COMPRESSION FIELD THEORY FOR SHEAR STRENGTH IN CONCRETE

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COMPRESSION FIELD THEORY FOR SHEAR STRENGTH IN
CONCRETE
Week 9
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Consider a truss under load. The moment is taken
by a couple formed by compression and tensile
forces applied at the upper and lower chords.
These forces include the chord forces and the
horizontal component of the diagonal force.
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The shear force is carried by the vertical
component of the diagonal force and the vertical
truss members.
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There is a theory that a concrete beam can be
treated as a truss. The uncracked block forms
the compression chord and the longitudinal
reinforcing steel forms the tensile chord. The
vertical members are the stirrups. The
compression diagonals are compression struts,
assumed to form in the concrete.
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Consider a cracked concrete beam
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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.
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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
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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 between the
  shear cracks (struts) is less than fc.

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Assume that the angle of the strut is ? and the
distance between the compressive and tensile
forces is jd were d is effective depth and jlt1.
Thus the horizontal distance is jd/tan?.
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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/tan? is (jd/tan?)/s The force per stirrup
is Avfy so Vs (Avfy jd) / (s tan?) (Avfy
jd) cot?/s (Note that if j 1 and ? 45o , we
get the old, familiar equation Vs (Avfy d) / s
)
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If a line is cut perpendicular to the cracks, it
has a length of jdcos?. It may cross several
struts. The total force in the struts will be
the concrete stress times the area.
Fc fc (jd cos?) bw where fc is the
concrete stress and bw web width.
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The force triangle shows that the force along 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?) sin? bw
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When ? 45o , the equations for Vc and Vs become
the familiar equations for shear in reinforced
concrete from ACI-318 and the AASHTO Standard
Specifications. 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.
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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
Modified Compression Field Theory.
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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 to
determine the angle of the struts. From the
crushing strength and the angle, the contribution
of the concrete, Vc , can be found.
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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 is applied, it changes
the apparent compressive strength. Lateral
compression holds cracks together and increases
compressive strength. Lateral tension pulls them
apart and decreases the compressive strength.
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Vecchio and Collins proposed that the compressive
strength of the strut is a function of both the
compressive stress along the strut and the
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), ?1 and
the angle of the strut, ?. To use MCFT,
values of ?1 and ? are assumed. It then takes 17
steps and 15 equations to recalculate ?1 and ?.
If these are not close to the assumed values, an
iteration is needed.
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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
1st Edition came out, there was controversy as
the MCFT often gave answers which were different
from the Std. Specs. The 2000 Interim of the 2nd
Edition of the LRFD uses new equations and
tables. These are given here.
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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 harped strands. Note
that there is a limit Vn lt 0.25fc bv dv
Vp bv web width dv effective depth for shear
d a/2
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The equation for Vs assumes the stirrups are
perpendicular to the longitudinal tensile
reinforcement. bv is the the minimum web width
within dv dv is the shear depth de a/2 gt
0.9de or 0.72 h s stirrup spacing Av stirrup
area. The factors ? and ? are unknown and must
be determined.
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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
The LRFD Tables use v/fc and ?x to find ? and ?.
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First change In the First Edition of the LRFD
Code
?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.
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In the 2nd Edition of the LRFD Code, 2000 Interim
?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).
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IMPORTANT DEFINITIONS The flexural tension side
of a beam is the ½ h on the flexural tension
side. In all the equations for shear which
require a value of the area of the longitudinal
tensile steel, As or Aps , ONLY the steel on the
flexural tension side counts. Tensile steel on
the flexural compression side (the ½ h on the
flexural compression side) or compression steel
is NOT counted for shear strength.
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The first term in the numerator, Mu / dv , is the
tensile force in the reinforcing steel due to the
moment. The dv is shear depth d a/2
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The second term in the numerator, Nu, is any
APPLIED axial force (not prestressing force). It
is assumed that ½ of the axial load is taken by
the steel. If the load is compressive, Nu is
negative.
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The third term in the numerator, (Vu Vp )cot?,
is the axial force component of strut force as
shown in the force triangle. Half the force is
assumed to be taken by the tensile steel, the
other half in the uncracked block.
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The last term in the numerator, Apsfpo corrects
for the strain in the prestressing steel due to
prestressing. The term fpo is the locked in
stress in the prestressing steel, usually taken
as 0.7fpu. In the 1st Edition, this was
defined as the stress in the prestressing steel
when the stress in the surrounding concrete is 0
ksi.
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The denominator is the stiffness of the
reinforcing steel. Notice that this equation
ASSUMES cracking. If the section doesnt crack
(?x lt 0), the effect of the uncracked concrete
must be considered
Ac is the area of concrete on the tension HALF of
the section.
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Once the values of v/fc and ?x are 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 ?
Vn Vc Vs Vp lt 0.25fc bv dv Vp Then Vu lt
? Vn
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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
As before, the section is assumed to be cracked.
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If the section is not cracked
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The second modification is that a crack spacing
parameter, sxe , is used in place of v in the
table.
sx lesser of dv or the spacing of longitudinal
steel placed in the web to control cracking. The
area must be at least 0.003 bvsx ag maximum
aggregate size inch.
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Once the values of sxe and ?x are 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 ?
Vn Vc Vs Vp lt 0.25fc bv dv Vp Then Vu lt
? Vn
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Minimum transverse reinforcing (stirrups) Needed
if Vu gt 0.5?(Vc Vp) smax If v lt 0.125
fc smax 0.8 dv lt 24 If v gt 0.125 fc smax
0.4 dv lt 12
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The critical section for shear is the section
near the support where the shear is
highest Larger of 0.5 dv cot ? or dv from the
face of the support IF the reaction at the
support is compressive. The values of ? and dv
are found at the critical section. If the
reaction is NOT compressive, the critical section
is the face of the support.
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• Some final notes
• The shear must be checked at the critical section
  and then at intervals along the beam, usually
  every 0.1L. The values of dv , ? and ? must be
  calculated at each section.
• As with all concrete members, minimum stirrups
  are required when Vu gt 0.5?(Vc Vp)
• For reinforced concrete members less than 16
  deep, ? and ? may be taken as 2 and 45o,
  respectively.

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Interface Shear

• The factored horizontal shear (Loading),
• Vh Vu / dv (LRFD C5.8.4.1-1)
• Vh ? ?Vn and ? 0.9
• Nominal resistance

• Acv area of concrete engaged in shear transfer

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Interface Shear

• Nominal resistance

• Acv area of concrete engaged in shear transfer
• c cohesion factor 0.1 ksi
• m friction factor 1.0
• Avf interface reinforcement
• Pc compressive force on Interface

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Interface Shear

• Limits on Vn
• Vn ? 0.2 fc Acv or 0.8 Acv
• Use the deck concrete strength for fc.

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Interface Shear

• Minimum steel requirement (LRFD 5.8.4.1-4)

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Interface Shear - Example
See PCA Example