Advance Design of RC Structure

Lecture 2

University of Palestine

Dr. Ali Tayeh

Seismic loads

There are two commonly used procedures for specifying

seismic design forcesThe equivalent static force procedureDynamic analysis

General Introduction

The seismic forces in a structure depend on a

number of factors including the:-Size & other characteristics of the earthquake Distance from the faultSite geology Type of lateral load resisting systemThe use & the consequences of failure of the structure

There are several analytical procedures to determine the magnitude of the base shear for which buildings must be designed, we will

only consider the equivalent lateral force procedure

The equivalent static force procedure In this method the inertia forces are specified as static force using empirical formulas.The formula developed to adequately represent the dynamic behavior of what are called “regular” structures.This method of analysis can be used if the structure is:

Regular structure under 240 feet (73meters) tall Irregular structure under 65 feet (20meters)

Regular structure means;A structure having reasonable

uniform distribution of a mass

and stiffness

►Uniform shape

►Uniform statical systemIrregular Structure Irregular Structure

Dynamic Analysis

A dynamic analysis can take a number of forms, but should account for the irregularities of the structure by modeling its “dynamic characteristics” including natural frequencies mode shapes and damping.

This method of analysis can be used if the structure is;Regular over 240 feet tallIrregular structure over 65 feetLocated on poor soils & have a period greater than 0.7 seconds.;

Design Base shear V

VC IV W

RT

T is the fundamental period of the structure in the direction under consideration

I is the seismic important factor

CV is a numerical coefficient dependent on the soil conditions at the site & the seismicity of the region

W is the seismic dead load

R is factor which accounts for the ductility & over strength of the structure system

Design Base shear VThe design base shear need not exceed :

2.5 aC IV WR

And cannot be less than

0.11 aV C IW

Where Ca is another seismic co-efficient depend on the soil conditions at the site & regional seismicity

Additionally in the zone of highest seismicity (zone 4) the design base shear must be greater than

0.8 vZN IV W

R

Where Nv is a near-source factor that depend on the proximity to & activity of known faults near the structure. Also Nv used in determining the seismic co-efficient Cv for building located in seismic zone 4

Seismic Zone FactorThe zone for a particular site is determined from a seismic zone map. The numerical value of Z are:

Zone12A2B34

Z0.0750.150.20.30.4

Important FactorThe importance factor I is used to increase the margin of safety for essential and hazardous facilities

See Table 1 in the handout sheets

Building Period3

4t nT C h

Where

Ct = 0.035 for steel moment frames

0.030 for concrete moment frames

0.030 for eccentric braced frames

0.020 for all other buildings

hn = The height of the building in feet

Fundamental natural period =

Structure System Coefficient RThe structural system coefficient, R is a measure of the ductility and over strength of the structural system, based primarily on performance of similar systems in past earthquakes.

See Table 2 in the handout sheets

Seismic Dead Load WAll the dead load of the structure including the partitions, total load not less than 10psf (0.48kN/m2), plus 25% of the floor live load in storage & warehouse occupancies Where design snow loads exceed 30 psf (1.44kN/m2) the design snow load shell be includedTotal weight of permanent equipment shall be included

Seismic Coefficients Cv & Ca

Depend on the expected ground acceleration at the site & that’s depend on the seismic zone & soil profile type

See Table 3 & 4

Soil Profile Type S

The effect of soil condition at the site on ground motion

See Table 5

Seismic Source Type A, B & C

It is used only in seismic zone 4 to specify the capacity & activity of faults in the immediate vicinity of the structure.

See Table 6

Near Source Factors Na & NvIt is used only in seismic zone 4 to determine the seismic coefficients Cv & Ca

See Table 7 & 8

Distribution of Lateral Force FXThe base shear V determined from previous equations are distributed over the height of the structure as a force at each level Fi, plus an extra force Ft at the top

1

n

t ii

V F F

The extra force at the top is

Ft = 0.07TV 0.25V if T 0.7 secFt = 0.0 if T 0.7 sec

The remaining portion of the total base shear (V-Ft) is distributed over the height including the top by the formula:

1

( )( ). x x

x t n

i ii

w hF V F

w h

Where W is the dead load of the (i) level including the partitions & 25% of the floor live load in storage & warehouse occupancies

h is the height of the (i) level above the shear base

Story ShearThe shear at any level x is the sum of all story forces at & above that level

1

n

t ii

V F F

Overturning Moment

The shear at any level x is the sum of all story forces at & above that level

1

( ) ( )n

x t n x i i xi

M F h h F h h

Resisting Moment = W B/2

Factor of safety = Re .1.5

. .

sisting Moment

OverTurning Moment

Force DiagramShear DiagramOver Turning Moment

Reliability/redundancy factor The seismic base shear determined from the previous equations must be multiplied by a reliability/redundancy factor for the later load resisting system

max

6.11 2 1.5

Br A

AB is the ground floor area of the structure in square meter

rmax is the maximum element-story shear ratio.

For shear walls rmax shell be taken as the maximum value of the product of the wall shear multiplied by 3.05/lw & divided by the total story shear, where lw is the length of the shear wall in meter.

For special moment-resisting frames, if exceeds 1.25, additional bays must be added.

Seismic Zone 0, 1 & 2 = 1

Displacement and DriftThe calculated story drifts are computed using the maximum inelastic response displacement drift (m), which is an estimate of the displacement that occurs when the structure is subjected to the design basis ground motion

sm R 7.0

S = design level response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design seismic forces.

Calculated story drift shall not exceed 0.025 times the story height for structures having a fundamental period of less than 0.70 seconds.

Calculated story drift shall not exceed 0.020 times the story height for structures having a fundamental period equal to or greater than 0.70 seconds.

Example 1Determine the UBC-97 design seismic forces for six story concrete shear wall office building. Located in seismic zone 3 on rock,. The story dead load is 850 kg/m2. live load 300 kg/m2

3m

6sto

ry

Elevation

7m

3

7m3

Plan

SolutionBase shear: VC I

V WRT

I = 1.0 Table 1 special occupancy structures

R = 5.5 Table 2 for shear wall-frame interaction system

Zone factor Z = 0.3 for seismic zone 3

Soil profile type SB Table 5 Rock ground

Cv = 0.3 Table 3 Z = 0.3 , SB

3

4t nT C h

Ct = 0.020 other building

►hn = 3m 6story = 18m 59ft

3

40.02 59 0.43secT onds

W = Dead load each floor area Number of story

►Presumed partitions and columns weight are accounted in the dead load per m2 . No live load would accounted with W

►W = 0.85 441 6 = 2249 ton

The Base shear

0.3 12249 285.3

5.5 0.43V ton

2.5 0.3 12249 306.7 285.3

5.5V

0.11 aV C IW

0.11 0.3 1 2249 74.2 285.3V

285.3V ton

2.5 aC IV WR

0.3aC

Vertical DistributionT 0.7 sec

Ft = 0.0

1

( )( ). x x

x t n

i ii

w hF V F

w h

1

(0.85 441).(18 15 12 9 6 3) 23615.6n

i ii

w h

( )285.3 0.012 ( )

23615.6x x

x x x

w hF w h

F1= 4.53 3 = 13.6 ton F4= 4.53 12 = 54.4 ton

F2= 4.53 6 = 27.2 ton F5= 4.53 15 = 68.0 ton

F3= 4.53 9 = 40.8 ton F6= 4.53 18 = 81.5 ton

If Ft 0. then the top floor will have two forces Ft + F6

Story ShearV6 = 81.5 ton V3 = 203.9 + 40.8 = 244.7 ton

V5 = 81.5 + 68 = 149.5 ton V2 = 244.7 + 27.2 = 271.9 ton

V4 = 149.5 + 54.4 = 203.9 ton V1 = 271.9 + 13.6 = 285.5 ton

Thus the shear force at the base = 285.5 ton

Overturning Moment

1

( ) ( )n

x t n x i i xi

M F h h F h h

M6 = 81.5 3 = 244.5 ton.m

M5 = 81.5 6 + 68.0 3 = 693 ton.m

M4 = 81.5 9 + 68.0 6 + 54.4 3 = 1304.7 ton.m

M3 = 81.5 12 + 68.0 9 + 54.4 6 + 40.8 3 = 2038.8 ton.m

M2 = 81.5 15 + 68.0 12 + 54.49 + 40.86 + 27.23 = 2854.5ton.m

M1 = 81.518+68.015+54.412+40.89+27.26+13.63 = 3711 ton.m

Thus the moment at the base = 3711 ton.m

Resisting MomentResisting moment = W Total B/2

= 0.85 441 6 21/2 = 23615,5 ton.m

Factor of safety=

Re . 23615,56,36 1.5

. . 3711

sisting Moment

OverTurning Moment

Summery

FloorWihiWihiFxVxMx

6374.85186747.381.581.5244.5

5374.85155622.7568.0149.5693.0

4374.85124498.254.4203.91304.7

3374.8593373.6540.8244.72038.8

2374.8562249.127.2271.92854.5

1374.8531124.5513.6285.53711.0

2249.1 23615.6 285.5

Elevation

13.6

27.2

40.8

54.4

68.0

81.5

3711

Shear Diagram Moment Diagram

2854.5

2038.8

1304.7

693.0

244.5

0.0