12745475 the flexural and shear design of deep beam

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10.1 INTRODUCTION Deep beam is a beam having large depth/thickness ratio and shear span depth ratio less than 2.5

for concentrated load and less than 5.0 for distributed load. Because the geometry of deep beams, they

behavior is different with slender beam or intermediate beam.

The structural element that might be classified as deep beam are :

Transfer Girder, is a girder that carry all the vertical load without any vertical element below the

girder.

Pile Cap, is a structural element that connect the vertical element with the deep foundation such

as bored pile.

Vertical Wall, wall slab under vertical load can be designed as deep beam.

10.2 BEHAVIOR OF DEEP BEAM The followings are the major different of deep beam element compared wth ordinary beam based on

the design assumption, as follows :

Two-Dimensional Action, because of the dimension of deep beam they behave as two-

dimensional action rather than one-dimensional action.

Plane Section Do Not Remain Plane, the assumption of plane section remain plane cannot be

used in the deep beam design. The strain distribution is not longer linear.

Shear Deformation, the shear deformation cannot be neglected as in the ordinary beam. The

stress distribution is not linear even in the elastic stage. At the ultimate limit state the shape of

concrete compressive stress block is not parabolic shape again.

The followings are the major behavior of deep beam element, as follows :

Cracking of deep beam will occur at c31 'f or c2

1 'f

The distribution of tensile stress at bottom fiber is constant over the span. In other word the

value of tensile stress at bottom fiber at support and at mid span is only little different, for

this reason in deep beam the tension reinforcement must be extend to the end of support

although that region is small bending moment region (in ordinary beam we can cut off the tension

reinforcement and not all of the tension reinforcement in mid span is extended to the end of

support, practically only two for anchor the stirrups.

The maximum tensile stress at the bottom fiber is far exceed the magnitude of compressive

stress.

CHAPTER

10 THE FLEXURAL AND SHEAR DESIGN OF DEEP BEAM


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SUPPORT STRESSMID SPAN SUPPORT

FIGURE 10.1 STRESS DISTRIBUTION OF DEEP BEAM

The cracks is vertical follows the direction of compression trajectory, in deep beam we must

provide both vertical stirrups and horizontal stirrups.

COMPRESSIVE ARC ACTION

FIGURE 10.2 CRACKS OF DEEP BEAM

10.3 FLEXURAL DESIGN OF DEEP BEAM 10.3.1 GENERAL

The flexural design for deep beam is not described in the ACI code, the method explained in this

section is from Euro – International Concrete Committee (CEB).


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10.3.2 CEB DESIGN OF DEEP BEAM

The flexural design procedure is for simply supported beams and for continuous beams.

TABLE 10.1 FLEXURAL DESIGN OF DEEP BEAM

TYPE SIMPLY SUPPORTED BEAMS CONTINUOUS BEAMS

Moment

Strength ( )jdfAM ysn = ( )jdfAM ysn =

Positive

Reinforcement ( )jdfMAy

us φ

=+

+ ( )jdfMAy

us φ

=+

+

Negative

Reinforcement – ( )jdf

MAy

us φ

=−

−

As

Minimum db

f4.1db

f4'f

A wy

wy

cmins ≥= db

f4'f

dbf4.1A w

y

cw

ymins ≥=

Lever

Arm

( ) 2hL1h2L2.0jd <≤⇒+=

1hLL6.0jd <⇒=

( ) 5.2hL1h5.1L2.0jd ≤≤⇒+=

1hLL5.0jd <⇒=

Positive

Reinforcement

Distribution

h20.0L05.0h25.0y <−= h20.0L05.0h25.0y <−=

Negative

Reinforcement

Distribution

– s1s A1

hL5.0A ⎟

⎠

⎞⎜⎝

⎛−=−

1ss2s AAA −=−

where :

jd = lever arm

Mu+ = positive ultimate flexure moment

Mu- = negative ultimate flexure moment

As+ = positive reinforcements area

As- = negative reinforcements area

h = beam depth

f’c = concrete cylinder strength (MPa)

fy = yield strength of reinforcements (MPa)

Where L is taken the minimum of effective span measured center to center of supports or 1.15 Ln.

In simply supported beams, the positive tension reinforcement is distributed in the lower of beam

section along the distance :

h20.0L05.0h25.0y <−= [10.1]


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POSITIVE REINFORCEMENTS

yh

LLn

FIGURE 9.9 DISTRIBUTION OF POSITIVE REINFORCEMENTS IN SIMPLY SUPPORTED BEAMS

In continuous beams the distribution of positive reinforcements is similar as in the simply supported

beam, the difference is the distribution of negative reinforcements.

hh3

0.2h

0.6h

As1As2

FIGURE 9.10 DISTRIBUTION OF NEGATIVE REINFORCEMENTS IN CONTINUOUS BEAMS

As1 is distributed along height h1=0.2h and As2 is distributed along h2=0.6h.

Reinforcements in zone h3 are come from the tension reinforcements that continued from the mid

span to the support section.

10.3.3 STEP – BY – STEP PROCEDURE The followings are the step – by – step procedure used in the flexural design for deep beam, as follows

:

Classified the structure as simply supported beam or continuous beam.

Calculate the approximate lever arm jd.


Lever

Arm

( ) 2hL1h2L2.0jd <≤⇒+=

1hLL6.0jd <⇒=

( ) 5.2hL1h5.1L2.0jd ≤≤⇒+=

1hLL5.0jd <⇒=


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Calculate the required positive or negative reinforcement As+, As

-.


Positive

Reinforcement ( )jdfMAy

us φ

=+

+ ( )jdfMAy

us φ

=+

+

Negative

Reinforcement – ( )jdf

MAy

us φ

=−

−

Check the required steel bars area with minimum steel bars area Asmin.


As

Minimum db

f4.1db

f4'f

A wy

wy

cmins ≥= db

f4'f

dbf4.1A w

y

cw

ymins ≥=

Choose the number of bars and the reinforcement is distributed as follows :


Positive

Reinforcement h20.0L05.0h25.0y <−= h20.0L05.0h25.0y <−=

Negative

Reinforcement –

s1s A1hL5.0A ⎟

⎠

⎞⎜⎝

⎛−=−

1ss2s AAA −=−

10.4 SHEAR DESIGN OF DEEP BEAM

10.4.1 GENERAL

The shear design of deep beam is similar as shear design of ordinary beam, the difference is only at

the concrete shear strength, limitation of ultimate shear force and horizontal and vertical

stirrups distribution.

10.4.2 BASIC DESIGN EQUATION

According to ACI code the design of deep beam due to shear force must follows the following condition

:

un VV ≥φ [10.2]

where :

Vn = nominal shear strength

nVφ = design shear strength

φ = strength reduction factor (0.85)

Vu = ultimate shear force, factored shear force


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As shear design of ordinary beam, the shear force is resisted by the concrete component and by the

shear reinforcement component, as follows :

scn VVV += [10.3]

where :

Vn = nominal shear strength

Vc = concrete shear strength without shear reinforcement

Vs = shear reinforcement (stirrup) shear strength

10.4.3 CONCRETE SHEAR STRENGTH

The concrete shear strength of deep beam is taken as :

dbM

dV120'f71

dVM5.25.3V w

u

uwc

u

uc ⎟⎟

⎠

⎞⎜⎜⎝

⎛ρ+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

50.2dV

M5.25.30.1u

u ≤⎟⎟⎠

⎞⎜⎜⎝

⎛−≤

[10.4]

where :

Vc = concrete shear strength (N)

Mu = ultimate flexure moment (Nmm)

Vu = ultimate shear force (N)

f’c = concrete cylinder strength (MPa)

d = effective depth

bw = width of beam web

ρw = longitudinal reinforcement ratio

Or the concrete shear strength can be determined as :

db'f61V wcc =

[10.5]

The maximum limit of concrete shear strength is :

db'f21V wcmaxc =−

[10.6]

The section must be enlarged if the ultimate shear force is not follows the condition below :

⎟⎠

⎞⎜⎝

⎛φ≤ db'f32V wcu

for 0.2d

Ln <

[10.7]

Or


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⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ +φ≤ db'fd

L10181V wc

nu

for 0.5d

L0.2 n ≤≤

[10.8]

10.4.4 STIRRUP SHEAR STRENGTH

The shear reinforcements must be provided in the deep beams follows the condition below :

cu VV φ≤ [10.9]

The strength of horizontal and vertical shear reinforcements is :

df12

dL11

sA

12d

L1

sAV y

n

h

vh

n

v

vs

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛−+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛+=

[10.10]

where :

Vs = horizontal and vertical stirrups shear strength (N)

Av = area of vertical stirrups

sv = spacing of vertical stirrups

Ln = clear distance of beam

d = effective depth

Avh = area of horizontal stirrups

sv = spacing of horizontal stirrups

fy = yield strength of stirrups

10.4.5 LIMITS OF SHEAR REINFORCEMENT

The minimum shear reinforcement area is :

( )vminv bs0015.0A =−

( )hminvh bs0025.0A =−

[10.11]

where :

Av-min = minimum vertical stirrups

Avh-min = minimum horizontal stirrups

b = width of beam

sv = spacing of vertical stirrups

sh = spacing of horizontal stirrups

The maximum spacing of shear reinforcement is :


10 - 8

TABLE 10.2 MAXIMUM SPACING OF SHEAR REINFORCEMENT

VERTICAL STIRRUPS HORIZONTAL STIRRUPS

5dsv ≤

mm500sv ≤

3dsh ≤

mm500sh ≤

10.4.6 CRITICAL SECTION IN DEEP BEAM

The critical section to determines the ultimate shear force in the deep beam is :

TABLE 10.3 CRITICAL SECTION OF DEEP BEAM DUE TO SHEAR

UNIFORM LOAD CONCENTRATED LOAD

( )nL15.0x = ( )a50.0x =

10.4.7 STEP – BY – STEP PROCEDURE

The followings are the step – by – step procedure used in the shear design for deep beam, as follows :

Determine the critical section to calculate the ultimate shear force Vu.

UNIFORM LOAD CONCENTRATED LOAD

( )nL15.0x = ( )a50.0x =

Check the ultimate shear force, enlarge the section if the condition is not achieved.

⎟⎠

⎞⎜⎝

⎛φ≤ db'f32V wcu for 0.2

dLn <

⎭⎬⎫

⎩⎨⎧

⎟⎠

⎞⎜⎝

⎛ +φ≤ db'fd

L10181V wc

nu for 0.5

dL0.2 n ≤≤

Calculate the concrete shear strength Vc

dbM

dV120'f71

dVM5.25.3V w

u

uwc

u

uc ⎟⎟

⎠

⎞⎜⎜⎝

⎛ρ+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

50.2dV

M5.25.30.1u

u ≤⎟⎟⎠

⎞⎜⎜⎝

⎛−≤

If cu V5.0V φ< then no shear reinforcements needed, but for practical reason provide minimum

shear reinforcement.

( )hminvh bs0025.0A =−

( )vminv bs0015.0A =−

If cu VV φ> then provide the shear reinforcements.


10 - 9

Calculate the ultimate shear force carried by the stirrups Vs.

cu

s VVV −φ

=

Choose the vertical and horizontal stirrups until the condition achieved.

df12

dL11

sA

12d

L1

sAV y

n

h

vh

n

v

vs

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛−+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛+=

Check the spacing of shear reinforcement sv and sh.

VERTICAL STIRRUPS HORIZONTAL STIRRUPS

5dsv ≤

mm500sv ≤

3dsh ≤

mm500sh ≤

If necessary check the chosen shear reinforcements for the basic design equation for shear

design.

scn VVV +=

df12

dL11

sA

12d

L1

sAV y

n

h

vh

n

v

vs

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛−+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛+=

The design procedure above is repeats until the basic design equation for shear design is

achieved.

10.5 APPLICATIONS 10.5.1 APPLICATION 01 – FLEXURAL DESIGN OF SIMPLY SUPPORTED DEEP BEAM

500

275

50470

PROBLEM

Design the flexural reinforcement of simply supported deep beam above.


10 - 10

MATERIAL Concrete strength = K – 300

Steel grade = Grade 400

Concrete cylinder strength = 9.243083.0'f c =×= MPa

85.01 =β

DIMENSION

b = 500 mm

h = 2750 mm

Concrete cover = 50 mm

d = 2700 mm

DESIGN FORCE

2625056000814.1qL

814.1Mu 22 =⎟

⎠

⎞⎜⎝

⎛ ××=⎟⎠

⎞⎜⎝

⎛= kgm

DEEP BEAM CHECKING

74.127004700

dLn == 0.574.10.1 ≤≤ Deep Beam Action

LEVER ARM

( ) ( )( ) 21002750250002.0h2L2.0jd =+=+= mm

POSITIVE REINFORCEMENT

262500000Mu = Nmm

( ) ( ) 34821004009.0

262500000jdf

MAy

us =

×=

φ=

++ mm2

421127005004004

9.24dbf4'f

A wy

cmins =×

×== mm2

47252700500400

4.1dbf4.1A wy

mins =×== mm2

4725As = mm2

Use 10D25, 4906254110D

4110A 22

s =⎟⎠

⎞⎜⎝

⎛ ×π=⎟⎠

⎞⎜⎝

⎛ π= mm2


10 - 11

DISTRIBUTION OF POSITIVE REINFORCEMENT

275

50

43.75

5D25

( ) ( ) ( ) 550275020.0h20.05.437500005.0275025.0L05.0h25.0y ==<=−=−=

5.437y = mm

The longitudinal positive reinforcement must be distributed at the lower base of the beam with a

distance 437.5 mm from the bottom fiber.

We place 5D25 at each face of the section.

10.5.2 APPLICATION 02 – SHEAR DESIGN OF SIMPLY SUPPORTED DEEP BEAM

500

1077

015

000

d=70.5

1077

0 DESIGN SHEAR FORCEDIAGRAM

DIAGRAMSHEAR FORCE

1500

0

275

50470

PROBLEM

Design the web reinforcement of simply supported deep beam above.

MATERIAL Concrete strength = K – 300

Steel grade = Grade 240

Concrete cylinder strength = 9.243083.0'f c =×= MPa

85.01 =β


10 - 12

DIMENSION

b = 500 mm

h = 2750 mm

Concrete cover = 50 mm

d = 2700 mm

DESIGN FORCE

705470015.0L15.0x n =×== mm

( ) 15078107704.1Vu == kg

150780Vu = N

LIMITATION CHECKING

381734227005009.243285.0db'f

32150780V wcu =⎟

⎠

⎞⎜⎝

⎛ ××=⎟⎠

⎞⎜⎝

⎛φ≤= N

The section is not enlarged.

CONCRETE SHEAR STRENGTH

88.12700150780

2625000005.25.3dV

M5.25.3u

u =⎟⎠

⎞⎜⎝

⎛×

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−

0036.02700500

4906db

A

w

sw =

×==ρ

( ) N20521432700500262500000

27001507800036.01209.247188.1V

dbM

dV120'f71

dVM5.25.3V

c

wu

uwc

u

uc

=⎭⎬⎫

⎩⎨⎧

×⎟⎠

⎞⎜⎝

⎛ ××+=

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

1744322205214385.0Vc =×=φ N

87216117443225.0V5.0 c =×=φ N

DESIGN OF STIRRUPS

872161V5.0150780V cu =φ<= Provide minimum web reinforcement

For horizontal and vertical stirrups we choose 2 legs φ10.

15710412

412A 22

v =⎟⎠

⎞⎜⎝

⎛ ×π=⎟⎠

⎞⎜⎝

⎛ πφ= mm2

HORIZONTAL STIRRUPS VERTICAL STIRRUPS

sh Avh-min sv Av-min

125 ( )hvh bs0025.0A =

( )1255000025.0Avh ×=

156Avh =

200 ( )vv bs0015.0A =

( )2005000015.0Av ×=

150Av =


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Ø10-125

Ø10-200

12745475 the flexural and shear design of deep beam

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