Seismic Analysis of Structures by TK Dutta, Civil Department, IIT Delhi, New Delhi.

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1 Seismic Analysis of Structures by Dutta, Civil Department, II Delhi, New Delhi. Module Response Analysis for Specified Ground Motion Exercise Problems:.. Find the effective mass and stiffness of the structure corresponding to the dynamic degree of freedom x shown in Figure.. All members are assumed as inextensible. Shear EI stiffness for the member AB is k. he shear stiffness of other members are given in l terms of k. If damping ratio is assumed as.5 and k m is assumed as ( rad s ), find the peak and rms displacements of x for El Centro earthquake by (i) Duhamel integral; (ii) frequency domain analysis using FF; and (iii) state space analysis in time domain. Figure... A rigid pipe embedded within non uniform soil is modeled as lumped mass system as shown in Figure.. he soil stiffness and damping are modeled with springs and dash pots. If the soil damping ratio is assumed to be.6 and k m is assumed as 5( rad s ), then find the peak and rms responses of y and θ to El Centro earthquake by (i) direct frequency domain analysis using FF, and (ii) direct time domain analysis using Newmark s- β -method. Figure..5. Figure. shows the stick model for a building shear frame. Shear stiffness coefficients are shown in the figure. he base of the frame is subjected to both rotational and translational ground motions. Using the direct frequency domain analysis, find the peak values of moment at the base and dusplacement of the top mass. ake the ground motion as El Centro earthquake and the rotational component of ground motion as translational k component divided by 5. ake = 5( rad s) and the damping ratio as.. m Figure..6. Figure.5 shows the two-storey shear frame which is subjected to different excitations at the three supports produced due to an arbitrarily assumed time lag in earthquake ground motion. Obtain the peak and rms values of the top displacement for El Centro ground motion by (i) direct time domain analysis using Newmark s- β -method, and (ii)

2 k 6 rad s m = as.5s; and modal damping ratio as.5. Figure.5 state space analysis in time domain. Assume ( ) ; time lag between supports.7. For the frame with a secondary system attached to the top floor as shown in Figure.6, find the peak value of the displacement of the mass of the secondary system for different support excitations as shown in the figure by (i) direct frequency domain analysis using FF, and (ii) state space analysis in frequency domain. Assume earthquake ground k motion as El Centro earthquake; time lag as 5s; 6( rad s) m = ; damping ratio for the frame and secondary system as.. Compare the result with that obtained for no time lag i.e., the same excitations at the supports. Figure.6.8. Using the modal time domain and frequency domain analyses, find the peak values of displacement of the top floor and the first storey drift of the frame shown in Figure.7 for El Centro earthquake excitation. Compare between the results obtained by considering contributions of the first two modes only and all modes. Also, find the times at which the peak values of the top storey displacement takes place when only two modes are considered and when all modes are considered. Assume modal damping ratio as.5 k and ( rad s) m =. Figure.7.9. Using the modal time domain analysis, find the rms and peak values of the dynamic degrees of freedom of a simplified model of a cable stayed bridge shown in Figure.8. Use El Centro earthquake with time lag between supports to 5 as.5s; 5 to 6 as 5s and 6 to 7 as.5s. Assume the following relationships: AE EI EI = ; m = m; = ( rad s) ; modal damping ratio as.5; l s sm (EI) tower =.5(EI) deck =.5EI ; (AE/) deck =.8(AE/L ) cable ; s=5; l is the length of the cable joining the top of the tower to the end of the bridge; m is the mass lumped at the top of the tower and m is the mass lumped at the centre of the bridge. Figure.8.. A six-storey shear frame, as shown in Figure.9, is subjected to El Centro ground motion. Using the mode acceleration method, find the rms and peak values of the top floor and the first floor displacements shown in the figure. For the analysis, consider contributions of the first three modes only. Compare the results of the mode acceleration method with those of the mode superposition method considering the contributions of all

3 k rad s m = modes. Assume modal damping ratio to be.5 and ( ) in terms of k. Figure.9. Obtain the moment.. A two-storey D frame with rigid slab, as shown in Figure., is subjected to El Centro earthquake. Using the mode acceleration method, find the peak values of the column moments at A. Lateral stiffness of the columns are the same in both directions. otal mass lumped at each floor is taken as m and the mass moment of inertia corresponding to the ml torsional degree of freedom is taken as. Assume modal damping ratio as.5; 6 k 8( rad s) m = ; m = kg and l = m. Consider first three modes for the response calculation and evaluate the accuracy of the results. Figure... A rigid slab is supported by three columns as shown in Figure.. Columns are rigidly connected to the slab and have the same lateral stiffness k in both directions. otal mass k of the slab is m and 8( rad s) m =. Find the time histories of displacements of the top of the columns when the base of the structure is subjected to two-component ground motions as shown in the figure. ake x xg as the El Centro earthquake and assume x yg = xg x. Use direct time domain analysis by employing Newmark s- β -method. Figure. ake the relevant figures from the slides or from the reference book

4 Module Response Analysis for Specified Ground Motion Exercise Solution: ERRAA FOR HE EX BOO pp,.. subheading: Absolute motion should be Absolute motions pp 8, nd para, nd line: the other two DOF locked should be the other DOF locked pp, Equation.5b: k and c should be k m and c m respectively. pp 5, Example.6: A matrix; th diaogonal term should be.5ρ n mrir λimrir r= r= pp, Equation.5: ρ i = should be ρ i = M M in which λ i is defined by Equation. for single support excitation. pp, Example.: (i) outside coefficient of n EI L rr will be.6 (ii) Due to computational mistakes, r is calculated wrongly. It should EI be r = ; = L AE AE Example.5: should be cos θ ; 7, but = cos θ ; l l, but = 7 Because of these changes, r matrix is changed to r =

5 ERRAA FOR HE EX BOO (Continued) Example.9: Since r matrix is wrong, the response without time delay for d.o.f 5 is not exactly zero. If the correct and r matrices (given as above for. (ii)) is used response for d.o.f 5 will be zero. he pattern of responses.shown in Figure.(a),(e),(d) remain the same with little changes in the values of ordinates, if the correct r matrix is used. Example.: Since r matrix is wrong, the ordinates of the figures (Figures.8-.9) change, if the correct r matrix (Example.5 as above) is used. he pattern of the responses remain nearly the same. With correct r matrix, peak values of z, z and z (given in able.) change to.77m,..5m and.m respectively. Refer to the exercise problem. and Figure.. he kinematic degrees of freedom areθ B, θ C and x. BC =.55ρ, CD =.5ρ, OB =.l and OC = l (θ is the instantaneous centre of rotation). he stiffness matrix corresponding to the degrees of freedom is given below.6l l θb k.l.7l θc.l.l.98 x = Condensed stiffness corresponding to the d.o.f x i.e., the equivalent stiffness is.79k. Equivalent mass corresponding to the d.o.f x is obtained by finding inertia force at x for unit acceleration given at x. he equivalent mass is determined as.m (see the method given in Appendix A)..79k n = =.rad s -.m

6 he peak and rms values of the displacement x are shown in able.8 and the time history of displacement x obtained by state space analysis is shown in Fig..8. able.8 Peak and rms values of x obtained by different methods Method Rms (m) Peak (m) Duhamel.,6 FF State space.6.68 Displacement (m) FF State Space ime (s) Figure.8 ime history of displacement x obtained by state space and frequency domain analysis. Refer to the exercise problem. and Figure.. he stiffness matrix corresponding to the degrees of freedom y and θ is given by 6.7 = k M =.75 m

7 he eigen value analysis provides k k λ, =.6,.9 m m k For 5 m = ; - =.9 rad s - =.96 rad s ( )( ) α = = β = = C =.6M = m k C =.5 m he equation of motion takes the form x+ + =.5. x 8 79 x.75 x g x = [ y θ ] he peak and rms values of the responses are shown in able.9. he time histories of responses obtained by both methods are shown in Figure.9

8 able.9 Peak and rms values of responses Method y(m) θ (rad) peak rms peak rms Direct Freq Newmark s ime Domain FF Analysis Displacement (m) ime (s) (a)

9 .8.6. FF Analysis ime Domain Rotation (rad) ime (s) (b) Figure.9 ime histories of responses (a) y-displacement (b) rotation-θ Refer to the exercise problem.5 and Figure. Absolute accelerations of the masses at d.o.f and are x = x + x + 6 θ a g x = x + x + θ a g Equations of motion for the system take the form Mx + + = a Cx x = m M C = α M + β = k Solution of the Eigen value problem provides - - = 5rads, = rad s ( )( ). 5. α = = =. β = = Equations of motion can be written as xg + 6 θ. x x x + = = 5 5 x xg + θ.6 g

10 Frequency response function matrix is given by H ( ) = i Using direct frequency domain analysis the responses may be obtained by Equation.97 in frequency domain. IFF of the responses provides the responses in time domain. Bending moment at the base is M () t = 6kx () t he peak value of x () t =-.m he peak value of M () ( ) t =.69 k =.69 k =.69 5m =.69m Refer to the exercise problem.6 and Figure.5. he stiffness matrix corresponding to the d.o.f including the support d.o.f ( x, x, x 5 numbered from left to right) 5k k k k k x k k x k k x k k x k k x5 5 r = xx xg = = M = m - - Natural frequencies of the system are obtained as = 6.988rads, =.8rad s he values of α and β are calculated as α =.77, β =.6 Equations of motion take the form

11 ( x g x ) g x g ( x g x g xg ) x+ + =.55. x x. + + State space equation is Y = AY + F A I M CM = = F.67 ( xg + x ) g + x g.( xg + xg + xg ) Because of the time lag of.5s between the supports, the total time history of excitation at each support is taken as 5s and is constructed as explained in Example.8. he time histories of the top displacement obtained by direct time domain () and state space time domain (S) analyses are given below. he peak value of top displacement =.68m ();.689 m(s) he rms value of top displacement =.m ();. m(s) Refer to the exercise problem.7 and Figure.6. he stiffness matrix for non support degrees of freedom xx and the coupling matrix xg between support and non support degrees of freedom are xx = k xg = k r = xx xg =.5.5 M = m.5 - Frequencies of the system are calculated as = 5. rad s, aking the first two frequencies α =. ; β =.9 - = 8.8rad s, - = 7.8rads

12 he equations of motion for the system take the form ( xg + x ) g ( xg x ) g ( x g xg ) x+ x + x = x g and g x have a time lag of 5 s. he time histories of the displacement of the top mass obtained by direct frequency domain (D) and state space frequency domain (S) are shown in Figure.. he peak value (time lag) =.878m (D);.9m (S) he peak value (no time lag) =.68m (D);.8m (S).. Direct Freq. Domain State Space Displacement (m) ime (s) Figure. ime history of the displacement of the secondary system Refer to the exercise problem.8 and Figure.7. Stiffness matrix for the system is

13 7 5 = k he natural frequencies of the system are - =.5rad s, - =.9rad s, M = m - =.5 rad s, - =.7rads = [ ] ; = [ ] = [ ] ; = [ ] λ =.67, λ =.55, λ =.7, λ = ime histories of top floor displacement obtained by considering the first two modes and all modes are: op story peak displacement =.67m (modes);.587m (all modes) Occurrence peaks of displacement = 6.8sec (modes); 6.85s (all modes) First story peak displacement =.565 m (modes);.55 (all modes) Refer to the exercise problem.9 and Figure.8. he length of the cable are computed as l 9.m = the cable joining the tower top to the centre of the bridge) =.6 m cosθ =.88 sinθ =.5 l (length of cosθ =.986 sinθ =.7; tower bases are fixed ( Figure.8 does not show them properly )..5EI AE AE = + cos θ + cos θ l l ( 8).75EI EI EI EI = 58m = + ( ) ( ) (.698 ) + ( ) (.7 )(.86 ) ( 5 ) = ; l = AE cosθ sinθ = AE cosθ sinθ = AE =. AE = 96. m ( )( ) l l l.6 l l AE AE AE AE = cos cos = = 9m ( ) ( ) θ θ l l l l

14 ( EI ) tower EI EI ( ) ( ) ( ) ( ) ( ) ( ) ( ) = = = = m= 95.m = 6 = 9 = = ; AE = cos θ =.8m 7 l = 6 = 5 7 = =.8m 5 = = 8 = 9 = = EI AE EI AE 9. = + sin θ = + (.7) = 8m( +.96) = 876.8m 9..6 ( 5) l ( 5) = 5 = 6 = 7 = 6EI = = 5m 9 = = 5m ( 5) 8 AE AE AE AE = + cos θ + cos θ =.8 +.7m+ 88m= 9m deck L L l cable 7 = 9m 5 = 6 = 8 = 9 = = 55 = 5 = 95.m 65 = 75 = 85 = 95 = 5 = 66 = 95.m 76 = 86 = 96 = 6 = 77 = = 9m 87 = 97 = 7 = ( 5) ( 5) EI EI EI EI = + = + = EI = EI = 5 m EI EI = = = m = EI = = 6 m 5 EI = = m 5 = 88

15 θθ ΔΔ θδ 5 = 6 m =.9..9 m = m.5 Condensing out the rotational stiffness to find and then partitioning it to identify the sub matrices for support and non support degrees of freedom, the matrix r is obtained as r = ss sn= where s and n refers to support and non support d.o.f ss 58 Sym = 58 m he mode shapes and frequencies are - =.6.6, =.55rads =, =.67, M = m 6 - = 5.8 rad s - = 5.57 rad s λ = ; λ ik i = k =

16 he time histories of excitations at the supports are of 5s duration and are constructed as explained in Example.8. he time histories of the vertical displacement of the deck and the top of the left tower are shown in Figure.. he rms and peak values of the three degrees of freedom are given as below: able.: Rms and peak values of responses rms peak d.o.f (m) (m) Displacement (m) ime (s) (a)

17 .8.6. Displacement (m) ime (s) (b) Figure. he time histories of displacements (a) top of the left tower (b) centre of the deck Refer to the exercise problem. and Figure.9. Stiffness and mass matrices for the frame are: Sym = k he natural frequencies of the frame are: - =.55rad s, - 6 =.rads - =.9 rad s, - = 7.85rad s, = = = = M = m - = 9.7 rad s, - 5 =.rad s,

18 5 = = λ =.58, λ =.7, λ =., λ =.5, λ 5 =.8, λ 6 =.76 Using Equations. and. (with the steps given to find out R( t ), the responses are obtained using mode acceleration approach considering the first three modes. For the mode superposition method, contributions of the first four modes are considered. he results are given below able.: Rms and peak responses Floor rms (m) peak (m) Mode accl. Mode sup. Mode accl. Mode sup. op st he time histories of the first two generalized displacements are shown in Figure.. he displacements of the top and the first floors obtained by both mode acceleration and mode superposition method are shown in Figure.. st Generalised Disp. (m) Mode Acceleration Mode Superposition ime (s) (a)

19 nd Generalised Disp. (m) Mode Acceleration Mode Superposition ime (s) op Floor Displacement (m) (b) Figure. ime histories of generalized displacements (a) first generalized displacement (b) second generalized displacement Mode Acceleration Mode Superposition ime (s) (a)

20 st Floor Displacement (m) Mode Acceleration Mode Superposition ime (s) (b) Figure. ime histories of displacements (a) top floor (b) first floor Refer to the exercise problem. and Figure.. he stiffness and mass matrices are of the frame are: = k M.5 = m.5

21 Frequencies of the system are - =.5rad s, - 6 = 69.6 rad s - = 5. rad s, - = 6.58rads, - = 8rad s, - 5 = 9.6 rad s, = =.68.. = = = = λ =.65, λ =.6, λ =.6, λ =.5, λ 5 =.657, λ 6 =.85 he results of the responses by considering first three modes are: Peak values : op floor y-displacement =.56m First floor x-displacement =.m First floor rotation =.9 rad he y-displacement of the top floor and the rotation of the first floor are shown in Figure.

22 ... Displacement (m) ime (s). (a).8 Rotation (rad) ime (s) (b) Figure. ime histories of (a) y-displacement of top floor (b) rotation of the first floor Refer to the exercise problem. and Figure..

23 Due to unit displacement given at the d.o.f one at a time, stiffness and the mass matrices are generated as explained in Appendix A. he stiffness and mass matrices corresponding to the d.o.f are 7 = 7 7 k α =.595, β = = m M I = [.5] C = Frequencies of the system are - = 9.66 rads, - = 5.5rad s, - = 8.77 rad s ime histories of displacements of the top of the columns are shown in Figure.5. Peak values of displacements are For d.o.f =.6m For d.o.f =.7m For d.o.f =.m

24 Displacement (m) ime (s) Displacement (m) (a) ime (s) (b)

25 .5.. Displacement (m) ime (s) (c) Figure.5 ime histories of (a) d.o.f- (b) d.o.f- (c) d.o.f-

Seismic Analysis of Structures by TK Dutta, Civil Department, IIT Delhi, New Delhi.

Seismic Analysis of Structures by TK Dutta, Civil Department, IIT Delhi, New Delhi. Seisic Analysis of Structures by K Dutta, Civil Departent, II Delhi, New Delhi. Module 5: Response Spectru Method of Analysis Exercise Probles : 5.8. or the stick odel of a building shear frae shown in