Earthquake Load
Determination
Earthquake Loads
Problem 1
Adjust Load Combination given by Equation 203-5.
Solution:
Simplified Static
Simplified Design Base Shear
W = total weight
R = numerical coeff. representative of
R the inherent overstrength and global
ductility
Vertical Distribution
Problem 2
Determine the minimum design lateral forces for a three-
story commercial building located in Lucena, Quezon. The
building is constructed using building frame systems with
ordinary reinforced concrete shear walls.
Deck
3rd
2nd
GF
3m 150T
3m 150T
3m 200T
Static Force Procedure
I. Method A
Ct = 0.035 for steel moment frames
= 0.030 for concrete moment frames ENGLISH SYSTEM
= 0.035 for eccentric moment frames
= 0.02 for all other buildings
hn hn hn
hn
n = no. of storey
II. Method B
Reyleighs Formula
Problem 3
Wihi Fx
950 3.6 950 (38.4) 78.487
1800 3.6 1800(34.8) 134.771
1800 3.6 1800(31.2) 120.829
1800 3.6 1800(27.6) 106.887
1800 3.6 1800(24) 92.94
2600 3.6 2600(20.4) 114.116
1850 3.6 1850(16.8) 66.869
1850 3.6 1850(13.2) 52.54
1950 3.6 1950(9.6) 40.276
2000 6 2000(6) 25.818
__________
387,420
Problem
Example. 4
Location: Cubao, Q.C.
Occupancy: Hospital
Structure: SMRF Concrete Special
Moment Resisting Force system
Soil Profile: Guadalupe Tuff - C
Seismic Source type A
Deck 3m 100T
4th 3m
3rd 3m
150T
2nd 3m 150T
GF 3m 200T
Cv= 0.56Nv
Nv=1.6(≤5km)
Cv= (0.56)(1.6) = 0.896
Ca= 0.4Na
Nv=1.2(≤5km)
Cv= (0.4)(1.2) = 0.48
Importance factor, I =1.15
W, Total weight = 600T
R= 0.85
T= Ct(hn)3/4
hn=12, Ct = 0.0731
T= 0.731(12)3/4 Wihi Fx
0.471sec < 0.7sec ; Ft=0
950 (38.4) 78.487
1800(34.8) 134.771
1800(31.2) 120.829
1800(27.6) 106.887
Σ=387,420