1. COMPRESSION FIELD THEORY FOR SHEAR STRENGTH IN CONCRETE

2. Consider a truss under load
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.

3. The shear force is carried by the vertical component of the diagonal force and the vertical truss members.

4. There is a theory that a concrete beam can be treated as a truss
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.

5. Consider a cracked concrete beam:

6. The “truss members” can be shown
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. 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. 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’.

9. 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.

10. 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 )

11. 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.

12. 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

14. 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.

15. 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”.

16. 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.

17. Why isn’t the crushing strength fc’
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.

18. 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.

19. 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.

20. 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

21. 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.

22. 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 .

23. 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.

24. 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).

25. 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.

27. 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

28. 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.

29. 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.

30. 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.

31. The denominator is the stiffness of the reinforcing steel
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:
Ac is the area of concrete on the tension HALF of the section.

33. 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 < 0.25fc’ bv dv + Vp
Then Vu <  Vn

34. 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.

35. If the section is not cracked:

36. 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 bvsx
ag = maximum aggregate size – inch.

37. 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 < 0.25fc’ bv dv + Vp
Then Vu <  Vn

38. Minimum transverse reinforcing (stirrups)
Needed if Vu > 0.5(Vc + Vp):
smax:
If v < fc’
smax = 0.8 dv < 24”
If v > fc’
smax = 0.4 dv < 12”

39. 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.

40. 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 > 0.5(Vc – Vp)
For reinforced concrete members less than 16” deep,  and  may be taken as 2 and 45o, respectively.