1. Design of Reinforced Concrete
Prof. Dr. Waleed A. Thanoon

2. Design of Flexural Members
A) Single Rectangular Reinforced Section
B) Single T Reinforced Section
C) Double Reinforced Concrete Beam

3. Introduction In design, the moment due to applied loading is known,
and it is required either to find:
the cross section dimensions with its steel reinforcement, or
the steel reinforcement and its detailing (in case the cross-section dimension is given).
In general, the design follows a trial and error procedure.
The dimension of different cross section in a structure are needed to calculate the distribution of bending moments and shear forces in the structure i.e. in moment distribution method the moment of inertia; “I” value is required to carry on the analysis.

4. Introduction
However initial cross section dimension are assumed to calculate the bending moment and shear forces.
The assumed cross section is next checked whether these dimensions can be used to resist the calculated moment and shear forces or not.

5. Concrete Cover
Concrete Cover
The concrete cover is required to ensure adequate resistance to fire and corrosion. It is defined as the distance between the outmost surface of steel reinforcement and the nearest concrete surface as shown in Fig. 1.
Fig. 1

6. (fire resistance are measured in hours).
Concrete Cover
Steel reinforcement loses its strength when subjected to high temperature.
Concrete is a good insulator with the result that it protects the steel and extends the time take for the temperature in the steel to reach dangerous level.
Hence greater cover means greater time for occupants to escape from brining building before collapse.
(fire resistance are measured in hours).

7. Concrete Cover
Another import role of concrete cover is to provides bonding layer to the steel reinforcement in order that both material act as a perfect integrated composite material.
The value of cover for grade 30 concrete is 25 mm in mild environmental condition.
Table 3.2 and 3.3 in BS 8110 part provides the suggested cover for other concrete grade and other environmental conditions.

8. Concrete Cover
Minimum cover to main reinforcement to resist different fire ratings in hour are shown in Table (2).
Fire Resistance Hour
Nominal Cover - mm
Beams
Floors
Ribs
Columns SS C 1 20
1.5 25 35 2 40 30 45
SS = Simply Supported
C = Continuous

9. A) Design of Single Rectangular Reinforced Section
Flexural Members
A) Design of Single Rectangular Reinforced Section

10. Section dimension are not given
Rectangular-section .
Design Case 1:
Cross section is given
Design Case 2:
Section dimension are not given
Design Case 3:
The percentage of reinforcing steel is known

11. Design case 1: Cross Section is given
In this case, the cross dimensions and the material strength (b, h, fcu, fy) are known and it is required to calculate the area of steel reinforcement and its detailing.
The design starts with assuming the number of steel layers to calculate d, the effective depth.
Table 1 shows different possibilities of steel arrangement in a reinforced concrete section

12. Table 1 The different possibilities of steel arrangement

13. Steel Arrangement

14. Rectangular-section
After deciding the number of steel layers, the effective depth is calculated as shown in Table (1).
However, since the steel amount is not known in the beginning of the design, the designer may assume the number of layer based on his experience and later check the assumption at the end of the design after calculating the steel area.

15. To do that the value of “K” need to be calculated as:.
Rectangular-section
Next the check whether the section need to be designed as single r.c. section or double r.c section has to be done.
To do that the value of “K” need to be calculated as:.
if K  0.156, the section is designed as singly reinforced concrete section (T.F)
if K > the section is designed as double reinforced concrete section.

16. In case of K  0.156, the lever arm z need to be calculated using:
Rectangular-section
In case of K  0.156, the lever arm z need to be calculated using:
The derivation of this equation can done as follows:
C = 0.45 fcu b . a
z = d – a/ or a = 2 (d - z)
M= C . z = 0.45 fcu b . a . z = 0.9 fcu b (d – z) . z

17. Rectangular-section
( BS8110-clause )

18. If As < As min use As = As min
Rectangular-section
The final step in design is to calculate the area of reinforcement necessary to resist the bending moment as:
If As < As min use As = As min

19. Design case 2: Section Dimension are not Given
Rectangular-section
Design case 2: Section Dimension are not Given
In this case, the width of the beam may be assumed to be similar to the width of columns in the frame or supporting walls.
The design can also starts by assuming b/d (width/effective depth) ratio. As a start b/d may be assumed equal to 0.5. Again the design must satisfy the ductility requirement (to ensure yielding of steel and hence T.F.).
C.F. Double r.c. section
Maximum limit axis
T.F. Single r.c section

20. Rectangular-section
To identify the depth of the section, design on maximum limit axis by equating the ultimate moment to the Mmax which is the maximum value allowed for a single reinforced section to fail in a ductile manner .
The above equation will yields the minimum effective depth required to satisfy ductility requirement as:
Lower than this depth with shift the design below the maximum limit axis and the failure will be controlled by compression failure (or steel in compression must be provided). While larger depth will leads to more ductile failure and less steel area.

21. zmax can be found using x=0.5d as explained in previous chapter as:
Rectangular-section
If minimum depth is used, the necessary steel reinforcement may be calculated using: or
zmax can be found using x=0.5d as explained in previous chapter as:
However, a section with higher depth can be used and the design will follow the procedure as explained in Case 1.

22. Rectangular-section
Example Case 2
Design a reinforced concrete section to resist a moment of Mu = 150 kNm, use fcu = 30 N/mm² and fy = 460N/mm².
Assume b = 200 mm and design the section on maximum condition.

23. Rectangular-section
use 3T20 = 942 mm2 > mm2
h = = mm
use h = 500 mm > OK

24. Check d = 500 – 45 = 445 mm  O.K. (T.F.)  O.K
Rectangular-section
Check
d = 500 – 45 = 445 mm
 O.K. (T.F.)
 O.K
Z = d – a/2 = 445 – 139/2 = mm
Mu = 0.95 x 460 x 942 x = kN.m > 150 kN  OK

25. Design case 3: The percentage of reinforcing steel is known
Rectangular-section
Design case 3: The percentage of reinforcing steel is known
Sometimes, the designer would like to fix the steel reinforcement ratio (r) to be laid between the min and max allowed by the code to control the amount of steel especially when the cost of steel reinforcement is high.
In such case, the depth of the beam cross required to resist the given ultimate moment need to be calculated for the specified steel ratio.

26. The design may be started using C = T:
Rectangular-section
The design may be started using C = T:
where a can be evaluated in terms of d :
Then the effective depth required to resist the given moment can be calculated using:
Finally the steel can found as:

27. Rectangular-section
Example Case 3
Design a rectangular beam having b = 250 mm, fcu = 25 N/mm2, fy = 460 N/mm2 . The steel ratio is approximately equal to 0.5%.
Check first the ductility requirement
> 0.5% T.F
= 0.13%
C = T
 O.K (T.F)

28. Rectangular-section
if Mu = 120 kNm,
d = mm
h = =
use h = 550 mm then
d = = 512 mm
As = r . b . d

29. The Design of Rectangular r.c. Beam
Flowchart 1
The Design of Rectangular r.c. Beam

30. Cont. The Design of Rectangular r.c. Beam

31. Design of Single T Reinforced Section
Flexural Members
Design of Single T Reinforced Section

32. The N.A. lies in the Flange
T-section
Design Case 1:
The N.A. lies in the Flange
Design Case 2:
The N.A. lies in the Web

33. T-section
The design of T-section will depends on the location of N.A. whether is located in the flange or in the web. In order to identify the location of the N.A, the moment capacity of the flange is calculated by considering the whole flange under compression i.e. a = hf as shown in figure:
Fig. 15 Stress-strain variation for T-beam section with N.A. lies in the flange

34. Mu < Mflange  the beam section is R.B.
T-section
The moment capacity of the flange is:
If the ultimate moment is less than or equal the flange moment capacity, then the N.A. lies in the flange since the moment can be resisted by smaller compression area. The design procedure are similar to the design of rectangular beam with b=bf.
Otherwise, the N.A will be located in the web and a larger compression area is required and the section will be treated as True T-beam.
Mu < Mflange  the beam section is R.B.
Mu > Mflange  the beam section is T.T.B.

35. Design case 1: The N.A. Lies in the flange
T-section
Design case 1: The N.A. Lies in the flange
if K  0.156, the section is designed as singly reinforced concrete section (T.F)
if K > the section is designed as double reinforced concrete section.

36. If As < As min use As = As min
T-section
In case of K  0.156, the lever arm z need to be calculated using:
then:
If As < As min use As = As min

37. Design case 2: The N.A. Lies in the web
T-section
Design case 2: The N.A. Lies in the web
Fig. 16 Stress-strain variation for T-beam section with N.A. lies in the web

38. T-section
Simplifying the above equation will yield a 2nd degree equation in terms of “x” the N.A. depth.
If x < xmax = 0.5d  T.F. the section is designed as single r.c. section
If x > xmax = 0.5d  a compression reinforcement will be required and the section must be designed as double r.c. section.

39. T-section
Alternatively, the ultimate may be compared with maximum moment calculated using:
if Mu  Mmax it is a T.F and the section is designed as singly reinforced concrete section
if Mu > Mmax it is a C.F and the section is designed as double reinforced concrete section.

40. The steel reinforcement can be found using:
T-section
The steel reinforcement can be found using:
where
and

41. Alternatively, the steel area can be calculated using:
T-section
Alternatively, the steel area can be calculated using:

42. T-section
Example 1
A simply supported beam having a span of 6 m and carries dead load equal to 14.8 kN/m and imposed load of 10 kN/m. The beam is monolithically cast with 100 mm thick floor slab. The width of the beam is 250 mm and the architecture fix the total depth to 350 mm. Using fcu = 30 N/mm2, fy = 460 N/mm2 find the steel reinforcement at the centre of the beam.

43. Mflange = 0.45 x 30x1450x100(304.5-100/2)x10-6 = 498.2 kN.m
T-section
Solution 1, cont.
Mflange = 0.45 x 30x1450x100( /2)x10-6 = kN.m
Mu = 165 < Mflange  the N.A. lies in the flange,
then:

44. T-section
K = which is < than so T.F and the beam is designed as single r.c. section
= 0.925d >0.95d
 use z = 0.95d = mm

45. T-section
Example
Mu = 260 kN.m
Mflange = 0.45 x 30 x 600 x 100( /2) x 10-6 = kN.m
Mu > Mflange  the N.A. lies in the web

46. Mu < Mmax  T.F., the section is considered as single r.c. section
T-section
= 0.156x30x250x x30( )100(340-50)x10-6
= kN.m
Mu < Mmax  T.F., the section is considered as single r.c. section
x = mm < 0.5 d = 170 mm  T.F.
2305 mm2

47. T-section
Alternatively, the area of the reinforcing steel may be found as follows:
Find the centroide of the concrete compression area by taking the moment of the compression area about the top edge of the flange:
As = mm2, use 5Y25 bars

48. Design of Singly Reinforced Section for T-beam
Flowchart 6
Design of Singly Reinforced Section for T-beam

49. Flowchart Design of Singly Reinforced Section for T-beam, cont.

50. Design of Double Reinforced Concrete Beam
Flexural Members
Design of Double Reinforced Concrete Beam

51. When the depth of the beam is restricted and the moment capacity of singly reinforced section is not adequate to resist the applied bending moment, compression reinforcement has to be added and the beam will be designed as a doubly reinforced section.
A doubly r.c. beam can be assumed to be made up of two beams A and B.
In beam A which is a singly r.c. section. The steel in tension As1 is required to balance the force of compression in concrete.
While in beam B, which imaginary, the tension steel As2 required to balance a compression force in the compression steel As’

52. Beam A will be designed to its maximum limit allowed by the code i. e
Beam A will be designed to its maximum limit allowed by the code i.e. x = xmax ; As1 = As max and M1 = Mmax etc.
The remaining moment M2=M-Mmax will be resisted by Beam B (which is imaginary) through the additional steel As’ and As2 shown in Fig.
Same equations presented in analysis chapter will be used in the design.

53. Double r.c. beam
Example 1
Design a r.c. beam for an effective span of 6m, ultimate load =128 kN/m and size of the beam is limited to b = 300 mm and h = 700 mm. use fcu = 20 N/mm2, fy = 460 N/mm2.
d = =652.5 mm
need steel in compression,
design as double reinforced section
= x 20 x 300 x x 10-6 = kN.m
Mu > Mmax  design as double reinforced section
M1 = Mmax = kN.m

54. Double r.c. beam
Or alternatively:
As1=1805 mm2
M2 = M – Mmax = 576 – = kN.m
this moment will be resisted by steel As’ and As2
then
the steel in compression is yielded

55. which is more than fy , use fs’ = 0.95 fy
Double r.c. beam
alternatively:
which is more than fy , use fs’ = 0.95 fy
M2 = x 106 = 0.95 x 460 x As2 ( )
As2 = mm2 = As’
Use for steel in compression 4Y16 which provides area = mm2
Total steel in tension will be:
As = = mm2
Use 6Y25 which provides As=2940 mm2 > o.k

56. Flowchart: Design of Design of Doubly Section for Rectangular beam

57. Flowchart Design of Design of Doubly Section for Rectangular beam

58. Design of Reinforced Concrete
Analysis of Double r.c. section
Updated 28 Oct 2009-Lv

59. If a specified single r. c
If a specified single r.c. section cannot resist the applied moment safely in a ductile manner.
Additional reinforcement may be provided in the compression zone to increase its flexural capacity of r.c section when the applied M exceeds the maximum capacity Mmax=0.156fcubd2 of a single reinforced concrete section.
Doubly r.c. section can be considered as a combination of singly r.c section and additional steel reinforcement at top and bottom of the section as shown in Fig. 21 .

60. Fig. 21

61. The singly reinforced section is designed to its limit capacity i. e
The singly reinforced section is designed to its limit capacity i.e. maximum limit condition discussed in previous sections. This means that x = xmax, As1 = As max and M1 = Mmax.
The remaining moment M2 = M – Mmax will be resisted by the additional steel A’s and As2 shown in Fig. 21.
However in analysis, it is required to check the capacity of a given section with known reinforcement and hence the following procedure is used.

62. Fig. 22

63. From the strain distribution shown in Fig. 22 
and the stress in compression steel is:
If the steel in compression yielded, e’s = 0.002, then: or
d’/x < 0.43 is the condition which ensure that the steel in compression is yielded.

64. However, since the depth of the N. A
However, since the depth of the N.A. (x) is not known, it is convenient to start the analysis assuming that all the steel reinforcements are yielded.
Using C=T equation to find the depth of N.A as:
here is assumed fs’ is assumed = fy then the depth of stress rectangle “a”:
and

65. Using the calculated x, check whether the reinforcement are yielded or not.
For steel in tension to be yielded, the calculated x must be less or equal to 0.5d:
For steel in compression to be yielded:
If the above two assumptions are satisfied proceed to calculate the moment using:

66. If the steel in compression is not yielded, then the exact value of the N.A. need to be evaluated as follows:
Using T=C1+C2 :
Putting:
Simplifying the above equation will lead to a 2nd degree equation with one positive root for x. The final moment will be:

67. For the case where x exceeds xmax:
If x > xmax = 0.5d then take x = 0.5d and calculate fs’
If fs’ is more than fy use fs’ = fy

68. Flowchart 3: Analysis of Doubly Reinforced Rectangular Beam Section

69. Example 4 Calculate the ultimate moment capacity of the doubly r.c.
section shown in figure if:
As = 5T25 and As’ = 5T16
As = 5T25 and As’ = 6T20

70. Solution (i) As = 5T25 and As’ = 5T16 As = 2454 mm2 As’ = 1000 mm2
Assume both steel are yielded and using C=T

71.  hence the steel in tension is yielded, o.k.
Check whether the steel in compression is yielded or not 
Hence the steel in compression is yielded, o.k.
The maximum bending capacity will be

72. Assume both steel are yielded and using C=T
(ii) As = 5T25 and As’ = 6T20
As = 2454 mm2 As’ = 1885 mm2
Assume both steel are yielded and using C=T
x = mm < 0.5d = 200 mm OK
the steel in compression not yield, not o.k.

73. =
= kNm, OK

74. Example 2 Not Updated Yet
Double r.c. beam
Example 2
Design the steel reinforcement for the T-beam section shown in Fig., to resist an ultimate moment of 500 kN.m. Use fcu = 15 MPa, fy = 460 MPa.
Not Updated Yet

75. Not Updated Yet = 0.156x15x300x4452 +0.45x15(1200-300)100(445-50)x10-6
Double r.c. beam
Mflange = 320 kN.m
Mu >Mflange  The N.A. lies in the web
= 0.156x15x300x x15( )100(445-50)x10-6
= kN.m
Not Updated Yet
Mu > Mmax  C.F. need a steel in compression design as a double r.c. section
M1 = Mmax = 796 kN.m
xmax = 0.5 d = 0.5 x 445 = mm
amax = 0.9 x = mm

76. Double r.c. beam
Not Updated Yet
As1=2526 mm2

77. Not Updated Yet Alternatively C = T
Double r.c. beam
Alternatively
C = T
0.45 x 15 x area = 0.87 x 460 x As1
area = 300 x x 100 = mm2
As1 = mm2
Not Updated Yet

78. Design of Reinforced Concrete
Analysis T-beam with Steel in Compression
Updated

79. The analysis of T- or L beam with steel in compression is dependent on the location of the N.A whether is located in the flange or in the web of the beam.
The analysis may be started assuming that:
the N.A lies in the flange and
the steels in compression and tension are yielded
These two assumptions need to be verified after finding the depth of the N.A (“x”) which can be calculated using C = T equation as follows (refer to figure.):

80. Fig.

81. the N.A lies in the flange (R.B) and assumption “a” is satisfied
C1 + C2 = T
if x ≤ hf
the N.A lies in the flange (R.B)
and assumption “a” is satisfied
if x > hf
the N.A lies in the web (T.T.B)
and assumption “a” is not satisfied

82. ≥ fy The steel is yielded and assumption “b” is satisfied.
Furthermore, the check for steel stresses need to be carried out. To check whether the steel in compression (A’s) yields or not: or
If :
≥ fy The steel is yielded and assumption “b” is satisfied.
< fy The steel is not yielded and fs’ need to be calculated.

83. Case 1: N.A lies in the flange Case 2: N.A lies in the web
For steel in tension, if x < 0.5 d, the steel in tension is yielded otherwise use x=0.5d.
In case that the steel in compression is not yielded, the solution will depends on the location of N.A. two cases will be explained in more details as follows:
Case 1: N.A lies in the flange
Case 2: N.A lies in the web

84. Case 1: N.A lies in the flange
Using C=T equation
Find x, by solving the above 2nd degree equation
If x < 0.5 d  (T.F.), ok
If x > 0.5 d  use x = 0.5 d
The moment capacity of the section is then calculated by taking moment of the forces about the centre of tension steel as:

85. Case 2: N.A lies in the web Refer to Fig. and using C=T equation or
Simplifying this equation will lead to a 2nd degree equation in “x”
If x < 0.5 d  (T.F.), ok
If x > 0.5 d  use x = 0.5 d

86. Table 1 shows the forces and their moment arms.
The moment capacity of the section is then calculated by taking moment of the forces about the centre of tension steel as:

87. Table 1 Compression forces in different parts of the doubly reinforced T-section
Web
Flange
Steel in
compression

88. Example 5 For the doubly reinforced T-beam shown in Fig. , find the
moment capacity for the following reinforcement cases:
Case a : As = 4T25 = mm2 and As’ = 2T10 = 157 mm2
Case b : As = 6T25 = mm2 and As’ = 2T16 = mm2
Case c : As = 8T25 = mm2 and As’ = 2T12 = 226 mm2
use:
fcu = 15 MPa
fy = 460 MPa
Updated

89. Solution Case a : As = 4T25 = 1963.5 mm2 and As’ = 2T10 = 157 mm2
assume: a) the steel reinforcements are yielded fs = fs’ = fy
b) the N.A. lies in the flange or x < hf
Using C=T
x = <  N.A. lies in the flange (o.k.)
< d/2 = 250mm  the steel in tension is yielded
 the steel in compression does not yield

90. Calculate fs’ using exact analysis

91. x = 88.79 < 100  N.A. lies in the flange
< d/2 = 250mm  the steel in tension is yielded
then
The moment capacity will be:
= kN.m

92. Solution Case b : As = 6T25 = 2945.2 mm2 and As’ = 2T16 = 402.12 mm2
assume: a) the steel reinforcements are yielded fs = fs’ = fy
b) the N.A. lies in the flange or x < hf
Using C=T
x = > 100  N.A. lies in the web and assumption “a”
is not satisfied and the section is T.T.B.

93. Calculate “x” using exact analysis
x = mm > hf  N.A. lies in the web (o.k.)
< d/2  T.F, the steel in tension is yielded, (o.k.)

94. M = kN.m

95. Solution Case c : As = 8T25 = 3927 mm2 and As’ = 2T12 = 226 mm2
The analysis can also start by assuming:
a) the steel reinforcements are yielded fs = fs’ = fy
b) the N.A. lies in the web or x > hf
x = mm > 100  N.A. lies in the web (o.k.)
> d/2  C.F steel in tension does not yielded (not o.k)
use x = 0.5 (500) = 250 > 100  T.T.B

96.  use fs’ = fy
M = kN.m

97. Flowchart 4: Analysis of Doubly Reinforced T-Beam Section

98. cont. Flowchart 4: Analysis of Doubly Reinforced T-Beam Section

99. Double r.c. beam
Example 2
Design the steel reinforcement for the T-beam section shown in Fig., to resist an ultimate moment of 500 kN.m. Use fcu = 15 MPa, fy = 460 MPa.

100. Double r.c. beam
Mflange = 320 kN.m
Mu >Mflange  The N.A. lies in the web
= 0.156x15x300x x15( )100(445-50)x10-6
= kN.m
Mu > Mmax  C.F. need a steel in compression design as a double r.c. section
M1 = Mmax = 796 kN.m
xmax = 0.5 d = 0.5 x 445 = mm
amax = 0.9 x = mm

101. Double r.c. beam
As1=2526 mm2

102. Alternatively C = T 0.45 x 15 x area = 0.87 x 460 x As1
Double r.c. beam
Alternatively
C = T
0.45 x 15 x area = 0.87 x 460 x As1
area = 300 x x 100 = mm2
As1 = mm2