Additional assumption in this state is the linear stress diagram at the compression concrete.The procedure in the analysis is given as follows:1. The depth of the compression block (c)is obtained using the equation below. Applying the equilibrium equation that ?F = 0. Where: Cs = As’fs’ = the compressive force for compression steelfs’ = 200,000?s’= compressive stress of steel based from Priestly et al where ?s’ is assumed to be less than 0.

00207 = the compressive force for compression concrete (volume of the compression block based from Hognestad Model, 1951) Ts = Asfs = the tensile force for tension steelfs = 200,000?s = tensile stress of steel based from Priestly et al where ?s is assumed to be less than 0.00207 = the tensile force for tension concrete (volume of the tension block based from linear stress diagram)All unknown quantities such strain and stress can be transformed in terms of the depth of compression block (c). An iterative procedure is adapted inorder to determine the compression block.

2. The modulus of rupture of concrete, fr is used to determine the strain of concrete and steel for both top and bottom. Once the strain of concrete at the extreme compression fiber (?cc) the curvature of the beam can now be calculated using the formula .3. The third and the last quantity to be determine is the nominal moment capacity. The nominal moment capacity at crack initiation is equal to the cracking moment. This moment is obtained by applying the summation of moment at the N.A.

A. After Flexural CrackingIn this state the tensile concrete is neglected, the assumption is that crack propagates up to the neutral axis as shown in Fig.3b. h d b N.A. c d-c fcc STRESS-FORCE DIAGRAM BEAM SECTION STRAIN DIAGRAM As Figure 3b. Cracked Beam subject to bending As’ ?cc ?s’ ?s Ts Cs Cc The moment capacity of the beam at this stage, the least possible scenario is still equal to the cracking moment obtained above.

Thus tasks now will only include the calculation of the depth of compression block by iteration such that ?F = 0 and ?M = Mcr and where Cs, Cc and Ts are calculated in the same manner above. After obtaining the depth of compression block, the curvature of the beam is to be calculated.1. Flexural Yielding StateFrom crack initiation, the stress of steel continuously increasing until it reaches it yield strength. The stage is known as flexural yielding. Using Priestly et al model for Grade 60 steel, the stress of steel fs is now equal to 60 ksi (420 MPa) see Fig.

3c. h d b N.A. c d-c fcc STRESS-FORCE DIAGRAM BEAM SECTION STRAIN DIAGRAM As Figure 3c. Flexural Yielding of Beam As’ ?cc ?s’ ?s= ?y T = Asfy Cs Cc The analysis of the beam at this state is almost similar to after cracking state. Here the tensile force of tension steel can be determined right away that is T = Asfy, leaving two unknowns in the equation ?F = 0.

The values of Cc and Cs once again can be express in terms of the depth of the compression block (c). By series of iteration the compression block’s depth is obtain and giving the curvature and the nominal moment capacity using the same procedure in crack initiation. 2. Ultimate StateAt ultimate state, ideally obtaining a ductile behaviour of failure is desired. Ductile failure happens when steel yields first before concrete reaches the rupture point. Meaning at this stage, the stress of tension steel must exceed its yield strength.

For analytical convenience, concrete is assumed to crush with the corresponding steel exceeds its yield strength. Using the Hognestad Model (1951) concrete crush at a strain of 0.003.

Unlike the previous stages where normally the compressive stress of concrete does not reach its peak where the compression block is based only on the ascending part of the stress-strain model, in the ultimate stage the entire stress-strain model is utilize as shown in Fig. 3d. h d b N.A. c d-c fcc STRESS-FORCE DIAGRAM BEAM SECTION STRAIN DIAGRAM As Figure 3d. Flexural Yielding of Beam As’ ?cc = 0.003 ?s’ ?s > ?y T = Asfy Cs Cc Applying the equation ?F = 0 and expressing the values of Cc, Cs and T in terms of the depth of the compression block (c) and by series of iteration the compression block’s depth is obtain.

Determine afterwards the curvature and the nominal moment capacity using the same procedure as of the previous stages. In expressing the stress of steel in terms of its strain the Priestley et al model is strictly followed. Since only the values vary not the concept, a spreadsheet program is used in the entire computations for the depth of compression block, nominal moment and curvature. The calculations are attached in the Appendices.After determining determine the depth of compression, nominal moment and curvature for each cases at corresponding stages, the results are presented in tables and graphs and thoroughly discussed. PART 2The second part of this paper is the presentation of the moment-curvature diagram for each cases using the control specimen. The each cases, the variable is the type of steel and the corresponding models and experimental data used as listed in Table 2.

The material property of concrete is maintained and this time the Kent and Park Model (1971) is used in the concrete analysis.Table 2. Material Properties CASES CONCRETE PROPERTY & MODEL STEEL PROPERTY & MODEL/DATA 1 C35-Kent and Park (1971) Model Grade 60-Priestley et al.

(1996) Model 2 C35-Kent and Park (1971) Model A1035 – ACI ITG 6 (2010) Model 3 C35-Kent and Park (1971) Model SD685 – Wang et al. (2009) Model 4 C35-Kent and Park (1971) Model Experimental Data 1 5 C35-Kent and Park (1971) Model Experimental Data 2 6 C35-Kent and Park (1971) Model Experimental Data 3 For simplicity in the analysis the moment-curvature at crack initiation, flexural yielding and ultimate stages were the points considered. The determination of moment and curvature is similar in procedure as of Part I in this paper. The computations and graphs are obtained by implementing a spreadsheet program for each cases. The results of moment and curvature are tabulated with the diagrams.