Design Stirrups Using F'c 4000 Psi and
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CE403 – REINFORCED CONCRETE DESIGN HANDOUT #3 CONTINUOUS BEAMS AND ONE-WAY SLABS CUA – FALL 2017 DR. LONG T. PHAN LONG.PHAN@NIST.GOV HANDOUT OBJECTIVES • ANALYZE CONTINUOUS RC BEAMS AND ONE-WAY SLABS USING: • APPROXIMATE METHODS (ACI MOMENT AND SHEAR COEFFICIENTS) • DESIGN CONTINUOUS RC BEAMS AND ONE-WAY SLABS FOR: • FLEXURE • SHEAR • SERVICEABILITY 2 OUTLINE • ANALYSIS METHOD FOR CONTINUOUS BEAMS • DESIGN REQUIREMENTS FOR CONTINUOUS RC BEAMS AND ONE-WAY SLABS • FLEXURE • SHEAR • DESIGN EXAMPLES 3 ARRANGEMENT OF LIVE LOAD FOR DESIGN 4 ARRANGEMENT OF LIVE LOAD FOR DESIGN 2nd Live Load 5 ARRANGEMENT OF LIVE LOAD FOR DESIGN Fixed (typ.) 2nd 6 ARRANGEMENT OF LIVE LOAD FOR DESIGN • Dead Load (D) on all spans; Live Load (L) on 2 adjacent spans: D+L A D B C D Loading pattern for max. negative moment at support B • Dead Load (D) on all spans; Live Load (L) on alternate spans: D+L A B C D+L D A D+L D B D D C D Loading pattern for max. negative moment at support A and positive moment in span AB Loading pattern for positive 7 moment in span BC APPROXIMATE ANALYSIS METHOD FOR MOMENT AND SHEAR IN CONTINUOUS BEAM AND SLAB DUE TO GRAVITY LOADS ACI Moment and Shear Coefficients Method Conditions for applicability: • • • • • Beams or slabs have two or more spans Loads are uniformly distributed. Live Load ≤ 3 Dead Load Beams are prismatic Spans are approximately equal, with difference in span length between two adjacent spans ≤ 20% Uniformly distributed load (L/D 3) 1.2n n Two or more spans Prismatic members 8 APPROXIMATE ANALYSIS METHOD ACI Moment and Shear Coefficients Method (Cont'd) Provide max. negative and positive moments and max. shears with consideration of alternative placements of live load • FOR CONTINUOUS BEAMS AND ONE-WAY SLABS: • MAXIMUM MOMENT AT VARIOUS LOCATIONS Cm (wu ln2 ) • MAXIMUM SHEAR AT VARIOUS LOCATIONS Cv wu ln Cm AND Cv = COEFFICIENTS DERIVED BASED ON ELASTIC ANALYSIS wu = TOTAL FACTORED LOAD PER UNIT LENGTH OF BEAM OR PER UNIT AREA OF SLAB ln = IS CLEAR SPAN FOR POSITIVE MOMENT AND SHEAR; OR THE AVERAGE OF THE TWO ADJACENT CLEAR SPANS FOR NEGATIVE MOMENT. 9 APPROXIMATE ANALYSIS METHOD ACI Moment and Shear Coefficients (Cont'd) wu Integral with support Spandrel support Cm Column support 1/24 n n n 1/14 1/16 1/11 1/10* 1/2 1/11 1/11 1/10* Positive Moment 0 *1/9 (2 spans) 1/16 Cv Simple support 1.15/2 1/2 1/2 1.15/2 1/2 ( wuln2 ) Negative Moment ( wuln2 ) ( wuln ) Shear 10 CONCRETE FLOOR SYSTEMS Flat Plate One-Way Joist • Short/medium spans (15 ft – 25 ft) • Moderate live loads • Standard • Long spans (30 ft – 50 ft) • Heavy live loads Flat Slab One-Way Joist • Short/medium span (20 ft – 30 ft) • Heavier live loads Two-Way Joist • Long spans (40 ft – 50 ft) • Heavy live loads • Wide-Module • Long spans (30 ft – 50 ft) • Heavy live loads Beam-Supported Slab • Long spans • Parking structures ONE-WAY SLABS La Lb 12 ONE-WAY SLABS Flexural Reinforcement Ratio for Slabs Lb 12" b=12" d As slab A nAs s s x d 12 x d Where : s As Area of one bar n average number of bars per foot of slab width ( s center - to - center spacing of bars 12" ) s 13 ONE-WAY SLABS Minimum Slab & Beam Thickness h (unless Deflections are Calculated) Min. Thickness h of non-prestressed Beams or One-way Slabs wc=145-150 pcf, fy=60 ksi 90 ≤wc≤ 120 pcf fy 60 ksi Simply supported: Solid one-way slabs Beams or ribbed slabs h l/20 h l/16 (1.65-0.005wc) 1.09 (0.4+fy /100,000) One end continuous: Solid one-way slabs Beams or ribbed slabs h l/24 h l/18.5 (1.65-0.005wc) 1.09 (0.4+fy /100,000) Both end continuous Solid one-way slabs Beams or ribbed slabs h l/28 h l/21 (1.65-0.005wc) 1.09 (0.4+fy /100,000) Cantilever Solid one-way slabs Beams or ribbed slabs h l/10 h l/8 (1.65-0.005wc) 1.09 (0.4+fy /100,000) 14 ONE-WAY JOIST Pan Joist (Ribbed Slab) Floor System Flexural Reinforcement hs h REQUIRED THICKNESS hs OF RIBBED SLAB : • hs 2 IN. FOR JOISTS FORMED WITH 20 IN. WIDE PANS • hs 2.5 IN. FOR JOISTS FORMED WITH 30 IN. WIDE PANS (ln /12) 15 Pan Joist (Ribbed Slab) Floor System Standard Form Dimensions • Widths: 20 in. & 30 in. (measured at bottom of ribs) • Depths: 6, 8, 10, 12, 14, 16 or 20 in. ONE-WAY JOIST Minimum Temperature & Shrinkage Reinforcement Ratios Minimum Ratios of Temperature & Shrinkage Reinforcement , min, in One-way Slabs (based on gross area b h) Slabs with Gr. 40 & 50 deformed bars 0.0020 Slabs with Gr. 60 deformed bars or welded wire fabric 0.0018 Slabs with reinforcement with fy > 60 ksi • Temperature & Shrinkage Reinforcement Ratios 0.0014 • Spacing of temperature and shrinkage reinforcing bars 5h or 18" 0.0018 x60,000 fy 17 DESIGN REQUIREMENTS FOR CONTINUOUS BEAMS & ONE-WAY SLABS • FLEXURE (STRENGTH REQUIREMENTS) • SHEAR • SERVICEABILITY REQUIREMENT (DEFLECTION) 18 DESIGN OF ONE-WAY SLABS – EXAMPLE 1 PROBLEM: DESIGN ONE-WAY FLAT SLAB FLOOR SYSTEM USING ACI MOMENT AND SHEAR COEFFICIENTS METHOD: • wl = 125 PSF; • wd= 10 PSF (FILL & SURFACE) + 10 PSF SUSPENDED LOADS + SELF WEIGHT • fy = 60,000 PSI; f'c = 3000 PSI 16'-0" 16'-0" 19 DESIGN OF ONE-WAY SLABS – EXAMPLE 1 (CONT'D) • MINIMUM THICKNESS OF ONE-WAY SLAB WITH ONE-END CONTINUOUS (TABLE ON SLIDE # 14): h l 16 x12 8"; assume d 7" 24 24 • SLAB SELF-WEIGHT: 150 x8 wo 100 psf 12 • FACTORED LOADS: wu 1.2wD 1.6wL 1.2(100 10 10) 1.6(125) 344 psf 20 • CLEAR SPAN: 16' – 9" – 9" = 14.5' DESIGN OF ONE-WAY SLABS – EXAMPLE 1 (CONT'D) 344 x14.52 wu l 72.3 ft - kips 1000 • MOMENT MULTIPLIER: • MOMENTS: A: AB: B: BC: C: 2 1 1 wu l 2 72.3 3.01 ft - kips 24 24 1 1 wu l 2 72.3 5.17 ft - kips 14 14 1 1 wu l 2 72.3 7.23 ft - kips 10 10 1 1 wu l 2 72.3 4.52 ft - kips 16 16 1 1 wu l 2 72.3 6.57 ft - kips 11 11 21 DESIGN OF ONE-WAY SLABS – EXAMPLE 1 (CONT'D) • FLEXURAL RESISTANCE FACTOR R: Mu R bd 2 3.01x12000 * 2 68 . 25 0.0012 A 0.1008 in /ft use #4@23" s 2 0.9 x12 x7 5.17 x12000 RAB 117 0.0020* As 0.168 in 2 / ft use #4@14" 0.9 x12 x 49 7.23x12000 RB 164 0.0028 As 0.24 in 2 / ft use #4@10" 0.9 x12 x 49 4.52 x12000 RBC 102 0.0017* As 0.14 in 2 / ft use #4@16" 0.9 x12 x 49 6.58 x12000 RC 149 0.0026 As 0.22 in 2 / ft use #4@10" 0.9 x12 x 49 RA • MIN. SLAB REINFORCEMENT FOR CONTROL OF TEMPERATURE AND SHRINKAGE CRACKING (SLIDE #17, ADJUSTED FOR EFFECTIVE DEPTH b d) : 0.0018 x12 x8 0.0021 (control for cases with * above) 12 x7 As ,min 0.0021bd 0.0018 x12 x7 0.18 in 2 per foot min 22 DESIGN OF ONE-WAY SLABS – EXAMPLE 1 (CONT'D) • CHECK BAR SPACING: smax = 3H = 3 8 = 24" OK • CHECK SHEAR (ACI COEFFICIENT METHOD, SLIDE #10): w l 1.15 x344 x14.5 Max. Vu 1.15 u n 2.87 kips 2 2 x1000 1 Vc (0.75)2 4000 x12 x7 x 9.03 kips Vu Shear is OK 1000 16'-0" 16'-0" 23 CE403 – REINFORCED CONCRETE DESIGN HANDOUT #4 - DESIGN FOR SHEAR Flexural Reinforcements Stirrup size ? Spacing s? s s Shear Reinforcements (U Stirrups) CUA – FALL 2017 DR. LONG T. PHAN LONG.PHAN@NIST.GOV SHEAR FORCES & SHEAR FAILURE IN RC MEMBERS • Shear (Vu): Forces that act perpendicular to longitudinal axis of members. • Vu is usually largest at the supports. Shear force at any distance x from the support (Vu/x ) decreases by the amount of load between the support and the distance x • Shear failure occurs abruptly (brittle failure, must be prevented!) • To prevent shear failure: Shear Strength (Vn) ≥ Design Shear Force Vu Load Point Where: Vn: Nominal Shear Strength; Vu: Factored Shear Force • Shear cracks (or diagonal tension cracks) are caused by principal tensile stresses acting perpendicular to the compressive stress trajectories, (typically inclined, between beam support and load point). Shear or diagonal tension crack 2 Beam support Typical shear (or diagonal tension) failure in beam SHEAR STRENGTH OF BEAM WITHOUT SHEAR REINF. (VN = VC) Shear strength Vc of a concrete beam without shear reinforcement is to be computed using either Equation 1 or 2 below (based on empirical data, 440 shear tests) Vc (1.9 f ' c 2500 V ud Mu )bwd 3.5 f ' c (Eqn. 1, detailed equation) , or Vn : Nominal shear force at a section where shear crack forms Mn : Nominal bending moment at same location Mn/(Vnd): Shear Span/Depth ratios ρ: Flexural Reinf. Ratio = As/bwd λ: Multiplier to account for use of lightweight concrete Stress Causing Shear Cracks, vcr = Vc 2.0 f ' c bwd (Eqn. 2, simplified equation) 3 COMPONENTS OF CONCRETE SHEAR STRENGTH VC Vcz = Shear Resistance of Uncracked Concrete C z Vi = Interface Shear Transfer Force T Vd = Dowel Force p Nominal Concrete Shear Strength Vn: Vn = Vc = Vcz + Viy + Vd (computed from Eqn. 1 or 2) 4 SHEAR STRENGTH OF BEAM WITH SHEAR REINFORCEMENT Vcz C s Avfy Avfy Vi T Vd z Standard U-stirrup for shear reinforcement (has two legs) Nominal Shear Strength Vn: Vn = Vcz + Viy + Vd + Vs p p: Horizontal projection of shear crack s: Spacing between stirrups Vc 5 Vn = Vc + Vs (Eqn. 3) SHEAR STRENGTH OF BEAM WITH SHEAR REINFORCEMENT (CONT'D) Vn = Vc + V s • SHEAR STRENGTH CONTRIBUTED BY CONCRETE VC (EQN. 1 OR 2): Vud Vc 1.9 f ' c 2500 bwd 3.5 f ' c bwd Mu • ASSUME p=d (CONSERVATIVE). NUMBER OF STIRRUPS BRIDGING THE CRACK n = d/s. SHEAR FAILURE OCCURS WHEN STRESS IN STIRRUPS REACHES fy. SHEAR STRENGTH CONTRIBUTED BY STIRRUPS VS: Vs nAv f y Av f y d s • DIVIDING EQN. 3 BY bwd TO OBTAIN NOMINAL SHEAR STRESS Vn: As f y Vn vn vc bwd bws (Eqn. 4, conservative) 6 DESIGN FOR SHEAR Total applied (factored) shear not exceeding shear capacity : Vu Vn ; where Vn Vc Vs Vu : Total applied shear force, acting at Critical Section, due to factored loads Vn : Nominal shear capacity Vc : Nominal shear strength provided by the concrete Vs : Nominal shear strength provided by the stirrups 0.75 for shear and torsion. Note: Ribbed floor slab, or joist slab system, does not need shear reinforcement in the joist ribs (Vc is sufficient for shear demand) 7 SHEAR DESIGN PROCEDURE Step 1: Locate Critical Section for Shear, where Vu is to be obtained • For non-prestressed members: Critical section is at d from face of support Vu d Vu Critical section (distance d from face of support) Vu d d • Exception Vu d Concentrated load acting within d Critical section (at face of support) Vu 8 Loaded on bottom edge SHEAR DESIGN PROCEDURE Step 2: Determine shear strength provided by the concrete Vc • For normal - weight, normal strengh rectangular member : wVu d bw d 3.5 f ' c bw d Vc 1.9 f ' c 2500 Mu or, Vc 2 f ' c bw d (for simplified calculation) • For " all - lightweight" concrete : • For " sand - lightweight" concrete : • Applicable for concrete with (Eqn. 1) (Eqn. 2) 0.75 0.85 f ' c 100 psi 9 SHEAR DESIGN PROCEDURE Step 3: Compute Vu – (Vc/2) at Critical Section • If, • If, 1 Vu Vc 0 : No web (shear) reinforcem ent needed 2 1 Vu Vc 0 : At least min shear reinf. must be provided 2 b s b s Av 0.75 f 'c w 50 w fy fy Ab s Longitudin al spacing of shear reinforcem ent, in. Av Total cross sectional area of shear reinforcem ent within distance s, in 2 Av = 2 Ab s f y Yield strength of shear reinforcem ent, psi. * Exceptions : M in Web Rein. is not required for slabs and footings, for concrete joist floor, and for beams with total depth not greater th an 10 in., or 2 1 times the thickness of the flange, 2 1 or the web width (whichever is greatest) 2 10 SHEAR DESIGN PROCEDURE Step 4: Check max. shear that can be provided by stirrups • If, Vu Vc Vs 0 : Shear reinf. must be provided, and if : Vu Vc Vs 8 f 'c bw d , increase the size of the section or the concrete compressive strength 11 SHEAR DESIGN PROCEDURE Step 5: Compute the distance from the face of the support beyond which the concrete can carry the total factored shear force • Vu Vu Vc Vc 2 Face of support Vc /2 12 Shear reinforcement not required SHEAR DESIGN PROCEDURE Step 6: Compute the distance from the face of the support beyond which min. shear reinforcement is required • Vu Vc Vu Face of support Vc Vc /2 13 Minimum Shear reinforcement Shear reinforcement not required SHEAR DESIGN PROCEDURE Step 7: Compute the required area Av and spacing s of shear reinforcement at the Critical Section Av f y d s Vu Vc • ACI limit on Maximum Stirrup Spacing s (so that each 45-degree shear crack is intercepted by at least one stirrup) For Vu Vc Vs 4 f 'c bw d : smax Av f y 0.75 f 'c bw Av f y 50bw ; d ; or 24 in. 2 3d ( for 45o longitudin al bent bars) 4 For Vu - Vc Vs 4 f 'c bw d : halved of above : smax d or 12 in. 4 14 SHEAR DESIGN PROCEDURE Step 8: Determine required shear reinforcement at a few additional controlling sections along the span Practical considerations: • Minimize number of different stirrup spacing (no more than 3 different spacings) • Use larger stirrup sizes at wider spacing (less costs for fabrication and placement) • Use preselected, practical stirrup spacing: s =d/2; d/3; and d/4 Vs Av f y d s For Gr. 60 steel : f y 60ksi s No. 3 U-stirrups No. 4 U-stirrups No. 5 U-stirrups d/2 20 kips 36 kips 56 kips d/3 30 kips 54 kips 84 kips d/4 40 kips 72 kips 112 kips 15 SHEAR DESIGN PROCEDURE Step 8: (cont'd) Example : Vu 110kips Vc 60kips Vs Vu Vc 110 60 50kips s No. 3 U-stirrups No. 4 U-stirrups No. 5 U-stirrups d/2 20 kips 36 kips 56 kips d/3 30 kips 54 kips 84 kips 16 d/4 40 kips 72 kips 112 kips DESIGN EXAMPLE 1 Select beam dimension based on required shear strength PROB. 4.1 A BEAM IS TO BE DESIGNED FOR FACTORED LOADS CAUSING A MAXIMUM SHEAR OF 44.0 KIPS. f'c=4000 PSI. SELECT THE BEAM DIMENSIONS BASED ON SHEAR STRENGTH REQUIREMENT FOR THE FOLLOWING CASES: (A) BEAM IS TO HAVE NO WEB REINFORCEMENT (B) BEAM IS TO HAVE ONLY MINIMUM WEB REINFORCEMENT (C) BEAM WITH WEB REINFORCEMENTS PROVIDING SHEAR STRENGTH VS = 2VC. ASSUME d = 2b. 17 DESIGN EXAMPLE 1 (CON'T.) SOLUTION: Vu = 44.0 KIPS; f'c= 4000 psi; Vu = Vn. (A) BEAM HAS NO WEB REINFORCEMENT: 1 1 Vu Vc (2 f 'c )bw d 2 2 Vu 44,000 bw d 928 in 2 f 'c 0.75 4,000 b 21.6 in, ; d 43.2 in. 18 DESIGN EXAMPLE 1 (CON'T.) SOLUTION: (B) BEAM HAS MINIMUM WEB REINFORCEMENT: Vu Vc (2 f 'c )bw d Vu 44,000 bw d 464 in 2 (2 f 'c ) 0.75(2 4,000 ) b 15.2 in, ; d 30.4 in. bw s bw s (determine Av 0.75 f 'c 50 with fy fy an assumed s or vice versa ) 19 DESIGN EXAMPLE 1 (CON'T.) SOLUTION: (C) BEAM HAS WEB REINFORCEMENT AND VS=2VC : Vu (Vc Vs ) (3Vc ) 3 (2 f 'c )bw d Vu 44,000 bw d 155 in 2 (6 f 'c ) 0.75(6 4,000 ) b 8.8 in, ; d 17.6 in. 20 DESIGN EXAMPLE 2 (SELF-READING) Determine shear strength of concrete in beam using detailed method Use detailed method, compute shear strength of normal weight concrete at the distance d from the face of support for the beam shown. the beam is loaded with uniformly distributed factored load wu=3.1 kips/ft. wu=3.1 kips/ft d h As Given: l = 20 ft; bw=12 in.; d=17 in. h=20 in; As=3.1 sq.in. f'c=3000 psi; fy=50,000 psi l=20' Center of support 8" 21 DESIGN EXAMPLE 2 (CON'T.) SOLUTION: STEP 1. (A) COMPUTE FACTORED MOMENT MU AT DISTANCE D FROM FACE OF SUPPORT (I.E., D + 4 IN. FROM CENTER OF SUPPORT): wu ( x)(l x) 3.1 17 4 17 4 Mu 20 49.5 ft - kips 2 2 12 12 (B) CALCULATE FLEXURAL REINFORCEMENT RATIO W w As 3.1 0.015 bw d (12)(17) 22 DESIGN EXAMPLE 2 (CON'T.) (C) Compute factored shear force Vu at distance d from face of support (i.e., d + 4 in. from center of support): l 20 17 4 Vu wu ( x) 3.1( ) 25.6 kips 2 2 12 (D) Calculate wVud/Mu wVu d Mu STEP 2. Compute Vc: 0.015(25.6)(17) 0.011 49.5(12) Vc vcbw d (1.9 f 'c 2500 wVu d Mu )bw d 0.75(1.9 3000 2500(0.011))(12)(17) 22.8 kips 23 CLASS EXERCISE 1 (SELF READING) Determine shear strength of concrete in beam using simplified and detailed method Compute factored shear force Vu and shear strength Vc of normal weight concrete beam at a point 3.5 ft from the center of support for the beam shown. use simplified and detailed methods. the beam is loaded with uniformly distributed factored load wu=3.1 kips/ft. check if web reinforcement is required at 3.5 ft from center of support. Given: l = 20 ft; bw=12 in.; d=17 in. h=20 in; As=3.1 sq.in. f'c=3000 psi; fy=50,000 psi wu=3.1 kips/ft d h As x=3.5 ft l=20' Center of support 24 CLASS EXERCISE 1 (CON'T.) SOLUTION: STEP 1. (A) COMPUTE FACTORED MOMENT MU AT 3.5 FT FROM CENTER OF SUPPORT: wu ( x)(l x) 3.1 3.520 3.5 89.5 ft - kips Mu 2 2 (B) CALCULATE FLEXURAL REINFORCEMENT RATIO W w As b d 3.1(12)(17) 0.015 w 25 CLASS EXERCISE 1 (CON'T.) (C) COMPUTE FACTORED SHEAR FORCE VU AT 3.5 FT FROM CENTER OF SUPPORT: l 20 Vu wu ( x) 3.1( 3.5) 20.2 kips 2 2 (D) CALCULATE WVUd/MU (DETAILED METHOD) wVu d Mu 0.015(20.2)(17) 0.0049 (89.5)(12) STEP 2. COMPUTE VC BY DETAILED METHOD: Vc vc bw d (1.9 f 'c 2500 wVu d Mu )bw d 0.75(1.9 3000 2500(0.0049))(12)(17) 20.2 kips 26 CLASS EXERCISE 1 (CON'T.) ALTERNATE STEP 2. COMPUTE VC BY SIMPLIFIED METHOD: Vc vcbw d (2 f 'c )bw d 0.75(2 3000 )(12)(17) 19.0 kips STEP 3. IS WEB REINFORCEMENT REQUIRED AT 3.5 FT FROM CENTER OF SUPPORT? 19.0 20.2 Since Vu 20.2 kips 0.5Vc kips or kips, 2 2 stirrups are required at 3.5 ft from center of supportof this beam 27 DESIGN EXAMPLE 3 (SELF READING) Design of vertical stirrups for beam with triangular shear diagram Prob. 4.2 A rectangular beam having b=12 in., d=22 in. and a 20 ft span length (face to face of simple supports). it is reinforced for flexure with three no. 11 bars and carries a uniform service dead load d=1.63 kips/ft (including self-weight) and a uniform service life load l=3.26 kips/ft. f'c=4000 psi, fy=60,000 psi. design the shear reinforcement, using no. 3 vertical u stirrups. the simplified equation for Vc may be used. 28 DESIGN EXAMPLE 3 (CON'T.) STEP 1. (A) FACTORED LOAD: WN = 1.2D+1.6L=1.2(1.63)+1.6(3.26)= 7.17 KIPS/FT wn L 7.17 x20 Vu 71.7 kips 2 2 71.7(120 22) (C) SHEAR FORCE AT DISTANCE D FROM FACE OF SUPPORT: Vu / d 58.6 kips 120 (B) SHEAR FORCE AT FACE OF SUPPORT: STEP 2. FIND VC USING SIMPLIFIED METHOD: 12 x22 Vc (2 f 'c )bwd 0.75(2 4000 ) 25 kips 1000 29 DESIGN EXAMPLE 3 (CON'T.) Steps 3 to 7. • Compute maximum shear to be carried by stir...
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Running head: RC design
1
RC design:
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RC design
2
Design wind loads at each floor level
Parameters & coefficients
Topographic factor K u = 1.00
Directionality factor K d= 0.85
Basic wind speed= 170m/hour
Gust effect factor= 0.85
I= 1.15
+GCpi Coefficient=0.18
-GCpi Coefficient = -0.18
C p= for windward wall= 0.8, leeward wall= -0.3, side wall= -0.7
p = q G C p – qi (GCpi)
Velocity pressure q z =0.00256 k z K u K d v2
Floor level
K h& k z
Velocity
Design wind
Wall area
Design wind
velocity
pressure (q
pressures (p)
Feet
loads l b
pressure
z)
l b/ f t2
squared
l b/ f t2
exposure
windward
coefficient
wall
1
0.85
53.45
26.416
4500
118872
2
0.932
58.61
29.305
4500
131872.5
3
1.016
63.89
31.945
4500
143752.5
K h & k z obtained through interpolation
I.e. at f t 20, k z = 0.90 & at f t 25 k z =0.94. Therefore, 25-20=5 & 0.94-0.90=0.04.
RC design
3
This implies that 5=0.04 how about 24-20=4
(4*0.04)/5= 0.032. Therefore k z at f t 24= 0.90+0.032=0....
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Design Stirrups Using F'c 4000 Psi and
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