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66

Seismic Response Coefficient

)/(/1.0 1

1

IRTS

IRSCIS DDS

sD =

Seismic Design Category E and F:

)/(//5.0 11

IRTS

IRSC

IRS DDS

s =

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67

Response Modification Factor, R

Basis:

Ductility Damping System Redundancy

Table 5.2.2

Special Moment Resisting Frame - Steel (SMRF) 8.0

Special Moment Resisting Frame - Reinforced Conc. (SMRF) 8.0

Ordinary Reinforced Shear Walls (ORSW) 5.5Ordinary Moment Resisting Frame Steel (OMRFS) 4.0

Ordinary Reinforced Concrete Moment Frames (ORCMF) 3.0

Ordinary Plain Masonry Shear Walls (OPMSW) 1.5

)/(/1

IRT

S

IR

SC DDS

s =

R R

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68

Response Modification Factor, R

Seismic Reponse Coefficient

0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

Period, Seconds

Seis

micResponseCoefficient,Cs

Elastic Spectrum

OPMSW (R=1.5)

ORCMF (R=3.0)

OMRFS (R=4)ORSW (R=5.5)

SMRF (R=8)

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69

Structural Framing Systems

Moment Resisting Frame.A structural system with anessentially complete space frame provides support forgravity loads. Moment Resisting frames provide resistanceto lateral load primarily by flexural action of members.

Bearing Wall System.A structural system without a complete vertical load-carryingspace frame. Bearing walls or bracing systems provide supportfor all or most gravity loads. Resistance to lateral load isprovided by shear walls or braced frames.

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70

Structural Framing Systems

Dual System

A structural system with the following features:

An essentially complete space frame which provides support

for gravity loads, Resistance to lateral load is provided by a especially

detailed moment-resisting frame (steel or concrete) which iscapable of resisting at least 25% of base shear.

The two system shall be designed to resist the total lateralload in proportion to their relative rigidities.

Building Frame SystemA structural system an essentially complete space frame providing

support for gravity loads. Resistance to lateral load isprovided by shear walls or braced frames.

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71

Fundamental Period, T

Method of calculation

Rayleigh

others

As T, Fundamental period ofvibration, increases, Csandtherefore, base shear, V, decreases

Design spectrum is constant at lowT

)/(/1

IRT

S

IR

SC DDS

s =

T

T

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72

Fundamental Period, T

)/(/1

IRT

S

IR

SC DDS

s =

T

CT = 0.035 for steel MRFCT = 0.030 for concrete MRF

CT = 0.030 for steel EBF

CT = 0.020 for other building systems

4/3

nTa hCT = 5.3.2.1-4

12 stories and10 ft story heightNTa 1.0=

5.3.3.1-2

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73

Measured Periods

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74

Fundamental Period, T

)/(/1

IRT

S

IR

SC DDS

s =

T If Tis calculated, Tused

to determine Cs is limitedby

auTCT

Design Spectral

Response Acceleration at 1 second, SD1Cu

0.40.3

0.2

0.150.1

0.05

1.21.3

1.4

1.51.7

1.7

Table 5.3.3

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75

Vertical Distribution of Seismic Forces

Fx

V

hx

VCF vxx=

==

n

i

k

ii

k

xx

vx

hw

hwC

1

T K

0.5 1.0

2.5 2.0Interpolate

5.3.4-1

5.3.4-2

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Example

Determine the NEHRP-97design seismic forces for a

nine-story ductile momentresisting steel-frameoffice building located inMemphis, Tennessee onSoil Site Class D. Thestory heights are all 13 ft;the plan area is 100 ft by170 ft. The total dead loadis 100 lb/ft2 at all levels.

170ft

100

ft

9

@ 13

ft

Elevation

Plan

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Example (contd)

40.1=SS

42.0=D

S

Table 4.1.2.4a 0.1=a

Table 4.1.2.4b 6.1=V

93.0)4.1(

3

2

3

2=== MSDS SS

45.0)67.0(3

2

3

211

=== MD

SS

4.14.10.1 === SaMS

SS

67.042.06.111 === SS vM

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R = 8 Special Steel Moment Resisting Frame

SecondshT na 25.1)117(035.0)(035.04/34/3

===Equation 5.3.3.1-1

Seismic Design Category

D from Tables 4.2.1a and 4.2.1b

)/(/1

IRT

S

IR

SC DDS

s =

05625.0)25.1/8(25.1

45.0093.0

25.1/8

93.0===

sC

ISISC

DS

/1.0

1 =

05625.025.145.01.0 ==S

C

05625.0=S

CTherefore,

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79

Example (contd)

kipsVCvxx

861)300,15(05625.0 ===

=

=

n

i

k

ii

k

xx

vx

hw

hwC

1

5.3.4-1

5.3.4-2

375.15.05.2

50.025.10.10.1 =

+=k

T K

0.5 1.02.5 2.0 Interpolate

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80

Example (contd)

Level h_x W_x hx^1.375 w_i*h-i Cvx Fx Vx Mx

9 117 1700 697.8 1186327.2 0.233 200.2 200.2 2603

8 104 1700 593.5 1008950.3 0.198 170.3 370.5 7419

7 91 1700 493.9 839712.9 0.165 141.7 512.2 14078

6 78 1700 399.6 679327.2 0.133 114.6 626.9 22227

5 65 1700 311.0 528694.5 0.104 89.2 716.1 31536

4 52 1700 228.8 389003.5 0.076 65.7 781.7 41698

3 39 1700 154.1 261916.4 0.051 44.2 825.9 52435

2 26 1700 88.2 149981.4 0.029 25.3 851.2 63501

1 13 1700 34.0 57825.7 0.011 9.8 861.0 74694

ummation 15300 3001.0 5101739.1 1.000 861.0

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81

Category Procedure

A N.A.B Neglect -

Seismic forces may be applied separately

in each of two orthogonal directions.C 5.2.6.3.1

Apply in critical direction

D 5.2.6.4.1 or

100% x-direction 30% y-direction

E,F 100% y-direction 30% x-direction

Orthogonal Loads

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83

Combination of Load Effects 5.2.7

DSQE DSE 2.0=

The effect of seismic load Edefined below as for loadcombinations in which the effects of gravity loads and seismicloads are additive (+) or for load combinations in which theeffects of gravity counteract seismic load (-)

The effects on the structure and its components due to gravityloads and seismic forces shall be combined in accordance with the

factored load combinations as presented in ASCE 7-95 except thatthe effect of seismic loads, E, shall be defined as:

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84

Combination of Load Effects 5.2.7

DSQE DSE 2.0=

E = the effect of horizontal andvertical earthquake induced forces,

SDS = the design spectral response accelerationat short periods,

D = the effect of dead load,

QE = the effect of horizontal seismic forces, and = the reliability factor.

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85

Reliability Factor,

Added to encourage redundancy in lateral load

resisting systems.

This added redundancy will provide multiple load

paths in structure prior to formation of failuremechanism

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86

Reliability Factor,

Seismic Categories A, B, and C

Seismic Category D

0.1=

xAr

xmax

202 =

5.2.4

5.10.1

Ax = the floor in square feet of the diaphragmlevel immediately above the story.

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87

rmax

rmax= the ratio of story shear resisted by the singleelement carrying the most shear force in the storyto the total story shear, for a given direction ofloading.

rmax= maximum element shear / story shear

For moment frames, maximum element shear is the summationof the shear in any two adjacent columns in the plane of a

moment frame. If a column is common to two bays withmoment resisting connections on opposite sides at the levelunder consideration, 70% of the shear in the common columnis used.

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88

rmax

For braced frames, the value of rmax is equal to thelateral force component in the most heavily loadedbrace element divided by the story shear.

For shear walls, maximum element shear is equal to

te maximum value of the the product of the wallshear multiplied by 10/lw, where lw is the wall lengthin feet.

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89

Example - Braced Frame

5.0max

=x

r

For a braced frame, the element-story shear ratio is determinedas shown below.

Assuming that each brace resists the seismic shear equally,the maximum element-story shear ratio is:

Brace

30

60

V/2 V/2

0.160305.0

202 =

=

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90

Example

30

60

V/6

V/6V/3

V/3

33.0max

=x

r

5.0603033.0

202 =

=

Special MomentResisting Frame

Use 1.0

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91

Example

100

Shear Wall

200

V/2

V/2

05.0100/5/105.0max === wlr x

mimmum1

83.22

20010005.0

202

=

=

=

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92

15

15

Load Combination - Ultimate Strength Design

75

2010 k

62.5 k

154.6 k

154.6 k

16.17 k 32.34 k 32.34 k 16.17 k

Dead

LoadLiveLoad

SeismicLoad

Determine the maximum and minimumultimate strength design loads acting onthe column footings of the special MRF

shown here. Frames are 25 feet oncenter, the floor live load is 125 lb/ft2

and the roof live load is 20 lb/ft2.Neglect any reduction of live load due tosupported area. Use SDs = 1.0.

21.56 k

10.78 k

32.34 k

33.0)56.213/(56.21max

==x

r

0.1

44.0

752033.0

202

=

=

=

use

DSQE DSE 2.0=

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94

Load Combination - Ultimate Strength DesignContd

= 2/)( DxWD

The force in one column due tothe effects of dead load is

WDx = dead load at level x

6.154

2/)2.1542(

=

=D

kips

= 2/)( LxWL

The force in one column due tothe effects superimposed floor load

WLx = floor live load at level x

25.31

2/)5.62(

=

=L

kips

2/rr WL =

The force in one column due tothe effects superimposed roof load

Wr = roof live load

52/)10( ==rL kips

10 k

62.5 k

154.6 k

154.6 k

Dead

Load

Live

Load

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95

Load Combinations

8.67)9.304.400.1(0.1)6.154(9.0

)2.0(0.19.00.19.0.6

288)25.31(0.1)9.304.400.1(0.1)6.154(2.1

0.1)2.0(0.12.10.12.1.5

2.138)5(5.0)25.31(6.1)6.154(2.1

5.06.12.1.2

1

=++=

=+++

+++=++

=++

++

DSQDED

LDSQDLfED

LLD

DSE

DSE

r

ASCE7-95, Sect 2.3.2

kips

kips

kips

Governs

Governs

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96

Inelastic Force-Deformation Curve

0

dR

R

dC

YV

SV

EV

Elastic Response of Structure

Successive

Yield Hinges

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97

Torsion

)05.0( DVeVTtat

+=+=

For category C, D, E, and F with Type 1 TorsionalIrregularities:

taxt

T +=

3.0Aavg

x

=

2

max

2.1

Mt= Torsional MomentMta= Accidental Torsional Momentmax= Maximum Displacement at level xavg= Average of displ at extreme pts of

the structure at level x

5.3.5.3

5.3.5.3

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98

Torsion

+

=2Rd

deV

RR

VRV Tx

CA

x

AL

Base shear of the moment frame

Rigidity of frames are designated by R.

For this figureRD =RB andRA =RC

Torsion Vx eTis resisted by the rigidity of all frames

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99

Overturning

Top

10floors

10

floors

0.1=

0.1=

8.0=

8.0=

= =n

xixiix hhFM )( 5.3.6

Foundationuse =0.75 for all floors

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100

Reason for

NEHRP 97 - Commentary - Section 5.3.6

The distribution of design story shears over height computed from the lateral forcesis intended to provide an envelope since the shear in all stories do not attain theirmaximum simultaneously. Thus, the overturning moments computed statistically fromthe envelope of story shears will be overestimated.

It is intended that the design shear envelope to be conservative. If the shear in aspecific story is close to the exact value, the shears in almost all other stories almostnecessarily overestimated. Hence, the overturning moments statically consistent withthe design story shears will be overestimated.

Under the action of overturning moments, one edge of the foundation may lift from

the ground for short duration of time. Such behavior leads to substantial reduction inseismic forces, and consequently, in the overturning moment.

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101

Overturning

Intended ductile responseWhy ductile detailing if you can use it?

Possible brittle failure

Rocking Response:Large displacements

Low energy dissipationCompatibility problems with

the rest of the structure

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102

Rayleighs Method for Determining Period

=

=

=n

iii

i

n

ii

Fg

wT

1

2

12

F1

F2

F3

Since the deflection has been determined for the preliminary design, a

more accurate T may be computed using Rayleigh's method as:

where:

n = number of floors

Wi = weight of each floor (kips)di = deflection of each floor relative

to base of structure

g = 32.2 ft/sec2

Fi = lateral force at each level (kips)

auTCT

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103

Drift

xBxT =

IC xed

x =

Cd = Deflection Modification Factor

xe = Deflection determined by an elastic analysis

EBF (Mom. Resisting Connections, at columns away from Links) 4Ordinary Reinforced Concrete Shear Walls 4

Special Masonry Moment Frames 5Special Moment Resisting Frame - Steel 5 1/2Special Moment Resisting Frame - Reinforced Conc. 5 1/2Ordinary Moment Resisting Frame Steel 3 1/2

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104

Drift

Verify that building elements can accommodate

interstory drift,

Verify that control joints and clearances to

adjacent structures can accommodate totaldisplacement,

Consider P-Delta if needed.

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105

Allowable Story Drift

Table 5.2.8

If Trather thanTa is used in calculating x, the limit of the T < CuTa doesnot apply. This implies that deflection is not as critical as strength.

)/(/

1

IRT

S

IR

SC DDS

s =

T auTCT N.A.

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106

Example (contd)

Allowable interstory displacements: Seismic Use

Group II

inhsxa

12.3)1213(020.0020.0 ===

xe

xexed

x I

C

4.425.1

5.5

=

==

5.5=d

C

in709.04.4

12.3=

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107

P-Delta Effects

if the stability coefficient, is greater than max, thestructure will be unstable

If 10.0 neglect

25.05.0

max =

dC

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109

P-Delta Effects

Ifmax

10.0

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Steps in Design

Select Preliminary Members

Calculate weight of structure, W

Calculate total seismic base shear, V, utilizing accelerationresponse spectrum, W and T

Distribute base shear to structure as a series of static forces andperform a stress analysis and determine drift

Check resulting stresses, displacements, and stability

Satisfy specific special design and detailing requirements based onseismic performance category

Overturning demand is based on lateral loads that are less thanultimate of lateral load resisting system