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5 CHAPTER 1 INTRODUCTION AND RESEARCH APPROACH The goal of this project was to develop proposed simpli- The concrete contribution is controlled by the value of the fied shear design provisions for the AASHTO LRFD Bridge coefficient as follows:. Design Specifications that would overcome perceived diffi- culties with using the current shear design provisions, which Vc = 0.0316 fcbv d v where f c is in ksi units (Eq. 2) are the provisions of the Sectional Design Model (A5.8.3). This Sectional Design Model constitutes the general shear The coefficient of 0.0316 is 1 / 1000 and is used to con- design requirements in the first three editions of the AASHTO vert the relationship for Vc from psi to ksi units. LRFD Bridge Design Specifications (1, 7, and 17). A variable angle truss model is used to calculate the con- Section 1.1 describes the problem that led to this project tribution of the shear reinforcement. See Equation 3 where and begins with a summary of the LRFD Sectional Design the angle of the field of diagonal compression, , is used in Model (A5.8.3), followed by a brief description of the calculating how many stirrups, [dvcot()/s], are included in basis of this model, and a discussion of the differences the transverse tie of the idealized truss. between the AASHTO LRFD and Standard Specifications (AASHTO, 2002) shear design provisions. Section 1.2 Av f y d v cot() summarizes the information that was available to develop Vs = s (Eq. 3) the proposed simplified provisions. This information con- sists of an overview of what is known about the mecha- where dv 0.9d or 0.72h, whichever is greater. (Eq. 4) nisms of shear resistance, a summary of code provisions, and descriptions of available experimental test data and The values for and are obtained from Table 1 for mem- analysis methods for shear. Section 1.3 defines project bers that contain at least the minimum required amount of objectives, the approach used for meeting these objectives, shear reinforcement (See Equation 5) and from Table 2 for and project tasks. members that contain less than that amount. bv s 1.1 THE AASHTO LRFD SHEAR Av,min 0.0316 fc where f c and fy are in ksi units (Eq. 5) DESIGN SPECIFICATIONS fy 1.1.1 Summary of the LRFD Sectional To obtain values for and from Table 1 (Av < Av,min), the Design Model (S5.8.3) designer selects the row in which to enter the table from the shear design stress ratio (v/f c) and the column by the longi- The AASHTO LRFD Section Design Model for Shear tudinal strain x at mid-depth, which may be taken as one-half (A5.8.3) is a hand-based shear design procedure derived of the strain in the longitudinal tension reinforcement, t. from the Modified Compression Field Theory (MCFT). This strain is equal to the force in the longitudinal tension Prior approaches focused on expressions for shear strength reinforcement divided by the axial stiffness of the tension that were then modified for the effect of other forces. This reinforcement. As shown in Equation 6 and illustrated in is a comprehensive design approach for structural concrete Figure 1, the effects of all demands on the longitudinal rein- members in which the combined actions of axial load, forcement are taken into account: flexure, and prestressing are taken into account when com- pleting the shear design of any section of any member. t Mu / d v + 0.5 N u + 0.5(Vu - Vp ) cot() - Aps f po In this approach, the nominal shear capacity is taken as a x = = sum of a concrete component, a shear reinforcement com- 2 2( Es As + E p Aps ) ponent, and the vertical (or transverse) component of the (Eq. 6) prestressing: Equation 6 assumes that the member is cracked and, there- Vn = Vc + Vs + Vp (Eq. 1) fore, only the axial stiffness of the reinforcement need be

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6 TABLE 1 Values of and for members with at least minimum shear reinforcement considered when evaluating t and x. If x is negative, then where ag is the maximum aggregate size in inches and taken the member is uncracked and the axial stiffness of the equal to 0 when f c 10 ksi. uncracked concrete needs to be considered per Equation 7. Table 2 shows that as sxe and x increase, the value of decreases and increases. The result is that, as the member becomes deeper and the value of the moment increases, t Mu / d v + 0.5 N u + 0.5(Vu - Vp ) cot() - Aps f po x = = the contributions of the concrete and shear reinforcement 2 2( Es As + E p Aps + Act Ec ) decrease. (Eq. 7) The LRFD Sectional Design Model introduced a new requirement into shear design provisions--the direct consid- where Act is the area of the concrete beneath mid-depth. eration of shear in determining the required capacity of the Alternatively, the designer can conservatively take x = 0 longitudinal reinforcement at any point along the length of if Equation 6 yields a negative value. the member (see Equation 9). Table 1 shows that as the longitudinal strain becomes larger, the values for decrease and the values for increase. Tmin 0.5 N u + 0.5Vu cot + Mu dv - Aps f ps (Eq. 9) This means that as the moment and longitudinal strain increase, both the magnitude of the concrete and shear rein- In the end regions of prestressed concrete members, the forcement contributions to shear resistance decrease. development length of the strands at the location of the first To obtain values for and when Av < Av,min, Table 2 is diagonal crack must be taken into consideration when satis- used. As for members containing at least minimum shear fying the requirements of Equation 9. reinforcement, the column by which the designer enters In the design of a member by the LRFD Sectional Design Table 2 is based on the value of the longitudinal strain at mid- Model, the member can be considered to be divided into depth, x. To determine the row, the spacing of the layers of design spans of length dvcot() as shown in Figure 3. Each crack control reinforcement is used, sxe (see Equation 8 and design span can be designed for the shear force midway Figure 2). along the length of the span. If the load is applied to the top of the member, then a staggered shear design concept may be 1.38sx used in which each design span is designed for the lowest sxe = 0.63 + ag (Eq. 8) value of shear occurring within the design span.

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7 TABLE 2 Values of and for members with less than minimum shear reinforcement The Sectional Design Model was developed for regions in designed by the Sectional Design Model for the shear force which engineering beam theory applies and there is a uni- at dvcot()/2 from the support. form flow of the diagonal compressive stresses. However, Figure 4 is a flowchart of the entire procedure for use of the LRFD specifications also permit the end region of mem- the LRFD Sectional Design Model. To further illustrate this bers (the distance between the support and dvcot()/2 from procedure, a brief example is given for the design of a sec- the support) that are subject to a complex state of stress to be tion of the 72-inch-deep bulb-tee girder in Figure 5. (This Figure 1. Effects of axial load, moment, shear, and prestressing on longitudinal strain in non-prestressed member.

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8 h/2 bw sz = dv sz x Act Flexural h/2 tension As Area 0.003bv sz side Figure 2. Evaluation of crack spacing parameter Sx. example was extracted from a design of a 120-foot single- 2. Assume x as -0.10 10 -3 x -0.05 10 -3, then span AASHTO-PCI bulb-tee beam bridge with no skew. The obtain example briefly illustrates the shear design procedure in = 22.8 and = 2.94 from Table 1 (S5.8.3.4-1). LRFD specifications. The critical section is taken at 0.06L from centerline of a support.) 3. Compute x Vu = 316.2 kips, Mu = 2134.0 ft-kips, N u = 0 kips, , Mu d v + 0.5 N u + 0.5(Vu - Vp ) cot - Aps f po x = Vp = 23.4 kips 2( Es As + E p Aps ) As = 0, Aps = 5.508 in 2 , f po = 189.0 ksi, fc = 6.5 ksi, = -1.091 10 -3 0.002 E p = 28500 ks Given that x is negative, recalculate bv = 6 in, dv = 73.14 in, = 0.9 Mu d v + 0.5 N u + 0.5(Vu - Vp ) cot - Aps f po x = 2( Ec Ac + Es As + E p Aps ) 1. Compute shear stress ratio vu fc = -0.080 10 -3 Vu - Vp vu = = 0.7473 ksi bv d v if x satisfies the assumed range, then = 22.8 and = vu fc = 0.115 2.94 are O.K. Design Section Design Section Design Section Design Section 0.5dvcot 0.5dvcot 0.5dvcot2 0.5dvcot2 dvcot dvcot2 dvcot dvcot2 Design Span Design Span Design Span Design Span Vu Vu Vr Vr Shea r Sh ea r Location Location Figure 3. Design regions and shear demand using the sectional design model.

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9 Start Assume value of x and take and from corresponding cell of Determine b v and dv Table 1. Eq. 5.8.2.9 Can longitudinal Yes Calculate x reinforcement resist End Calculate Vp required tension? Eq. 5.8.3.4.2-1 Eq.5.8.3.5 Calculate shear stress ratio v/fc, Eq. 5.8.2.9-1 Is No No calculated x less than assumed If the section is within value? Can you the transfer length of use excess shear any strands, calculate Yes capacity to reduce the No Provide additional the average effective longitudinal steel longitudinal value of fpo Is requirements in reinforcement assumed x too Eq.5.8.3.5-1? Yes conservative? If the section is within ( too high?) the development length Yes of any reinforcing bars, No calculate the effective Choose values of value of As Determine transverse and corresponding reinforcement to to large x , Table 1 ensure V < Vn Eq. 5.8.3.3 Figure 4. Flowchart for LRFD design procedure. 4. Determine shear reinforcement Av fy dv (cot + cot )sin From Vs = , and set fy = 60.0 s Vc = 0.0316 fcbv dv = 103.9 kips, ksi, then Av Vs Vu = = 0.021 in 2 /in Vs = - Vc - Vp = 224.0 kips s fy dv cot Av Use #4 bar double legs @12 in., = 0.033 in 2 /i s /in > 0.021 in 2 /in Av fy dv (cot + cot )sin This provides Vs = = 344.6 kips s 5. Compute maximum limit check: Vc + Vs 0.25 fc bv dv Vc + Vs = 448.5 kips 0.25 fc bv dv = 713.1 kips, O.K. 6. Compute longitudinal reinforcement check at the end of beam Mu N V As fy + Aps f ps + 0.5 u + ( u - 0.5 Vs - Vp ) cot dv Mu N V Figure 5. Design example + 0.5 u + u - 0.5 Vs - Vp cot = 456.4 ki ips dv implementing the LRFD sectional design model. As fy + Aps f ps = 460.1 kips 456.4 kips, O.K.