NSCP Seismic Provisions SMRSF Horizontal Forces

8/11/2019 NSCP Seismic Provisions SMRSF Horizontal Forces

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SEISMICLATERAL FORCES COMPUTATION

STATIC LATERAL FORCE PROCEDUREBASED ON NSCP

2001/2010

Learning OutcomesUpon completion of this lecture,

students will be able to

Describe the seismic coefficients forbase shear computation Compute the base shear based onthe NSCP 2001/2010 Compute the static seismic lateralforces vertically and horizontally on amulti-story building


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Governing Concepts in Earthquake ResistantGoverning Concepts in Earthquake Resistant

Design Design

Resist minor earthquake without damage Resist moderate earthquake without structural

damage, but possibly experience some non-structural damage

Resist major earthquake without collapse, butpossibly with some structural as well asnonstructural damage.

To become earthquake resistant, the structureTo become earthquake resistant, the structuremust possess : must possess :

Strength - the structure must have the ability tocarry or resist the earthquake forces withoutfailure

Ductility - the structure must have the ability todeform past the elastic range without failure todissipate the energy induced by the earthquake.


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Basis for Seismic Design

Seismic zoning Site characteristics Occupancy Configuration Structural system Height Limits

Effect of Earthquake on a Structure 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

a V

F = maV = FV = maV = (a/g) WV = (Seimic Coefficients) W


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Static Lateral Force Procedure: Design

Base Shear

1992 NSCP

CvCv IIV =V = WW

R TR TNote: Cv / T Z CNote also that the 2001 NSCP specifies lower values

of R.

Z I CZ I CV =V = WW

RR

2001/2010 NSCP

SEISMIC LOAD CALCULATION


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Design Base Shear

V= Cv I W Eq. 208-4RT

Where:V = is the total design seismic forceCv = Seismic coefficient in Table 208-8 I = Importance factor in Table 208-1W = Total dead load per Sect. 208.5.1.1R = Numerical coefficient in Table 208-11T = Fundamental period of vibration

Design Base Shear (Contn.)

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


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Design Base Shear ( Contn.)

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 208-3Nv = Near Source Factor in Table 208-5

Seismic Parameters

Na = Near source Factor,from Table 208-4

Nv = Near source Factor,from Table 208-5Ca = Seismic Coefficient from

Table 208-7 Cv = Seismic Coefficient from

Table 208-8


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Seismic Parameters

Table 208-1 - Seismic Importance Factors

OccupancyCategory 1

SeismicImportance

Factor, I

SeismicImportance 2

Factor, I pI . Essential

Facilities 3 1.50 1.50

II . HazardousFacilities 1.25 1.50

III . SpecialOccupancyStructures 4

1.00 1.00

IV .StandardOccupancyStructures 4

1.00 1.00

V.Miscellaneousstructures 1.00 1.00

Seismic Parameters

Table 208-3 Seismic Zone Factor Z

ZONE 2 4

Z 0.20 0.40


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Seismic Parameters

Table 208-6 - Seismic Source Types 1

Seismic Source DefinitionSeismicSource

TypeSeismic Source Description Maximum Moment

Magnitude, M

A

Faults that are capable of producing large magnitude eventsand that have a high rate of seismicactivity

M 7.0

B All faults other than Types A andC 6.5 = M < 7.0

C

Faults that are not capable of producing large magnitudeearthquakes and that have a

relatively low rate of seismicactivity

M < 6.5

1 Subduction sources shall be evaluated on a site-specific basis .

Seismic Parameters

Table 208-4 Near-Source Factor N aClosest Distance To

Known SeismicSource 2

SeismicSource

Type 5km

10 km

A 1.2 1.0B 1.0 1.0C 1.0 1.0


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Seismic Parameters

Table 208-5 Near-Source Factor, N V Closest Distance To

Known Seismic Source 2 SeismicSource

Type 5 km 10 km 15 kmA 1.6 1.2 1.0B 1.2 1.0 1.0C 1.0 1.0 1.0

Table 208-2 - Soil Profile TypesAverage Soil Properties For Top 30 m Of

Soil Profile

Soil ProfileType

Soil ProfileName/ Generic

Description

Shear Wave

Velocity,

V s(m/s)

SPT, N (blows /

300 mm)

Undrained Shear Strength, (kPa)

S A Hard Rock > 1,500

S B Rock 760 to1,500

S C Very Dense Soil ad

Soft Rock 360 to 760 > 50 > 100

S D Stiff Soil Profile 180 to 360 15 to 50 50 to 100S E

1 Soft Soil Profile < 180 < 15 < 50

S F Soil Requiring Site-specific Evaluation.

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 s u =

n

i it

n

i ii

x xt x

t

t

F F V

hw

hw F V F

sT F

sT TV F

1

1

)(

7.0if 0.0

7.0if 07.0i=3

i=2

i=1

h x

wi

w x

wn

F x

F i

F n F t

V

hnhi


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4. Story Shear (Vx) and Overturning Moment (Mx)

General Procedure Static lateral forceprocedure 208.4.8.2

M x h x

F x

F i

F n F t

Vx

hn

hi

)()( xin

xi i xnt x

n

xi it x

hh F hh F M

F F V

+=

+=

=

=

Sample Problem

Calculate the seismic design loads for a three-story reinforced concrete frame building under therequirements of NSCP 2001 edition. The proposedbuilding would be used as an office andapproximately located 5Km away from Valley FaultSystem (See Figure 208-2B of NSCP 2001).


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Sample Problem (Contn.)Design Criteria

A typical floor plan, roof plan and elevation of thebuilding are shown below. The columns, beams andslabs have constant cross sections throughout theheight of the building. Though the uniformity andsymmetry used in this example have been adapted

primarily for simplicity, the structure itself is ahypothetical one, and has been chosen mainly forillustrative purposes. Other pertinent design data areas follows:

Sample Problem (Contn.)Material Properties:

Concrete fc' = 28MPa (4ksi)

Concrete weight, wc = 24 kN/m^3

Concrete CHB weight, 150mm thick solidgrouted = 2.62 KN/m^2 ( vertical surface)

Rebar fy = 413 MPa


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Sample Problem (Contn.)

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 (Contn.)Other Superimposed Dead Load:Concrete Toppings at Roof = 0.075mConcrete Toppings at Typ.Flr. = 0.05mInterior 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 wallsolid grouted (2.62 kN/m^2)


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Sample Problem (Contn.)

Seismic Parameters:Importance factor, I = 1.00 (Assumed

Occupancy Category IV,See Section 103)

Seismic Zone factor, Z = 0.40 (Zone 4, see Section208.4.4.1)

Soil Profile Types = S D (see Section208.4.3.1)

Seismic Source Type = A (See Fig. 208-2B,Valley Fault System isType A)

Typical Floor Plan

C1 C1 C1

C1 C1 C1

C2 C2 C2

1 2 3

C

B

A

10 m 8 m

6 m

6 m.4x.6 typ.

C1 = 0.4x0.6C2 = 0.4x0.4Typ. Beam Size (incl. intermediate beams) = 0.4x0.6


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Frame Elevation Along Grids A, B & C

10 m 8 m

1 2 3

3.5 m

3.5 m

3.5 m

Base

Frame Elevation Along Grids 1, 2 & 3

6 m 6 mA B C

3.5 m

3.5 m

3.5 m

Base


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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


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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


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Step 2: Calculate Total Mass of theBuilding

Wt. of Roof level = 2296kNWt. of 3rd Floor = 2473kNWt. 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 WIn Zone 4, Base Shear shall not be < the ff:V= 0.8 Z Nv I W

R


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Fundamental Period ofVibration 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 SeismicSource Type A and distance of

5km from Fault Line.

Nv = 1.6 Near source Factor,from Table 208-5, for SeismicSource Type A and distance of

5km from Fault Line.


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Seismic Parameters

Ca = 0.44Na Seismic Coefficient fromTable 208-7, for Soil profiletype S D and Seismic Zone 4

Ca = 0.44(1.2) = 0.53

Cv = 0.64Nv Seismic Coefficient fromTable 208-8, for Soil profile

type S D and Seismic Zone 4Cv = 0.64(1.6) = 1.02

Seismic Parameters

R = 8.5 Numerical coefficientrepresentative of the globalductility capacity of structuralsystem from Table 208-11

I = 1.00 Seismic Importance Factor fromTable 208-1


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Step 4: Vertical Distribution

Fx = (1129-0)(24108) = 543 kN at Roof level50075

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 StoreyHt.

hx wx wxhx Wxhx wihi

Fx

Roof 3.5 10.5 2296 24108 .481 543

3 rd 3.5 7.0 2473 17311 .346 391

2nd 3.5 3.5 2473 8656 .173 195

Base

= 7242 50075 1.00 1129


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Horizontal Distribution (Contn.)

For fc = 28mPa

Ec = 24.84 x 10 6 kN/m^2

G = 9.92 x 10 6 kN/m^2

P = 1000 kN (assumed)


Deflections & Stiffness Along X-Direction

Frame Col.

b x d(m)

Heigh

th

(m)

Are

aA

m 2I

m 4

f =

Ph3

12E c I

v =

1.2Ph GA

Total Stiffness

k i =1/

A A1 .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

=137.6


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rame Col.b x d(m)

Heighth

(m)

AreaAm 2

Im 4

f =Ph 3

12E c I

v = 1.2Ph GA

Total Stiffness

k i =1/

B B1 .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

=42.81



Frame Col.

b x d(m)

Heigh

th

(m)

Are

aA

m 2I

m 4

f =

Ph3

12E c I

v =

1.2Ph GA

Total Stiffness

k i =1/

C C1 .4x.6 3.5 0.24 .0072 .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

=137.6


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Deflections & Stiffness Along Y-Axis

Frame Col.b x d(m)

Heighth

(m)

AreaA

m 2

Im 4

f =Ph 3

12E c I

v = 1.2Ph GA

Total Stiffness

k i =1/

1 1A .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37

1B .4x.4 3.5 0.16 .00213 0.0675 .0026 .0701 14.27

1C .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37 =57.01


Deflection & Stiffness Along Y-Axis

Frame Col.

b x d(m)

Heigh

th(m)

Are

aAm 2

Im 4

f =

Ph3

12E c I

v =

1.2Ph GA

Total

Stiffnessk i =1/

2 2A .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37

2B .4x.4 3.5 0.16 .00213 0.0675 .0026 .0701 14.27

2C .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37

=57.01


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Deflections & Stiffness Along Y-Axis

Frame Col.b x d(m)

Heighth

(m)

AreaA

m 2

Im 4

f =Ph 3

12E c I

v = 1.2Ph GA

Total Stiffness

k i =1/

3 3A .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37

3B .4x.4 3.5 0.16 .00213 0.0675 .0026 .0701 14.27

3C .6x.4 3.5 0.24 .0032 0.0450 .0018 .0468 21.37

=57.01


Center of MassCenter of mass Xm and Ym can be calculated bytaking statistical moment about a chosen location

Xm = m x m

Ym = m y m


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Center of Mass

The center of mass, by inspection:

Xm = 9 m from grid 1

ym = 6 m from grid C


Center of RigidityCenter of rigidity Xr and Yr can be calculated bytaking statistical moment about a chosen location

Xr = ky x ky

Yr = kx y kx


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Center of RigidityCenter 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


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


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Calculation of force Px Along Grid A:P roof = (137.61) (543) + ( 137.61x6)(543x0.6)

318.03 (19181.55)= 248.98 kN

P 3rd = (137.61) (391) + (137.61x 6) (391 x 0.6)318.03 (19181.55)

= 179.28 kNP

2nd= (137.61) (195) + ( 137.61x6)195x0.6)

318.03 (19181.55)= 89.41 kN

Horizontal Dist. Along X DirectionFrame

Stiffnessk i

Relat.Stiff.

k i k i

Direct

Force

d

(m)

d 2 k id

2 k id2

k id2

TorsionForce

Direct+

TorsionA 137.61 0.433 0.433 6 36 4953.9 0.258 0.0258 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

k i =318.03

1.00 1.00

1 57.01 0.333 - 9.33 87.05 4962.7 0.259 0.0167 0.0167

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 -

k i =171.03

1.00 19181.73


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ASEP

Direct and Torsional Shear

Assume Fx = 1Direct Shear CoefficientVDx = (k x d y ) (Fx ) = (k x d y ) (1 )

k x k xMt = F x (e y ) = 1 (e y )Torsional Shear CoefficientV

Tx= (k

xd

y) (M

t) = 1 (k

xd

y

2 ) (1)( ey)

J r d y J r J r = (k x d y2 + k y d x2 )

Seismic Forces Along Grids A

10 m 8 m1 2 3

195x.4588=89.47

Base

2nd/F

3rd/F

Roof 543x.4588=249.13

391x.4588=

179.39


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Horizontal Dist. along Y Direction

Apply the forces along the Y-direction Consider the eccentricity e x in computing the

torsional moment Repeat the same procedure Determine the lateral forces for frames 1, 2 and 3

NSCP Seismic Provisions SMRSF Horizontal Forces

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