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Lecture 34 - Elastic Flexural Analysis for Serviceability November 20, 2001 CVEN 444 Lecture Goals • Serviceability • Crack width • Moments of inertia Introduction Recall: Ultimate Limit States Lead to collapse Serviceability Limit States Disrupt use of Structures but do not cause collapse Introduction Types of Serviceability Limit States - Excessive crack width - Excessive deflection - Undesirable vibrations - Fatigue (ULS) Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc.. Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc.. Crack Width Control Bar crack development. Crack Width Control Temperature crack development Crack Width Control Reasons for crack width control? • Appearance (smooth surface > 0.01 to 0.013 public concern) • Leakage (Liquid-retaining structures) • Corrosion (cracks can speed up occurrence of corrosion) Crack Width Control Corrosion more apt to occur if (steel oxidizes rust ) • Chlorides ( other corrosive substances) present • Relative Humidity > 60 % • High Ambient Temperatures (accelerates chemical reactions) • Wetting and drying cycles • Stray electrical currents occur in the bars. Limits on Crack Width ACI Code’s Basis max.. crack width = 0.016 in. for interior exposure 0.013 in. for exterior exposure Cracking controlled in ACI code by regulating the distribution of reinforcement in beams/slabs. Limits on Crack Width Gergely-Lutz Equation = 0.076fs 3 dc A = Crack width in units of 0.001 in. = Distance from NA to bottom (tension) fiber, divided by distance to reinforcement. =(h-c)/(d-c) fs = Service load stress in reinforcement in ksi Limits on Crack Width Gergely-Lutz Equation = 0.076fs 3 dc A dc = Distance from extreme tension fiber to center of reinforcement located closest to it, (in.) A = effective tension area of concrete surrounding tension bars (w/ same centroid) divided by # bars. (for 1 layer of bars A = (2dc b)/n Limits on Crack Width ACI Code Eqn 10-5 ( limits magnitude of z term ) Note: = 0.076 z z = f s 3 dc A ( =1.2 for beams) Interior exposure: critical crack width = 0.016 in. ( = 16 ) z = 175k/in Exterior exposure: critical crack width = 0.013 in. ( = 13 ) z = 145k/in Limits on Crack Width Tolerable Crack Widths Tolerable Exposure Condition Crack Width Dry air or protective membrane - 0.016 in. Humidity, moist air, soil - 0.012 in. Deicing chemicals - 0.007 in. Seawater and seawater spray - 0.006 in. wetting and drying Water-retaining structures - 0.004 in. (excluding nonpressure pipes) Limits on Crack Width Thin one-way slabs: Use =1.35 z = 155 k/in (Interior Exposure) z = 130 k/in (Exterior Exposure) fs = service load stress may be taken as fy f s = 0.60 f y 1.55 – average load factor 1.55 - strength reduction 0.90 . factor for flexure Example-Crack Given: A beam with bw= 14 in. Gr 60 steel 4 #8 with 2 #6 in the second layer with a #4 stirrup. Determine the crack width limit, z for exterior and interior limits (145 k/in and 175 k/in.). Example-Crack Compute the center of the steel for the given bars. As = 4 #8 bars 2 # 6 bars = 4 0.79 in 2 ) 2 0.44 in 2 ) = 4.04 in 2 Example-Crack The locations of the center of the bars are db y1 = cover d stirrup 2 1.0 in. = 1.5 in. 0.5 in. 2 = 2.5 in. db d b2 y2 = 2.5 in. d b 2 2 1.0 in. 0.75 in. = 2.5 in. 1.0 in. 2 2 = 4.375 in. Example-Crack Compute the center of the steel for the given bars. y= y A i i A i 2.5 in.) 4 0.79 in 2 ) 4.375 in.) 2 0.44 in 2 ) = 4.04 in 2 = 2.91 in. Example-Crack Compute number of equivalent bars, n. Use the largest bar. n= Ai = 4.04 in 2 = 5.11 2 Abar 0.79 in Compute the effective tension area 2 yb 2 2.91 in.)14 in.) A= = = 15.93 in 2 n 5.11 Example-Crack The effective service load stress is fs = 0.60 f y = 0.6 60 ksi ) = 36 ksi Compute the effective tension area z = f s 3 d c A = 36 ksi ) 3 2.5 in.) 15.93 in 2 ) = 122.9 k/in. Example-Crack The limits magnitude of z term. 122.9 k/in. < 145 k/in. - Interior exposure 122.9 k/in. < 175 k/in. - Exterior exposure Crack width is w = 0.076 z = 0.076 1.2 )122.9 ) = 11.2 or = 0.0112 in. Deflection Control Reasons to Limit Deflection (1.) Visual Appearance 1 * l are generally visible 250 ( 25 ft. span 1.2 in. ) (2.) Damage to Non-structural Elements - cracking of partitions - malfunction of doors /windows Deflection Control (3.) Disruption of function - sensitive machinery, equipment - ponding of rain water on roofs (4.) Damage to Structural Elements - large ’s than serviceability problem - (contact w/ other members modify load paths) Allowable Deflections ACI Table 9.5(a) = min. thickness unless ’s are computed ACI Table 9.5(b) = max. permissible computed deflection Allowable Deflections Flat Roofs ( no damageable nonstructural elements supported) l LLinst ) 180 Allowable Deflections Floors ( no damageable nonstructural elements supported ) l LLinst ) 180 Allowable Deflections Roof or Floor elements (supported nonstructural elements likely damaged by large ’s) l 480 Allowable Deflections Roof or Floor elements ( supported nonstructural elements not likely to be damaged by large ’s ) l 240 Allowable Deflections Deflection occurring after attachment of nonstructural elements allow Need to consider the specific structures function and characteristics. Moment of Inertia for Deflection Calculation For I cr I e I gt (intermediate values of EI) Brandon M a a Ie = cr * I gt 1 - M cr * I derived cr Ma Ma fr Ig Mcr = Cracking Moment = yt Igt = Moment of inertia of transformed cross-section fr = Modulus of rupture = 7.5 f c Moment of Inertia for Deflection Calculation M a a I e = cr * I gt 1 - M cr * I cr Ma Ma yt = Distance from centroid to extreme tension fiber Ma = maximum moment in member at loading stage for which Ie ( ) is being computed or at any previous loading stage Ig = Moment of inertia of concrete section neglect reinforcement Moment of Inertia for Deflection Calculation M 3 3 cr * I 1 - M cr Ie = *I g cr Ma Ma or 3 M I e = I cr I g - I cr ) cr Ma Moment Vs curvature plot M M = slope = = EI EI “Moment Vs Slope” Plot The cracked beam starts to lose strength as the amount of cracking increases Moment of Inertia For wc = 90 to 155 lb/ft3 Ec = c 33 f c 1.5 psi) For normal weight concrete Ec = 57000 f c psi) (ACI 8.5.1) Deflection Response of RC Beams (Flexure) A- Ends of Beam Crack B - Cracking at midspan C - Instantaneous deflection under service load C’ - long time deflection under service load D and E - yielding of reinforcement @ ends & Note: Stiffness (slope) decreases midspan as cracking progresses Deflection Response of RC Beams (Flexure) wl 2 wl 2 M = 2 M = wl M = 12 24 12 The maximum moments for distributed load acting on an indeterminate beam are given. Deflection Response of RC Beams (Flexure) For Continuous beams ACI 9.5.2.4 I e avg) = 0.50 I e m id) 0.25 I e1 I e2 ) ACI Com. 435 2 ends continous : Weight Average I eavg) = 0.70I emid) 0.15I e1 I e2 ) 1 end continous : I e avg) = 0.85I emid) 0.15I e1 ) I e m id) = I e@ midspan I e1 = I e@ end 1 I e2 = I e@ end 2 Uncracked Transformed Section Part (n) =Ej /Ei Area n*Area yi yi*(n)A Concrete 1 bw*h bw*h 0.5*h 0.5*bw*h2 A’s n A’s (n-1)A’s d’ (n-1)*A’s*d’ As n As (n-1)As d (n-1)*As*d n* A yi * n * Ai y= y *n A i i * i Note: (n-1) is to remove area of concrete n A i * i Cracked Transformed Section Finding the centroid of singly Reinforced Rectangular Section y y by nAs d by nAs y = by nAs d 2 y= yi Ai = 2 2 Ai by nAs b 2 y nAs y - nAs d = 0 2 Solve for the quadratic for y 2nAs 2nAs d y 2 y- =0 b b Cracked Transformed Section Singly Reinforced Rectangular Section 2nAs 2nAs d y 2 y- =0 b b I cr = by nAs d - y ) 1 3 2 3 Es Note: n= Ec Cracked Transformed Section Doubly Reinforced Rectangular Section 2n - 1)As 2nAs 2n - 1)As 2nAs d y 2 y- =0 b b y - d )2 nAs d - y )2 I cr = by n - 1)As 1 3 3 Note: Es n= Ec Uncracked Transformed Section Moment of inertia (uncracked doubly reinforced beam) 2 1 h I gt = bh bh y - 3 12 2 concrete n - 1)As y - d ) n - 1)As y - d ) 2 2 steel Note: 1 Ig = bh 3 12 Cracked Transformed Section Finding the centroid of doubly reinforced T-Section 2t be - bw ) 2n - 1)As 2nAs y 2 y bw - be - bw ) t 2 2n - 1)As 2nAs d =0 bw Cracked Transformed Section Finding the moment of inertia for a doubly reinforced T-Section 2 t 1 1 I cr = be y bet 3 y - bw y - t )3 12 2 3 flange beam n - 1)As y - d ) nAs d - y ) 2 2 steel Reinforced Concrete Sections - Example Given a doubly reinforced beam with h = 24 in, b = 12 in., d’ = 2.5 in. and d = 21.5 in. with 2# 7 bars in compression steel and 4 # 7 bars in tension steel. The material properties are fc = 4 ksi and fy= 60 ksi. Determine Igt, Icr , Mcr(+), Mcr(-), and compare to the NA of the beam. Reinforced Concrete Sections - Example The components of the beam As = 2 0.6 in 2 ) = 1.2 in 2 As = 4 0.6 in 2 ) = 2.4 in 2 1k Ec = 57000 f c = 57000 4000 1000 lb = 3605 ksi Reinforced Concrete Sections - Example The compute the n value and the centroid, I uncracked Es 29000 ksi n= = = 8.04 Ec 3605 ksi n area (in2) n*area (in2) y i (in) y i *n*area (in2) I (in4) d (in) d2*n*area(in4) A's 7.04 1.2 8.448 2.5 21.12 - -9.756 804.10 As 7.04 2.4 16.896 21.5 363.26 - 9.244 1443.75 Ac 1 288 288 12 3456.00 13824 -0.256 18.89 313.344 3840.38 13824 2266.74 Reinforced Concrete Sections - Example The compute the centroid and I uncracked y= yi ni Ai = 3840.38 in 3 = 12.26 in. ni Ai 313.34 in 2 I = I i di2 ni Ai =13824 in 4 2266.7 in 4 = 16090.7 in 4 Reinforced Concrete Sections - Example The compute the centroid and I for a cracked doubly reinforced beam. 2n - 1)As 2nAs 2n - 1)As 2nAs d y 2 y- =0 b b 2 7.04 ) 1.2 in 2 ) 2 8.04 ) 2.4 in 2 ) y2 y 12 in. 2 7.04 ) 1.2 in 2 ) 2 8.04 ) 2.4 in 2 ) 21.5 in.) - s =0 12 in. y 2 4.624 y - 72.664 = 0 Reinforced Concrete Sections - Example The compute the centroid for a cracked doubly reinforced beam. y 2 4.624 y - 72.664 = 0 -4.624 4.624 ) 4 72.664 ) 2 y= 2 = 6.52 in. Reinforced Concrete Sections - Example The compute the moment of inertia for a cracked doubly reinforced beam. I cr = by n - 1)As y - d ) nAs d - y ) 1 3 2 2 3 1 I cr = 12 in.) 6.52 in.) 3 3 7.04 ) 1.2 in ) 6.52 in. - 2.5 in.) 2 2 8.04 ) 2.4 in ) 21.5 in. - 6.52 in.) 2 2 = 5575.22 in 4 Reinforced Concrete Sections - Example The critical ratio of moment of inertia 4 I cr 5575.22 in = 4 = 0.346 I g 16090.7 in I cr 0.35I g Reinforced Concrete Sections - Example Find the components of the beam Cc = 0.85 f cba = 0.85 4 ksi )12 in.) 0.85 ) c = 34.68c c - 2.5 in. 0.0075 s = 0.003) = 0.003 - c c = 29000 0.003 - 0.0075 217.5 fs = Es s = 87 - c c 2 217.5 Cs = As fs - 0.85 f c ) = 1.2 in ) 87 - c 261 = 100.32 - c Reinforced Concrete Sections - Example Find the components of the beam T = 2.4 in 2 ) 60 ksi ) = 144 k T = Cc Cs 261 144 k = 34.68c 100.32 - 34.68c 2 - 43.68c - 261 = 0 c The neutral axis 43.68 43.68) 4 261) 34.68 ) 2 c= 2 34.68 ) = 3.44 in. Reinforced Concrete Sections - Example The strain of the steel 3.44 in. - 2.5 in. = s 0.003) = 0.0008 0.00207 3.44 in. 21.5 in. - 3.44 in. s = 0.003) = 0.0158 0.00207 3.44 in. Note: At service loads, beams are assumed to act elastically. c = 3.44 in. y = 6.52 in. Reinforced Concrete Sections - Example Using a linearly varying and s = E along the NA is the centroid of the area for an elastic center My s =- I The maximum tension stress in tension is f r = 7.5 f c = 7.5 4000 = 474.3 psi 0.4743 ksi Reinforced Concrete Sections - Example The uncracked moments for the beam My sI s= M = I y fr I 0.4743 ksi 16090.7 in 4 ) M cr ) = = = 650.2 k-in. y 24 in. - 12.26 in.) fr I 0.4743 ksi 16090.7 in 4 ) M cr - ) = = = 622.6 k-in. y 12.26 in. Homework-12/2/02 Problem 8.7