Lecture-05 Serviceability Requirements & Development of Reinforcement

Lecture-05 Serviceability Requirements & Development of Reinforcement By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com 1 Section 1: Deflections Topics Addressed Deflection in RC One-way Slabs and Beams, Deflection in RC Two-way Slabs, Examples Section 2: Cracking in RC Members Crack Formation, Equations for Maximum Crack Width, Reasons for Crack Width Control, Crack Control Reinforcement in Deep Members, ACI Provisions for Crack Control, Example Section 3: Development & Splices of Reinforcement Bond Strength, Bond failure, ACI provisions for Development Length, Splices of Reinforcement 2 1

Section 1: Deflections 3 Topics Addressed Introduction to deflections Deflection in RC One-way Slabs and Beams Deflection in RC Two-way Slabs Example 1 Example 2 4 2

Introduction to Deflections Deflection Effects It is important to maintain control of deflections so that members designed mainly for strength at prescribed overloads will also perform well in normal service. Excessive deflections can lead to cracking of supported walls and partitions, ill-fitting doors and windows, poor roof drainage, misalignment of sensitive machinery and equipment, and visually offensive sag etc. 5 Introduction to Deflections Deflection Types Immediate (short-term) Due to applied loads on the member. Long-term Due to creep and shrinkage of concrete. 6 3

Introduction to Deflections w Deflection history of fixed ended beam subjected to uniformly distributed load w Yielding of reinforcement 5 4 3 Due to shrinkage and creep 3 Full service load 2 Cracking at midspan 1 Ends of beam crack Mid-span deflection 7 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Immediate deflection at a given load in a structure is calculated using equations of elastic deflection. Δ = f(loads, spans, supports)/ei EI is the flexural rigidity. f(loads, spans, supports) is a function of particular load, span and support arrangement. 8 4

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Elastic equation in a general form for deflection of cantilever and simple and continuous beams be written as: subjected to uniform loading can Δ i = K(5/48)M a l 2 /E c I e Where, M a = the mid-span moment (when K is so defined) for simple and continuous beams. For cantilever beams, M a will be the support moment. l = Span length. 9 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of K: Theoretical values of deflection coefficient K for uniformly distributed loading w: Reference: PCA Notes. 10 5

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Alternatively, deflections can be directly computed for different conditions of loading and end conditions as below: 11 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of l: In elastic deflection formulae, l is the span length and is least of: l n + h c/c distance between supports. Slab h Support l n l n c/c distance c/c distance 12 6

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of E c (modulus of elasticity of concrete) Unless stiffness values are obtained by a more comprehensive analysis, immediate deflection shall be computed with the modulus of elasticity (E c ) of section as given in ACI 19.2.2.1. E c = w c 1.5 33 f c (in psi) Note: The formula is used for values of w c between 90 and 160 lb/ft 3. E c strongly influences the behavior of reinforced concrete members under short-term loads, particularly showing more variation with concrete quality, concrete age, stress level, and rate or duration of load. 13 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Effective Moment of Inertia M I cr Moment, M a M cr I g I e Graph shows the portion of load deflection curve up to stage 3 as discussed earlier D cr D e D t Idealized short term deflection of RC beam Deflection D 14 7

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : In elastic deflection equation, the effective moment of inertia I e is calculated as (ACI 24.2.3.5a): Where, M a = maximum service load moment at the stage for which deflections are being considered. M cr = Cracking moment = f r I g /y t (where f r = 7.5 f c ) I g = Gross moment of inertia I cr = Moment of inertia of cracked section 15 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Un-cracked section moment of Inertia (I g ) [about centroid]: For an un-cracked concrete section, full section will be effective so that: I e = I g = bh 3 /12 h b 16 8

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Un-cracked section moment of Inertia (I g ) [about base]: I g = I c.a = bh 3 /12 is the moment of inertia of rectangular section about the centroid. For determination of moment of inertia about base of the section, transfer formula may be used: I base = I c.a + Ae 2 = bh 3 /12 + Ae 2 Where, A = bh ; e = h/2 (for given case) I base = bh 3 /3 e h b 17 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ) [about neutral axis]: Similarly, moment of inertia of cracked section about neutral axis can be determined through transfer formulae (I = bh 3 /3 + Ae 2 ) For the section shown: I cr = [b w (kd) 3 /3] + na s (d kd) 2 kd kd in above equation is unknown. n = E s / E c e N.A d h na s b w 18 9

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ): Determination of kd: Taking moments of area about the neutral axis: kd b w kd/2 = na s (d kd) kd 2 b w /na s = d kd kd Let B = b w /na s, then kd 2 B = (d kd) N.A d h Simplification through quadratic formula gives, na s kd = { (1 + 2Bd) 1}/B b w 19 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ): Alternatively, gross and cracked moment of inertia of rectangular and flanged sections can be calculated from the given tables: 20 10

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ): 21 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ): Note: Effective width of T-section as per ACI 6.3.2.1 22 11

Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Determination of I e : Cracked section Moment of Inertia (I cr ): Note: Effective width of T-section as per ACI 6.3.2.1 23 Deflection in RC One-way Slabs and Beams Immediate (short-term) Deflection Application of I e for simple and continuous members We have seen that I e depends on M a, which will be different at different locations. Therefore according to (ACI R 24.2.3.7); for Simply supported Beams: I e = I e,midspan Beams with one end continuous: I e, average = 0.85I e,midspan + 0.15 (I e,continuous end ) Beams with both ends continuous: I e, average = 0.70I e,midspan + 0.15 (I e1 + I e2 ) Where, I e1 and I e2 refer to I e at the respective beam ends. (ACI R 24.2.3.7) 24 12

Deflection in RC One-way Slabs and Beams Long-term Deflections Shrinkage and creep due to sustained loads cause additional longterm deflections over and above those which occur when loads are first placed on the structure. Such deflections are influenced by: Temperature, Humidity, Curing conditions, Age at the time of loading, Quantity of compression reinforcement, and Magnitude of the sustained load. 25 Deflection in RC One-way Slabs and Beams Long-term Deflections (ACI 24.2.4) Additional Long-term deflection resulting from the combined effect of creep and shrinkage is determined by multiplying the immediate deflection caused by the sustained load with the factor l D. x is a time dependent factor for sustained load. Δ (cp +sh) = λ Δ (Δ i ) sus x l D = 1 + 50r r = A s /bd Time x 3 months 1.0 6 months 1.2 12 months 1.4 5 years 2.0 26 13

Deflection in RC One-way Slabs and Beams Long-Term Deflections It is important to note here that long term deflections are function of immediate deflections due to sustained load only i.e., Δ (cp +sh) = λ Δ (Δ i ) sus Sustained loads are loads that are permanently applied on the structure e.g., dead loads, superimposed dead loads and live loads kept on the structure for long period. 27 Deflection in RC One-way Slabs and Beams Deflection Control according to ACI code: Two methods are given in the code for controlling deflections. Deflections may be controlled DIRECTLY by limiting computed deflections [see Table 24.2.2] or INDIRECTLY by means of minimum thickness [Table 7.3.1.1] for one-way systems. 28 14

Deflection in RC One-way Slabs and Beams Deflection Control according to ACI code: In DIRECT approach, the deflections are said to be within limits if the combined effect of immediate and long term deflections does not exceed the limits specified in ACI table 24.2.2. 29 Deflection in RC One-way Slabs and Beams Deflection Control according to ACI code: In the INDIRECT approach, the deflections for beams and one-way slabs are controlled indirectly by the minimum requirements given in ACI table 9.3.1.1 and 7.3.1.1. However, this method is applicable only to the cases of loadings and spans commonly experienced in buildings and cannot be used for unusually large values of loading and span. 30 15

Deflection in RC Two-way Slabs Immediate (short-term) Deflection The calculation of deflections for two-way slabs is complicated even if linear elastic behavior can be assumed. For immediate deflections 2D structural analysis is required in which the values of E c and I e for one-way slabs and beams discussed earlier may be used. 31 Deflection in RC Two-way Slabs Long-Term Deflections The additional long-term deflection for two-way construction is required to be computed using the multipliers given in ACI 24.2.4.1. 32 16

Deflection in RC Two-way Slabs Deflection Control according to ACI code Deflections may be controlled directly by limiting computed deflections [see Table 24.2.2] or indirectly by means of minimum thickness, table 8.3.1.1 and table 8.3.1.2 for two-way systems 33 Example 1 Analyze the simply supported reinforced concrete beam for shortterm deflections and long-term deflections at ages 3 months and 5 years. Data: f c = 3000 psi (normal weight concrete); f y = 40,000 psi E s = 29,000,000 psi Superimposed dead load (not including beam weight) = 120 lb/ft Live load = 300 lb/ft (50% sustained) ; Span = 25 ft 34 17

Example 1 Data: A s = 3 #7 = 1.80 in 2 ρ = A s /bd = 0.0077 A s = 3 #4 = 0.60 in 2 (A s not required for strength) ρ = A s /bd = 0.0026 35 Example 1 Solution: Step 01: Minimum beam thickness, for members not supporting or attached to partitions or other construction likely to be damaged by large deflections: h min = (l/16) {multiply by 0.8 for f y = 40000 psi} h min = (25 12/16) 0.8 = 15 22 is the given depth, therefore O.K. 36 18

Example 1 Solution: Step 02: Moments w d = 0.120 + 0.150 12 22/144 = 0.395 kips/ft M d = w d l 2 /8 = 0.395 25 2 /8 = 30.9 ft-kips M l = w l l 2 /8 = 0.300 25 2 /8 = 23.4 ft-kips M d+l = 54.3 ft-kips M sus = M d + 0.5M l = 30.9 + 0.5 23.4 = 42.6 ft-kips 37 Example 1 Solution: Step 03: Modulus of rupture, modulus of elasticity, modular ratio f r = 7.5 (f c ) = 7.5 (3000) = 411 psi E c = w 1.5 c 33 (f c ) = 150 1.5 33 (3000) = 3.32 10 6 psi n = E s /E c = 29 10 6 / 3.32 10 6 = 8.7 38 19

Example 1 Solution: Step 04: Gross and cracked section moments of inertia I g = bh 3 /12 = 12 22 3 /12 = 10650 in 4 B = b/na s = 12/{8.7 1.80} = 0.766 r = (n 1)A s /na s = {(8.7 1)0.60/(8.7 1.80) = 0.295 kd = [ {2dB(1 + rd /d) + (1+r 2 )} (1 + r)]/b = 5.77 I cr = {b(kd) 3 /3} + na s (d kd) 2 + (n 1) A s (kd - d ) 2 = 3770 in 4 I g /I cr = 2.8 39 Example 1 Step 05: Effective moment of inertia M cr = f r I g /y t = [411 10650/11]/12000 = 33.2 ft-kips a. Under dead load only: M cr / M d = 33.2/30.9 > 1, hence (I e ) d = I g = 10650 in 4 b. Under sustained load: (M cr /M sus ) 3 = (33.2/42.6) 3 = 0.473 (I e ) sus = (M cr /M a ) 3 I g + {1 - (M cr /M a ) 3 }I cr I g (where M a = M sus ) = 0.473 10650 + (1 0.473) 3770 = 7025 in 4 c. Under dead+ live load: (M cr /M d+l ) 3 = (33.2/54.3) 3 = 0.229 (I e ) d+l = (M cr /M a ) 3 I g + {1 (M cr /M a ) 3 }I cr I g (where M a = M d+l ) = 0.229 10650 + (1 0.229) 3770 = 5345 in 4 40 20

Example 1 Step 06: Initial or short term deflection (Δ i ) d = K(5/48)M d l 2 /E c (I e ) d = (1)(5/48)(30.9)(25) 2 (12) 3 /(3320)(10650) = 0.098 K = 1 for simply supported spans (Δ i ) sus = K(5/48)M sus l 2 /E c (I e ) sus = (1)(5/48)(42.6)(25) 2 (12) 3 /(3320)(7025) = 0.205 (Δ i ) d+l = K(5/48)M d+l l 2 /E c (I e ) d+l = (1)(5/48)(54.3)(25) 2 (12) 3 /(3320)(5345) = 0.344 (Δ i ) l = (Δ i ) d+l (Δ i ) d = 0.344 0.098 = 0.246 41 Example 1 Step 06: Initial or short term deflection Allowable deflection: For flat roofs not supporting and not attached to nonstructural elements likely to be damaged by large deflections: (Δ i ) l l/180 l/180 = 300/180 = 1.67 (Δ i ) l =0.246 < l/180, OK Allowable deflection: Floors not supporting and not attached to nonstructural elements likely to be damaged by large deflections: (Δ i ) l l/360 l/360 = 300/360 = 0.83 (Δ i ) l = 0.246 < l/360, OK 42 21

Example 1 Step 07: Long-term deflections at ages 3 months and 5 yrs Duration ξ λ = ξ/(1 + 50ρ ) Δ isus Δ (cp +sh) =λδ isus Δ il 5 years 2.0 1.77 0.205 0.363 0.246 3 months 1.0 0.89 0.205 0.182 0.246 Allowable deflection: From Table 24.2.2, the most stringent requirement has been placed on Roof and floor construction supporting or attached to nonstructural elements likely to be damaged by large deflections (sum of the long-term deflection due to all sustained loads and the immediate deflection due to any additional live load)= l/480 l/480 = 300/480 = 0.63 inch 43 Example 1 Step 07: Long-term deflections at ages 3 months and 5 yrs. Which computed deflection should be considered for this case? The code says that this should include Sum of the long-term deflection due to all sustained loads and the immediate deflection due to any additional live load As the partition will be installed after the immediate deflection due to dead load has occurred, total deflection which would affect the partition would include 1. Only long term dead load deflection (not including dead load immediate deflection) 2. Both Short and long term sustained live load portion 3. Additional live load immediate deflection. This is immediate deflection due to total live load minus sustained live load 44 22

Example 1 Step 07: Long-term deflections at ages 3 months and 5 yrs. Therefore total deflection for this case would be equal to λ Δ d + Δ i sus live + λ Δ i sus live + Δ i additional live (λ Δ d + λ Δ i sus live ) + ( Δ i sus live + Δ i additional live ) λ (Δ d + Δ i sus live ) + Δ il total (Δ d + Δ i sus live ) is total sustained load and Δ il total would be denoted by Δ il Finally we have λ (Δ isus ) + Δ il 45 Example 1 Step 07: Long-term deflections at ages 3 months and 5 yrs. The deflection [ λ Δ isus + Δ il ] is given as follows: Duration λδ isus Δ il λ Δ isus + Δ il 5 years 0.363 0.246 0.61 3 months 0.182 0.246 0.428 [ λ (Δ isus ) + Δ il ] should be l/480 ; where l/480 = 300/480 = 0.63 [ λ (Δ isus ) + Δ il ] = 0.61 < 0.63 ; OK 46 23

Example 2 Find the deflection under full service load for the beam of the hall shown below. Data: f c = 3000 psi (normal weight concrete); f y = 40,000 psi Dead load (w d ) = 2.26 kip/ft; Live load (w l ) = 0.40 kip/ft Full service load (w s ) = w d + w l = 2.66 kip/ft Note: No live load is sustained. 47 Example 2 Solution: Summary of Strength Design of Beam: According to ACI 9.3.1 {table 9.3.1.1} = h min = l/16 = 36.9 Taken h = 5 = 60 ; d = h 3 = 57 ; b w = 18 (assumed) Also, l = 61.5 Flexural Design = 12 #8 bars 48 24

Solution: Example 2 Immediate deflection (under full service load): Δ i = K(5/48)M a l 2 /E c I e K = 1 (for simply supported beam) l = 61.5 = 738 M d = w d l 2 /8 =12 (2.26 61.5 2 /8) = 12821 in-kip M l = w l l 2 /8 =12 (0.40 61.5 2 /8) = 2269.35 in-kip M d+l = w d+l l 2 /8 =12 (2.66 61.5 2 /8) = 15091 in-kip E c = 57 f c = 57 (3000) = 3122 ksi 49 Example 2 Solution: Immediate deflection (under full service load): For calculation of I e, various parameters are required: Note: Effective width as per ACI 6.3.2.1 50 25

Example 2 Solution: Immediate deflection (under full service load): Effective width of T-beam (ACI 6.3.2.1) T - Beam 1 b w + 16h 2 b w + s w s w s w 3 b w + l n /4 Least of the above values is selected This effective width of T-section is different from that used in design for torsion and direct design method. 51 Example 2 Solution: Immediate deflection (under full service load): n = E s /E c = 29000/3122 = 9.289 C = b w /(na s ) = 18/ (9.289 12 0.79) = 0.204 b = 114 (ACI 6.3.2.1) f = h f (b b w )/ (na s ) = 6.54 y t = h ½ [ {b b w }h f2 + b w h 2 ]/ [{b b w }h f + b w h] = 39.39 I g = (b b w )h f3 /12+b w h 3 /12+(b b w )h f (h h f /2 y t ) 2 +b w h(y t h/2) 2 = 599578 in 4 M cr = 7.5 f c I g /y t = 6252 in-kip < M d (therefore section is cracked for dead load) kd = [ {C(2d + h f f) + (1+f)2} (1 + f)]/c = 9.05 I cr = (b b w )h f3 /12+b w (kd) 3 /12+(b b w )h f (kd h f /2) 2 + na s (d kd) 2 = 229722 in 4 52 26

Example 2 Solution: Immediate deflection (under full service load): For Dead load (M a = M d = 12821 in-kip): I e(d) = (6252/12821) 3 599578 + [1 (6252/12821) 3 ] 229722= 272608 in 4 For Dead + Live loads (M a = M d+l = 15091 in-kip): I e(d+l) = (6252/15091) 3 599578 + [1 (6252/15091) 3 ] 229722= 256030 in 4 53 Example 2 Solution: Immediate deflection (under full service load): Deflection due to dead load: Δ i(d) =K(5/48)M d l 2 /E c I ed = 1 (5/48) 12821 (61.5 12) 2 /(3122 272608)= 0.85 Deflection due to dead + live loads: Δ i(d+l) =K(5/48)M d+l l 2 /E c I e(d+l) =1 (5/48) 15091 (61.5 12) 2 /(3122 256030)= 1.07 Deflection due to live load: Δ il =Δ i(d+l) Δ i(d) = 1.07 0.85 = 0.22 Note: In this case, Δ i(d) is Δ i(sus), the deflection due to sustained load. 54 27

Solution: Example 2 Long term deflection (under full service load): Δ (cp +sh) = λ Δ (Δ i ) sus l D = x 1 + 50r r = A s /bd Duration ξ λ Δ isus λδ isus Δ il λδ isus + Δ il 5 years 2 2 0.85 1.7 0.22 1.92 3 months 1 1 0.85 0.85 0.22 1.07 Allowable deflection (case 1): With no false ceiling attachments, the case for Roof and floor construction supporting or attached to nonstructural elements not likely to be damaged by large deflections applies: λδ isus + Δ il l/240 l/240 = 61.5 12/240 = 3.07 λδ isus + Δ il = 1.92 ; As 1.92 < 3.07, OK 55 Example 2 Solution: Long term deflection (under full service load): Allowable deflection (case 2): With false ceiling attachments, the case for Roof and floor construction supporting or attached to nonstructural elements likely to be damaged by large deflections applies: λδ isus + Δ il l/480 l/480 = 61.5 12/480 = 1.54 λδ isus + Δ il = 1.92 ; As 1.92 > 1.54, N.G. (The deflection should not have exceeded 1.54 inch required by ACI code) 56 28

Example 2 Solution: Long term deflection (under full service load): Now for the beam in question, table below displays the live load deflections due to various depths of beam (ranging from minimum depth as per ACI code requirement up to 60 ). Depth (inches) λδ isus + Δ il (inches) ACI Limitation Remarks 36.9 (h min of ACI) 5.42 3.07 Not Governing 40 4.60 3.07 Not Governing 50 2.91 3.07 OK 60 1.92 3.07 OK From the above table, it is concluded that minimum depth requirements of ACI does not govern for unusually large values of loading and/or span. 57 Section 2: Cracking in RC Members 58 29

Topics Addressed Crack Formation Reasons for Crack Width Control Parameters Affecting Crack Width ACI Code Provisions for Crack Control Maximum Spacing Requirements Crack Control Reinforcement in Deep Flexural Members Example 1 Equations for Maximum Crack Width Example 2 59 Crack Formation All RC beams crack, generally starting at loads well below service level, and possibly even prior to loading due to restrained shrinkage. In a well designed beam, flexural cracks are fine, so-called hairline cracks, almost invisible to a casual observer, and they permit little if any corrosion to the reinforcement. 60 30

Crack Formation As loads are gradually increased above the cracking load, both the number and width of cracks increase, and at service load level a maximum width of crack of about 0.016 inch (0.40 mm) is typical. If loads are further increased, crack widths increase further, although the number of cracks do not increase substantially. 61 Crack Formation In ACI code prior to 1995, the limitation on crack width for interior and exterior exposure was 0.016 and 0.013 inch respectively. Research in later years has shown that corrosion is not clearly correlated with surface crack widths in the range normally found with reinforcement stresses at service load levels. For this reason, the former distinction between interior and exterior exposure has been eliminated in later codes. 62 31

Crack Formation Therefore, the limiting value of crack width both for interior and exterior exposures is now taken as 0.016 inch. 63 Crack Formation Because of complexity of the problem, present methods for predicting crack widths are based primarily on test observations. Most equations that have been developed predict the probable maximum crack width, which usually means that about 90 % of the crack widths in the member are below the calculated value. 64 32

Reason for Crack Width Control Appearance is important for concrete exposed to view such as wall panels. Corrosion is important for concrete exposed to aggressive environments. Water tightness may be required for marine/sanitary structures. 65 Parameters Affecting Crack Width Concrete cover Experiments have shown that both crack spacing and crack width are related to the concrete cover distance (d c ), measured from center of the bar to the face of concrete. Increasing the concrete cover increases the spacing of cracks and also increases crack width. w 1 (> w) 66 33

Parameters Affecting Crack Width Distribution of reinforcement in tension zone of beam Generally, to control cracking, it is better to use a larger number of smaller diameter bars to provide the required A s than to use the minimum number of larger bars, and the bars should be well distributed over the tensile zone of the concrete. This portion may crack Smaller diameter bars distributed over tension zone 67 Parameters Affecting Crack Width Stress in Reinforcement (crack width f sn ) Where f s is the steel stress and n is an exponent that varies in the range of 1.0 to 1.4. For steel stress in the range of practical interest, say from 20 to 36 ksi, n may be taken equal to 1.0. Steel stress may be computed based on elastic cracked section analysis. Alternatively, f s may be taken equal to 0.60f y (ACI 24.3.2.1) 68 34

Parameters Affecting Crack Width Bar Deformations Beams with smooth round bars will display a relatively small number of wide cracks in service, while beams with bars having proper surface deformations will show a larger number of very fine, almost invisible cracks. 69 ACI Code provision for crack control Crack width is controlled in the ACI Code: (ACI 24.3.2) s (inches) = (540/f s ) 2.5c c OR 12(36/f s ) (whichever is less) The center-to-center spacing between the bars in a concrete section shall not exceed s as given by the above equation. f s is the bar stress in ksi under service condition. s b c c is the clear cover in inches from the nearest surface in tension to the surface of the flexural tension reinforcement. 70 35

ACI Code provision for crack control The stress f s is calculated by dividing the service load moment by the product of area of reinforcement and the internal moment arm. n.a d a For equilibrium, M s = A s f s (d a/2) f s = M n /A s (d a/2) Where, A s A s f s M s = service load moment b 71 ACI Code provision for crack control Alternatively, the ACI code permits f s to be taken as 60 percent of the specified yield strength f y (because seldom will be reinforcing bars stressed greater than 60 % of f y at service loads). The condition f s = 0.60f y is for full service load condition. For loading less than that, f s shall be actually calculated. 72 36

Maximum Spacing Requirement Maximum spacing requirement corresponding to concrete cover as per ACI equation {s = (540/f s ) 2.5c c }: 7.5 15 4.5 For various values of concrete cover and f s of 24, 36 and 45 ksi, the maximum center-tocenter spacing between the reinforcing bars closest to the surface of a tension member to control crack width can be plotted as shown. 73 Crack Control Reinforcement in Deep Flexural Members For deep flexural members with effective depth d exceeding 36 inches, additional longitudinal skin reinforcement for crack control must be distributed along the side faces over the full depth of the flexural tension zone. s sk d/6, 12 or 1000A b /(d 30) A sk = A s /2 74 37

Example 1 Figure shows the main flexural reinforcement at mid span for a T girder in a high rise building that carries a service load moment of 7760 in-kips. The clear cover on the side and bottom of the beam stem is 2 ¼ inches. Determine if the beam meets the crack control criteria in the ACI Code. 75 Example 1 Solution: Since the depth of the web is less than 36 inches, skin reinforcement is not needed. To check the bar spacing criteria, the steel stress can be estimated closely by taking the internal lever arm equal to distance d h f /2. f s = M s / {A s (d h f /2)} = 7760/ {7.9 (32.25 6/2)} = 33.6 ksi 76 38

Solution: Example 1 Alternately, the ACI code permits using f s = 0.60f y = 0.60 60 = 36 ksi. Therefore, according to ACI 24.3.2, the maximum center-to-center spacing between reinforcing bars required to control crack width is: s = (540/fs) 2.5cc = {540/ (33.6)} 2.5 2.25 = 10.4 inches OR s = 12 (36/fs) = 12 {36/ (0.60 60)} = 12 (10.4 inches governs) Therefore, required spacing is 10.4. The provided spacing shall be less than or equal to 10.4. From given figure, c/c s provided = 5.375. Therefore the crack control criteria of ACI code is met. 77 Equations for Maximum Crack Width As alternate to ACI crack width equation, crack width can be calculated using following equations. These equations have been used in deriving the ACI Code equation on crack control. Gregely and Lutz equation: w = 0.076βf s (d c A) 1/3 Frosch equation: w = 2000(f s /E s )β [d c2 + (s/2) 2 ] 78 39

Equations for Maximum Crack Width Where, w = maximum width of crack, thousandth inch f s =steel stress at loads for which crack width is to be determined, ksi E s = modulus of elasticity of steel, ksi d c = thickness of concrete cover measured from tension face to center of bar closest to that face, inch s = maximum bar spacing, inch. 79 Equations for Maximum Crack Width β = ratio of distance from tension face to neutral axis to distance from steel centroid to neutral axis = e/f f e 80 40

Equations for Maximum Crack Width A = concrete area surrounding one bar which is equal to total effective tension area of concrete surrounding total reinforcement divided by number of bars, in 2. A = 2d cg b/n (where n = number of bars) 81 Example 2 Check the maximum crack width of the beam shown below. The beam is subjected to service load moment of 7760 in kips. The clear cover on the side and bottom of the beam stem is 2 ¼ inches. The material strengths are f c = 4 ksi and f y = 60 ksi. Use: Gregely & Lutz equation, Frosch equation, 36 32 ¼ 6 10 #8 3 ¾ 27 82 41

Example 2 Solution: Gregely & Lutz equation: a=3.68 6 w = 0.076βf s (d c A) 1/3 e = 32.32 f = 28.57 Determine the location of neutral axis for the determination of β. a = A s f s / (0.85 f c b) = (10 0.8) 0.6 60/ (0.85 4 27) = 3.13 27 c = a/β 1 = 3.13/ 0.85 = 3.68 β = e/f = 32.32/ 28.57 = 1.13 83 Example 2 Solution: Gregely & Lutz equation: w = 0.076βf s (d c A) 1/3 Determine f s. f s = 0.6f y = 0.6 60 = 36 ksi d cg = 2.75 + 1 = 3.75 A = 2d cg b/n = 7.5 27/10 = 20.25 in 2 d c = 2.75 2d cg = 7.5 w = 0.076 1.13 36(2.75 20.25) 1/3 27 = 11.80 thousandth inch OR 0.012 (0.30 mm) 84 42

Example 2 Solution: Frosch equation: w = 2000(f s /E s )β [d c2 + (s/2) 2 ] c/c spacing between bars (s) = 5.375 E s = 29000 ksi w = 2000 (36/29000) 1.13 [(2.75) 2 +(5.375/2) 2 ] = 10.78 thousandth inch OR 0.011 (0.28 mm) s = 5.375 27 85 Section 3: Development & Splices of Reinforcement 86 43

Topics Addressed Definition of Development Length Bond failure ACI provisions for Development of Tension Reinforcement ACI provisions for Development of Standard Hook in Tension Dimensions & Bends for Standard Hooks Various Scenarios where l dh must be satisfied Splices of Deformed Bars 87 Definition of Development Length Length of embedded reinforcement required to develop the design strength of the reinforcement at critical section. If the actual length (l) is equal to or greater than the development length (l d ), no premature bond failure occurs. 88 44

Bond Failure There are two types of Bond Failure 1. Direct pullout of reinforcement: Direct pullout of reinforcement occurs in members subjected to direct tension. 2. Splitting of concrete: In members subjected to tensile flexural stresses, the reinforcement causes splitting of concrete as shown. 89 ACI Provision for Development of Tension Reinforcement Basic Equation (ACI 25.4.2.3) For deformed bars or deformed wire, l d shall be: f y l d = 3 40 λ f c Ψ z Ψ e Ψ s c b + Kt r d b (25.4.2.3a) d b In which the confinement term (c b + K tr )/d b shall not be taken greater than 2.5. 90 45

ACI Provision for Development of Tension Reinforcement Basic Equation (ACI 25.4.2.3) For the calculation of ld, modification factors shall be in accordance with Table 25.4.2.4. 91 ACI Provision for Development of Tension Reinforcement Simplified Equations (ACI 25.4.2.2) Section 25.4.2.2 recognizes that many current practical construction cases utilize spacing and cover values along with confining reinforcement, such as stirrups or ties, that result in a value of (c b + K tr )/d b of at least 1.5. For other cases, (c b + K tr )/d b shall be taken equal to 1.0 Based on these values of 1.5 and 1.0, equation (25.4.2.3a) can be simplified as given on next slide. 92 46

ACI Provision for Development of Tension Reinforcement Simplified Equations (ACI 25.4.2.2) l d determined shall not be less than 12. 93 ACI Provision for Development of Tension Reinforcement l d for grades 40 and 60 (αβλ = 1 and f c = 3000 psi ) (Clear spacing of bars being developed or spliced not less than db, clear cover not less than db, and stirrups or ties throughout ld not less than the code minimum OR Clear spacing of bars being developed or spliced not less than 2db and clear cover not less than db.) Bar No Grade 40 Grade 60 #3 12 16 #4 15 22 #5 18 27 #6 22 33 #7 32 48 #8 37 55 #9 41 62 l d for grades 40 and 60 (αβλ = 1 and f c = 3000 psi ) (Other cases) Bar No Grade 40 Grade 60 #3 16 25 #4 22 33 #5 27 41 #6 33 49 #7 48 72 #8 55 82 #9 62 92 94 47

ACI Provision for Development of Tension Reinforcement Table: Generalized formulae for development lengths for usual cases of reinforcement. αβλ = 1 f c = 3000 psi l d for No. 6 and smaller bars (inches) l d for No. 7 and larger bars (inches) Grade 40 Grade 60 Grade 40 Grade 60 Clear spacing of bars being developed or spliced not less than db, clear cover not less than db, and stirrups or ties throughout ld not less than the code minimum OR Clear spacing of bars being developed or spliced not less than 2db and clear cover not less than db. 32d b 45d b 36d b 55d b Other Cases 45d b 66d b 55d b 82d b 95 ACI Provision for Development of Tension Reinforcement Further reduction in development length (ACI R25.4.2.3) The development length can be reduced by increasing the term (c + K tr )/d b in the denominator of equation 25.4.2.3a c is the smaller of either the distance from the center of the bar to the nearest concrete surface or one-half the center-to-center spacing of the bars being developed. Bars with minimum clear cover not less than 2d b and minimum clear spacing not less than 4d b and without any confining reinforcement would have (c + K tr )/d b value of 2.5. Therefore l d can be further reduced by substituting the value of 2.5 instead of 1.5 in equation 25.4.2.3a 96 48

ACI Provision for Development of Standard hook in Tension (ACI 25.4.2) Development length l dh, in inches, for deformed bars in tension terminating in a standard hook shall be the greater of (a) through (c). 97 ACI Provision for Development of Standard hook in Tension (ACI 25.4.2) For the calculation of l dh, modification factors shall be in accordance with Table 25.4.3.2. Factors ψ c and ψ r shall be permitted to be taken as 1.0. At discontinuous ends of members, 25.4.3.3 shall apply. 98 49

ACI Provision for Development of Standard hook in Tension (ACI 12.5) For bars being developed by a standard hook at discontinuous ends of members with both side cover and top (or bottom) cover to hook less than 2-1/2 in., (a) through (c) shall be satisfied: (a) The hook shall be enclosed along l dh within ties or stirrups perpendicular to l dh at s 3d b (b) The first tie or stirrup shall enclose the bent portion of the hook within 2d b of the outside of the bend (c) ψ r shall be taken as 1.0 in calculating l dh in accordance with 25.4.3.1(a) where d b is the nominal diameter of the hooked bar. 99 ACI Provision for Development of Standard hook in Tension (ACI 25.4.2) l dh (inches) for grades 40 and 60 (αβλ = 1 and f c = 3000 psi ) l dh = (f y /50 f c )d b Bar No. Grade 40 Grade 60 #3 6 8 #4 7 11 #5 9 14 #6 11 16 #7 13 19 #8 15 22 #9 16 25 100 50

ACI Provision for Development of Standard hook in Tension (ACI 25.4.2) l dh (inches) for grades 40 and 60 (generalized formulae) Grade 40 Grade 60 16d b 22d b 101 Dimensions and bends for standard hooks (ACI 25.3.2) Minimum inside bend diameters for bars used as transverse reinforcement and standard hooks for bars used to anchor stirrups, ties, hoops, and spirals shall conform to Table 25.3.2. Standard hooks shall enclose longitudinal reinforcement. 102 51

Various scenarios where l dh must be Beam Column Joint satisfied Development of beam reinforcement in column shall be > l dh Development of column reinforcement in beam shall be > l dh Beam Column 103 Various scenarios where l dh must be satisfied Development of column reinforcement in foundation 104 52

Splices of Deformed Bars Introduction Splice means to join. In general, reinforcing bars are stocked by supplier in lengths upto 60. For this reason, and because it is often more convenient to work with shorter bar lengths, it is frequently necessary to splice bars. Splices in the reinforcement at points of maximum stress should be avoided. Splices should be staggered. 105 Splices of Deformed Bars Types Bar splicing can be done in three ways: Lap Splice Mechanical Splice Welded Splice 106 53

Splices of Deformed Bars Lap Splice: Splices for #11 bars and smaller are usually made simply lapping the bars by a sufficient distance to transfer stress by bond from one bar to the other. The lapped bars are usually placed in contact and lightly wired so that they stay in position as the concrete is placed. According to ACI 12.14.2.3, bars spliced by noncontact lap splices in flexural members shall not be spaced transversely farther apart than one-fifth the required lap splice length, nor 6 inches. 107 Splices of Deformed Bars Lap Splice: According to ACI 25.5.2.1, minimum length of lap for tension lap splices shall be as required for Class A or B splice, but not less than 12 inches, where: Class A splice... 1.0l d Class B splice... 1.3l d Where l d is as per in ACI 25.4 (discussed earlier). 108 54

Splices of Deformed Bars Lap Splice: Lap splices in general must be class B splices according to ACI 25.5.2.1, except that class A splice is allowed when the area of the reinforcement provided is at least twice that required by analysis over the entire length of the splice and when ½ or less of the total reinforcement is spliced within the required lap length. The effect of these requirements is to encourage designers to locate splices away from regions of maximum stress to a location where the actual steel area is at least twice that required by analysis and to stagger splices 109 Splices of Deformed Bars Lap Splice: According to ACI 25.5.5.2, tension lap splices shall not be used for bars larger than #11 (Because of lack of adequate experimental data on lap splices of No. 14 and No. 18 bars). 110 55

Splices of Deformed Bars Mechanical Splice: In this method of splicing, the bars in direct contact are mechanically connected through sleeves or other similar devices. According to ACI 25.5.7.1, a full mechanical splice shall develop in tension or compression, as required, at least 125 percent of specified yield strength f y of the bar. This ensures that the overloaded spliced bar would fail by ductile yielding in the region away from the splice, rather than at the splice where brittle failure is likely. 111 Splices of Deformed Bars Welded Splice: Splicing may be accomplished by welding in which bars in direct contact are welded so that the stresses are transferred by weld rather than bond. According to ACI 25.5.7.1, A full welded splice shall develop at least 125 percent of the specified yield strength f y of the bar. This is for the same reason as discussed for mechanical splices. 112 56

Splices of Deformed Bars Lap splice location The splicing should be avoided in the critical locations, such as at the maximum bending moment locations and at the shear critical locations. 113 References Control of Deflection in RC Structures, Reported by ACI Committee 435 ACI 318-14 Design of Concrete Structures by Nilson, Darwin and Dolan [Chapter 5, 6]. 114 57

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