Lecture 6a -Earthquake Load by guzalg

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									CE-4102: Earthquake Load –
UBC 97
By Dr. Alberto Guzmán
Basic Definitions
Earthquake Load
 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

								
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