The residential building of the sketch includes a basement of area 12x18 m^{2} and height 3 m, a ground floor and four storeys of identical dimensions and a top floor of area 4x6 m^{2} and height 2.5 m. The masses at levels 0, 1, 2, 3, 4 are equal to M_{G}=220 t and M_{Q}=44 t, at level 5 to M_{G}=180 t and M_{Q}=44 t, while at the top level to M_{G}=20 t and M_{Q} =4 t. The building is situated in the seismic area Z_{1} and the distribution of seismic accelerations is triangular. The design seismic acceleration of magnitude 0.12g is applied at the center of mass of the building. The calculation of the seismic and wind forces as well as a comparison between them is asked.
Figure 2.51: The geometrical and loading model of the building's wind and earthquake actions Figure 2.51: The geometrical and loading model of the building's wind and earthquake actions
M_{i} [t] : masses, w [kN/m^{2}]: wind loads, a_{i}[m/sec^{2} ]: seismic accelerations Since the building is residential ψ_{2}=0.30 and consequently during an earthquake the dynamic masses are evaluated asM=M_{G}+0.30·M_{Q}. Thus, the dynamic masses at levels 0, 1, 2, 3 and 4 are equal to M _{G+0.30Q,i=04}=220+0.30x44=233 t, at level 5 is equal to M_{G+0.30Q,5}=180+0.30x44=193 t, while at the top level is equal M_{G+0.30Q,6}=20+0.30x4=21 t.
Figure 2.52: Wind forces Fw are less significant comparing to earthquake forces Fs. Figure 2.52: Wind forces Fw are less significant comparing to earthquake forces Fs.
W [kN]:gravity loads F_{w} [kN]: windforces F_{s} [kN]: seismic forces Assessment of seismic forces The total mass of the building during earthquake is M=4x233+193+21=1146 t, while the CM (mass center) is located at distance z_{ο}from the ground floor basis:
The design acceleration at the CM is 0.12g. Therefore acceleration at the 1^{st} level is equal to a_{1} =(3.0/9.0)x0.12g=0.04g. Respectively, at the rest levels: a _{2} =0.08g , a_{3}=0.12g, a_{4}=0.16g, a_{5}=0.20g and a_{6} =0.24g. The seismic force imposed on level 'i' is: F _{s,i} =M _{G+0.30Q,i}· a_{i} F _{s,1} =233t · (0.04x9.81)m/sec ^{2} =91 kN F _{s,2} =233t · (0.08x9.81)m/sec ^{2} =183 kN F _{s,3} =233t · (0.12x9.81)m/sec ^{2} =274 kN F _{s,4} =233t · (0.16x9.81)m/sec ^{2} =366 kN F _{s,5} =193t · (0.20x9.81)m/sec ^{2} =379 kN F _{s,6} =21t · (0.24x9.81)m/sec ^{2} =49 kN The total seismic force (or the total seismic base shear force) F_{s,tot} is equal to the sum seismic forces at different levels and its center of application is located at about 2/3 of the building height. F _{s,tot} =1146t · (0.12x9.81)m/sec ^{2} =1349 kN Assessment of wind forces For maximum wind load w=1.50 kN/m^{2}, the corresponding force towards the most unfavourable vertical direction of the 18.0 m on each storey is F_{w}=18.0m·3.0m·1.50kN/m^{2}=81 kN, while on the top floor is F_{w}=6.0m·2.5m·1.50kN/m^{2}=23 kN. The wind load on each storey is equally shared between the two levels defining it. Therefore, 6^{th} level: F_{w,6}=23/2=12 kN 5^{th} level: F_{w,5}=23/2+81/2=52 kN Rest levels: F_{w,i=04}=81/2+81/2=81 kN The total wind force is N_{w,tot}=12+52+4x81=388 kN and its point of application is located at about 1/2 of the building height. Comparison of wind and seismic forces
