#### Installation — business terrible - 1 part

September 8th, 2015

As was noted above, the general method and the simplified method in the 2004 CSA code are in fact now the same thing. The key to the simplified method is the appreciation that the value of the longitudinal strain at the mid-depth of the beam, ex, is in fact a “constrained” parameter for shear failures. Consider that for 400 MPa steel the flexural reinforcement will yield and cause a flexural failure rather than a shear failure when the strain in this steel equals fy/Es = 400/200,000 = 0.002. With the strain at mid-depth taken as one half of the strain in the bottom chord of the member for simplicity, the ex strain associated with flexural yield will equal 0.001, or 1 x 10-3 for this reinforcement. Thus for 400 MPa steel, if the strain ex exceeds 1 x 10-3, the member will not fail in shear, but will fail in flexure and none of the equations in this paper will govern.

The simplified method is derived by assuming that the strain term, ex, is equal to 0.85 x 10-3, or slightly less than the value associated with the yield of 400 MPa reinforcement. Substituting this value into Equation (5) produces a constant angle в = 35°. Substituting ex into Equation (4) will produce a constant value of в for members with stirrups, and a size effect equation for members without stirrups. Thus the shear strength of members with stirrups can be determined by the simplified method as:

Av

bwdv + 1.430* — fydv, s

which is no more complicated than the traditional ACI shear strength provisions. The major difference from the ACI provisions is that the Vc term is slightly lower and the beneficial effect of adding stirrups is estimated to be higher in Canada. As the Canadian provisions allow the design of members without stirrups up to the full value of Vc whereas the ACI code requires stirrups for members loaded in excess of Vc /2, these differences seem acceptable.

For members without stirrups, the same assumptions about the value of ex are made and the final version of Equation (1) with all simplifications reduces to:

This equation indicates that when stirrups are not provided that a size effect remains which must be taken into consideration at design time. At the same time, similar to the 1994 code, this equation is not iterative as all the terms in this equation will be known once the flexural design has been completed.

A significant benefit of basing the simplified method on the general method is that it becomes clear what to do as the limits of the simplified method are exceeded. As an example, Equations (8) and (9) are only appropriate when used with 400 MPa reinforcement. For designs based on, for example, 500 MPa reinforcement, the code specifies that Equations (8) and (9) may not be used. In this situation, however, it is clear how to generate new simplified equations for use with this higher strength steel. For this case, the strain at mid-depth associated with flexural yield will be fy/2Es = 1.25 x 10-3. Taking the same factor of 85% of this yield strain produces a value of ex = 1.06 x 10-3 for use with 500 MPa reinforcement. This too can be substituted into the previous equations to produce safe equations for use with this steel. The value of в, for example, would be taken as 36.4°, and the numerator in Equation (9) would be replaced by the value of 200 rather than 230. Design by this newly derived simplified method could then proceed. Note that the use of this technique implies that members with high strength reinforcement will be weaker in shear than

members with normal strength reinforcement. This is not necessarily the case, however. What is predicted is that when higher strength reinforcement is used, that the shear strength will be lower at flexural failure and so this lower, safe value must be used for all designs if a simplified type equation is desired.