SEISMIC LATERAL FORCES COMPUTATION STATIC LATERAL FORCE PROCEDURE BASED ON NSCP 2001/2010
Learning Outcomes Upon completion of this lecture, students will be able to • Describe the seismic coefficients for base shear computation • Compute the base shear based on the NSCP 2001/2010 • Compute the static seismic lateral forces vertically and horizontally on a multi-story building
Governing Concepts in Earthquake Resistant Design
• Resist minor earthquake without damage • Resist moderate earthquake without structural damage, but possibly experience some nonstructural damage • Resist major earthquake without collapse, but possibly with some structural as well as nonstructural damage.
To become beco me earthquake earthq uake resistant, resistant, the structure become earthquake must possess :
• Strength Strength - the structure structure must must have the ability to carry or resist the earthquake forces without failure • Ductility Ductility - the structure structure must must have the ability ability to deform past the elastic range without failure to dissipate the energy induced by the earthquake.
Governing Concepts in Earthquake Resistant Design
• Resist minor earthquake without damage • Resist moderate earthquake without structural damage, but possibly experience some nonstructural damage • Resist major earthquake without collapse, but possibly with some structural as well as nonstructural damage.
To become beco me earthquake earthq uake resistant, resistant, the structure become earthquake must possess :
• Strength Strength - the structure structure must must have the ability to carry or resist the earthquake forces without failure • Ductility Ductility - the structure structure must must have the ability ability to deform past the elastic range without failure to dissipate the energy induced by the earthquake.
Basis for Seismic Design • Seismic zoning • Site characteristics • Occupancy • Configuration • Structural system • Height Limits
Effect of Earthquake on a Structure
• Earthquake causes ground motion • Structures accelerate during ground motion • Acceleration causes inertial force • Inertial force is proportional to the mass (F=ma)
F = ma
F = ma V=F V = ma V = (a/g) W V = (Seimic (Seimic Coeff Coefficien icients) ts) W
a
V
Static Lateral Force Procedure: Design Base Shear • 1992 NSCP
• 2001/2010 NSCP
V= V=
ZIC R
W
Cv I RT
Note: Cv / T <=> Z C Note also that the 2001 NSCP specifies lower values of R.
SEISMIC LOAD CALCULATION
W
Design Base Shear V= Cv I W
Eq. 208-4
RT Where: V
= is the total design seismic force
Cv = Seismic coefficient in Table 208-8 I
= Importance factor in Table 208-1
W = Total dead load per Sect. 208.5.1.1 R
= Numerical coefficient in Table 208-11
T
= Fundamental period of vibration
Design Base Shear (Cont’n.) Base Shear need not exceed the ff: V= 2.5 Ca I W
Eq. 208-5
R Where: Ca
= Seismic Coefficient in Table 208-7
I, W, & R = as previously defined
Design Base Shear ( Cont’n.) Base Shear shall not be < the ff: V= 0.11 Ca I W
Eq. 208-6
In Zone 4, Base Shear shall not be < the ff: V= 0.8 Z Nv I W
Eq. 208-7
R Where: Z = Seismic Zone Factor in Table 208-3 Nv = Near Source Factor in Table 208-5
Seismic Parameters Na =
Near source Factor, from Table 208-4
Nv =
Near source Factor, from Table 208-5
Ca =
Seismic Coefficient from Table 208-7
Cv =
Seismic Coefficient from Table 208-8
Seismic Parameters Table 208-1 - Seismic Importance Factors Seismic Seismic Occupancy 2 Importance Importance 1 Category Factor, I Factor, I p I. Essential 1.50 1.50 3 Facilities II. Hazardous 1.25 1.50 Facilities Special III. Occupancy 1.00 1.00 4 Structures IV.Standard Occupancy 1.00 1.00 4 Structures V.Miscellaneous 1.00 1.00 structures
Seismic Parameters Table 208-3 Seismic Zone Factor Z ZONE
2
4
Z
0.20
0.40
Seismic Parameters 1
Table 208-6 - Seismic Source Types Seismic Source Definition Seismic Source Seismic Source Description Maximum Moment Type Magnitude, M Faults that are capable of producing large magnitude events A M ≥ 7.0 and that have a high rate of seismic activity All faults other than Types A and B 6.5 = M < 7.0 C Faults that are not capable of producing large magnitude C earthquakes and that have a M < 6.5 relatively low rate of seismic activity 1 Subduction sources shall be evaluated on a site-specific basis .
Seismic Parameters Table 208-4 Near-Source Factor N a Closest Distance To Known Seismic Seismic 2 Source Source Type ≤5 ≥ 10 km km 1.2 1.0 A 1.0 1.0 B 1.0 1.0 C
Seismic Parameters Table 208-5 Near-Source Factor, N V Closest Distance To Seismic 2 Source Known Seismic Source Type 5 km 10 km 15 km A 1.6 1.2 1.0 B 1.2 1.0 1.0 C 1.0 1.0 1.0
Seismic Parameters Table 208-2 - Soil Profile Types
Soil Profile Type
Soil Profile Name/ Generic Description
S A
Hard Rock
S B
Rock
Average Soil Properties For Top 30 m Of Soil Profile Shear Wave SPT, N Undrained Shear Velocity, (blows / Strength, (kPa) 300 mm) V s (m/s) > 1,500 760 to 1,500
Very Dense Soil ad 360 to 760 > 50 > 100 Soft Rock S D Stiff Soil Profile 180 to 360 15 to 50 50 to 100 1 S E Soft Soil Profile < 180 < 15 < 50 Soil Requiring Site-specific Evaluation. S F See Section 208.4.3.1 1 Soil Profile Type S E also includes any soil profile with more than 3.0 meters of soft clay defined as a soil with plasticity index, Pl >20, w mc ≥ 40 percent and su < 24 kPa. The Plasticity Index, PI , and the moisture content, w mc , shall be determined in accordance with approved national standards. S C
Seismic Parameters Table 208-7 - Seismic Coefficient, Ca Soil Profile Seismic Zone Factor , Z type Z= Z = 0.4 0.2 0.16 0.32 N a SA 0.20 0.40 N a SB 0.24 0.40 N a SC 0.28 0.44 N a SD 0.34 0.36 N a SE See Footnote 1 of Table SF 208-8
Seismic Parameters Table 208-8 - Seismic Coefficient, C v Soil Seismic Zone Factor , Z Profile Z = 0.2 Z = 0.4 type 0.16 0.32 N V SA 0.20 0.40 N V SB 0.32 0.56 N V SC 0.40 0.64 N V SD 0.64 0.96 N V SE See Footnote 1 SF 1 Site-specific geotechnical investigation and dynamic site response analysis shall be performed to determine seismic coefficients.
Seismic Parameters Table 208-11 Basic Structural 2 System
3.MomentResisting Frame Systems
Structural Systems Lateral-Force-Resisting System Description
1.Special Moment-Resisting Frame (SMRF) a. Steel 4 b.Concrete 2. Masonry Moment-Resisting Wall Frame (MMRWF) 3. Concrete Intermediate Moment5 Resisting Frame (IMRF) 4. Ordinary Moment-Resisting Frame (OMRF) 6 a. Steel 7 b.Concrete 5. Special Truss Moment Frames of Steel (STMF)
Height Limit for Zones 4 (m)
R
8.5 8.5
2.8 2.8
N.L. N.L.
6.5
2.8
50
5.5
2.8
--
4.5 3.5
2.8 2.8
50 --
6.5
2.8
75
Fundamental Period of Vibration of the Structure, T T= Ct (hn)3/4 Ct = 0.0853 for Steel Moment Resisting Frame Ct = 0.0731 for RC Moment Resisting Frame Ct = 0.0488 for all other buildings hn = height of structure above base
General Procedure – Static lateral force procedure 208.4.8.2 1. Seismic Dead Load Computation: W V 2. Base Shear Computation V = V ≤
V
C v IW RT 2 . 5 C a IW
R V ≥ 0 . 11 C a IW V ≥
0 . 8 ZN
v
IW
R
General Procedure – Static lateral force procedure 208.4.8.2 3. Vertical Distribution of Forces
F t = 0.07TV if T > 0.7 s F t = 0.0 if T ≤ 0.7 s F x =
(V − F t ) w x h x n
∑i 1 wi hi
wn
n
F t
i=3 w x
F i
wi
F x
i=2 i=1 h x
=
V = F t + ∑i =1 F i
F n
V
hn hi
General Procedure – Static lateral force procedure 208.4.8.2 4. Story Shear (Vx) and Overturning Moment (Mx) F n n
V x = F t + ∑ i = x F i
F t
F i n
M x = F t ( hn − h x ) + ∑ i = x F i ( hi − h x )
F x
Vx
Mx
hn hi
h x
Sample Problem Calculate the seismic design loads for a threestory reinforced concrete frame building under the requirements of NSCP 2001 edition. The proposed building would be used as an office and approximately located 5Km away from Valley Fault System (See Figure 208-2B of NSCP 2001).
Sample Problem (Cont’n.) Design Criteria A typical floor plan, roof plan and elevation of the building are shown below. The columns, beams and slabs have constant cross sections throughout the height of the building. Though the uniformity and symmetry used in this example have been adapted primarily for simplicity, the structure itself is a hypothetical one, and has been chosen mainly for illustrative purposes. Other pertinent design data are as follows:
Sample Problem (Cont’n.) Material Properties: Concrete fc' = 28MPa (4ksi) Concrete weight, wc = 24 kN/m^3 Concrete CHB weight, 150mm thick solid grouted = 2.62 KN/m^2 ( vertical surface) Rebar fy = 413 MPa
Sample Problem (Cont’n.) Member sizes: Column C1
=0.40mx0.60m
Column C2
=0.40mx0.40m
Beam sizes
=0.40m wide x 0.60m deep
Slab Thickness =0.15 m
Sample Problem (Cont’n.) Other Superimposed Dead Load: •Concrete Toppings at Roof
= 0.075m
•Concrete Toppings at Typ.Flr. = 0.05m •Interior Partition at Typ. Flr.
= 1.00 kN/m^2
• Assume 1.5m high parapet RC wall along the perimeter of the roof level. • Assume the perimeter of the typical floor is covered with 150mm thk. CHB wall solid grouted (2.62 kN/m^2)
Sample Problem (Cont’n.) Seismic Parameters: •Importance factor, I
=
1.00 (Assumed Occupancy Category IV, See Section 103)
•Seismic Zone factor, Z = 0.40 (Zone 4, see Section 208.4.4.1) •Soil Profile Types =
SD (see Section 208.4.3.1)
•Seismic Source Type =
A (See Fig. 208-2B, Valley Fault System is Type A)
Typical Floor Plan 1
2
3
10 m C
8m
C1
C1
C1
C2
C2
C2
C1
C1
6m B
.4x.6 typ.
6m A
C1
C1 = 0.4x0.6 C2 = 0.4x0.4 Typ. Beam Size (incl. intermediate beams) = 0.4x0.6
1
2 10 m
3 8m
3.5 m 3.5 m 3.5 m
Base
Frame Elevation Along Grids A, B & C
A
B 6m
C 6m
3.5 m 3.5 m 3.5 m
Base
Frame Elevation Along Grids 1, 2 & 3
Step 1:Compute the Mass At Roof Level: • Slab weight= 0.15 x24
= 3.60kPa
• Topping wt = 0.075x24
= 1.80kPa
• Parapet wt.= 0.15x1.5x60x24 = 1.50kPa (12x18) • Beam wt.= 0.40x0.60x126x24 = 3.36kPa (12x18) • Col. wt.1= .4x.6x3.5/2(6)(24) = 0.28kPa (12x18) • Col. Wt.2= .4x.4x3.5/2(3)(24) =0.09kPa (12x18)
Mass Computation at Roof Level Total Unit Load at Roof Level: = 3.60 + 1.80 + 1.50 + 3.36 + 0.28 + 0.09 = 10.63 kPa Total weight at Roof Level: = 10.63 x 12 x 18
= 2296 kN
Step 1:Compute the Mass At Typ Floor: • Slab weight= 0.15 x24
= 3.60kPa
• Topping wt = 0.05x24
= 1.20kPa
• Peri.Wall wt.=3.5x60x2.62
=2.54kPa
(12x18) • Beam wt.= Same as in Roof
= 3.36kPa
• Interior Partition
= 1.00kPa
• Col. wt.1= .4x.6x3.5(6)(24)
= 0.56kPa
(12x18) • Col. Wt.2= .4x.4x3.5(3)(24)
=0.18kPa
(12x18)
Mass Computation at Typ.Floor Total Unit Load at Typ Floor: = 3.60 + 1.20 + 2.54 + 3.36 + 0.56 + 0.18 = 11.45 kPa Total weight at Typ.Floor: = 1145 x 12 x 18
= 2473 kN
Step 2: Calculate Total Mass of the Building Wt. of Roof level
= 2296kN
Wt. of 3rd Floor
= 2473kN
Wt. of 2nd Floor
= 2473kN
Total Mass
= 7242 kN
Step 3: Calculate Total Base Shear V= Cv I W RT Base Shear need not exceed the ff: V= 2.5 Ca I W R Base Shear shall not be < the ff: V= 0.11 Ca I W In Zone 4, Base Shear shall not be < the ff: V= 0.8 Z Nv I W R
Fundamental Period of Vibration of the Structure, T T= Ct (hn)3/4 Ct = 0.0731 for RC Moment Resisting Frame hn = 10.50 m , height of structure above base T = 0.0731(10.5)3/4 =
0.43 sec.
Seismic Parameters Na = 1.2
Near source Factor, from Table 208-4, for Seismic Source Type A and distance of 5km from Fault Line.
Nv = 1.6
Near source Factor, from Table 208-5, for Seismic Source Type A and distance of 5km from Fault Line.
Seismic Parameters Ca = 0.44Na
Seismic Coefficient from Table 208-7, for Soil profile type SD and Seismic Zone 4
Ca = 0.44(1.2) = 0.53 Cv = 0.64Nv
Seismic Coefficient from Table 208-8, for Soil profile type SD and Seismic Zone 4
Cv = 0.64(1.6) = 1.02
Seismic Parameters R = 8.5
Numerical coefficient representative of the global ductility capacity of structural system from Table 208-11
I = 1.00
Seismic Importance Factor from Table 208-1
Total Base Shear V= Cv I W = 1.02(1)(7242) = 2021 kN RT 8.5(.43) Base Shear need not exceed the ff: V= 2.5 Ca I W = 2.5(.53)(1)(7242) = 1129 kN R 8.5 Base Shear shall not be < the ff: V= 0.11 Ca I W = 0.11(.53)(1)(7242) = 422 kN In Zone 4, Base Shear shall not be < the ff: V= 0.8 Z Nv I W = .8(.4)(1.6)(1)(7242) = 436kN R 8.5
Step 4: Vertical Distribution V = Ft + Σ Fi
NSCP 2001 Eq. 208-13
Ft = 0.07 T V
NSCP 2001 Eq.208-14
Ft = 0 for T ≦ 0.7 sec Fx = (V-Ft)wxhx NSCP 2001 Eq 208-15 Σ wihi
V = 1129 kN T = 0.43 sec.
Step 4: Vertical Distribution Fx = (1129-0)(24108) = 543 kN at Roof level 50075 Fx = (1129-0)(17311) = 391 kN at 3rd flr. 50075 Fx = (1129-0)(8656) = 195 kN at 2nd flr. 50075
Step 4: Vertical Distribution Levels Storey hx wx wxhx Wxhx Fx Ht. Σwihi Roof 3.5 10.5 2296 24108 .481 543 rd
3.5
7.0
2473 17311 .346
391
nd
3.5
3.5
2473 8656
195
3 2
.173
Base Σ
=
7242 50075 1.00
1129
Step 5: Horizontal Distribution Lateral force will be distributed horizontally in direct proportion to the relative rigidities of the resisting elements. Rigidity of an element is the force required to deflect one end of an element a unit distance relative to the other end. Rigidity is therefore inversely proportional to deflection. Relative rigidity is defined as the ratio of the rigidity of an element to the total rigidities of all the elements within a floor to floor distance.
Horizontal Distribution (Cont’n.) δ
= δf + δv = Ph3
+ 1.2Ph
12Ec I
GA
Displacement for vertical members with both ends fixed
Where: δ
= total displacement
δf = δv
displacement due to bending
= displacement due to shear
P = Lateral force on pier or column h = height of pier or column I = moment of inertia of pier in direction of bending Ec = modulus of elasticity of concrete G = shear modulus
Horizontal Distribution (Cont’n.) For fc’ = 28mPa Ec = 24.84 x 106 kN/m^2 G = 9.92 x 106 kN/m^2 P = 1000 kN (assumed)
Horizontal Distribution (Cont’n.) Deflections & Stiffness Along X-Direction Frame Col.
A
A1
Heigh Are Total δf = δv = 3 bxd t a I Ph 1.2Ph Stiffness δ 4 (m) h A m 12Ec I GA k i =1/ δ 2 (m) m .4x.6 3.5 0.24 .0072 .0200 .0018 .0218 45.87
A2
.4x.6
3.5
0.24 .0072
.0200
.0018 .0218
45.87
A3
.4x.6
3.5
0.24 .0072
.0200
.0018 .0218
45.87 Σ=137.6
Horizontal Distribution (Cont’n.) Deflections & Stiffness Along X-Direction rame Col.
B
B1
Heigh Area Total δf = δv = 3 bxd t A I Ph 1.2Ph Stiffness δ 2 4 (m) h m m 12Ec I GA k i =1/ δ (m) .4x.4 3.5 0.16 .00213 0.0675 .0026 .0701 14.27
B2
.4x.4
3.5
0.16 .00213 0.0675 .0026 .0701
14.27
B3
.4x.4
3.5
0.16 .00213 0.0675 .0026 .0701
14.27 Σ=42.81
Horizontal Distribution (Cont’n.) Deflections & Stiffness Along X-Direction Frame Col.
C
C1
Heigh Are bxd t a I (m) h A m4 2 (m) m .4x.6 3.5 0.24 .0072
δf =
.0200
.0018 .0218
45.87
C2
.4x.6
3.5
0.24 .0072
.0200
.0018 .0218
45.87
C3
.4x.6
3.5
0.24 .0072
.0200
.0018 .0218
45.87
3
δv =
Ph 1.2Ph 12Ec I GA
Total δ
Stiffness k i =1/ δ
Σ=137.6
Horizontal Distribution (Cont’n.) Deflections & Stiffness Along Y-Axis Frame Col.
1
1A
Heigh Are bxd t a I 4 (m) h A m 2 (m) m .6x.4 3.5 0.24 .0032
1B
.4x.4
1C
.6x.4
δf = 3
δv =
Total
Ph 1.2Ph 12Ec I GA
δ
Stiffness k i =1/ δ
0.0450 .0018 .0468
21.37
3.5
0.16 .00213 0.0675 .0026 .0701
14.27
3.5
0.24 .0032
21.37
0.0450 .0018 .0468
Σ=57.01
Horizontal Distribution (Cont’n.) Deflection & Stiffness Along Y-Axis Frame Col.
2
2A
Heigh Are bxd t a I 4 (m) h A m 2 (m) m .6x.4 3.5 0.24 .0032
2B
.4x.4
2C
.6x.4
δf = 3
δv =
Ph 1.2Ph 12Ec I GA
Total δ
Stiffness k i =1/ δ
0.0450 .0018 .0468
21.37
3.5
0.16 .00213 0.0675 .0026 .0701
14.27
3.5
0.24 .0032
21.37
0.0450 .0018 .0468
Σ=57.01
Horizontal Distribution (Cont’n.) Deflections & Stiffness Along Y-Axis Frame Col.
3
3A
Heigh Are bxd t a I 4 (m) h A m 2 (m) m .6x.4 3.5 0.24 .0032
3B
.4x.4
3C
.6x.4
δf = 3
δv =
Total
Ph 1.2Ph 12Ec I GA
δ
Stiffness k i =1/ δ
0.0450 .0018 .0468
21.37
3.5
0.16 .00213 0.0675 .0026 .0701
14.27
3.5
0.24 .0032
21.37
0.0450 .0018 .0468
Σ=57.01
Horizontal Distribution (Cont’n.) Center of Mass Center of mass Xm and Ym can be calculated by taking statistical moment about a chosen location Xm =
mx m
Ym =
my m
Horizontal Distribution (Cont’n.) Center of Mass The center of mass, by inspection: Xm = 9 m from grid 1 ym = 6 m from grid C
Horizontal Distribution (Cont’n.) Center of Rigidity Center of rigidity Xr and Yr can be calculated by taking statistical moment about a chosen location Xr =
ky x ky
Yr = kx y kx
Horizontal Distribution (Cont’n.) Center of Rigidity Center of rigidity Yr and Xr; Yr = (137.6x12)+(42.81x6)+(137.6x0) (137.6+42.81+137.6) = 6 m from grid C Xr = (57.01x0)+57.01x10)+(57.01x18) (57.01x3) = 9.33 m from grid 1
Horizontal Distribution (Cont’n.) Calculation of eccentricity ex & ey: ex = (Xr-Xm) + 5% (18m) = (9.33 - 9) + 0.05(18) = 1.23 m ey = (Yr-Ym) + 5% (12m) = (6-6) + .05(12)
= 0.60 m
Horizontal Distribution (Cont’n.) Calculation of force Px at Roof level: Px = (kx ) (Fx ) + (kx dy ) (Mt ) Σ
k x
Jr
Mt = Fx (ey ) Torsional Moment Jr = Σ (k x dy2 + ky dx2 ) Rotational Stiffness
Horizontal Distribution (Cont’n.) Calculation of force Px at Roof level: Mt = Fx (ey ) Torsional Moment = 543(0.6) = 325.8 kN-m Jr = Σ (k x dy2 + ky dx2 ) Rotational Stiffness = (137.61x 62 + 42.81x 02 + 137.61x 62 + 57.01x 9.332 + 57.01x 0.672 + 57.01x 8.672 ) = 19181.55 kN-m
Horizontal Distribution (Cont’n.) Calculation of force Px Along Grid A: Proof = (137.61) (543) + ( 137.61x6)(543x0.6) 318.03 (19181.55) =
248.98 kN
P3rd = (137.61) (391) + (137.61x 6) (391 x 0.6) 318.03 (19181.55) =
179.28 kN
P2nd = (137.61) (195) + ( 137.61x6)195x0.6) 318.03 =
(19181.55)
89.41 kN
Horizontal Dist. Along X Direction Fr Stiffnes Relat Direc am s .Stiff. t d e k i k i Force (m) Σk i A 137.61 0.433 0.433 6
d2
k id2
k id2 2 Σk id
Torsion Force
36
4953.9
0.258
0.0258
Direct + Torsion 0.4588
B 42.81
0.135 0.135
0
0
0
0
0
0.135
C 137.61
0.433 0.433
6
36
4953.9
0.258
-.0258
0.433
0.0167
0.0167
Σk i
1.00
1.00
1
=318.03 57.01 0.333
-
9.33 87.05 4962.7
0.259
2
57.01
0.333
-
0.67 0.45
25.65
0.0013 -0.0012
-
3
57.01
0.333
-
8.67 75.17 4285.4
0.2234 -0.0155
-
Σk i
=171.03
1.00
19181. 73
Direct and Torsional Shear Assume Fx = 1 Direct Shear Coefficient VDx = (kx dy ) (Fx ) = (kx dy ) (1 ) Σ
Σ
k x
k x
Mt = Fx (ey ) = 1 (ey ) Torsional Shear Coefficient VTx = (kx dy ) (Mt ) = 1 (kx dy2 ) (1)( ey) Jr
dy
Jr
Jr = Σ (k x dy2 + ky dx2 ) ASEP
1 543x.4588= 249.13
2 10 m
3 8m Roof
391x.4588= 179.39
3rd/F
195x.4588= 89.47
2nd/F Base
Seismic Forces Along Grids A
1
2 10 m
543x.135= 73.31
3 8m Roof
391x.135= 52.79
3rd/F
195x.135= 26.33
2nd/F Base
Seismic Forces Along Grids B
1 543x.433= 235.12
2 10 m
3 8m Roof
391x.433= 169.3
3rd/F
195x.433= 84.43
2nd/F Base
Seismic Forces Along Grids C
