Step 1:
The building height is H=6·3.0=18.0 m.
The structure is classified as wall system in both directions, thus C_{t,x}=C_{t,y}=C_{t}=0.050. The structure is also regular in both plan and elevation, thus q=3.60.
T_{1,x}= T_{1,y}=Τ_{1}=C_{t}·H^{3/4}=0.050·18.0^{3/4}=0.44 sec
Step 2:
Since 0.15=T_{B}≤0.44=T_{1}<0.50=T_{C} (see §6.1.6)
S_{d}(T_{1})=γ_{I}·a_{gR}·S·(2.5/q)=1.0·0.15·1.2·2.5/3.6= 0.125g.
Since T_{1}=0.44≤1.00=2T_{C} à λ=0.85, thus a_{CM}=λ·S_{d}(T_{1})=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 M_{6}=165.0 t, therefore Σ(M_{i})=165.0+5·186.3=1096.5 t.
The elevation of the centre of mass is
Z_{CM}=Σ(M_{j}·Z_{j})/Σ(M_{j})= =(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 Z_{CM}=10.4 m
The triangular distribution gives the accelerations by means of the expression
a_{j}=(Z_{j}/Z_{CM})·a_{CM}=(0.10625g/10.4)·Z_{j} à a_{j}=0.0102g·Z_{j}
which yields
a_{1}=0.031g, a_{2}=0.061g, a_{3}=0.092g, a_{4}=0.122g, a_{5}=0.153g, a_{6}=0.184g and seismic forces
H_{1}=a_{1}·M_{1}=0.031·10m/sec^{2}·186.3·10^{3}kg=58 kN, H_{2}=114, H_{3}=171, H_{4}=227, H_{5}=285 and
H_{6}=0.184·10m/sec^{2}·165·10^{3}kg =304 kN.