Question 14.2: Generate a moment-curvature curve for a concrete section sho...
Generate a moment-curvature curve for a concrete section shown in Figure 14.15 using the Mander model for confined concrete.
Axial load in column = 1000 kN (assumed); axial load capacity = 4100 kN
Section properties:
Width = 500 mm, height = 500 mm, clear cover to ties = 40 mm
Concrete properties:
f_{ck} = 30 MPa, E_{c} = 27400 MPa
Reinforcing steel properties:
f_{y} = 415 MPa, E_{c} = 200000 MPa, f_{u} = 550 MPa, \in_{u} = 0.09, \in _{sh} = 0.0105, E_{sh}= 10000 MPa, f_{yh}= 415 MPa, diameter of longitudinal bars \phi _{L} = 25 mm, diameter of stirrups \phi _{T} = 12 mm and stirrup spacing s = 100 mm.
Mander concrete model
Unconfined strain = 0.002; Unconfined stress = 25 MPa (Cylinder); Ultimate strain = 0.02051
Ultimate stress = 25 MPa; Spalling strain = 0.005; Confined stress = 35.7 MPa
If A_{shx} is the total amount of lateral reinforcement in a hoop layer crossing a section perpendicular to y-axis, then the maximum effective confining stress that can be developed in that direction is given by
f_{lx}^{\prime}=f_{lx}k_{e} (14.13a)
f_{lx}=p_{tx}f_{yh} (14.21a)
∴ f_{lx}^{\prime}=p_{tx}f_{yh}k_{e} (14.25)
For the square cross-section considered in this example:
Effective cover to centre line of the tie (hoop) = 40 + 12/2 = 46 mm
Effective cover to centre of longitudinal bars = 40 + 12 + 25/2 = 64.5 mm
s = 100 mm c/c
Distance between two longitudinal bars = (500 − 40 − 12 − 40 − 12 − 25)/3 = 123.7 mm c/c
∴ w_{i} = 123.7 − 25 = 98.7 mm clear
A_{i} = 12 × 98.72 /6 = 19480 mm²
A_{core}= x_{cl}\times y_{cl} = (500 – 46) × (500 – 46) = 206116 mm²
Net area of concrete A_{cc}=A_{core} – A_{t} = 206116 – 12 × 490 = 200236 mm²
A_{e}=(A_{core}-A_{i})\left\lgroup1-\frac{s-\phi _{T}}{2x_{cl}} \right\rgroup \left\lgroup1-\frac{s-\phi _{T}}{2y_{cl}} \right\rgroup
A_{e} = (206116 – 19480) × (1 – (100 – 12)/(2 × 454)) × (1 – (100 – 12)/(2 × 454)) = 152240 mm
K_{e}=\frac{A_{e}}{A_{cc}} = 0.76
Total area of lateral steel in x-direction = 4 × 113 = 452 mm²
p_{tx}=\frac{A_{tx}}{sy_{cl}} =\frac{425}{100\times (500-46)}
The effective confining lateral pressure can be computed using Equation (14.25).
f_{lx}^{\prime}=p_{tx}f_{yh}k_{e} = 0.00996 × 415 × 0.76 = 3.14 MPa
The M-f curve is shown in Figure 14.16.
Idealized plastic moment = 591 kNm
Yield curvature = 0.0000109; ultimate curvature = 0.000113
