividing As by the area of one bar
to be used. Round off this number to the next integer.
Problems :
Design a rectangular beam for a 6 m simple span to support a uniform
dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3.
Design a rectangular beam reinforced for tension only to support a
service dead load moment of 85 kN.m ( including its weight) and service
live load moment of 102 kN.m. Use p =0.6pb, d/b = 1.75 , fc’=28 MPa,
fy = 276 MPa.
)
2
1
1
(
1
y
u
f
mR
m
p 



95. Problems :
Design a rectangular beam for a 6 m simple span to support a uniform
dead load of 15 kN/m and uniform live load of 24 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 350 MPa,Wc=23.5 kN/m3
25
.
3
]
2
)
59
.
20
(
0104
.
0
1
)[
350
(
0104
.
0
)
2
1
(
59
.
20
)
20
(
85
.
0
350
'
85
.
0
0104
.
0
350
)
350
600
(
)
600
(
85
.
0
)
20
(
85
.
0
4
.
0
4
.
0
.
4
.
327
8
)
6
(
24
7
.
1
8
)
6
(
8
.
22
4
.
1
/
8
.
22
8
.
7
15
8
.
7
)
15
24
(
20
.
0
2
2





















pm
pf
R
fc
fy
m
p
p
m
kN
M
m
kN
W
w
y
u
b
u
D
B

96. mm
b
R
M
d
R
M
bd
u
u
u
u
570
350
)
25
.
3
(
9
.
0
)
10
(
24
.
327 6
2






Trial section 350 mm x 570mm effective depth, total depth 650 mm
Actual weight
2
.
3
)
570
)(
350
(
9
.
0
)
10
(
24
.
327
/
8
.
7
/
35
.
5
5
.
23
)
65
.
0
(
35
.
0
2
6
2






bd
M
R
m
kN
m
kN
w
u
u
B


97. 7
4
)
20
(
2035
20
2035
570
)
350
(
0102
.
0
102
.
0
)
350
)
2
.
3
)(
59
.
20
(
2
1
1
(
59
.
20
1
)
2
1
1
(
1
2
2














N
bars
mm
try
mm
pbd
A
p
f
mR
m
p
s
y
u

98. Problems :
Design a rectangular beam for a 5 m simple span to support a uniform
dead load of 12 kN/m and uniform live load of 20 kN/m applied along it’s
entire length. fc’=20 MPa, fy = 400 MPa,Wc=23.5 kN/m3.
Design a rectangular beam reinforced for tension only to support a
service dead load moment of 65 kN.m ( including its weight) and service
live load moment of 80 kN.m. Use p =0.45pb, d/b = 1.5 , fc’=20 MPa,
fy = 300 MPa.
Plate # 3 Design of beams reinforced for tension
A reinforced concrete T- beam spaced at 3.0 m on centers has a
span of 4.0 m with a slab thickness of 75 mm. The effective depth
is 750 mm and the width of the web is 300 mm. The beam is
reinforced with steel of area 4200 mm2. If fc’ = 20.7 MPa and
fy= 345 MPa, calculate the ultimate moment capacity.

99. 2
bd
M
R u
u


)
2
1
1
(
1
y
u
f
mR
m
p 


'
85
.
0 fc
f
m
y

Design of reinforcement of a beam with given moment and
cross sectional dimension
Given : b,d, Mu ,fc’ , fy,
Required : Number of steel bars
General Procedure :
1. Solve for Ru and m
2. Solve for p
3. Check for yielding of tension steel and pmin if required.
4. Solve for As : As = pbd

100. Problem :
Determine the required tension steel area for a rectangular beam with
b =250 mm, d =330mm, fc’ =20.7 MPa, fy = 414 MPa. The beam is
required to support a factored moment of 110 kN.m.
49
.
4
)
330
)(
250
(
90
.
0
)
10
(
110
2
6
2




bd
M
R u
u
53
.
23
)
7
.
20
(
85
.
0
414
'
85
.
0



fc
f
m
y
)
2
1
1
(
1
y
u
f
mR
m
p 


01276
.
0
)
414
49
.
4
)
53
.
23
(
2
1
1
(
53
.
23
1




p
Note :
If p > 0.006 no need to
check for pmin
p < 0.02 no need to
check for pmax
As = pbd
As = 0.01276(250)330
As = 1052.7 mm2
Solution

101. DESIGN OF CONTINUOUS BEAM
In the design of continuous beam, the cross section is determined
by the maximum moment obtained by any structural analysis
method or by its equivalent NSCP coefficients.
The reinforcements are designed from the moment obtained at the
different sections of maximum positive and negative moments.
Problem:
Figure shows a continuous beam of three spans with the left and right
ends discontinuous and integral with the support . Design the section
and reinforcements at critical sections using the given service uniform
loading. Given dead loads includes the weight of the beam. fc’= 28
MPa, fy =350 MPa . Use NSCP coefficients to determine the moments.

102. DL = 12 kN/m DL = 15 kN/m DL = 20 kN/m
LL =16 kN/m LL = 18 kN/m LL = 24 kN/m
4 m 5 m 6 m
A B C D E F G
Factored loads
W1 = 1.4(12)+1.7(16)= 44 W2=1.4(15)+1.7(18)= 51.6 W3 = 1.4(20)+1.7(24)=68.8
m
kN
L
w
M A .
44
16
)
4
(
44
16
2
2
1
1






Design moments by NSCP coeffecients
m
kN
L
w
MB .
28
.
50
14
)
4
(
44
14
2
2
1
1




103. m
kN
L
w
M n
C .
49
.
104
10
)
2
5
4
(
6
.
51
10
2
2
1
2







m
kN
L
w
MD .
63
.
80
16
)
5
(
6
.
51
16
2
2
2
2



m
kN
L
w
M n
E .
12
.
208
10
)
2
5
6
(
8
.
68
10
2
2
2
3







m
kN
L
w
MF .
91
.
176
14
)
6
(
8
.
68
14
2
2
3
3



m
kN
L
w
MG .
9
.
157
16
)
6
(
8
.
68
16
2
2
3
3







104. Proportioning of uniform beam size
Note: use the biggest computed design Moment Mu =208.12 kN.m
MPa
Ru 53
.
5
)
2
)
71
.
14
(
01825
.
0
1
)(
350
(
01825
.
0 


71
.
14
)
28
(
85
.
0
350
'
85
.
0



fc
f
m
y
0365
.
0
350
)
350
600
(
)
600
(
85
.
0
)
28
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

01825
.
0
0365
.
0
)
5
.
0
(
5
.
0 

 b
p
p
)
2
1
(
pm
pf
R y
u 


105. mm
b
R
M
d
u
u
410
)
250
)(
53
.
5
(
9
.
0
)
10
(
12
.
208 6




Try b = 250 mm
004
.
0
350
4
.
1
4
.
1
min 


y
f
p
Section A m
kN
Mu .
44


16
.
1
)
410
)(
250
(
9
.
0
)
10
(
44
2
6
2



bd
M
R u
u

0034
.
0
)
350
71
.
14
)
16
.
1
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

106. Use p = 0.004
2
410
410
)
250
(
004
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 2
)
20
(
)
4
(
410
2



Section B m
kN
Mu .
28
.
50

32
.
1
)
410
)(
250
(
9
.
0
)
10
(
28
.
50
2
6
2



bd
M
R u
u

0034
.
0
)
350
71
.
14
)
32
.
1
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p
Top bars

107. Use p = 0.004
2
410
410
)
250
(
004
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 2
)
20
(
)
4
(
410
2



bottom bars
Section C m
kN
Mu .
49
.
104


74
.
2
)
410
)(
250
(
9
.
0
)
10
(
49
.
104
2
6
2



bd
M
R u
u

00834
.
0
)
350
71
.
14
)
74
.
2
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

108. 2
86
.
854
410
)
250
(
00834
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 3
)
20
(
)
4
(
86
.
854
2



top bars
Section D m
kN
Mu .
63
.
80

11
.
2
)
410
)(
250
(
9
.
0
)
10
(
63
.
80
2
6
2



bd
M
R u
u

00632
.
0
)
350
71
.
14
)
11
.
2
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

109. 2
06
.
648
410
)
250
(
00632
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 3
)
20
(
)
4
(
06
.
648
2



bottom bars
Section E m
kN
Mu .
12
.
208


44
.
5
)
410
)(
250
(
9
.
0
)
10
(
12
.
208
2
6
2



bd
M
R u
u

018
.
0
)
350
71
.
14
)
44
.
5
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

110. 2
1845
410
)
250
(
018
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 6
)
20
(
)
4
(
1845
2



Top bars
Section F m
kN
Mu .
91
.
176

62
.
4
)
410
)(
250
(
9
.
0
)
10
(
91
.
176
2
6
2



bd
M
R u
u

0148
.
0
)
350
71
.
14
)
62
.
4
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

111. 2
1517
410
)
250
(
0148
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 5
)
20
(
)
4
(
1517
2



bottom bars
Section G m
kN
Mu .
9
.
157


12
.
4
)
410
)(
250
(
9
.
0
)
10
(
9
.
157
2
6
2



bd
M
R u
u

013
.
0
)
350
71
.
14
)
12
.
4
(
2
1
1
(
71
.
14
1
)
2
1
1
(
1







y
u
f
mR
m
p

112. 2
5
.
1332
410
)
250
(
013
.
0 mm
pbd
As 


Try 20 mm bars
pcs
N 5
)
20
(
)
4
(
5
.
1332
2



top bars
Gross moment of inertia using 100 mm covering
4
6
3
3
10
5400
12
)
600
(
300
12
mm
x
bh
Ig 


mm
N
x
y
I
f
M
t
g
r
cr .
10
79
.
55
300
10
)
5400
(
1
.
3 6
6




113. A B C D E F G
2 of 20
2 of 20
3 of 20
3 of 20
6 of 20
5 of 20
5 of 20
A B C D E F G
2 of 20
2 of 20
1 of 20 1 of 20 3 of 20 3 of 20 3 of 20
2 of 20 2 of 20
2 of 20 2 of 20
Placement of bars
Requirement
Layout
3 of 20

114. 250 mm
500 mm
250 mm
500 mm
250 mm
500 mm
Section at A and B Section at C Section at D
250 mm
500 mm
250 mm
500 mm
250 mm
500 mm
Section at E Section at F Section at G

115. DL = 9 kN/m DL = 12 kN/m DL = 15kN/m
LL =12 kN/m LL = 14 kN/m LL = 18 kN/m
3.6 m 4.0 m 4.5 m
Plate # 4: Design of continuous beams
Problem
Design the uniform size and reinforcements at critical section of
the continuous beam shown above. fc’= 20 MPa, fy = 300 MPa.
Given dead loads includes the weight of the beam.

116. 2
bd
M
R u
u


)
2
1
1
(
1
fy
mR
m
p u



y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



b
p
p 75
.
0
max 
DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS
Given :b,d,d’, Mu ,fc’,fy
Req’d : As, As’
General Procedure
1. Solve for Ru
2. Solve for p
3. Check if the beam needs compression reinforcement
if p > pmax then compression reinforcement is necessary

117. bd
p
As max
1 
b
fc
f
A
a y
s
'
85
.
0
1

)
2
(
1
1
a
d
f
A
M y
s
u 
 
1
2 u
u
u M
M
M 

5. Solve for a
6. Solve for Mu1
7. Solve for Mu2
4. Solve for As1
8. Solve for As2
)
'
(
2
2
d
d
f
M
A
y
u
s



9. Solve for fs
’
a
d
a
fs
)
(
600 '
1
' 


Case 1
If fs’ ≥ fy then fs’ = fy
( compression steel yields at failure)
2
'
s
s A
A 
Case 2
If fs’ < fy then use fs’
( compression steel does not yield at failure)
'
2
'
s
y
s
s
f
f
A
A 
10. Solve for As
2
1 s
s
s A
A
A 


118. Problem:
Design the reinforcement of a rectangular beam to carry a factored
moment of 272 kN.m. The beam width is 250 mm,effective depth
400mm. Use fc’ = 20.7 MPa, fy = 345 MPa, d’ =60 mm.
55
.
7
)
400
)(
250
(
9
.
0
)
10
(
272
2
6
2



bd
M
R u
u

0317
.
0
)
345
55
.
7
)
61
.
19
(
2
1
1
(
61
.
19
1




p
61
.
19
)
7
.
20
(
85
.
0
345
'
85
.
0



fc
f
m
y
0275
.
0
345
)
345
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

0206
.
0
)
0275
.
0
(
75
.
0
max 

p
p > pmax compression reinforcement is necessary
Solution

119. 2
max
1 2060
400
)
250
(
0206
.
0 mm
bd
p
As 


mm
b
fc
f
A
a
y
s
57
.
161
250
)
7
.
20
(
85
.
0
)
345
(
2060
'
85
.
0
1



m
kN
a
d
f
A
M y
s
u .
18
.
204
10
)
2
57
.
161
400
(
345
)
2060
(
9
.
0
)
2
( 6
1
1 




m
kN
M
M
M u
u
u .
82
.
67
18
.
204
272
1
2 




2
6
2
2 42
.
642
)
60
400
)(
345
(
9
.
0
)
10
(
82
.
67
)
'
(
mm
d
d
f
M
A
y
u
s 





MPa
a
d
a
fs 6
.
410
57
.
161
)])
60
(
85
.
0
[
57
.
161
(
600
)
(
600 '
1
'






Compression
Steel yields
at failure
2
2
'
42
.
642 mm
A
A s
s 
 2
2
1 42
.
2702
42
.
642
2060 mm
A
A
A s
s
s 





120. Plate # 5 :
DESIGN FOR FLEXURE :DOUBLY REINFORCED BEAMS and T BEAMS
Design the reinforcement of a rectangular beam to resist a dead load
moment of 200 kN.m(including its own weight) and a live load moment of
300 kN.m . The beam is limited in size to 350 mm by 600mm overall
depth. Steel covering ( from centroid of bars to outermost fiber is 100 mm
for both tension and compression reinforcement. Use fc’ = 27.5 MPa, fy =
414 MPa
A reinforced concrete T-beam with d = 550 mm, bw = 300 mm , slab
thickness =100 mm is 4.8 m long and spaced 3 m on centers. The beam
support a service dead load moment of 400 kN.m (including its weight) and
service live load moment of 500 kN.m. If fc’=27.5 MPa, fy = 414 MPa ,
determine the required steel area.
Design the reinforcement of a T- beam to support a uniform service dead
load of 25 kN/m and service live load of 30 kN/m on a simple span of 8 m .
Properties of the T- beam are as follows : bf = 1500 mm,
bw = 250 mm, d = 600 mm, t = 100 mm, fc’ = 20.7 MPa, fy = 345 MPa.

121. DESIGN FOR BENDING : T-BEAMS (z > t)
Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical
interior beam), L, fc’,fy
Required: As
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Solve for Asf
y
w
f
sf
f
t
b
b
fc
A
)
(
'
85
.
0 

3. Solve for Muf
)
2
(
t
d
fy
A
M sf
uf 
 
If Mu > Muf then design as T- beam

122. uf
u
uw M
M
M 

)
2
(
'
85
.
0
a
d
ab
fc
M w
uw 
 
y
w
sw
f
ab
fc
A
'
85
.
0

4. Solve for Muw
5. Solve for a
6. Solve for Asw
sw
sf
s A
A
A 

d
b
A
p
d
b
A
p
w
sf
f
w
s
w


7. Solve for As
8. Check for yielding of tension steel
y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



w
f
b
w p
p
p
p 

 )
(
75
.
0
max

123. Design the reinforcement of a T- beam to resist a factored moment of
750 kN.m. Properties of the T- beam are as follows : bf = 550 mm,
bw = 300 mm, d = 600 mm, t = 110 mm, fc’ = 20.7 MPa, fy = 345 MPa.
2
5
.
1402
345
110
)
300
550
)(
7
.
20
(
85
.
0
)
(
'
85
.
0
mm
f
t
b
b
fc
A
y
w
f
sf 




solution
m
kN
t
d
fy
A
M sf
uf .
33
.
237
10
)
2
110
600
(
345
)
5
.
1402
(
9
.
0
)
2
( 6





Mu > Muf design as T- beam
m
kN
M
M
M uf
u
uw .
67
.
512
33
.
237
750 





124. )
2
(
'
85
.
0
a
d
ab
fc
M w
uw 
 
2
6
.
3370
345
)
300
(
3
.
220
)
7
.
20
(
85
.
0
'
85
.
0
mm
f
ab
fc
A
y
w
sw 


2
1
.
4773
6
.
3370
5
.
1402 mm
A
A
A sw
sf
s 




)
2
600
(
300
)
7
.
20
(
85
.
0
)
9
.
0
(
)
10
(
67
.
512 6 a
a 

2
600
8
.
107915
2
a
a 

0
6
.
215831
1200
2


 a
a
mm
mm
a 110
3
.
220
2
)
6
.
215831
(
4
)
1200
(
1200 2






125. 008
.
0
)
600
(
300
5
.
1402
0265
.
0
)
600
(
300
1
.
4773






d
b
A
p
d
b
A
p
w
sf
f
w
s
w
Check for yielding of tension steel
0275
.
0
345
)
345
600
(
)
600
(
85
.
0
)
7
.
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

w
f
b
w p
p
p
p 




 0266
.
0
)
008
.
0
0275
.
0
(
75
.
0
)
(
75
.
0
max
Tension steel yields at failure

126. DESIGN FOR BENDING : T-BEAMS (z < t)
Given: bw,t,Mu,d,center to center spacing of beams(assuming symmetrical
interior beam), L, fc’,fy
Required: As
1. Determine the effective flange width bf
bf = L/4
bf = bw + 16t
bf = center to center spacing of beams
Use the smallest value of bf
Note: if bf is given omit step 1
2. Solve for Asf
y
w
f
sf
f
t
b
b
fc
A
)
(
'
85
.
0 


127. )
2
(
t
d
fy
A
M sf
uf 
 
3. Solve for Muf
If Mu < Muf then z < t ,design as wide rectangular beam
4. Solve for a
)
2
(
'
85
.
0
a
d
ab
fc
M f
u 
 
5. Solve for As
y
f
s
f
a
b
fc
A
'
85
.
0

d
b
A
p
f
s

y
y
b
f
f
fc
p
)
600
(
600
'
85
.
0 1



p
p
p b 
 75
.
0
max
6. Check for yielding of tension steel

128. Problem :
A reinforced concrete T-beam with bf = 810 mm, d = 300 mm,
bw = 200 mm , t = 100 mm, fc’=20.7 MPa, fy = 414 MPa is to be
designed to carry an ultimate moment of 221 kN.m Determine the
required steel area.
2
2
.
2683
414
100
)
200
810
)(
7
.
20
(
85
.
0
)
(
'
85
.
0
mm
f
t
b
b
fc
A
y
w
f
sf 




m
kN
t
d
fy
A
M sf
uf .
9
.
249
10
)
2
100
300
(
414
)
2
.
2683
(
9
.
0
)
2
( 6





solution
Design as wide rectangular beam
)
2
(
'
85
.
0
a
d
ab
fc
M f
u 
 

129.   )
2
300
(
810
)
7
.
20
(
85
.
0
9
.
0
)
10
(
221 6 a
a 

2
300
6
.
17229
2
a
a 

0
2
.
34459
600
2


 a
a
mm
a 33
.
64
2
)
2
.
34459
(
4
)
600
(
600 2




2
5
.
2214
414
33
.
64
)
810
)(
7
.
20
(
85
.
0
'
85
.
0
mm
f
a
b
fc
A
y
f
s 


0091
.
0
)
300
(
810
5
.
2214



d
b
A
p
f
s   0213
.
0
414
)
414
600
(
600
85
.
0
)
7
.
20
(
85
.
0



b
p
0091
.
0
016
.
0
)
0213
.
0
(
75
.
0
max 


p Tension steel yields at failure

130. Plate # 6: Flexural design of T-beams
A reinforced concrete T-beam with d = 550 mm, bw = 300 mm , slab
thickness =100 mm is 4.8 m long and spaced 3 m on centers. The beam
support a service dead load moment of 400 kN.m (including its weight)
and service live load moment of 500 kN.m. If fc’=27.5 MPa, fy = 414 MPa
, determine the required steel area.
Design the reinforcement of a T- beam to support a uniform service
dead load of 25 kN/m and service live load of 30 kN/m on a simple span
of 8 m . Properties of the T- beam are as follows : bf = 1500 mm,
bw = 250 mm, d = 600 mm, t = 100 mm, fc’ = 20.7 MPa, fy = 345 MPa.

131. Reinforced concrete slab are large flat plates that are supported at its
sides by reinforced concrete beams, walls, column, steel beams or by
the ground. If the slabs are supported on opposite sides only they are
called one way slab since bending will occur on one direction only.
A one way slab is considered as a wide & shallow rectangular beam.
Reinforcing steel is usually spaced uniformly over its width. One way
slabs are analyzed by considering a one meter strip, which is assumed
independent of the adjacent strips.
Maximum spacing of reinforcement
Flexural reinforcement shall not be spaced farther apart than 3 times
the slab thickness nor 450 mm.
Minimum size of flexural reinforcement = 12 mm
ONE WAY SLAB

132. The area of shrinkage reinforcement shall be
where Grade 275 deformed bars are used… 0.002bt
where Grade 415 deformed bars are used… 0.0018bt
where reinforcement with fy > 415 MPa measured
at yield strain of 0.35% are used …. 0.0018(400)bt/fy
Shrinkage and temperature reinforcement
Shrinkage reinforcement shall not be spaced farther apart than 5
times the slab thickness nor 450 mm.
Minimum size of shrinkage and temperature bars = 10 mm

133. Design of one way slab
Given: Loads, type of slab, fc’,fy,Wc
Req’d: t, size and spacing of main bars and
temperature bars
General Procedure
1. Determine the minimum slab thickness t using table for minimum
thickness of non prestressed beams and one way slab. This
thickness should be at least 75 mm
2. Compute the weight of the slab ( this is to be added to the given
dead load)
3. Calculate the design moment Mu
4. Compute the effective depth d
d = t – covering – ½ bar diameter ( minimum of 12 mm)
5. Compute design constants
2
bd
M
R u
u


'
85
.
0 fc
f
m
y
 )
2
1
1
(
1
y
u
f
mR
m
p 



134. 1000
1
s
A
A
S 
1000
1
t
t
A
A
S 
p > pmin
5. Solve for As
As = pbd
6. Solve for the spacing of bars
where : A1 = area of 1 bar
Use the smallest of the following
a)S b)3t c) 450 mm
7. Solve for area of temperature bars
At = 0.002bt , At = 0.0018bt, At =0.0018(400)bt/fy
8. Solve for the spacing of temperature bars
where : A1 = area of 1 temperature bar ( minimum of 10mm dia.)
Use the smallest of the following
a) St b) 5t c)450 mm

135. Problem:
Design a one way slab having a simple span of 3.0 m. The slab is to
carry a uniform dead load of 2.5 KPa and uniform live load of 4.2 kPa.
fc’ = 27.6 MPa , fy = 276 MPa for main bars and temperature bars.
Concrete weighs 23.5 kN/m3
)
700
4
.
0
(
20
y
f
L
t 

Solution
mm
mm
t 75
120
)
700
276
4
.
0
(
20
3000




Slab thickness
Weight of Slab (assuming 1 m wide strip)
m
kN
Ws /
82
.
2
12
.
0
)
1
(
5
.
23 

m
kN
WD /
32
.
5
82
.
2
5
.
2 


Total dead load

136. Factored uniform load
m
kN
W
W
W L
D
u /
59
.
14
)
2
.
4
(
7
.
1
)
32
.
5
(
4
.
1
7
.
1
4
.
1 




Design Moment
m
kN
L
W
M u
u .
41
.
16
8
)
3
(
59
.
14
8
2
2



Effective depth assuming 12 mm bar
mm
t
d 94
2
12
20
120
2
1
20 





 
MPa
bd
M
R u
u 06
.
2
)
94
)(
1000
(
9
.
0
)
10
(
41
.
16
2
6
2




76
.
11
)
6
.
27
(
85
.
0
276
'
85
.
0



fc
f
m
y

137. 00782
.
0
)
276
)
06
.
2
)(
76
.
11
(
2
1
1
(
76
.
11
1




p
00507
.
0
276
4
.
1
4
.
1
min 


y
f
p
2
4
.
735
94
)
1000
(
00782
.
0 mm
pbd
As 


mm
t
mm
say
A
A
S
s
360
3
150
7
.
153
)
1000
(
4
.
735
4
)
12
(
1000
2
1






Using 12 mm bars
2
240
120
)
1000
(
002
.
0
002
.
0 mm
bt
At 


mm
t
mm
say
A
A
S
t
t 600
5
320
327
)
1000
(
240
4
)
10
(
1000
2
1






Using 10 mm bars
Temperature bars

138. 12 mm main bars @ 150 mm o.c
10 mm temp bars @ 320 mm o.c
120 mm

139. Problem:
Design a 4 m long one way slab with one end discontinuous as shown in
the figure . The slab is to carry a uniform dead load of 3.6 KPa
and uniform live load of 4.0 kPa. fc’ = 20.7 MPa , fy = 415 MPa for
main bars and fy = 276 MPa for temperature bars.
Concrete weighs 22.56 kN/m3.
4m 4m
column column column
C B A
mm
say
L
t 170
67
.
166
24
4000
24




140. Weight of Slab (assuming 1 m wide strip)
m
kN
Ws /
83
.
3
17
.
0
)
1
(
56
.
22 

m
kN
WD /
43
.
7
83
.
3
6
.
3 


Total dead load
Factored uniform load
m
kN
W
W
W L
D
u /
2
.
17
)
0
.
4
(
7
.
1
)
43
.
7
(
4
.
1
7
.
1
4
.
1 




Design Moments
m
kN
L
W
M u
A .
2
.
17
16
)
4
(
2
.
17
16
2
2




m
kN
L
W
M u
B .
66
.
19
14
)
4
(
2
.
17
14
2
2



m
kN
L
W
M u
C .
58
.
30
9
)
4
(
2
.
17
9
2
2




Effective depth assuming 12 mm bar
mm
t
d 144
2
12
20
170
2
1
20 





 
MPa
Ru 92
.
0
)
144
)(
1000
(
9
.
0
)
10
(
2
.
17
2
6


MPa
Ru 05
.
1
)
144
)(
1000
(
9
.
0
)
10
(
66
.
19
2
6


MPa
Ru 635
.
1
)
144
)(
1000
(
9
.
0
)
10
(
58
.
30
2
6



141. 58
.
23
)
7
.
20
(
85
.
0
415
'
85
.
0



fc
f
m
y
00337
.
0
00228
.
0
)
415
)
92
.
0
)(
58
.
23
(
2
1
1
(
58
.
23
1
use
pA 



00337
.
0
415
4
.
1
4
.
1
min 


y
f
p
00337
.
0
00261
.
0
)
415
)
05
.
1
)(
58
.
23
(
2
1
1
(
58
.
23
1
use
pB 



00415
.
0
)
415
)
635
.
1
)(
58
.
23
(
2
1
1
(
58
.
23
1




C
p
2
min 28
.
485
144
)
1000
(
00337
.
0 mm
bd
p
A
A sB
sA 



mm
mm
say
A
A
S
s
450
230
233
)
1000
(
28
.
485
4
)
12
(
1000
2
1





Using 12 mm bars

142. 2
6
.
597
144
)
1000
(
00415
.
0 mm
bd
p
A c
sC 


mm
mm
say
A
A
S
s
450
180
189
)
1000
(
6
.
597
4
)
12
(
1000
2
1





Using 12 mm bars
2
340
170
)
1000
(
002
.
0
002
.
0 mm
bt
At 


mm
mm
say
A
A
S
t
t 450
230
9
.
230
)
1000
(
340
4
)
10
(
1000
2
1





Using 10 mm bars
Temperature bars

143. C B A
12 mm continuous bent bars at 230 mm 0.C
12 mm extra bars at 230 mm 0.C
10 mm temperature
bars at 230 mm 0.C
170mm

144. Design a 4.5 m long one way slab with one end discontinuous as shown in
the figure . The slab is to carry a uniform dead load of 4.2 KPa
and uniform live load of 4.5 kPa. fc’ = 20.7 MPa , fy = 345 MPa for
main bars and fy = 276 MPa for temperature bars. Concrete weighs
23.5 kN/m3. Draw layout of bars.
4.5m 4.5m
column column column
Plate # 7: Design of one way slab
Design a one way cantilever slab of 2.0 m span . The slab is to
carry a uniform dead load of 2.4 KPa and uniform live load of 3.6 kPa.
fc’ = 27.6 MPa , fy = 415 MPa for main bars and temperature bars.
Concrete weighs 22.56 kN/m3. Draw layout of bars.

145. Prelim Exam
A rectangular beam has b =300 mm, d = 500 mm, As = 6 of 32 mm,
fc’ =27.6 MPa,fy =414 MPa. If the beam is simply supported on a span of 6 m,
determine the concentrated live load that could be applied at the third points on
the beam if steel covering is 80 mm and concrete weighs 23.5 kN/m3.
A doubly reinforced rectangular concrete beam has b =350 mm,d =600mm,
fc’=27.5 MPa, fy = 345 MPa, As =3625 mm2, As’ = 775mm2 ,covering for tension
and compression bars 80 mm and 63 mm respectively. If the beam is an interior
span of a three span continuous beam supporting a service dead load of 20 kN/m
(weight included) determine the maximum uniformly distributed live load it can
support on an average clear span of 5.0 m. Use NSCP moment coeffecients.
400 mm
75 mm
400 mm
3 of 16 mm
Determine the ultimate moment capacity
of the triangular beam shown in figure.
fc’ =20.7 MPa, fy = 345 MPa.

146. mm
b
fc
f
A
a
y
s
18
.
120
350
)
5
.
27
(
85
.
0
)
345
(
2850
'
85
.
0
1



y
s f
MPa
a
d
a
f 




 65
.
332
18
.
120
)])
63
(
85
.
0
[
18
.
120
(
600
)
(
600 '
1
' 
Compression steel does not yields at failure
2
1 2850
775
3625
' mm
As
As
As 




∑Fx =0 0.85fc’ab + As’fs’ = Asfy
0.85(27.5)350a + 775fs’ = 3625(345)
10.56a +fs’ = 1613.7
fs’ =1613.7 – 10.56 a EQ.1
Solution to #2

147. a
a
a
d
a
fs
)])
63
(
85
.
0
[
(
600
)
(
600 '
1
' 




2
.
)
55
.
53
(
600
'
EQ
a
a
fs 


2
.
1
. EQ
EQ 
a
a
a
)
55
.
53
(
600
56
.
10
7
.
1613



32130
600
56
.
10
7
.
1613 2


 a
a
a
0
32130
7
.
1013
56
.
10 2


 a
a
0
6
.
3042
96
2


 a
a
mm
a 12
.
121
2
)
6
.
3042
(
4
)
96
(
96 2




12
.
121
)
55
.
53
12
.
121
(
600
' 

s
f
y
s f
MPa
f 
 7
.
334
'
)
2
(
1
1
a
d
f
As
Mu y 
 
)
'
(
'
'
2 d
d
fs
As
Mu 
 
6
1
10
)
2
12
.
121
600
)(
345
)
2850
(
9
.
0 

Mu
m
kN
Mu .
36
.
477
1 
6
2
10
)
63
600
(
7
.
334
)
775
(
9
.
0 

Mu
m
kN
Mu .
36
.
125
2 
m
kN
Mu
Mu
Mu .
72
.
602
2
1 


2
'
max 6
.
6500
'
75
.
0 mm
f
fs
A
bd
p
A
y
s
b
s 


0365
.
0
345
)
345
600
(
)
600
(
7
.
334
)
5
.
27
(
85
.
0



b
p

148. #1
p=0.03217
pmax = 0.021 tension steel does not yield at failure
a = 260.22 mm
fs =379.92 MPa
Mu = 610.31 kN.m
WD = 4.1 kN/m
P = 171.9 kN
#2
a = 121.12 mm compression steel does not yield at failure
fs’=334.7 MPa
Mu = 602.72 kN.m
WL = 125.34 kN.m

149. 400mm
75 mm
400 mm
3 of 16 mm
Ac
0.003
s

c
375 - c
x
a
C=0.85fc’Ac
T=Asfy
375 - 2a/3
Solution
2
2
2
.
603
4
)
16
(
3
mm
As 


Assume that steel yields at failure( subject to checking)
C=T 0.85fc’Ac=Asfy 0.85(20.7)Ac =603.2(345) Ac =11,827.45 mm2

150. ax
Ac
2
1

475
400

a
x
1
.
842
.
0 EQ
a
x 

)
842
.
0
(
2
1
45
.
11827 a
a
 mm
a 6
.
167

mm
a
c 17
.
197
85
.
0
6
.
167
1




From the strain diagram
c
c
s 

400
003
.
0

2000000
s
s
f


17
.
197
17
.
197
400
003
.
0
)
200000
(


s
f
y
s f
MPa
f 
 22
.
617 tension steel yields at failure
By similar triangles
m
kN
a
f
A
M y
s
u .
91
.
53
10
)
3
6
.
167
]
2
[
400
(
345
)
2
.
603
(
9
.
0
)
3
2
375
( 6






151. SHEAR AND DIAGONAL TENSION
Another type of beam failure other than bending is shear failure.
Shear failure are very dangerous if it happens before flexure failure
because they can occur without warning
BASIC CODE REQUIREMENT
Factored shear strength Vu shall be equal or less than design shear ФVn
n
u V
V 

s
c
n V
V
V 

where:
Vc = shear carried by concrete
Vs = shear carried by the stirrups
Vu = factored shear strength

152. SHEAR STRENGTH PROVIDED BY CONCRETE
d
b
'
fc
6
1
V w
c 
d
b
'
fc
3
.
0
d
b
7
M
d
V
p
120
'
fc
V w
w
u
u
w
c 








0
.
1
M
d
V
u
u

Shear strength provided by concrete subject to shear and flexure
only;
or in more detailed calculation
where : '
fc is in MPa and shall not exceed 0.7 MPa
bw = width of the beam web for T-beams,
width of the beam for rectangular beams
d =effective depth of the beam
pw = As/bwd

153. Spacing limits of shear reinforcement
Spacing S of shear reinforcement placed perpendicular to the axis
of the member shall not exceed d/2 for nonprestressed members
and 3/4 h for prestressed members, nor 600mm. When d
b
'
fc
33
.
0
V w
s 
maximum spacing given by the above limits shall be reduced by
one half.
2
V
V c
u


Minimum shear reinforcement
When
except in the following conditions:
a)slabs and footings
b)concrete joist construction
c)beams with total depth not greater than 250 mm,2.5 times flange
thickness or half the width of the web whichever is greatest.
, minimum area of shear reinforcement
shall be provided in all reinforced concrete flexural members

154. Where shear reinforcement is required, the minimum area of shear
reinforcement shall be computed by:
y
w
v
f
3
S
b
A 
Where:
Av =cross sectional area of the stirrups
taken twice for u-shaped stirrups
d
b
'
fc
3
2
S
d
f
A
V w
y
v
s 

Shear strength provided by reinforcement
a) When shear reinforcement perpendicular to
the axis of the member is used
b) When inclined stirrups are used as shear reinforcement
d
b
'
fc
3
2
S
)
cos
(sin
d
f
A
V w
y
v
s 





155. CRITICAL SECTION FOR BEAM SHEAR
Maximum factored shear force Vu maybe computed in accordance
with the following provided that:
a) the support reaction ,in the direction of the applied shear
introduces compression on the end regions of member
no concentrated load occur between the face of the support and
location of the critical section.
b) For non prestressed members,sections located less than a distance
of d from the face of the support maybe designed for the same shear
Vu as that computed at a distance of d
Size of stirrups
Main bars smaller than or equal to 32 mm diameter: 10 mm
Main bars greater than 32 mm diameter : 12 mm
Shear carried by stirrups
c
u
s V
V
V 



156. PROBLEMS
Determine the minimum cross section required for a rectangular
beam to satisfy the condition that web reinforcement be neglected
Vu = 72 kN,fc’ =27.6MPa. Assume d = 1.6b.
2
V
V c
u


2
)
6
(
bd
'
fc
Vu


2
)
6
(
)
b
6
.
1
(
b
'
fc
Vu


2
)
6
(
)
b
6
.
1
(
6
.
27
85
.
0
)
1000
(
72
2

)
6
.
1
(
6
.
27
86
.
0
)
12
(
72000
b 
mm
350
b 
mm
560
350
)
6
.
1
(
d 


157. A rectangular beam with b = 270 mm, d = 500 mm is provided with 10
mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa;
a) Determine the required spacing if Vu = 40 kN
b) Determine the required spacing if Vu = 92 kN
c) Determine the required spacing if Vu = 236 kN
d) Determine the required spacing if Vu = 473 kN
Solution
Shear carried by concrete
d
b
'
fc
6
1
V w
c 
N
108
,
103
500
)
270
(
21
6
1
Vc 

N
43821
2
)
103108
(
85
.
0
2
Vc




158. 2
V
N
42000
V c
u



a) Stirrups not necessary
N
92000
Vu 
b)
c
u
s V
V
V 


N
3
.
5127
103108
85
.
0
92000
Vs 


2
2
v mm
1
.
157
4
)
10
(
2
A 


mm
3
.
4228
3
.
5127
500
)
276
(
1
.
157
V
d
f
A
S
s
y
v




159. N
206216
)
500
)(
270
(
21
3
1
d
b
'
fc
3
1
w 

mm
250
2
500
2
d
S 


N
236000
Vu 
c)
c
u
s V
V
V 


N
174539
103108
85
.
0
236000
Vs 


mm
120
mmsay
124
174539
500
)
276
(
1
.
157
V
d
f
A
S
s
y
v



d
b
'
fc
3
1
V w
s 
Use S = 250 mm

160. N
d
b
fc w 206216
'
3
1

mm
d
S 250
2
500
2



d
b
fc
V w
s '
3
1

Use S = 120 mm
N
Vu 473000

d)
c
u
s V
V
V 


N
Vs 5
.
453362
103108
85
.
0
473000



N
d
b
fc w 206216
'
3
1

N
d
b
fc w 412432
'
3
2

d
b
fc
V w
s '
3
2

Beam size is inadequate for shear

161. d
b
'
fc
6
1
V w
c 
2
V
V c
u 

2
V
V c
u 

Design of vertical stirrups
Given :bw or b,d,fc’,fy, beam loading & span,
Required: size and spacing of stirrups
General Procedure
1.Calculate factored shear force VU at the critical section.
2.Calculate shear strength of concrete:
provide stirrups
stirrups not necessary

162. 2
c
u
V
V 

c
u
s
V
V
V 


d
b
fc
V w
s '
3
2

s
y
v
V
d
f
A
S 
d
b
fc w
'
3
1
d
b
fc
V w
s
'
3
1

d
b
fc
V w
s
'
3
1

Assuming that
3. Calculate the shear strength provided by the stirrups
Note: if
4. Calculate the required spacing of stirrups
Spacing is the smallest of:
a)
Calculate
b ) S = d/2 when
c ) S = d/4 when
5. Check for minimum required area of stirrups
y
w
v
f
S
b
A
3

Note: Av must be less than or equal to
the actual area of Stirrups
adjust the beam size

163. Problem:
A simply supported reinforced concrete beam 230 mm wide with an effective
depth of 500 mm has a span of 6m. The beam carries a dead load of 9
kN/m ( including its own weight) and live load of 18 kN/m applied throughtout
its entire span. Determine the required spacing of 10 mm stirrups. fc’ = 28
MPa ,fy= 345 MPa.
N
d
b
fc
V w
c 101420
500
)
230
(
28
6
1
'
6
1



wu
6 m
Wu =1.4(9)+1.7(18)=43.2 kN/m R =Wu(3)=43.2(3)=129.6kN
R

164. 43.2 kN/m
R=129.6kN
0.5
VU
Shear force at crtical section
Vu = 129.6-0.5(43.2)=108kN =108000N
N
Vc
5
.
43103
2
)
101420
(
85
.
0
2



2
c
u
V
V


101420
85
.
0
108000



 c
u
s V
V
V

N
Vs 25639

2
2
1
.
157
4
)
10
(
2
mm
Av 


mm
V
d
f
A
S
s
y
v
1056
25639
500
)
345
(
1
.
157




165. N
d
b
fc w 202840
'
3
1

mm
d
S 250
2
500
2



d
b
fc
V w
s '
3
1

mm
S 250

Use
y
w
v
f
S
b
A
3

2
2
1
.
157
56
.
55
)
345
(
3
)
250
(
230
mm
mm
Av 


Minimum required area
Use 10 mm u shaped stirrups spaced at 250 mm on centers

166. PLATE # 7 : SHEAR
A simply supported reinforced concrete beam 250 mm wide with an
effective depth of 600 mm has a span of 7.5m. The beam carries a
dead load of 12 kN/m ( including its own weight) and live load of 24
kN/m applied throughtout its entire span. Determine the required
spacing of 10 mm stirrups. fc’ = 28 MPa ,fy= 345 MPa.
A rectangular beam with b = 300 mm, d = 550 mm is provided with 10
mm vertical stirrups with fy = 276 MPa. Assuming fc’ = 21 MPa;
a) Determine the required spacing if Vu = 50 kN
b) Determine the required spacing if Vu = 220 kN
c) Determine the required spacing if Vu = 360 kN
d) Determine the required spacing if Vu = 500kN

167. Design of beams for bending shear and deflection
General procedure
1. Design section and reinforcement by bending
2. Design stirrups by shear
3. Check adequacy of design by deflection
Problem :
Design a rectangular beam for a 6 m simple span to support a
uniform dead load of 18 kN/m(weight included) and uniform live load
of 12 kN/m applied along it’s entire length. fc’=20 MPa, fy = 345 MPa
for main bars and stirrups,Wc=23.5 kN/m3 ,p=0.6pmax Consider
immediate deflection due to live load only with an allowable of 1/360 of
span length. fr=3.1 MPa n=10,Ec=20000 MPa. Use 100 mm covering.

168. 29
.
20
)
20
(
85
.
0
345
'
85
.
0



fc
f
m
y
Solution
m
kN
Wu /
6
.
45
)
12
(
7
.
1
)
18
(
4
.
1 


m
kN
L
W
M u
u .
2
.
205
8
)
6
(
6
.
45
8
2
2



02659
.
0
345
)
345
600
(
)
600
(
85
.
0
)
20
(
85
.
0
)
600
(
600
'
85
.
0 1





y
y
b
f
f
fc
p

012
.
0
02659
.
0
)
75
.
0
(
6
.
0
)
75
.
0
)(
6
.
0
( 

 b
p
p
MPa
pm
pf
R y
u 635
.
3
)
2
]
29
.
20
[
012
.
0
1
)(
345
(
012
.
0
)
2
1
( 





169. 6
.
3
)
500
)(
250
(
9
.
0
)
10
(
5
.
202
2
6
2




bd
M
R u
u
mm
b
R
M
d
u
u
500
250
)
635
.
3
(
9
.
0
)
10
(
5
.
202 6




01186
.
0
)
345
6
.
3
)
29
.
20
(
2
1
1
(
29
.
20
1
)
2
1
1
(
1







y
u
f
mR
m
p
2
5
.
482
,
1
500
)
250
(
01186
.
0 mm
pbd
As 


Try b = 250 mm
pcs
4
4
)
25
(
5
.
1482
N 2




170. N
d
b
fc
V w
c 93169
500
)
250
(
20
6
1
'
6
1



wu
6 m
R =Wu(3)=45.6(3)=136.8kN
R
45.6 kN/m
0.5
VU
Shear force at critical section
Vu = 136.8- 0.5(45.6)=114kN =114000N
R=136.8

171. N
Vc
8
.
39596
2
)
93169
(
85
.
0
2



2
c
u
V
V


93169
85
.
0
114000



 c
u
s V
V
V

N
Vs 40949

2
2
1
.
157
4
)
10
(
2
mm
Av 


mm
V
d
f
A
S
s
y
v
661
40949
500
)
345
(
1
.
157



Using 10 mm u shape stirrups
N
d
b
fc w 186338
'
3
1


172. mm
d
S 250
2
500
2



mm
S 250

Use
y
w
v
f
S
b
A
3

2
2
1
.
157
56
.
55
)
345
(
3
)
250
(
230
mm
mm
Av 


Minimum required area

173. Gross moment of inertia
4
6
3
3
10
4500
12
)
600
(
250
12
mm
x
bh
Ig 


m
kN
mm
N
x
y
I
f
M
t
g
r
cr .
5
.
46
.
10
5
.
46
300
10
)
4500
(
1
.
3 6
6




b=250 b
x
500-x
nAs
As
d=500
2
2
1570
4
)
20
(
5
mm
As 


Transformed section
100

174. )
(
2
2
x
d
nA
x
b s 
 )
500
)(
1570
(
10
2
250
2
x
x


x
x 6
.
125
62800
2

 0
62800
6
.
125
2


 x
x
mm
x 6
.
195
2
)
62800
(
4
)
6
.
125
(
6
.
125 2





2
3
2
3
)
6
.
195
500
)(
1570
(
10
3
)
6
.
195
(
250
)
(
3





 x
d
nA
bx
I s
cr
4
6
10
3
.
078
,
2 mm
x
Icr 
a) m
kN
L
W
M L
a .
54
8
)
6
)(
12
(
8
2
2




175. 6
3
6
3
10
3
.
2078
54
5
.
46
1
10
4500
54
5
.
46
x
x
Ie























4
6
10
6
.
624
,
3 mm
x
Ie 
Effective moment of inertia
cr
a
cr
g
a
cr
e I
M
M
I
M
M
I

























3
3
1
Instantaneous deflection due to live load
mm
I
E
L
W
e
c
L
79
.
2
)
10
(
6
.
3624
)
20000
(
384
)
6000
)(
12
(
5
384
5
6
4
4




allowable deflection due to live load
mm
79
.
2
mm
67
.
16
360
6000
360
L
allowable 




Section is adequate

176. Plate #8:Design of beams for bending shear and deflection
Make a complete design of a rectangular beam reinforced for tension
only for a 7.5 m simple span to support a uniform dead load of
24 kN/m(weight included) and uniform live load of 18 kN/m applied
along it’s entire length. fc’=20 MPa, fy = 300 MPa for main bars and
stirrups,Wc=23.5 kN/m3 ,p=0.18(fc’/fy). Consider immediate deflection
due to total load with an allowable of 1/360 of span length.
Es = 200000MPa,Ec=18500 MPa, . Use 100 mm covering.
'
7
.
0 fc
fr 

177. BOND AND DEVELOPMENT LENGTH
Bond
In reinforced concrete, concrete and steel act as a unit. For
this to happen, there must be absolutely no slippage of the
bars in relation to the surrounding concrete. The steel and
concrete must stick or bond together so that there will be
transfer of stress from steel to concrete and vice-versa.
Failure of transfer of stress makes the concrete an
unreinforced member thus it will be subject to collapse.
Development Length
Bar development length is the embedment necessary to
assure that the bar can be stressed to its yield point with
some reserved to ensure member toughness.

178. T = Abfy
F
Ld
db
Basic concept of development length
ΣF = 0
F = T
y
2
b
d
b f
d
4
L
d



b
y
d d
4
f
L


Where : μ = average bond stress
Ld = minimum development length
The code however provides the basic development length
Ldb for various conditions. The values provided are modified
for different conditions. Thus the minimum development length
provided by the code is;
Ld = Ldb(applicable modification factors) ≥ 300 mm

179. '
fc
f
A
02
.
0
L
y
b
db 
'
fc
f
25
L
y
db 
'
fc
8
f
d
3
L
y
b
db 
Basic Development Length of bars in tension
For 32 mm bar & smaller and deformed wire
For 36 mm bar
For deformed wire
but not to be taken less than 0.6dbfy
Basic Development Length of bars in compression
but not to be taken less than 0.04dbfy
'
fc
f
d
24
.
0
L
y
b
db 
Where
Ldb = basic development length ( mm)
Ab = area of one bar (mm2)
db = diameter of one bar (mm)
fc’,fy in MPa

180. Modification Factors for bars in tension
provided
A
required
A
s
s
Condition Modification
Factor,m
a) For bars in beams or column with a minimum cover not less than specified by the code 1.0
b) For bars in beams or column with transverse reinforcement satisfying the requirement of the code 1.0
c) Bars in beams or column with a clear spacing not less than 3db 1.0
d) Bar in the inner layer of slab or wall reinforcement and with a clear spacing not less than 3db 1.0
e) Any bars with cover of not less than 2db and with a clear spacing not less than 3db 1.0
f) For bars with cover of db or less with a clear spacing of 2db or less 2.0
g) For bars not included in items a to f 1.4
h) For 32 mm bar and smaller with clear spacing not less than 5db and with cover from face of the
member to edge bar, measured in the plane of the bar, not less than 2.5db, the factors from items
a to g may be multiplied by 0.8
0.8
i) Top reinforcement 1.3
j) Lightweight aggregate concrete 1.3
K)Lightweight aggregate concrete when fct is specified
l) For reinforcement enclosed within special reinforcement not less than 6 mm diameter and not
more than 100 mm pitch, within 12 mm or larger circular ties spaced at not more than
100 mm on center or larger ties or stirrups spaced not more than 100 mm on center and arranged
such that alternate bars shall have supported by the corner of a tie hoop with an included angle
not more than 1350,the factors in items a through g maybe multiplied by 1.8
1.8
m) Excess Reinforcement. Development length maybe reduced where reinforcement in a flexural
member is more than required by analysis
ct
f
fc
8
.
1
'

181. Modification Factors for bars in Compression
provided
A
required
A
s
s
Condition
Modification
Factor,m
a) Excess reinforcement. Reinforcement more than
that required by analysis
b) Spiral and Ties. Reinforcement enclosed within
spiral reinforcement not less than 6 mm diameter
and not more than 100 mm pitch or within 10 mm
ties and spaced not more than 100 mm on center.
0.75
Problems
A rectangular beam 200 mm wide and 400 mm deep is reinforced with
3 of 22 mm tension top bars with fc’ = 20.7 MPa and fy = 275 MPa.
Calculate the required development length.

182. '
02
.
0
fc
f
A
L y
b
db 
Solution
mm
Ldb 460
7
.
20
275
4
)
22
(
02
.
0
2



mm
Ldb 363
)
275
)(
22
(
06
.
0 
 mm
Ldb 460

Modification Factor
Top bar = 1.3
Required development length
Use
mm
L
L db
d 598
)
460
(
3
.
1
3
.
1 


y
b
db f
d
L 06
.
0


183. A rectangular beam 250 mm wide and 500 mm deep is reinforced with 4 of 25 mm
with fc’ = 27 MPa and fy = 345 MPa. If the member is made up of lightweight
aggregate with fct = 2.88 MPa ,Calculate the required development length.
'
02
.
0
fc
f
A
L y
b
db 
Solution
mm
Ldb 652
27
345
4
)
25
(
02
.
0
2



mm
Ldb 518
)
345
)(
25
(
06
.
0 

mm
Ldb 652

Use
y
b
db f
d
L 06
.
0

Modification Factor
Lightweight concrete
With specified fct
00
.
1
)
88
.
2
(
8
.
1
27
8
.
1
'


ct
f
fc
db
d L
L 0
.
1

)
652
(
0
.
1

d
L
mm
Ld 652


185. A cantilever beam 320 mm wide and 500 mm deep is reinforced with 3 of 36 mm
straight top bars with fc’ = 27 MPa and fy = 345 MPa. Calculate required
development length.
Solution
Modification Factor
Top bar = 1.3
Required development length
mm
L
L db
d 2158
)
1660
(
3
.
1
3
.
1 


'
25
fc
f
L y
db 
mm
Ldb 1660
27
)
345
(
25



186. A rectangular beam 250 mm wide and 410 mm deep is reinforced with
3 of 20 mm compression bars with fc’ = 20.7 MPa and fy = 275 MPa.
Calculate the required development length.
'
24
.
0
fc
f
d
L y
b
db 
mm
Ldb 254
7
.
20
275
)
20
(
24
.
0


mm
Ldb 220
)
275
)(
20
(
04
.
0 

mm
Ldb 254

Use
y
b
db f
d
L 04
.
0

No applicable modification factor mm
Ld 254


187. Plate # 9 Development Length
A rectangular beam 300 mm wide and 450 mm deep is reinforced with 4 of 25 mm
tension top bars with fc’ = 20.7 MPa and fy = 345 MPa. Calculate the required
development length.
A rectangular beam 200 mm wide and 350 mm deep is reinforced with 3 of 20 mm
with fc’ = 30 MPa and fy = 415 MPa. If the member is made up of lightweight
aggregate with fct = 2.4 MPa ,Calculate the required development length.
A cantilever beam 300 mm wide and 450 mm deep is reinforced with 6 of 22 mm
straight top bars with fc’ = 20.7 MPa and fy = 415 MPa. Calculate required
development length.
A rectangular beam 350 mm wide and 600 mm deep is reinforced with 4 of 32 mm
compression bars with fc’ = 20.7 MPa and fy = 275 MPa. Calculate the required
development length.

188. Development Length of Flexural Reinforcement
Tension reinforcement in flexural members maybe developed by :
a) bending across the web to be anchored
b) made continues with reinforcement on opposite face ofthe member
Critical points for development length in flexural members are at points
of maximum stress and at points where the adjacent reinforcements
terminates or is bent. Reinforcement shall extend beyond the point at
which it isno longer required to resist flexure for a distance equal to the
effective depth of member or 12db whichever is greater, except at
supports of simple beam and free end of cantilevers. Continuing
reinforcement shall have an embedment length not less than the
development length beyond the point where bent or terminated tension
reinforcement is no longer required to resist flexure

189. a
u
n
l
V
M
Ld 

Development of positive moment bars
The code stipulates that at least one third the positive
reinforcement in simple members and one fourth the positive
reinforcement in continuous members shall extend along the
same face in the member into the supports. In beams such
reinforcement shall extend into the support at least 150mm.
At simple supports and at points of inflection,positive
moment tension reinforcement shall be limited to a diameter
such that Ld computed by
Where:
Mn = nominal moment strength assuming that all reinforcement
at section are stressed to specified yield strength fy
Vu = factored shear force at supports for simple beams and at
points of inflection for continuous beams

190. la = end anchorage ;at support shall be the embedment length beyond
the center of support; at point of inflection limited to the effective depth
or 12db whichever is greater.
Value of Mn/Vu maybe increased by 30% when the ends of the
reinforcement are confined by a compressive reaction such as a
column below but not when the beam frame into the girder.
a
u
n
l
V
M
Ld 
 3
.
1
a
u
n
l
V
M
Ld 
 3
.
1 a
u
n
l
V
M
Ld 

When or Use smaller bar
Size or increase
End anchorage la
)
2
(
a
d
f
A
M y
s
n 

b
fc
f
A
a
y
s
'
85
.
0

Where:

191. Max Ld Max Ld
la 1.3Mn/Vu Mn/Vu la
tension bars
at least 1/3 of positive reinforcement
VU

192. la is the larger
Max Ld
or 12db la Mn/Vu tension bars
at least ¼ of positive moment reinforcement
.
CL
Point of inflection
150 mm min
Value of d

193. Development length for Negative moment Reinforcement
Negative moment reinforcement should have an embedment length into the span to
develop the calculated tension in the bar,or a length equal to the effective depth of the
member or 12db whichever is the greatest. At least one third of the total negative
reinforcement should have an embedment length beyond the point of inflection not less
than the effective depth of the member or 12 db or 1/16 of the clear span whichever is
greatest.
Development length for Negative moment
Ld
` the larger value of 12db,d or Ln/16
at least 1/3 of the total Negative moment reinforcement
Clear span =Ln
Point of inflection

194. Ln1/4 0.3 Ln1 0.3 Ln2
Ln1/8 Ln1/8 Ln2/8
Ln1 Ln2
150 mm min 150 mm min
Recommended bar details for continuous beams

195. 12db
r r
D
900 hook 4db
db
bend diameter,D =2r 1800 hook
D =6db for 10 mm through 25 mm bars
D =8db for 28 mm through 32 mm bars
D =10db for 36 mm bars
Hooks
If sufficient space is not available to anchor tension bars by running
them straight for the required development length as required by the
code, hooks maybe used.

196. Ldb STANDARD HOOKS
65 mm min
12db
6db for 36 mm bar
4db 4db for 10 mm through 25 mm bars
5db for 28 mm through 32 mm bars
Development of standard hooks
Basic Development Length of standard hooks
'
100
fc
d
l b
hb 
Actual Development Length of standard hooks
ldh = lhb(applicable modification factors) ≥8db nor 150 mm

197. Modification Factors
1. If the reinforcing bar has an fy other 415 MPa, lhb is to be multiplied by fy/415
2. When 900 hooks and 32 mm or smaller bar are used and when 60 mm or
more of side cover normal to the hook is present, together with at least 50 mm
cover for the bar extension, lhb is to be multiplied by 0.7.
3. When hooks made of 32 mm and smaller are enclosed vertically and
horizontally within ties or stirrups ties spaced no farther apart than 3db, lhb is to be
multiplied by 0.8
4. Where the amount of flexural reinforcement exceeds the theoretical amount
required and where the specifications being used do not specifically require that
development lengths be based on fy the value of lhb is to be multiplied by
Asrequired/As provided.
5. When light weight concrete are used, apply a modification factor of 1.3
6. For bars being developed by standard hook at discontinuous end with side
cover and top or bottom cover over hook less than 60 mm,hooked bar shall be
enclosed within ties or stirrups spaced along the full development length ldh not
greater than 3db where db is the diameter of hooked bar. For this case, the
factor mentioned in item 3 shall not apply.

198. Splices of Reinforcement
Splicing maybe done by welding, by mechanical connections or most
frequently by lapping bars. Lapped bars are usually tied in contact.
Lap splice must not be used for bars larger than 32 mm.
Splices in tension
The minimum length of lap for tension lap splice shall be as required for
class A or class B, but shall not be less than 300 mm, where
Class A splice 1.0Ld
Class B splice 1.3Ld
Lap splices of deformed bars and wires in tension shall be class B splice
except that class A splice are allowed when
a)the area of the reinforcement provided is at least twice than that
required by analysis.
b)One half or less of the total reinforcement is spliced within the
required lap strength.
Splices of deformed bars in compression
Compression bars maybe spliced by lapping,end bearing, welding or
mechanical devices. The minimum length of such bars should be the
development length Ld but may not be less than 0.07dbfy for fy of 415
MPa or less, or (0.13fy -24)db for fy greater than 415 MPa.

199. Problems
A simply supported beam is reinforced with three of 28 mm
bars with fc’= 27.6 MPa and fy = 275 MPa. Assuming that side, bottom and top
cover to be greater than 60 mm, determine the following:
a) the required development length if a 900 hook is used
b) the required development length if a 1800 hook is used
ldh
ldh
Solution
Using a 900 hook
'
100
fc
d
l b
hb 
mm
lhb 533
6
.
27
)
28
(
100


Modification factor for fy other than 415 MPa
6626
.
0
415
275
415



y
f
m
Required development length
Ldh =0.6626(533)0.7=247.2 say 250 mm
Modification factor for 900 hook = 0.7

200. ldh
Solution
Using a 1800 hook
'
100
fc
d
l b
hb 
mm
lhb 533
6
.
27
)
28
(
100


Modification factor for fy other than 415 MPa
6626
.
0
415
275
415



y
f
m
Required development length
Ldh =0.6626(533)=353.17 say 355 mm
Problem
For the simply supported beam shown in figure below, investigate whether the
bars size is satisfactory for the required development length. The beam is
reinforced with 4 of 25 mm bars . fc’ = 20.7 MPa and fy = 414 MPa, Vu = 270 kN.
The beam is made up of normal sand concrete and the reaction produces
compression on concrete.

201. 300mm
600 mm
175 mm
Solution
Basic development length
'
02
.
0
fc
f
A
L y
b
db 
mm
Ldb 894
7
.
20
414
4
)
25
(
02
.
0
2



mm
Ldb 621
)
414
)(
25
(
06
.
0 

y
b
db f
d
L 06
.
0

mm
Ldb 894

Use
Since there is no applicable
modification factor
mm
Ld 894


202. )
2
(
a
d
f
A
M y
s
n 

b
fc
f
A
a
y
s
'
85
.
0

mm
a 154
300
)
7
.
20
(
85
.
0
414
4
)
25
(
4
2



)
2
154
600
(
414
4
)
25
(
4
2



n
M
mm
N
x
Mn .
10
14
.
425 6

a
u
n
l
V
M

3
.
1
mm
x
2
.
2221
175
)
1000
(
270
10
14
.
425
3
.
1
6


mm
mm 2
.
2221
894 
a
u
n
d l
V
M
L 
 3
.
1
bars are adequate