COMPRESSION FIELD THEORY FOR SHEAR STRENGTH IN CONCRETE

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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 openedThis 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 jhttps://slideplayer.com/slide/248938/1/images/19/MODIFIED+COMPRESSION+FIELD+THEORY+AND+THE+LRFD+CODE.jpg
20 The shear strength of the beam is: 
Vn = Vc + Vs + VpVc = contribution of the concreteVs = contribution of the stirrupsVp = 
vertical component of the force in harped strands.Note that there is a limit:Vn 0.9de or 0.72 hs = stirrup spacingAv = 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 0.5dvThis 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. 
https://slideplayer.com/slide/248938/1/images/20/The+shear+strength+of+the+beam+is%3A.jpg
https://slideplayer.com/slide/248938/1/images/21/The+equation+for+Vs+assumes+the+stirrups+are+perpendicular+to+the+longitudinal+tensile+reinforcement..jpg
https://slideplayer.com/slide/248938/1/images/22/For+sections+with+at+least+the+minimum+amount+of+transverse+steel+%28stirrups%29%3A.jpg
https://slideplayer.com/slide/248938/1/images/23/First+change%3A+In+the+First+Edition+of+the+LRFD+Code%3A.jpg
https://slideplayer.com/slide/248938/1/images/24/In+the+2nd+Edition+of+the+LRFD+Code%2C+2000+Interim%3A.jpg
https://slideplayer.com/slide/248938/1/images/25/REALLY+IMPORTANT+DEFINITIONS%3A.jpg
https://slideplayer.com/slide/248938/1/images/27/The+first+term+in+the+numerator%2C+Mu+%2F+dv+%2C+is+the+tensile+force+in+the+reinforcing+steel+due+to+the+moment..jpg
https://slideplayer.com/slide/248938/1/images/28/The+second+term+in+the+numerator%2C+Nu%2C+is+any+APPLIED+axial+force+%28not+prestressing+force%29..jpg
https://slideplayer.com/slide/248938/1/images/29/The+third+term+in+the+numerator%2C+%28Vu+%E2%80%93+Vp+%29cot%EF%81%B1%2C+is+the+axial+force+component+of+strut+force+as+shown+in+the+force+triangle..jpg
https://slideplayer.com/slide/248938/1/images/30/The+last+term+in+the+numerator%2C+Apsfpo+corrects+for+the+strain+in+the+prestressing+steel+due+to+prestressing.+The+term+fpo+is+the+locked+in+stress+in+the+prestressing+steel%2C+usually+taken+as+0.7fpu..jpg
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 0.5(Vc + Vp):smax:If v fc’smax = 0.4 dv 0.5(Vc – Vp)For reinforced concrete members less than 16” deep, 
 and  may be taken as 2 and 45o, respectively. 
 
https://slideplayer.com/slide/248938/1/images/31/The+denominator+is+the+stiffness+of+the+reinforcing+steel.jpg
https://slideplayer.com/slide/248938/1/images/33/Once+the+values+of+v%2Ffc%E2%80%99+and+%EF%81%A5x+are+calculated%2C+use+the+table+in+the+LRFD+Code+to+find+%EF%81%B1+and+%EF%81%A2.+If+the+value+of+%EF%81%B1+is+close+to+the+original+assumption%2C+use+the+%EF%81%A2+given.+If+not%2C+use+the+table+value+of+%EF%81%B1+as+the+next+estimate+and+repeat+the+calculations+of+%EF%81%A5x+.+Iterate.+After+finding+the+value+of+%EF%81%A2+and+%EF%81%B1+%3A.jpg
https://slideplayer.com/slide/248938/1/images/34/If+the+section+does+NOT+have+at+least+the+minimum+required+transverse+steel+%28stirrups%29%2C+two+modifications+are+made.+First%2C+the+strain%2C+%EF%81%A5x+%2C+is+the+maximum+longitudinal+strain+in+the+web.+It+can+be+calculated+by%3A.jpg
https://slideplayer.com/slide/248938/1/images/35/If+the+section+is+not+cracked%3A.jpg
https://slideplayer.com/slide/248938/1/images/36/The+second+modification+is+that+a+crack+spacing+parameter%2C+sxe+%2C+is+used+in+place+of+v+in+the+table..jpg
https://slideplayer.com/slide/248938/1/images/37/Once+the+values+of+sxe+and+%EF%81%A5x+are+calculated%2C+use+the+table+in+the+LRFD+Code+to+find+%EF%81%B1+and+%EF%81%A2.+If+the+value+of+%EF%81%B1+is+close+to+the+original+assumption%2C+use+the+%EF%81%A2+given.+If+not%2C+use+the+table+value+of+%EF%81%B1+as+the+next+estimate+and+repeat+the+calculations+of+%EF%81%A5x+.+Iterate.+After+finding+the+value+of+%EF%81%A2+and+%EF%81%B1+%3A.jpghttps://slideplayer.com/slide/248938/1/images/38/Minimum+transverse+reinforcing+%28stirrups%29.jpg
https://slideplayer.com/slide/248938/1/images/39/The+critical+section+for+shear+is+the+section+near+the+support+where+the+shear+is+highest%3A.jpg
https://slideplayer.com/slide/248938/1/images/40/Some+final+notes%3A.jpg