Below are the first 10 and last 10 pages of uncorrected machine-read text (when available) of this chapter, followed by the top 30 algorithmically extracted key phrases from the chapter as a whole.
Intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text on the opening pages of each chapter. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.
Do not use for reproduction, copying, pasting, or reading; exclusively for search engines.
OCR for page 12
12 · Further Iteration Required for Capacity Evaluation In the LRFD Sectional Design Model, x and thus and are functions of Vu. Thus, the shear design force must be known in order to evaluate Vc, Vs, and the nominal shear strength. As a result, the procedure for evaluating capacity is iterative and requires the engineer to guess the capacity, evaluate model parameters and Vn, and then check that the calculated capacity is close to the factored load. · Empirical versus Model-Based Justification The Standard Specifications justify the relationship for Vc by experimental test data (23) which indicates that the mea- sured shear capacity of prestressed and non-prestressed test beams is conservatively predicted by the sum of Vc (lesser of Vci and Vcw) and the contribution of the shear reinforcement, Figure 8. Shear demands on longitudinal reinforcement Vs, as calculated using a 45-degree parallel chord truss at end of prestressed girder. model. The LRFD Sectional Design Model shear provisions are derived from a comprehensive behavioral model (the MCFT); the member being overly reinforced in shear and failing by therefore, the basis of this model is the MCFT. The calculated diagonal crushing of the concrete or another means before capacities by the LRFD Sectional Design Model were illus- yielding of the shear reinforcement. According to the MCFT, trated by experimental test data (24) to provide conservative and based on the results of shear tests on elements (21, 22), such estimates of shear capacity. failure mechanisms do not occur until design shear stresses are · Difference in Shear Reinforcement Requirements and in excess of 0.25 f c. The difference between these limits is Capacity Ratings shown in Figure 9. The LRFD shear design requirements different consider- · LRFD Requires an Iterative Shear Design Procedure ably from those of the Standard Specifications. This leads to The LRFD shear design procedure requires the evaluation significant differences in required amounts of shear rein- of the longitudinal strain at mid-depth, x, in order to obtain forcement and rated capacities of existing structures. Because values for and from Table 1 and Table 2. Because x is a the structure of the design provisions is so different, it cannot function of (see Equations 1-6 and 1-7), the design proce- be readily said when one set of provisions will be more con- dure is iterative. The angle is first assumed and then x is servative than the other. Further, with use of the Standard evaluated for the given value of . The value of is obtained Specifications it is easy to perform independent checking of from Table 1 or Table 2, and then x is checked to confirm designs. However, the opposite is true with use of the LRFD that is not significantly changed by using the new value of . Specifications. If it is, then it may be necessary for a different column to be used for obtaining and . 1.2 INTRODUCTION TO SHEAR BEHAVIOR AND DESIGN PRACTICES This section summarizes the resources considered and used to develop the proposed simplified provisions. This subsection presents the development of U.S. code provisions and compression field approaches for shear design and dis- cusses the factors that influence the primary mechanisms of shear resistance; lists other code provisions warranting consideration; and presents an overview of available experi- mental test data, analysis tools, and design data. 1.2.1 Development of Traditional U.S. Code Provisions for Shear The basic model for how shear is carried in structural concrete is the parallel chord truss model that was first proposed by Ritter Figure 9. Maximum allowable design shear stress. in 1899 (25). In this model, the load is carried in reinforced con-
OCR for page 13
13 crete in the same manner as load flows in a truss with the load When the 45-degree parallel chord truss model was intro- zigzagging its way to the support. The load flows down the duced in the United States in the early 1900s, researchers at the concrete diagonal struts and then is lifted to the compression University of Illinois (26) and the University of Wisconsin (27, chord by transverse tension ties on its way to the support. 28) observed through experimental research that the shear Equilibrating the flow of forces puts tension in the bottom chord capacity of beams was greater than that predicted by this truss and compression in the top chord of the truss. Although the model by nearly a constant amount (see Figure 11). Thus, the model is traditionally shown as one truss with stirrups at a longi- idea of a concrete contribution to shear resistance was intro- tudinal spacing of "d," such as given in Figure 10a, it was cor- duced. This contribution was originally taken as equal to a rectly understood by Ritter that there was a continuous band of shear stress of between 2 and 3 percent of f c multiplied by the diagonal compression carried up and over cracks by a band of shear area (b × d). However, over time that contribution stirrups, Figure 10b. For a 45-degree truss, the capacity provided became linked to the diagonal cracking strength because this by the shear reinforcement is equal to the capacity of an individ- provided a better fit with test data. The most commonly used ual stirrup multiplied by the number of stirrups over the length, relationship in U.S. design practice for the diagonal cracking "d " which is approximately equal to "d/s." See Equation 10. load, and thus the concrete contribution to shear resistance in reinforced concrete members, is given by Equation 11: Av fy d Vs = s (Eq. 10) Vc = 2 fcbv d where f c is in psi units (Eq. 11) Figure 10. Parallel chord truss model. Figure 11. Shear strength of RC beams with shear reinforcement.