Learn : Stress Strain relationship for concrete, Stress Strain relationship for steel,Design Strength Values for Steel Design Stresses at Specified Strains. fe 415 and fe 500
Stress Strain relationship for concrete and Stress Strain relationship for steel
Stress Strain relationship for concrete
Stress Strain relationship for concrete : The experimental or actual stress strain curve for concrete is very difficult to use in design Therefore, IS code 456:2000 has simplified or idealized it as shown in Fig. 4.1.
- For design purpose, the compressive strength of concrete in the structure in taken as 0.67 times the characteristic strength. The 0.67 factor is introduced to account for the difference in the strength indicated by a cube test and the strength of concrete in actual structure.
- The partial safety factor (r_{mc}), equal to 1.5 is applied in addition to this 0.67 factor.
- The initial portion of the curve is parabolic. After a strain of 0.002 (0.2%), the stress becomes constant with increasing load, until a strain of 0.0035 is reached and here the concrete is assumed to have failed.
Stress Strain relationship for steel
Stress Strain relationship for steel : The stress-strain curve for mild steel, Fe 415 and Fe 500 are shown in Fig. 4.2 and 4.3.
For mild steel, the value of characteristic stress is taken as yield stress and the design curve is obtained after applying a factor of safety of 1.15 to yield stress. The design stress for steel is equal to \[\frac {f_{y}}{1.15}\] i.e., 0.87 f_{y}. For high strength bars (Fe 415 and Fe 500), the yield point is not distinct, hence yield stress is taken as 0.2 percent proof stress and factor of safety is applied to it.
Design Strength Values for Steel
(i) For mild steel \[f_{d}=\frac{f_{y}}{1.15}=0.87f_{y}\]
and \[(f_{y}=250 N/mm^{2}) f_{d}=0.87\times250=217.5 N/mm^{2}\]
(ii) For Fe 415, \[f_{d}=0.87\times415=361N/mm^{2}\]
(iii) For Fe 500, \[f_{d}=0.87\times500=435N/mm^{2}\].
TABLE 4.1. Design Stresses at Specified Strains.
Fe 500 steel : \[f_{y}=500(N/mm^{2})\] | Fe 415 steel : \[f_{y}=415N/mm^{2}\] | |||
Strain | Stress (N/mm^{2}) | Strain | Stress (N/mm^{2}) | |
0.00174 | 347.8 | 0.00144 | 288.7 | |
0.00195 | 396.6 | 0.00163 | 306.7 | |
0.00226 | 391.3 | 0.00192 | 324.8 | |
0.00277 | 413.0 | 0.00241 | 342.8 | |
0.00312 | 423.9 | 0.00276 | 351.8 | |
0.00417 | 434.8 | 0.00380 | 360.9 |
(a) Fe 500 steel (b) Fe 415 steel
One of the most critical points in the theory of steel fibre reinforced concrete (SFRC) is quantifying the residual stresses in tension. Due to concrete interaction with fibres, a cracked section is able to carry a significant portion of tensile stresses, called the residual stresses. Because of a great diversity in the shape and aspect ratio of fibres and, consequently, varying bond characteristics, there are no currently available reliable constitutive models. In present practices, residual stresses needed for strength, deflection and crack width analysis are quantified by means of standard bending tests. However, such tests require relatively sophisticated and expensive equipment based on the displacement-controlled loading. Besides, the test results are highly scattered. This paper investigates an alternative approach for defining the residual stresses. The approach aims at deriving equivalent stress–strain relations of cracked tensile concrete using test moment–curvature relationships of flexural concrete members with ordinary reinforcement and steel fibres. Tests on eight lightly reinforced beams (reinforcement ratio 0.3%) with different contents of steel fibres (0%, 0.5%, 1.0%, and 1.5% by volume) have been carried out. Based on the proposed technique, equivalent stress–strain relations were defined for each of the beams and further used for curvature and crack width analyses.