Design and Detailing of RC Deep beams as per IS 456-2000

Module 3
Design of Reinforced Concrete Deep Beams
Introduction – Minimum thickness -Steps of Designing Deep beams – design by IS 456 - Detailing of Deep beams.
Two Span Continuous
Deep Beam with Two
Point Loading

1. Deep Beam Definition - IS 456
RC deep beams have useful applications in
• tall buildings,
• offshore structures and foundations
2. Deep Beam Application
A deep beam is having a depth comparable to the span
TRANSFER GIRDER
RIBBED MAT FOUNDATION
3. Deep Beam – Types
• Simply Supported or Continuous
• Rectangular or Flanged Beams
• Top or Bottom or Side Loaded
• with or without openings

4. Behaviour of Deep Beams
• The elementary theory of bending for simple beams may not be applicable to deep beams even
under the linear elastic assumption.
• A deep beam is in fact a vertical plate subjected to loading in its own plane. The strain or stress
distribution across the depth is no longer a straight line, and the variation is mainly dependent on
the aspect ratio of the beam.
• The analysis of a deep beam should therefore be treated as a two dimensional plane stress
problem, and two-dimensional stress analysis methods should be used in order to obtain a realistic
stress distribution in deep beams even for a linear elastic solution.

2m
4m
COMPRESSION
TENSION
4m
4m
TENSION
COMPRESSION
FE RESULTS

L/D  2
L/D < 1
• Following approximations are suggested for design
purposes to compute the lever arm Z
Z
Z

5. Compressive force path concept
• The load-carrying capacity of an RC structural member is
associated with the strength of concrete in the region of the
paths along which compressive forces are transmitted to the
supports.
• The path of a compressive force may be visualized as a flow
of compressive stresses with varying sections perpendicular
to the path direction and with the compressive force,
representing the stress resultant at each section.

• Failure is considered to be related to the development of
tensile stresses in the region of the path that may develop
due to a number of causes, the main ones being ;
• Changes in the path direction
• Varying intensity of compressive stress field along
path Stress increase at the tip of inclined cracks
• Bond failure

6. Arch and tie action
• Mode of failure is not associated with beam
action.
• The variation in bending moment along the
beam span is mainly effected by a change of
the lever arm rather than the magnitude of the
internal horizontal actions.
• Such behaviour has been found to result from
the fact that the force sustained by the tension
reinforcement of a deep beam at its ultimate
limit state is constant throughout the beam
span.
• RC deep beam at its ultimate limit state cannot
rely on beam action to sustain the shear
forces, it would have to behave as a tied arch.

COMPRESSION
TENSION
TENSION
COMPRESSION

7. Deep beam behaviour at ultimate limit state
Behaviour of a deep RC beam with a rectangular cross section and without shear
reinforcement may be divided into two types of behaviour depending on
• either a/d, for beams subjected to two-point loading,
• or L/d, for beams under UDL
• The Figure indicates that the mode of failure is
characterized by a deep inclined crack which
appears to have formed within the shear span
independently of the flexural cracks.
• The inclined crack initiates at the bottom face of the
beam close to the support, extends towards the top
face of the beam in the region of the load point.
Case 1:Deep beam without web reinforcement subjected to two-point loading with a / d=1.5
• Eventually causes failure of the compressive zone in the middle zone of the beam.
• The causes of failure should therefore be sought within the middle, rather than the shear
span of such beams

• Failure is associated with a large reduction of the size of the compressive
zone of the cross-section coinciding with the tip of the main inclined crack.
• This type of failure may be prevented either by providing transverse
reinforcement that would sustain the tensile stresses that cannot be
sustained by concrete alone, or by reducing the compressive stresses.
• Transverse reinforcement only within the shear span can be equally effective.
• Such reinforcement reduces the compressive stresses that develop in the
cross-section which coincides with the tip of the inclined crack, as it sustains
a portion of the bending moment developing in that section.
• However, the presence of transverse reinforcement beyond the critical
section is essential, as with stirrups only to the critical section does not
safeguard against brittle failure.
• This type of failure may be prevented by extending the transverse
reinforcement beyond the critical section to a distance approximately equal
to the depth of the compressive zone.

Case 2 ; Deep beam, without shear reinforcement, under two point loading with a/d = 1.0
• This mode of failure is characterized by a deep inclined
crack which appears to have formed within the shear
span independently of the flexural cracks.
• Inclined crack almost coincides with the line joining
the load point and the support.
• It usually starts within the beam web, almost half way
between the loading and support points, at a load
level significantly lower than the beam load-carrying
capacity, and propagates simultaneously towards
these points with increasing load.
• Eventually, collapse of the beam occurs owing to a
sudden extension of the inclined crack towards the top
and bottom face of the beam in the regions of the load
point and support, respectively, within the shear span.
• Such a mode of failure is usually referred to as
‘diagonal- splitting’
Diagonal Splitting

• Due to the large compressive forces carried by deep beams, it is unlikely
that, the presence of conventional web reinforcement in the form of
vertical stirrups considerably improves load-carrying capacity.
• Such reinforcement may delay the cracking process but may give only a
small increase in load-carrying capacity.
• Web reinforcement is provided in order to prevent splitting of the
inclined portion of the compressive force path (diagonal splitting).

8. Rebar Detailing
A. SS BEAMS

NOTE:
Anchorage of positive reinforcement may be achieved by bending of the bars in a horizontal
plane

C. Web Reinforcement in Deep Beams
CASE 1: TOP LOADED DEEP BEAMS
• Load is resisted by ARCH action as it is stiffer than
Truss action
• Stirrups are not necessary as they do not cross the
cracks
• A minimum reinforcement placed in both vertical
and horizontal directions as in RC walls is adequate.
Failure
Pattern in
TOP LOADED
DEEP BEAMS

• In continuous beams half
the flexural reinforcement
(horizontal) provided over
the supports may be part
of this
• Near the supports,
additional bars of the
same size used for web
reinforcement should be
introduced as shown

CASE 2: BOTTOM LOADED DEEP BEAMS
Failure
Pattern in
BOTTOM
LOADED
DEEP BEAMS
• Load is resisted mainly by vertical or inclined tension
towards the supports
• To enable the compression arch to develop, the whole
of the suspended load must be transferred by means
of vertical reinforcement into the compression zone of
the beam.
• Suspender Stirrups should completely surround the
bottom flexural reinforcement and extend into the
compression zone of the beam.
• Spacing should not exceed 150 mm

Example 1 – Simply Supported Deep Beam
A transfer girder 5.25 m length supports two columns located at 1.75 m from each end.
Column loads = 3750 kN . Total depth of the beam = 4.2m and width of support = 520mm.
Concrete Grade = M40, Fe 415 steel.
Design and Detail the girder.
Leff
1. C/C distance between supports
2. 1.15 x clear span whichever is less

Step 1: Check for bearing capacity at support
• Let B = Beam width
• Allowable stress = 0.45 x 40 = 18MPa
• Support width = 520 mm
• Effective width of support = 0.2x Clear Span
= 0.2x(5250 - 2x520) = 842 mm
• Adopt 520 mm
• 18 = 1.5 x 3750 x 103 / (520 x B); B = 600 mm
• Leff = 5250 - 520 = 4730 mm or 1.15 x (5250 – 2 x520) = 4841 mm ; Leff = 4730 mm CL 29.2
• D/b < 25 or L/b < 50 : 4200/600 = 7 ; 4730/600 = 7.88
• L/D = 4730/4200 = 1.13 > 1 and < 2 ; Deep Beam category CL 29.1

Step 2 : Factored Moments, Ast
• Mu = 1. 5 x ( 3750 x 1.49) = 8382 kNm
• Lever arm Z = 0.2 (Leff + 2D) = 0.2 x (4730 + 2 x 4200) = 2626mm CL 29.2 (a)
• Ast = Mu/(0.87fy Z) = 8382 x106 /(0.87 x 415 x 2626) = 8841 mm2
• Ast min = 0.85 x b x D /415 = 0.85 x 600 x 4200 /415 = 5161 mm2 cl 26.5.1.1
• Adopt 18 - #25 , Ast = 8836 mm2
Step 3: Detailing of Rebars
CL 29.3.1
• Tension Zone Depth = 0.25 D – 0.05Leff = 0.25 x 4200 –
0.05 x 4730 = 814 mm
Assume Clear bottom and side cover = 40 mm
• Arrange bars in 6 rows in a depth = 814 mm
814
39
155
600

Step 4: Detailing of Vertical Rebars: CL 32.5
• Ast min / m length = 0.0012 x 600 x 1000 = 720 mm2/m
• Provide on each face : 360 mm2/m
• Spacing of #12 rebars = 1000 x 113/360= 313 mm < 450 mm
• Adopt #12@300 mm c/c
Step 5: Detailing of Horizontal Rebars
29.3.4 Side Face Reinforcement
Side face reinforcement shall comply with requirements of minimum
reinforcement of walls
• Spacing of #16 rebars = 1000 x 201/600 = 335 mm < 450 mm

Dia 50 Dia
Anchorage value
Ld 0.8Ld
With 900 Bend
8dia*
With 1800 Hook
16 dia** * ** * **
25 1250 -200 -400 1050 850 840 680
16 800 -128 -256 672 544 540 435
End Anchorage as per CL 29.3.1(b)
40
520
355
5dia
(125)
485
• For #25 - middle bars Corner bars
ELEVATION
40
520
3555dia
(125)
Support Face
PLAN
40
40
600(BeamWidth)
325
1
1
2
2

814
Tension Zone
CL 29.3.1
Compression Zone
3386 mm
5250
520
#12@300 ; cl 32.5 (a)
Vertical Stirrups
#16@300;cl32.5(c) 18-#25 in
6 Rows
#12@300
Vertical additional rebars
near support CL 32.5 (a)
#16@300 additional rebars near support (horizontal)
adequately anchored as per CL 32.5(c)
1260
2100
TYP

Example 2 – Simply Supported Deep Beam ; M20, Fe415
• Allowable stress = 0.45 x 20 = 9MPa
= 0.2 x 5000 = 1000 mm
• Adopt 500 mm
• Total Load W = 0.25 x 3.5 x 6 x 25 + 200 x 6 = 1332 kN
• Reaction at each support = W/2 =666 kN
• Bearing Pressure = 1.5 x 666 x 103 / (500 x 250) = 8 MPa < 9 MPa OK
• Leff = 5500 mm or 1.15 x 5000 = 5750 mm ; Leff = 5500 mm CL 29.2
• D/b = 14 ; < 25 or L/b=22; < 50
• L/D = 5500/3500 = 1.57 > 1 and < 2 ; Deep Beam category CL 29.1

• w(kN/m) = 1332/6 = 222 kN/m
• Mu = 1. 5 x 222 x 5.52/8 = 1260 kNm
• Lever arm Z = 0.2 (Leff+ 2D) = 0.2 x (5500 + 2 x 3500) = 2500mm CL 29.2 (a)
• Ast = Mu/(0.87fy Z) = 1260 x 106 /(0.87 x 415 x 2500) = 1396 mm2
• Ast min = 0.85 x b x D /415 = 0.85 x 250 x 3500 /415 = 1792 mm2 CL 26.5.1.1
• Adopt 10 - #16 , Ast = 2010 mm2
Step 3: Detailing of Rebars
CL 29.3.1
• Tension Zone Depth = 0.25 D – 0.05Leff = 0.25 x 3500 – 0.05 x 5500 = 600 mm
• Assume Clear bottom and side cover = 40 mm
• Arrange bars in 5 rows in a depth = 600 mm
600
40
140
140
140
250
140

• Spacing of #10 rebars = 1000 x 78.5/150= 523 mm > 450 mm
• Spacing of #12 rebars = 1000 x 113/250 = 452 mm > 450 mm

Dia 50 Dia
With 1800 Bend
Anchorage
value = 16dia
Ld 0.8Ld
12 600 -192 408 326
16 800 -256 544 435
• For #16 in all rows
40
500
3805dia
(80)
Support Face
PLAN
40
40
250(BeamWidth)
64 (min = 4dia)
1
1
2
2
Rebars are embedded into the support by extending it to
a maximum possible length and then providing 1800
hook which project along the width of the beam

600
Tension Zone
CL 29.2.1
Compression Zone
2900 mm
6000
500
#10@450 ; cl 32.5 (a)
Vertical Stirrups
#12@450
cl32.5(c)
10 -#16 in
5 Rows
#10@450
Vertical additional rebars
near support CL 32.5 (a)
#12@450 additional rebars near support (horizontal)
adequately anchored as per CL 32.5(c)
1050
1750
TYP
1050

Example 3 : Fixed ends and continuous Deep Beam
• Concrete Grade = M35
• Allowable stress = 0.45 x 35 = 15.75 MPa CL 34.4
= 0.2 x 5000 = 1000 mm
• Adopt 500 mm
• Total Load W = 0.25 x 3 x 11.5 x 25 + 200 x11.5 = 2515.625 kN
• Reaction at Interior support = W = 2515.625/2 =1257.81 kN
• Bearing Pressure = 1.5 x 1257.81 x 103 / (500 x 250)
= 15 MPa < 15.75 MPa OK
• Leff = 5500 mm or 1.15 x 5000 = 5750 mm
• Adopt Leff = 5500 mm CL 29.2
• D/b = 12 < 25 or L/b=22; < 50
• L/D = 5500/3000 = 1.83 > 1 and < 2.5
• Deep Beam category CL 29.1

w(kN/m) = 2515.625 /11.5 = 218.75 kN/m
Span Moment
• Mu = 1. 5 x 218.75 x 5.52/24 = 413.6 kNm
• Lever arm Z = 0.2 (Leff + 1.5D) = 0.2 x (5500 + 1.5 x 3000) = 2000mm CL 29.2 (b)
• Ast = Mu/(0.87fy Z) = 413.6 x 106 /(0.87 x 415 x 2000) = 573 mm2
• Adopt 8 - #16 , Ast = 1608 mm2
Support Moment
• Mu = 1. 5 x 218.75 x 5.52/12 = 827 kNm
• Ast = Mu/(0.87fy Z) = 827 x 106 /(0.87 x 415 x 2000) = 1145 mm2
• Adopt Ast = 1536 mm2

Step 3: Detailing of Rebars in Span region
CL 29.3.1
• Arrange bars in 4 rows in a depth = 475 mm 475
40
145
145
145
250

Step 4: Detailing of Rebars in Support region CL 29.3.2
• Clear Span / D = 5000 / 3000 = 1.67 > 1 and < 2.5
• Rebars are placed in two zones CL 29.3.2 (b)
• Ast = 1536 mm2
• Zone1
• Depth = 0.2D = 0.2 x 3000 = 600 mm
• Ast1 = 1536 x 0.5 x (1.67 – 0.5) = 900 mm2
• Adopt 6 - #16 – 1206 mm2 in three rows
• Zone2
• Depth = 0.3D = 0.3 x 3000 = 900 mm on both sides of mid depth
• Ast1 = (1536 – 900) = 636 mm2

• Spacing of #10 rebars = 1000 x 78.5/150= 523 mm > 450 mm
• Adopt #10@450 mm c/c (stirrups)
• Spacing of #12 rebars = 1000 x 113/250 = 452 mm > 450 mm

900 900 0.3D
900
900
(0.5D)
1500
8 - #16 in
4 Rows
475
(0.2D) 600
(0.6D)
1800
6 - #16 in 3 Rows (Zone 1)
1500* 1500*1500* 1500*
#10@450 ; cl 32.5 (a)
Vertical Stirrups
6 - #12 in
3 Rows
(Zone2)
5500 5500
3000
* Curtailment position measured
from support face
Additional Rebars in Support
Regions (A,B,C) on both faces
#10@450 (vertical)
+
#12 @450 (Horizontal)
A
B C
#12 @450
CL 32.5 (c)
250
8 - #16 in
4 Rows
6 - #12 in
3 Rows (Zone2)
6 - #16 in
3 Rows (Zone 1)
#10@450

Example 4 : Fixed ends and continuous Deep Beam
A reinforced girder 4.5 m deep is continuous over two spans 9 m c/c, resting on column supports 900 mm
width is to be designed to support a total load of 200 kN/m including its own weight. M20 and Fe415
• Concrete Grade = M20; Allowable stress = 0.45 x 20 = 9 MPa CL 34.4
• Support width = 900 mm; Effective width of support = 0.2x Clear Span = 0.2 x 8100 = 1620 mm
• Adopt 900 mm
• Total Load W = 200 x 18.9 = 3780 kN
• Reaction at Interior support = W = 3780/2 =1890 kN
• 1.5 x 1890 x 103 / (900 x B) = 9 ; B = 350 mm
• Leff = 9000 mm or 1.15 x 8100 = 9315 mm
• Adopt Leff = 9000 mm CL 29.2
• D/b = 12.8 < 25 or L/b=25.7 < 50
• L/D = 9000/4500 = 2 > 1 and < 2.5
• Deep Beam category CL 29.1

Span Moment
• Mu = 1. 5 x 200 x 92/24 = 1012.5 kNm
• Ast = Mu/(0.87fy Z) = 1012.5 x 106 /(0.87 x 415 x 3150) = 890 mm2
• Adopt 8 - #25 in 4 rows
Support Moment
• Mu = 1. 5 x 200 x 92/12 = 2025 kNm
• Ast = Mu/(0.87fy Z) = 2025 x 106 /(0.87 x 415 x 3150) = 1780 mm2
• Ast min = 3226 mm2 CL 26.5.1.1
• Adopt Ast = 3226 mm2

Step 3: Detailing of Rebars in Span region
CL 29.3.1
• Arrange bars in 4 rows in a depth = 675 mm 675
45
210
210
210
350

Step 4: Detailing of Rebars in Support region CL 29.3.2
• Clear Span / D = 8100/ 4500 = 1.8 > 1 and < 2.5
• Rebars are placed in two zones CL 29.3.2 (b)
• Ast = 3226 mm2
• Zone1
• Depth = 0.2D = 0.2 x 4500 = 900 mm
• Ast1 = 3226 x 0.5 x (1.8 – 0.5) = 2097 mm2
• Adopt 8 - #20 – 2512 mm2 in Four rows
• Zone2
• Depth = 0.3D = 0.3 x 4500 = 1350 mm on both sides of mid depth
• Ast1 = (3226 – 2097) = 1129 mm2

• Spacing of #10 rebars = 1000 x 78.5/210= 373 mm < 450 mm
• Adopt #10@300 mm c/c (stirrups)
• Spacing of #12 rebars = 1000 x 113/350 = 322 mm < 450 mm

Dia 50 Dia
With 1800 hook
Anchorage
value = 16 dia
Ld 0.8Ld
12 600 -192 408 326
16 800 -256 544 435
20 1000 -320 680 544
25 1250 -400 850 680
40
900
7355dia
(125)
Support Face
PLAN
40
40
350(BeamWidth)
100 (min = 4dia)
1
1
2
2 Rebars are embedded into the support by extending it to
a maximum possible length and then providing 1800
hook which project along the width of the beam
For # 25 bars

1350 1350 0.3D
1350
1350
(0.5D)
2250
8 - #25 in
4 Rows
675
(0.2D) 900
(0.6D)
2700
8 - #20 in 4 Rows (Zone 1)
2250* 2250*2250* 2250*
#10@300 ; cl 32.5 (a)
Vertical Stirrups
6 - #16 in
3 Rows
(Zone2)
9000 9000
4500
*0.5D - Curtailment position
measured from support face
Additional Rebars in Support
Regions (A,B,C) on both faces
#10@300 (vertical)
+
#12 @300 (Horizontal)
A B C
#12 @300
CL 32.5 (c)

Design and Detailing of RC Deep beams as per IS 456-2000

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