The Strength of Members Subjected to Shear

The 2004 CSA shear provisions are based on the shear resisting mechanism shown in Figure 4. This free body diagram shows the end portion of a beam, cuts through the longitudinal reinforcement and stirrups, follows the diagonal shear crack and then cuts the top flexural compression region. Flexural moments (Mf) and axial tension (Nf) are resisted by the force couple between the flexural compressive force C and the tension in the reinforcement Fit, here calculated at the crack. Shear forces (Vf) are resisted by three mechanisms. Shear stresses (vc) on the crack surface itself are resisted by aggregate interlock, which is the primary method of strength resistance observed for members without stirrups (Fenwick, 1968; Taylor, 1970; Kani, 1979). Stirrups that cross the diagonal crack provide a steel contribution, Vs, and any vertical component to prestressing forces, Vp, will also resist shear forces. Even from such a simple diagram as that in Figure 4, useful deductions may still be made. In this case, note that the shear on the crack, vc, has a vertical component that resists shear, but also has a horizontal component that must be balanced by additional strain in the longitudinal reinforcement. This is one of the causes of shear-moment interaction in that shear force cause stress in the flexural reinforcement.

If the contribution due to Vp is ignored, the following equations are the basic shear strength terms used in the CSA code for both the general and simplified methods:

V, = Vc + Vs < 0.25^>cf’bwdv,

і— Ay

Vr = фсХр^/f’ ■ bwdv + <fis — fydyCOtO, (1) where Vr is the factored shear strength, bwdv is the shear area taken as the web width multiplied by the shear depth shown in Figure 4 and taken simply as 0.9d. The term X relates to the use of lightweight concrete. The equation is written in metric notation (MPa, mm, Newtons), with фс =

0. 65 and ф5 = 0.85 as the material reduction factors for concrete and steel. The ability of the member to resist aggregate interlock stresses is represented by the variable в, and the angle of the crack, в, indicates how many stirrup legs will cross the crack, and is necessary to determine Vs.

Development of shear provisions can therefore be boiled down to finding a method to quantify the parameters в and в. The 1994 provisions used tables to estimate these values, but the 2004 code uses simple equations to do this. As these two variables are conceptually independent, they will be treated in separate sections.