ARCE 412: STRUCTURAL DYNAMICS. Homework 1 (Due ) (a) (b) (c)

Transcription

1 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS March, 9 Problem Homework (Due 4--9) m (a) (b) (c) m m 5 m 5 m 5 m For each of the frame structures show, calculate the stiffessk of the frame i the directio of the force. Assume uiform flexural stiffess EI ad eglect axial deformatio. Solutio: Structure (a) (b) (c) k [ ] EI m Problem EI =, k-ft EA = 5 k 8 ft 7 ft EA = 5 k EI =, k-ft 5 ft (a) (b) 6 ft 6 ft momet coectio! For the two structures above, calculate the lateral stiffessk. Solutio: Structure (a) (b) k [k/ft] //9 :4 PM C:\calpoly\arce4\homework\sprig_9\hw.doc

2 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April, 9 Problem Homework -Solutio m (a) (b) (c) m m 5 m 5 m 5 m For each of the frame structures show, calculate the stiffessk of the frame i the directio of the force. Assume uiform flexural stiffess EI ad eglect axial deformatio. Momet Diagrams for Uit Force (a) (b) (c).9.9 Displacemet Calculatio (a) EI Δ= (. + 5.) = 4 k = =.47 EI Δ (b) EI Δ=.5 (. +.5) = 8.5 k = =. EI Δ (c) EI Δ = ( ) =.86 k = =.58 EI Δ Deflected Shape (for illustratio oly) (a) (b) (c) 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

3 Sprig Quarter 9 Problem EI =, k-ft EA = 5 k 8 ft 7 ft EA = 5 k EI =, k-ft 5 ft (a) (b) 6 ft 6 ft momet coectio! For the two structures above, calculate the lateral stiffessk. (a) Δ = + + = + = 5 k = =.46 k/ft( as).78 7 (7 8) ft/k 7 N =.5 M (b) - 5 α = ta = 68. cos α =.74 cos α =.79 6 EI EA 5 k = + cos α = +.79 = = 6.46 k/ft( as) L L 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

4 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April, 9 Homework (Due 4-6-9) Readig: Chopra Sectios. ad., Review mass momet of iertia Problem (a) (b) (c) (d) W EI u ft 4 ft 6 ft 6 ft 4 ft ft (e) (f) W =.5 k C (g) (h) EI =, k-ft EA = 5 k D 8 ft (i) B A 6 ft 6 ft EI W (j) A B C u ft ft Fid the atural circular frequecy ω = k/ m of the above systems/structures. Commets: (d) For torsioal vibratio of the disk of massm (circular shaft massless). The shear modulus of the shaft isg. (h) For vibratio i the x -ory directio. The platform of weightw is braced laterally i each side by two steel cables. The cables have axial stiffess EA. Due to high prestressig, the compressio cables cotribute to the structural stiffess. (i) Cosider axial deformatio oly i cable BD. (j) Use both a flexibility (apply uit force i directio ofu ) ad a stiffess approach. I the stiffess approach, start with a x stiffess matrix correspodig to the three dofs Δ B or u, ϕb, ϕc, the elimiate the two rotatios to get a scalar stiffess relatio ivolvigu oly (this is the static codesatio techique leared i ARCE 6). Solutio: Structure (a) (b) (c) (d) (e) (f) (g) ω [rad/sec] k + k kk ( k + k) k 4 Gπd 48EI 9EI EI. m ( k + k ) m ( k + k + k ) m 6mLR L m L m m Note: m = W / g ω [rad/sec] (h) (i) (j).89 EA hm.57.7 EI m 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw.doc

5 Sprig Quarter 9 Problem (a) Plot three cycles of a free vibratio respose of a SDF system with a mass of. k-s /i ad a stiffess of 5 k/i subjected to a iitial displacemet of iches ad zero iitial velocity. LabelT, f, u(), u ()ad u. (b) Double the stiffess of the SDF system i (a) ad replot the respose. LabelTu, (), u ()ad u. (c) Plot the respose of the system i (b) for a iitial displacemet of iches ad a iitial velocity of i/sec. Label Tu, (), u ()ad u. 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw.doc

6 Sprig Quarter 9 ARCE 4: Structural Dyamics April 6, 9 Homework -Solutio (a) (b) (c) (d) W EI u ft 4 ft 6 ft 6 ft 4 ft ft (e) (f) W =.5 k C (g) (h) EI =, k-ft EA = 5 k D 8 ft (i) B A 6 ft 6 ft EI W (j) A B C u ft ft Fid the atural circular frequecy ω of the above systems/structures. Commets: (d) For torsioal vibratio of the disk of massm (circular shaft massless). The shear modulus of the shaft isg. (h) For vibratio i the x -ory directio. The platform of weightw is braced laterally i each side by two steel cables. The cables have axial stiffess EA. Due to high prestressig, the compressio cables cotribute to the structural stiffess. (i) Cosider axial deformatio oly i cable DE. (j) Use both a flexibility (apply uit force i directio ofu ) ad a stiffess approach. I the stiffess approach, start with a x stiffess matrix correspodig to the three dofs Δ B or u, ϕb, ϕc, the elimiate the two rotatios to get a scalar stiffess relatio ivolvigu oly. 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

7 Sprig Quarter 9 Problem k + k k k k as m (a) = + ω = ( ) k k kk k = = ω = as + k + k ( k + k) m k k b ( ) ( ) ( ) ( ) k + k k k + k k () c k = = = = ω = ( as) k + k + k + + ( k + k + k) ( k + k + k) m k k k + k k ( k + k ) k 4 4 GJ Gπd k Gπd (d) k = =, MMI = mr ω = = ( as) L L MMI 6LmR 48EI 48EI (e) k = ω = ( as) L Lm EI 9EI 9EI (f) k = = ω = ( as) L L Lm EI EI EI Δ + + = k = = EI ω = as m (g) = ( ) EA k = α s) L EA EA EA EA EA k = 4 cos 45 = 4.5 = ω = =.89 ( as) h h h hm hm (h) cos (stiffess of each of the 4 cable 5.76., 5,.97.5 (i) Δ= =.97 ft/k k = = 5.76 k/ft ω = =.57 rad/sec( as) (j) EI Δ= ( ) +.5 = 7.97 EI.7EI EI k = =.7 k/ft ω = =.7 ( as) 7.97 m m work for g,i,j show o ext page 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

8 Sprig Quarter 9 work for g i P = C P = + M [k-ft] M 6 N =.5 BD EI =, k-ft EA = 5 k D 8 ft B A 6 ft 6 ft work for j A P = B ft ft M [k-ft] stiffess approach for j stiffess matrix K.4.6 =.8. use EI = dofs are,, = Δ, ϕ, ϕ.6..4 A A B e = [ ](elimiate) r = (retai) stiffess submatrices correspodig to dofs to be elimiated ad sigle dof to be retaied.8. K = ee rr =..4 K.4 Ker = = [ ] re.6 K K ee = Kco = Krr KreKee Ker =.4 [.6] = k =.7 EI Flexibility ad stiffess approaches lead to the same scalar stiffessk ad hece to the same atural frequecy. 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

9 Sprig Quarter 9 Problem (a) Plot three cycles of a free vibratio respose of a SDF system with a mass of. k-s /i ad a stiffess of 5 k/i subjected to a iitial displacemet of iches ad zero iitial velocity. LabelT, f, u(), u ()ad u. (b) Double the stiffess of the SDF system i (a) ad replot the respose. LabelTu, (), u ()ad u. (c) Plot the respose of the system i (b) for a iitial displacemet of iches ad a iitial velocity of i/sec. Label Tu, (), u ()ad u. 4 u() Displacemet [i] u () = slope of ut () at t= (a) (b) T T (c) T Time [sec] u u u 4//9 7: AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc 4

10 Sprig Quarter 9 ARCE 4: Structural Dyamics April 6, 9 Readig: Chopra Sectio. Homework (Due 4--9) Problem : m=ksec /i ut () k=5 k/i The structure above with give lateral stiffessk is set ito free vibratio with a iitial displacemet of.5 i. ad a iitial velocity of i/sec. Make the followig assumptios regardig dampig: (a) udamped (b) % damped ( ζ=. ) (c) % damped ( ζ=. ) Fid the solutio for the displacemet ut () ad use MATLAB to plot three cycles of vibratio. Problem : (a) What is the amplitude of motio of the system i Problem for the udamped case? (b) What is the maximum displacemet of the system i Problem for the damped cases? (c) What is the required dampig ratio ζ to reduce the displacemet at.5 sec to? For (b) ad (c) use MATLAB to calculate the respose ut () for closely spaced time poits. The use the max commad for (b) ad trial ad error for (c). Problem (Chopra.): For a system with dampig ratio ζ, determie the umber of free vibratio cycles required to reduce the displacemet amplitude to % of the iitial amplitude; the iitial velocity is zero. l().66 Solutio: j% = = πζ ζ Problem 4 (Chopra.4): The vertical suspesio system of a automobile is idealized as a viscously damped SDF system. Uder the -lb weight of the car the suspesio system deflects i. The suspesio is desiged to be critically damped. (a) Calculate the dampig ad stiffess coefficiets of the suspesio. Solutio: (b) With four l6lb passegers i the car, what is the effective dampig ratio? ( a) k = 5 lb/i (c) Calculate the atural vibratio frequecy for case (b). c = ccr = 5.9 lb-sec/i (b) ζ =.98 Problem 5 (Chopra.5): (c) ωd = 5.8 rad/sec The stiffess ad dampig properties of a mass-sprig-damper system are to be determied by a free vibratio test; the mass is give as m =. lb-sec /i. I this test the mass is displaced i. by a hydraulic jack ad the suddely released. At the ed of complete cycles, the time is sec ad the amplitude is. i. Determie the stiffess ad dampig coefficiets. Solutio: k = 75.5 lb/i, c =.7 lb-sec/i 4//9 :54 PM C:\calpoly\arce4\homework\sprig_9\hw.doc

11 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April 6, 9 Homework -Solutio Problem Use MATLAB to plot three cycles of displacemet respose ut () of the structure above (rigid girder, colum fixed at the base). The structure is set ito free vibratio with a iitial displacemet of.5 i. ad a iitial velocity of i/sec. Make the followig assumptios regardig dampig: (a) udamped (b) % damped ( ζ=. ) (c) % damped ( ζ=. ) m=ksec /i ut ().5 k=5 k/i.5 ζ = ζ =. Displacemet [i].5.5 ζ =..5 Problem : (a) What is the amplitude of motio of the system i Problem for the udamped case? (b) What is the maximum displacemet of the system i Problem for the damped cases? (c) What is the required dampig ratio ζ to reduce the displacemet at.5 sec to? u () (a) u = [ u() ] + =.5 + =.6 " ω 5 (b) umax =.4 " at t =.7 sec ( ζ =.) u =.7" at t =.4 sec( ζ =.) max Time [sec] (c) try ζ =. ut ( =.5) =.989 (good eough) Problem (Chopra.) MOVED TO HW #4 For a system with dampig ratio ζ, determie the umber of free vibratio cycles required to reduce the displacemet amplitude to % of the iitial amplitude; the iitial velocity is zero. u l().66 l l j % j πζ πζ u = = = = j. πζ ζ j + % 4//9 :4 PM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

12 Sprig Quarter 9 Problem 4 (Chopra.4) The vertical suspesio system of a automobile is idealized as a viscously damped SDF system. Uder the -lb weight of the car the suspesio system deflects i. The suspesio is desiged to be critically damped. (a) Calculate the dampig ad stiffess coefficiets of the suspesio. (b) With four l6lb passegers i the car, what is the effective dampig ratio? (c) Calculate the atural vibratio frequecy for case (b). (a) The stiffess coefficiet is k = = 5 lb/i The dampig coeffciet is c = ccr = km = 5 = 5.9 lb-sec/i 86 (b) With passegers, weight isw = 64 lb. The dampig ratio is ζ = c c c = cr km = 64 = 5 86 (c) The atural vibratio frequecy for case (b) is 5 86 ω = ω D ζ = = rad/sec Problem 5 (Chopra.5) MOVED TO HW #4 The stiffess ad dampig properties of a mass-sprig-damper system are to be determied by a free vibratio test; the mass is give as mm =. lb-sec /i. I this test the mass is displaced i. by a hydraulic jack ad the suddely released. At the ed of complete cycles, the time is sec ad the amplitude is. i. Determie the stiffess ad dampig coefficiets. () Determie ζ ad ω u ζ = = = = πj. l l.8.8% u π j + Therefore, assumptio of small dampig i the above equatio is valid. T.5 sec.5 sec 4.89 rad/sec D D T T π = = = ω = =.5 () The stiffess coefficiet is k = ω m = = 75.5 lb/i () The dampig coefficiet is c = mω = = 8.77 lb-sec/i c = ζc = =.7 lb-sec/i cr cr 4//9 :4 PM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc

13 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April, 9 Readig: Chopra Sectio.,. Homework 4 (Due 4-7-9) Problem (Chopra.): For a system with dampig ratio ζ, determie the umber of free vibratio cycles required to reduce the displacemet amplitude to % of the iitial amplitude; the iitial velocity is zero. Solutio: l().66 j% = = πζ ζ Problem (Chopra.5): The stiffess ad dampig properties of a mass-sprig-damper system are to be determied by a free vibratio test; the mass is give as m =. lb-sec /i. I this test the mass is displaced i. by a hydraulic jack ad the suddely released. At the ed of complete cycles, the time is sec ad the amplitude is. i. Determie the stiffess ad dampig coefficiets. Solutio: k = 75.5 lb/i, c =.7 lb-sec/i Problem : A SDF structure is excited by a siusoidal force. At resoace the amplitude of displacemet was measured to be i. At a excitig frequecy of oe-teth the atural frequecy of the system, the displacemet amplitude was measured to be. i. Estimate the dampig ratio of the system. Problem 4: I a forced vibratio test uder harmoic excitatio it was oted that the amplitude of motio at resoace was exactly four times the amplitude at a excitatio frequecy % higher tha the resoat frequecy. Determie the dampig ratio of the system. Problem 5: The displacemet respose of a SDF structure to harmoic excitatio ad iitial coditios u() ad u () is give by ζωt u( t) = e [ Acos ω t + Bsi ωdt] + C si ωt + Dcos ωt D trasiet vibratio steady-stae vibratio Fid expressios for the costats of itegratio A ad B i terms of CDu,, (), u (), ω, ωd, ω, ad ζ. Solutio: A= u() D u() + ωζ [ u() D] Cω B = ω D 4//9 7:9 AM C:\calpoly\arce4\homework\sprig_6\hw4.doc

14 Sprig Quarter 9 Problem 6: W =4 k pt ( ) = 5 si( t) [k] E I = 9 ksi 4 = 75 i 5 ft ft Determie ad plot the respose of the frame above ( t 5sec). Assume the girder is rigid ad eglect the mass of the colums. (a) Assume at rest iitial coditios ad zero dampig. (b) Assume 5% dampig ad iitial coditios u () = i ad u () = 5 i/sec. Submit two figures, oe for (a) ad oe for (b), cotaiig three plots each (the trasiet, stead-state ad total resposes). Write the umerical values for costats ABC,,, Do the figures. Problem 7: k = k/i W = 5 k pt () = p si( ωt) (a) Calculate the vertical displacemet of the catilever tip due to gravity. Assume 5% dampig ad cosider steady state motio oly: (b) For p = k ad ω = 5,,5 rad/sec calculate the amplitude of motio. Which of the three forcig frequecies causes the largest displacemet? Explai. (c) For ω = 5 rad/sec calculate the maximum allowable amplitude p of the forcig fuctio such that the deflectio of the catilever due to gravity plus dyamic actio is dowward at all times. 4//9 7:9 AM C:\calpoly\arce4\homework\sprig_6\hw4.doc

15 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April 7, 9 Homework 4-Solutio Problem For a system with dampig ratio ζ, determie the umber of free vibratio cycles required to reduce the displacemet amplitude to % of the iitial amplitude; the iitial velocity is zero. u l().66 l l j% j πζ πζ u = = = = j+ j%. πζ ζ Problem (Chopra.5) The stiffess ad dampig properties of a mass-sprig-damper system are to be determied by a free vibratio test; the mass is give as mm =. lb-sec /i. I this test the mass is displaced i. by a hydraulic jack ad the suddely released. At the ed of complete cycles, the time is sec ad the amplitude is. i. Determie the stiffess ad dampig coefficiets. () Determie ζ ad ω u ζ = l l.8.8% πj u = = = j+ π. Therefore, assumptio of small dampig i the above equatio is valid. T.5 sec.5 sec 4.89 rad/sec D D T T π = = = ω = =.5 () The stiffess coefficiet is k = ω m = = 75.5 lb/i () The dampig coefficiet is c = mω = = 8.77 lb-sec/i c = ζc = =.7 lb-sec/i cr cr Problem A SDF system is excited by a siusoidal force. At resoace, the amplitude of displacemet was measured to be i. At a excitig frequecy of oe-teth the atural frequecy of the system, the displacemet amplitude was measured to be. i. estimate the dampig ratio ζ of the system. At ω = ω (a) u = ( u st ) = ζ At ω =.ω u ( ust ) =. ( β =. excitatio is so slow, that it ca be cosidered static) Substitutig( u ) =.i (a) gives st ζ =.5 If we do t wat to make the assumptio of static respose for. β =, we ca use the approach followed i Problem 4 (see below). 4/7/9 :7 AM C:\calpoly\arce4\homework\sprig_9\hw4_sol.doc

16 Sprig Quarter 9 Problem 4 I a forced vibratio test uder harmoic excitatio it was oted that the amplitude of motio at resoace was exactly four times the amplitude at a excitatio frequecy % higher tha the resoat frequecy. Determie the dampig ratio of the system. We assume that dampig is small eough to justify the approximatio that the resoat frequecy is ω ad the resoace amplitude is/ζ. The give data the implies: (a) ( u) ω= ω = ( u ) st ζ (b)( u ) ω=. ω = ( u ) ( ) st = ust β + ζ β. + ζ. Combiig Eqs. (a) ad (b) ( ) ( ) ( ) ( ) ( u) ω= ω.576( ) = = ζ = as ( ζ) ( u) ω=.ω ( ζ) 4 (.44) + (.4 ζ) Assumptio of small dampig is reasoable. 4/7/9 :7 AM C:\calpoly\arce4\homework\sprig_9\hw4_sol.doc

17 Sprig Quarter 9 Problem 5 The displacemet respose of a SDF structure to harmoic excitatio ad iitial coditios u() ad u () is give by ζωt u( t) = e [ Acos ω t + Bsi ωdt] + C si ωt + Dcos ωt D trasiet vibratio steady-stae vibratio Fid expressios for the costats of itegratio A ad B i terms of CDu,, (), u (), ω, ω, ω, ad ζ. u() = A+ D A= u() D( as) ωζt ωζt ut () = e ( AωD si ωdt + BωD cos ωdt) ωζe ( Acos ωdt + Bsi ωdt) + Cωcos ωt Dωsi ωt u () = BωD ωζa+ Cω u () + ωζ A Cω u () + ωζ [ u() D] Cω B = = ( as) ω ω D D Problem 6 Determie ad plot the respose of the frame above ( t 5sec). Assume the girder is rigid ad eglect the mass of the colums. Idetify the steady state ad the trasiet portio of the respose (both i the equatios ad o the plot). (a) Assume at rest iitial coditios ad zero dampig. (b) Assume 5% dampig ad iitial coditios u () = i ad u () = 5 i/sec. Submit two figures, oe for (a) ad oe for (b), cotaiig three plots each (the trasiet, stead-state ad total resposes). Idicate the umerical values for costats ABC,,, Do the figures. (a) W =4 k k pt = t 4EI = = =.88 k/i L (5 ) ω = = 4. rad/sec E = 9 ksi 4 4 I = 75 i 5 ft p = 5 k ω = rad/sec β ω = = =.74 ω 4. ft D ( ) 5 si( ) [k] C ζ = C D A β p ζβ D p = = k ( β ) + ( ζβ) k ( β ) + ( ζβ) p 5 = = =.45 k = = ( β ).88(.74 ) ω B = C = Cβ = =.6 ω p β p ut () = siω si t + wt k β k β trasiet respose steady state respose 4/7/9 :7 AM C:\calpoly\arce4\homework\sprig_9\hw4_sol.doc

18 Sprig Quarter 9.5 u() =, u () =, ζ = A =, B =.6 C =.45 D = total trasiet steady-state.5 Displacemet [i] Time [sec] (b) ωζt u() t = uc() t + up() t = e ( AcosωDt + Bsi ωdt) + C siωt + Dcosωt trasiet respose steady state respose p β p ζβ u () + ωζ [ u() D] Cω C = D = A= u() D B = k ω ( β ) + ( ζβ) k ( β ) + ( ζβ) A=.95 i B =.6 i C =.97 i D =.95 i D 4 u() = " u () = 5 "/ sec ζ = 5% A =.95 B =.6 C =.97 D =.95 Displacemet [i] total trasiet steady-state Time [sec] 4/7/9 :7 AM C:\calpoly\arce4\homework\sprig_9\hw4_sol.doc

19 Sprig Quarter 9 Problem 7 k = k/i W = 5 k pt () = p si( ωt) (a) Calculate the vertical displacemet of the catilever tip due to gravity. Assume 5% dampig ad cosider steady state motio oly: (b) For p = k ad ω = 5,,5 rad/sec calculate the amplitude of motio. Which of the three forcig frequecies causes the largest displacemet? Explai. (c) For ω = 5 rad/sec calculate the maximum allowable amplitude p of the forcig fuctio such that the deflectio of the catilever due to gravity plus dyamic actio is dowward at all times. W 5 (a) ust = = =.5 "( as) k 86.4 (b) ω = =.4 rad/sec 5 5 k ω= 5 rad/sec β = =.4 u =.9 =.596 i( as).4 k/i k ω= rad/sec β = =.84 u =.76 =.8 i( as).4 k/i 5 k ω= 5 rad/sec β = =.6 u =. =.6 i( as).4 k/i Excitatio frequecy ω is closest to resoace largest amplitude p k p (c) u = ust Rd =.5. > p <.6 k( as) 4/7/9 :7 AM C:\calpoly\arce4\homework\sprig_9\hw4_sol.doc

20 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April 7, 9 Problem Homework 5 (Due 4--9) W 6 x5 W = 4 k ut () W 6 x7 W 6 x7 ft ζ=% 5 ft u () g t The sigle bay/sigle story momet frame above with floor weight of 4 kips above is subject to te cycles of harmoic groud motio π ug () t = i siωt < t < T T = ω whose frequecy ω is twice that of the atural frequecy ω. Determie the costats A, B, C, D i the expressios below for the displacemet of the floor relative to the movig groud ζω π u() t = e t ( AcosωDt + BsiωDt) + C siωt + Dcosωt < t < T T = ω After cycles the groud motios stops ad the structure vibrates freely. Determie the costats A, B i the expressio below for secods of free vibratio after the groud has stopped movig. ζω ut ( + T) = e t ( A cosω t + B siω t) < t < sec D D Plot ut () ad u() t as a fuctio of time. g Partial solutio: T T =.5sec =.75 sec A =.747 i, B = 6.5 i, C =. i, D =.747 i A =.85 i, B = 5.6 i Displacemet [i] groud displacemet ug( t) displacemet ut ( ) of floor (mass) relative to groud 4 Time [sec] C:\calpoly\arce4\homework\sprig_9\hw5.doc 4/7/9

21 Sprig Quarter 9 April, 9 ARCE 4: STRUCTURAL DYNAMICS Homework 5-Solutio Problem W 6 x5 W = 4 k ut () W 6 x7 W 6 x7 ft ζ=% 5 ft u () g t The sigle bay/sigle story momet frame above with floor weight of 4 kips above is subject to harmoic groud motio π ug () t = i siωt < t < T T = whose frequecy ω is twice that of the atural frequecy ω. ω Determie the costats A, B, C, D i the expressios below for the displacemet of the floor relative to the movig groud for cycles of groud motio ζω π u() t = e t ( AcosωDt + BsiωDt) + C siωt + Dcosωt < t < T T = ω After cycles the groud motios stops ad the structure vibrates freely. Determie the costats A, B i the expressio below for secods of free vibratio after the groud has stopped movig. ζω ut ( + T) = e t ( A cosω t + B siω t) < t < sec D D Plot ut () ad u() t as a fuctio of time. g groud displacemet ug( t) displacemet ut ( ) of floor (mass) relative to groud Displacemet [i] Time [sec] 4 C:\calpoly\arce4\homework\sprig_9\hw5_sol.doc 4/7/9 7:9 AM - -

22 Sprig Quarter 9 Vibratio properties M for uit lateral load 7 k" 7 k" EI EI β f k C G =, 5 9, = 4.5 k-i 6 = 7, 8 9, = 6. k-i ω = ω = 6 = =.94 i/k = = =.9 k/i f.94 i/k T ω T ω u g m 4/ 86.4 = π = π =.5sec k.9 π π = = = 7.9 rad/sec T.5 = T =.5.5 =.75 sec β π π = = = 5.84 rad/sec T.75 ( t) = i si 5.84 g ( ) ( ) u = 5.8 si 5.84t =, 84 i/sec si 5.8t p = m u = = g 4 k / 86.4 i/sec, 84 i/sec 9 k C:\calpoly\arce4\homework\sprig_9\hw5_sol.doc 4/7/9 7:9 AM - -

23 Sprig Quarter 9 Costats i respose equatio (see Hw #4) ωζ t u() t = uc() t + up() t = e ( A cosωdt + B si ωdt) + C siωt + Dcosωt trasiet respose steady state respose C p = k β ( β ) + ( ζβ) D p = k ζβ ( β ) + ( ζβ) A = u() D B C D A u () + ωζ [ u() D] Cω = ω D 9 = =. i.9. ( ) + ( ) 9. = =.747 i.9. ( ) + ( ) =.747 =.747 i B = = 6.5 i 7.9. I order to describe the free vibratio respose after the groud motio has stopped, we eed to fid the displacemet ut ( = T) ad the velocity ut ( = T) at the ed of the groud excitatio. These values serve as iitial coditios for the free vibratio. u( t) ζωt = e ( Acos ωdt + Bsi ωdt) + C si ωt + Dcos ωt < t < T =.75 sec u( t = T) =.85 i ωζt ut () = e ( AωD si ωdt + BωD cos ωdt) ωζe ( Acos ωdt + Bsi ωdt) + Cωcos ωt Dωsi ωt ut ( = T) = i/sec ωζt A B = u() =.85 i u () + ωζ u() = = = 5.6 i ω D 7.9. ζωt ut ( + T) = e ( A cosω t+ B siω t) < t< sec D D plot see page C:\calpoly\arce4\homework\sprig_9\hw5_sol.doc 4/7/9 7:9 AM - -

24 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April, 9 Homework 6 (Due 4-4-9) Readig: Chopra Sectios 5., 5.4 Problem A SDF system has the followig properties: m=.5 k sec /i, k= k/i, ζ=.5. Use the NEWMARK average acceleratio method to calculate ad plot the displacemet respose ut () of this system to a force pt () = si( πt) (at rest iitial coditios). (a) Use had calculatio with Δ t =.sec for as may timest i that you eed to uderstad the procedure. (b) Implemet the NEWMARK average acceleratio method for a SDF system i MATLAB by writig a fuctio fuctio u = ewmark(m,c,k,p,delt) where mckare,, the mass, viscous dampig coefficiet, ad stiffess, respectively of the SDF structure, p is a row vector cotaiig the loadig fuctio ad Δt is the time step. The output argumet u is a vector (say row vector) cotaiig the displacemet respose of the SDF system. (c) Test your algorithm by calculatig the displacemet respose ut () for t sec. Use time steps Δ t =.5,.,. sec. Also plot the exact solutio. The exact solutio is (see HW 4) u( t) = exp( ζωt) ( Acos ω t + Bsi ω t) + Csi ωt + Dcos ωt D D Submit: (a) Had calculatios. (b) A sigle plot cotaiig the solutio with Δ t =.5,.,. sec ad the exact solutio. Use differet colors or differet liestyles to distiguish solutios for differet Δ t. Problem Use the NEWMARK average acceleratio method to calculate ad plot the displacemet respose (the lateral deformatio) of two SDF systems with atural periodst = secadt =.5 sec, respectively to the 989 Loma Prieta earthquake (recorded at Gilroy). Assume 5% dampig. The record lomaprieta.dat (o blackboard) cotais groud acceleratio data poits i cm/sec at a spacig of. sec. 6 4 Acceleratio [cm/s ] 4 Loma Prieta Groud Acceleratio Time [s] Submit: Sigle page with two plots (groud acceleratio i uits of g ad relative displacemet respose i [i] (the output of the NEWMARK fuctio), sigle plot for the two periods). 4//9 8: AM C:\calpoly\arce4\homework\sprig_9\hw6.doc

25 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April, 9 Homework 6-Solutio Problem A SDF system has the followig properties: m=.5 k sec /i, k= k/i, ζ=.5. Use the NEWMARK average acceleratio method to calculate ad plot the displacemet respose ut () of this system to a force pt () = si( πt) (at rest iitial coditios). (a) Use had calculatio with Δ t =.sec for as may timest i that you eed to uderstad the procedure. (b) Implemet the NEWMARK average acceleratio method for a SDF system i MATLAB by writig a fuctio fuctio u = ewmark(m,c,k,p,delt) where mckare,, the mass, viscous dampig coefficiet, ad stiffess, respectively of the SDF structure, p is a row vector cotaiig the loadig fuctio ad Δt is the time step. The output argumet u is a vector (say row vector) cotaiig the displacemet respose of the SDF system. (c) Test your algorithm by calculatig the displacemet respose ut () for t sec. Use time steps Δ t =.5,.,. sec. Also plot the exact solutio. The exact solutio is (see HW 4) u( t) = exp( ζωt) ( Acos ω t + Bsi ω t) + Csi ωt + Dcos ωt D D Submit: (a) Had calculatios. (b) A sigle plot cotaiig the solutio with Δ t =.5,.,. sec ad the exact solutio. i t pt () ut () ut () ut () Displacemet [i] Δ t =.5 Δ t =. Δ t =., exact( virtually idetical) Time [sec] Numerical (Newmark Average Acceleratio Method usig differet time steps) ad aalytical respose. 4/9/9 :58 AM C:\calpoly\arce4\homework\sprig_9\hw6_sol.doc

26 Sprig Quarter 9 Problem Use the NEWMARK average acceleratio method to calculate ad plot the acceleratio, velocity ad displacemet respose of two SDF systems with atural periodst = secadt =.5 sec, respectively to the 989 Loma Prieta earthquake (recorded at Gilroy). Assume 5% dampig. The record (\asgar\lomaprieta.dat o the ARCE shared drive) cotais groud acceleratio data poits i cm/sec at a spacig of. sec. 6 4 Acceleratio [cm/s ] 4 Loma Prieta Groud Acceleratio Time [s] Submit: Sigle page with two plots (groud acceleratio i uits of g ad relative displacemet respose i [i] (the output of the NEWMARK fuctio), sigle plot for the two periods)..8 Groud Acceleratio [g] max u =.665 g g t [sec] Displacemet [i] ζ =.5 T = sec T =.5 sec 5 5 Time [s] u u max mx a = =.6 i.5 i Loma Prieta groud acceleratio ad displacemet respose of two SDF structures (displacemet is relative to the groud) with 5% dampig. 4/9/9 :58 AM C:\calpoly\arce4\homework\sprig_9\hw6_sol.doc

27 Sprig Quarter 9 structor: Asgar Neuehofer ARCE 4: STRUCTURAL DYNAMICS April 4, 9 Homework 7 (Due 4-9-9) Readig: Chopra, Chapter 6 Problem Calculate the 5% damped displacemet (i cm) ad acceleratio respose spectra (i g) (also called the spectral displacemet ad spectral acceleratio, respectively) of the Loma Prieta ad Northridge earthquakes (o Blackboard). Cosider 5 atural periodst equally spaced betwee T =. sec ad T = sec. Submit: Two plots (spectral displacemet ad acceleratio), sigle plot for the two EQ motios ad table with spectral values (see below) Hit: Cut ad paste spectral values A ad D from MATLAB ito EXCEL Solutio: Spectral Displacemet [cm] ζ =.5 Northridge Loma Prieta T [sec] Spectral Acceleratio [g] ζ =.5 Loma Prieta Northridge T [sec] # T [sec] D [cm] LP A [g] LP D [cm] NR A [g] NR Problem A -ft-log vertical catilever made of a steel pipe supports a -lb weight attached at the tip as show. The properties of the pipe are: outside diameter=6.65, iside diameter=6.65 i. Fid () the peak deformatio, () the equivalet static force ad () the bedig stress resultig from the Loma Prieta ad Northridge earthquakes. Igore the weight of the pipe ad assume 5% dampig. CAUTION: Earthquake records are i cm/sec Partial Solutio: Loma Prieta: σ = 5. ksi( as) Northridge: σ = 6. ksi( as) 4/7/9 7:9 AM C:\calpoly\arce4\homework\sprig_9\hw7.doc

28 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April 9, 9 Homework 7-Solutio Problem Calculate the 5% damped displacemet (i cm) ad acceleratio respose spectra (i g) for the Loma Prieta ad Northridge earthquakes (i directory asgar o the ARCE shared drive). Cosider 5 atural periodst equally spaced betwee T =. sec ad T = sec. Submit: Two plots (spectral displacemet ad acceleratio), sigle plot for the two motios. Table (see below) with spectral values. Hit: Cut ad paste spectral values A ad D from MATLAB ito EXCEL 8 Spectral Displacemet [cm] ζ =.5 Northridge Loma Prieta T [sec].5 Spectral Acceleratio [g].5.5 ζ =.5 Northridge Loma Prieta T [sec] 4/9/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw7_sol.doc

29 Sprig Quarter 9 MATLAB statemets %HW 7, NEWMARK METHOD, groud acceleratio, respose spectrum clear all close('all') width =.5; width =. size=6; size=; load c:\calpoly\arce4\eqrecords\lomaprieta.dat; load c:\calpoly\arce4\eqrecords\orthridge.dat; iput = [lomaprieta orthridge]; t = [:.:]; zeta =.5; m=;delt =.; delt = ( -.)/5; T = [.:delt:]; for i=: for j=:legth(t); j TT = T(j); k = 4*pi^./ TT^; c = zeta * * sqrt( k * m); [u,udot,uddot] = ewmark(m,c,k,-iput(:,i)'*m,delt); dis(i,j) = max(abs(u)); acc(i,j) = dis(i,j).* k / 98; ed ed col = ['b';'r']; st=['- ';'--'] figure() hold o for i=: plot(t,dis(i,:),col(i,:),'liewidth',width,'liestyle',st(i,:)); ty = [col(i,:) '*'] plot(t,dis(i,:),ty,'liewidth',width); ed xlabel('\itt_ \rm[sec]','fotsize',size); ylabel('spectral Displacemet [cm]','fotsize',size); set(gca,'plotboxaspectratio',[.5 ]); set(gca,'fotsize',size); prit -depsc -r -tiff c:\calpoly figure() hold o for i=: plot(t,acc(i,:),col(i,:),'liewidth',width,'liestyle',st(i,:)); ty = [col(i,:) '*'] plot(t,acc(i,:),ty,'liewidth',width); ed xlabel('\itt_ \rm [sec]','fotsize',size); ylabel('spectral Acceleratio [g]','fotsize',size); set(gca,'plotboxaspectratio',[.5 ]); set(gca,'fotsize',size); stop %build table res(:,[ 4 5]) = [ T' dis' acc']; %prit -depsc -r -tiff c:\calpoly 4/9/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw7_sol.doc

30 Sprig Quarter 9 Numerical results T [sec] D [cm] LP A [g] LP D [cm] NR A [g] NR /9/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw7_sol.doc

31 Sprig Quarter 9 Problem A -ft-log vertical catilever made of a 6-i omial-diameter stadard steel pipe supports a -lb weight attached at the tip as show. The properties of the pipe are: outside diameter=6.65, iside diameter=6.665 i. Fid () the peak deformatio, () the equivalet static force ad () the bedig stress resultig from the Northridge ad Loma Prieta earthquakes. Igore the weight of the pipe. I k W T π = ( ) = 8.4 i 64 EI 9, 8.4 = = =.47 k/i L = lbs = π =.465sec Spectral displacemet,force, momet,stress 6.4cm from Newmark Loma Prieta: DT ( =.465, ζ =.5) = =.5 i( as).54 cm/i F =.5.47 =.54 k ( as) M =.54 = 44 k-i 44 σ = 6.65 / = 5. ksi( as) cm Northridge: DT ( =.465, ζ =.5) = =.8 i ( as).54 cm/i F =.8.47 =.56 k ( as) M =.56 = 7 k-i 7 σ = 6.65/ = 6. ksi( as) 8.4 Note: Sectio properties are calculated based o omial dimesios. Use of steel maual results i smaller sectio properties ad correspodigly larger stresses. The results are σ = 58. ksi(loma Prieta) σ = 4. ksi(northridge) 4/9/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw7_sol.doc

32 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS April 9, 9 Homework 8 (Due 5-4-9) Readig: Chopra Sectio 9., 9. Problem A uiform rigid bar of total massm is supported o two sprigsk adk at the two eds ad subjected to dyamic forces as show i the figure. Formulate the equatios of motio with respect to the two DOFs defied at the left ed of the bar. Partial Solutio: ml k + k k m L k = kl kl m = ml ml Problem Fid the x (a) mass, (b) flexibility (usig PVF had calculatio) (c) stiffess (use beam elemet i MATLAB, elimiate uwated DOFs) matrices of the beam for the two DOFs show. Check whether the stiffess ad flexibility matrices are the iverse of each other. Partial Solutio: L 8 7 f = 486EI 7 8 Problem A rigid bar is supported by a weightless colum as show. Fid the x (a) mass (b) flexibility (usig PVF had calculatio) (c) stiffess (by ivertig the flexibility matrix) matrices of the system defied for the two DOFs show. Fid the stiffess matrix by ivertig the flexibility matrix. L 5 f = 6EI 5 4 4/9/9 8: AM C:\calpoly\arce4\homework\sprig_9\hw8.doc

33 Sprig Quarter 9 Problem 4 Solutio for (4.) 6 EI K = k 6 b,with k = h b b m M = m b m 6 The figure shows a uiform slab supported o four colums rigidly attached to the slab ad clamped at the base. The slab has a total mass m ad is rigid i plae ad out of plae. Each colum is of circular cross sectio with momet of iertia I or I. With the DOFs selected as ux, uy, u θ () at the ceter of mass of the slab (as show) () at the ceter of stiffess of the slab (ot show) fid the mass ad stiffess matrices i terms ofm ad the lateral stiffess k = EI / h of the smaller colum. Problem 5 ' ' y z x 6 ' EA 6 ' t =.5 ft γ C = 5 #/ft ' 6 ' Solutio K M W slab.64 = EA [ft,k] =.68 [ft,k,sec] 644. = 86.4 k Fid the mass ad stiffess matrices with respect to the three i plae degrees-of-freedom ux, uy, ϕ z. I-plae stiffess is provided by the three diagoal braces. 4/9/9 8: AM C:\calpoly\arce4\homework\sprig_9\hw8.doc

34 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May 6, 9 Homework 8-Solutio Relative displacemet, velocity ad acceleratio respose time history for Loma Prieta Earthquake( ζ = 5%). 4/8/9 8:9 PM C:\calpoly\arce4\homework\sprig_9\hw8_sol.doc - -

35 Sprig Quarter 9 4/8/9 8:9 PM C:\calpoly\arce4\homework\sprig_9\hw8_sol.doc - -

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37 Sprig Quarter 9 4/8/9 8:9 PM C:\calpoly\arce4\homework\sprig_9\hw8_sol.doc - 4 -

38 Sprig Quarter 9 Problem c Usig the beam elemet with four degrees-of-freedom the stiffess matrix with respect to the 8 free degrees-of-freedom (4 rotatios ad traslatios) is K free = use E, I, L = = = Sice zero mass is associated with the rotatios, we elimiate the four rotatios to geerate a stiffess matrix with respect to the two traslatios oly. EI Kfial = ( as) L Below are the correspodig MATLAB statemets free = [ ]; Kfree = K(free,free); r = [ 4]; e = [ 5 6]; Krr = Kfree(r,r); Kre = Kfree (r,e); Ker = Kfree (e,r); Kee = Kfree (e,e); %Fial x st. matrix Kfial = Krr - Kre*iv(Kee) * Ker Problem 4. Whe we apply the degrees-of-freedom at the ceter of rigidity all off-diagoal terms of the stiffess matrix a zero (that is the defiitio of the ceter of rigidity). Furthermore, the diagoal elemets related to the displacemets are the same as before. Hece we oly eed to calculate elemet K of the stiffess matrix. The mass matrix is o loger diagoal, couplig exists betwee the acceleratio i they -directio ad the momet (ad also, by symmetry) betwee the agular acceleratio ad they -force). Elemet M of the mass matrix also chages. 5 d = b, d = b K kd kd k b k b b k = + = 4 + = = + + = ( ) M m b b m b mb EI K = k 6 = k 6 ( as),withk = h 7 b.8b 6 M = b m =.667 b m( as) 6 7b.667b.944b b 6 6 4/8/9 8:9 PM C:\calpoly\arce4\homework\sprig_9\hw8_sol.doc - 5 -

39 Sprig Quarter 9 Problem 5 t =.5 ft γ C = 5 #/ft y z x ' ' EA 6 ' 6 ' ' 6 ' Fid the mass ad stiffess matrices with respect to the three i plae degrees-of-freedom ux, uy, ϕ z. I-plae stiffess is provided by the three diagoal braces. k k k = EA EA α = L = = k = cos.8.64 = k = EA EA EA k = cos α =.77 =.95 EA L 6.97 k = k =.95 EA 4 =.77EA k = EA = 6.9 EA K = EA k, ft( as) W slab 48 ' 4 '.5 '.5 k/ft 86.4 k = = m 86.4 k-sec = =.68. ft (4 48 ) k-sec -ft MMI = + =.68 M =.68 [ft,k,sec]( as) 644. u =, u =, u = k. EA k k. EA ft ft.95 EA u =, u =, u = k k k ft ft 6 ft 6 ft 6 ft 6 ft 6 ft 6 ft EA 6.97 =.77EA.77 4 EA =.84 EA.8 k k k ft ft u =, u =, u =.84 EA 6 ft 6 ft 6 ft 4/8/9 8:9 PM C:\calpoly\arce4\homework\sprig_9\hw8_sol.doc - 6 -

40 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May 8, 9 Readig: Chopra Sectio.,.,. Homework 9 (Due 5--9) Problem Determie the atural vibratio frequecies ad mode shapes of the structure fork = k adk = k. Sketch the mode shapes. Use the mass ad stiffess matrices determied i HW 8. m (total mass) Solutio: k k ω =.59 ω =.76 φ = m m.64 / L φ =.66 / L Problem Determie the atural vibratio frequecies ad mode shapes of the structure i terms of meiad, L. Sketch the mode shapes. Use the mass ad stiffess matrices determied i HW 8. Solutio: EI EI ω = ω 4 = 8.84 φ 4 = ml ml = φ Problem : Determie the atural vibratio frequecies ad mode shapes of the structure (rigid beams) i terms of meiadh,. Sketch the mode shapes. Use MATLAB s eig fuctio to calculate the eigevalues (atural frequecies) ad eigevectors (mode shapes). Normalize the mode shapes such that the value at the roof level has uit value. I Matlab, use E =, I =, h =, m =. Model the frame i RISA or ETABS, perform a frequecy aalysis to fid the mode shapes ad frequecies. Compare the results. A correct model should give you the exact same frequecies ad mode shapes. Solutio: EI EI EI ω =.4 ω = ω = 7.4 φ =.686 φ =.5 φ =.86 mh mh mh 5/8/9 7:5 AM C:\calpoly\arce4\homework\sprig_9\hw_9.doc

41 Sprig Quarter 9. May 4, 9 ARCE 4: STRUCTURAL DYNAMICS Homework 9-Solutio 5/4/9 :7 PM C:\calpoly\arce4\homework\sprig_9\hw9_sol.doc - -

42 Sprig Quarter 9 Problem. 5/4/9 :7 PM C:\calpoly\arce4\homework\sprig_9\hw9_sol.doc - -

43 Sprig Quarter 9 5/4/9 :7 PM C:\calpoly\arce4\homework\sprig_9\hw9_sol.doc - -

44 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May 5, 9 Readig: Chopra Sectios.7-.8 Problem Homework (Due 5--9) For the structure of HW 8, P 4 (rigid slab o four colums), fid the three atural periods T ad correspodig mode shapes Φ for i-plae vibratios. Defie the degrees-of-freedom () at the ceter of mass of the slab (as show) () at the ceter of stiffess of the slab (ot show) ad show that both frequecies ad mode shapes are idepedet of where we apply the degrees-of-freedom. Order the periods i descedig order ad sketch the mode shapes. Use MATLAB. () Model the structure i ETABS or RISA D ad calculate atural periods ad mode shapes. Iclude plots of the mode shapes i your submittal. Model the colums as fixed-fixed. UseW =. k (total weight of slab), b = ft, k = EI / h = k/ft. Solutio: T =.67 sec, T =.56 sec T =.46 sec Problem 5 k u W 4 x 48 W 4 x 9 W 4 x 5 k 5 k ft ft ft u u girders assumed rigid Shape 4 I [i ] W 4x W 4x /5/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw_.doc

45 Sprig Quarter 9 W 4x 8 () Determie the atural periods T ad modes φ of the three-story frame above. Sketch the mode shapes ad T idetify the associated atural frequecies. Normalize each mode so that the modal mass M = φ m φhas uit value. Solutio: Φ = for mi k, sec, ft T = [ ] sec Φ = for m i k, sec, i () Determie the free vibratio respose of the frame for the followig three vectors of iitial displacemets (the iitial velocity is zero). Neglect dampig. Cosider t.sec. u T T T I = II =.5 III = [i] u u Commet o the relative cotributio of the three vibratio modes to the respose produced by the three iitial displacemets. Submit: () Summary of aalysis. Sketch of mode shapes. () Oe plot for each vector of iitial displacemets showig the roof respose i modes,, ad ad the total roof respose (i.e. 4 lies per plot). () Use had calculatio to calculate at t = sec for the three iitial coditios:. The total (all modes combied) first floor displacemet.. The roof displacemet i mode.. The secod floor displacemet i mode. Partial Solutio displacemet i [i]... IC IC IC Roof displacemet for IC mo de mode mode t otal Importat arrays ad their dimesios: N NT umber of degrees of freedom (dof) =umber of modes umber of time poits u structure respose ( N x NT matrix) q modal respose ( N x NT matrix) Φ mode shapes ( N x N matrix) ϕ( ij, ) value of mode shape jat dof i u = Φ q matrix product = row times colum = modal supoerpositio 5/5/9 :5 AM C:\calpoly\arce4\homework\sprig_9\hw_.doc

46 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May, 9 Homework -Solutio Problem For the structure of HW 8, P 4 (rigid slab o four colums) fid the three atural periods T ad correspodig mode shapes Φ for i-plae vibratios. Defie the degrees-of-freedom () at the ceter of mass of the slab (as show) () at the ceter of stiffess of the slab (ot show) ad show that both frequecies ad mode shapes are idepedet of where we apply the degrees-of-freedom. Order the periods i descedig order ad sketch the mode shapes. UseW =. k (total weight of slab), b = ft, k = EI / h = k/ft. Ceter of mass (compare HW 8 solutio) k 6 b 6 6, m K = = = M = = =, 7, b b b Ceter of stiffess (compare HW 8 solutio) K = k 6 = 6 = 6 M = m b = = , b 7b b MATLAB statemeets. [phi lam] = eig(k,m); ΦCM = %frequecies.57.8 omega = sqrt(diag(lam)) CR CM %sort. [omega j] = sort(omega); phi = phi(:,j); ΦCR = Mode Mode T = *pi./ omega.57.8 T =.67 sec, T =.56 sec T =.46 sec Sice.98 = ad.96 = , the two mode shape describe idetical motios. 5/8/9 : AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc - -

47 Sprig Quarter 9 Sketch of mode shapes Mode Mode Mode Problem 5 k u W 4 x 48 5 k u W 4 x 9 5 k u W 4 x Shape 4 I [i ] W 4x W 4x9 999 W 4x 8 girders assumed rigid () Determie the atural periods T ad modes φ of the three-story frame above. Sketch the mode shapes ad T idetify the associated atural frequecies. Normalize each mode so that the modal mass M = φ m φhas uit value. Solutio: T = [ ] sec Φ = () Determie the free vibratio respose of the frame for the followig three vectors of iitial displacemets (the iitial velocity is zero). Neglect dampig. Cosider t.sec. u T T T I = II =.5 III = [i] u u Commet o the relative cotributio of the three vibratio modes to the respose produced by the three iitial displacemets. 5/8/9 : AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc - -

48 Sprig Quarter 9 Submit: () Summary of aalysis. Sketch of mode shapes. () Oe plot for each vector of iitial displacemets showig the roof respose i modes,, ad ad the total roof respose (i.e. 4 lies per plot). () Use had calculatio to calculate at t = sec for the three iitial coditios:. The total (all modes combied) first floor displacemet.. The roof displacemet i mode.. The secod floor displacemet i mode. MATLAB code %-story frame (rigid girders) clear all close('all') %story mass ad stiffesses m = 5/86.4; E = 9; k = 4*E*8 /44/^; k = 4*E* 999 /44/^; k = 4*E* 485 /44/^; K = [k+k -k ; -k k+k -k; -k k]/; M = [m ; m ; m/]; %frequecies [phi lam] = eig(k,m); omega = sqrt(diag(lam)) %sort frequecies ad modes [omega j] = sort(omega); phi= phi(:,j);t= *pi./ omega N = legth(t); %ormalize phi for i=:n M(i) = phi(:,i)' * M * phi(:,i); phi(:,i) = phi(:,i)./ sqrt(m(i)); ed %iitial coditios %u = [ ]'; u = [-.5 ]'; %u = [ - ]'; q = iv(phi) * u; t = [:.:]; NT= legth(t); umode=zeros(n,nt,n); %modal displacemets for i=: q(i,:) = q(i).* cos(omega(i).* t); ed %structure displacemets i idividual modes for i=:n umode(:,:,i) = phi(:,i) * q(i,:); ed %total structure displacemets (all modes combied) utot = phi * q; col = ['r';'g';'b';'k']; figure hold o for i=: plot(t,umode(,:,i),col(i,:)); ed plot(t,utot(,:),col(4,:)); axis([ - ]); 5/8/9 : AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc - -

49 Sprig Quarter 9 Summary of matrices ad had calculatios K = k/i M =.94 k sec /i Φ= Φ = q q q q - =Φ u -, u, =Φ = -, u, =Φ = -, u, =Φ = = = = cos ω -.8 cos ω =.85 cos ω.755 For IC. u = (.8) +.4 (.) =.6 i. u =.795 (.).85 =.6 i,. u = (.8) =.574 i other IC aalogous, Summary displacemet i [i]... IC IC IC /8/9 : AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc - 4 -

50 Sprig Quarter 9 Plots roof displacemet [i] Roof displacemet for IC mode mode mode total roof displacemet [i] time [sec] Roof displacemet for IC mode mode mode total Sketch of mode shapes time [sec] roof displacemet [i] Roof displacemet for IC mode mode mode total time [sec] 5/8/9 : AM C:\calpoly\arce4\homework\sprig_9\hw_sol.doc - 5 -

51 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May, 9 Homework (Due 5-9-9) 48. k u k = 65 k/ft 96.6 k u k = 65 k/ft 96.6 k u k = 5 k/ft Calculate the respose of the three-story frame above to the Loma Prieta groud motio. Select 5% dampig. Respose quatities of iterest are: (-) The iter-story drift ratio i % for the three stories. Story heights are ft for stories ad ad 5 ft for story. (4) The overturig momet i k-ft. Plot the respose i modes ad ad the total respose for < t < sec. Do't iclude the third mode (its cotributio is virtually zero ad the plots would become too busy). Distiguish the three lies by color or lie-style. O your plot, clearly mark the maximum absolute value of the respose i each mode as well as the maximum of the total respose. Why is the maximum value of the sum (sum of all modes) ot equal to the sum of the maximum modal values? Develop a table summarizig the maximum value of each result quatity i the first mode ad the total maximum value. Commet o the effects of the secod mode o the total maximum. Which of the four respose quatities is most affected by the secod mode? Some help o MATAB implemetatio %Total structure respose u = phi * q girders assumed rigid k = story stiffess %Structure respose (floor displacemets) i idividual modes u = phi(:,) * q(,:) u = phi(:,) * q(,:) u = phi(:,) * q(,:) %st story drift i three modes ad total d = u(,:); d = u(,:); d = u(,:); d = u(,:); %d story drift i three modes ad total 5//9 : PM C:\calpoly\arce4\homework\sprig_9\hw_.doc

52 Sprig Quarter 9 d = u(,:) - u(,:); d = u(,:)- u(,:); d = u(,:)-u(,:); d = u(,:)-u(,:); %do't forget to covert drift ito drift ratio %rd story drift i three modes ad total your tur... %forces actig at floor level (three modes ad total) F = K * u; F = K * u; F = K * u; F = K * u; %overturig momet(three modes ad total) your tur... Some Solutios: T =.5 sec T =.6 sec T =. sec Γ =.698 Γ =.466 Γ =.97 Note: Γ 's are obtaied with mass-orthoormal mode shape usig uits of feet. 5//9 : PM C:\calpoly\arce4\homework\sprig_9\hw_.doc

53 Sprig Quarter 9 Iterstory Drift [%] t [s] Mode Mode Total 5//9 : PM C:\calpoly\arce4\homework\sprig_9\hw_.doc

54 Sprig Quarter 9 ARCE 4: STRUCTURAL DYNAMICS May, 9 Homework -Solutio 48. k u k = 65 k/ft 96.6 k u k = 65 k/ft 96.6 k u k = 5 k/ft Calculate the respose of the three-story frame above to the Loma Prieta groud motio. Select 5% dampig. Respose quatities of iterest are: (-) The iter-story drift ratio i % for the three stories. Story heights are ft for stories ad ad 5 ft for story. (4) The overturig momet i k-ft. Plot the respose i modes ad ad the total respose for < t < sec. Do't iclude the third mode (its cotributio is virtually zero ad the plots would become too busy). Distiguish the three lies by color or lie-style. O your plot, clearly mark the maximum absolute value of the respose i each mode as well as the maximum of the total respose. Why is the maximum value of the sum (sum of all modes) ot equal to the sum of the maximum modal values? Develop a table summarizig the maximum value of each result quatity i the first mode ad the total maximum value. Commet o the effects of the secod mode o the total maximum. Which of the four respose quatities is most affected by the secod mode? Some help o MATAB implemetatio %Total structure respose u = phi * q girders assumed rigid k = story stiffess %Structure respose (floor displacemets) i idividual modes u = phi(:,) * q(,:) u = phi(:,) * q(,:) u = phi(:,) * q(,:) %st story drift i three modes ad total d = u(,:); d = u(,:); d = u(,:); d = u(,:); %d story drift i three modes ad total 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - -

55 Sprig Quarter 9 d = u(,:) - u(,:); d = u(,:)- u(,:); d = u(,:)-u(,:); d = u(,:)-u(,:); %do't forget to covert drift ito drift ratio %rd story drift i three modes ad total your tur... %forces actig at floor level (three modes ad total) F = K * u; F = K * u; F = K * u; F = K * u; %overturig momet(three modes ad total) your tur... Some Solutios: T =.5 sec T =.6 sec T =. sec Γ =.698 Γ =.466 Γ =.97 Note: Γ 's are obtaied with mass-orthoormal mode shape usig uits of feet. Solutio: Table : Modal ad combied maximum values of four respose quatities cosidered. Mode Mode Mode Total DR [%] DR [%] DR [%] OTM [k-ft] Note: Mode ad Mode colums ot required i HW. I time history aalysis we first calculate the sum of the idividual modal time histories ad the take the maximum of this summatio. We thus calculate the maximum of the sum. Sice the modal maximum values usually occur at differet time, the maximum of the sum is commoly smaller tha the sum of the maximum values. The sum of the maximum values gives us a overly coservative estimate of the maximum respose. The maximum of the sum over all modes ca also be smaller tha the maximum i mode, sice higher modes may cotribute with differet sigs. The combied drift ratio i story (.4), for example, is smaller tha that i mode (.4%) sice the effects of modes ad is egative. Higher modes icrease the drift ratio i story by about 7% (.7/.6=.7) ad the drift ratio i story by about 66%. As already metioed higher modes reduce the drift ratio i story. Higher modes oly have a small effect o the overturig momet, which is clearly domiated by mode (6/96=.8). 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - -

56 Sprig Quarter 9 Iterstory Drift [%] t [s] Mode Mode Total 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - -

57 Sprig Quarter 9 Iterstory Drift [%] t [s] Mode Mode Total 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - 4 -

58 Sprig Quarter 9 Iterstory Drift [%] t [s] Mode Mode Total 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - 5 -

59 Sprig Quarter 9 Overturig Momet [k ft] t [s] Mode Mode Total 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - 6 -

60 Sprig Quarter 9 ETABS model: A frequecy aalysis i ETABS yields the followig results: MATLAB.499 sec.6 sec.8 sec A time history aalysis i ETABS yields the followig results for the iter-story drift ratio (MATLAB results are from Table ): MATLAB The results are reasoably close to those from MATLAB.` 6//9 :58 PM C:\calpoly\arce4\homework\sprig_9\hw sol.doc - 7 -