6. Shear and Diagonal Tension in Beams 4-1 ~ 4-3 20121022

7/23/2019 6. Shear and Diagonal Tension in Beams 4-1 ~ 4-3 20121022

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Design of Reinforced Concrete 101-1

Chapter 4. Shear and Diagonal Tension in Beams

Review

1. Introduction

2. Bending of Homogeneous Beams. e n orce oncre e eam e av or

4. Design of Tension-Reinforced Rectangular Beams

.

6. Rectangular Beams with Tension and Compression

Reinforcement

7. T-Beams


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Review

Assume rebars yielded; Whitney stress block

' s y s yA f A f

1

11

0.85 '

(0.85 ' )( ) ' ( ')

c

n c s

f b

cM f bc d A f d d

0.003 cun

c c


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Review

(III) Cracked section Nonlinear Behavior Strength Behavior

Assumption: rebars yielded; Whitney stress block


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(III) Cracked section Nonlinear Behavior Strength BehaviorReview

ssume re ars y e e ; tney stress oc

: ?Note check if rebars yielded

'' ?

s cu y

c d

c

,

'

if not yielded

c d

'' '

s s s s cuc

c d1

1

.

' ' ' '

s y c s s cuc o ve c

c

c1.

2

n c s s


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Review

Scheme 1


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Review

s y

'0

s y s yA f A fF c

Scheme 2

1.

c

''s u

c d

c

'

' 's s sf E110.85 ' ( ) ' ( ')

2

n c s y

cM f cb d A f d d

1

'0 0.85 ' 's y c s s u

c dF A f f cb A E

c

Solve for c

110.85 ' ( ) ' '( ')

2n c s s

cM f cb d A f d d


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3.6.3 Reinforcement ratios of doubly reinforced beam

Review

max permitted by ACI codemax1.

0.004 u u u

0.004 uc d


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Review

max1.

.

0.004

u u u

u

c dc d

max 10 0.85 ' ' '0.004

uy c s

u

F bdf f db bdf

c u s s1 max max

f ' f ' f '

= 0.85 + ' '+u y y

!! is required by codemax


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Review

cy ensures yielding of the compression steel at failure2. cy

Ifcy , , ,As

10 0.85 ' ' '

ucy y c yu y

F bdf f d b bdf

'

d

dc u1cy

y u y

f ' = 0.85 + 'f -


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3.6.4 Design of Doubly Reinforced Beams

Review

' ' s y s yAssume f f check if adjustment

= =. . .

2

2 2 2 2; ( );2 0.85 ' s ys s y

c

A faA bd M A f d a

f b

. a cu a e 1 res s e y s ,

3. Assume f =f

1 2 uM M

1' M

4. Calculate As=As2+As

( ')yf d d

( ')

;'

s s yA A f a

a c

5. Calculate cy use As

1.

''

c

s s u

c df E

c

6. If


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Review

Strength Analysis Case 1: a < hf

1. f =f This will nearl alwa s be the case because of thelarge compressive concrete area provided by the flange

2. a


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Review

Strength Analysis Case 2: a > hf

1

.

2. 0.85 '( )

'

s y

c w fC f b b h

2

1

. .

0.85 '( )4.

0.85 ' 0.85 '

c w

s y c w f A f f b b hT Ca

b b

1 1 2 2 1 25. ( ); ( );

2 2

c w c w

f

n n n n nh aM C d M C d M M M


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Example:

15cm15cm

12 2' 280 ( 0.85) 4200

c y

kgf kgf f f

cm cm

1. Mn

2. d

sb,

3.


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Example:

1. Mn,assume steel yield

0.85 ' 0.85 280 8.6 4200

c

c c c s y

A

C T

C f A A T A f

2

151.8

151.8

c

c

A cma

A

. .30

( ) 1353416 13.5

2n s y

a cm cmb

aM A f d kgf cm ton m

1

0.002

5.065.95 0.0

0.85

s y

s

check

ac cm

03 0.017 !d c

okc

13.5n

so M ton m


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Example:0.003cu

2. Asb1

.0.003 0.002

0.85 24 20.4

0.85 ' 15 7.5 0.85 ' 2 7.5 20.4 99603

cu y

a c cm

C C C k

299603 23.7

c c

sb y

sb

T A f C

CA cm

y


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Example:

3. Mn-Asb

1 2

7.5 20.4( ) ( )

2 2

7.5 20.4

nM C d C d

0.85 '(15 7.5)(40 ) 0.85 '(2 7.5 20.4)(40 )

2 23140868 31.3

c cf f

kgf cm ton m


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Chapter Outline

1. Introduction

2. Diagonal Tension in Homogeneous Elastic Beams

3. Reinforced Concrete Beams without Shear

4. Reinforced Concrete Beams with Web Reinforcement

6. Effect of Axial Forces

7. Truss Model8. Strut and Tie Model

9. Shear Friction


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4.1 Introduction

Brittleness in nature

Shear failure of RC, more properly called diagonal tension

failure, is one example of sudden failure mode.

Strength hierarchyFlexural strength < shear strength

(ductile) (brittle)

Capacity design of shear strength in seismic design RC beams are generally provided with special shear

reinforcement to ensure that flexural failure would occur


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4.1 Introduction


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4.1 Introduction

2 p

MwLV

2

=

L


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4.2 Diagonal Tension in Homogeneous Elastic Beams

VQ

v

I


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Shear in homo eneous rectan ular beams

VQIb

effective adhesiveness3h

delaminated

3

22 ( )12 48 bh

I12

bh

I


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Stress Combination and Stress Tra ectories


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Stress Combination and Stress Tra ectories

Position 1

1 v1

v

t1=v1t2=-v1

2

2 -v1


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Position 2

t1 t2

t1t2

Orientation of principal stress Principal tensile stress

r en a on o crac

2 22 v v

2

1 22 4

t v

2f

f

2

22 2

2 4 f ft v


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4.2 Diagonal Tension in Homogeneous Elastic Beams Stress Combination and Stress Trajectories


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f f C


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D i f R i f d C t 101 1


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Design of Reinforced Concrete 101 1


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4.3 Reinforced Concrete Beams without Shear Reinforcement

1. Idealized uncrackedElastic RC section



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2. Idealized cracked Elastic RC Section

: constant shear because of

1. ,normal stress,shear stress2. Aggregate interlocking

dT vbdx

1

dT

v b dx



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4.3 Reinforced Concrete Beams without Shear Reinforcement Shear Behavior

2. Idealized cracked Elastic RC Section

M T j d( )

. .

dM d T j d

beam in linear elastic range N A position kept the same

1( ) dTdM j d dT and vb dx

1

Vjdv

b dx bjd

For convenience, the ACI adopted as an index of shear intensity.

The simplified expression

v bd



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3. Shear crack zone

vc: s ear res s ance rom compress on

concrete

i

interlockingvd: shear resistance from dowel action



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g



dM Vdx

( )

w x

dM d T jd V

( ) =

x x

dT d jd jd T

3,

a

if d

=beam action + arch action

homogeneous isotropic material within elastic range.



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g


Beam action

where d(T)/dx is the shear flow across any

horizontal plane between the reinforcement

,

Fig. 6-5c. For beam action to exist, this

shear flow must exist.



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Arch action

This occurs if the shear flow cannot be transmitted, because the steel is

unbonded, or if the transfer of shear flow is disrupted by an inclined crack

extending from the load to the reactions.

In such a case, the shear is transferred by arch action rather than beam

action, as illustrated above. In this member, the compression force C in the

inclined strut and the tension force T in the reinforcement are constant over

e eng o e s ear span.



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4.3 Reinforced Concrete Beams without Shear Reinforcement Formation of Diagonal Crack



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Large V, small M web-shear crack

23.5 '( ) 0.93 '( ) crcr c c

V kgfv f psi f

bd cm

Large M Flexural shear crack

21.9 '( ) 0.50 '( ) crcr c cv f psi f

bd cm

Note: same order of tensile strength of concrete'cf



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In between, weighting parameter v/f

1( )v Kbd

2

cM f bkd jd

22 2

c

f Kk j b d bd

11

( )

, ( ) K

Kv VdbdthereforeM K M

2 2bd


Ch t 4 Sh d Di l T i i B


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In between, weighting parameter v/f d/a

1

( ) K M

1

K d

2K a


Ch t 4 Sh d Di l T i i B


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, , ua

arch action vd


Chapter 4 Shear and Diagonal Tension in Beams


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6. Shear and Diagonal Tension in Beams 4-1 ~ 4-3 20121022