ML20040B743

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Computer Re-Evaluation of Reinforced Concrete Masonry Walls Program:Block Wall.
ML20040B743
Person / Time
Site: Grand Gulf Entergy icon.png
Issue date: 01/19/1982
From:
MISSISSIPPI POWER & LIGHT CO.
To:
Shared Package
ML20040B716 List:
References
PROC-820119-01, NUDOCS 8201260377
Download: ML20040B743 (17)


Text

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l APPENDIX J COMPUTER RE-EVALUATION OF REINFORCED CONCRETE MASONRY WALLS 4

PROGRAM: " BLOCK WALL" m

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l TABLE OF C0!!TD;TS I

1. Introdoction 1.1 Determination of Section State (Cracked vs. Uncracked) 1.2 Seismic Analysis 1.3 Stress and Deflection Calculations 1.4 Governing Codes

. 2. Analytical Procedure 2.1 Elock Wall Stress Calculation 27 Eigenvalue Solution and Response Calculation

3. Computer Program 3.1 Flow Chart-of the " Block Wall" Progran

'3.2 Hand Calculation for Conputer Verification

, 3.3 Co=puter Calculation 3.4 Comparison Between Hand Calculation and Computer Calculation 9

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. 1. INTROD'JCTION ,

A f ortran co= peter code " Block Walls" has been developed to analyze . block walls for axial load and flexural effects due to external and/or seismic loading. The block wall is analyzed as a simplified three degree of freedom beam model. The modal analysis technique is used in conjunction with the response sp'ectrum method to obtain the seismic response of the wall model. An iterative method is used to

' determine the actual stress and section properties (ef fective moment of inertia) of a vall section. Convergence criteria is established to verify that the assumed section condition results in the same inertial loading for two seccessive iterations.

j The working stress method for concrete analysis is used for stress calculations.

Finally, the calculated stresses are checked against the established a110wables.

I Determination of Section State (Cracked vs. Uncracked) 1.1 Iteration Procedure.

1. For the first iteration, the wall is assumed uncracked.

I 2. As a result of Step 1, and based on the calculated inertial forces, the section is checked for cracking. .

3. If cracked conditions exist, an effective moment of inertia is determined

, using the following AC1 Formula: 4 IH er ) 3 IMer I 3 l e

- It + I ~ l er (Ha j , (M a j ,

,' n,.

g g, k# j

~

where,

- Hre = Uncracked moment capacity.

t

. M, = A'pplied maximum moment on the wall.

i lt = Moment of inertia of transformed uncracked section.

I cr - Moment of inertia of the cracked section.

fr = Modulus of rupure.

1 y = Distance of neutral. plane from tension face. i s 4. A new iteration is initiated tesrecompute the frequencies, mode shapes and modal participation factors.

5. The procedure is repeated until convergence is achieved.

1.2 Seismic Analysis ,

The wall is represented by a three degree of freedom simplified bean model. ,

A response spectrum analysis is performed yielding the ine'rtial loading to be toposed on the system.

4-Four types of end conditions are allowed for the beam model used to perf orn the analysis as shown schematically below:

Mass Point *M y My M ,

. _ _3 p

//// ////

f. L/.4 _f_.L/4 _. j _.. ..L./4 .f.__.L/4 .f

. M M M 2 3

_ , , 4

/\

jf riusuu M M M g

_I 2 _3 y d, . - ,

d y M M 2 3 3 _ _ 4 g

j .L/3 J /3 L/3 y

. ./ . _[

1.3 Stress and Deflection Calculations The stress calculations are perf omed f or the final configration of the ,

section using working stress methods. Based on inertial loads, applied external loads, and the computed section .stif fness, the beam model deflection is determined.

_ 1.4 Governing Codes

1. ACI 531-79 and commentary.
2. Uniform Building Code,1970 edition.
3. Other codes as specified.
2. ANALYTICAL PROCEDURE 2.1 Block Wall Stress Calculation The governing equations for block wall stress calculations are developed using a working stress approach.

=

~fifective Width fffective fidth .

SP P N.

., \.

s - .

h C

4 M,_ L ,

g

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1 y_

1 A

- e

., YICR +- -AS *

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,_S Cracked Section  ; Uncracked Section ps

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Idealized Section for Analysis

_a_ l 1

The section properties are calculated based on a transformed section with the

.- block material as a base. Using the standard concrete analysis equilibrium concept namely:

)[ FORCES =0 or Tension = Compression

-M = Section Internal Moment Moment The folloying equations for stress calculation for bending are obtained: Case A : Uncracked section fMB = (M/1UCR) x YCU - f ST = NS:t x (M/10CR) x (YTU-DS) fSC = NS!! x (M/IUCR) x (YCU-DP) Case B: Cracked section fMB = (M/ICR) x YCCR fST = NSM x (M/ICR) x (YTCR-DS) fSC = NSM x (M/ICR) x (YCCR-DP) Note 1. For both Casc A and Case B the axial compression stresses are calculated and interaction is checked. (fMA/Tm) + (fMB/T!!B)51.0

2. For axial tension it is assumed that the reinforcing steel only
    ~

carries the tension. Definition of variables used in the above equations: 21.* = Bending moment T:!B = Allowable masonry compressive stress due to bending FMA = Allowable masonry compressive stress due to axial force fMB = Masonry compressive stress due to bending fMA = Masonry _ compressive stress due to axial force IUCR = Uncracked moment of inertia

                                 .1CR = Cracked moment of inertia YCU = Distance to extreme fiber in compression (uneracked)

YTV = Distance to extreme fiber in tension (uneracked) YCCR = Distance to extreme fiber in compression (cracked) YTCR = Distance to extreme fiber in tension (cracked) AC = Transformed, compressive area of section NS:1 = Modular ratio for steel-2.2 Eigen Value Solution and Response Calculation The following two matrices are determined based upon boundary conditions and structural properties. Flexibility matrix = IF] Mass Matrix =NN

1) Calculate transformation matrix NidN] = N1V3 /2 ] .
2) Using Causs elimination technique with column pivoting, calculate the structural stiffness matrix.

[k] = [F~I] ,

3) Calculate transformed stiffness carrix [k] such that:

Ik] = N%) Ik1 N%17

4) Tridiagonalize [k] using Householder's method and evaluate the characteristic
                 .       value equation:

t ik) (Q 1) + W2 1 (Q3 ) = 0

5) Calculate eigenvalues using Sturn sequence on the tridiagonal matrix.
6) Calculate eigenvectors using Wilkinson's method on the tridia;;onal matrix.
7) (Wi ) are the eigenvalues . for the untransformed stif fness matrix [k).

Calculate the frequencies: f 1 = W1 /2K .

8) Eigenvectors { Qil must betransformedintothevectors((ilofthe untransforced matrix:

{ li l - twx1 1 o1 l .

9) Co=pute codal participation factors:

n - - - - 1

                                        ~

1 ij yih . . .

10) The modal values of the inertia forces {Plg at the dynamic degrees of freedom for the ith mode are given by:

{Plg=(R)(ai)[${hgl i

                              'R g    = Participation factor for the i th mode th mode a3 = Acceleration for the i

{fgl = Mode Shape for the i th mode ll) Using the calculated inertial loads and the seismic monents, shear and the corresponding deflection are calcula,ted using the SRSS method since the modes are not closely spaced. 69 l

t 1.0 COMPUTER PROGRAM 3.1 Flow Chart of the " Block Wall" Procram Start , Read Problem Title Define Initial Conditions of Section No External Load Type O External Load Applied Type 1 If KType EQ. 0 YES 102 t Read Axial Load P

 ~

Read Shear Force V Read Bending }!oment M

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Read Section Properties: 102 Read }!aterial Properties: Read Properties for Stress Calculations If Seis=ic Consideration is Not Required YES 103 Input Floor Response Spectrum as a 2 D Array FRS (I, J) Frequency versus . Acceleration -

e 103 Input Boundary Conditions for Sicplified Beam Element (~

1) , Calculate the first three Frequencies of the beam model 500 11) Ext ra ct the "C" values iii) Use SRSS to co=pute final inertia loads.

Determine Total Bending Compare with moment from previous step. Is convergence satisfied? YES 600 Calculate -

                                                                                  . Stresses I                                                       .

Co to 500, next iteration. Maximum number of iterations = 10 600 Calculate masonry co=pression stress, tensile steel stress, co=pression steel stress. Print Stresses. Compare with allowables. Flag overstressing. Stop m

'               3.2 Mand Calculation for Computer Verification

! Assume two core masonry units, 44% solid by volume with running bond. Nominal thickness is 12 inches, with two number 5 vertical reinforcing bars at 16 inches spacing. Exa ct dimensions are: . 11 5/8" x 7 5/8" x 15 5/B", t, = 1.25,- tw - 1.12" A. Uncracked section properties

                /

Transfore all caterials to block material:

     ,                             steel E,    29x10 6

grout E c 1.4x10 6 n1=

                                             =                 = 29        ;     n2 "                 "

6

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  • 6 block E n 1.0x10 block I , 1x10 Tensile steel area As = 0.31 sq. inches Tension steel cover D's = 3.37 5 inches Thickness of the wall H = 11.625 inches Ef fective width of beam beff = 15.625 inches .
     . '                                                                                                         15.625                        i I                                             I s

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                                                    .- '.~ '>                                                  !            . g '-    :

in a Ei E( .g y s - 7 .

                   .                                   .s-                               o                   a 6

p - g g a . n 4 , .

                                                                *-                         U                   l s. 3, a L                                                                              .                                      :

g 3.36 8.59 . 15.625 l- '~ ~8 D i Assumed Section Transformed Section 4 Uncracked moment of Inertia = tI = 1096.2 in &

3. Cracked sect!on properties 15.625"  !

I I I L 0

                   -                                     n                                             d i

l l l 11.95 .l Cracked co=ent of inertia = l er

                                                                                       = 326.7 in' C. Calculation of Effective Area (Axial & Shear)

Reference ACI 531-79 Codt AAXIA1. - 2(1.12 + 6.1325 + 1.12) 1.25 + (9.125x1.12)3 + (6.1325x9.125)x1.4 -

                                + 2 x (6.1325 + 1.12) = 144.4 in2 D.      Calculation of Shest Area                                         f Reference ACI 531-79 Code ASHEAR = (11.625 - 3.875) 1.12x3
                                + 2x (6.1325+1.12) + 1.4 (11.625 - 3.875 - 1.25) 6.1325
                                - 26.04 + 15.33 + 55.8 = 97.2 in2-b
                                           ,                     eff                              1 y.

g .v '9 , .,s ; v d,  :...

                                                                      .    .c.. ...:Q
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                                  .          y.N                  Q                  .       ,

n. 3 l'

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                                                                            *".'". * ,       Q, e,

nl I 6.1325 . D Note: The above calculations are for uniform inertia loadings. 8 O i .

         . .      g' I. Dyna =ic Inertia Loading ,

For a 12 inch wall grouted at 16 inches on center, the average weight of i a completed wall is 111 lb/ft2

                                                                                                    ~

wt/ unit length . Illx16 = 12.3 lb/in 144

   '                             '~         7T n                7T                   6 tir                  1.6x10 x1096.2x3ss.4 I

2(L)2' Ay 12.3

                                                                                              =   5.98 cps 2(240) d[    .
                       ~

l Acceleration = 0.28g Inertia loading intensity k'i = Acceleration x vt/ unit length

                                                                       = 0.28 x 0.0123 Scismic mo=ent . (0.28 x 0.0123) (240)2 = 24.79 in-kips                     '

8 F. Deter =ine the maxi =ue bendinr stress i Tension = 29 x 24.79 x (7.846 - 2.62) = 11.5 kai

      .                                                  326.75 t

Co=pression - (24.79 x 2.528) = 0.192 kai 326.75 . . . w I e l . . I

                            '. * ! L _. . _ _ _ _          ._~...

3.3

  • Computer Calculation
          -                                    ***eBLOCK UALLS PROGRAM *A.
                                               *e**  VERSIDH 6 2/12/01 see*0UESTIONS SHOULD BE ADDRESSED 10 eses sees E. AKK0USH    GPD X 3196        ****

esse S. CLOSE GPD X 3196 esse

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  • UNITS XIPS INCHES
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INPUT PROBLEM TITLE (UP 10 10 CHARACTERS)

     > EXAMPLE
!        DEFINE INITIAL CONDITION OF SECTION f

NO EXTERNAL LOAD APPLIED TYPE O

     ~ I
            . EITERNAL LDAD IS APPLIED TYPE 1
     >0 INPUT SECTIONS PROPERTIES AS, ASP,DS,DP,H,L,BEFF, HEIGHT IN.             .
     >.31,,2.62,,12.,2/0.,15.6,240.

INPUT IUCR,1CR,YCU,Y10,YCCR,YTCR, AAXI AL, ASHEAR, AC - UHERE: IUCR=UNCRACKED INERTIA ICR= CRACKED IHERTIA YCU=DIST. TO EXTREME FIBER IN COMP.(UNCRACHED) YTUvDIST. TO EXTREME, FIBER IN TENSION (UNCRACKED) , YCCR=DIST. TO EXTREME FILER IN COMP.(CRACKED) YTCR=DIST. TO EXTREME FIBER IN TENSIOH(CRACKED) AAXIAL= EFFEC 11VE AXIAL AREA ASHEAR= EFFECTIVE SHEAR AREA AC= TRANSFORMED COMPRESSIVE AREA 0F SECTION ' .

>1096.22,326.74,4.84,5.535,2.528,7.846,144.4,97.2,34.8 3

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  • INPUT YOUNG MODULUS , AVERAGE UT. PER UNIT LENGTH AND MODULAR RATIDS
             >1400.,.0123,29.,1.4
                                                       , COMP. STRENGTH OF GROUT ll                 INPUT' COMP. STRENGTH OF MASONRY AND       Y1 ELD STRENGTH OF REINFORCING STEEL                               j
              >1.,1.8,40.                                                                       (

DEF AULT AllouAILE STRESSES ARE ACI 531-79** IF ACCEP1ABLE TYPE 0 IF UNACCEPTABLE TYPE 1

               >0                                                                               ;

4 CNECK IF SEISHIC LOADING IS TO BE CONSIDERED

       .                IF OBE SEISHIC CONSIDERATION IS REQUIRED TYPE 1
        .(              IF SSE    SEISMIC CONSIDERA110N IS REQUIRED TYPE 2 IF SEISn1C CONSIDERATION'IS NOT REQUIRED TYPE O
                  >2 4

INPUT FLOOR RESPONSE SPECTRUM SPECTRUM INPUT IS A 2-D ARRAY DEFINING FREQUENCY INCPS V TYPE *No NunLER OF P01HT USED TO DESCRIBE THE CURVE T

                   >B                    ,

1 INPUT 8 SET OF FREQUENCY VS ACCELERATIONS ENTRIES EACH CN A NEU

                    ).2,.12
                     >1.2,.34                                              .
                     >2.,2.45
                     >2.6,2.45                                                        ,
                     >2.8,.75
                     >3.5,.75
                     >5.99,.28
                     >1000.,.28 INPUT ALD1110HAL UEIGHTS AT MASS PTS. 1.2,3
30. ,0. ,0. t * .

O

l BOUNDARY CONDITIONS ASSUMED FOR SIMPLIFIED BEAM MODEL S.S BOTH ENDS TYPE 1 S.S ONE END FIXED THE OTHER TYPE 2 BOTH ENDS FIXED TYPE 3 1 SIMPLE CANTILEVER TYPE 4 1

    >1

+ cesseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeese e*ee DATA FROM INTERNAL STORAGEeee* ese* BLOCK UALLS PROGRAMe**

                                         **ee VERSION 6 2/12/81 i

esse 00ESTIONS SHOULD BE ADDRESSED T0eese

                     -                   see E. AKKDUSH        GPD X 3196          *es*

t e*** S. CLOSE GPD X 3196 se**

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see. T. JOSEPH GPD I 3192 e**e

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  • UNITS KIPS INCHES *
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                                        **** PROB. TITLE: EXAMPLE         *ess
                                          **** SECTION PROPERTIES ***e AS= .31   ASP = .00      DS= 2.62        DP=   .00
  • H= 12.0 L=240.0 B= 15.6 D= 7.8 J
           *=                                                                             *
                                             *=*INPU1 FOR STRESS CALCULATICNe*e IUCR=UNCRACKED INERTIA =      1096.22 ICR= CRACKED INERTIA       326.74 YCU = DIST . 10 EXTREME FIBER IN COMP.(UNCRACKED)= 4.840 YTU=DIST.10 EI1REME FI)ER IN TENSION (UNCRACKED)= 5.535 YCCR=DIST.10 EX1kEME FIBER IN COMP.(CRACKED)= 2.528 Yith=DIST. 10 EXTREME FILER IN 1ENSI0d(CRACKED)= 7.846
   .      AAZIAL= EFFEC 11VE AXIAL AREA =      144.40 ASHEAk= EFFECTIVE SNEAR AREA =         97.20 AC=1RANSFORMED COMPRESSIVE AREA 0F SECTION=

34.B0

                                              **** MATERIAL PROPERTIES *ise YOUNG MODULUS =        1400.00 AVERAGE UT. PER UNIT LENG1H= .01230000 MODULAR RATIOS = 29.0 1.4 COMPRESSIVE STRENG1H OF MASONRY =          1.0              g COMPRESSIVE STRENGTH OF GROU1=          1.S                 ,

YIELD OF REINFORCING STEEL = 40.0 k

           ** SSE SEISMIC CONSIDERATION FOR THIS PROBLEM **

f FLODR RESPONSE SPECTRUM DEFINITION F G

                                                      .20          i          .12 1.20                     .36 2.00                    2.45 2.'60                   2.45 2.80                     .75 3.50                .
                                                                          ' .75
  • 5.99 .28 1000.00 .29 ADDITIONAL UEIGHTS Al MASS P15. ARE:

ADDul= .000 ADDU2= .000 ADDU3= .000

ocPEAM MODEL IS S.S AT BOTH ENDSee ce* FREDEHCIES ARE *** 5.989 23.790 50.511 ce* MODAL PARTICIPATION FACTORS ARE*** .07 .00 .01 ces ACCELERATIONS ARE *** .200 .280 .280 ( c8* SEISMIC HOMENT= 25.9 KIPS.IH

                                                *****RESULTS OF ANALYSISessee MASONRY COMPRESSIVE BENDING STRESS =           .2007KSI               ALLOUABLE =   .825KSI MASONRY AXIAL COMPRESSIVE STRESS =           .0000KS1                 ALLOUAPLE =   .394KS!

TENSILE STEEL STRESS = 12.030$KSI ALLOUABLE = 36.000KSI EOMTRESSIVE STEEL STRESS = .0000 KS! ALLOUABLE = 36.000KSI HA50NRY SHEAR STRESS = .0031KSI ALLOUABLE = .05BKSI mat! MUM DEFLECTION = .093572 IN. PO YOU UANT TO RUN PLDCK UALL AGAIN YES TYPE 1 40 TYPE 0

        ~> 0 r,

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                             ,b "

3.4 comparison Between 11and calculation and computer calculation Block Wall Protram Hand calculation Natural Frequencies (CPS) 5.99, 23.79, 50.51 5.98 Seismic Accelerations (g's) 0.28, 0.28, 0.28 0.28 Seismic Moment * (in-kips) 25.9 24.79 - l:

  ,                                           Masonry Compressive Stress * (psi)               201                           192 Reinforcing Steel Stress * (psi) ,             12030                          11500
  • Note: The program calculates a higher value due to the inclusion of the second and third modes.

I  ! k g h d e 4 f l e e I J.

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