EXAMPLE 11. MASONRY WITH THE EQUAL FRAME METHOD

PARADEIGMA 11: " MASONRY M.I.P." 16 current. These sizes are the final step sizes at each end. Of course they are not recalculated λ because then we would have an endless process. From all the λ of each step, the largest is selected and at the end plastic joint (internal release) is activated in NEXT STEP, i.e. the corresponding moments are released at the ends of the member. If the option is activated: then the moments at the other end are also released, regardless of λ. Depending on the range λ given to the parameters, the plastic joints are activated in the respective limbs. These limbs are the "candidates" for plastic joints to be implemented in the next step. In the next step, the analysis is performed on the modified beam and the known dots are placed at the ends of the previously "candidate" members, which are painted in a colour determined by comparing the measure of the rotation of the plastic joint and the limits as defined below for in-plane and out-of-plane respectively: ▪ θDL=0 (once it is exceeded, i.e. once it is activated, it turns blue) ▪ θSD=θu/γRd (once it is exceeded, it turns yellow) ▪ θNC=4/3*θu/γRd (once it exceeds it, it turns red) for in-plane γRd=1.5 for out- of-plane γRd=2 The way of calculating θu is different for in-plane and out-of-plane failure. PDF Refer to the pdf file of the presentation entitled ¨Part 3 o : Valuation of an existing building made of load-bearing masonry with linear finite elements (L.F.I.P.); and to KADET for more.