Question 14.1: Generate moment-curvature curve for a reinforced concrete un...
Generate moment-curvature curve for a reinforced concrete unconfined section is shown in Figure 14.13.

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Section properties:
breadth = 300 mm, depth = 500 mm, clear cover to stirrups = 25 mm
Concrete properties:
f_{ck} = 30 MPa, f^{\prime}_{c} = 25 MPa,\varepsilon_{0} = 0.002, \varepsilon _{cu} = 0.0035 and E_{c}= 27400 MPa
spalling strain = 0.0050
Reinforcing steel properties (hot rolled steel):
f_{y} = 415 MPa, E_{s} = 200000 MPa, f_{u} = 460 MPa, \varepsilon _{u}= 0.12, E_{sh}= 10000 MPa (5% ofE_{s}), diameter of longitudinal bars \phi _{L} = 25 mm, diameter of stirrups \phi _{T} = 10 mm and stirrup spacing x = 150 mm.
The basic relations for generating M-f curve were discussed earlier. These equations were used to derive the moment-curvature curve shown in Figure 14.14. The iterations were done in Excel sheet. The final values of M and f for a given strain level are illustrated in Table 14.6.
Idealized plastic moment = 326 kNm
Strain | Depth of Neutral Axis mm | Moment kNm | Curvature |
0 | 0 | 0 | 0 |
0.0005 | 174 | 123.3 | 0.000003 |
0.001 | 180 | 231.1 | 0.0000055 |
0.0015 | 183 | 315.9 | 0.000008 |
0.002 | 152 | 322.9 | 0.000013 |
0.0025 | 137 | 325.7 | 0.000018 |
0.003 | 128 | 326.2 | 0.000023 |
0.0035 | 123 | 326.4 | 0.000028 |
0.004 | 122 | 327.9 | 0.000033 |
0.0045 | 124 | 332.7 | 0.000036 |
0.005 | 126 | 336.4 | 0.00004 |
