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Concrete In Australia : March 2014
Concrete in Australia Vol 40 No 1 33 The effective aggregate size, age , is the maximum aggregate size except it is taken as zero for concrete strengths greater than 70 MPa. The first term in Equation 2 for β accounts for the detrimental effect of increasing longitudinal strain on shear capacity (the strain effect), while the second term accounts for the detrimental effect of increasing crack spacing on shear strength (the size effect). Note that if high-strength reinforcement or low stiffness reinforcement (FRP) is used, the longitudinal strains will likely be considerably higher resulting in lower shear capacities. To estimate the longitudinal strain at mid-depth, the following general equation is used: ) (2 5.0 / 0 p p s s p p f f v f x A E A E f A N V d M (4) Where the terms Mf, Vf, and Nf are the moment and shear (both taken as positive) and the axial force (tension positive) at the section of interest. While As ,E s , Ap, Ep, are the areas and Young’s moduli of the regular and bonded prestressed reinforcement respectively. The term fp0 is the stress in the prestressing reinforcement when loaded to an extent that the strain in the surrounding concrete is equal to zero. Note that axial load and prestressing are considered in a compatible way in the CSA code shear provisions. While the MCFT equations indicate that a variable angle truss model could be used for the shear design of members with stirrups, the following equation is provided in the CSA code to directly calculate the appropriate value of θ: x 7000 29 (5) To demonstrate the use of the above design equations, consider the high-strength concrete large member without stirrups (L-10H) in Figure 3. It has the following parameters: bw = 300mm,d =1400mm,f c ’ = 75MPa,age =0mm, As = 3500 mm2, E s = 200 GPa, and a shear span of 4050 mm. The critical section for shear for this beam is indicated by the CSA code to be a distance of dv = 0.9d = 1260 mm away from the edge of the loading plate. At this location, a shear of 1 kN will result in a moment of (4.050-0 .075-1 .26) = 2.720 kNm. This M/V ratio can be taken as constant. The crack spacing term in this case is sxe = 1260 × 35/(16+0) = 2760 mm. While design of a section for shear is a direct calculation, predicting the shear strength of a given section requires some iteration. It is usually best to assume a value of εx to allow Equations 1 and 2 to be evaluated, and then check the assumed value of εx using Equation 4. With εx assumed equal to 1.0×10 -3 , Equation 2 gives a value for β of 0.0553. Substituting this into Equation 1 indicates a shear strength estimate of 167 kN. This will have a corresponding moment of 167 × 2.720 = 454 kNm. Substituting these force values into Equation 4 produces a calculated strain of 0.377x10-3 . Thus the assumption of a strain of 1.0×10 -3 was conservative. An appropriate next iteration could be taken as εx = (2.0×0.377×10 -3+1.0×10 - 3)/3.0 = 0.585 × 10 - 3 producing a shear strength of 223 kN and a calculated strain from Equaton 4 of εx = 0.50×10 -3 . Figure 9: Reduction of compressive strength with principal tensile strain. Figure 10: Levels of approximation in MCFT analysis. 28-38 - Cover story.indd 33 28-38 - Cover story.indd 33 28/01/14 3:09 PM 28/01/14 3:09 PM