1. Elastic and inelastic relations..... mx+cx+Q(x)= -ma x Q x Q Q=kx elasticinelastic

2. Exercise 1 (A hysteretic energy dissipation index E h ) A hysteretic energy dissipation index E h corresponds to equivalent viscous damping factor h. Derive the equation (3.1) by calculating energy dissipation ΔU by viscous damping per cycle under the ‘resonant steady-state’ and putting ΔU = ΔW. E h = ΔW/2πF m D m (3.1) c: damping coefficient(=2hm ), m: mass of the system : natural circular frequency(= ) k=F m /D m y: amplitude, a: maximum amplitude(=D m ) p: input frequency, : phase difference ‘Resonant steady-state’ means p=.

3. Example 2 (A hysteretic energy dissipation index E h in the case of elasto-plastic model) Derive the equation shown below by calculating energy dissipation ΔW under the ‘resonant steady-state’ in the case of elasto-plastic model.

4. Excercise 2 (A hysteretic energy dissipation index E h in the case of Clough model) Derive the equation (3.8) by calculating energy dissipation ΔW under the ‘resonant steady-state’ in the case of modified Clough model. (3.8)

5. Simple method to get elastic period of SDOF system w: a unit weight(=12000N/m 2 ), Σ Ａ f : sum of whole floor area of the building(m 2 ), g: gravity(cm/s 2 ), F c : compressive strength of concrete (N/mm 2 ), b: width of a column(cm), D: depth of a olumn(cm), h: story height(cm), n: number of story stiffness of concrete (N/mm 2 ) moment of inertia of a column (cm 4 )

6. Simple method to get base shear coefficient of SDOF system τ c : ultimate shear strength of columns(N/mm 2 =F c /15), Ａ c : sum of 1st story column section area(cm 2 ), w : a unit weight(=12000N/m 2 ), Σ Ａ f : sum of whole floor area of the building(m 2 )

7. Example 3 (Simple method to get fundamental parameters of SDOF system ) 3-story building: column size: 60cm x 60cm Fc= 24(N/mm2) plan elevation story height: 3.6m ⇔ longitudinal direction

8. Exercise 3 (Simple method to get fundamental parameters of SDOF system ) 12-story building: column size: 95cm x 95cm Fc= 48(N/mm 2 ) plan story height: 3.5m elevation ⇔ longitudinal direction

9. Newmark’s design criteria ・ property of energy conservation: For short period systems (T<0.5s) the energy dissipation is constant. ・ property of displacement conservation: For long period systems (T>0.5s) the response displacement is constant. Newmark’s design criteria was used to decide strength of the system for each input strong ground motion.

10. Property of energy conservation For short period systems (T<0.5s) The area of trapezoid OBCE = the one of △ OAD (μ δ y +(μ-1) δ y )Q y /2= δ L* Q L /2 δ y = Q y /k, δ L = Q L /k

11. Property of displacement conservation For long period systems (T>0.5s) inelastic response displacement is the same as elastic response displacement. μ δ y = Q L /k δ y = Q y /k

12. Example 5 (Newmark’s design criteria) Calculate response displacement in the case that 1) elastic period=0.3 sec., base shear coefficient= 0.4, elastic response acceleration=0.8g at 0.3 sec. 2) elastic period=1.0 sec., base shear coefficient= 0.1, elastic response acceleration=0.2g at 1.0 sec.

13. Exercise 5 (Newmark’s design criteria) Calculate response displacement under the input of El-Centro NS using elastic spectra (damping factor=0.05, Fig.4.3,(b), p.139) and Newmark’s design criteria in the case that 1) 7-story reinforced concrete building (base shear coefficient= 0.3, story height=2.8(m)) 2) 20-story steel building (base shear coefficient= 0.05, story height=3.5(m)) T=0.02H (T: period of the building (s), H: height of the building(m)) for reinforced concrete building T=0.03H (T: period of the building (s), H: height of the building(m)) for steel building

14. Tripartite response spectra Response spectra which show response acceleration, velocity and displacement simultaneously using the relations: S V =ωS D S A =ωS V =ω 2 S D S D : response displacement S V : response pseudo-velocity S A : response pseudo-acceleration

15. Ductility factor Ductility factor μ is defined as the ratio of the maximum response displacement x to the yielding displacement x y. Ductility factor is used as the index of representing the damage level.

16. Ductility factor of short and long period system Short period systemlong period system μ=dm/dy=5μ=dm/dy=2

17. Response ductility factor spectra Ductility factors were calculated in the case of bilinear Takeda model under the input of El-Centro NS and Fukiai (Kobe EQ.) changing the base shear coefficient.

18. Response ductility factor spectra Input: El-Centro NS Bilinear Takeda model

19. Response ductility factor spectra Input: Fukiai (Kobe EQ.) Bilinear Takeda model

20. Comparison of response ductility factor spectra Input: Fukiai (Kobe EQ.) Bilinear Takeda model Input: El-Centro NS Bilinear Takeda model

21. Required strength Required strength which give maximum response ductility factors within constant values is important, because we need design base shear coefficient in the structural design. ↑ Allowable ductility factor μa

22. Required strength spectra Input: El-Centro NS Bilinear Takeda model

23. Required strength spectra Input: Fukiai (Kobe EQ.) Bilinear Takeda model

24. Comparison of required strength spectra Input: El-Centro NS Bilinear Takeda model Input: Fukiai (Kobe EQ.) Bilinear Takeda model

25. Example 8 (Calculation of required strength spectrum from response ductility factor spectra) Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of El-Centro NS. Response ductility factor spectra changing base shear coefficient of the system input motion: El-Centro NS T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 99.999 38.270 8.309 1.826 1.174 0.200 99.999 99.999 99.999 99.999 65.070 19.300 6.894 2.350 1.557 0.500 99.999 99.999 99.999 45.490 13.500 6.573 3.564 2.734 1.987 1.000 99.999 73.020 24.880 13.080 4.469 2.071 1.474 1.285 1.030 1.500 74.490 33.440 13.380 5.099 1.954 0.947 0.632 0.474 0.379 2.000 47.700 21.080 7.728 3.817 1.831 0.887 0.592 0.444 0.355 3.000 21.190 9.574 3.388 1.865 1.114 0.570 0.380 0.285 0.228 99.999: greater than 100

26. Example 8 (Calculation of required strength spectrum from response ductility factor spectra) Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of El-Centro NS. Response ductility factor spectra changing base shear coefficient of the system input motion: El-Centro NS T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 99.999 38.270 8.309 1.826 1.174 0.200 99.999 99.999 99.999 99.999 65.070 19.300 6.894 2.350 1.557 0.500 99.999 99.999 99.999 45.490 13.500 6.573 3.564 2.734 1.987 1.000 99.999 73.020 24.880 13.080 4.469 2.071 1.474 1.285 1.030 1.500 74.490 33.440 13.380 5.099 1.954 0.947 0.632 0.474 0.379 2.000 47.700 21.080 7.728 3.817 1.831 0.887 0.592 0.444 0.355 3.000 21.190 9.574 3.388 1.865 1.114 0.570 0.380 0.285 0.228 99.999: greater than 100

27. Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS required T(s) strength (base shear coefficient) 0.1 0.366 0.2 0.364 0.5 0.286 1.0 0.120 1.5 0.067 2.0 0.049 3.0 0.019

28. Exercise 8 (Calculation of required strength spectrum from response ductility factor spectra) Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of Taft EW and compare it with El-Centro NS. Response ductility factor spectra changing base shear coefficient of the system input motion: Taft EW T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 84.430 1.109 0.720 0.540 0.432 0.200 99.999 99.999 99.999 74.450 23.240 4.561 1.577 1.079 0.859 0.500 99.999 99.999 32.930 13.910 5.297 1.396 1.168 0.864 0.691 1.000 91.840 37.000 10.040 3.978 1.734 0.793 0.529 0.396 0.317 1.500 51.110 18.410 7.052 2.825 1.389 0.655 0.437 0.327 0.262 2.000 31.780 11.420 4.222 1.352 0.855 0.428 0.285 0.214 0.171 3.000 15.710 6.619 2.097 0.957 0.479 0.239 0.160 0.120 0.096 99.999: greater than 100

29. Exercise 8 (Calculation of required strength spectrum from response ductility factor spectra) Calculate required strength (base shear coefficient) spectrum from response ductility factor spectra in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of Taft EW and compare it with El-Centro NS. Response ductility factor spectra changing base shear coefficient of the system input motion: Taft EW T(s) Cy=0.005 0.01 0.02 0.05 0.1 0.2 0.3 0.4 0.5 0.100 99.999 99.999 99.999 99.999 84.430 1.109 0.720 0.540 0.432 0.200 99.999 99.999 99.999 74.450 23.240 4.561 1.577 1.079 0.859 0.500 99.999 99.999 32.930 13.910 5.297 1.396 1.168 0.864 0.691 1.000 91.840 37.000 10.040 3.978 1.734 0.793 0.529 0.396 0.317 1.500 51.110 18.410 7.052 2.825 1.389 0.655 0.437 0.327 0.262 2.000 31.780 11.420 4.222 1.352 0.855 0.428 0.285 0.214 0.171 3.000 15.710 6.619 2.097 0.957 0.479 0.239 0.160 0.120 0.096 99.999: greater than 100

30. Required strength spectra (allowable ductility factor=4 ) by El-Centro NS and Taft EW required strength (base shear T(s) coefficient) El-Centro Taft NS EW 0.1 0.366 0.197 0.2 0.364 0.219 0.5 0.286 0.133 1.0 0.120 0.050 1.5 0.067 0.042 2.0 0.049 0.022 3.0 0.019 0.016

31. Equivalent linear system Inelastic responses can be estimated by equivalent linear systems with equivalent period and equivalent viscous damping. x Q x Q inelastic with equivalent viscous damping elastic equivalent period

32. Method to calculate inelastic responses by equivalent linear systems How to decide equivalent period and damping equivalent period: period corresponding maximum displacement equivalent damping: equivalent viscous damping

33. Example 9 (Required strength spectrum by equivalent linear system) Calculate required strength (base shear coefficient) spectrum (T=0.1, 0.2, 0.5, 1.0, 1.5, 2.0 and 3.0 s) by equivalent linear system in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of El-Centro NS from elastic response spectra (h=0.05)→ using the damping reduction factor in Equation shown below and compare it with required strength spectrum by inelastic system and Newmark’s design criteria. F h : reduction factor from h=0.05, h: damping factor response T(s) acceleration (cm/s 2 ) 0.1 560.7 0.2 640.7 0.3 695.9 0.4 601.8 0.5 820.5 1.0 507.4 1.5 186.7 2.0 174.9 3.0 112.3 4.0 45.4 6.0 31.0

34. Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS using equivalent linear system comparing with actual spectrum Required strength (base shear coefficient) Inelastic Equivalent Newmark ’ s T(s) system linear design system criteria 0.1 0.366 0.386 0.216 0.2 0.364 0.363 0.217 0.5 0.286 0.306 0.316, 0.209 1.0 0.120 0.105 0.129 1.5 0.067 0.068 0.048 2.0 0.049 0.027 0.045 3.0 0.019 0.019 0.029

35. Required strength spectrum (allowable ductility factor=4 ) by El-Centro NS using equivalent linear system comparing with actual spectrum Required strength (base shear coefficient) Inelastic Equivalent T(s) system linear system 0.1 0.366 0.386 0.2 0.364 0.363 0.5 0.286 0.306 1.0 0.120 0.105 1.5 0.067 0.068 2.0 0.049 0.027 3.0 0.019 0.019

36. Exercise 9 (Required strength spectrum by equivalent linear system) Calculate required strength (base shear coefficient) spectrum (T=0.1, 0.2, 0.5, 1.0, 1.5, 2.0 and 3.0 s) by equivalent linear system in the case that allowable ductility factor=4, model: bilinear Takeda model (α=0.5, β=0.01 ） under the input of Taft EW from elastic response spectra (h=0.05)→ using the damping reduction factor in Equation shown below and compare it with required strength spectrum by inelastic system and Newmark’s design criteria. F h : reduction factor from h=0.05, h: damping factor response T(s) acceleration (cm/s 2 ) 0.1 211.9 0.2 422.5 0.3 398.4 0.4 387.4 0.5 340.3 1.0 156.4 1.5 129.3 2.0 84.9 3.0 47.2 4.0 27.1 6.0 22.7

37. Required strength spectrum (allowable ductility factor=4 ) by Taft EW using equivalent linear system comparing with actual spectrum Required strength (base shear coefficient) Inelastic Equivalent T(s) system linear system 0.1 0.197 0.255 0.2 0.219 0.233 0.5 0.133 0.094 1.0 0.050 0.051 1.5 0.042 0.028 2.0 0.022 0.016 3.0 0.016 0.014

38. Required strength spectrum (allowable ductility factor=4 ) by Taft EW using equivalent linear system comparing with actual spectrum Required strength (base shear coefficient) Inelastic Equivalent Newmark ’ s T(s) system linear design system criteria 0.1 0.197 0.255 0.082 0.2 0.219 0.233 0.163 0.5 0.133 0.094 0.131, 0.087 1.0 0.050 0.051 0.040 1.5 0.042 0.028 0.033 2.0 0.022 0.016 0.022 3.0 0.016 0.014 0.012

39. Example & Excersise10 (Prediction of response ductility factor from elastic response using equivalent linear system and comparison with actual structural damage) Calculate response ductility factor in the case of longitudinal direction of Kuoshing National Elementary School Building B under the input of recorded strong ground motion of 1999 Chi-chi, Taiwan earthquake using equivalent linear system and compare with actual structural damage.

40. Building A (Shikang Natl. Elem. Sch.)

41. Elastic response spectrum at Shikang National Elementary School Building A

42. Building B (Kuoshing Natl. Elem. Sch.)

43. Elastic response spectrum at Kuoshing National Elementary School Building B

44. Shikang N.E.S. Kuoshing N.E.S. Building A Building B unit of weight 1.2 tonf/m2 Fc 154.7 216.5 kgf/cm2 No. of story 3 3 column1 depth(longitudinal) 33.0 50.0 cm column1 depth(transvers) 46.8 58.0 cm column2 depth(longitudinal) 52.3 50.0 cm column2 depth(transvers) 52.4 58.0 cm column3 depth(longitudinal) 40.0 50.0 cm column3 depth(transvers) 65.0 58.0 cm span length1(transvers) 783.2 500.0 cm span length2(transvers) 261.3 cm span length(longitudinal) 300.0 400.0 cm story height 343.0 360.0 cm No. of span(longitudinal) 16 5 No. of span(transvers) 2 5 column1 transvers column2 column3 span length (longitudinal) span length1 (transvers) span length2 (transvers) longitudinal Information of the buildings

45. Simple method to get elastic period of SDOF system w: a unit weight(=12000N/m 2 ), Σ Ａ f : sum of whole floor area of the building(m 2 ), g: gravity(cm/s 2 ), F c : compressive strength of concrete (N/mm 2 ), b: width of a column(cm), D: depth of a olumn(cm), h: story height(cm), n: number of story stiffness of concrete (N/mm 2 ) moment of inertia of a column (cm 4 )

46. equivalent period T’

47. Simple method to get base shear coefficient of SDOF system τ c : ultimate shear strength of columns(N/mm 2 =F c /15), Ａ c : sum of 1st story column section area(cm 2 ), w : a unit weight(=12000N/m 2 ), Σ Ａ f : sum of whole floor area of the building(m 2 )

48. Equivalent viscous damping and damping reduction factor Decide μso that response acceleration is equal to base shear coefficient at equivalent period T’ and equivalent viscous damping factor E h from elastic response spectrum of damping factor 0.05.