Question 14.9: Re-do Example 14.6 and locate the performance point using th...

For T = 1 sec, stiffness = 3944 N/m
It is a medium period system having a period T >T_{C} = 0.5 sec
The calculations are same as in Example 14.6. The reduction factor R = 1.87 and the ductility demand is also 1.87.
Step 5: Compute R, S_{a} and S_{d} coordinates for ductility µ = 1.87 using Equation (14.32) and (14.33). This curve will intersect the capacity curve at the same point B as the point of intersection of the vertical line. The target displacement is 0.0933 m, which is same as obtained by the Newmark–Hall method.

S_{\text{d inelastic}} =µ S_{\text{d inelastic}}/R    for     T < T_{C}     (14.32a)

S_{\text{d inelastic}} =S_{\text{d inelastic}}      for     T > T_{C}        (14.32b)

R = (µ − 1)T/T_{C} + 1     for        T < T_{C}    (14.33a)

R = µ      for        T >T_{C}      (14.33a)

Typical calculations are shown in Table 14.13. The double lines in the table show change in reduction factor R region. The performance point in the A–D response spectra is shown in Figure 14.27

It can be seen that the Newmark–Hall spectra required a ductility of 1.87 and target
displacement of 9.33 cm which are the same as required by the EC8-Part 1. The demand curves for a ductility of 1.87 are different in both the methods. Nevertheless, both the demand curves do intersect at the same performance point.