Presentation on reinforced concrete deflections attached. That said, there are plenty of free beam analysis spreadsheets out there that can easily be set up with subdivided beams, so you can vary the flexural stiffness along the length of the beam, including: So if you are going to refine the beam analysis, you really need to refine the section stiffness analysis as well (unless you are confident that deflections 2-3 times greater than predicted will not have any significant adverse effects). Differential temperature effects can also greatly reduce the cracking moment. Shrinkage greatly reduces the cracking moment, which greatly increases deflections, even with symmetrical reinforcement. Significant loss of tension stiffening happens in weeks or days, rather than years. The loss of stiffness immediately after cracking is much greater than given by the ACI formula. I don't know the Canadian code, but if the deflection provisions are similar to ACI 318 the problem with doing a more detailed analysis to sharpen your pencil is that the calculated deflections are already way over-sharpened, especially for lightly reinforced sections: RE: Concrete Beam Effective Inertia Shotzie (Structural) If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it. I like to debate structural engineering theory - a lot. And it's the chunk of the span contributing most to deflections. I'd think it even more prudent for a uniformly loaded simple span as the moments tend to stay pretty flat over a good chunk of the span. The advice given above regarding midspan/end moment of inertias was mostly developed for continuous beams rather than simple spans. I'd be happy to run some things for you if you're inclined to attempt some bench marking in the future. I bet one could get this done in a page or two of programming.įor my precast work, I use a program called ConciseBeam that basically does what you've described. My vision was a MathCAD worksheet that would basically do old school double integration method, using a variable I. In the context of spreadsheet making, doing it once isn't so much different from doing it a thousand times. ![]() Would it be possible/correct to discretion the beam into shorter sections and compute the deflection cumulatively (going back to my structural analysis notes from University) Finding Z-section moment of inertia around x axisĪs a result, the moments of inertia of the each sub-area A, B and C around the axis x are: Indeed, centroidal axis x, lies at a distance equal to h/2-t_f/2 from the parallel axis through the sub-area A centroid (and the same distance from sub-area C centroid too). Therefore, the parallel axes theorem should be employed for these two sub-areas (later in text, a short introduction on the parallel axes theorem is given). This axis happens to also be centroidal for the rectangular sub-area B, however the same is not true for sub-areas A and C. Let's consider here, the moment of inertia around the centroidal axis x, of the z-section, which is parallel to the two flanges. See our article on finding the moment of inertia of compound shapes for more details on the methodology. It also determines the maximum and minimum values of section modulus and radius of gyration about x-axis and y-axis. It is important however that the individual moments of inertia I_A, I_B, I_C, have to be determined around the same axis, before such a summation can be made. This calculator uses standard formulae and parallel axes theorem to calculate the values of moment of inertia about x-axis and y-axis of angle section. However, the end result should be the same if the methodology is followed consistently. It is perfectly possible to split the given z-section in many different ways. ![]() The wanted moment of area I of the entire z-section, around a specific neutral axis, can be considered as the additive combination of I_A+I_B+I_C, of the individual moments of the sub-areas, over the same axis. Sub-areas B, C are identical in terms of sectional area, each having a width equal to b_f=b-t_w. ![]() Sub-area B accounts for the entire web (including the two small areas where flanges and web interesect), while sub-areas A and C account for the remaining flange parts, (A for the upper flange and C for the lower one). The moment of inertia of a z-section can be found if the total area is divided into three, smaller ones, A, B and C as shown in figure below.
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