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 ApplicationA 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

4mTENSION

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 stateBehaviour 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

B. CONTINUOUS BEAMS

C. Web Reinforcement in Deep BeamsCASE 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 supports2. 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 RebarsCL 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/cStep 5: Detailing of Horizontal Rebars29.3.4 Side Face ReinforcementSide face reinforcement shall comply with requirements of minimum reinforcement of walls



• Spacing of #16 rebars = 1000 x 201/600 = 335 mm < 450 mm

• Adopt #16@300 mm c/c

Dia 50 Dia

Anchorage valueLd 0.8Ld

With 900 Bend8dia*

With 1800 Hook16 dia** * ** * **

25 1250 -200 -400 1050 850 840 68016 800 -128 -256 672 544 540 435

End Anchorage as per CL 29.3.1(b)

40

520

3555dia(125)

485

• For #25 - middle bars Corner bars

ELEVATION

40

520

3555dia(125)

Support Face

PLAN

40

40600

(Bea

m W

idth

)

325

1

12

2

814Tension ZoneCL 29.3.1

Compression Zone3386 mm

5250

520

#12@300 ; cl 32.5 (a)Vertical Stirrups

#16@

300;

cl

32.

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

Step 1: Check for bearing capacity at support

• 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

Step 2 : Factored Moments, Ast• 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 RebarsCL 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 mm600

40

140

140

140

250

140




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


Step 5: Detailing of Horizontal Rebars

29.3.4 Side Face ReinforcementSide face reinforcement shall comply with requirements of minimum reinforcement of walls



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


Dia 50 DiaWith 1800 Bend

Anchorage value = 16dia

Ld 0.8Ld

12 600 -192 408 32616 800 -256 544 435


• For #16 in all rows

40

500

3805dia(80)

Support Face

PLAN

40

40250

(Bea

m W

idth

)

64 (min = 4dia)

1

12

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

600Tension ZoneCL 29.2.1

Compression Zone2900 mm

6000

500


#12@

450

cl

32.

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 Step 1: Check for bearing capacity at support

• 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

Step 2 : Factored Moments, Ast

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

• Ast min = 0.85 x b x D /415 = 0.85 x 250 x 3000 /415 = 1536 mm2 CL

26.5.1.1

• 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


26.5.1.1

• 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


Dia 50 DiaWith 1800 Bend

Anchorage value = 16dia

Ld 0.8Ld

12 600 -192 408 32616 800 -256 544 435


• For #16 in all rows

40

500

3805dia(80)

Support Face

PLAN

40

40250

(Bea

m W

idth

)

64 (min = 4dia)

1

12

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

900 900 0.3D900

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*


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 @450CL 32.5 (c)

250

8 - #16 in 4 Rows

6 - #12 in 3 Rows (Zone2)

6 - #16 in3 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

Step 1: Check for bearing capacity at support

• 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

Step 2 : Factored Moments, Ast

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


26.5.1.1

• 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 DiaWith 1800 hook

Anchorage value = 16 dia

Ld 0.8Ld

12 600 -192 408 32616 800 -256 544 43520 1000 -320 680 54425 1250 -400 850 680


40

900

7355dia(125)

Support Face

PLAN

40

40350

(Bea

m W

idth

)

100 (min = 4dia)

1

12

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 0.3D1350

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*


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 @300CL 32.5 (c)

Education

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