7.8 Use of Response Spectra for Designing MDF Systems
291
Mass z
(1)
i = φ
(1)
i B 1 S
(1)
d
z
(2)
i = φ
(2)
i B 2 S
(2)
d
z
(3)
i = φ
(3)
i B 3 S
(3)
d
z i
1
1.000 × 1.240 × 47.5
= 58.9
1.000 × (−0.317) × 9.16
= 2.904
1.000 × 0.082 × 5.55
= 0.446
58.976
2
0.759 × 1.240 × 47.5
= 44.705
(- 0.804) (−0.317) 9.16
= 2.3346
(−2.462) (0.082) 5.44
= −1.098
44.782
3
0.336 × 1.24 × 47.5
= 19.790
(−1.157) (−0.317) 9.16
= 3.3596
(2.580) (0.082) 5.44
= 1.15
20.15
The values of z i given in the last column have been calculated on the basis of root
mean square value of each mode.
Let us calculate the storey shear on the basis of Eqs. (7.54a–b) in the following
table:
Mode
Top storey
Second storey
1st story
φ
(r )
1 − φ
(r )
2
V
(r )
1
kN
φ
(r )
2 − φ
(r )
3
V
(r )
2
kN
φ
(r )
3
V
(r )
3
kN
1
0.241
51.218
0.423
89.898
0.336
107.112
2
1.804
−18.89
0.353
−3.901
−1.157
18.184
3
−3.462
−5.572
−5.042
−8.115
2.580
6.229
rms of maximum
storey shear (kN)
54.767
90.360
108.930
A look into the tables will immediately reveal that both in the case of displacements
and the storey shear, the first mode always plays the most dominant part.
The base shear = 108.930 kN
Let us check the base shear from Eq. (7.58)
V
(1)
3 = B 1 S
(1)
a
m 1 φ
(1)
1 + m 2 φ
(1)
2 + m 3 φ
(1)
3
= 1.24 × 1.15 × [1.00 + 0.759 + 0.336] × 36000
= 107.548 kN
V
(2)
3 = B 2 S
(2)
a
m 1 φ
(2)
1 + m 2 φ
(2)
2 + m 3 φ
(2)
3
= −0.317 × 1.65 × [1.000 − 0.804 − 1.157] × 36000
= 18.095 kN
V
(3)
3 = B 3 S
(3)
a
m 1 φ
(3)
1 + m 2 φ
(3)
2 + m 3 φ
(3)
3
= 0.082 × 1.9 × [1.000 − 2.462 + 2.58] × 36000
= 6.271 kN
Thus, the base shear in each mode calculated from Eqs. (7.58) agreed exactly with
Eqs. (7.54a–b) [the difference between them is due to round-off errors].