Seismic Analysis

Seismic Analysis of Proposed 4-Storey Multi-Purpose Building (based on NSCP 2015)

1. Basis for Design using Section 203.3
1.2D +1.0E + f1L (203-5)
where: f1 = 1.0 for floors in places of public assembly, for live loads in excess of
4.8kPa, and for garage live load, or
= 0.5 for other live loads
2. Occupancy Categories using Table 208-1
III. Special Occupancy Structures
I = 1.00 and Ip =1.00
3. Seismic Design Consideration
Table D1
Recommended Seismic Design Parameters
Seismic Design Parameter Reference Recommended Value
Soil Profile Type NSCP Table 208-2 SD
Seismic Zone Factor Z NSCP Table 203-3 0.40
Seismic Source Type NSCP Table 208-4 A/B
4. Seismic Zone 4 Near-Source Factor
Using Table 208-5 and 208-6
Near source factor, Na = 1.0 and Near source factor, Nv = 1.0
5. Seismic Responds Coefficients
Using Table 208-7 and 208-8
Seismic Coefficient, Ca = 0.44Na = 0.44(1.0) = 0.44
Seismic Coefficient, Cv = 0.64Nv = 0.56(1.0) = 0.64
6. Structural System Coefficient
R= a measure of the ductility and over strength of the structural system,
based primarily on performance of similar past earthquakes
Using Table 208-11A,
Basic Seismic-Force Resisting System: C. Moment-Resisiting Frame
System (Special Reinforced Concrete)
R = 8.5 and Ωo = 2.8
7. Static Force Procedure
Using section 2.5.2.1 the total design base shear in a given direction shall
be determined from the following equations:
𝑪𝒗 𝑰
V= W (208-8)
𝑹𝑻
The total base shear need not exceed the following:
𝟐.𝟓𝑪𝒂 𝑰
V= W (208-9)
𝑹
The total design base shear shall not be less than the following:
V = 0.11CaIW(208-10)
 In addition, for seismic Zone 4, the total base shear shall also not be less than the
following:
𝟎.𝟖𝒁𝑵𝒗 𝑰
V= W (208-11)
𝑹
8. Structure Period
Using Method A. For all buildings, the value of T may be approximated
from the following equation:
T = Ct(hn)3/4 (208-12)
where: Ct = 0.0731 for reinforced concrete moment-resisting frames and
eccentrically braced frames

9. Vertical Distribution of Force

V = Ft + ∑𝒏𝒊=𝟏 𝑭𝒊 (208-15)
The concentrated force Ft at the top, which is in addition to Fn shall be
determined from the equation:
Ft = 0.07TV (208-16)
Ft = 0.07TV ≤0.25V if T > 0.7 sec
Ft = 0.0 if T ≤ 0.7 sec
(𝑽−𝑭𝒕 )𝒘𝒙 𝒉𝒙
Fx= ∑𝒏
(208-17)
𝒊=𝟏 𝒘𝒊 𝒉𝒊
where, w is the weight at a particular level and h is the height of a particular level
above the base shear. At each floor, the force is located at the center of mass.
10. Storey Shear and Overturning Moment
The storey shear at level x is the sum of all the storey forces at and above
that level:
Vx= Ft + ∑𝒏𝒊=𝒙 𝑭𝒊
The overturning moment at particular level Mxis the sum of the moments of
the storey forces above, about that level. Hence,
Mx = Ft(hn – hx) + ∑𝒏𝒊=𝒙 𝑭𝒊 (hi – hx)
11. Torsion and P-Delta Effect
Storey shear x 5% of the floor dimension, perpendicular to the direction of
force.
 Table D
Estimated Weight of the Building (4-Storey- Multipurpose Building)
Roof Deck Beam + Column
Weight
Length (m) Width (m) Depth (m) Pieces γconcrete(kN/m3)
(kN)
2 0.30 0.40 14 23.50 78.96
3 0.30 0.40 6 23.50 50.76
4 0.30 0.40 76 23.50 857.28
5 0.30 0.40 16 23.50 225.6
0.5 0.5 1.5 68 23.50 599.25
TOTAL 1811.85
Third Floor Beam + Column
Weight
Length (m) Width (m) Depth (m) Pieces γconcrete(kN/m3)
(kN)
2 0.30 0.40 14 23.50 78.96
3 0.30 0.40 6 23.50 50.76
4 0.30 0.40 76 23.50 857.28
5 0.30 0.40 16 23.50 225.6
0.5 0.5 3 68 23.50 1198.5
TOTAL 2411.1
Second Floor Beam + Column
Weight
Length (m) Width (m) Depth (m) Pieces γconcrete(kN/m3)
(kN)
2 0.30 0.40 14 23.50 78.96
3 0.30 0.40 6 23.50 50.76
4 0.30 0.40 76 23.50 857.28
5 0.30 0.40 16 23.50 225.6
0.5 0.50 3 68 23.50 1198.5
TOTAL 2411.1
Ground Floor Beam + Column
Weight
Length (m) Width (m) Depth (m) Pieces γconcrete(kN/m3)
(kN)
2 0.30 0.40 14 23.50 78.96
3 0.30 0.40 6 23.50 50.76
4 0.30 0.40 76 23.50 857.28
5 0.30 0.40 16 23.50 225.6
0.5 0.50 3.5 68 23.50 368.36
TOTAL 1398.25
TOTAL BEAM + COLUMN WEIGHT 8032.3
Third Floor Slab (Typical to Ground Floor Slab)
Length (m) Height (m) Thickness (m) Pieces γconcrete(kN/m3) Weight (kN)
3.00 2.00 0.210 1 23.50 21.15
4.00 2.00 0.210 8 23.50 225.6
5.00 2.00 0.210 3 23.50 105.75
4.00 3.00 0.210 4 23.50 169.2
4.00 4.00 0.210 21 23.50 1184.4
5.00 4.00 0.210 10 23.50 705
TOTAL 2411.1
Ground Floor wall
Length (m) Height (m) Thickness (m) γCHB(kN/m3) Weight (kN)
126 3.00 0.15 21.20 1202.04
139 3.00 0.10 21.20 884.04
TOTAL 2086.08
Second Floor wall
Length (m) Height (m) Thickness (m) γCHB(kN/m3) Weight (kN)
122 3.00 0.15 21.20 1163.88
201 3.00 0.10 21.20 1278.36
TOTAL 2422.24
Third Floor wall
Length (m) Height (m) Thickness (m) γCHB(kN/m3) Weight (kN)
129 3.00 0.15 21.20 1230.66
194 3.00 0.10 21.20 1233.84
TOTAL 2464.5
 Table D
Calculation of Seismic Forces
Level hx (m) Wx (kN) Wxhx Fx+Ft Vx Mx
Roof 11 1,811.85 19,930.35 454.3661 454.3661 1,363.0983
3rd 8 7,286.70 58,293.60 1,328.9599 1,783.3260 6,713.0763
2nd 5 7,244.44 36,222.20 825.7828 2,609.1088 14,540.4027
1st 2 5,895.43 11,790.86 268.8045 2,877.9133 22,719.7765
∑ - 22,238.42 126,237.01 2,877.9133 - -

Where:
T = Ct(hn)3/4
T = (0.0731)(113/4
T = 0.4415< 0.7sec, therefore Ft = 0

𝐶𝑣 𝐼 (0.64)(1.0)
V= W = (8.5)(0.4415)(22,238.42kN) = 3792.5758kN
𝑅𝑇
2.5𝐶𝑎 𝐼 2.5(0.44)(1.0)
V= W= (22,238.42kN) = 2877.9132kN (governs)
𝑅 8.5
V = 0.11CaIW = 0.11(0.44)(1.0)( 22,238.42kN = 1076.3395kN
0.8𝑍𝑁𝑣 𝐼 0.8(0.4)(1)(1)
V= W= (22,238.42kN) = 837.2111kN
𝑅 8.5

(𝑉−𝐹𝑡 )𝑤4 ℎ4 (2877.9132−0)(19,930.35)

F4= = = 454.3661kN
∑𝑤𝑥 ℎ𝑥 126,237.01
(𝑉−𝐹𝑡 )𝑤3 ℎ3 (2877.9132−0)(58,293.60)
F3= = = 1328.9599kN
∑𝑤𝑥 ℎ𝑥 126,237.01
(𝑉−𝐹𝑡 )𝑤2 ℎ2 (2877.9132−0)(36,222.20)
F2= = = 825.7828kN
∑𝑤𝑥 ℎ𝑥 126,237.01
(𝑉−𝐹𝑡 )𝑤1 ℎ1 (2877.9132−0)(11,790.86)
F1= = = 268.8045kN
∑𝑤𝑥 ℎ𝑥 126,237.01

V4= Ft + ∑4𝑖=4 𝐹𝑖 = 0 + 454.3661kN = 454.3661kN

V3= Ft + ∑4𝑖=3 𝐹𝑖 = 0 + 454.3661kN + 1328.9599kN = 1,783.3260kN
V2= Ft + ∑4𝑖=2 𝐹𝑖 = 0 + 454.3661kN + 1328.9599kN + 825.7828kN = 2,609.1088kN
V1= Ft + ∑4𝑖=1 𝐹𝑖 = 0 + 454.3661kN + 1328.9599kN + 825.7828kN + 268.8045kN
= 2,877.9133kN
 M4 = Ft (hn – hx) + ∑4𝑖=4 𝐹𝑖 (hi– hx) = 454.3661kN (3.0)m = 1,363.0983kN-m
M3 = Ft (hn – hx) + ∑4𝑖=3 𝐹𝑖 (hi– hx) = 454.3661kN (6.0)m + 1328.9599kN (3.0)m
= 6,713.0763kN-m
M2 = Ft (hn – hx) + ∑4𝑖=2 𝐹𝑖 (hi– hx) = 454.3661kN (9.0)m + 1328.9599kN (6.0)m +
825.7828kN (3.0)m = 14,540.4027kN-m
M1 = Ft (hn – hx) + ∑4𝑖=1 𝐹𝑖 (hi– hx) = 454.3661kN (11.0)m + 1328.9599kN (9.0)m +
825.7828kN (6.0)m + 268.8045kN (3.0)m = 22,719.7765kN-m