So far, all the calculations depended on the structure geometry and were not affected by the magnitude of the external loading. For instance the centre of stiffness, the structural eccentricities, and the torsional radii, are independent of the seismic force magnitude.
Next, the displacements and stress resultants of the structure, due to external seismic loading H, will be calculated.
The relevant seismic force H is applied at the centre of mass C_{M} of the diaphragm. This force is resolved in two forces H_{x} and H_{y} parallel to the two axes of the principal system. In order to apply the previous analysis, the forces H_{x}, H_{y} are transferred to the centre of stiffness C_{T} together with the moment M_{CT} according to the expression:
(9)
Example C.7-1: Given a horizontal force in x direction equal to is H=90.6 kN, carry-out the analysis of the structure.
Calculation of moment M_{CT} :
The structural eccentricities are e_{ox} =-0.626 and e_{oy}= y_{CM}=-2.334m
For a given force Η=H_{X}=90.6 kN in the initial system X0Y, the equivalent loading in the principal system (a=22.44^{ο}) is: H_{x}=H_{X}·cosa=90.6·0.924=83.72 kN and H_{y}=-H_{X}·sina=-90.6·0.382=-34.64 kN and
Μ_{CT}=-H_{x}·e_{oy}+H_{y}·e_{ox}=-83.72kN·(-2.334m)+[-34.64kN·(-0.626m)]=195.4 kNm+21.7kNm=217.1 kNm
The following quantities are calculated using the H_{x}, H_{y}, M_{CT}:
· the displacements δ_{xo}, δ_{yo} and θ_{z} of C_{T} using the expressions:
, , (10)
where , ,
Example C.7-2:
Quantities K_{xx}=Σ(K_{xxi})= 161.2·10^{6}N/m, K_{yy}=Σ(K_{yyi})= 161.2·10^{6} N/m and Κ_{θ}=3134·10^{6} N·m are already calculated therefore:
δ_{xo}=H_{x}/Κ_{xx }=83.72·10^{3}N/(274.41·10^{6}N/m)=0.305 mm,
δ_{yo}=H_{y}/Κ_{yy }=-34.46·10^{3}N/(161.2·10^{6} N/m)=- 0.214 mm and
θ_{z}=M_{CT}/K_{θ}=217.1kNm/(3134·10^{3}kNm)=0.692·10^{-4}
· the displacements δ_{xi}, δ_{yi} of the top of each column using the expressions:
, (11)
· and the displacements δ_{ζ}_{i}, δ_{η}_{i} by transferring δ_{xi}, δ_{yi} to the local system of each column using the expressions:
, where (12)
Example C.7-3:
C_{1}: δ_{x1}= δ_{xo}- θ_{z}·y_{1}=0.305mm-0.692·10^{-4}·(-3.50·10^{3}mm)=(0.305+0.242)mm=0.547 mm
δ_{y1}= δ_{yo}+ θ_{z}·x_{1}=-0.214mm+0.692·10^{-4}·(-4.35·10^{3}mm)=(-0.214-0.301)mm=-0.515 mm
By transferring to the local system where φ’_{1}=0.0-22.44°=-22.44°.
δ_{ζ}_{1}= δ_{x1}·cosφ’_{1}+δ_{y1}·sinφ’_{1}=0.547·0.924-0.515·(-0.382)=0.505+0.197=0.702 mm
δ_{η}_{1}=-δ_{x1}·sinφ’_{1}+δ_{y1}·cosφ’_{1}=-0.547·(-0.382)+(-0.515)·0.924=0.209-0.476=-0.267 mm
C_{2}: δ_{x2}= δ_{xo}- θ_{z}·y_{2}=0.305mm-0.692·10^{-4}·(-5.79·10^{3}mm)=(0.305+0.400)mm=0.705 mm
δ_{y2}= δ_{yo}+ θ_{z}·x_{2}=-0.214mm+0.692·10^{-4}·1.19·10^{3}mm=(-0.214+0.082)mm=-0.132 mm
By transferring to the local system where φ’_{2}=0.0-22.44°=-22.4°.
δ_{ζ}_{2}= δ_{x2}·cosφ’_{2}+δ_{y2}·sinφ’_{2}=0.705·0.924+(-0.291)·(-0.132)=0.652+0.038=0.701 mm
δ_{η}_{2}=-δ_{x2}·sinφ’_{2}+δ_{y2}·cosφ’_{2}=-0.705·(-0.382)+(-0.132)·0.924=0.269-0.122=0.147 mm
C_{3}: δ_{x3}= δ_{xo}- θ_{z}·y_{3}=0.305mm-0.692·10^{-4}·1.12·10^{3}mm=(0.305-0.078)mm=0.227 mm
δ_{y3}= δ_{yo}+ θ_{z}·x_{3}=-0.214mm+0.692·10^{-4}·(-2.44·10^{3}mm)=(-0.214-0.169)mm=-0.383 mm
By transferring to the local system where φ’_{3}=30.0-22.44°=7.56°.
δ_{ζ}_{3}= δ_{x3}·cosφ’_{3}+δ_{y3}·sinφ’_{3}=0.227·0.991+(-0.383)·0.132=0.225-0.050=0.175 mm
δ_{η}_{3}=-δ_{x3}·sinφ’_{3}+δ_{y3}·cosφ’_{3}=-0.227·0.132+(-0.383)·0.991=-0.030-0.380=-0.410 mm
C_{4}: δ_{x4}= δ_{xo}- θ_{z}·y_{4}=0.305mm-0.692·10^{-4}·(-1.17·10^{3}mm)=(0.305+0.081)mm=0.386 mm
δ_{y4}= δ_{yo}+ θ_{z}·x_{4}=-0.214mm+0.692·10^{-4}·3.10·10^{3}mm=(-0.214+0.215)mm=0.001 mm
By transferring to the local system where φ’_{4}=45.0-22.44°=22.56°.
δ_{ζ}_{4}= δ_{x4}·cosφ’_{4}+δ_{y4}·sinφ’_{4}=0.386·0.923+0.001·0.384=0.355+0.000=0.355 mm
δ_{η}_{4}=-δ_{x4}·sinφ’_{4}+δ_{y4}·cosφ’_{4}=-0.386·0.384+0.001·0.923=-0.148+0.001=-0.147 mm
In the example considered, for seismic action in X direction, the deformation due to rotation at column C_{2} gives δ_{x2,}_{θ}=0.400 mm, higher than the deformation due to translation δ_{xo}=0.305 mm. The total displacement is therefore δ_{x2}=0.305+0.400=0.705 mm.
Figure C.7: Moment distribution (Mji,1- Mji,2=Vji·h)
· shear forces and bending moments of each column in its local system based on the relationships
, (13)
, (14)
,
Example C.7-4:
Calculation of shear forces and bending moments :
The moment distribution factor is assumed the same in both directions a_{ζ}_{i}=a_{η}_{i}=0.50
C_{1}: V_{ζ}_{1}=δ_{ζ}_{1}·K_{ζ}_{1}=0.702mm·31.1·10^{6} N/m =21.8 kN
V_{η}_{1}=δ_{η}_{1}·K_{η}_{1}=-0.267mm·31.1·10^{6} N/m =-8.3 kN
M_{ζ}_{1,1} =V_{ζ}_{1}·h·0.50=21.8·3.0·0.50=32.7 kNm M_{ζ}_{1,2}=- M_{ζ}_{1,1}=-32.7 kNm
M_{η}_{1,1}=V_{η}_{1}·h·0.50=-8.3·3.0·0.50=-12.5 kNm M_{ηι}_{,2}=- M_{η}_{1,1}=12.5 kNm
C_{2}: V_{ζ}_{2}=δ_{ζ}_{2}·K_{ζ}_{2}=0.701mm·31.1·10^{6} N/m =21.8 kN
V_{η}_{2}=δ_{η}_{2}·K_{η}_{2}=0.147mm·31.1·10^{6} N/m =4.6 kN
M_{ζ}_{2,1}=V_{ζ}_{2}·h·0.50=21.8·3.0·0.50=32.7 kNm M_{ζ}_{2,2}=- M_{ζ}_{2,1}=-32.7 kNm
M_{η}_{2,1}=V_{η}_{2}·h·0.50=4.6·3.0·0.50=6.9 kNm M_{η}_{2,2}=- M_{η}_{2,1}=-6.9 kNm
C_{3}: V_{ζ}_{3}=δ_{ζ}_{3}·K_{ζ}_{3}=0.175mm·186.6·10^{6} N/m =32.7 kN
V_{η}_{3}=δ_{η}_{3}·K_{η}_{3}=-0.410mm·26.2·10^{6} N/m =-10.8 kN
M_{ζ}_{3,1}=V_{ζ}_{3}·h·0.50=32.7·3.0·0.50=49.1 kNm M_{ζ}_{3,2}=- M_{ζ}_{3,1}=-49.1 kNm
M_{η}_{3,1}=V_{η}_{3}·h·0.50=-10.8·3.0·0.50=-16.2 kNm M_{η}_{3,2}=- M_{η}_{3,1}=16.2 kNm
C_{4}: V_{ζ}_{4}=δ_{ζ}_{4}·K_{ζ}_{4}=0.355mm·19.7·10^{6} N/m =7.0 kN
V_{η}_{4}=δ_{η}_{4}·K_{η}_{2}=-0.147mm·78.7·10^{6} N/m =-11.6 kN
M_{ζ}_{4}=V_{ζ}_{4}·h·0.50=7.0·3.0·0.50=10.5 kNm M_{ζ}_{4,2}=- M_{ζ}_{4,1}=-10.5 kNm
M_{η4,1}=V_{η4}·h·0.50=-11.6·0·0.50=-17.4 kNm M_{η4,2}=- M_{η4,1}=17.4 kNm