Question 14.8: Re-do Example 14.5 and locate the performance point using th...

For T = 0.3 sec, stiffness = 43820 N/m
This is a short period system having a period T < T_{C} = 0.5 sec.

The calculations are same up to Step 3 of Example 14.5. The reduction factor R = 3.79.

Step 4: The period 0.3 sec lies in the transition zone. Now compute ductility from

R = (T/T_{C}) (µ − 1) + 1
or,    µ = (R − 1)/ (T/T_{C}) + 1
or,     µ = (T_{C}/T) (R − 1) + 1
∴      µ = (0.5/0.3) (3.79 − 1) + 1 = 5.65

Step 5: Compute R, S_{a} and S_{d} coordinates for ductility µ = 5.65 using Equation (14.32) and (14.33) and plot. This demand curve will intersect the capacity curve at B. The point of intersection B gives the target displacement equal to 0.0226 m. Again, target displacement corresponds to a ductility equal to

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)

µ = 0.0226/0.004 = 5.65          OK

Step 6: If this point of intersection B is joined with the point of intersection A between the elastic demand curve and period T = 0.3 sec as obtained in Step 1, an inclined line is obtained. It means the elastic and inelastic displacements are not equal in this range.
Typical calculations are shown in Table 14.12. The double lines in the table show change in reduction factor R region. The location of performance point in the A–D response spectra is shown in Figure 14.26. The ductility curve for 5.65 does pass through the performance point.

Table 14.12 Calculations for R, s_{a} and s_{d} coordinates

It can be seen that the Newmark–Hall spectra required a ductility of 6.32 and target displacement of 2.53 cm. A slight difference in the two methods is expected.