reinforced concrete design beam complete lecture. a complete lecture

1. Concrete Floor Systems
Analysis and Design of Slabs
Basic Design Steps
Example: Design of 90′ x 60′ Hall
References
At the end of this lecture, students will be able to
Recall basic design steps of shear and flexure
Classify Concrete Floor Systems
Choose different framing systems for structures.
Analyze and Design one-way slabs for flexure by
ACI Approximate method
Discuss effect of beam spacing on slab moments
RCD

2. CONCRETE FLOOR
SYSTEMS
STRUCTURAL MEMBER
RCD

3. Concrete Floor Systems
RCD

4. Concrete Floor Systems
RCD

5. Concrete Floor Systems
RCD

6. Concrete Floor Systems
RCD

7. Concrete Floor Systems
RCD

8. RCD
Two possible arrangement of reinforcement
are shown a) straight bar b) bent bars are
used

9. Concrete Floor Systems
RCD

10. Concrete Floor Systems
RCD
Beam Supported Slabs

11. Beam Supported slab
RCD
Beam supported slab (One way)
Beam supported slab
(One way)
Beam supported slab (Two way)

12. Concrete Floor Systems
RCD
Punching shear is a typical
problem in flat plate
Flat Plate
Flat plate (Two way
slab)

13. RCD
Flat Slab
Flat slab
(Two way
slab)
Drop Panel: Thick part of slab in the
vicinity of columns
Column Capital: Column head of
increased size
Punching shear can be reduced by
introducing drop panel
and column capital

14. RCD
One way and two way joist (Beam used to support
floors or roofs) system (grid or waffle slab)
Joist: T-beams called joists are formed by creating
void spaces in what otherwise would be a solid
slab.
Peshawar University Auditorium
ribs

15. RCD
One-way Joist construction consists of a
monolithic combination of regularly spaced
ribs and a top slab arranged to span in one
direction or two orthogonal directions.
Peshawar University Auditorium
ribs
structural system
will be however
called as joist
system if the pan
width (clear spacing
between ribs) is less
than or equal to 30
inches (ACI 8.11.3).

16. RCD
Two way joist system (grid or waffle slab)
A two-way joist system, or waffle slab, comprises evenly spaced concrete joists
spanning in both directions and a reinforced concrete slab cast integrally with
the joists.
joist

17. RCD
Two way joist system (grid or waffle slab)
Solid slab

18. Two way joist
RCD
One way and two way joist system (grid
or waffle slab)
One way joist slab system

19. Types of Slab
One way slab
RCD
Two way Slab

20. Introduction
Types of Slab
One way slab
Two way Slab
A reinforced concrete slab is a broad, flat plate, usually horizontal, with top and
bottom surfaces parallel or nearly so.
It may be supported by RC beams, by masonry or reinforced walls, by structural
steel members, directly by columns or continuously by the ground
Flexural stresses are dominated in slab therefore slabs are called flexural
members
Shear forces are mostly controlled by thickness of slab
Torsional reinforcement are provided at the corner of slab (two way slab)
The reinforcement are provide parallel to surface of slab in both direction
depends on flexural behavior
RCD

21. Torsional reinforcement are provided at the corner of slab (two way
slab)
RCD

22. The slab which is supported only on two sides or supported on four sides with
longer to shorter side ratio equal to or greater than two i.e
aspect ratio M=L/S ≥ 2 (lb/la = ≥ 2)
In one way slab, loads travel along the shorter span of slab
Therefore main bars are provided along the deflection profile as shown in the
figure
In the other direction temperature/shrinkage/distribution reinforcement are
provided
RCD
Temperature
reinforcement

23. RCD
The slab which is supported on four sides, with longer to shorter side ratio
less than two i.e aspect ratio M=L/S < 2 (lb/la <2)
In two way slab, loads travel along the both span of slab
Therefore main bars are provided along both direction as shown in the
figure

24. RCD
Design of one way slab
Flexural design of slab is same as design of rectangular beam with 12 inches
width (unit strip of 1 feet)
The minimum thickness of one ay slab is calculated as per ACI code, the table
is shown below
Simply supported (pin-pin)
One-end continuous (pin-fixed)
Both-end continuous (fixed-fixed)
Cantilever (fixed-free)
l/16
l/18.5
l/20
l/8
and Beam

25. RCD
Design of slab
Analysis
Unlike beams and columns, slabs are two dimensional members. Therefore
their analysis except one-way slab systems is relatively difficult.
Design
Once the analysis is done, the design is carried out in the usual manner. So no
problem in design, problem is only in analysis of slabs.
Analysis Methods
Analysis using computer software (FEA)
SAP 2000, ETABS, SAFE etc.
ACI Approximate Method of Analysis
Strip Method for one-way slabs
Moment Coefficient Method for two way slabs
Direct Design Method for two way slabs

26. RCD
Analysis and Design of One-Way Slab Systems
Strip Method of Analysis for One-way Slabs
For purposes of analysis and design, a unit strip of one way slab, cut out at
right angles to the supporting beams, may be considered as a rectangular
beam of unit width, with a depth h and a span la as shown.
The strip method of analysis and design of
slabs having bending in one direction is
applicable only when:
Slab is supported on only two sides on stiff
beams or walls,
Slab is supported on all sides on stiff beams
or walls with ratio of larger to smaller side
greater than 2.
Note: Not applicable to flat plates etc., even
if bending is primarily in one direction.

27. RCD
Analysis and Design of One-Way Slab Systems
Strip Method of Analysis for One-way Slabs
Basic Steps for Structural Design
Step No. 01: Sizes:-Sizes of all structural and non structural elements are
decided.
Step No. 02: Loads:-Loads on structure are determined based on occupational
characteristics and functionality (refer Appendix C at the end of this lecture)
Step No. 03: Analysis:-Effect of loads are calculated on all structural elements
Step No. 04: Design:-Structural elements are designed for the respective load
effects following the code provisions.

28. RCD
Basic Steps for Structural Design

29. RCD
Basic Steps for Structural Design

30. RCD
Basic Steps for Structural Design
Loads:
According to ACI 8.2 — Service loads shall be in accordance with the general
building code of which this code forms a part, with such live load reductions as
are permitted in the general building code.
BCP SP-2007 is General Building Code of Pakistan and it refers to ASCE 7-10
for minimum design loads for buildings and other structures.
One-way slabs are usually designed for gravity loading
(U = 1.2D + 1.6L).

31. RCD
Basic Steps for Structural Design
Loads:

32. RCD
Basic Steps for Structural Design
Analysis:
Chapter 8 of the ACI addresses provisions for the analysis of concrete
members.
According to ACI 8.3.3, as an alternate to frame analysis, ACI approximate
moments shall be permitted for analysis of one-way slabs with certain
restrictions, which are as follows:

33. RCD

34. RCD

35. RCD

36. Uniaxial
Column Axial
Column
Biaxial
Column
RCD

37. RCD
spandrel beam - when wall can not take weight of slab or floor , In such cases,
the beams are provided exterior walls at each floor level to support the wall load
and perhaps some roof load also , known as spandrel beam.
It is not necessary that under the beam always be wall, there can be column and
window but spandrel beam will provide only at exterior wall.

38. RCD
interior support
Exterior support
Mid span
Mid span
interior support
Exterior support
Mid span
Mid span

39. RCD
b
a
Shear for a slab

40. RCD
Basic Steps for Structural Design
Design:
Capacity ≥ Demand
Capacity or Design Strength = Strength Reduction Factor (ϕ) x Nominal
Strength
Demand = Load Factor xService Load Effects
Bar spacing (in inches) = Ab/As × 12
(Ab = area of bar in in2, As = Design steel in in2/ft)

41. RCD
Maximum spacing for main steel reinforcement in one way slab according to
ACI 7.6.5 is minimum of 3hf or 18’’
The shrinkage/ temperature reinforcement is calculated as per ACI code 7.12.2
as shown in below table
Maximum spacing for shrinkage steel in one way slab according to ACI 7.6.5 is
minimum of 5hf or 18’’
In slab Minimum Steel reinforcement as per ACI code = shrinkage/ temperature
reinforcement
Maximum steel reinforcement is same as for rectangular beam #3, #4 and #5
bars are commonly used in slabs

42. RCD
Design of slab
The following Example illustrates the design of a one-way slab. It will be noted
that the code (7.7.1.c) COVER REQUIREMENT for reinforcement in slabs (#11 and
smaller bars) is ¾ in. clear, unless corrosion or fire protection requirements are
more severe

43. RCD
Assume structural
configuration. Take time
to reach to a
reasonable arrangement
of beams, girders and
columns. It
depends on experience.
Several alternatives are
possible.
One way and Two way slab

44. RCD
Assume structural
configuration. Take time to
reach to a reasonable
arrangement of beams,
girders and columns. It
depends on experience.
Several alternatives are
possible.
One way and Two way slab

45. RCD
One way and Two way slab

46. RCD
Wall width = 18 in, given
Assume beam width = 18 in
Slab
Beam
One way and Two way slab

47. RCD
Two way Slab
100’x60’

48. RCD
Two way Slab

49. RCD
Two way Slab SLAB, BEAMS, COLUMN AND FOOTING
S1
S1 S1
S1
S2
S2
S2
S2
S3 S3
S4 S4
B1 B1
B2 B3 B2
B4
B5
B5
B4
C3
C1
C2
19’
20’
19’
20’
B2

50. DESIGN EXAMPLES
RCD

51. Design slab and beams of a 90′ × 60′ Hall. The height of Hall is 20′. Concrete
compressive strength (fc′) = 3 ksi and steel yield strength (fy) = 40 ksi. Take 3″
mud layer and 2″ tile layer above slab. Take Live Load equal to 40 psf.
RCD

52. RCD
Assume structural
configuration. Take time
to reach to a
reasonable arrangement
of beams, girders and
columns. It
depends on experience.
Several alternatives are
possible.

53. RCD
Wall width = 18 in, given
Assume beam width = 18 in

54. RCD

55. RCD

56. RCD
Mu = coefficient × wu × ln
2

57. RCD
Or from table of areas of bars
In next slide
OR other way around bar spacing (in inches) = Ab/As × 12
Using # 3 bar (Ab = 0.11 and As = 0.144 so
Bar spacing (in inches) = Ab/As × 12 =9.166 c/c’’ #3@ 9.166 c/c’’

58. RCD
Design of slab
A table of areas of bars in slabs such as Appendix A, Table A.6 is very useful in
such cases for selecting the specific bars to be used.
Temperature Steel
Primary Flexural
reinforcement

59. RCD

60. RCD

61. RCD
Slab Design
Placement of positive reinforcement:
Positive reinforcing bars are placed in the direction of flexure stresses
and placed at the bottom (above the clear cover) to maximize the “d”,
effective depth.

62. RCD
Slab Design
Placement of negative reinforcement:
Negative reinforcement bars are placed perpendicular to the direction of
support (beam in this case). At the support these rest on the reinforcement
of the beam.
beam

63. RCD
Slab Design
Placement of negative reinforcement:
At the far end of bars, the chairs are provided to support the negative
reinforcement. As each bar will need a separate chair therefore to reduce the
number of chairs supporting bars are provided perpendicular to the direction of
negative reinforcement.

64. RCD
Slab Design
Reinforcement at discontinuous support:
At the discontinuous end, the ACI code recommends to provide
reinforcement equal to 1/3 times the positive reinforcement provided at
the mid span.

65. RCD
Slab Design
Step No 05: Drafting
Main reinforcement = #3 @ 9″ c/c (positive & negative)
Shrinkage reinforcement = #3 @ 9″ c/c
Supporting bars = #3 @ 18″ c/c
Chairs or supporting bars

66. RCD
Beam Design
Step No 01: Sizes
Minimum thickness of beam (simply
supported) = hmin = l/16
l = clear span (ln) + depth of member (beam) ≤
c/c distance between supports
Let depth of beam = 5′
ln + depth of beam = 60′ + 5′ = 65′
c/c distance between beam supports = 60 + 2
× (9/12) = 61.5′
Therefore l = 61.5′
Depth (h) = (61.5/16)×(0.4+fy/100000)×12=
36.9″ (Minimum by ACI 9.3.1.1).
Take h = 5′ = 60″
d = h – 3 = 57″
bw = 18″ (assumed)

67. RCD
Beam Design
Step No 02: Loads
Load on beam will be equal to
Factored load on beam from slab +
factored self weight of beam web
Factored load on slab = 0. 214 ksf
Load on beam from slab = 0. 214 ksf x 10
= 2.14 k/ft
Factored Self load of beam web =
=1.2 x (54 × 18/144) × 0.15 =1.215 k/ft
Total load on beam = 2.14 + 1.215 = 3.355 k/ft
10’

68. RCD
Shear Force & Bending Moment Diagrams
Beam Design
Step No 03: Analysis
Vu = 87.23 kip
Mu = 19034 in-kip

69. RCD
Beam Design
Step No 04: Design
Design for flexure
Step (a): According to ACI 6.3.2.1, beff for T-beam is minimum of:
16hf + bw = 16 × 6 + 18 =114″
(clear length of beam)/4 =(60′/4) ×12 + 18 =198″
clear spacing between beams + bw =8.5′ × 12 + 18 =120″
So beff = 114″
Step (b): Check if beam is to be designed as rectangular beam or T-beam.
Assume a = hf = 6″ and calculate As:
As =Mu/ {Φfy (d–a/2)} =19034/ {0.9 × 40 × (57–6/2)} = 9.79 in2
Re-calculate “a”:
a =Asfy/ (0.85fc′beff) =9.79 × 40/ (0.85 × 3 × 114) = 1.34″ < hf
Therefore design beam as rectangular beam.
After trials As = 9.38 in2 (Asmax = 20.83 in2 ;Asmin = 5.13 in2)
Therefore As = 9.38 in2 (12 #8 bars)

70. RCD
Beam Design
(Asmax = 20.83
in2 ;Asmin = 5.13
in2)
Therefore As =
9.38 in2 (12 #8
bars)

71. RCD
Beam Design
Step No 04: Design
Design for flexure
Skin Reinforcement : (ACI 9.7.2.3)
As the effective depth d of a beam is greater than 36 inches, longitudinal skin
reinforcement is required as per ACI 9.7.2.3.
Askin, = Main flexural reinforcement/2 = 9.60/2 = 4.8 in2
Range up to which skin reinforcement is provided:
d/2 = 56.625/2 = 28.3125″

72. RCD
Beam Design
Step No 04: Design
Design for flexure
Skin Reinforcement
For #8 bar used in skin reinforcement,
ssk (skin reinforcement spacing)
is least of:
d/6 = 56.625/6 = 9.44″, 12″, or
1000Ab/(d – 30) = 1000×0.80/(56.625 – 30) = 30.05″
Therefore ssk = 9.44″ ≈ 9″
With this spacing, 3 bars on each face are required.
And for # 8 bar, the total area of skin reinforcement is:
Askin = 6 × 0.80 = 4.8 in2
skin reinforcement

73. RCD
Beam Design
Step No 04: Design
Design for Shear
Vu = 87.23 kip
ΦVc = Φ2 √ fc′bwd = (0.75 × 2 × √3000 × 18 × 56.625)/1000 = 83.74 kip
ΦVc < Vu (Shear reinforcement is required)
sd = ΦAvfyd/(Vu – ΦVc)
Using #3, 2 legged stirrups with Av = 0.11 × 2 =0.22 in2}
sd = 0.75 × 0.22 × 40 × 56.625/(87.23 – 83.74) = 107″

74. RCD
Beam Design
Step No 04: Design
Design for Shear
Maximum spacing and minimum reinforcement requirement as permitted by ACI
9.7.6.2.2 and 10.6.2.2 shall be minimum of:
Avfy/(50bw) =0.22 × 40000/(50 × 18) ≈ 9.5″
d/2 =56.625/2 =28.3″
24″
Avfy/ 0.75 √fc’ bw = 0.22 × 40000/ {(0.75 × √3000 × 18} = 11.90″
Therefore, smax = 9.5″
ΦVc /2 = 83.74/2 = 41.87 kips at a distance of 17.5 ft from face of the support.
Therefore no reinforcemnt is required in this zone, however, we will provide #3,
2-legged vertical stirrups at 12 in. c/c

75. RCD
Beam Design
Step No 04: Design
Design for Shear
Other checks:
Check for depth of beam:
ΦVs ≤ Φ8 √fc’ bwd (ACI 22.5.1.2)
Φ8 √fc’ bwd = 0.75 × 8 × √3000 × 18 × 56.625/1000 = 334.96 k
ΦVs = (ΦAvfyd)/smax
= (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.3 k < 334.96 k, O.K.
So depth is O.K. If not, increase depth of beam.

76. RCD
Beam Design
Step No 04: Design
Design for Shear
Other checks:
Check if “ ΦVs ≤ Φ4 √ fc′bwd ” (ACI 10.7.6.5.2):
If “ΦVs ≤ Φ4 √ fc′ bwd”, the maximum spacing (smax) is O.K.
Otherwise reduce spacing by one half.
Φ4√ fc′bwd = 0.75 × 4 × √3000 × 18 × 56.625/1000= 167.47 k
ΦVs = (ΦAvfyd)/sd
= (0.75 × 0.22 × 40 × 56.625)/9.5 = 39.33 k < 167.47 k, O.K.

77. RCD
Beam Design
Step No 05: Drafting

78. RCD
In this design example, the beams were supported on walls.
This was done to simplify analysis.
For practical reasons, however, the beams must be supported
on columns and hence the structural analysis will be that of a
frame rather than simply supported beam.
In the subsequent slides, the analysis and design results for
beams supported on columns have been provided.

79. Design slab, beams, columns, and footing of a 90′ × 60′ Hall. The height of Hall
is 20′. Concrete compressive strength (fc′) = 3 ksi and steel yield strength (fy)
= 40 ksi. Take 3″ mud layer and 2″ tile layer above slab. Take Live Load equal
to 40 psf.
RCD

80. RCD
Frame Analysis
3D model of the hall showing beams supported on columns.

81. RCD
Frame Analysis
A 2D frame can be detached from a 3D system in the following manner
Column size = 18’’ × 18’’

82. RCD
Frame Analysis
Various methods can be used for frame analysis. Using moment distribution
method, the following results can be Obtained:

83. RCD
Frame Analysis
Slab Design
Slab design will remain the same as in case of beams supported on
walls.
Main reinforcement = #3 @ 9″ c/c (positive & negative)
Shrinkage reinforcement = #3 @ 9″ c/c
Supporting bars = #3 @ 18″ c/c

84. RCD
Frame Analysis
Beam Design
Beam flexure design will be as follows:
Mu (+ve) = 17627 in-kips
Mu (-ve) = 1407 in-kips
As (+ve) = 8.68 in2
Use 6 #8 in 1st layer & 2 #8 + 4 #7 bars in 2nd layer)
As = (8)(0.80) + (4)(0.60) = 8.80 in2 (As,max = 0.0203bd = 20.83 in2 OK)
As (-ve) = 0.69 in2 (As,min = 0.005bd = 5.13 in2, so As,min governs)
Use 7 #8 bars (5 bars in 1st layer and 2 bars in 2nd layer)
As = (7)(0.80) = 5.60 in2
Beam shear design will be same as in previous case.

85. RCD
Beam Design
(As,max = 0.0203bd =
20.83 in2 )
As (-ve) = 0.69 in2
(As,min = 0.005bd =
5.13 in2)

86. RCD
Frame Analysis

87. RCD
DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH
UNIAXIAL BENDING using DESIGN AIDS

88. RCD
The ACI column interaction diagrams are used to design or analyze
columns for different situations. In order to correctly use these diagrams, it
is necessary to compute the value of γ (gamma), which is equal to the
distance from the center of the bars on one side of the column to the center
of the bars on the other side of the column divided by h,
Column cross sections for normalized
interaction curves in Appendix A,
DESIGN OF RC MEMBERS SUBJECTED TO COMPRESSIVE LOAD WITH
UNIAXIAL BENDING using DESIGN AIDS

89. RCD
Frame Analysis
Column Design: Using ACI Design Aids
Main Reinforcement Design
Size:
18 in. × 18 in.
Loads:
Pu = 103.17 kips
Mu = 1407 in-kips
Calculate the ratio γ , for 2.5 in. cover (c): g = (h – 2c) / h
= (18 – 5)/18 = 0.72
Calculate Kn, Kn = Pu/(Ø fc′Ag) = 103.17/(0.65 × 3 × 324) = 0.16
Calculate Rn, Rn = Mu/(Øfc′Agh) = 1407/(0.65 × 3 × 324 × 18) = 0.12
fc′ = 3 ksi, fy = 60 ksi

90. RCD
Column Design
Main Reinforcement Design
For given material strength,
the column strength interaction
diagram gives the
following reinforcement ratio:
ρ = 0.01
Ast = 0.01 × 324 = 3.24 in2
Using 8 #6 bars

91. RCD
Column Design
Tie Bars:
Using 3/8″ Φ (#3) tie bars for 3/4″ Φ (#6) main bars (ACI 9.7.6.4.2),
Spacing for Tie bars according to ACI 9.7.6.4.3 is minimum of:
16 × dia of main bar =16 × 6/8 =12″ c/c
48 × dia of tie bar = 48 × (3/8) =18″ c/c
Least column dimension =18″ c/c
Finally use #3, tie bars @ 12″ c/c

92. RCD
Column Design
12″ c/c
12″ c/c

93. RCD
Footing Design
Isolated column footing; square or rectangular

94. RCD
Footing Design
Data Given:
Column size = 18″ × 18″
fc′ =3 ksi
fy = 40 ksi
qa = 2.204 k/ft2
Factored load on column = 103.17 kips (Reaction at the support)
Service load on column = 81.87 kips (Reaction at the support due to service
load)

95. RCD
Footing Design
Sizes:
Assume h = 15 in.
davg = h – clear cover – one bar dia
= 15 – 3 – 1(for #8 bar) = 11 in.
Assume depth of the base of footing from ground level (z) = 5′
Weight of fill and concrete footing, W= γfill(z - h) + γch
=100 × (5 – 1.25) +150 × (1.25) = 562.5 psf = 0.5625 ksf

96. RCD
Footing Design
Sizes:
Effective bearing capacity, qe = qa – W
= 2.204 – 0.5625 = 1.642 ksf
Bearing area, Areq = Service Load/ qe
= 81.87/1.642 = 49.86 ft2
Areq = B x B = 49.86 ft2 => B = 7 ft.
Critical Perimeter, bo = 4 x (c + davg)
= 4 × (18 + 11) =116 in
7 ft
7 ft
5.5 ft
5.5 ft

97. RCD
Footing Design
Loads:
qu (bearing pressure for strength design of footing):
qu = factored load on column / Areq
= 103.17 / (7 × 7) = 2.105 ksf
Analysis:
Beam Shear
Beam shear failure at a distance “d” from face of column
27.014k
2.105*(7/2-1.5/2)-11/12)*7 quB

98. RCD
Footing Design
Analysis:
Punching shear:
Punching shear failure at a
distance “d/2” from face of column
Vup = quB2 – qu(c + davg)2
Vup = (2.105 × 49) –2.105 × {(18+11)/12)}2
103.145-12.29
= 90.85 kip
qu(c + davg)2
quB2

99. RCD
Footing Design
Analysis:
Flexural Analysis:
Flexural Failure at face of column
Mu = quBk2/2
k = (B – c)/2 = (7 x 12 –18)/2
= 33 in = 2.75´
Mu = 2.105 × 7 × 2.75 × 2.75/2
= 55.72 ft-k = 668.60 in-kip
quB

100. RCD
Footing Design
Design:
Design for Punching Shear:
Vup = 90.85 kip
Punching shear capacity (ΦVcp)
= Φ4√fc’ bodavg
= 0.75 × 4 × 3000 × 116 × 11/1000
= 209.66 k >Vup, O.K

101. RCD
Footing Design
Design:
Design for Flexure:
Mu = 668.60 kip-in
a = 0.2davg = 0.2 × 11 = 2.2″
As = Mu/ {Φfy(davg – a/2)} = 668.60/ {0.9 × 40 × (11 – 2.2/2)} = 1.87in2
a = Asfy/ (0.85fc′B) = 1.83 × 40/ (0.85 × 3 × 7 × 12) = 0.35″
After trials, As = 1.71 in2 (Asmin = 0.005Bdavg = 4.62 in2 so Asmin governs)
Now, the spacing can be calculated as follows:
Using #8 bars: No. of bars = 4.62/0.79 ≈ 6 bars.
Spacing = 6.5 × 12 /5 = 15 in. c/c
Hence 6 bars can be provided in the foundation if they are placed 15 in. c/c
(Max. spacing should not exceed 3h or 18 in.)

102. RCD
Drafting

104. RCD
Drafting

105. RCD
IMPORTANT TABLES

106. RCD
IMPORTANT TABLES

107. RCD
IMPORTANT TABLES

108. RCD
IMPORTANT TABLES

109. RCD

110. RCD

111. RCD

112. RCD
Bar spacing (in inches) = Ab/As × 12
Using # 4 bar (Ab = 0.2 and As = 0.27 so
Bar spacing (in inches) = Ab/As × 12 =8.8 c/c’’ #4@ 8.8 c/c’’
#4@ 14 c/c’’
#4@ 24 c/c’’
#4@ 14 c/c’’
#4@ 17 c/c’’
#4@ 8.8 c/c’’

113. RCD

114. RCD
Two possible arrangement of reinforcement
are shown a) straight bar b) bent bars are
used

115. RCD

116. RCD
Two way Slab
The slab which is supported on four sides, with longer to shorter side ratio
less than two i.e aspect ratio M=L/S < 2 (lb/la <2)
In two way slab, loads travel along the both span of slab
Therefore main bars are provided along both direction as shown in the
figure

117. RCD
Two way Slab
100’x60’

118. RCD
Two way Slab

119. RCD
Two way Slab SLAB BEAMS COLUMN AND FOOTING
S1
S1 S1
S1
S2
S2
S2
S2
S3 S3
S4 S4
B1 B1
B2 B3 B2
B4
B5
B5
B4
C3
C1
C2

120. RCD

121. RCD

122. RCD

123. RCD

124. RCD

125. RCD

126. RCD

127. RCD
1 2 3 4 5 6 7 8 9

128. RCD

129. 1 2 3 4 5 6 7 8 9
RCD

130. RCD
COEFFICIENTS FOR NEGATIVE
MOMENTS IN SLABs

131. RCD
COEFFICIENTS FOR DEAD
LOAD POSITIVE ATIVE
MOMENTS IN SLABs

132. RCD
COEFFICIENTS FOR LIVE
LOAD POSITIVE MOMENTS IN
SLABs

133. RCD
RATIO OF LOAD W IN la AND lb
DIRECTIONS FOR SHEAR IN
SLAB AND LOAD ON
SUPPORTS

134. RCD

135. RCD

136. RCD
la
Mb (-ve)
Ma (-ve)
Ma (-ve)
M
b
(+ve)
Ma (+ve)
lb
Mb (-ve)

137. RCD
Bar spacing (in inches) = Ab/As × 12
Using # 4 bar (Ab = 0.2 and As = 0.288 so
Bar spacing (in inches) = Ab/As × 12 =7 c/c’’ #4@ 7 c/c’’
#4@ 7 c/c’’
#4@ 5 c/c’’

138. RCD
#4@ 12 c/c’’
#3@ 6c/c’’

139. RCD
#3@ 4.5c/c’’
#3@ 12c/c’’

140. RCD
la
Mb (-ve)
Ma (-ve)
Ma (-ve)
M
b
(+ve)
Ma (+ve)
lb
Mb (-ve)

141. RCD

142. RCD

143. RCD

144. RCD

145. RCD

146. RCD

147. RCD