Any input concerning assignment of hinges to beams, columns and shear walls is highly appreciated. I would like to know what would be a better estimate for relative distances for plastic hinges in case of beams, columns. While assigning plastic hinges, I have an option of using ASCE 41-17 (Seismic Evaluation and Retrofit of Existing buildings". I will be using ETABS 2017 for performing Pushover Analysis. Until now, I have decided to go with "Displacement Co-efficient method". I'll be performing "Non-Linear Static Pushover Analysis" for my model. I am interested in performing "Performance Based Design" for a 20 story building. LRFD allows proper, balanced statistical probabilities of failure while ASD has a single safety factor independent of how well you know your cow. I've used stress in my LRFD example, but most LRFD specifications use moment, shear, or force variables instead of stress. While flexible, ductile failure modes can have higher factors. Abrupt, highly variable failures need lower resistance factors Live loads vary all over the place.Īlso, the resistance factor can be varied depending on the variability of the strength estimates that exist with different materials and different failure modes. What LRFD gives you that ASD doesn't is the ability to use different load factors (the 1.6) depending on the variability of the load. So you find a log big enough so that it will theoretically break when a 1600 lb cow is placed on it and you only count on 90% of its strength. Also, you know that the 1000 psi limit varies a bit with different logs so you apply a little safety factor to it - say reduce the 1000 psi to 900 psi just to be safe. So because of this uncertainty you apply a safety factor to the weight of the cow -say 1.6. You also know that the weight of the cow used to find the stress in the log might be plus or minus a few pounds. You still know that the log can be stressed to 1000 psi before it breaks. So you have a safety factor against failure of 1/0.6 = 1.67. your maximum allowable stress is 0.6 x 1000 = 600 psi.) So you find a log large enough such that the stress is limited to only 600 psi. In fact, you want to limit the stress with some safety factor to make sure that you are OK. So you need to find a log large enough so that the wood is not stressed to 1000 psi. You know that the wood of your log can be stressed to 1000 psi before it breaks. The log needs to be strong enough not to break under the weight of your cow and it needs to be stiff enough so the log won't sag too much under her weight. You have a small stream and need a log to drop over the stream so your ballet cow can dance across it. It really boils down to how you apply a safety factor to the design of members.
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