COMPRESSION FIELD
THEORY FOR SHEAR
STRENGTH IN CONCRETE
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.
The shear force is carried by the vertical
component of the diagonal force and the vertical
truss members.
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.
Consider a cracked concrete beam:
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.
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
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’.
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 j<1.
Thus the
horizontal
distance is
jd/tan.
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 )
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.
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
Vc  f c  jd cos  sin  b w
Not e t hat if j  1,   45
o
and f c  4 f c '
Vc  2 f c ' b w d
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.
In the 1980’s, 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”.
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.
Why isn’t 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.
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.
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.
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 < 0.25fc’ bv dv + Vp
bv = web width
dv = effective depth for shear = d – a/2
Vc  0.0316
Vs 
fc ' bvd v
A v f y d v cot
s
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 > 0.9de or 0.72 h
s = stirrup spacing
Av = stirrup area.
The factors  and  are unknown and must be
determined.
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:
v
Vu   Vp
 bvd v
The LRFD Tables use v/fc’ and x to find  and .
First change: In the First Edition of the LRFD Code:
Mu
 0.5N u  0.5Vu cot  A ps f po
dv
x 
 0.002
E s A s  E p A ps
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.
In the 2nd Edition of the LRFD Code, 2000 Interim:
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
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).
REALLY 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.
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
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
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
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.
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
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.
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
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.
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps 
The denominator is the stiffness of the reinforcing
steel. Notice that this equation ASSUMES cracking.
If the section doesn’t crack (x < 0), the effect of the
uncracked concrete must be considered:
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps  E c A c 
Ac is the area of concrete on the tension HALF of the section.
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  :
Vc  0.0316
Vs 
fc ' bvd v
A v f y d v cot
s
Vn = Vc + Vs + Vp < 0.25fc’ bv dv + Vp
Then Vu <  Vn
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:
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
E s A s  E p A ps 
As before, the section is assumed to be cracked.
If the section is not cracked:
Mu
 0.5N u  0.5Vu  Vp cot  A ps f po
dv
x 
 0.002
2E s A s  E p A ps  E c A c 
The second modification is that a crack spacing
parameter, sxe , is used in place of v in the table.
s xe
1.38
 sx
 80in.
a g  0.63
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.
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  :
Vc  0.0316
Vs 
fc ' bvd v
A v f y d v cot
s
Vn = Vc + Vs + Vp < 0.25fc’ bv dv + Vp
Then Vu <  Vn
Minimum transverse reinforcing (stirrups)
Needed if Vu > 0.5(Vc + Vp):
smax:
If v < 0.125 fc’
smax = 0.8 dv < 24”
v
Vu   Vp
If v > 0.125 fc’
smax = 0.4 dv < 12”
bv s
Av,min  0.0316 f c '
fy
 bvd v
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.
Some final notes:
1) 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.
2) As with all concrete members, minimum
stirrups are required when Vu > 0.5(Vc – Vp)
3) For reinforced concrete members less than
16” deep,  and  may be taken as 2 and 45o,
respectively.