Lecture 6a -Earthquake Load by guzalg

VIEWS: 1,472 PAGES: 33

• pg 1
```									CE-4102: Earthquake Load –
UBC 97
By Dr. Alberto Guzmán
Basic Definitions
Three procedures to determine seismic load:
◦ Static Force Procedure
◦ Simplified Design Base Shear
◦ Dynamic Analysis Procedure
For any of those procedure is needed:
◦ The Seismic Zone
◦ The Importance Factor
◦ The Soil Type
Static Force Procedure
Static Force Procedure (Cont.)
This is stated on UBC-97 section 1630.2

T is the fundamental structure period
W is the total structure weight
Ca, Cv are Seismic Coefficients.
R is the Response modification factor
Static Force Procedure (cont)
Structure Period, T:
Simplified Procedure
Lateral Distribution
One the base shear have been obtained:

◦ Horizontal Distribution: Base shear distribution
on all directions and for all Lateral Force Resisting
System.

◦ Vertical Distribution: Distribution for each Lateral
Force Resisting System through elevations.
Lateral and Vertical Distribution
Vertical Distribution: Static Force Procedure
Vertical Distribution (Cont.)

Vertical Distribution: Simplified Procedure
Horizontal Distribution
1      2        3

A                       1        V/2

Vtotal

B                       2        V/2
Horizontal Distribution
1         2            3

A

Interior Frames have double
tributary area than exteriors

B

1         2   V/2      3

Vtotal       V/4
V/4
Horizontal Distribution
Vtotal

VB = 2VA

VA              VB          VA

Vtotal = 2VA + VB
Vtotal = 2VA + 2VA
Vtotal = 4VA
Vtotal        Vtotal
VA =        → VB =
4             2
Seismic Coefficients, Cv and Ca
◦ Seismic coefficient Cv can be obtained on Table 16-R

◦ Seismic coefficient Ca can be obtained on Table 16-Q
Design Response Spectra
Response Modification Factor, R
The RMF, R is set for on Table 16-N

1.

2.

3.

4.
Seismic Zone Factor, Z and
Soil Profile, S
Seismic zone factor, Table 16-I

Seismic Soil Profile, Table 16-J
Seismic Importance Factor, I
Seismic Zone, UBC-97
Table N: Response Modification
Factor, R
Table N: Response Modification
Factor, R
Table N: Response Modification
Factor, R
Table N: Response Modification
Factor, R
Simplified Method
The structure shown bellow will be the housing for a new toxic
and explosive chemical plant facility located in San Juan, Puerto
Rico and was built over a very dense sand. Determine the
earthquake load distribution for frame at axis 3.
A             B            C
30'- 0"       30'-0"

23'-0"
4

18'-0"
3

23-0"
18'-0"
2

23'-0"
18'-0"

1

Story         Weight
First         900 kips

Second        1300 Kips
Roof          650 Kips
Base Shear
According to 1630.2.3

From table 16-I~ Z = 0.3

From table 16-J ~ Soil profile = Sc
Base Shear (Cont.)
From table 16-Q ~ Ca = 0.33

From table 16-N ~ R = 8.5
Base Shear (cont).
3.0 xCa    3.0 x0.33
V=         W=           W
R          8.5
V = 0.116W
Horizontal Distribution
V = 2Va + 2Vb
V = Vb + 2Vb
V
Vb =
3
0.116 xW
Vb =
3
Vb = 0.0387W
Base Shear (cont)
Vertical Distribution:
◦ According to section 1630.2.3.3

◦ Thus:                             25.15k

◦ F1=0.0387x900 = 34.83 kips
50.31k
◦ F2=0.0387x1300=50.31 kips
◦ F3 =0.0387x650=25.15 kips         34.83k
Static Method
The following parameters are the same:
◦   Z = 0.3
◦   Soil profile = Sc
◦   Ca = 0.33
◦   R = 8.5
According to UBC: 1630.2
Vmin ≤ V ≤ Vmax
Cv I      2.5Ca I
0.11Ca I W ≤      W≤             W
RT          R
From table 16-K, the importance factor, I = 1.25
From table 16-R, the velocity coefficient , Cv = 0.45
From UBC:1630.2.2,
T = 0.03(69) 3 / 4 = 0.718 sec
Static Method
Thus,
Cv I     0.45(1.25)
V=       W=             W = 0.0922W             Controls
RT      8.5(0.718)
0.11Ca I W = 0.11(0.33)(1.25)W = 0.0454W
2.5Ca I      2.5(0.33)1.25
W=                 W = 0.121 W
R              8 .5
The total shear will become:
V = 0.0922 (650 + 1300 + 900) = 262.77 k
Them, the shear that need to be distributed on our
frames is (from previous example):
262.77 k             Horizontal distribution
V=             = 87.59 k
for frame at axis 3
3
Static Method
Once the horizontal distribution is obtained, the shear
have to be distributed vertically.           According to
UBC:1630.5 V = F + n F
t   ∑
i =1
Ft = 0 if T ≤ 0.7
i

Ft = 0.07 T V
Ft ≤ 0.25V

Ft = 0.07 (0.718)(87.59) = 4.40 k             Controls

0.25V = 0.25 (87.59) = 21.90 k
The shear should be distributed over the structure
height:           (V − Ft )Wx hx
Fx =      n

∑W h
i =1
i   i
Static Method
Thus, for V = 87.59k and Ft = 4.40k, the following
distribution could be found:
Wx h x                       (V − Ft )Wx hx
Floor     Weight        hi     Wi• hi     n
(Kips) Fx =       n
(Kips)

(kips)                         ∑W h                            ∑W h
(ft)             i =1
i   i
i =1
i   i

Roof       650         69     44,850           0.3578                    29.76
Second     1,300        46     59,800           0.4771                    39.69
First      900         23     20,700           0.1651                    13.73
∑        2,850               125,350                                    83.19

Simplified                            Static
Practice Problem

```
To top