Concrete Cracking. ε ctr = f ctr / E c = 0.6 / 4400 = x 10-3

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1 Concrete Cracking Concrete is known as a sensitive material for cracking. The code defines concrete modulus of elasticity (E c ) and concrete cracking-limit tensile stress (f ctr ) as: E c = 4400 (f cu ) 1/2 (in N/mm 2 ) (Code eq. 2-1 page 2-15) f ctr = 0.6 (f cu ) 1/2 (in N/mm 2 ) (Code eq b page 4-49) Hence; the cracking- limit tensile strain (ε ctr ) is; ε ctr = f ctr / E c = 0.6 / 4400 = x 10-3 This means that concrete cracking is expected to take place if concrete tensile strains (ε c ) exceed the cracking- limit value (ε ctr ). Tensile strains develop in concrete because of different reasons, the main ones may be: 1- Shrinkage of constrained concrete at its early stages. 2- Applied loads. 3- Thermal effects. 4- Rusting (corrosion) of steel: Rusting is; simply, the oxidization of steel. The volume of oxidized steel is greater than its original volume (4 to 7 times). The increase in the steel volume creates an internal pressure on concrete cover, which leads to its cracking due to the development of local tensile strains (these cracks develop parallel to the bars). 5- Excessive Deformations of the structural: Such as deformations due to differential settlement of supports, excessive deflections. Prof.Dr.Hany.M.El-Hashimy Crack control - 1 -

2 Philosophy of Control of Concrete Cracking Distribution and width of concrete cracks depend on the level of introduced tensile strains (ε c ) [i.e. more intense cracking takes place with the increase in ε c ]. Therefore control of concrete cracking can be achieved by keeping (ε c ) as low as possible; specially in the exposure zones 3 & 4 defined in Table (4-11) in the code. Minimizing concrete cracking requires fulfilling the following measures: 1- Appropriate concrete dimensions: Code clause (page 4-59) The smaller the section, the greater tensile strains (ε c ) for the same loading configurations. 2- Appropriate thickness of concrete cover: (c c ) a) To insure sufficient bond between steel & concrete. b) To form a good protection for the reinforcing bars against corrosive attacks from the ambient surroundings. Table 4-13 in the code defines the minimum thickness of concrete cover for different structural elements function of the exposure zone, concrete characteristic strength and the type of structural element. Definition: The thickness of concrete cover in table 4-13 is to be measured OUT-SIDE the bar diameter. Prof.Dr.Hany.M.El-Hashimy Crack control - 2 -

3 3- Good quality concrete: a) Concrete characteristic strength (f cu ): f cu should be > 25 N/mm 2. b) Aggregates & mixing water should conform with the standard specifications and the code; specially: Size and strength of aggregates. Content of salts & minerals. Amount of fine particles. c) Cement dosage: in the exposure zones 3 and 4; a minimum dose of 400 kg cement/cubic meter of concrete is highly recommended. d) Cement type: should be appropriate for the amount of salts in the environment (soil, water, atmosphere, etc). e) Minimum Porosity: Lowest possible water/cement ratio ( < 0.42) Good concrete casting & compaction. f) Appropriate additives: (such as plasticizers & superplasticizers) in the concrete mix. These products improve (f cu ) and the workability; i.e. help in minimizing the mixing water. g) Good concrete casting and compaction. h) Good concrete curing: Especially within the first 2 to 3 weeks after casting to avoid cracking at early ages. Prof.Dr.Hany.M.El-Hashimy Crack control - 3 -

4 4- Expansion (movement), construction and shrinkage joints: Refer to Code clause 7-4 (page 7-12). 5- Employing the steel reinforcement in controlling the section tensile strain: The width and distribution of concrete cracks are influenced by the details of steel reinforcement as follows: a) Steel type: Using ribbed bars of High Grade Steel (360/520) or (400/600) is preferable than using smooth bars of Normal Mild Steel (240/350). The mechanical interlock of bars ribs greatly improves its bond with the surrounding concrete. b) Bar diameter: It is always preferable to use bars of small diameters (as much as possible) to increase their surface area and hence improve the bond conditions. The minimum diameter of bars in exposure zones 3 & 4 is 12 mm. c) Minimum bar spacing: Code clause (page 7-9) Respecting the specified min. distance between bars insures efficient concrete surrounding to the bars and hence best bond conditions. Prof.Dr.Hany.M.El-Hashimy Crack control - 4 -

5 d) Maximum bar spacing: Code clause (page7-10) Maximum distance between bars in the structural elements should satisfy the code requirements. If the distance between bars exceeds the specified ranges, cracks can develop and propagate between the bars. e) Splices & anchorage length: Code clause (page 4-37) Providing sufficient length and distribution of bar splices and anchorages insures the presence of good bond conditions along the whole bar. f) Steel tensile strains & stresses: Code clause (page 4-52) Reducing the tensile strains in well bonded steel reinforcement leads to reduce the tensile strains in the surrounding concrete. This can play an important rule in reducing the intensity of cracking. The code provides 2 alternative methods for controlling steel tensile strains: i- Either Satisfying equation (4-66). Code page 4-52 ii- or Satisfying the maximum steel stresses given in code Tables (4-14) & (4-15) Code page 4-59 Refer also to the provisions for concrete durability in clause (page 2-18) in the Code. Prof.Dr.Hany.M.El-Hashimy Crack control - 5 -

6 Tensile concrete stresses (Code clause ) EXAMPLE 1: (Simple Bending) M = 45 KN.m = 45 x 10 6 N.mm b = 1000 mm t = 400 mm fcu=25 N/mm 2 Tensile stresses due to applied forces: f ct (N) = 0.0 f ct (M) = (M / I).Y = 6 M /b t 2 = 6 * 45 x 10 6 /(1000 x ) = 1.65 N/mm 2 Allowable tensile stress: f ctr = 0.6 (25) 1/2 = 3 N/mm 2. tv = 400 [ 1 + (0/1.5)] = 400 mm. [i.e. t v = t for N=0.0] η = 1.6 (from Table 4-16) f ctr / η = 3 / 1.6 = N/mm 2 > f ct O.K. EXAMPLE 2: (Ecc. Tension) N= + 60 KN M = 30 KN.m b = 1000 mm t = 350 mm fcu=25 N/mm 2 Tensile stresses due to applied forces: f ct (N) = N / A = / (1000 * 350) = 0.17 N/mm 2. f ct (M) = M.I/ Y = 6 M /b t 2 = 6 * 30 x 10 6 / (1000*350 2 ) = 1.47 N/mm 2. f ct = f ct (N) + f ct (M) = = 1.64 N/mm 2. Allowable tensile stress: f ctr = 0.6 (25) 1/2 = 3 N/mm 2. tv = 350 [ 1 + (0.17 / 1.47)] = 390 mm. η = 1.6 (from Table 4-16) f ctr / η = 3 / 1.6 = N/mm 2 > f ct O.K. Prof.Dr.Hany.M.El-Hashimy Crack control - 6 -

7 EXAMPLE 3: (T-section subjected to ecc. tension) N= + 60 KN M = - 40 KN.m fcu=25 N/mm 2 b = 300 mm t = 600 mm t f = 100 mm Properties of gross-section: (i.e. pre-cracking) Since we are dealing with uncracked sections, Always take into account the flange(s); even if the flange is at the tension side. B = b + 6 t f = (6 x 100) = 900 mm. A sec = (300 x 500) + (900 x 100) = = mm 2 Y = ( x 350) + ( x 50) = mm I = (900 x ) + [ x ( ) 2 ] 12 + (300 x ) + [ x ( ) 2 ] = 8195 x 10 6 mm 4 12 Z top = I = 8195 x 10 6 = mm 3 Y Prof.Dr.Hany.M.El-Hashimy Crack control - 7 -

8 Cont. EXAMPLE 3: Tensile stresses due to applied forces: f ct (N) = N A sec = = 0.25 N/mm 2. f ct (M) = M Z top = (40x10 6 ) = 1.16 N/mm 2. f ct = f ct (N) + f ct (M) = = 1.41 N/mm 2. Allowable tensile stress: f ctr = 0.6 (25) 1/2 = 3 N/mm 2. tv = 600 [ ] = 729 mm η = 1.7 (from Table 4-16) f ctr = 3 = 1.76 N/mm 2 > f ct O.K. η 1.7 Prof.Dr.Hany.M.El-Hashimy Crack control - 8 -