CEE 461: Reinforced Concrete Design, slide 8-1

Module 8:

Shear and Diagonal Tensionin Beams

CEE 461: Reinforced Concrete Design, slide 8-2

Shear and Diagonal Tension

Figure 4.1 Shear failure of reinforced concrete beam.

CEE 461: Reinforced Concrete Design, slide 8-3

Shear and Flexural Stress

neutralaxis

h

b shearstress

flexuralstress

Ib

VQv =

I

Myf =

CEE 461: Reinforced Concrete Design, slide 8-4

Principal stress per Mohr’s circle

22

v4

f

2

ft +±=

v

f

2

f

2

f

v

t1 t2

tension compression

Ib

VQv =

I

Myf =

CEE 461: Reinforced Concrete Design, slide 8-5

Plain Concrete Beam

V

M

Ib

VQv =

I

Myf =

1

2

1

3

f

2v

3 vf

CEE 461: Reinforced Concrete Design, slide 8-6

Principal Stresses

Figure 4.3: Stress trajectories in plain concrete beam.

!

t =f

2±

f 2

4+ v 2

CEE 461: Reinforced Concrete Design, slide 8-7

Diagonal Tension Cracking

(b) Flexure-shear cracking

(a) Web-shear cracking

Figure 4.5. Diagonal tension cracking in reinforced concrete beams.

!

vcr =Vcr

bd= 3.5 f 'c

CEE 461: Reinforced Concrete Design, slide 8-8

Concrete Shear Strength

Figure 4.6. Correlation of Eq 4.3a with test results.

!

vcr =Vcr

bd= 2.0 f 'c

!

vcr =Vcr

bd= 1.9 f 'c + 2500

"Vd

M f 'c# 3.5 f 'c

CEE 461: Reinforced Concrete Design, slide 8-9

Internal Shear Forces

Figure 4.10. Redistribution of internal shear forces.

Figure 4.9

Vs

Viy

Vd

Vcz

!

Vint = Vcz + Vd + Viy + Vs

!

Vint

= Vc

+ Vs

CEE 461: Reinforced Concrete Design, slide 8-10

Shear Design

As

d

b

Av

Av

s

!

Vu = " (Vc + Vs )

!

" = 0.75 for shear

!

Vc

bd= 1.9 f 'c + 2500

"Vd

M f'c# 3.5 f 'c

!

Vs = Av fyd

s

!

smax

=d

2

CEE 461: Reinforced Concrete Design, slide 8-11

Shear Demand, VuFigure 4.12: Locations of critical design shear

face ofsupport

face ofsupport

Vu

VuVu

end supported beambeam-column framing

CEE 461: Reinforced Concrete Design, slide 8-12

Shear Demand, VuFigure 4.12: Locations of critical design shear

Vu Vu

load near bottom of beamconcentrated loadwithin d distance

face ofsupport

face ofsupport

CEE 461: Reinforced Concrete Design, slide 8-13

Shear Demand, VuFigure 4.12: Locations of critical design shear

beam supported bygirder of similar depth hanging beam

Vu

CEE 461: Reinforced Concrete Design, slide 8-14

Minimum Web ReinforcementFor regions of light shear force:

!

if Vu " #Vc then no Av required, but....

!

Av min = 0.75 f 'cbws

f y

" 50bws

f y

!

if Vu "#Vc

2 then no Av required

!

Av min

!

Vu

= "Vc

!

Vu

="V

c

2

CEE 461: Reinforced Concrete Design, slide 8-15

22”

16”

wu = 9.4 k/ft

10’

Example

!

"Vc = 0.75(2 f 'c bd ) = 33.4k

!

"Vs

= 43.4k

CEE 461: Reinforced Concrete Design, slide 8-16

ExampleDesign stirrups at “d” out from support:

!

"Vs

= Vu# "V

c= 76.8

k # 33.4k

= 43.4k

!

Av

s=Vs

fyd=

43.4k

0.75

60ksi (22")

= 0.0438"

!

s =0.11

in 2

x 2 legs

0.0438in

= 5.0"!

smax =Av f y

50bw=0.22(60 ,000 )

50(16")= 16.5"

!

smax

=d

2= 11"

Minimum stirrups:

CEE 461: Reinforced Concrete Design, slide 8-17

Example

!

Vc

bd= 1.9 f 'c + 2500

"Vd

M f'c# 3.5 f 'c

!

note M

V varies along span

!

"Vs

= Vu# "V

c= 43.4

k

!

Av

s=Vs

fyd