Question 14.5: A SDOF system has the following data: Mass = 100 kg, Damping...

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

Spectral acceleration corresponding to the ground motion with peak acceleration 1 g in Example 14.4 was 2.71 g. In the present case, the peak ground acceleration is 0.25 g.
∴ peak spectral acceleration = 0.25 × 2.71 × 9.81 = 6.65 m/sec²
Elastic force = 100 × 6.65 = 665 N
Yield displacement = 0.004 m

The steps for locating the performance point are as follows:
Step 1: Plot the elastic spectral acceleration – spectral displacement curve for 5% damping using Equation (14.26) along with radial lines for T = 0.3 and 0.5 sec.

S_{di}=\frac{T^2_i}{4π^2}S_{ai}g        (14.26)

Step 2: Next, the capacity curve is plotted. The initial stiffness line is extended up to the demand curve with µ = 1, which meets at point A.
Step 3: Knowing the spectral accelerations in elastic demand and capacity curves, estimate the reduction factor R.

R = 6.65/1.75 = 3.79

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

R = (T/T_{C})µ

or,                                                  µ = R/(T/T_{C})

or,                                                µ = (T_{C}/T)R

∴                                                    µ = (0.5/0.3) 3.79 = 6.32

Step 5 : Compute R, S_{a} and S_{d} coordinates for ductility µ = 6.32 using Equation (14.31) and (14.32). Now plot  S_{a} vs S_{d}  for m = 6.32. This demand curve will intersect the capacity curve at B. The point of intersection B is the target displacement equal to 0.0253 m. Again, target displacement corresponds to a ductility equal to

µ = 0.0253/0.004 = 6.32            OK

R = 1       for         T < T_{A}               (14.31a)

R=(2\mu -1)^{y/2}                for T_{A} < T <T_{B}      (14.31b)

R=(2\mu -1)^{0.5}  for     T_{B}  < T <T_{C}^{\prime}      (14.31c)

R=(T/T_{C})\mu  for   T_{C}^{\prime} < T < T_{C}      (14.31d)

R = µ          for              for  T >T_{C}      (14.31e)

\gamma =\frac{\log \left\lgroup\frac{T}{T_{A}} \right\rgroup }{\log \left\lgroup\frac{T_{B}}{T_{A}} \right\rgroup }      (14.31f)

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)

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 confirms that the inelastic and elastic displacements in the short period range are not equal.
Typical calculations are shown in Table 14.9. The double lines in the table show change in reduction factor R region. The performance point B is shown in Figure 14.23.
Table 14.9 Calculation for R, s_{a} and s_{d} coordinates