B U I L D I N G D E S I G N

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Damian Spencer
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1 B U I L D I N G D E S I G N

2 10.1 DESIGN OF SLAB P R I O D E E P C H O W D H U R Y C K DESIGN BY COEFFICIENT METHOD Loads: DL = 150 pc LL = 85 pc Material Properties: c = 3000 psi = 60,000 psi Design the panel (S 4 ) as two-wa slab with beam b coeicient Method.

3 C K S1 S S3 S4 14 t 15 t 8 t 15 t 15 t Top Floor Plan 19 t 0 t 3 Figure 10.: Top loor plan

4 C K Step1: Determination o Minimum Thickness p 840 h = 180 = 4.67 in. 180 whwre, P = (15+0) = 840 Select h = 5 in. as trial depth Step : Calculation o Factored Load DL = w c * DL = 150 x LL = 85 ps h ps ps ; where w c = 150 pc W = 1.4D L = (1.4 * * 85) = 3.7 ps Step 3 : Determination o Moment Coeicient la 15 Length ratio, m =. 75 lb 0 From the end condition case tpe is Case 4 From Appendix D-1 C a, neg = 0.076; C b, neg = 0.04 From Appendix D- C a,dl,pos = 0.043; C b,dl,pos = From Appendix D-3 C a,ll,pos = 0.05; C b,ll,pos = Step 4 : Calculation o Moment Middle Strip Moment: Positive Moments at Midspan M a, pos = C a, dl Wl a + C a, ll Wl a 4

5 C K M a, pos = * 3.7 * * 3.7 * 15 = 4974 t-lb M b, pos = C b, dl Wl b + C a, ll Wl b M b, pos = * 3.7 * * 3.7 * 0 = 699 t-lb Negative Moments at Continuous Edge M a, neg = C a, neg W l a = * 3.7 * 16 = 457 t lb M b, neg = C b, neg W l b = 0.04 * 3.7 * 0 = 34 t lb Negative Moment at Discontinuous Edge M a, neg, discontinuous = 3 1 * Ma, pos = 3 1 * 457 = 1509 t lb M b, neg, discontinuous = 3 1 * Mb, pos = 3 1 * 34 = 745 t lb Column Strip Moment: Column strip moments are /3 o corresponding middle strip s moments in respective direction. Step 5 : Check the Design Thickness d = M u ( Here = max = 0.75ρ b = 0.75 * 0.85*β 1 c' ) / c * = *1 d = *0.016*60,000*1( *0.016* ) 3 h required = (d + clear cover= 1 in ) = 3.66 in. =.66 in. h required h design, design is ok 5

6 C K Step 6 : Calculation or Reinorcement or Middle Strip In Short Direction: Minimum reinorcement: As = *b*d = *1*4 = =0.09in /t Spacing, S max = *h= 10 Midspan M u = 4974 *1 lb- in. B Iteration process Trial No. Trial, a As M u As = a = d a 0.85 cb Remark not ok not ok nearl ok ok Using # 3 bar required spacing: 1x0.11 Spacing = 4.4 4c / c 0.30 Continuous Edge M u = 457 *1 lb- in. B Iteration process As = 0.7 in /t. Using # 3 bar required spacing: 1x0.11 Spacing = 4.88". 4.5c / c 0.7 Discontinuous Edge As = in /t <As (min) 6

7 C K Using # 3 bar required spacing: Spacing = 1x " 10" c / c In long Direction: As = *b*d = *1*3.5 = =0.08in /t Midspan M u = 699 *1 lb- in. B Iteration process As = 0.15 in /t. Using # 3 bar required spacing: 1x0.11 Spacing = " c / c 0.15 Continuous Edge M u = 34 *1 lb- in. B Iteration process As = 0.1 in /t. Using # 3 bar required spacing: 1x0.11 Spacing = 11 10" c / c 0.1 Discontinuous Edge As = 0.04 in /t <As (min) Using # 3 bar required spacing: Spacing = 1x " c / c = 10 in.c/c

8 C K Cornar reinorcement # 4 in c/c upto L/5 in respective directions. Step 7 : Calculation or Reinorcement or Column Strip The average moments in columns being two-third o the corresponding moments in the 3 middle strips, adequate steel will be urnished i the spacing o this steel is times that in the middle strip. Using # 3 bar spacing or column strip Midspan = 4 x 3 = 6 c/c Continuous edge = 4.5x 3 = 7.5 c/c Discontinuous edge = 10x 3 = 15 c/c But maximum allowable spacing = h = 10 in. Use10 c/c Step 8: Check or shear Total load = 3.7*0*15 = lb From Appendix D-4 m=.75 case-4 Wa=.76 Wb=.4 V u,a = 69810*.76 *(015) 748lb/ t V u,a = 69810*.4 *(015) 36lb/ t 8 Slab strength = ØV c = Ø** c *b*d = 0.85** 3000*1*4 = 4470 lb >>V u ; Ok.

P R I O D E E P C H O W D H U R Y C E @ K 8.

9 P R I O D E E P C H O W D H U R Y C K Whole Slab Reinorcement Step 9: Detailing Figure 10.3: Detailing o slab 9

10 C K B A B A 0 t (Long Direction) 15 t (Long Direction) Figure 10.3: Detailing o slab (continued) 10

11 10 in c/c 8.5 in c/c 4 in c/c 10 in c/c 5in 10 in c/c Section A-A 10 in c/c P R I O D E E P C H O W D H U R Y C K

C E @ K 8. 0 1 7 6 9 4 4 1 8 3 Figure 10.

12 C K Figure 10.3: Detailing o slab (continued) 10. DESIGN OF BEAM B 5 B 4 B 3 B B 1 Figure10.4: Beam or Design 1

13 C K Step 1: Determination o load on the Beam (we have used E-TABS data, but conventional approach is like this) For beam B : a * b a * a Slab area ( ) 4 14*19 14*14 15*19 15*15 ( ) +( ) 170.5st 4 4 Slab load = 170.5*(5/1)*150= lb = 10641/14= 760 lb/t Choosing o beam is a matter o trial and error or experience For this beam let assume a 1 inch X 16 inch section Weight o beam lb/ t 1 1 Wall load = 1000 lb/t Deadload ( ) 1960lb/ t Live load=85 ps Total live load= lb/t Factored load U=1.4D+1.7L= lb/t So or this case, Total load, w = 4.5 k/t u 13

14 C K Beam Lenth (t) Section (in*in) Area contributing (t ) DL (slab) (lb/t) DL (beam) (lb/t) Wall load (lb/t) Total DL (lb/t) Total LL (lb/t) Total actored load (lb/t) B * B 15 1* B 3 8 1* B * B * Step : Determination o approximate moment (we have used E-TABS data, but conventional approach is like this) Approximate Moment M u M u ( ) 0.08 wl ( ) wl The Beam should be designed or highest negative moment, that is 81.7 t kips. Step 3: Selection o the cross sectional dimension o the (we have used ETABS data, but conventional approach is like this) Minimum thickness l " in = 1 in. 1 1 (Keeping the assumed size) Width rom equation M u bd c 14

15 C K M u b d Here, c (1.5) = in = 14 in can be taken. M u = in-kips max b 0.75 b c b max 1 Using ETABS the analsis this step can be done. A beam section o 1in x16in and column o 18in x 18 in is chosen and analzed. The ollowed moments were ound. 15

16 C K Figure 10.5: BMD o B rom ETABS (top loor) Step 4: Determination whether the Beam has to be designed as Singl or Doubl reinorced I we add the slab thickness Then we get total depth, h=1 in. Eective depth=17in. And web width, b=1 in. Assuming singl reinorced beam, Steel area, A s = bd A s = A s = 3.64 in h =5 in with the Beam thickness 16 in, Here, max where, 0.85 b 0.75 b 1 c b = Now, As a c b a =6.4 in And M n = A s (d-a/ ) M n = 3.64*60*(17-6.4/) M n = 70.5 in-kips > Imposed moment = 185.5*1 k-t. So the beam is to be design as singl reinorced beam. 16

17 C K Step 5: Determination whether the beam has to be designed as Rectangular Beam or a T Beam Eective lange width b Smallest o the ollowing three condition will determine the eective lange width b. Span/4= 15*1/4= 45in. b 16h b in. w c/c distance/= 10in. So eective lange width b=45 in. Let a= h = 5 in Now, A s M u a d A s =.83 in.83 So, = 4517 =.0037 d Now, a as a h c in. So the beam will be designed as rectangular beam which would have the ollowing cross section. Step 6: Rectangular Beam Design Now A s M u a d 17

18 C K And a As 0.85 c b Trial No. Assumed a As = M u d a As a = 0.85 b Remark not ok ok c Here M u = k-t (considering ETABS data) Beam M u(+) Reinorcement(in ) M u(-) Reinorcement(in ) B ; # ; 4#6 B ; # ; 4#8 B ; # 3 # ; #5 B ; # ; 4#8 B ; # ; 4#6 Step 7: Shear Reinorcement V c c b w d Vu = kips kips kips. V c Now wherever o the beam the Vu exceeds Vc we need to provide shear reinorcement. For this case nowhere o the beam the shear exceeds Vc V where shear orce exceeds c kips. but minimum reinorcement required 18

19 Av d s V V u c P R I O D E E P C H O W D H U R Y C K Using # 3 bar s in up to 5 t rom support on each side. The rest with minimum spacing. As V s 4 c b w d d 17 S max 8. 5in s 4.5 First stirrup at.5. 5 in. Step 8 & 9: Design or Torsion and Serviceabilit Normall in buildig design torsion and Serviceabilit is not considered because the tortional eect is normall countered b shear reinorcement. 19

20 0 P R I O D E E P C H O W D H U R Y C K

21 A B C A B C Step 10: Detailing P R I O D E E P C H O W D H U R Y C K

C E @ K 8. 0 1 7 6 9 4 4 1 8 3 47 in 5 in 1 in 4 no # 8 Bar 17 in # 3 @ 4.

22 C K in 5 in 1 in 4 no # 8 Bar 17 in # 4.5in c/c Section B-B 1 in Section A-A 47 in 5 in 16 in # 4.5in c/c no # 7 Bar 1 in 1 in Section B-B

23 C K in 1 in 17 in 4 no # 8 Bar # 4.5in c/c 1 in no # 7 Bar Section C-C Figure 10.6: Detailing o beam (continued) 3

24 C K DESIGN OF COLUMN Design o column C subject to axial load as a tied column with ollowing data: ' c = 4 ksi (as column requires more importance in design and better qualit) = 60 ksi Step 1: Determination o Factored Load P u = 1.4 D L These values were collected rom E TABS. Step : Steel Ratio Assumption Since value o axial load is low, take g = 0.04 Step 3: Determination o Concrete Gross Area Pu = 0.80 A g 0.85 (1 ) For tied column 0.70, p 380 kips c 686=.80 A g * 0.70 [0.85 * 4(1-0.04) * 60] A g = 16 in Step 4: Selection o Column Size g u g Let us choose a square column o size = 14in. x 18 in. A g = 5 in 4

25 C K Step 1,, 3 & 4 is done b ollowing table (Data From ETABS) Combination P u (kips) M x ip in) M (kip in) Section * * * * * * * * * *14 Step 5 : Check or Steel Ratio P = 0.80 A g 0.85 (1 ) u c g g = (0.80 * 5 * 0.70).85*4(1 ) *40 0 g g g = 0.06 =.6%; Limit o g is ok. Step 6: Calculation o Reinorcement A st = A g * g = 5 * 0.06 = 6.55 in Let us choose # 8 bar (A b = 0.79 in ) Ast 6.55 No o bar, N 8nos. A 0.79 Step 7 : Selection o Ties b Use # 3 bar or ties Step 8 : Determination o Vertical Spacing o Ties 16 d b o longitudinal reinorcement = 16 x 1 = 16in (dia # 8 = 1 in) 48 d b o tie bar = 48 x = 18 in Least dimension o column section = 14 in Choose vertical spacing o ties = 14 in 5

26 C K Step 9: Arrangement o Ties Clear spacing between longitudinal bars in direction =[Column dimension ( x clear cover) ( x dia o ties) (3 x dia o bar)] = [18 ( x 1.5) (* ) (3* 1)] = 5.65 in < 6 in No additional ties are required DESIGN OF COLUMN FOR BIAXIAL BENDING USING LOAD CONTOUR METHOD Step 1 : Determination o Factored Load and Moment P u = 1.4D + 1.7L = 686kip (ACI Code-00) M nx = M ux =845kip-in and M n = M u =78kip-in Step : Selection o Chart rom Appendix F or Bending About X- Axis h d 18.5 Find: = h 18 Selecting the design chart o Appendix F chart corresponding to Step 3 : Determination o Moment Parameter or Bending About X Axis Find: P A g n 686 = and g =0.04 Intersection o these two points on the chart given the value o Determine M nxo = kip in M A g nxo h =0.45 Step 4 : Selection o Chart rom Appendix F or Bending About Y- Axis h d 14.5 Find = h 14 Selecting chart corresponding to 6

27 C K Step 5 : Determination o Moment Parameter or Bending About Y Axis Pn 686 Find:. 7 and A 5 g g =0.04 (same as step 3) M Intersection o these two points give A h Determine M no = g no = kip in Step 6 : Checking o Design Strength M M nx nxo 1.15 M M n no 1.15 Replacing all the values in equation 845 = = (Design is ok) 7

28 C K Step 7 : Detailing #3 Vertical spacing 18in. 14 in. 1 o 8 # 8 bar 8

29 C K Figure 10.7: Detailing o column 10.4 DESIGN OF FOOTING Rectangular Footing Data: Unactored column dead load, DL =666.8 K Unactored column live load, LL = K Column size = 18 in x 14 in. Allowable soil pressure, q a = 4400 ps c = 3000 psi = 60,000 psi Step 1 : Determination o Eective Bearing Pressure (q a ) Let, bottom o the ooting is at 3 depth rom grade. So, D = 3 q e = (q a 15 x D ) ps = ( x 3) = 3775 ps Step : Determination o Area o Footing A= DL LL q e ( ) t With width 14 t, Size o the ooting = 16 t x 14 t Step 3 : Determination o Bearing Pressure or Strength Design q u = 1.4DL 1.7LL ps * b 1.4* * *14 = 563ps 3 9

30 C K Step 4 : Determination o Eective Depth rom Punching Shear Consideration 4 b d q c o u * b c d * 4 * 3000* 416 d Step 5 : Check or Flexural Shear 16 d d d = inch. 1 Long Direction V c (1bd) c = * 0.85 * 3000(1*14*33) = lb V u c / 1 d qu * b = 563 x = lb V u < V c, design is ok. Short Direction V c (1ld) c = * 0.85 * 3000* (1 * 16* 33) = lb b c/ 1 d Vu qu * lb / 1 33 = 563 x 16 1 = lb 30

31 C K V u < V c design is ok. Step 6 : Check or Bending Moment max = 0.75 b * c 87,000 87,000 = Long Direction Moment qu M u c1/ 1 * b lb-t = 16 * = lb-t = 337 k-in d = max 1M b u max 1 c inch 337 = *0.016*60 x *0.016 x 1 3 = in. d < d (provided), design is ok. Short Direction Moment M qu b / 1 * lb-t 8 u, s C = 14 * = lb t = 0803 k-t 0803 d = *0.016*60* *0.016 x 1 3 = 1.43 in 31

32 d < d (provided0, design is ok. P R I O D E E P C H O W D H U R Y C K Step 7 : Estimation o Thickness t min = d + 3 in + 1 db = = [Using #8 bar] Provide, t = 37 in. Step 8 : Determination o Steel Area or Long Direction (A s, l ) For Iteration Method assuming a = 4 inch Trial Assumed, a A s, l = M u, l ( d a / ) (in ) a = A s, l 0.85 b c (in) Remark [M u = 337k-in] Not ok Not ok ok A s = 13.4 in Check or Minimum Steel 3 c *1 33 As, min b *1d 15 in 60,000 But not less than in 60,000 A s, min = b *1d 14* A s,l < A s, min 3

33 So A s = 18.5 in P R I O D E E P C H O W D H U R Y C K Number o Bar Select # 9 bar (A b = 1 in ) As, l 18.5 N = A 1 b Spacing bx1 14x1 S = N 19 in c/c Use 19 # in c/c Step 9 : Determination o Steel Area or Short Direction (A s,s ) Required Steel Area A s, s = A s,s = M u, s, *1 d a / * / 1 in Check or Minimum Steel = 1.1 in 60,000 A s, min = *1d 16*133 A s, s < A s, min Use A s, s = A s,min = 1.1 in Steel or Band Width 33 Steel Area

34 A s,bw = A s,s x % 1 P R I O D E E P C H O W D H U R Y C K = A s,bw = 17.8 x = 19.7 in Band width = 14 t. No o Bar Select # 9 bar (A b = 1 in ) As, bw 19.7 No o bar, N = A 1 Spacing b bx1 14x1 S = N 0 in c/c Use 0 # 9 bar 8 in c/c Steel or Outside o Band Steel Area A s, s As, bw A s, out o band = in No o Bar 34 Select # 6 bar (A b = 0.44 in ) 0.54 No o bar, N= 1.3 on each side 0.44 Spacing

35 1x1 Spacing, S = 6 in c/c Use 3 # 6 bar 8 in c/c on each side. P R I O D E E P C H O W D H U R Y C K

36 C K Step 10: Reinorcement Detailing 18 in 37 in 14 t 16 t 18 in # 6 19 # 9 0 # 9 # 6 4in 16 t Figure 10.8: Detailing o ooting 36

37 C K DESIGN OF WATER TANK DESIGN OF A ROOF TOP WATER TANK Step 1: Water requirement Water consumption rate = 40 gpcd. Number o persons= no o lats X 6 =1 X 6 =7 persons (considering six persons per lat) Total water requirement=7 X 40=88 gal/da 880 Dail requirement = ct 46 ct/da 6.4 Step : Tank dimension Let inside Dimension L=15 t =4.57 m B =7 t =.1336 m So L/B=15/7> 46 Height o water level= t Free board = 0.5 t. Step 3: 37 So inal height = =4.9 t 5 t Part 1 Here, h =H/4 or 1 m (larger height to be considered) Part (H-H/4) or (H-1) m to be considered. For L/B> We have, h = H/4=5/4=1.5 t m <1m h =1 m =3.8 t

38 C K Step 4: Design or long wall 3 wh Moment M = 6 Here w= 9.81 kn/ m 3 10 kn/ m 3 H=5t=1.54 m M = 5. 9 kn-m = lb t =4.351 kt =5.3 kin [1 kn-m= lb-t] Check or, 1 M max ckjbd d d 4.98inch Overall depth= x4/8=6.75 inch (considering # 4 bars) Here, c s r ksi 0.5 E n E s c s c ' c 30ksi psi. n 9 k 0.88 n r 9. k j Let us take overall thickness o wall=7.0 inch So eective depth = X4/8=5.5 inch 38

39 C K Then, A s M in / jd A (min) 0.3% s s in / t 100 t (o cross sectional area) So, A s 0.63in / t (provide # 4 4 inch c/c) Direct tension in the wall, 10(1.541) 5.59kN / m.1336 T L B w( H h) lb/ t A TL F A (min) 0.5in / t s s s in / t So # 3 5 inch c/c to be provided Since steel is provided on both aces thereore steel to be provided on both aces as # 3 10 inch c/c. Step 5: Design or short wall Force P=w(H-h)=10 X(1.54-1)= 5.4 kn/ m = lb/t (per m run) Eective span in horizontall spaced slab=7+6.5/1=7.54t=.3 m Pl Bending moment at end, M= 1 w( H h) B 1 M KN-m (per m run) 1 = lb t=0.66k in (per t run) Reduction in moment due to tensile steel = T x = x 1.5/1=47.88 lb t=0.575 k in 39

40 C K Design moment Design moment = M-T x = = 0.1 k-in Steel requirement A s A (min) 0.5in / t s A M Tx in / jd s s 0.5in / t t We will use # 3 5 inch c/c. At mid section PL M 10.33kin A 0.07in / t So s A s (min) A s will be provided as # 3 5 in c/c at mid section. Step 6: Cantilever eect on short column Maximum moment whh M max kn. m.54kn. m 6 6 = lb t =.5 k-in Steel requirement A s M jd s in / t As (min) 0.5in / t So use # 3 5 inch c/c. 40

41 C K Step 7: Design o base slab L/B>, so we will design or one wa slab Minimum thickness L 151 t 9in (For 60 grade steel) 0 0 Let thickness = 9 inch 9 Total weight o base slab= 150 ( ) ks 1 9 Eective width, B= t Moments, M max kt=38.9 k-in 8 Depth d 38.9 =3.17 inch (OK) d in / inch 8 A s t Use # 3 4.5inch c/c. So # 3 9 inch c/c should be used at each ace. 41

42 C K Step 8: Detailing L=15 t 7 in 5 t 18 t 10 t Figure 10.9: Detailing o over head water tank 4

43 C K A A # 3 10 inch c/c # 4 4 in c/c # 3 9 inch c/c Section A-A Figure 10.9: Detailing o over head water tank (continued) 43

44 C K DESIGN OF UNDERGROUND WATER TANK General data Volume to be stored= 46 94ct (For two das store dail requirement 46 ct) Angle o repose 30, 6 dr Unit weight o soil=w=15 pc=0 kn/ m Most critical condition: Empt water tank and wet soil. wet Step 1: Tank dimension Let inside dimension, L=15 t=4.51 m. B=7 t=.1336 m. 94 So height o water level= 8.8 t 15 7 Free board=0.3 t Final height= =9.3t.896 m. Step : Design o long walls 1 sin Pressure exerted b wet soil = wh 1 sin p 46.96kN / m 1 sin6 = kN / m 1 sin6 ph Tension near the water ace= kn.m k in (Per inch run) So tension near water ace/ t run= X0.3048=31.7 kip inch 44

45 C K Tension awa rom water ace M max ph 6.6 kn/m =3.8 k-in (per inch run) =70.96 kin (per t run) From cracking consideration the thickness o wall is determined. M D F ct bd et 6 6M in b D=9.8 in 9.5 in Here 8 ' Let, M max ct 6 c ct 7.5 ' psi = k-in D= Total thickness c Eective depth = =8 inch d=8 inch Step 3: Vertical reinorcement (long walls) M Steel requirement, A 0.33in t s jd / s A (min).003bt in A s s 0.33in / t / t So use # 4 7 inch c/c (inner orce) Steel requirements or M=31.7 kip in 31.7 As 0.15in A 0.34in / t s / t A (min) s So use # 4 7 inch c/c (Outer orce) 45

46 C K Step 4: Horizontal reinorcement (long walls) Minimum steel requirements A s (min).003bt 0.34in / t Use # 4 7 inch c/c Step 5: Design o short wall Earth pressure at the bottom P=46.96 kn/ m PL Max moment at the center M= M 1 =.07 kin (per m length) =195.35k-in (per in length) =59.55 k-in (per t length) 9.5 L t.375m 1 Now M max d c kjbd inch 8inch Step 6: Vertical reinorcement (short wall) M A 0..75in s jd / s t Use # 4 8 inch c/c. 46

47 C K Step 7: Horizontal reinorcement (short wall) A s (min) 0.34in / t Use # 4 7 inch c/c that is 14 inch c/c both side. Step 8: Design o base slab Thickness provided=9.5 inch (let) Minimum reinorcement=0.003bt=0.34 in / t Use # 4 7 inch c/c. 47

48 C K Step 7: Detailing 7 t 15 t 9.5 t 9.5 in 18t 10 t 48

49 C K # 4 7 inch c/c # 4 8 inch c/c. # 4 7 inch c/c # 4 14 inch c/c Figure 10.10: Detailing o under ground water tank (continued) 49

50 C K DESIGN OF SHEAR WALL 9 t 8 t 9 t 6.5 t 39 t Figure 10.11: Plan o a loor Given data: 6 storied building. Height o each loor 10 t. Maximum wind pressure per st= 10 lbs Zone 1 R w =4 Total seismic load 0 kips/t. I= Importance actor =

51 C K Step 1: Horizontal loads on the structure Wind loads F w b windpressure lb/t =1.3 kips/t V Seismic loads ZIC R w W Where, W= Total seismic load 0 kips/t R w 4 Z=0.075 I= S Now C T C 4 t hn T T C [s=1 or sti soil] =.19 V F 0.5V F t e wxhx w h x x ( ) (101) 10 (w x =w x ) Step : Calculation o moment or vertical lexural reinorcement 1 M u t-kips 6 1 M u kip 51

52 C K Step 3: Design o vertical lexural reinorcement 6.5 t 10 in 8 t 10 in Figure 10.1: Shear Wall Cross Section Balanced steel ratio 3 87 b max Area o steel i singl reinorced A s maxbd (96 5) in 1 As 1 Then a 0.85 c in b Maximum moment that can be developed a 34.7 M n A s d in-kips =41944 t-kips > M u [So no compression steel.] Using minimum 5 A b d 00 (6.51) 91 w s min in in

53 C K A s 3.66 maximum moment that can be developed is Now using in M n kip in As, a 7. 14in t- kips > M u Thereore A s 3.66in which can be provided b 11#14 bars in back ace and with 4.75 square inch o steel area. Step 4: Shear reinorcement Values o V c : Vc c hd Kip. Shear Vu in no case exceed V c kips. Thereore minimum shear reinorcement should be provided in the lange. Horizontal shear reinorcement: Maximum spacing, l w S, 3h or 18 in 5 S 15.6, 88 or 18 in So, S 18 inch (can be provided b # 3 bars) Vertical shear reinorcement: Maximum spacing l w S1, 3 h or 18 in in, 88 in or 18 in S1 So, S 1=18 in (can be provided b # 3 bars) 53

54 Step 5: detailing P R I O D E E P C H O W D H U R Y C K t 10 in 8 t # 3 18 in 11# 14 bars 54

55 C K # 3 bars o 15. inches (Vertical bars) 10 t x 6 store Figure 10.13: Detailing o shear wall (continued) 55

56 C K DESIGN OF DOGLEGGED STAIR CASE Step 1: General arrangement LANDING PASSAGE Figure 10.14: Dog legged stair case (general arrangement) 8t The igure above shows the plan o the stair hall. Let the rise be 6 inch and trade be 6.5inch. The width o each light is 7.5/=3.75t 10 Height o each light = = 5 t. 5 1 No o risers required = = 10 risers in each light. 6 No o tread in each light = 10-1 = 9. Space occupied be trades = 9 10=90in= 7.5 t. Width o landing =6.5 t. Width o passage =6.5t. Size o stair hall = 7.5 t 0 t. 56

57 C K Step : Design constants For steel = 60,000 psi And or concrete c = 3000 psi Step 3: Determination o loading The landing slab acts together with the going as a single slab. The bearing o the slab into the wall ma be considered 8 inch. 8 Then the eective span = t. 1 l Considering one-wa slab with both end continuous minimum thickness is. 8 So, t = l inches 8in inches Sel weight o the slab = 1501= 100 pl. 1 1 Tread Riser Tread Sel weight o the steps = = =50 pl. Floor inish = 0 pl. Total dead load = =170pl Live load = 100 pl. So, Design actored load = =408 pl. 57

58 C K Step 4: Bending Moment Calculation Maximum Moment 1 M max wl lb-t =19.8 k-in. 8 8 Check or depth 3 87 max 0.75b M max d b c 19.8 = d =4.8 inch And t = 4.8+1=5.8 inch (with 1 inch clear cover) t =5.8 inch < 8 inch (Ok) d available = 8-1 = 7 inch Step 5: Reinorcement Calculation Distribution Bar. Minimum reinorcement is provided as temperature and shrinkage reinorcement. Temperature and shrinkage reinorcement, A st b t in / t # 3 bar can be used. The spacing will be, S = =7.5 inch c/c

59 C K Longitudinal Steel. This is selected b trial. Trial No Trial-1 Trial- Assumed a (inch) a=1.0 a=1.07 M As Steel Area, As a a d 0.85c (inch) (inch ) b Comments nearl Ok Ok So, A s 0.55inch is provided. It can be urnished b using # 4 bar. 0.1 Spacing = inch center to center

$5 inch # 4 bar @5 inch c\c 6 inch #$

60 C K Step 6: detailing # 4 inch c/c 11.5 inch # 4 inch c\c 6 inch # 3 inch c/c 4.5 t 7.67 t 4 t 60

Design of a Multi-Storied RC Building

Design of a Multi-Storied RC Building 16 14 14 3 C 1 B 1 C 2 B 2 C 3 B 3 C 4 13 B 15 (S 1 ) B 16 (S 2 ) B 17 (S 3 ) B 18 7 B 4 B 5 B 6 B 7 C 5 C 6 C 7 C 8 C 9 7 B 20 B 22 14 B 19 (S 4 ) C 10 C 11 B 23