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Content SEISMIC ANALYSIS OP A WALL-BEARING REINFORCED CONCRETE BUILDING A Thesis Presented to the Faculty of the School of Engineering University of Southern California In Partial Fulfillment of the Requirements for the Degree Master of Science in Civil Engineering by George Min-shui Lew June 195^ UMI Number: EP41921 All rights reserved INFORMATION TO ALL USERS The quality of this reproduction is dependent upon the quality of the copy submitted. In the unlikely event that the author did not send a complete manuscript and there are missing pages, these will be noted. Also, if material had to be removed, a note will indicate the deletion. UMI' Dissertation Publishing UMI EP41921 Published by ProQuest LLC (2014). Copyright in the Dissertation held by the Author. Microform Edition © ProQuest LLC. All rights reserved. This work is protected against unauthorized copying under Title 17, United States Code ProQuest* ProQuest LLC. 789 East Eisenhower Parkway P.O. Box 1346 Ann Arbor, Ml 481 0 6- 1346 Q L ' 5 ¥ J - ( e > H This thesis, w ritte n by / ......George. . M i n - s h u l L e w ................................................... under the guidance ofhl.S .F aculty Committee and approved by a ll its members, has been presented to and accepted by the School of E ngineering in p a rtia l fu lfillm e n t o f the re quirements fo r the degree of Master of Science in Civil Engineering Date. Faculty Committee TABLE OP CONTENTS CHAPTER PAGE I. INTRODUCTION......... 1 Purpose ............ 1 Description of the building ....... 1 Basis of design...................... 2 II. DETERMINATION OF EARTHQUAKE FORCES........ 4 Gravity inertia loading method ...... 4 Seismic coefficients .•.••...••• 5 Increase in stresses for seismic design . 5 III. DESIGN FOR VERTICAL LOADS ........ 8 Design of roof slab .......... 8 Design of floor slab .•••....••• 13 Design of walls ............... 14 Design of wall footing 21 IV. SEISMIC ANALYSIS .................. 24 Earthquake forces In north-south direction 25 Parapet wall 25 Earthquake force normal to Walls AD, BC, EH, and FG . . . ............. 26 Earthquake force normal to Walls DE and C F ..................... 28 iv CHAPTER PAGE Distribution of lateral force to shear Walls AB, CD, EF, and G H .. 30 Distribution of lateral force among piers in Wall AB ........... 37 Stresses in Wall AB.......... 38 Distribution of lateral force between piers in Wall C D .......... 42 Stresses in Wall CD.... ... 43 Distribution of lateral force among piers in Wall E F .......... 46 Stresses in Wall EF ............... 47 Distribution of lateral force among piers in Wall GH ............... 31 Stresses in Wall G H .............. 53 Wall footing pressure due to seismic forces •••••••..•••.•• 56 Earthquake forces in east-west direction 57 Earthquake force normal to Wall AB . . 57 Earthquake force normal to Wall CD . . 60 Earthquake force normal to Wall EF . . 62 Earthquake force normal to Wall GH . . 64 Distribution of lateral forces to Walls AH and B G ............ 66 CHAPTER PAGE Distribution of lateral force among piers in Walls AH and B G ......... 67 Stresses in Walls AH and BG*.*..* 69 Shearing stress at intersection of roof slab and shear wall ....... 72 V. CONCLUSIONS........................... 73 BIBLIOGRAPHY.................... 76 LIST OF TABLES TABLE PAGE I. Widths of Wall Footings............ 22 II. Weight "W" for Seismic Analysis ...... 24 III. Deflection of Piers of Wall A B ......... 31 IV. Deflection of Piers of Wall CD ••*... 32 V. Deflection of Piers of Wall EF ...... 32 VI. Deflection of Piers of Wall G H ......... 33 VII. Relative Rigidities of Walls ....... 33 VIII. Location of Center of Mass........ 35 IX. Deflection Due to Bending and Shear In Wall A B ............................. 37 X. Distribution of Lateral Force among Piers in Wall A B .................. 38 XI. Moment of Inertia of Net Section of Wall AB 38 « XII. Deflection Due to Bending and Shear In Wall C D ............................. 42 XIII. Distribution of Lateral Force between Piers in Wall CD ••••.••••••. 42 XIV. Moment of Inertia of Net Section of Wall CD 43 XV. Deflection Due to Bending and Shear in Wall E F ............................. 46 vii TABLE PAGE XVI* Distribution of Lateral Force among Piers in Wall E F .......... ......... 47 XVII* Moment of Inertia of Net Section of Wall EF 47 XVIII. Deflection Due to Bending and Shear in i i Wall G H .............. ... .......... 51 XIX. Distribution of Lateral Force among Piers in Wall G H ....................... . 52 XX. Moment of Inertia of Net Section of Wall GH 52 XXI. Deflection Due Bending and Shear in Wall AH or B G ................ .............. 67 XXII. Distribution of Lateral Force among Piers in Wall AH or B G ............... .. . 68 XXIII. Moment of Inertia of Net Section of Wall AH or B G ............ 68 I I LIST OF FIGURES FIGURE PAGE 1. Plan of Building ................... 6 j 2. Elevations of Building ..... .......... 7 ' 3» Lateral Load Normal to Walls AD, BC, EH, and : I . , . i ! F G 26 I i I 4. Moment Diagrams for Walls AD, BC, EH, and FG 27 j j 5. Lateral Load Normal to Walls DE and CF . . . 28 ' I I j 6. Moment Diagram for Walls DE and CF . . . . . 29 I 7. Wall Deflections 35 i 8. Shear Wall AB . . . . . 9. Stresses for Wall AB • 10. Stress Diagram for Pier 1, Wall AB ... 11. Shear Wall CD ....... 12. Stresses for Wall CD • 13. Stress Diagram for Pier 1, Wall CD ... 14. Shear Wall EF . . . . . 15. Stresses for Wall EF . 1 6. Stress Diagram for Pier 1, Wall EF ... 17. Shear Wall GH ....... 18. Stresses for Wall GH • 19. Stress Diagram for Pier 1, Wall GH . . . r o o • Plan of Footing along Wall AB • FIGURE PAGE 21. Lateral Load Normal to Wall AB ....... 57 22. Moment Diagrams for Wall AB......... 58 23. Lateral Load Normal to Wall GD ....... 60 24. Moment Diagrams for Wall CD......... 6l 25. Lateral Load Normal to Wall EF . ........ 62 26. Moment Diagrams for Wall EF . . . ........ 63 27. Lateral Load Normal to Wall G H ........... 64 28. Moment Diagrams for Wall G H ............... 65 29. Shear Wall AH or B G ...................... 67 30. Stresses for Wall AH or B G ........... 71 31. Stress Diagram for Pier 6, Wall BG . . . . . 72 I CHAPTER I INTRODUCTION It is obvious that severe earthquakes are highly destructive to life and property. The importance of designing buildings and structures to resist the stresses produced by lateral forces due to earthquakes can hardly be overstated. The first concerted action, taken by engineers and building departments in California to de sign structures to resist earthquake forces, was that which followed the Long Beach earthquake in 1933* To day, every building code in California requires build- ings and structures to be designed to resist earthquake forces. Purpose. The purpose of this thesis is (1) to analyze a one-story wall-bearing reinforced concrete building for seismic forces by a simple and practical method which is widely used in Californiaj (2) to indi cate that when an ordinary one-story wall-bearing re inforced concrete building is adequately designed for ! vertical load, the building usually can safely resist the seismic forces. i » i Description of the building. The building under I 2 consideration is a one-story reinforced concrete build- ; ing with dimensions as follows: length, 42’-4"; width, | 20*-8”; height, 13'-7 1/2”; thickness of exterior walls, j 8"; interior walls 6". Plan and elevation views of the building are shown in Figures 1 and 2, Pages 6 and 7» t Basis of design. The building is designed accord- : i ing to the 1954 Los Angeles Building Code. The ultimate compressive strength of concrete used in the analysis is f ( l = 2,500 psi. The allowable unit stresses are as l follows: j fc = Extreme fiber stress of concrete in ; compression = 1,1 25 psi v = Shearing stress of concrete = 75 psi u = Bond stress of concrete = 250 psi fg = Tensile stress of reinforcing steel = j 18 ,00 0 psi Soil pressure = 2,000 psf The building is first designed for vertical load, j Then it is analyzed to determine whether the wall thick- ! nesses and the reinforcement are adequate to resist the ] lateral force due to earthquake. j i The roof slab is designed as a two-way slab. j Method 2 of Section 710 of the American Concrete Institute’s "Building Code Requirements for Reinforced Concrete (AC1 318-51)'% for designing two-way slabs is used. To determine the concrete and steel stresses due to combined bending and axial load, Diagram 23, Page 434 of "Principles of Reinforced Concrete Construction, by P. E. Turneaure and E. R. Maurer is used. CHAPTER II DETERMINATION OP EARTHQUAKE FORCES i I I Gravity inertia loading method. The seismic forces applied to buildings and structures are determined by the gravity inertia loading method. Seismic forces are applied horizontally at the roof level for one-story ! buildings and at the roof level and each floor level above f I the foundation for multi-story buildings* These forces I acting on a building or structure are caused by the build-! ! 1 i i ing or structure itself. If the building or structure I had no weight, there would be no forces acting. Because ! of the inertia of its mass, the building or structure resists motion of the ground. From the principles of mechanics, it is known that a force is directly proportional to the mass and to the acceleration. Expressed as a formula: I F = g ! where F = force (in this case, the seismic force). i i W = weight of the building above the plane under j consideration. i a = acceleration of the building due to earthquake. ! g « acceleration due to gravity, 32 ft. per sec. 5 per sec. The ratio a/g in the above formula is known as the seismic coefficient. Representing this by C, the general seismic force formula becomes F = CW Seismic coefficients. In the Los Angeles Build ing Code the seismic coefficient C for a building acting as a unit is < ! 0 .6 0 j N + 4.5 1 | where N is the number of stories above the story under consideration. For parapet and other cantilever walls, C = 1.00. For bearing or non-bearing walls, C = 0.20. Increase in stresses for seismic design. When designing for seismic forces, a 33 1/3 per cent increase in the normal working stresses is allowed for the com bined dead, live, and earthquake loads. I I ■n ~r N i V —| —£ £ D £ :.| 1-1 I. M • T T77 rT H Pane/ I i i 1 1 < 1 H 1 1 1 1 Panel JE Panel ffl O ' « 0 b I 1 1 3 8" 14'-O" 6" - p a i r ~ ? r 12 -O ' ' 42''4" F /4'0" 8 / v y . / . P la n o f B 7 ' . Q . 9 V ,«*y . J. “ 1 b-1 - ! __ 44“ 4 '- O ' b-Z 4‘-0\ 20'-8 4-0“ Waff AB b 3 ______ I — I 4 '-4 ‘4 v -IV). r y b-i 5-0 " , 3 0 . hi /2'-8 ' 2 0 ' - S ' Wa/t CD § 0 N b-l b-1 4-4, 3r O' b-O 20 -8 ' b-3 3-0 W all EF . 4 - 4 ' ' Waft GH O ' " 5 . -mm ■ ir i i m i i i i n m i t J L nr K T ‘ M ii i i n i i n i i 1 1 ■ ■ s < M N 5 . § bf 5 -2' S'-O' 6 -6 ‘ 16 k 3 '-o " f: 6 l: 6 6'-6 5-0' 42'-4“ Wa/t AH or BG S '- 2 " N F ig . 2. F /e v a fio n s o f Boi/oiincj CHAPTER III DESIGN FOR VERTICAL LOADS Design of Roof Slab-1 - Panel I t = Perimeter 2 x 20.00 x 12 + 2 x 14.58 x 12 jgo = 155 = 4.6” Try t = 5 1/2”, di = 4 1/2", & d2 - 4" Live load = 20 Roofing = 10 5 1/2” slab = 69 Plaster ceiling = 10 Total w 109 #/sq. ft. m = *58 « o.7 20.00 Short-span (East-West) Middle 1/2 of span + M * = CwS2 - 0.062 x 109 x (14.58)2 x 12 = + 17,270" #/ft. 1 Frank Kerekes, Building Code Requirements for Reinforced Concrete (AG1 381-511, (Detroit: American Concrete institute, 1$5I), pp. 618-9. A - M - 17,270 - s fsJdi 18,000 x 0.857 x 4.5 0 .2 5 sq. In./ft. Use #4 @ 9 1/2 c.c. West 1/4 of span - M = CwS2 « 0.041 x 109 x (14.58)2 x 12 - 11,400” #/ft. As - M » ______11,400_______ = fgjdi 18,000 x 0.857 x 4.5 0 .1 6 sq. in./ft. Use #3 @ 8 1/2” c.c. East 1/4 of span - M = CwS2 = 0.082 x 109 x (14.58)2 x 12 - 22, 800” #/ft. A - M _ 12,800 , s fsjc3i 16,000 x 0.857 x 4.5 0.33 sq. in./ft. Use #4 @ 7” c.c. Long-span (North-South) Middle 1/2 of span + M = CwS2 = 0.044 x 109 x (14.58)2 x 12 + 12,220” #/ft. 10 A M 12,220 S ~ fsJd2 18,000 x 0.857 x 4 0.20 sq. in./ft. #4 @ 12” c.c. North 1/4 of span and south l/4 of span - M * CwS2 = 0.029 x 109 x (14.58)2 x 12 * - 8,060” #/ft. a = M = 8 ,0 6 0 — s fsjd2 18,000 x 0.85? x 4 0.13 sq. in./ft. #3 @ 10” c.c. Use 5 1/2” slab Shear per ft. width in the short-span middle strip Shear per ft. width in the long-span middle strip wS = 109 x 14.58 = 53Q# 11 v = ..X = _____ 53° .. = 13 psi bjd2 12 x 0.857 X 4 ^ H u « -2— = g— 53Q— _ « no psi jd20 0.057 x 4 x 1.4 Panel II Same thickness as Panel I. 12.5 n A “ = 20^ ' °-6 Short-span (East-West) Middle 1/2 of span + M = CwS2 = 0.059 x 109 x (12.5)2 x 12 + 12,100” #/ft. « _ M _ 12,100 8 fsJ^i ~ 1 8 ,000 x 0.557 x 4.5 0.17 sq. in./ft. Use #3 @ 8” c.c. East 1/4 of span and west 1/4 of span - M - CwS2 = 0.078 x 109 x (12.5)2 x 12 15*930” #/ft. Ac = M = 15,9^0_______ = s fsjdi l'8,000 x 0.857 x 4.5 0 .2 3 sq. in./ft. Use #4 @ 10 1/2” c.c. Long-span (North-South) Middle 1/2 -of span + M • CwS2 = ■ 0.03T x 109 x (12.50)2 x 12 = 7,580" #/ft. As = * - 7,580 fsjd2 18,000 x 0.857 x 4 0*12 sq* in./ft* Use #3 @ 11” c.c. North 1/4 of span and south 1/4 of span -M = CwS2 = 0.025 x 109 x (12.50) 2 x 12 - -5,110" #/ft. A. = M = 5,110_______ = s fsJd2 1 8 ,0 0 0 x 0.857 x 4 O.O83 sq. in./ft. Min. steel reg’d: As = 0.0025 hd = 0.0025 12 x 4 = 0.12 sq. in./ft. Use #3 @ 11" c.c. Shear per ft. width in the short-span middle strip w| x _ 109 X la-5 x _ 600# f = v = ______ 292,______= 13 psi bjdx 12 X 0.8 57 X 4 .5 13 u = — 3L_ = ___ --------- « 86 psi jdiO 0.85T x 4.5 x 1.8 Shear per ft. width in the long-span middle strip wS _ 109 x 12.5 _ ~~3 T v = V = ____i55_____ = ii psi bjd2 12 x 0.857 x 4 u “ Jd20 = 0.857^x^4 x 1.3 = 102 psl Panel III Identical to Panel I except the reinforce ment in the short-span is opposite hand. Design of Floor Slab The 1st floor slab is on fill, therefore, there is no bending stress need to be designed for. Use 4" slab. Reinforcement for shrinkage and temperature stresses: As = 0.002 x 12 x 4 = 0.096 sq. in./ft. Use #3 @ 12” c.c. each way. Place bars at center of slab. I 14 Design of Walls For Walls AB, GH, AH and BG, use thickness « = 8H. ! Minimum reinforcement required for shrinkage and temperature stresses: As = 0.002 x 12 x 8 = 0.19 sq. in./ft. I | Use one curtain of #4 bars spaced @12” c.c. vertically and horizontally with minimum 2" clearance from outside face of wall: j As = 0.20 sq. in./ft. j For Walls CD and EF, use thickness = 6”. ! ' Minimum reinforcement required for shrinkage I ' and temperature stresses: I j As = 0.002 x 12 x 6 * 0.14 sq. in./ft. Use #3 bars spaced @ 9 1/2” c.c. vertically and horizontally at center of wall: As = 0.14 sq. in./ft. Place 2 additional #5 bars around all window j and door openings. Extend these bars at least 24” beyond the corner of the openings. i i ( Check walls for axial load: ; i The critical wall is the 6” wall. | I i i ' Load from roof = 666 + 600 = 1,266 ! ! ! Wt. of 1 ft. width of wall = 10 x 75 = 750 i I 7 I Load per ft._of.wall________ ___ _= 2,pi6#_J 15 Axial stress = 2>Ol6_ = 28 psi 6 x 12 Allowable axial stress for 6" wall: fa - (0.20 - 0.007 x f) f£ = (0.20 - 0.007 x IP- £ 2,500 = 150 psi Allowable axial stress for 8" wall: fa - (0.20 - 0.007 x IP_iL_i£)x 2500 - 238 psi 8 Parapet Wall As = 0.002 x 8 x 12 = 0.19 sq. in./ft. For outer curtain, extend #4 vertical bars @ 12" c.c. from wall below. For inner cur tain, use #3 @ 12" c.c. vertically and extend them 12" below top of roof slab. Use #4 @ 12" c.c. horizontally for both curtains. Check walls for combined bending and direct stress: Diagram 23 on Page 434 of "Principles of Reinforced Concrete Construction" by Turneaure and Maurer will be used for investigating the concrete and steel stresses: For Walls AD, BC, EH and FG j I M * 8, 060" #/ft. from roof : 16 Since the moment is resisted by only the piers between the openings of the wall, the moment must be increased by the ratio of the length of wall to the total length of piers* This is also true for axial loading. M = 8, 060“ #/ft. x M = 12,530" #/ft. 9 Load from roof = 530 Wt. of wall above window head = 6 .2 9 x 100 = 629 1>159 N = 1,159 x ~ - 1800 #/ft. y el = | + d" = + 1.75 « 6.97 + 1.75 - 8.72" Ml = uei - 1 ,8 0 0 x 8*72 = 15*700“# pn = » 0*20 x 12 = 0.035 12 x 5.75 li= W * - °-66 o J«l 6.8 * 15.700 . ^ psl bd2 12 x (5.75) M fs - no. JL _ 12 X U « 1^^700 . g 650 t bd8 12 x (5*75) Allowable concrete stress: 17 fa 238 ° " 0.4"5f* ' 1125 " °#211 c r _ - ( J L + J & s - — Q' B + 6 x 6.97 fP ~ fa^t + CDe' 3 '8 + 0.211 x 6 x 6.97 696 psi For Walls DE and CF M » 5, 110” #/ft. x = 1 0, 220” #/ft. 6 Load from roof = 455 Wt. of wall above window head = 629 N = 1,084 #/ft x - 2 ,1 6 8 #/ft. el = f + d" = + 1.75 - 4.71 + 1*75 = x N 2,168 Ml = Nei « 2 ,1 6 8 x 6.46 - 14,000" # pn - 0.035 d - lilj = 0 .8 9 ©i 6.46 f# - Cc J * - ^ J L 1 ±.°P° .. 198 p3l bd2 12 x (5* 7 5) fs » n0s - H — ..8 x l i *>000 = 3 ,3 9 0 psi bd 12 x (5. 7 5 )2 i 12 T ! 6.46” 18 238^8 + 0.211 x 6 x ^ .71^ 8 + 6 x 4.71 619 psi For Walls AB and GH Walls AB & GH have the same loading condi tions. However, Wall GH is the weaker of the two, therefore, it is only necessary to investi gate the stresses for Wall GH. x . i k ? . » . 3 3 = 2,240 #/ft. 10.33 e-, = M + d" = > 1 ™ + 1.75 = 9.50 + 1.75 = 11.25“ 1 N 2240 = Nei = 2,240 x 11.25 = 25, 200” # pn = 0.035 M * 11,400" #/ft. x ffi-f| - 21, 300" #/ft io* a Load from roof 666 Wt. of wall above window head = 5 .2 9 x 100 = 529 N = For Wall CD 19 M - 2 2 ,8 0 0 - 15*930 » 6, 870" #/ft. x 3 ^ 3 3 = 8130" #/ft. Load from roof « 666 + 600 = 1,266 Wt. of wall above door = 3 x 75 8 8 225 N = - 1,491 #/ft. x iii33 = 1.765 #/ft. 20 i fs = ncs Jil = 12 x 11 x 8,130 = 9 ,9 3 0 psi ; bd? 12 x {3)2 Allowable concrete stress: C = ~ 0 * g*133 0A5fc 1125 f_ = fa(U_le_) - 150(_A.1J5 JE j/lO . ) - p a t + CDe 6 + 0.133 x 6 x 4.66 521 psi For Wall EF M = 6, 870° #/ft. x lii33 „ 9, 960° # 13.33 N = 1,491 x * 2,l6o # 13.33 e - i = — = = 4.60” H 2 ,1 6 0 Mi * 9^960” # pn » 0.047 £ = 0.65 el - Ml 6 ,3 x 9>96o m f :o - « e a r - "ii/x '(3 ) ~ " 581 psl * 521 psl allowable 21 f . nCs J h . ia x 11 x 9960 . s s bd* 12 x (3 ) 2 Since the concrete is overstressed, It is necessary to increase the steel, if the wall thickness remained the same. Use #4 bars spaced @ 11 1/2” c.c. As = 0.21 sq. in./ft. p . -Or!1 - . S . . . ! 2 = 0.07 12 x 3 fc = cc - = 5*6 x 9*960 _ 5^6 pSi < 521 psi *> d2 12 x (3 )2 Design of Wall Footing In order to minimize the differential settlement of the building, It is deemed advisable to design the footings so that the soil pressure due to dead load (the constant load) Is approximately uniform. This is accom- j plished, as shown in Table I, by using a design load which 1 is the XD.L. (col. 5) divided by the smallest value of --- or O.895 (col. 7)» This method gives a ZD.L. + L.L. ; constant ratio of the design load to the dead load. Therefore, a uniform soil pressure for dead load Is I • obtained. 22 TABLE I WIDTHS OF WALL FOOTINGS q . § ■ v i .g ' " ' S V * " n / n J ' - s $ - " ■ > v S p' 4 ^ N ^ N> KfO 'vi Kl AB 1,563 543 123 2,106 2 ,2 3 0 0.945 2 ,3 6 0 15 CD 925 1,033 233 1 ,9 6 0 2 ,1 9 0 0.895 2 ,1 9 0 14 EF 925 1,033 233 1 ,9 6 0 2,190 0.895 2 ,1 9 0 14 GH 1,563 5^3 123 2,106 2 ,2 3 0 0.895 2 ,3 6 0 15 AD 1,563 432 98 1,995 2 ,0 9 0 0.954 2 ,2 3 0 14 DE 1,563 371 84 1,934 2 ,0 2 0 0.958 2 ,1 6 0 14 EH 1,563 432 98 1,995 2 ,0 9 0 0.954 2 ,2 3 0 14 BC 1,563 432 98 1,995 2 ,0 9 0 0.954 2 ,2 3 0 14 CF 1,563 371 84 1,934 2 ,0 2 0 0.958 2 ,1 6 0 14 FG 1,563 432 98 1,995 2 ,0 9 0 0.954 2 ,2 3 0 14 t The allowable soil pressure is 2,000 psf for | dead load and live load. A 10" footing is used, thus, | the soil pressure minus the pressure due to the weight of the footing = 2 ,0 0 0 - 10/12 x 1 x 150 = 1,875 psf per lineal ft. In Table I, each footing width is ob- ; tained by multiplying the design load by 12/1, 875. i i j 23 Net soil pressure = ZD»L» = 1*995 = 1,710 psf ftg. width 1.167 Footing pressure 125 psf Balance soil pressure 1*835 psf Max. overhang = 4" * 0.333' M - 1,875 x (0.333ig . 104'# = 1248"# 2 As = -Ji—- * . x. .--------- ■ = 0.0125 sq. in. s fJ35 1 8 ,00 0 x 0.857 x 0 .5 (insignificant) No steel is required for overhang due to bending. Longitudinal steel for footing = As « 0.002 x 10 x 15 - 0.30 sq. in. Use 2-#4 space 3” from the bottom and sides of the footing. CHAPTER IV SEISMIC ANALYSIS The following weights of the various parts of the building are taken from mid-way between roof slab and floor slab to the top of the parapet wall. TABLE II WEIGHT "V" FOR SEISMIC ANALYSIS Wt. of Net Wt. Part Total Wt. mm Opening (lb.) Roof Slab 19.33 X 41 x 89 70,500 Wall AB 19-33 X 8.29 x 100 - 8 x 3 x 90 13.870 Wall CD 19.33 X 5 X 75 - 3 x 2 x 65 = 6 ,8 6 0 Wall EF 19.33 X 5 x 75 - 6 x 2 x 63 = 6,470 Wall GH 19*33 X 8.29 x 100 - 9 x 3 x 90 a s 13,600 Wall AH 42.33 X 8.29 x 100 - 16 x 2 x 90 = 32,220 Wall BG 42.33 X 8 .2 9 x 100 - 16 x 2 x 90 s 32 ,22 0 Total W = 175.740 The seismic force to be resisted by the building as a whole is: F = CW 25 Where C * * O.6 0 = 0.60 - 0.133 N + 4.5 0 + 4.5 F = 0.133 x 175.740 = 23.500# Earthquake forces in north-south direction: The earthquake forces act normal to Walls AD, DE, EH, BC, CF, FG, and parallel to Walls AB, CD, EF, GH. Parapet wall♦ F = CW » 1 x 100 x 2.83 - 283 #/ft. M = iFh = I x 283 x 2 .8 3 x 12 = 4,800”# 2 2 p - = 0 *11 , - 0.00158 bd 12 x 5.Si = V(pn ) 2 + 2pn - pn = \j{0.00158 x 12 )2 + (2 x 0.00158 x 12) 0.00158 x 12 - 0.177 j=l--|=l - 2liZI - 0.941 f. = 2 x 4,800_____ _ - J J 2 psi kjbd2 0.177 x 0.941 x 12(5. 8 1 )2 f = — !L -----k * . ? 1 ° . c! _____ - 7 .9 8 0 psi s As« J d 0.11 x 0. 9^1 x 5.81 26 Earthquake force normal to Walls AD, BC, EH, and FG. ' , r r , I . , , 4 5 \ „ 5.0\ T .4.5\ /4 0 * # £,-2,27* \ / - - O ' F, - o. 20 x /oo x /4 - 280 * / f Y. M F2 ' - 0.20 * /0OX9 - /80 #/fth£ F3 0.20 X /OO X /4 - 280 */&. Rz -1,048* Fig. 3» Lateral Load Normal to Walls AD, BC, EH, and FG. x 10 = 280 x 6 .2 9 x 10.14 + 180 x 4 x 5 + 280 x 3 x 1.5 RX - 2,273# R2 X 10 + 280 X 3.29 X 1.645 = 280 x 3 x 1.5 + 180 x 4 x 5 + 280 x 3 x 8 .5 R2 * 1,048# M @ Ri = 280 x 3.29 x 1.645 = 1,515# M @ V = 0 * 1,048 x 4.16 - 280 x 3 x 2.66 - 180 x 1.1 6 x 0 .5 8 « 2,004^ 27 Dead load moment from roof = l40wS2 = 14 x 0.029 x (14.58) 2 = 7,690*# \ Max. M- +9,2OS'#- | X s\ C om jb/rrect M o m e n t' Seism ic Moment Deere/ L o a d M o m en t Fig. 4. Moment Diagrams for Walls AD, BC, EH, and FG. Combined moment per linear foot of wall, M - = 1,023*# = 12,2801 '# 9 Dead load from roof = H§. 89 x^l4*58 _ 433 Wt. of wall above window = 6.29 x 100 = 629 N = s 1 ,0 6 2 x ■ —^ - l,650#/ft. 3.0' 4 , 0 ,3.0,3.29 * Fig. 28 el - Ml - pn = d = «1 1 5 + d " = l 2 ? , 2 .§ - ° + 1.75 = 7.44 + 1.75 N 1 . 6 5 0 Nei = 1650 x 9.19 - 1 5. 160"# 0.035 5.75 = 0.63 9.19 = 9.19 r v f l fc - Cc J!i - 6.9 X 15.160 » 264 psi bd2 12 x (5*75)2 fa = nCs J*i - 12 x 14 x 15,i6q - 6,420 psi bd2 12 x (5.75)2 r „ f /t + De \ T 1 03 = o-3« / 8 + 6 x 7.44 ^ fP t”+ "CDe X 1#33 23d 8 + 0.211 x 6 x 7.4V X 1.33 = 719 X 1.33 - 956 psi 1 Earthquake force normal to Walls EE and OF. 1 .5 1 .5 / . S l 3.0. 15 f - 5 IS 1 2 . . O' Rr-1853’ * v-ot, R .2 - (>71# F, - 0.20 X 100 x/2 - 240 #/ft. h+ F2 - 0 20 X /OOxG - 120 f3 - 0.20* /OOX? - /QO^/fF hi 5. Lateral Load Normal to Walls DE and CF. 29 Ri x 10 - 240 x 6.29 x 10.14 + 120 x 4 x 5 + 180 x 3 x 1.5 Hi » 1,853# R2 x 10 + 240 x 3.29 x 1.645 = 240 x 3 x 1.5 + 120 x 4 x 5 + ISO x 3 x 8 .5 r2 = 677# M @ Ri = 240 x 3.29 x 1.645 = 1,300'# M @ V = 0 = 677 x 4.14 - 180 x 3 x 2.64 - 120 x 1.14 x 0.57 = 1,295’ # Dead load moment from roof = 12CwS2 = 12 x 0.025 x 89 x (12.5) 2 - 4,170'# \ M ax. M - + 5 ,4 1 0 '# | / / / Seismic Moment D ead Loarct Moment ComJb/ ned Moment Fig. 6. Moment Diagram for Walls DE and CF Combined moment per linear foot of wall M - - 912'# - 10,940"# 30 Dead load from roof = E| = §?. « = 371 Wt. of wall above openings = 6.29 x 100 - 629 N = 1 ,0 0 0 x ~ = 6 200C#/f t. e - . = H + d” = ■ 1. 0/ . 9 . 4. 2 + I.75 = 5 .4 7 + 1.75 - 7.22” x N 2000 Mi = Nex = 2 0 0 0 x 7 .2 2 = 14,440"# pn = 0.035 £ = 5-Z5 = 0 .8 0 el 7 .2 2 fc - Cc -5i - — M Alg. = 222 psi C 1 2 x (5* 7 5) fs . ncs * = 12- * a . 4,810 psi S 8 bd2 12 x (5*75) 2 ,t + De v o, 8 + 6 x 5.47 . fp = fa(t + CDe^ x 1 ,3 3 ~ 238 + 0.211 x 6 x 5.47^ x 1 .3 3 = 651 x 1 .3 3 - 866 psi Distribution of lateral force to shear Walls AB, CD, EP, and GH. These walls are parallel with the direction of the seismic force. The deflections of the piers are calculat* 1 31 p formula^ A= \ t0.333{ § ) 3 +f] Where A * deflection of pier in inches t = thickness of pier in inches H = height of pier in inches d = width of pier in inches TABLE III DEFLECTION OF PIERS OF WALL AB Pier H 3 A Rigidity 1 A Deflec. of wall sections 1 A, A z ^ 3 a 0.097 0.0122 0.0122 b-1 1.153 0.208 0 4.81 b-2 1 .2 5 0 0.238 0 4.20 0.0724 b-3 1.153 0 .2 0 8 0 4.81 c 0.145 0.0183 0.0183 Total A 0.1029 2 Analysis of Small Reinforced Concrete Buildings for Earthquake Forces (Chicago: Portland Cement" Associa Hon7~IFbTTp.*3^r” f 32 TABLE IV DEFLECTION OF PIERS OF WALL CD Pier H A Rigidity 1 A Deflec* of wall sections 1 !_ + !. A, A 2 a 0.145 0.0243 0.0243 b-1 1.400 0.3860 2.59 b-2 0.553 0.1 02 0 9 .8 0 0.0807 Total A 0.1050 TABLE V DEFLECTION OF PIERS OF WALL EF Pier H 3 A Rigidity 1 A Deflec* of wall sections 1 + L. + A, A 2 A 3 a 0.145 0.0243 0.0243 b-1 1.617 0.5040 1.984 b-2 1.167 0 .2 83 0 3.540 0.1332 b-3 1.617 0.5040 1.984 Total A 0.1575 I i TABLE VI DEFLECTION OF PIERS OF WALL GH 33 Pier H S Rigidity A 1 A Deflec. of wall sections 1 i + i + i + i A 2 A3t a 0.097 0.0122 0.0122 b-1 1 .5 0 0 0 .3 28 0 3 .0 5 0 b-2 2 .0 0 0 0.5830 1.713 b-3 2 .0 0 0 0.5830 1.713 0.1048 b-4 1 .5 0 0 0 .3280 3 .0 5 0 e 0.145 0.0183 0.0183 Total A 0.1333 TABLE VII RELATIVE RIGIDITIES OF WALLS Wall Deflection A Rigidity Relative Rigidities 1* 4- A a T i AB 0.1029 9.72 0.295 CD 0 .1050 9.53 O.289 EF 0.1575 6.35 0.192 GH 0.1353 7.39 0.224 I Total A 32.99 1.000 The lateral force F * 23,500# Is distributed to the different walls in proportion to their relative rigidities! j This can be done if the lateral force F is placed at the ! center of rigidity to produce equal deflections in the walls, as shown in Figure 7 (a), Page 3 5. The center of rigidity is located by using the relative rigidities as weights and taking moments about the center line of Wall ! AB: 1 I x * 0 .2 8 9 X 1 4 .5 8 + 0.192 X 2 7 .0 8 + 0.244 X 41.67 ; « 18.77' 1 1 j The lateral force is distributed to the walls thus: 1 j Wall AB * 23,500 x 0.295 - 6930# j Wall CD - 23,500 x 0.289 - 6790# 1 Wall EF = 23,500 x 0.192 = 451C# i Wall GH - 23,500 x 0.224 - 5270# The lateral force, however, is not applied at the center of rigidity, but at the center mass. Therefore, in addition to the forces Just determined, there are j forces acting on the four walls due to torsion. ! | The distance from Wall AB to the center of mass 1 equals: x » .3,2.652,700 _ 2 0. 8 0* 175,720 e = 2 0 .8 0 - 18.77 - 2.03' TABLE VIII LOCATION OF CENTER OF MASS 35 Part Weight (lb.) Moment Arm from Wall AB (ft.) Moment (ft. lb.) Roof Slab 73,000 20.83 1, 520,000 Wall AB 13,760 0 0 Wall CD 7 ,1 1 0 14.58 103,700 Wall EF 6 ,7 2 0 2 7 .0 8 182,000 Wall GH 13,490 41.67 563,000 Wall AH 3 0 ,80 0 20.83 642,000 Wall BG 3 0 ,8 0 0 20.83 642,000 Total 175,720 3, 6 5 2,0 00 /4.S6' /2.5\t. M.5&\ -K = 20.80' D Ce/rfer of C 3- /8-77‘ * / 4 f 4 . / 9 8.3/\ t F H 22.89' . r_ . _ 4 3 ( o r ) Cenre/- of r/giafify GH Fig* 7* Wall Deflections Torsion = 23,500 x 2.03 = 47,700'# In resisting this torsion, the walls will deflect as shown in Figure 7 (b) above. The reactions on the walls due to the torsion of the lateral force about the center of rigidity equal EF X ---- - X 0.295 18.77 x 9-A92 x 8.31 _ 0.295 18.77 r 0.224 r 22.89 „ 0.295 18.77 1,020# Design load for walls: Wall AB - 6,930# Wall CD = 6,790# Wall EF - 4,510 + 318 = 4,828# Wall GH - 5>270 + 1,020 - 6,290# The total deflection of a pier is the sum of the deflections due to bending and shear and is calculated by the formula A = Dm + Dv = -ll- + I 36lg A Where A = total deflection of a pier due to bend ing and shear in inches Dm = deflection due to bending in inches Dv = deflection due to shear in inches H = height of pier in inches A = cross-sectional area of pier in sq. Inches 37 Ig = * cross-sectional moment of inertia of pier in direction of bending in inches4 The lateral load applied to a wall will be distribu ted among the individual piers in proportion to their rigidities* Distribution of lateral force among piers in Wall AB. 6 930 / 98 I . - 98' 0 4.93 4.0 c9 4.0' 4.0 4.33 20.67 Fig. 8. Shear Wall AB DEFLECTION DUE TO TABLE BENDING IX AND SHEAR IN WALL AB Pier d H A Ig Dm Dv A 1 52 60 4l6 93.700 0.0640 0.144 0.208 2 48 60 384 73.700 0.0813 0.156 0.237 3 52 60 4l6 93.700 0.0640 0.144 0.208 J TABLE X DISTRIBUTION OF LATERAL FORCE AMONG PIERS IN WALL AB 38 Pier _ l _ I A Farce A 1 4.81 0.347 2,405 2 4.22 0.306 2,210 3 4.8l 0.3 47 2,405 TABLE XI MOMENT OF INERTIA OF NET OF WALL AB SECTION Pier A (in.2) L L2 (in.) (in.2) AL2 (in.1 *) (S.*) AL2 + Ig (in.4) 1 4l6 98 9,604 4,000,000 93.700 4,093,700 2 384 0 0 0 73.700 73,700 3 416 98 9.604 4,000,000 93,700 4,093,700 Total I 8, 261,100 Stresses in Wall AB, Shearing stresses Piers 1 & 3 = M Q 5 = 6 psi 4l6 Pier 2 2,120 <- “ " 584" = 6 psi Flexure stresses I I f Me P x 2 x 2 x d _ PHd 1 Ig ' 4lg Piers 1 & 3 f = 2*405 x SO x 52 _ 20 psi 4 x 93*700 Pier 2 f -'.2*120 x 60 x 48 „ 21 ± 4 x 73*700 Stress due to vertical load Piers 1, 2, & 3 f . I . 1^25 . = 18 psi A 8 x 12 Stresses due to overturning Overturning moment = 6,930 x 7*23 x 12 60 1, 000”# Piers 1 & 3 Pier 2 601,0 00 x 24 8, 261,100 « * + 2 psi © © '3 f t s > ' s ? > ( a ) S tre s s e s Due to V e rH c o tt L o a d ( b ) S tre s s e s D u e to F /e x u re It — I 4. N Cc) S tre s s e s D u e to O v e rtu rn in g (d ) C o m b in e d S tre s s e s Fig. 9* Stresses for Wall AB 41 I < 0 ' ■ > 1 * 2’ 4 £ •I < g > /2" r 4Q"______ S2"___________ -O v fs ic /e farce o f ’p /e r \- # 4 B a rs /V 2 # N ii 5 ? « S 1 O n tz Fig. 10. Stress Diagram for Pier 1, Wall AB 1st bar from left = . f t • * • 7^. x 8.25 - 8,420 psi 2 0.2 2nd bar from left * = —^ + ^5 x AfL = 8,130 psi 2 0.2 42 Distribution of lateral force between piers in Wall CD. s. o' 20.& T Fig. 11. Shear Wall CD TABLE XII DEFLECTION DUE TO BENDING AND SHEAR IN WALL CD Pier d H A Ig Dm DV " ■ - P f l A 1 60 84 360 108,000 0.1524 0.233 0.3854 2 152 84 912 1, 75 6,0 00 0.0094 0.092 0.1014 TABLE XIII DISTRIBUTION OF LATERAL FORCE BETWEEN PIERS IN WALL CD Pier 1 _L A -£r A Force 1 2 2.59 9.86 0.208 0.792 1,410 5*380 TABLE XIV MOMENT OF INERTIA OF NET SECTION OF WALL CD Pier A (in,2) L (in.) L2 (in.2) AL2 (in.*) AL2 + I c r (in.*) 1 360 99.2 9*840 3,5*0,000 108,000 3,648,000 2 912 42.8 1,834 1, 67 2 ,0 0 0 1, 756,000 3,428,000 Total I 7, 076,000 Stresses in Wall CD Shearing stresses Pier 1 = = 4 psi 360 Pier 2 * 5. 1 , 3 80 = 6 pSi 912 Stress due to vertical load Piers 1 & 2 f = f - « f- — ~ = 28 psi A 6 x 12 Flexure stresses Pier 1 44 Pier 2 f . 5 ,3 8 0 x , 84 x, 152 = + 1Q 4 x 1, 756,0 00 Stresses due to overturning Overturning moment = 6,790 x 10.23 x 12 833,000" # Pier 1 f = 833,000 x 69.2 _ . 8 al 7,076,000 f = 833,000 x 1 2 9 .2 - m _ 2,5 psi 7,076,000 Pier 2 f = 833,000 x 33*2 = _ 4 pSi 7,076,000 f = 833,000 x 1 1 8 .8 = + 14 psi 7, 076,000 © © S i N ‘ N 5 « 0 N ? < * > 0 * Cat) S tre s s e s D u e to V e r t t e a t L o a d fi> ) S tre s s e s D u e to O v e rtu rn in g Fig. 12. Stresses for Wall CD (c ) S tresses Doe fo F/ext/re 45 • s ( d ) C om bined S tre s s e s £ < n Pig, 12. Stresses for Wall CD (continued) *3 B a rs fv T 0 ■ *. * * • « . * I •' * . 4 ; . * j • . A . .4 . . • _ - . * <j , ’ ; 2W 2 ' * 4[ 5€>9, d'-4 7i" 4" &o" V Pig. 13. * Stress Diagram for Pier 1, Wall CD \ 46 ! I 1st bar from left = + ^ 6 x = - 5 ,5 0 0 psi 2 0.14 2nd bar from left * A f f i . +, ^-L x ^ = 8 ,1 1 0 psi j 2 0.14 i j 3rd bar from left = Aft.A + ^39 x . , , i L * - 5 = 1 0 ,1 8 0 psi 2 0.14 Distribution of lateral force among piers in Wall EF. t 4,828# £ - .1 L ‘ 9a" j * 1 ) 1 c>q. 1 0 © © 0 s . ^.33 \ 3.0 . 6. O' 3.0 4.33 20.67' I Fig. 14. Shear Wall EF TABLE XV DEFLECTION DDE TO BENDING AND SHEAR IN WALL EF Pier d H A H Om Dv A 1 52 84 312 7 0 ,3 0 0 0.2340 0.269 0.503 2 72 84 432 186,600 0.0884 0.194 0.283 3 52 84 312 7 0 ,30 0 0.2340 0.269 0.503 47 TABLE X V I } DISTRIBUTION OF LATERAL FORCE AMONG PIERS IN WALL EF Pier 1 A >1 ->l_ Force 1 1.99 0.265 1,280 2 3.53 0.470 2,268 3 1.99 O.265 1,280 TABLE XVII MOMENT OF INERTIA OF NET SECTION OF WALL EF Pier A (in.2) L (in.) L2 AL2, (in.2) (in.4) Ig (in?4) AL2 +, Ig (in.*)8 1 312 98 9,604 2,995,000 70,300 3,065,300 2 432 0 0 0 186,600 186,600 3 312 98 9,604 2,995,000 70,300 3,065,300 Total I 6,317,200 Stresses in Wall EF. Shearing stresses i j Piers 1 & 3 * ~ — - 4 psi ! 312 i 48 Stress due to vertical load Piers 1, 2, & 3 f « 2*03-6 = 28 psi 6 x 12 Flexure stresses Piers 1 & 3 Stresses due to overturning Overturning moment = 4,828 x 10.23 x 12 ® f , 1*280 x 84 x 52 4 x 70,300 + 20 psi Pier 2 f = 2*268 x 84 x 72 4 x 186,600 = + 18 psi 592, 000"# Piers 1 & 3 f = 592,000 x 72 6,317*200 = + 7 psi f 592,000 x 124 _ = + 12 psi 6, 317*200 Pier 2 ■ f _ 592 J 0 0 0 X 36 _ , O na4 f “ 6,317,200 ' ± 3 PS1 ( ^7 ^ S tre s s e s D u e to V e rt/e a t I oar <s t ' I \ 5 \ ^ \ N N 5 ? N (t> ) S tre s s e s D u e to Ftexune f c ^ l l » * ) I N . v > • N ( c ) S tre s s e s D u e to O v e rtu rn in g Cct) ComJbineot S tre s s e s Pig. 15. Stresses for Wall EP 50 • d . • . * . • • , • ^ . * ; . • , 2 J£+ 9'z" z 4 7 2 " 5-z" /5" 37 Pig. 16. Stress Diagram for Pier 1, Wall EF 1st bar from left = 192 + 156 x 7 = 8,710 psi 2 oTPT 2nd bar from left = 156 + 106 x 9«5 = 8 ,8 9 0 psi 2 0.14 51 Distribution of lateral force among piers in Wall GH L ■ - /<?<?" L - f 0 4 ' A-33 6,290# -+■ 1-33" © 3 33 k - - a © 3.0 2.5 3.0 © 2.5 3.0 © 3.33 20.67' Pig. 17. Shear Wall GH TABLE XVIII DEFLECTION DUE TO BENDING AND SHEAR IN WALL GH Pier d H A Dv A 1 40 60 320 42,700 0.1404 0.1875 0.3 28 2 30 6o 240 18,000 0.3333 0.2500 0.583 3 30 60 240 18,000 0.3333 0.2500 0.583 4 40 60 320 42,700 0.l4o4 0.1875 0.328 52 TABLE X IX DISTRIBUTION OF LATERAL FORCE AMONG PIERS IN WALL GH Pier 1 A l A T — ^ A Force 1 3.050 0 .3 2 0 2,013 2 • 1.715 0 .1 8 0 1 ,1 3 2 3 1.715 0 .1 8 0 1 ,1 3 2 4 3.050 0 .3 2 0 2,013 TABLE XX MOMENT OF INERTIA OF NET SECTION OF WALL GH Pier A (in.2) L iP (in.) (in.2) AL2h Ig, (In.') (i n ) AL2 + Ig (in.4) 1 320 104 10,816 3,460,000 42,700 3,502,700 2 240 33 1,089 262,000 18,000 280,000 3 240 33 1,089 262,000 18 ,00 0 280,000 4 320 104 10,816 3,460,000 42,700 3,502,700 Total I 7,565,400 53 Stresses in Wall GH. Shearing stresses Piers 1 & 4 = fLiPll = g psi 320 Piers 2 & 3 * - 5 psi 240 Stress due to vertical load Piers 1, 2, 3, 8e 4 f « , = 18 psi 8 x 12 Flexure stresses Piers 1 8s 4 f « 2,013 x 60 x 40 _ + 28 psi 4 x 42,700 ~ Piers 2 8s 3 f . = + 28 psi 4 x 18,000 — K Stresses due to overturning Overturning moment = 6,290 x 7»23 x 546,000”# Piers 1 8s 4 f = ^ 6-iQQQ...x 84 . + 6 7, 565,000 12 - e „ 546,000 X 124 _ j n f '"'''7/5B5/006'” 1 9 PSi Piers 2 & 3 f » 546,000 x 18 7»565/000 f = 546,000 x 48 . + 1 psi I 7.565.000 0 ® * s . S > ! V . ( a ) S tre s s e s D u e to V e r tic a l L o a d ( b ) S tre s s e s D u e fo F /e x u re $1 -A ^ V 1A— C c) S tre s s e s D o e to O v e rtu rn in g (o f) C o m b /n e a f S tre s s e s 54 Fig, 18, Stresses for Wall GH 55 Or/As/c/e farce of/>/ee r #4 Bars 4 0 " /2" Pig* 19* Stress Diagram for Pier 1, Wall GH 1st bar from left = . ^2^ - t x * 1 1 ,36 0 psi 2nd bar from left = f f f i L . . ! —3§. x _i£ = 15,240 psi 2 0.2 The tensile stresses of the reinforcement in the piers of the cross walls are well within the allowable steel stress of 24,000 psi (1*33 x 18,000) for combined dead load and seismic force. 56 Wall footing pressure due to seismic forces. The critical footing is the portion under Wall BC adjacent to Wall AB. Fig. 20. Plan of Footing Along Wall AB I = [-^ x 5.^6 x (1.167) 3 + 5*46 x 1.167 x (10)23 x 2 + i x 1 .2 5 x (1 8. 8 3 )3 12 - 1,971 ft.4 S = I . 1*971 = 186 ft#3 c 1 0 .5 8 Overturning moment * M 0 T = 6930 x 13.40 = 92,900’ # 57 Soil pressure due to seismic « Mg> T' * = 22 ’229 = 498 S 186 Dead load soil pressure = 1,835 Total soil pressure = 2,333# /ft. 2 Allowable soil pressure for combined dead load and seismic force = 1.33 x 1,835 = 2,440 #/ft.^ Footings computed in Table I, Page 22 are to be used. Earthquake forces in East-West direction The earthquake forces act normal to Walls AB, CD, * EF and GH. Earthquake force normal to Wall AB. /> s 0.20 x fOO* 79.33 - 38'7*yff./ii #o< A,;2,977* * O ■ O. 20 X/OOX 1I.33 - 227 // 0 •Qt J - f3 - 0.20X/OOX/9.33 - 387 , « > 3.67 t 4 . o \ 4.0' +o\ 3.67 * ■< — /9 . 3 3' --- * ■ Fig. 21. Lateral Load Normal to Wall AB 58 Rl x 10 = 387 x 5.29 x 10.64 + 227 x 5 x 5*5 + 387 x 3 x 1 .5 % * 2,977# R2 x 10 + 387 X 3.29 X 1.645 ■ 387x2x1+ 227 x 5 x 4.5 + 387 x 3 x 8.5 R2 = 1>366# M @ RX - 387 x 3.29 x i.645 = 2,094'# M @ V = 0 = 1,366 x 3.90 - 387 x 3 x 2.40 - 227 x 0.90 x 0.45 = 2,453*# Dead load moment from roof ® 19*33 CwS2 = 19*33 x 0.041 x 89 x (14.5 8 )2 - 15>000 '# M ax. M -+ n ,o 9 4 ‘# S eism ic M o m en t D e a d L o a d M o m e n t C o m b in e d M o m e n t Fig. 22. Moment Diagrams for Wall AB Combined moment per linear foot of wall, 59 M » - 1,5 0 7’ # = 18,080"# 11.33 Dead load from roof « = —§2 x 666 = 543 109 ¥t* of wall above window = 5*29 x 100 = 529 N - 1,072 x 19.•33, = l,830#/ft. 11.33 e - i - 15 + d" - + 1*75 - 9.88 + 1.75 * 11.63" N lo30 Ml = Nei = 1,830 x 11.63 - 21,300"# pn = 0.035 fc - Cc _J!i = 7.5 x 21,300 - 403 psi bd2 12 x (5*75)2 fg = nCs J?1 12 x 18 x 21,300 = 1 1 ,60 0 psi bd2 12 x (5.75)2 ■ p _ - p /1 + De \ v n _ o-aA / 8 + 6 x 9*88 ^ fP - fa(t + CDe ,33 3 V + 0.211 x S~ T J M ) X 1 .3 3 = 781 X 1 .3 3 * * 1,040 psi 6o Earthquake force normal to Wall CD. Kr-/,34Q* F, - 0.20X 75x 19.33 - 290 h t . E C > ' » ^v-o J F7 i - 0.20x . 75* /6- 33 - 245 ■ < 0 Q V 4 0 K Fgz(,24*4 4.33 3.0 /2-0' '9.33' ^ Fig. 23* Lateral Load Normal to Wall CD Hi x 10 = 290 x 3 x 8 .5 + 245 x 7 x 3*5 Hi = 1,3^0# R2 x 10 = 245 x 7 x 6 .5 + 290 x 3 x 1.5 R2 - 1,24S# H @ V = 0 = 1245 x 5.08 - 245 x 5.08 x 2.54 - 3,170'# Dead load moment from roof = (19*33 CwL)u - (19*33 CwL)* = 19*33 x 0.0 82 x 89 x (14.58) 2 - 19*33 x 0 .0 7 8 x 89 x (12.5) 2 = 9,040'# 61 Seismic M o m e n t ¥ 9, 040' # Max. M * % 332 ' 4 f C o m b ined M o m e n t D e a d L o a d M o m en t Fig* 24. Moment Diagrams for Wall CD Combined moment per linear foot of wall, M = %332 = 571*# = 6,850"# 16.33 Dead load from roof = x 1266 = 1033 iuy Wt. of wall above door = 3 x 75 N = 225 1 ,2 5 8 x ^ 3 , 3 1 6 .3 3 - l,490#/ft. N 1,490 Mx - 6,850"# pn = 0.047 1 = .J— = 0 .6 5 ei 4.60 7O' , 3.0 62 fo = Oo _“i = 6 .3 X 6 ,8 5 0 . 400 psi 0 C bd2 12 X (3)2 fs = nCs J!l = 12 x 11 x 6 ,8 5 0 . 8 ,3 7 0 psi b d 5 12 x ( 3)2 fp - fa(v~— ' ■ %!■■■) x 1.33 = 150 (----6 + 6 xJi.6o ) p avt + CDe '6 + 0.133 x 6 x 4.60 X 1.33 « 521 x 1.33 = 693 psi Earthquake force normal to Wall EF. 230# v--o- 3.67 3.0 6.0 /9.33‘ 30 3.6 7 F , - 0.20x75* /9.33 = 290 # / f f . At F2 - - 0.20* 75 x /3. 33 - 200 */ft At Fig. 25. Lateral Load Normal to Wall EF Rl x 10 = 290 x 3 x 8 .5 + 200 x 7 x 3.5 R! - 1,23C# R2 x 10 = 200 x 7 x 6 .5 + 290 x 3 x 1.5 R2 = 1,040# 63 M @ V = 0 = 1,040 x 5.20 - 200 x 5.20 x 2.60 = 2,704 »# Dead load moment from roof = 9*04o! # (Same as for Wall GD) Max. M - +%22/'# S eism ic M om ent D ecrat L o o rai M o m e n t C o m b in ed M om ent Pig. 2 6. Moment Diagrams for Wall EP Combined moment per linear foot of wall, M = » 691'# = 8,290"# 13.33 N = l,490#/ft. (Same as for Wall CD) ei - M - = 5#56« X N 1,490 Mi - 8, 290"# pn = 0.07 ©1 5.56 0.54 3.0, 5.0' 7 . 0 3 . 2 9 . 64 f0 - Cc -rS - 5, »8. Jk|90 = 453 psi bd2 12 x (3)2 fs * * nCs 12 x 9 x 8,290 = 8 ,2 9 0 psi bdZ 12 x (3)2 fn = fa(t + De ) x 1 .3 3 * 150 ( • ?----.±, , § .. £ _J) p avt + GDe 3 '6 + 0.133 x 6 x 5.56' x 1.33 - 565 x 1.33 = 752 psi Earthquake force normal to Wall GH. 2 .6 7 3.0' 25 3.0 2 5 30 2.67 Z9.33' 3 * % -2 ,921 ViO R2-/,32> * F, - £7.2<? x /<?<? f/9.33 - 387 */ff. A/. Fz -0.20)(/OOx/0.33 - 207 * / f f . A t. F3 - 0.20x/OOX /9.33 - 387 A/. Fig. 27. Lateral Load Normal to Wall GH Rl x 10 = 387 x 5.29 x 10.64 + 207 x 5 x 5*5+ 387 x 3 x 1.5 R! - 2,921# R2 x 10 + 387 x 3.29 x 1.645 = 387 x 2 x 1 + 207 x_5-X_4-.5—+- 387—x. 3 x . 8 .5________________ 65 r2 = 1,321# M @ Ri = 387 X 3 .2 9 X 1.645 = 2,094’ # H @ V * 0 - 1321 x 3 .7 8 - 387 X 3 X 2.28 - 207 X 0.78 x 0.39 » 2,287'# Dead load moment from roof = 19*33 CwS2 = 19*33 x 0.041 x 89 x (14.58)2 - 15, 0 0 0’ # Se/s/n/c M o m e n t D e n rc / lo c r a t M o m e n t May. M- +/7,Q94 '# C o m b /n e at M o m e n t Pig. 28. Moment Diagrams for Wall GH Combined moment per linear foot of wall, M = = 1,652’ # = 19,830"# 10.33 Dead load from roof = —§2 x 666 = 543 109 Wt. of wall above window = 5 .2 9 x 100 = 529 N " 1,027 x Iof33 6 6 2,000#/ft. ei - § + d" - ^§Vb" §§ + 1-75 - 9.92 + 1.75 » 11.67” Mx = Nex = 2,000 x 11.67 = 23,340”# £ = -5.75 „ o.49 ei 11.67 pn = 0.035 f0 = Co J& - I^_x„„g_3,j40 . „ 2 psl bd 12 x (5.75)2 fs = nCs _5i = ,^23,340 _ 12>700 ± bd2 12 x (5.75)s fP - x ^ = 238 (f l A t a n V ^ x 9.92} X 1.33 - 782 X 1.33 « l,04o psi Distribution of lateral forces to Walls AH and BG. These two walls are parallel with the direction of the seismic force. Since these two walls are identi cal, their relative rigidities are equal and their centers of rigidity coincide with the center of mass, therefore, no torsion is produced. The seismic force acts on each wall = 1/2 x 23,500 = 11,75C#« Pig. 29. Shear Wall AH or BG Distribution of lateral force among piers in Wall AH or BG. TABLE XXI DEFLECTION DUE BENDING AND SHEAR IN WALL AH OR BG Pier d H A JS Dm Dy A 1 62 48 496 159,000 0.0193 0.0968 0.1161 2 78 48 624 316,000 0.0097 0.0769 0.0866 3 18 48 144 3,890 0.7890 0.3333 1.1220 4 18 48 144 3,890 0.7890 0.3333 1.1220 5 78 48 624 316,000 0.0097 0.0769 0 " . 0866 6 62 48 496 159,000 0.0193 0.0968 0.1161 6 8 TABLE XXII DISTRIBUTION OP LATERAL FORCE AMONG PIERS IN WALL AH OR BG Pier 1 A 1 A *A Force 1 8.61 0.205 2,410 2 11.52 0.274 3*220 3 0.89 0.021 247 4 0.89 0.021 247 5 11.52 0.274 3,220 6 8.61 0.205 2,410 TABLE XXIII MOMENT OF INERTIA OF NET OF WALL AH OR BG SECTION Pier A (in.2) L L2 (in.) (in.2) AL2 (in.*) (S.*j AL2 +, Ig (in.*) 1 496 223 49,729 24,650,000 159*000 24,809,000 2 624 93 8,649 5,400,000 316,000 5,716,000 3 144 27 729 105*000 3*890 108,890 4 144 27 729 105*000 3*890 108,890 5 624 93 8,649 5,400,000 316,000 5,716,000 6 496 223 49,729 24,650,000 159*000 24,809*000 Total I 6l,267»780 Stresses In Wall AH and BG. Shearing stresses Piers 1 & 6 = = 5 psl 496 Piers 2 & 5 = = 5 psi 024 Piers 3 & 4 = - 2 psi Stress due to vertical load Piers 1, 2, 3, 4, 5, & 6 f = l695 = 18 psi 8 x 12 Flexure stresses Piers 1 & 6 r 2,410 x 48 x 62 f = —r ;■ 1 — .- = + 11 psi 4 x 159,000 — Piers 2 & 5 ^ 3,220 x 48. x 78 . _ f = 4 X 3 1 6,0 00 i 10 psl Piers 3 & 4 „ 247 x 48 x 18 f = — ------- = + 14 psi 4 x 3890 “ Stresses due to overturning Overturning moment = 11,750 x 7*23 x 12 - 1,020,000” # Piers 1 & 6 f * s 1, 020,0 00 x 25^ = i [ psi 6 1,267,780 ” f = 1>020,000 x 192 = + 3 nSi 61,267,780 Piers 2 & 5 , * » 1,020,000 x 132 _ , o nsi 61,267,780 " - v f = 1,020,000 x 54 _ ^ q o pSi 61,267,780 ~ Piers 3 & 4 ^ 1,020,000 x 36 _ ^ ^ ^0, f “ 61,267,780 ” i °* P 1,020,000 x 18 f = 6 1, 2b7 ,7 8 0 ~ ± ° * 3 psl 71 0 0 0 0 © © 5 . s 3 j 5 0 f a ) S tre s s e s D u e to V e r t /c a t I o n e / Z1 R. ^ f T 7 Ct>) S tre s s e s D u e to F /e x u re A 5 L -J'2s ' £ 3 i. n ( c ) S tre s s e s D u e to O v e rtu rn in g / I V 9 ^ * / f ^ J'2) £ S \ ^ / Q Q S < * 01 / c r » cn k J V ^ h?i % « / ^ ( c t) C o m h '/n ect S tre s s e s Fig. 30. Stresses for Wall AH or BG / S c / / / / S e t t ' tSc/S/ 72 9'z 0t//s/afa /htcf ey/'/y/er- Fig. 31. Stress Diagram for Pier 6, Wall BG 1st bar from right = -22 x « 3*830 psl c U«u 2nd bar from right - 2ff-±.l§ x = 4,700 psi 2 0.2 Shearing stress at intersection of roof slab and shear wall. The solid roof slab acts as a horizontal girder when a seismic force is applied on the roof. The roof slab must transmit the seismic force to the shear Walls AB, CD, EF and GH. The maximum shearing stress thus developed is at the intersection of the roof slab and Wall AB. F = 6,930 K 19.33 x 12 x 5.5 = 5.4 psi CHAPTER V CONCLUSIONS Results of this analysis indicate that when a one-story wall-bearing reinforced concrete building is adequately designed for vertical load and adequately provided with reinforcing steel for shrinkage and temperature stresses in the walls, the building can safely resist seismic forces. The tensile stresses developed in the reinforce- i | rnent of the shear walls due to combined axial and earth- ! quake loads are well within the allowable stress of 18 ,00 0 psi without taking advantage of the 33 1 /3 per cent increase in design stress or 24,000 psi permitted | when earthquake forces are considered. The bar stressed i the greatest is the second one from the left in Pier 1, j Wall GH. This bar is stressed only to 15,240 psi. ! Since Wall GH has the greatest area of open space, it | is therefore stressed the greatest. One must bear in mind that the method used in 1 this analysis to distribute the lateral forces to the 1 I j shear walls is only an approximate method. This may j be shown most easily by a comparison of Piers No. 2 and I No. 3 in Figure 2 9, Page 6 7. At Pier No. 3 the spandrel 74 above and the wall section below are each evidently much stiffer than the pier itself; the assumption of fixed ends is Justified. At Pier No. 2 this condition is obviously reversed; the pier is stiffer than the elements which restrain its ends. Although the latter condition partially invalidates the primary assumption, one should consider the method as the best available practical one and the best aid to his judgment. It is generally agreed that If the building is not overstressed when the stresses are determined by the principles of mechanics the design is satisfactory. The method of distributing the total lateral force to the shear walls in accordance with their rigidi ties is limited to wall-bearing buildings with rigid diaphragms, namely, reinforced concrete slabs. If the diaphragms are flexible, such as wood diaphragms, the total lateral force distributed to the shear walls should be in accordance with tributary areas. When buildings have more than four stories, it is more economical to use skeleton framing, which consists of beams, girders, and columns. In this case the beams, girders and columns are designed as resisting elements to seismic forces, while the exterior walls serve only as an enclosure. To analyze structural frameworks for 75 seismic stresses* the most practical method is moment distribution. The actual deformations and stresses caused by an earthquake are dynamic ones. The vibrations which radiate out through the ground from the epicentre are exceedingly complex* These vibrations and the response of buildings subjected thereto defy a complete dynamic- stress analysis with our present knowledge of earthquake characteristics. Because of this lack of earthquake knowledge, it is justified to resort to some simpler and more practical method such as the gravity inertia loading method. This method is being used with success in California, one of the great earthquake regions of the United States. No doubt, the gravity inertia load ing method will continue to be valid until such time that scientists will devise methods to predict the intensity, duration, and character of an earthquake. When such time comes, no doubt seismic analysis can be based upon a complete dynamic-stress analysis. BIBLIOGRAPHY A. BOOKS Los Angeles City Building Code and Other Ordinances. VST." 17 edition? ~L6s IHgeTeil- CoITing Pub- llshing Company# 1954. Turneaure, Frederick E. and Edward E. Mauer, Principles of Reinforced Concrete Construction. Fourth Edition; Wew' York: John Wiley and Sons", IrTc., 1935• B. PERIODICAL ARTICLES Bolin, Harry W., "Earthquake Resistant Design for New School Buildings; Experience in Applying California’s Structural Law of 1933>" Engineering News Record, 118:415-17* March, 1937. Dale, E. P., "Long Beach Earthquake a Triumph for Con crete Construction," Concrete, 4l:10, June, 1933* "Earthquake Again Proves Efficiency of Concrete Con struction," Concrete, 41:9, April, 1933. Green, N. B., "Reinforced Concrete in the Long Beach Earthquake," Engineering News-Record, 110:560, May, 1933. "Reinforced Concrete Wins Approval as Shockproof Con struction," Concrete, 41:5, May, 1933* C. PUBLICATIONS OF ENGINEERING ORGANIZATIONS Analysis of Small Reinforced Concrete Buildings for Earthquake Forces. Chicago: Portland Cement Associa tion, 1946. 78 Kerekes, Frank, Building Code Requirements for Reinforced Concrete (ACI Betroit's American Concrete l' nstTtuTe,~I^5T:----- Oaiasrersiry at bcwtbarn GoHlNaAi

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