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33 Theta However x shall not be taken as less than -0.2 10-3. f pc cot = 1.0 + 3 1.8 For members having less than minimum shear reinforce- fc ment, as required by Equation 32, the equivalent crack spac- when Vcw < Vct (where stress is in ksi units) (Eq. 24) ing parameter, sxe, is calculated as: which is equivalent to: 1.38sx f pc sxe = (in. units) (Eq. 30) cot = 1.0 + 0.095 1.8 0.63 + ag fc (where stress is in psi units) (Eq. 25) where ag is the maximum aggregate size (in.). Then, the fac- tor accounting for the shear resistance of cracked concrete, This expression was selected so that cot() was equal to 1.0 , can be computed from: ( = 45 degrees) when fpc = 0 (i.e. non-prestressed member). The slope of the influence of fpc on provides a good 4.8 51 correlation with test data. = (in. units) (Eq. 31) (1 + 1500 x )(39 + sxe ) The complete design procedure is shown in Figure 21. The minimum area of shear reinforcement is: 3.2 CHANGE PROPOSAL 2: MODIFICATION OF LRFD SECTIONAL DESIGN MODEL (S5.8.3) bv s Av ,min = fc (in., psi) (Eq. 32) fy The shear design provisions in the 1994 Canadian Stan- dards Association code for the Design of Concrete Structures It should be noted that minimum shear reinforcement is (6) were essentially the same as the Sectional Design Model required when the factored shear force exceeds Vc, rather in the first three editions of the LRFD Bridge Design Specifi- than Vc/2 as required by the ACI 318-02 code. Furthermore, cations (1, 7, 17). In order to simplify the CSA shear design the minimum amount of shear reinforcement is greater than provisions, the 2004 code introduced equations for evaluating the minimum amount required by ACI 318-02 and the and that replaced the tables. Furthermore, a new equation AASHTO Standard Specifications. for x was introduced by assuming that was 30 degrees when For members having at least minimum transverse rein- evaluating the influence of shear on the longitudinal strain, x. forcement, the angle of the diagonal compression field, , is Change proposal 2 is the adoption of the CSA relationships calculated as: for , , and x. These provisions are herein referred to as the CSA Method. This method is presented below. = 29 + 7000 x (Eq. 33) Vn = Vc + Vs + Vp 0.25 fc bv d v + Vp (Eq. 26) and the coefficient, , is obtained from Equation 31 with the where: Vc = fcbv d v (in., psi) : concrete equivalent crack spacing parameter, sze, set to 12 inches. contribution (Eq. 27) In the modified LRFD design provisions presented in Appendix F, the contractor proposes that when the member is Av f y d v (cot + cot )sin not continuous, or cast integrally with the support, the end and Vs = : steel s contribution (Eq. 28) region is designed by the strut-and-tie method in LRFD Arti- cle 5.6.3 (7) when the design shear stress exceeds 0.18f c at the As shown in Figure 1, the longitudinal strain, x, is com- first critical section from the support. This is to guard against puted at mid-depth of the cross-section by: a diagonal compression failure that could occur due to fun- neling of the diagonal compression above a simple support. Mu d v + 0.5 N u + Vu - Vp - Aps f po The complete CSA design procedure is presented in x = Figure 22. 2( Es As + E p Aps ) (Eq. 29a) When x is negative, it is taken as either zero or recalcu- 3.3 DISCUSSION OF DESIGN EXAMPLES lated by changing the denominator of Equation 29a such that the equation becomes: To illustrate the use of these two proposed methods in dif- ferent design situations, eight design examples were pre- Mu d v + 0.5 N u + Vu - Vp - Aps f po pared. These examples were selected from existing PCI x = 2( Es As + E p Aps + Ec Act ) (Eq. 29b) examples, suggestions from project panel members, and new examples selected by the contractor. Each example begins where Act is the area of concrete in tension. with the completed flexural design at a section with specified

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34 Figure 21. Flowchart for use of proposed simplified provisions.

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35 Figure 22. Flowchart for shear design in accordance with CSA.