Step 1:
The building height is H=6·3.0=18.0 m.
The structure is classified as wall system in both directions, thus Ct,x=Ct,y=Ct=0.050. The structure is also regular in both plan and elevation, thus q=3.60.
T1,x= T1,y=Τ1=Ct·H3/4=0.050·18.03/4=0.44 sec
Step 2:
Since 0.15=TB≤0.44=T1<0.50=TC (see §6.1.6)
Sd(T1)=γI·agR·S·(2.5/q)=1.0·0.15·1.2·2.5/3.6= 0.125g.
Since T1=0.44≤1.00=2TC à λ=0.85, thus aCM=λ·Sd(T1)=0.85·0.125g=0.10625g
Step 3:
The mass of each floor is M=186.3 t, except from the mass of the last floor which is M6=165.0 t, therefore Σ(Mi)=165.0+5·186.3=1096.5 t.
The elevation of the centre of mass is
ZCM=Σ(Mj·Zj)/Σ(Mj)= =(186.3·3.0+186.3·6.0+186.3·9.0+186.3·12.0+186.3·15.0+165.0·18.0)/1096.5=11353.5/1096.5 à ZCM=10.4 m
The triangular distribution gives the accelerations by means of the expression
aj=(Zj/ZCM)·aCM=(0.10625g/10.4)·Zj à aj=0.0102g·Zj
which yields
a1=0.031g, a2=0.061g, a3=0.092g, a4=0.122g, a5=0.153g, a6=0.184g and seismic forces
H1=a1·M1=0.031·10m/sec2·186.3·103kg=58 kN, H2=114, H3=171, H4=227, H5=285 and
H6=0.184·10m/sec2·165·103kg =304 kN.