7 Design of Deep Flexural Member

8/10/2019 7 Design of Deep Flexural Member

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Design of Deep Flexural Member

Revised : 08-Oct-2013

By : LB3


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The code gives two definitions for deep members.For flexure, members with overall depth-to-clear-span ratios

greater than 2/5 for continuous spans (Ln/d > 2/5) or 4/5 for

simple spans (Ln/d > 4/5) are defined as deep(SNI 2847-

2002 Ps.12.7.1).

For shear, a deep member is one with an effective

depth-to-clear-span ratio larger than 1/5 (Ln/d > 1/5) (SNI

2847-2002 Ps.13.8.1).


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Ln

d

Simple Deep Beam

Continous Deep Beam

d

Ln Ln


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Ln

Clear Span (Ln)

Column

Deep Beam


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Deep Beam


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Problems need to be solved in DeepMember :

1. Lateral Buckling !!!

2. Non-Linear distribution of Strain !!!


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Fortunately, most walls andbeams receive lateral support

from supported floor or roof

members, so lateral buckling of

the compression flange is rarelya problem.


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Some form of lateral support isrequired at intervals not

exceeding 50 times the least

width of the compression flange

(SNI 2847-2002 12.4.1), even ifthe member is free-standing .


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Lateral buckling in a verticaldirection, particularly near

concentrated loads and at supports,

can be checked by the moment

magnifier method for columns, or bynumerical or energy methods.

A simplified procedure for wall-like

beams (tilt-up panels) is provided in

Reference. If the height-to-thicknessratio of a member is limited to 25,

buckling should not be a problem.


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The special shear strength provisions of 13.8 areintended to apply mainly to simply supported deep beams.

Tests of continuous deep beams have indicated that the

special shear provisions of 13.8 are inadequate for

continuous members.Therefore, the design of continuous deep flexural

members for shear strength must be based on the regular

beam design procedures of 13.1 through 13.5 with 13.8.5

substituted for 13.1.3, and must also satisfy the provisions of

13.8.4, 13.8.9 and 13.8.10. (Codes = SNI 2847-2002)


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The code requires that nonlinear distribution ofstrain be taken into account in flexural design of deep

members.

The elastic analyses by Dischinger and others (Refs.

19.2-19.4) have shown that the shape of the elastic stresscurve can be quite different from the linear distribution

usually assumed.

At midspan, the neutral axis moves away from the

loaded face of the member as the span-to-depth ratio

decreases (see Fig. 19-3).Over the supports, the resultant elastic tensile forces

can be within a third of the member depth from the top fiber.


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Reference 19.3, however, recommends that tensilereinforcement be distributed throughout the tensile area and

centered at or near the resultant of the tensile forces, so that,

when cracking occurs, there will not be a sudden shift in the

location of the resultant tensile force. Both methods of sizingand placing reinforcement are illustrated in Example 19.1

and it is left to the judgment of the designer to choose the

more appropriate method.


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Development of horizontal tensile reinforcement insingle-span simply-supported deep members requires

special consideration. Since moments increase rapidly from

zero at the face of the support, the reinforcement may not

have sufficient anchorage length to develop the required

moment strength near the support. Tensile bars may be

anchored by development length (if available), standard

hooks, or by special anchorage devices.

Ln

d

Simple Deep Beam

Tensile ReinforcementAnchorage


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The most radical departure from a linear strain andstress distribution is in compression areas at or near

supports of continuous members. Compressive forces may

be confined to the bottom 5 or 10 percent of the member

depth and compressive stresses may be as high as 14 times

those indicated by linear strain and stress distribution. 19.2

In these cases, reinforcing details require special

consideration. If service load compressive stresses approach

about 0.45 fc, it may be necessary to treat the compression

area as an axially loaded member, using laterally tiedreinforcement to carry the compressive forces as the

moment strength is approached.


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The special shear strength provisions for deepflexural members apply only to members having a clear-

span-toeffective-depth ratio ( ln /d) less than 5. Section 13.8

(SNI 2847-2002) deals with deep members loaded on the top

face as shown in Fig. 19-4. Since the principal tensile forces

in deep members are primarily horizontal (vertical

cracking), horizontal shear reinforcement is effective in

resisting the tensile forces. Truss bars are, therefore, not

recommended as shear reinforcement in deep members.


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Different shear design procedures are prescribed forsimply supported and continuous deep flexural members.

Design of simply supported members for shear must be

based on the special provisions of 13.8. Design of continuous

members for shear must be based on the regular beam

design procedures of 13.1 through 13.5 as well as 13.8.4,

13.8.9 and 13.8.10. Also, when loads are applied through the

sides or bottom of the member, simply supported or

continuous, the shear design provisions of 13.1 through 13.5

must be used. (SNI 2847-2002)


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The factored shear force Vu must not exceed the shearstrength Vn = (Vc + Vs ), where Vc is the shear strength

provided by the concrete and Vs is the shear strength

provided by the shear reinforcement, both horizontal and

vertical. Vc may be computed from either the more complex

Eq. (13-29), which takes into account the effects of the tensile

reinforcement and Mu/Vud at the critical section, or may be

determined from the simpler Eq. (13-2) . Equation (13-29) is

illustrated in Fig. 19-5.

)2.13....(6

1' EqdbfV wcc


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For design convenience, the required area of shearreinforcement Av and Avh in terms of the factored shear

force Vu can be computed using Eq. (11-30) as follows:


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The first step in the design is to check if Vu is less thanVc , with Vc equal to 1/6 fc bwd. If the shear strength

provided by the concrete is not adequate to carry the

factored shear force Vu, calculate Vs for minimum shear

reinforcement and add to Vc . Using the minimum shear

reinforcement requirements of 13.8.9 (Av = 0.0015bws) and

13.8.10 (Avh = 0.0025bws2), shear strength Eq. (11-30)

reduces to :


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Substituting Vc from Eq. (13-28) and Vs from above,the shear strength with minimum shear reinforcement

becomes :

If shear strength with minimum shear reinforcement is

still not adequate, the more complex Eq. (13-29) can be used

to calculate a higher concrete shear strength, or additional

shear reinforcement Av and Avh may be added to increasethe shear strength of the section. Shear reinforcement

required at the critical section must be provided throughout

the span in all cases (13.8.11).

12001.0029.0 yw

nc

fbldVVu


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The shear strength Vn = Vc + Vs must not be takengreater than:

2

3

2'

d

LnifdbfV wcn

5210

18

1'

d

lfordbf

d

lV nwc

nn


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A strict reading of 13.8.8 would appear to suggest thatno shear reinforcement is needed in a simply supported

deep beam unless Vu exceeds Vc , which may be as high

as 1/2 fc bwd. However, a deep beam without shear

reinforcement is not recommended. It would be more

appropriate to conform with 13.8.9 and 13.8.10 for the

design of simply supported deep beams. (SNI 2847-2002)


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Sections 14.11.4 and 14.12.4 were added to the 95Code concerning reinforcement detailing of deep flexural

members. The sections require that at interior supports of

deep flexural members, positive moment tension

reinforcement and negative moment tension reinforcement

shall be continuous with that of adjacent spans in order to

develop proper anchorage. (Codes = SNI 2847-2002)

Continous Deep Beam

d

Ln Ln

Positive

Negative

Positive


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Example 1 : Determine the require simple supportedtransfer girder supporting the single column below. Fc =

4000 psi (27.585 MPa_), fy = 60000 psi (413.774 MPa).

400

500

12008750

1800

3600400 400

Vd = 889.84kN

Vl = 1112.30kN


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Step 1 : (SNI 2847-2002 Ps. 13.8.1)Determine if deep beam provisions for shear design apply.

Deep beam provisions apply.

mmd 10751251200

5348.310753600

dln


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Step 2 : (SNI 2847-2002 Ps. 13.8.5)Determine critical section for shear (neglect uniform dead

load since it is small compared to the concentrated loads) :

Step 3 : (SNI 2847-2002 Ps. 13.8.6)

Determine shear strength without shear reinforcement.

1075900180050.050.0 da

kNV

dbfV

c

wcc

89.3521075500585.276

175.0

6

1'


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Step 4 :Determine ultimate shear forces acting on the support :

Shear strength provided by concrete Vc is not adequate to

carry the factored shear force Vu.

Step 5 : (SNI 2847-2002 Ps.13.8.4)

Check maximum shear permitted.

kN

VVV ldu 77.1423

2

3.11126.184.8892.1

2

6.12.1

348.3d

lfor n


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Step 5 : (SNI 2847-2002 Ps.13.8.4)Check maximum shear permitted.

kNV

kNV

V

dbfd

lV

n

n

n

wcn

n

17.1570562.209375.0

562.2093

1075500585.271075

360010

18

1

1018

1'


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Step 6 : (SNI 2847-2002 Ps.13.8.9 & Ps.13.8.10)Determine shear strength with minimum reinforcement.

Substituting minimum Av and Avh into Eq.(11-30).

kNkNVsVc

kNVsVs

fybldVs wn

77.142344.70955.35689.352

55.35612/77.4135003600001.01075029.075.0

12/001.0029.0


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Step 7 : (SNI 2847-2002 Ps.13.8.7)Determine shear strength of Vc using more complex Eq. (11-

29) at critical section.

At critical section :

71205.25.3 'db

M

dV

fdV

M

Vc w

u

u

wcu

u

837.0107577.1423

90077.1423

dV

M

u

u


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Step 7 : (SNI 2847-2002 Ps.13.8.7)Determine shear strength of Vc using more complex Eq. (11-

29) at critical section.

5.2406.1837.05.25.35.25.3 dV

M

u

u

0163.01075500

8750

db

A

w

sw

kN

dbMdVf w

u

uwc

94.817

7

1075500

83.0

0163.0120585.274.1

71204.1 '


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Step 8 : (SNI 2847-2002 Ps.13.8.8)Determined required shear reinforcement.

Use minimum horizontal reinforcement :

12

11

12

1

2

d

l

s

Ad

l

s

A

df

VVn

vh

n

v

y

cu

mmmm

df

VV

y

cu /429.2107574.41375.0

457.61374.1423 2

mmmmd

s

mmmsbA wvh

500333.358

3

1075

3

3.0/3753005000025.00025.0

2

2

2


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Step 8 : (SNI 2847-2002 Ps.13.8.8)Use No.5 (15.7)@300mm (each face), Avh = 400 mm2/(0.3)m

mmmm

s

Avh /333.1300

2002 2

2

mmmmd

s

mmms

A

mmmm

s

A

mmmmsA

v

v

v

5002155

1075

5

09.0/39290357.4

/357.4

/429.212

348.311333.112348.31

2

2

2


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Step 8 : (SNI 2847-2002 Ps.13.8.8)Use No.5 @90mm (each face), Avh = 400 mm2/(0.09)m

Alternatively, decrease the spacing of the horizontal bars to

No.5 @ 200mm. (each face).

mmmm

s

Avh /2200

2002 2

2

mmmms

A

mmmms

A

v

v

/183.3

/429.212

348.311

212

348.31

2

2


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Step 8 : (SNI 2847-2002 Ps.13.8.8)

Use No.5 @120mm (each face), Avh = 400 mm2/(0.12)m

mmmm

d

s

mmms

Av

5002155

1075

5

12.0/382120183.3 2


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Step 9 : (SNI 2847-2002 Ps.13.8.8)Check shear strength provided using No.5 @120mm (each

face) for horizontal and No.5 @200mm (each face) for

vertical shear reinforcement.

kNkN

VuVV

kNVVkNkNV

c

sc

s

77.1423649.1423

649.1423199.81045.613199.810266.108075.0


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Step 10 : (SNI 2847-2002 Ps.13.8.11)Both horizontal and vertical shear reinforcement required at

the critical section must be provided throughout the span.

See reinforcement detail below.

Use No.5 @120mm (each face) for horizontal and No.5

@200mm for vertical shear reinforcement.

Note : The main flexural reinforcement must be anchored to

develop the specified yield strength fy in tension at the faceof the support.


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No.5 (15.7mm D16) @ 200 mm

No.5 (15.7mm D16) @ 120 mm

No.5 (15.7mm D16) @ 200 mm


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For shear design of continuous deep members, thedesign procedure is the same as for ordinary beams. The

maximum factored shear force Vu is calculated at the critical

section defined in 13.8.5. The factored shear force Vu must

not exceed the shear strength provided by the section (Vc

+ Vs), where Vc may be computed from either the more

complex Eq. (13-5), or the simpler Eq. (13-3), Vc = 2 fcbwd.

(SNI 2847-2002)


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Section 13.8.3 also specifies that the design ofcontinuous deep flexural members must also satisfy 11.8.4,

13.8.9 and 13.8.10. Section 13.8.4 sets an upper limit to Vn.

Sections 13.8.9 and 13.8.10 specify minimum vertical and

horizontal shear reinforcement, respectively. (SNI 2847-2002)


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If the shear strength with minimum shearreinforcement is still not adequate, the more complex Eq.

(13-5) can be used to calculate a higher concrete shear

strength, or additional shear reinforcement Av may be

added to increase the shear strength of the section. Using

Eq. (13-15), the required shear reinforcement is :


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The first step in the design is to check if Vu is less thanfVc, with Vc equal to 2 fc bwd. If the shear strength

provided by the concrete is not adequate to carry the

factored shear force Vu, calculate Vs for minimum shear

reinforcement and add to Vc. Using the minimum shear

reinforcement of 13.8.9 (Av = 0.0015bws), the shear strength

Eq. (13-15) reduces to :


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Note that the minimum shear reinforcement of 13.8.9is greater than that required by Eq. (13-13). The shear

strength with minimum shear reinforcement becomes :

If the shear strength with minimum shear

reinforcement is still not adequate, the more complex Eq.

(13-5) can be used to calculate a higher concrete shear

strength, or additional shear reinforcement Av may beadded to increase the shear strength of the section. Using

Eq. (11-15), the required shear reinforcement is :


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Shear reinforcement may be varied along the lengthof span as for ordinary beams; however, a minimum area of

both vertical and horizontal reinforcement, Av andAvh, in

accordance with 13.8.9 and 13.8.10 must be provided

throughout the full span length. Note that the spacing s of the

vertical shear reinforcement Av must not exceed d/5 nor 18

in. (somewhat closer maximum spacing than that permitted

for ordinary beams). Note also that the horizontal shear

reinforcement Avh does not contribute to the shear strength

Vs for continuous deep members.


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As for simply supported deep members, incontinuous deep members, the shear strength Vn must not

be taken greater than :

23

2'

d

Lnifdbf wc

521018

1'

d

l

fordbfd

l nwc

n

7 Design of Deep Flexural Member

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