Question 13.7: (a) Plot the ratio of peak deformation in an inelastic syste...

(a) The relation between the peak deformations in a SDOF inelastic system and the corresponding elastic system were derived in Equation (13.35). The same are shown in Table 13.5 for different ductility and plotted in Figure 13.34. Although the calculation are shown for period T upto 100 sec, the plot is shown only upto 1 sec for clarity. Similar calculations can be done for other values of ductility

F_y=F_O \ \text{and} \ x_m=\mu x_0        (13.35a)

F_y=\frac{F_0}{R}=\frac{F_0}{\sqrt{2\mu -1}x_0}      (13.35b)

x_m=\frac{\mu }{R}x_0=\frac{\mu }{\sqrt{2\mu -1}}x_0      (13.35c)

F_y=\frac{F_0}{\mu } \ \text{and} \ x_m= x_0           (13.35d)

Table 13.5 Relation between Maximum Elastic and Inelastic Displacements with Period T for Different Ductility μ

(b) For the ground motion of Example 13.5, period T_{C} = 0.5 sec.

∴ Inelastic acceleration = \frac{A}{\sqrt{2\mu -1} }             (13.34c)

Elastic acceleration A = 1 g × 2.71 = 2.71 g
∴ Inelastic acceleration = \frac{2.71 g}{\sqrt{2\mu -1} }

and
Yield strength              F_{y}=\frac{100\times 2.71g}{\sqrt{2\mu -1} }

Also, the maximum deformation is given by Equations (13.35c) for T < T_{C}

 x_{m}=\frac{\mu }{R} x_{0}=\frac{\mu }{\sqrt{2\mu -1} } x_{0}           (13.35c)

Maximum elastic deformation  x_{0}=\frac{A}{\omega ^{2}} =\left\lgroup\frac{T}{2\pi } \right\rgroup^{2}A

∴                        x_{m}=\frac{\mu }{\sqrt{2\mu -1} } \left\lgroup\frac{T}{2\pi } \right\rgroup^2 A

=\frac{\mu }{\sqrt{2\mu -1} } \left\lgroup\frac{0.4}{2\pi } \right\rgroup^{2} 2.71 g

=\frac{0.1077\mu }{\sqrt{2\mu -1} } m

The maximum strength and maximum deformation are given as follows:
For µ = 1, maximum strength = 271 g = 2658.51 N, maximum deformation x_{0} = 0.1077 m
For µ = 4, maximum strength = 1004.8 N, maximum deformation x_{m} = 0.1628 m