advance design of rc structure lecture 2 university of palestine dr. ali tayeh seismic loads
TRANSCRIPT
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