EXAMPLE 11. MASONRY WITH THE EQUAL FRAME METHOD

PARADEIGMA 11: " MASONRY M.I.P." 27 l1= 0.154268 VRd_f= 4.444897 (scale=1.000000) l2= 2.873666 VRd_v1= 82.798276 (scale=1.000000) l3= 1.661277 VRd_v2= 47.865983 (scale=1.000000) l4= 0.137181 VRd_2= 1.859916 (scale=1.000000) l1= 10000.000000 VRd_f= 0.000000 (scale=1.000000) l2= 2.699420 VRd_v1= 77.777778 (scale=1.000000) l3= 1.457687 VRd_v2= 42.000000 (scale=1.000000) l4= 10000.000000 VRd_2= 0.000000 (scale=1.000000) The first 4 l's represent the beginning and the next 4 the end. So we see that the smallest λ is λ4=0.137181 which indeed corresponds to the out-of-plane bend. The value 10000 at the end means that the corresponding strength was not calculated for this end because end has tensile strength and the end actually failed in tension (red square). The strengths that are not calculated when there is tensile force is the two in-plane and out-of-plane bending forces (1) and (4). 3.5 CONTROL AT THE LEVEL OF THE INSTITUTION This check is for the whole vector and compares the movement dm which is the movement corresponding to the last step of the pushover With the targeted movements corresponding to the performance levels. In this example the value is dm=4.21 cm. This is the maximum displacement that the carrier can withstand before it collapses. This is compared to the targeted