APPENDIX 1 MIX DESIGN

Transcription

1 185 APPENDIX 1 MIX DESIGN A1.1 TARGET STRENGTH The majority of the construction works in study area are residential buildings with multiple floors and they are not more than three floors. The minimum grade of concrete recommended is M30 (IS: ) and a target strength of MPa is fixed in this work. Also, this work is carried out to find the influence of brackish water in strength of concrete in the study area. In order to account, for some of the possible variations, the target strength is increased to 30 MPa. Hence, the mix design was carried out for characteristic design strength of 30 MPa. (Krishnaraju 2008). A1.2 DESIGN OF CONCRETE MIX M30 Grade designation = M30 Characteristic strength (f ck ) = 30 N/mm 2 Type of cement = OPC 53 Grade confirming to IS : Maximum size of coarse aggregate = 20 mm (angular) Degree of workability = 0.90 compaction factor Type of exposure = Mild Specific gravity of cement = 3.15 Specific gravity of water = 1.00 Chemical admixture = Super plasticizer Conplast SP430 Mineral admixtures = Flyash and Silicafume

2 186 Specific gravity of coarse aggregate = 2.67 Specific gravity of fine aggregate = 2.62 Water absorbtion of coarse aggregate = 0.15% Water absorbtion of fine aggregate = 0.10% As per codal provisons, Sand content = 36% of total aggregates Water cement ratio for M30 concrete = 0.45 ( IS: 456, 2000) From Figure.2 of IS: , select water content = 172 kg For 20 mm size of aggregates, maximum water content = 172 kg/m 3 Standard deviation (SD) for M30 Concrete = 5.00 MPa Target mean strength = SD = *5 = = N/mm 2 From IS: , Aggregate type = Crushed broken granites From IS: for 20mm size of coarse aggregates Max. Water content = 186 kg/m 3 Hence, Cement content = 172/0.45 = 382 kg/m 3 (say 380 kg/m 3 ) Minimum cement content required = 380 kg Formula for mix proportion of fine and coarse aggregate, 1000 (1-a 0 ) = [(Cement content / Sp.gr.of cement) + water content + (F a / sp.gr.* P f )] 1000(1-a 0 ) = [(Cement content / Sp.gr. of cement)+water content +(C a /sp.gr. *P c )]

3 187 Where, C a = Coarse aggregate content, F a = Fine aggregate content, P f = Sand content as percentage of total aggregates = 0.40 P c = Coarse aggregate content as percentage of total aggregates = 0.64 a 0 = Percentage of air content, As per IS:10262, for 20mm normal size aggregate, Entrapped air content is 2% = 0.02 Hence, 1000 (1-0.02) = [186+ (380/3.15) + (F a /2.62*0.40)] F a = 706 kg/m 3 (say 710 kg/m 3 ) 1000 (1-0.02) = [186+(380/3.15)+ (C a /2.67*0.71)] C a = 1277kg/m 3 (say 1280 kg/m 3 ) A1.3 MIX RATIO Cement : Sand: Coarse Aggregate: w/c = (380: 710: 1280: 0.45) follows, The design mix proportions for the required target strength is as Cement : Sand: Coarse Aggregate: w/c = (1 : 1.87 : 3.37 :0.45)

4 188 APPENDIX 2 A2.1 NEURAL NETWORK DESIGN AND TRAINING The network architecture or features such as number of neurons and layers are very important factors that determine the functionality and generalization capability of the network. For this model, standard multilayer feed forward back propagation neural networks are designed with neural fitting tool software. The network consists of three layers, they are the input layer, hidden layer and out put layer. In order to determine the optimal architecture, different network with different number of layers and neuron in the hidden layer are designed and tested. The ANN models developed for the prediction of 5 outputs from 11 inputs. The number of neurons is adjusted in steps and the various stages of training of the model are summarized in Table A2.1 The first stage, 10 neurons have been considered for training, testing and validation. First stage has given a very high value of MSE for training, testing and validation. Regression coefficient R was not close to 1 for training, testing and validation. In first set of trails, the percentage of samples considered for training the network is 80%, the percentage of samples for testing is 10 % and the percentage of samples for validation is 10 %. Each stage of training is carried out by 3 trials. The number of neurons was increased by 5 up to 25 in four steps. Then percentage of samples selected was adjusted to 70 % for training, 15 % for testing and 15 % for validation. The training is repeated until minimum MSE values and R values close to 1 are arrived. The expected results are arrived after 6 set of trials. The

5 189 various stages of training are summarized in Table A3.1 with number of neurons, MSE and R values of the training results. Table A2.1 Training stages of ANN model S.No Percentage of samples No. of neorons Training Testing Validation Training MSE Testing R value Testing Validation Training Validation 6.54E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E

6 190 APPENDIX 3 A3.1 MODEL CALCULATION OF SERVICE LIFE The flexural members in a structure are beam and slab. The steel reinforcements are subjected to corrosion, the service life of beam and slab will deteriorate faster than any other elements of the structure. The methodology in calculation of service life is explained here, by taking beam as the case. A3.2 METHOD TO FIX PERCENTAGE LOSS IN DIAMETER The permissible percentage of loss can be fixed by following the procedure given below. Bending due to actual load = M Bending moment to factored load = FM Moment carrying capacity ( M u ) of the beam as per IS: M u = 0.87 * f y *A st * d [1- ( f y * A st / f ck * bd) ] where, f y = Charateristic strength of steel in N / mm 2 f ck = Charateristic strength of concrete in N / mm 2 A st = Area of steel in mm 2 b = Breadth of the beam in mm d = Effective depth of the beam.

7 191 For safe design, FM < M u (A3.1) Due to corrosion, the area of steel will get reduced. Let A stc be area of steel after corrosion. Now, the moment carrying capacity (M uc ) of the beam after corrosion M u = M uc = 0.87 * f y *A stc * d [1- ( f y * A stc / f ck * bd) ] For safety against corrosion, M uc > M (A3.2) The value of M, FM, M u and M uc depends on the extent to which area of steel is reducing. Hence, A stc cannot be allowed to reduce more than certain limit so that, the condition in equation (A3.2) should be satisfied. Percentage of loss in area of steel due to corrosion Percent loss = [A stc / A st ] * 100 Percentage of loss that can be permitted Permissible percentage of loss = [ 1 - (Percent loss ) ] A3.3 Case study A rectangular beam of cross section 230 mm x 450 mm is reinforced with 4 numbers of 12 mm RTS bars. The grade of concrete in M20 and the grade of steel is Fe 415. The permissible loss in diameter of steel is 2.39 mm. The member is subjected to uniform rate of corrosion due to carbonation. The rate of corrosion is 0.25 mm/year. Estimate the service life of the structural member. Ignore the effect of bondage. A3.3.i Procedure The procedure to calculate the service life can be understood from the model calculation worked based on the case study.

8 192 A3.3.ii Given Data Size of the beam = 230 mm x 450 mm Reinforcements = 4 x 12 mm RTS bars Grade of concrete = M20 Grade of steel = Fe 415 Permissible loss in diameter = 2.39 mm Rate of corrosion = 0.25 mm/year = 250 µm/year Corrosion nature = Carbonation So, Value of K = 0.5 A3.3.iii Solution The service life of structure is given in equation (4.4) as, Service life of structure (t) = D / [ K * (2 x 10-3 ) * R(t)] Where, t = Service life of structure in years. D = Permissible loss in diameter of reinforcement in mm K = Correction factor R(t) = Corrosion rate at time ( µm/year) Substituting the given values in equation for service life, t = D / [ K * (2 x 10-3 ) * R(t)] t = 2.39 / [ 0.5 * (2 x 10-3 ) * 250] t= 9.56 years A3.3.iv Result Safe Service life of structure under given Corrosion environment = 9.56 years