Question 14.6: A SDOF system has the following data: Mass = 100 kg, Damping...

For T = 1 sec, stiffness = 3944 N/m
It is a medium period system having T > 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 maximum 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 = 0.05 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 = 1 and 0.5 s.

S_{di}=\frac{T_i^2}{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: From the point of intersection A, draw a vertical. It intersects the capacity curve at B. The point of intersection gives the target displacement equal to 0.0933 m.
Step 4: Knowing the spectral accelerations in elastic and capacity curves, estimate the reduction factor R.

R = 3.68/1.97 = 1.87

Step 5: The period 1.0 sec lies beyond T_{C} = 0.5 sec. Now compute ductility from

R = µ
or,           µ = R = 1.87

Step 6: Compute R, S_{a} and S_{d} coordinates for ductility µ = 1.87 using Equation (14.31) and (14.32). Now plot S_{a} vs S_{d} for µ = 1.87. This curve will intersect the capacity curve at the same point B as the point of intersection of the vertical line. It confirms that the elastic and inelastic displacements in this period range are equal.

Again, target displacement corresponds to a ductility equal to

µ = 0.0933/0.050 = 1.866 ≈ 1.87        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)

Typical calculations are shown in Table 14.10. The double lines in the table show change in reduction factor R region.
The location of performance point is shown in Figure 14.24.

Table 14.10 Calculation for R, s_{a} and s_{d} coordinates