week 9 shear lrfd

8/12/2019 Week 9 Shear LRFD

1/40

COMPRESSION FIELD

THEORY FOR SHEAR

STRENGTH IN CONCRETE


2/40

Consider a truss under load. The moment is

taken by a couple formed by compression and

tensile forces applied at the upper and lowerchords. These forces include the chord forces

and the horizontal component of the diagonal

force.


3/40

The shear force is carried by the vertical

component of the diagonal force and the vertical

truss members.


4/40

There is a theory that a concrete beam can be

treated as a truss. The uncracked block forms

the compression chord and the longitudinalreinforcing steel forms the tensile chord. The

vertical members are the stirrups. The

compression diagonals are compression struts,

assumed to form in the concrete.


5/40

Consider a cracked concrete beam:


6/40

The truss members can be shown. Notice that the

compression struts are formed by the concrete

between the shear cracks. Also notice that the angle

of the struts can vary.


7/40

The shear force, V, can be assumed to be the sum

of two forces, the forces in the stirrups and the

vertical component of the force in the concrete.

This leads to the basic equation:

V = Vc+ Vs


8/40

ASSUMPTIONS ABOUT

SHEAR STRENGTH The beam fails when the concrete in the

struts reaches its crushing strength.

At failure, the beam has shear cracks and

the cracks have opened

This would cause the stirrups to yield.

The compressive strength of concrete betweenthe shear cracks (struts) is less than fc.


9/40

Assume that the

angle of the strut

is and the

distance between

the compressive

and tensile

forces is jd wered is effective

depth and j


10/40

The stirrup contribution is:

Force per stirrup X number of stirrups.

If the stirrups are spaced at s, the number of

stirrups in the length jd/tanis:

(jd/tan)/s

The force per stirrup is Avfyso:

Vs= (Avfyjd) / (s tan) = (Avfyjd) cot/s

(Note that if j = 1 and = 45o, we get the old,

familiar equation: Vs

= (Av

fy

d) / s )


11/40

If a line is cut

perpendicular to

the cracks, ithas a length of

jdcos. It may

cross several

struts. The totalforce in the

struts will be the

concrete stress

times the area.

Fc

= fc

(jd cos) bwwhere fcis the concrete stress and bw web width.


12/40

The force triangle

shows that the forcealong the struts is

V / sin.

Substituting into the previous equation and

assuming V is the shear force carried by the

concrete:

Vc= fc(jd cos) sinbw


13/40

db'f2V

'f4fand

45,1jifthatNote

bsincosjdfV

wcc

cc

o

wcc


14/40

When = 45o, the equations for Vcand Vsbecome

the familiar equations for shear in reinforced

concrete from ACI-318 and the AASHTO StandardSpecifications. It has long been known that the

actual angle varies along the beam and that the

angle can be anywhere from 25 to 65 degrees.

While it is possible to calculate the angle, it is

difficult. In the days before computers or

calculators, it was nearly impossible. Therefore, the

value of 45 degrees was chosen for simplification.

The value of the crushing strength was also chosen

as a simplification.


15/40

In the 1980s, Vecchio and Collins proposed a

method for finding the shear strength of a beam.

This method required the calculation of the actual

angle, , and the crushing strength of the concrete

struts. The crushing strength is a function of the

tensile strain perpendicular to the strut.

The original theory was called Compression Field

Theory. Later the theory was improved to account

for additional mechanisms, such as aggregate

interlock, and was renamed ModifiedCompression Field Theory.


16/40

MODIFIED COMPRESSION FIELD

THEORY

The basis of the Modified Compression Field

Theory (MCFT) is to determine the point at which

the diagonal compressive struts fail and todetermine the angle of the struts. From the

crushing strength and the angle, the contribution

of the concrete, Vc, can be found.


17/40

Why isnt the crushing strength fc ? The value of fc

is for uniaxial load. The concrete fails by cracking

parallel to the load. If a lateral (biaxial) force isapplied, it changes the apparent compressive

strength. Lateral compression holds cracks together

and increases compressive strength. Lateral tension

pulls them apart and decreases the compressivestrength.


18/40

Vecchio and Collins proposed that the

compressive strength of the strut is a function of

both the compressive stress along the strut andthe tensile stress perpendicular to the strut. They

wrote several equations in terms of the applied

average shear stress, v = V/bd, the principal

tensile strain (perpendicular to strut), 1and theangle of the strut, .

To use MCFT, values of 1and are assumed. It

then takes 17 steps and 15 equations to

recalculate 1and . If these are not close to the

assumed values, an iteration is needed.


19/40

MODIFIED COMPRESSION FIELD

THEORY AND THE LRFD CODE

Obviously, no one wanted to use an iterative

procedure involving 17 steps and 15 equations. As

a result the LRFD Code simplified the method to use

a table.

Unfortunately, soon after the 1stEdition came out,

there was controversy as the MCFT often gaveanswers which were different from the Std. Specs.

The 2000 Interim of the 2ndEdition of the LRFD

uses new equations and tables. These are given

here.


20/40

The shear strength of the beam is:

Vn= Vc+ Vs+ Vp

Vc= contribution of the concrete

Vs= contribution of the stirrups

Vp= vertical component of the force in harpedstrands.

Note that there is a limit:

Vn< 0.25fc bvdv+ Vp

bv= web width

dv= effective depth for shear = da/2


21/40

s

cotdfA

V

db'f0316.0V

vyv

s

vvcc

The equation for Vs assumes the stirrups are

perpendicular to the longitudinal tensile

reinforcement.

bvis the the minimum web width within dv

dvis the shear depth = dea/2 > 0.9deor 0.72 h

s = stirrup spacingAv= stirrup area.

The factors and are unknown and must be

determined.


22/40

For sections with at least the minimum amount of

transverse steel (stirrups):

A value of is assumed.

Next, the average shear stress, carried by the

concrete and stirrups, is found:

vv

pu

db

VV

v

The LRFD Tables use v/fc and xto find and .


23/40

002.0AEAE

fAcotV5.0N5.0d

M

pspss

popsuu

v

u

x

x= longitudinal strain at the level of the tensile

reinforcement.

This equation ASSUMES cracked section. Also,the contribution of the harped strand is ignored.

This equation was used for all beams, regardless

of the amount of stirrups.

First change: In the First Edition of the LRFD Code:


24/40

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x

x= longitudinal strain at 0.5dv

This equation ASSUMES cracked section and is

only for beams with at least the minimum amount

of transverse reinforcing (stirrups).

In the 2nd Edition of the LRFD Code, 2000 Interim:


25/40

REALLY IMPORTANT DEFINITIONS:

The flexural tension side of a beam is the h on

the flexural tension side.

In all the equations for shear which require a valueof the area of the longitudinal tensile steel, Asor

Aps, ONLY the steel on the flexural tension side

counts. Tensile steel on the flexural compression

side (the h on the flexural compression side) orcompression steel is NOT counted for shear

strength.


26/40

M


27/40

The first term in the numerator, Mu/ dv, is the

tensile force in the reinforcing steel due to the

moment. The dvis shear depth = da/2

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x

M


28/40

The second term in the numerator, Nu, is any

APPLIED axial force (not prestressing force). It is

assumed that of the axial load is taken by the

steel. If the load is compressive, Nuis negative.

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x


29/40

The third term in the numerator, (VuVp)cot, is

the axial force component of strut force as shown in

the force triangle. Half the force is assumed to betaken by the tensile steel, the other half in the

uncracked block.

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x


30/40

The last term in the numerator, Apsfpocorrects for the

strain in the prestressing steel due to prestressing.The term fpois the locked in stress in the

prestressing steel, usually taken as 0.7fpu.

In the 1stEdition, this was defined as the stress inthe prestressing steel when the stress in the

surrounding concrete is 0 ksi.

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x

M


31/40

The denominator is the stiffness of the reinforcing

steel. Notice that this equation ASSUMES cracking.

If the section doesnt crack (x< 0), the effect of the

uncracked concrete must be considered:

Acis the area of concrete on the tension HALF of the section.

002.0

AEAE2

fAcotVV5.0N5.0d

M

pspss

popspuu

v

u

x

002.0AEAEAE2

fAcotVV5.0N5.0d

M

ccpspss

popspuuv

u

x


32/40


33/40

Once the values of v/fc and xare calculated, use

the table in the LRFD Code to find and . If the

value of is close to the original assumption, use

the given. If not, use the table value of as the

next estimate and repeat the calculations of x.

Iterate. After finding the value of and :

s

cotdfAV

db'f0316.0V

vyv

s

vvcc

Vn= Vc+ Vs+ Vp < 0.25fc bvdv+ Vp

Then Vu< Vn


34/40

If the section does NOT have at least the minimum

required transverse steel (stirrups), two

modifications are made. First, the strain, x

, is the

maximum longitudinal strain in the web. It can be

calculated by:

002.0

AEAE

fAcotVV5.0N5.0dM

pspss

popspuu

v

u

x

As before, the section is assumed to be cracked.


35/40

If the section is not cracked:

002.0AEAEAE2

fAcotVV5.0N5.0d

M

ccpspss

popspuu

v

u

x


36/40

The second modification is that a crack spacing

parameter, sxe, is used in place of v in the table.

.in8063.0a

38.1ss

g

xxe

sx= lesser of dvor the spacing of longitudinal

steel placed in the web to control cracking.

The area must be at least 0.003 bvsx

ag= maximum aggregate sizeinch.


37/40

Once the values of sxeand xare calculated, use

the table in the LRFD Code to find and . If the

value of is close to the original assumption, use

the given. If not, use the table value of as the

next estimate and repeat the calculations of x.

Iterate. After finding the value of and :

s

cotdfAV

db'f0316.0V

vyv

s

vvcc

Vn= Vc+ Vs+ Vp < 0.25fc bvdv+ Vp

Then Vu< Vn


38/40

Minimum transverse reinforcing (stirrups)

Needed if Vu> 0.5(Vc+ Vp):

smax:

If v < 0.125 fc

smax= 0.8 dv< 24

If v > 0.125 fc

smax= 0.4 dv< 12

vv

pu

db

VVv

y

vcv

f

sbfA '0316.0min,


39/40

The critical section for shear is the section near

the support where the shear is highest:

Larger of 0.5 dvcot or dvfrom the face of the

support IF the reaction at the support is

compressive. The values of and dvare found

at the critical section.

If the reaction is NOT compressive, the critical

section is the face of the support.


40/40

Some final notes:

1) The shear must be checked at the criticalsection and then at intervals along the beam,

usually every 0.1L. The values of dv, and

must be calculated at each section.

2) As with all concrete members, minimumstirrups are required when Vu> 0.5(VcVp)

3) For reinforced concrete members less than

16 deep, and may be taken as 2 and 45o,

respectively.

week 9 shear lrfd

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