Proposed Three Storey Commercial Bldg. Concrete Design
A project study Presented to ENGR. Cesario A. Bacosa, Jr. MPA, Ph.D Professor College of Engineering Holy Trinity University Puerto Princesa City Palawan
In partial fulfillment of the Requirements for the degree of Bachelor of Science in Civil Engineering CE – CE – 524 524 (Geotechnical Earthquake Engineering)
Presented by: GARRY B. GUARDACASA
DESIGN OF A THREE-STOREY REINFORCED CONCRETE BUILDING
DESCRIPTION OF THE STRUCTURE: The design involves a three-storey reinforced concrete commercial building to be located in a commercial area, restaurant area, assembly, and telescoping area. Located in wind zone 3 and seismic zone 2. Figures present both the structural and architectural plans for the structure. A summary of the functional requirements are given below: Roof Deck: Usage – Telescoping – Telescoping Framing – Framing – Reinforce Reinforce concrete slab and girders Flooring – Flooring – Concrete Concrete rd
3 Floor: Usage – Usage – Commercial Commercial space Framing – Framing – Reinforce Reinforce concrete slab and girders/beams Flooring – Flooring – Terrazzo Terrazzo tiles (25mm) Ceiling – Ceiling – Concrete Concrete nd
2 Floor: Usage – Usage – Commercial Commercial space Framing – Framing – Reinforce Reinforce concrete slab and girders/beams Flooring – Flooring – Terrazzo Terrazzo tiles (25mm) Ceiling – Ceiling – Concrete Concrete Ground Floor: Usage – Usage – Commercial Commercial space Framing – Framing – Reinforce Reinforce concrete slab and girders/beams Flooring – Flooring – Terrazzo Terrazzo tiles (25mm) Ceiling – Ceiling – Concrete Concrete
STRUCTURAL DESIGN CRITERIA The following sections presents the design criteria, structural idealization, an analytic procedures and methods used in the structural design of the building. The design criteria include the codes and specifications, materials strength, and loadings deemed applicable for this structure.
CODES AND SPECIFICATIONS The following structural codes and specifications were used in the design de sign of the building. 1. NATIONAL STRUCTURAL CODE OF THE PHILIPPINES 2001 (NSCP 2001) 2. PHILIPPINE NAT’L STANDARDS (PNS) SPECIFICATIONS 3. BUILDING CODE REQUIREMENTS FOR REINFORCED CONCRETE, ACI 318 – 05, – 05, AMERICAN CONCRETE INSTITUTE (ACI) 4. AMERICAN SOCIETY FOR TESTING AND MATERIALS (ASTM) STANDARD SPECIFICATIONS.
MATERIALS STRENGTHS The following materials strengths were used in the structural investigation of the building. 1. CONCRETE – CONCRETE – The The minimum concrete compressive strength for the structural elements, f’c = 27MPa. 2. REINFORCING STEEL STEEL - Rebar’s shall have a grade fy = 414MPa for slab, beams, columns and footings. 3. SOIL – SOIL – Allowable Allowable bearing pressure = 150 kPa.
LOADS Design loads and forces are those resulting from dead and live loads acting in the most critical combination, using the appropriate load factors recommended by the governing codes. The basic load types and their corresponding magnitudes are taken as follows: DEAD LOADS The loads include loads of constant magnitude that remain in one position. This consists mainly of the weight of the structure and other permanent attachments to the frame and supported thereby:
6” CHB wall
--------
3.534kPa
4”CHB wall
--------
2.356kPa
Ceiling
--------
0.24kPa
Floor finish
--------
1.53kPa
Concrete
--------
23.56kN/m3
LIVE LOADS The live loads include the loads that may change in magnitude and position. Live loads that move under their own power are called moving loads. The minimum values of these loads depend on the occupancy and are normally specified by the governing building code (NSCP 2001). Commercial(Stores Retail)
--------
4.8kPa
Hallway
--------
4.8kPa
Toilet
--------
2.4kPa
Telescoping
--------
4.8kPa
WIND LOADS – Refer to NSCP 2001
EARTHQUAKE FORCES Seismic forces were determined based on the equivalent static force procedure and computed following the provisions of NSCP 2001. The total design base shear in a given direction shall be determined from the following equation: V
C v I
W
R T
Need not Exceeds:
V
2.5 C a I
W
R
But shall not be less than:
V
0.11Ca IW
For Seismic Zone 4, V shall not be less than:
V
0.8 Z N v I
W
R
LOAD COMBINATIONS Reinforced concrete sections shall be designed using the “strength design” methods. Using the load factors and critical load combination from the following: U = 1.4 D + 1.7 L U = 0.75 (1.4 D + 1.7 L + 1.87 E)
Analysis of Structural: 50m x 15m = 750m² A. Windload (WL) See Table 207-1, NSCP-01 WindLoad Force Resisting Frame F = qzGcfAf (other structure) Component & Cladding: When: Exposure c = G = 0.85 Maintain Force: F = qzGcfAf G – see @ 207-1 Cf – see @ table 207-1 to 207-10 Af – Projected Area Normal to wind Gf = Obtained by rational analysis
= 1.03, = 1.08,
= 1.16, = 1.0
V = 125 (Palawan Zone)
= 5m,
= 10m,
Lw = 1.0 Cf = for structure: Table 207-7 Qh = Height above NGL (Natural Ground Line) Table 207-1 (Building) h/D = 15/10 = 1.5 Normal Wind Force @ Cf = 1.5 Short Section (Assume Closed Window) Af1 = 3(15) = 45 Af2 = 3(15) = 45 Af3 = 3/2 (15) = 22.5
= 15m
Long section (Closed Window) Af1 = 3(50) = 150 Af2 = 3 (50) = 150 Af3 = 3/2 (50) = 75
Wind Force: Long section
= 0.761 kN/m² = 0.761(0.85)(1.5)(150) = 145.54 kN = 0.798 kN/m² = 0.798(0.85)(1.5)(150) = 152.62 kN = 0.857 kN/m² = 0.857(0.85)(1.5)(75) = 81.951 kN Wind Force: Short Section
= 0.761 (0.85)(1.5)(45) = 43.66 kN = 0.798 (0.85)(1.5)(45) = 45.79 kN = 0.857 (0.85)(1.5)(22.5) = 24.59 kN
SHORT SECTION F3 = 24.59 kN
F2 = 45.79 kN
F1 = 43.66 kN
V
2V
2V
V
N k 5 9 . 1 8 = 3 F
N k 2 6 . 2 5 1 = 2 F
N k 9 5 . 5 4 1 = 1 F
V
V 2
N O I T C E S G N O L
V 2
V 2
V 2
V
Short section: A. V+2V+2V+V = F1 + F2 + F3 6V = 43.66 + 45.79 +24.59 6V = 113.98 V = 18.99 kN (Exterior) 2V = 37.99 kN (Interior) B. V+2V+2V+V = F2 + F3 6V = 45.79 + 24.59 6V = 70.38 V = 11.73 kN (Exterior) 2V = 23.46 kN (Interior) C. V+2V+2V+V = F3 6V = 24.59 kN V = 4.1 kN (Exterior) 2V = 8.2 kN (Interior) Long Section: A. V+2V+2V+2V+2V+V = F1 + F2 + F3 10V = 145.59 + 152.62 + 81.951 V = 38.02 kN (Exterior) 2V = 76.03 kN (Interior) B. 10V = F2 + F3 10V = 152.62 + 81.51 V = 23.46 kN (Exterior) 2V = 46.91 kN (Interior) C. 10V = F3 10V = 81.95 V = 8.2 kN (Exterior) 2V = 16.4 kN (Interior)
Assume the size of the Beam @ Roof Deck B1 = 300mm x 500mm = 10 pcs B2 = 300mm x 500mm = 6 pcs
10m
5m
B3 = 300mm x 500mm = 24 pcs
10m
B4 = 300mm x 500mm = 13 pcs
5m
B5 = 300mm x 500mm = 1 pc
8m
rd
@ 3 floor B1 = 300mm x 500mm = 10 pcs B2 = 300mm x 500mm = 6 pcs
10m
5m
B3 = 300mm x 500mm = 24 pcs
10m
B4 = 300mm x 500mm = 13 pcs
5m
B5 = 300mm x 500mm = 1 pc
6m
nd
@ 2 floor B1 = 300mm x 500mm = 10 pcs B2 = 300mm x 500mm = 6 pcs
10m
5m
B3 = 300mm x 500mm = 24 pcs
10m
B4 = 300mm x 500mm = 13 pcs
5m
B5 = 300mm x 500mm = 1 pc
6m
Assume all column size is 500mm x 500mm
Slab thickness = 150mm Walling (CHB 0.1 x 0.2 x 0.4 )
COMPUTATION FOR DEAD LOAD @ Roof Deck Beam
B
H
Length
Concrete
Quantity
B – 1
0.3
0.5
10m
23.56 = 35.34
5
176.7kN
B – 2
0.3
0.5
5m
23.56 = 17.67
3
53.01kN
B – 3
0.3
0.5
10m
23.56 = 35.34
12
424.08kN
B – 4
0.3
0.5
5m
23.56 = 17.67
5
88.35kN
B – 2
0.3
0.5
8m
23.56 = 28.27
1 Total =
Ceiling
1
15
50
Wall
1
0.1
130
Column
1
0.5
0.5
1.2 = 900
Total
28.27 kN 770.41kN
1
900 kN
23.56 = 306.28
1
306.28 kN
23.56 = 5.89
16
94.24kN
Total =
2070.9kN
Quantity
Total
rd
@ 3 floor Beam
B
H
Length
Concrete
B – 1
0.3
0.5
10m
23.56 = 35.34
5
176.7kN
B – 2
0.3
0.5
5m
23.56 = 17.67
3
53.01kN
B – 3
0.3
0.5
10m
23.56 = 35.34
12
424.08kN
B – 4
0.3
0.5
5m
23.56 = 17.67
5
88.35kN
B – 2
0.3
0.5
6m
23.56 = 21.20
1
21.20kN
Total = Ceiling
1
15
50
Wall
5
0.1
130
Column
5
0.5
0.5
1.2 = 900
763.34kN
1
900 kN
23.56 = 1531.4
1
1531.4kN
23.56 = 29.45
16
471.2kN
Total =
3665.9kN
nd
@ 2 floor Beam
B
H
Length
Concrete
Quantity
B – 1
0.3
0.5
10m
23.56 = 35.34
5
176.7kN
B – 2
0.3
0.5
5m
23.56 = 17.67
3
53.01kN
B – 3
0.3
0.5
10m
23.56 = 35.34
12
424.08kN
B – 4
0.3
0.5
5m
23.56 = 17.67
5
88.35kN
B – 2
0.3
0.5
6m
23.56 = 21.20
1
21.20kN
Total = Ceiling
1
15
50
1.2 = 900
Wall
5
0.1
130
23.56 = 1531.4
Column
5
0.5
0.5
23.56 = 57.72
The following data give from NSCP Cv = 0.69 S = 1.0 I = 1.0 R = 5.6 Ct = 0.0731 V = CvIw / RT = 0.69(1.0)(9402.7) / 5.6(0.56) = 2068.83kN F4 = (2068.83 – 0) 31050 / 86038.5 = 746.61kN F3 = (2068.83 – 0) 36659 / 86038.5 = 881.48kN F2 = (2068.83 – 0) 18329.5 / 86038.5 = 440.7kN Total =
2068.83kN
T = Ct (hn)3/4; T<0.70 (second) T = 0.0731 (15) ¾ = 0.56 < 0.7 Moment at the base M = 746.61(15) + 881.48 (10) + 440.7(5) = 22217.45kN-m
Total
763.34kN
1
900 kN
1
1531.4 kN
16
471.2kN
Total =
3665.9kN
Level
Wx
hx
Wxhx
Fx
Roof
2070.9
15
31050
746.61
3
3665.9
10
36659
2
3665.9
5
18329.5
440.7
86038.5
2068.83
9402.7
Floor Shear and Moment Level
Vx
Roof 3 2
Mn
An
746.61
11199.15
3.54
1628.09
8814.8
2.36
2203.5
1.18
2068.8
22217.45 A4 = F4(g) / W4 = 746.61(9.81) / 2070.9 = 3.54 A3 = F3(g) / W4 = 881.48(9.81) / 3665.9 = 2.36 A2 = F3(g) / W4 = 440.7(9.81) / 3665.9 = 1.18 Spectral Acceleration M4 = hx(Fx) = 15(746.61) = 11199.15 M3 = hx(Fx) = 10(881.48) = 8814.8 M2 = hx(Fx) = 5(440.7) = 2203.5 Over Turning and Resisting moment RM = R(L) ;
∑R = 0.85(w)
3R = 0.85(2070.9 + 3665.9 + 3665.9) = 2664.1kN Resisting moment (RM) = 2664.1(3) = 7992.3kN-m
881.48
For Longitudinal Section st
Column Shear at 1 floor V+2V+2V+2V+2V+V = 2068.8 10V = 2068.8 V = 206.8kN (Exterior) 2V = 413.6kN (Interior) nd
Column Shear at 2 floor 10V = 1628.09 V = 162.8kN (Exterior) 2V = 325.62kN (Interior) rd
Column Shear at 3 floor 10V = 746.61 V = 74.661kN (Exterior) 2V = 149.32kN (Interior) Moment at column
= 206.8(5/2) =517kN-m = 413.6(5/2) = 1034kN-m = 162.8(5/2) =407kN-m = 325.62(5/2) = 814.1kN-m = 74.661(5/2) =186.65kN-m = 149.32(5/2) = 373.3kN-m Moment at Beam (Exterior)
=517 kN-m = 407 kN-m = 186.65 kN-m Moment at Beam (Interior)
= 1034 kN-m = 814.1 kN-m = 373.3 kN-m
Shear at Beam (Exterior) V1 = 2(517) / 10 = 103.4 kN V2 = 2(407) / 10 = 81.4 kN V3 = 2(186.65) / 10 = 37.33 kN Axial Force (Load of Column Exterior) P1 = 222.13 kN P2 = 118.73 kN P3 = 37.33 kN Shear at Beam (Interior) V1 = 2(1034) / 10 = 206.8 kN V2 = 2(814.1) / 10 = 162.82 kN V3 = 2(373.3) / 10 = 74.66 kN Axial Force (Load of Column Interior) P1 = 444.28 kN P2 = 237.48 kN P3 = 74.66 kN
For Transverse Section st
Column Shear at 1 floor V = 506.04kN (Exterior) 2V = 1012.07kN (Interior) nd
Column Shear at 2 floor V = 398.4kN (Exterior) 2V = 796.8 kN (Interior)
rd
Column Shear at 3 floor V = 183.11 kN (Exterior) 2V = 366.22 kN (Interior)
Moment at column
= 506.04(5/2) = 1265.1kN-m = 1012.07(5/2) = 2530.175kN-m = 398.4(5/2) = 996kN-m = 796.8(5/2) = 1992kN-m = 183.11(5/2) = 457.775kN-m = 366.22(5/2) = 915.55kN-m Moment at Beam (Exterior)
= 1265.1 kN-m = 996 kN-m = 457.775 kN-m Moment at Beam (Interior)
= 2530.175 kN-m = 1992 kN-m = 915.55 kN-m
Shear at Beam (Exterior) V1 = 2(1265.1) / 5 = 506.04 kN V2 = 2(996) / 5 = 398.4 kN V3 = 2(457.775) / 5 = 183.11 kN Axial Force (Load of Column Exterior) P1 = 1087.55 kN P2 = 581.51 kN P3 = 183.11 kN
Shear at Beam (Interior) V1 = 2(2530.175) / 5 = 1012.07 kN V2 = 2(1992) / 5 = 796.8 kN V3 = 2(915.55) / 5 = 366.22 kN Axial Force (Load of Column Interior) P1 = 2175.1 kN P2 = 1163.02 kN P3 = 366.22 kN
SEISMIC LOAD TABULATION (TRANSVERSE)
LEVEL
BEAM EXTERIOR V
INTERIOR
M
V
M
RD
183.11
457.775
366.22
915.55
3
398.4
996
796.8
1992
2
506.04
1265.1
1012.07
2530.175
LEVEL
COLUMN EXTERIOR
INTERIOR
V
M
P
V
M
P
RD
183.11
457.775
183.11
366.22
915.55
366.22
3
398.4
996
581.51
796.8
1992
1163.02
2
506.04
1265.1
1087.55
1012.07
2530.175
2175.1
SEISMIC LOAD TABULATION (LONGITUDINAL)
LEVEL
BEAM EXTERIOR
INTERIOR
V
M
V
M
RD
37.33
186.65
74.66
373.3
3
81.4
407
162.82
814.1
2
103.41
517
206.8
1034
LEVEL
COLUMN EXTERIOR V
M
INTERIOR P
V
M
P
RD
74.661
457.775
37.33
149.32
915.55
74.66
3
162.8
996
118.73
325.62
1992
237.48
2
206.8
1265.1
222.13
413.6
2530.175
444.28
Vact = 0.40 (1) (9402.7) / 8.5 (0.70) = 632.11 kN Vmax = 2.5 (0.28)(1)(9402.7) / 8.5 = 774.34 kN Vmin = 0.11(0.28)(1)(9402.7) = 289.6 kN Vmax>Vact>Vmin
OK!!
Stiffness of Structures and Displacement Column (500mm x 500mm) Fc` = 27 Mpa E = 4700√fc` = 4700√27 = 24421.92 N/mm² I = 1/12 (500)(500)² = 10416666.67 mm⁴ K = EI/L = 24421.92 N/mm²(10416666.67 mm⁴) / 5(1000) = 50879000.02 L1 = (746.61 + 881.48 + 440.7)1000 / 50879000.02 = 0.0406 L2 = [(746.61 + 881.48)1000 / 50879000.02]+0.0406 = 0.0726 L3 = [(746.61+ 881.48)1000 / 50879000.02]+0.0726= 0.105 mRD = w / g = 2070.9 / 9.81 = 211.1 WRD = √k / m = √50879000.02 / 211.1 = 33.79 TRD = 2pie / W = 0.186 sec m3 = w / g = 3665.9 / 9.81 = 373.7 W3 = √k / m = √50879000.02 / 373.7 = 19.1 T3 = 2pie / W = 0.33 sec
m2 = w / g = 3665.9 / 9.81 = 373.7 W2 = √k / m = √50879000.02 / 373.7 = 19.1 T2 = 2pie / W = 0.33 sec
DYNAMIC LATERAL FORCES st
1 MODE LEVEL RD 3 2 Σ
E
Hi 5 5 5
2
Wi 2070.9 3665.9 3665.9 9402.7
2
Фi
2.86 1.959 1.0
Wiфi 5922.77 7181.5 3665.9 16770.17
2
Wiфi 16939.12 14068.56 3665.9 34673.58
2
W = (ΣWiфi) / ΣWiфi = (16770.17) / 34673.58 = 8111.03 E
V = (W /g) Sa From response spectrum With TRD = 0.186 sec say 0.2 sec Sv = 307.25 mm /sec (from table Earthquake Resistance page 28 table 3 – 1) Fi = (Wiфi / ΣWiфi) V Sa = (2pie / T) Sv = (2pie / 0.186)(0.30725) = 10.38 m / sec
2
V = (8111.03)(10.38) / 9.81 = 8582.31 FiRD= (Wiфi / ΣWiфi) V = (5922.77 / 16770.17)(8582.31) = 3031.04 Fi3= (Wiфi / ΣWiфi) V = (7181.5/ 16770.17)(8582.31) = 3675.21 Fi2= (Wiфi / ΣWiфi) V = (3665.9/ 16770.17)(8582.31) = 1876.06
Fi 3031.04 3675.21 1876.06 8582.31
nd
2 MODE LEVEL RD 3 2 Σ
E
Hi 5 5 5
2
Wi 2070.9 3665.9 3665.9 9402.7
2
Фi
- 0.657 0.725 1.0
Wiфi - 1360.6 2657.78 3665.9 4963.1
2
Wiфi 893.91 1926.9 3665.9 6486.71
2
W = (ΣWiфi) / ΣWiфi = (4963.1) / 6486.71 = 3797.36 E
V = (W /g) Sa From response spectrum With T = 0.33 sec say 0.4 sec Sv = 614.25 mm /sec (from table Earthquake Resistance page 28 table 3 – 1) Fi = (Wiфi / ΣWiфi) V Sa = (2pie / T) Sv = (2pie / 0.33)(0.61425) = 11.7 m / sec
2
V = (3797.36)(11.7) / 9.81 = 4527.14 FiRD= (Wiфi / ΣWiфi) V = (- 1360.6/ 4963.1)(4527.14) = -1241.1 Fi3= (Wiфi / ΣWiфi) V = (2657.78/ 4963.1)(4527.14) = 2424.32 Fi2= (Wiфi / ΣWiфi) V = (3665.9/ 4963.1)(4527.14) = 3343.88
Fi -1241.1 2424.32 3343.88 4527.1
rd
3 MODE LEVEL RD 3 2 Σ
E
Hi 5 5 5
2
Wi 2070.9 3665.9 3665.9 9402.7
2
Фi
0.378 -1.610 1.0
2
Wiфi 782.8 -5902.1 3665.9 -1453.4
Wiфi 295.9 9502.38 3665.9 13464.18
Fi -100.78 759.87 -471.97 187.12
2
W = (ΣWiфi) / ΣWiфi = (-1453.4) / 13464.18 = 156.89 E
V = (W /g) Sa From response spectrum With T = 0.33 sec say 0.4 sec Sv = 614.25 mm /sec (from table Earthquake Resistance page 28 table 3 – 1) Fi = (Wiфi / ΣWiфi) V Sa = (2pie / T) Sv = (2pie / 0.33)(0.61425) = 11.7 m / sec
2
V = (156.89)(11.7) / 9.81 = 187.12 FiRD= (Wiфi / ΣWiфi) V = (782.8/ -1453.4)(187.12) = -100.78 Fi3= (Wiфi / ΣWiфi) V = (-5902.1/ -1453.4)(187.12) = 759.87 Fi2= (Wiфi / ΣWiфi) V = (3665.9/ -1453.4)(187.12) = -471.97
Most probable design base shear 2
2
2 1/2
2
2
2 1/2
Using the SRSS method, V = (V 1 +V2 +V3 ) = (8582.31 +4527.1 +187.12 ) V = 9704.93kN
DYNAMIC PROCEDURES FROM EARTHQUAKE RESISTANCE From a moment-resisting steel frame, the response modification factor is obtained from UBC Table 16-N as Rw = 12 3/4
T = Ct(hn)
Ct = 0.035 hn = 15m 3/4
T = 0.035(15) = 0.2668 sec say 0.3 sec 2
Sa = 9650 mm / sec (from table Earthquake Resistance page 28 table 3 – 1) LEVEL RD 3 2 Σ
E
Wi 2070.9 3665.9 3665.9 9402.7
2
Фi
1.00 0.575 0.252 -
2
Wiфi 2070.9 2107.89 923.8 5102.6
2
Wiфi 2070.9 1212.04 232.8 3515.74
2
W = (ΣWiфi) / ΣWiфi = (5102.6) / 3515.74 = 7405.7 E
V = (W /g) Sa= (7405.7/9.81) ( 9.65) = 7284.91 FiRD= (Wiфi / ΣWiфi) V = (2070.9/ 5102.6)(7284.91) = 2956.6 Fi3= (Wiфi / ΣWiфi) V = 2107.89/ 5102.6)(7284.91) = 3009.4 Fi2= (Wiфi / ΣWiфi) V = (923.8/ 5102.6)(7284.91) = 1318.9 S = 1.2 from UBC table 16-J, for soil type 2 CA = 1.25 S / T
2/3
2/3
= 1.25(1.2) / (0.2668) = 3.62
C / Rw = 3.62 / 12 = 0.3 VA = (ZIC/Rw)W Z = 0.4 for zone 4 from UBC Table 16-I I = 1.0 as given W = 9402.7 (total loads)
Fi 2956.6 3009.4 1318.9 7284.9
Vi 2956.6 5966 7284.9
VA = (0.4)(1.0)(3.62/12)(9402.7) = 1134.6 TB = 1.3TA = 1.3(0.2668) = 0.3468 secs> 0.9 secs TB = 0.35 secs 2/3
C = 1.25S / TB = 1.25(1.2) / 0.35
2/3
= 3
VB = (ZIC/Rw)W = (0.4)(1.0)(3/12)(9402.7) = 940.27 V = 0.90VB = 0.90(940.27) = 846.24 Or V = 0.80V A = 0.80 (1134.6) = 907.68 , which governs!! Then, the design lateral force at each level is given by: Fn = VWiфi / ΣWiфi = 907.68(W iф) / 5102.6 = 0.1779 Wiф FRD = 0.1779 Wiф = 0.1779(2070.9) = 368.38 kN F3 = 0.1779 Wiф = 0.1779(2107.89) = 374.96kN F2 = 0.1779 Wiф = 0.1779(923.8) = 164.33kN
STORY SHEAR FORCE st
1 Mode LEVEL RD 3 2 Σ
Hi 15 10 5 -
Wi 2070.9 3665.9 3665.9 9402.7
Фi
1.00 0.675 0.252 -
WiФi 2070.9 2474.48 923.8 5469.18
2
Fi 372.76 445.41 166.28 984.45
Vi 372.76 818.17 984.45
2
Fi 390.99 173.03 -238.78 325.24
Vi 391 564.03 325.25
Wi Ф i 2070.9 1670.27 232.8 3973.97
T = 1.5 secs> T = 0.7 secs Sv = 300 mm / sec Sa = 2pieSv / T = 2pie(0.3) / 1.5 = 1.3 / 9.81 = 0.13g E
2
W = 5469.18 / 3973.97 = 7526.96 V = 7526.96 (0.13g /g) = 978.5 Fn = (WiФi / 5469.18)(978.5) = 0.18WiФi FRD = 0.18WiФi= (0.18)(2070.9) = 372.76kN F3 = 0.18WiФi= (0.18)(2474.48) = 445.41kN F2 = 0.18WiФi= (0.18)(923.8) = 166.28kN nd
2 Mode LEVEL RD 3 2 Σ
Hi 15 10 5 -
Wi 2070.9 3665.9 3665.9 9402.7
Фi
1.00 0.25 -0.345 -
WiФi 2070.9 916.475 -1264.74 1722.64
T = 0.56 secs< T = 0.7 secs Sa = 0.3g E
2
W = 1722.64 / 2736.35 = 1084.47 V = 1084.47 (0.3g /g) = 325.34 Fn = (WiФi / 1722.64)(325.34) = 0.1888WiФi FRD = 0.1888WiФi = (0.1888)(2070.9) = 390.99 kN F3 = 0.1888WiФi = (0.1888)(916.475) = 173.03 kN F2 = 0.1888WiФi = (0.1888)(-1264.74) = -238.78 kN
Wi Ф i 2070.9 229.12 436.33 2736.35
COMBINED STORY FORCE E
E
100(W1 + W2 ) / W = 100(7526.96+1084.47) / 9402.7 = 91.58% > 90% OK!!! LEVEL Roof 3 2 Base
F1i 372.76 445.41 166.28 984.45
F2i 390.99 173.03 -238.78 325.24
Where F1i = lateral force at level I for the first mode F2i = lateral force at level I for the second mode 1/2
Fci = (F1i + F2i)
For Longitudinal Section st
Column Shear at 1 floor V+2V+2V+2V+2V+V = 1309.02 10V = 1309.02 V = 130.9 kN (Exterior) 2V = 261.8 kN (Interior) nd
Column Shear at 2 floor 10V = 1018.05 V = 101.81 kN (Exterior) 2V = 203.61kN (Interior) rd
Column Shear at 3 floor 10V = 540.21 V = 54.021kN (Exterior) 2V = 108.042kN (Interior)
Fci 540.21 477.84 290.97 1036.78
Moment at column
= 130.9(5/2) = 327.25kN-m = 261.8(5/2) = 654.5kN-m = 101.81(5/2) = 254.525kN-m = 203.61(5/2) = 509.025kN-m = 54.021(5/2) = 135.0525kN-m = 108.042(5/2) = 270.11kN-m Moment at Beam (Exterior)
= 327.25 kN-m = 254.525 kN-m = 135.0525 kN-m Moment at Beam (Interior)
= 654.5 kN-m = 509.025 kN-m = 270.11 kN-m
Shear at Beam (Exterior) V1 = 2(327.25) / 10 = 65.45kN V2 = 2(254.525) / 10 = 50.905kN V3 = 2(135.0525) / 10 = 27.04kN Axial Force (Load of Column Exterior) P1 = 143.4kN P2 = 77.945kN P3 = 27.04kN Shear at Beam (Interior) V1 = 2(654.5) / 10 = 130.9kN V2 = 2(509.025) / 10 = 101.805kN V3 = 2(270.11) / 10 = 54.022kN
Axial Force (Load of Column Interior) P1 = 286.727kN P2 = 155.827kN P3 = 54.022kN
For Transverse Section st
Column Shear at 1 floor V = 218.17kN (Exterior) 2V = 436.34kN (Interior) nd
Column Shear at 2 floor V = 169.68kN (Exterior) 2V = 339.35kN (Interior)
rd
Column Shear at 3 floor V = 90.035kN (Exterior) 2V = 180.07kN (Interior)
Moment at column
= 218.17(5/2) = 545.425kN-m = 436.34(5/2) = 1090.85kN-m = 169.68(5/2) = 424.2kN-m = 339.35(5/2) = 848.375kN-m = 90.035(5/2) = 225.1kN-m = 180.07(5/2) = 450.175kN-m
Moment at Beam (Exterior)
= 545.425kN-m = 424.2kN-m = 225.1kN-m Moment at Beam (Interior)
= 1090.85kN-m = 848.375kN-m = 450.175kN-m
Shear at Beam (Exterior) V1 = 2(545.425) / 5 = 218.17kN V2 = 2(424.2) / 5 = 169.68kN V3 = 2(225.1) / 5 = 90.04kN Axial Force (Load of Column Exterior) P1 = 477.89kN P2 = 259.72kN P3 = 90.04kN
Shear at Beam (Interior) V1 = 2(1090.85) / 5 = 436.34kN V2 = 2(848.375) / 5 = 339.35kN V3 = 2(450.175) / 5 = 180.07kN Axial Force (Load of Column Interior) P1 = 955.76kN P2 = 519.42kN P3 = 180.07kN
SEISMIC LOAD TABULATION (TRANSVERSE)
LEVEL
BEAM EXTERIOR V
INTERIOR
M
V
M
RD
90.035
225.1
180.07
450.175
3
169.68
424.2
339.35
848.375
2
218.17
545.425
436.34
1090.85
LEVEL
COLUMN EXTERIOR V
M
INTERIOR P
V
M
P
RD
90.035
225.1
90.04
180.07
450.175
180.07
3
169.68
424.2
259.72
339.35
848.375
519.42
2
218.17
545.425
477.89
436.34
1090.85
955.76
