A five
story reinforce concrete building is to be designed as MRFS and is located in
Kathmandu having very soft soil. Determine the seismic base shear according to NBC105:2077
L.L=3kN/m2
Floor Finish= 1kN/m2 Roof Live load=1.5kN/m2 importance
factor =1.25 M25 concrete Fe500 steel
Slab
Thickness: 125 mm Beam: 300*450 Column: 450*450 Story Height =3.2m
Solution:
Calculation
of Total DL and LL
Weight
of Slab: 5*20*12*0.125*25
=3750kN
Weight
of beam= (20*4+12*5)*5*0.3*0.45*25
=2362.5kN
Weight
of Column=0.45*0.45*20*(3.2-0.45)*5*25
=1392.1875kN
Floor
Finish =20*12*5*1
=1200kN
Live
load =20*12*3*4
=2880kN
Roof
Live Load =1.5*20*12
=360kN
Total
Dead load = weight of (Slab + beam + column+ floor Finish)
=3750+2362.5+1392.1875+1200
=8704.6875kN
Total
Live Load = 2880+360
=3240kN
Seismic
weight
The seismic weight at each level, Wi
shall be taken as the sum of the dead loads and the factored seismic live loads
between the mid-heights of adjacent stories.
The seismic live load is taken as 0.3LL
|
Floor |
Column(kN) |
Beam(kN) |
Slab(kN) |
Floor
Finish(kN) |
Live
Load(kN) |
Seismic
Weight(DL+0.3LL) |
|
1floor |
278.4375 |
472.5 |
750 |
240 |
720 |
1965.9375 |
|
2nd
Floor |
278.4375 |
472.5 |
750 |
240 |
720 |
1965.9375 |
|
3rd
Floor |
278.4375 |
472.5 |
750 |
240 |
720 |
1965.9375 |
|
4th
Floor |
278.4375 |
472.5 |
750 |
240 |
720 |
1965.9375 |
|
Top
Level |
139.21875 |
472.5 |
750 |
240 |
0 |
1601.7185 |
|
Sum= |
=9465.4685kN |
|||||
Seismic Weight =9465.4685kN
The approximate fundamental period of vibration, T1,
in seconds is determined from following empirical equation
T1 = kt *H
¾
=0.075*160.75
=0.6sec
The
approximate fundamental time period calculated using empirical equation shall
be increased by a factor of 1.25.
T=1.25*0.6
=0.75sec
For Kathmandu Very Soft Soil (Type D soil)
Z=0.35 Importance Factor I=1.25
For T=0.75 and D Type soil the
The Spectral Shape Factor Ch(T)=2.25
Elastic Site Spectra for horizontal loading is given
by C(T)=Ch(t)*Z*I
=2.25*0.35*1.25
=0.984375
Reinforced Concrete Moment Resisting
Frame
Ductility Factor(Rμ)=
4 Over strength Factor Ultimate limit State(Ωu)=
1.5
For
the ultimate limit state, the horizontal base shear co-efficient shall as given
by:
Cd(𝑇𝑖)=C(𝑇𝑖)/Rμ x Ωu
=0.9843/4*1.5
=0.1640
Base Shear Force= base shear
co-efficient*seismic Weight
=0.1640*9465.4685kN
V=1552.34kN
VERTICAL
DISTRIBUTION OF SEISMIC FORCES
The
lateral seismic force (Fi) induced at each level ‘i’ shall be calculated as:
Fi=(Wihik/ΣWihik
)x V
K=1.125 for 0.75 Sec
|
Floor |
Height
|
Weight |
Wi*hi1.125 |
Seismic
Force |
Shear
Force |
|
Top
Level |
16 |
1601.7185 |
36242.75 |
472.6468 |
472.6468 |
|
4th
Floor |
12.8 |
1965.9375 |
34608.35 |
451.3324 |
923.9792 |
|
3rd
Floor |
9.6 |
1965.9375 |
25039.45 |
326.543 |
1250.522 |
|
2nd
Floor |
6.4 |
1965.9375 |
15868 |
206.9368 |
1457.459 |
|
1st
Floor |
3.2 |
1965.9375 |
7275.51 |
94.88095 |
1552.34 |
|
|
Sum=119034.1 |
|
|||
