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.

Annotation 2022-10-19 120402
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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
Annotation 2022-10-19 121316

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