inköpings Universitet, Hållfasthetslära, IEI/IKP Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, TMMI09 2007-03-16 kl 14-18 Ö S N I N G A R ---- SOUTIONS 1. Ange sambanden mellan vinkelfrekvens ω, cyklisk frekvens f och svängningstid (periodtid) T för en svängning. English: Give the relationships between the angular frequency ω, the cyclic frequency f and the period time T of a vibration. ösning: ω=2πf and f = 1 T 2. En massa M hänger i en fjäder med styvhet k. Vad blir egenvinkelfrekvensen för systemet? English: A mass M is hanging in a spring with spring stiffness (spring constant) k. What is the (angular) eigenfrequency of this system? ösning/solution: ω e = k M 8
Tekniska Högskolan i inköping, IKP Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, 070316 kl 14-18 DE 1 - (Teoridel utan hjälpmedel) 3. Rita (skissa) ett Wöhler-diagram och förklara dess användning. English: Draw (sketch) a Wöhler diagram and explain how it is used. ösning/solution: spänning/stress (MPa) a amplitude log N A Wöhler diagram is shown in the figure. The diagram gives a relation between the loading stress amplitude and the expected fatigue life. In the figure, a given stress amplitude σ ai gives the expected fatige life N i at that loading. 4. Ange Neubers hyperbel: skriv upp ekvationen och förklara de ingående storheterna. Förklara även hur den används. English: Give the Neuber hyperbola: write down the equation and explain the different factors in it. Also, explain how it is used. ösning/solution: The Neuber hyperbola reads ε σ= K 2 2 f σ E where ε and σ is strain and stress (at point of stress concentration), K f is the fatigue notch factor (K t, the stress concentration factor, is sometimes used here), and σ is the stress far away from the stress concentration). The intersection of the Neuber hyperbola and the material relationship (for example the Ramberg-Osgoods relation) gives the stress and the strain at the point of stress concentration. 9
Tekniska Högskolan i inköping, IKP Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, 070316 kl 14-18 DE 2 - (Problemdel med hjälpmedel) 5. En massa M hängs upp i två fjädrar (styvhet k k 1 = k och k 2 =2k). 1 k1 (a) Bestäm egenvinkelfrekvensen för systemet M k om massan monteras enligt figur (a). 2 k2 M (b) Vad blir egenvinkelfrekvensen om massan (a) (b) monteras enligt figur (b)? English: 5. A mass M is mounted with two springs (stiffness k 1 = k and k 2 = 2k). (a) Determine the (angular) eigenfrequency of the system if the mass is mounted as shown in figure (a). (b) What will the eigenfrequency be if the mass is mounted as in figure (b). ösning/solution: (a) The equation of motion of the mass is Mẍ = F 1 F 2 (a) where F 1 = k 1 x and F 2 = k 2 x This gives Mẍ +(k 1 + k 2 )x = 0 which gives (b) The equation of motion of the mass now becomes Mẍ = F For the two springs in series one obtains (same force F in the two springs) Enter this into (b). It gives and the eigenfrequency becomes ω e = k M = (k 1 + k 2 ) M = 3 k M = 1.732 k M x = x 1 + x 2 = F k 1 + F k 2 giving k 1k 2 k 1 + k 2 x = F Mẍ + k 1k 2 k 1 + k 2 x = 0 Thus, the eigenfrequency in case (b) goes down to approximately half the eigenfrequency in case (a). (b) ω e = k 1k 2 (k 1 + k 2 )M = 2 3 k M = 0.816 k M 10
Tekniska Högskolan i inköping, IKP /Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, 040312 kl 8-12 DE 2 - (Problemdel med hjälpmedel) c 6. En masslös axel bär ett svänghjul, radie R, som utsätts för ett vridande moment M v enligt figur. Bestäm amplituden av momentet i axeln vid stationärtillstånd om M v (t) =M 0 sin Ωt, där Ω = 0,90 ω 0 och ω 0 är systemets odämpade egenvinkelfrekvens. En dämpare enligt den nedre figuren dämpar svänghjulets rörelse. Dämparen har dämpfaktor c. Välj relativ dämpning ζ = 0,1, där cr 2 /J =2ζω 0. English: 6. A shaft (massless) carries a flywheel with radius R. The flywheel is loaded by a moment M v according to figure. Determine the amplitude of the momentet (the torque) in the shaft at stationary conditions if M v (t) =M 0 sin Ωt, where Ω = 0,90 ω 0 and ω 0 is the undampded eigenfrequency of the system. A damper, as shown in the lower figure, will damp the motion. The damper has the damping factor (damping stiffness) c. Choose relativ damping ζ = 0,1, where cr 2 /J =2ζω 0. Solution: Fd, GK J M v ( t) R M v ( t) M axel M v ( t) Study the shaft, damper, and flywheel separately. Dissect and enter the torque M axel (axel = shaft) between the shaft and the flywheel. Enter the force F d between the flywheel and the damper. Also, enter the rotation angle ϕ of the flywheel. The angle ϕ is positive in the same direction as the moment M axel is applied on the shaft. For the flywheel one obtains ( φ = ϕ) J φ=m v (t) M axel F d R (1) The moment-deformation relationship for the shaft is The force-deformation relationship for the damper is φ= M axel GK (2) F d = cr φ (3) Eliminate the moment M axel and the force F d from equations (1), (2) och (3). It gives the differential equation J φ+cr 2 φ+ GK φ=m (t)=m v 0 sin Ωt (4a) 11
This equation can be written on the form φ+ cr2 J GK φ+ J φ=m 0 sin Ωt J (4b) Entering the undamped eigenfrequency ω 0 / J and using cr 2 /J =2ζω 0 the differential equation is obtained as φ+2ζω 0 φ+ω 0 2 φ= M 0 J sin Ωt (4c) The solution to equation (4a) can be written as one particular solution and one homogeous solution: φ = φ part + φ hom. The homogeneous solution φ hom can be neglected because it will be damped out. Note, however, that the homogeneous part gives the undamped eigenfrequency ω 0 = GK / J (this was used above). The righ hand side contains the frequency Ω. Therefore also the solution will contain this frequency. Assume that the particular solution φ part has the form φ part = A sin Ωt + B cos Ωt (We need both the sinus and the cosinus term because due to the damping the angle φ will not be in phase with the load.) Enter φpart into (4a). It gives J ( AΩ 2 sin Ωt BΩ 2 cos Ωt)+cR 2 ( AΩ cos Ωt BΩ sin Ωt) Split up in terms containing sinus and terms containing cosinus. It gives JΩ2 A cr 2 Ω B + GK A sin Ωt = M 0 sin Ωt This is a system of equations for the constants A och B. On matrix form it can be written JΩ 2 + GK cr 2 Ω A M 0 cr 2 Ω JΩ 2 + GK = (7a,b) B 0 The constants A and B become + GK ( A sin Ωt +B cos Ωt)=M 0 sin Ωt (5) (6a) JΩ2 B + cr 2 Ω A + GK B cos Ωt = 0 (6b) 12
The amplitude becomes A = B = M 0 ( JΩ 2 + GK / ) ( JΩ 2 + GK / ) 2 + c 2 R 4 Ω 2 (8a) M 0 cr 2 Ω ( JΩ 2 + GK / ) 2 + c 2 R 4 Ω 2 (8b) ( JΩ 2 + GK / ) 2 +(cr 2 Ω) 2 C = A 2 + B 2 = M 0 ( JΩ 2 + GK / ) 2 + c 2 R 4 Ω 2 (9) The torque M axel in the shaft can now be determined. One obtains, using (2), M axel φ=gk (A sin Ωt + B cos Ωt) (10) The amplitude M axel0 of the torque becomes M axel0 C A 2 + B 2 M 0 ( JΩ 2 + GK / ) 2 +(cr 2 Ω) 2 ( JΩ 2 + GK / ) 2 + c 2 R 4 Ω 2 M 0 (11) ( JΩ 2 + GK / ) 2 +(cr 2 Ω) 2 This amplitude can be written, using ω 0 / J and cr 2 /J =2ζω 0, M axel0 = M 0 (GK / ) 2 (1 Ω 2 /ω 2 0 ) 2 + 4ζ 2 ω 2 0 J 2 Ω 2 (GK / ) 2 (1 Ω 2 /ω 0 2 ) 2 + 4ζ 2 ω 0 2 J 2 Ω 2 M 0 (1 Ω 2 /ω 2 0 ) 2 + 4ζ 2 Ω 2 2 /ω 0 Enter the relative damping ζ = 0,10 (i.e. 10 % of critical damping) and Ω = 0,9ω 0. It gives M axel0 = M 0 (15) = 3, 82M (1 0, 90 2 ) 2 0 (17b) + 4 0, 10 2 0, 90 2 For the undamped system one obtains M 0 M axel0 = (1 0, 90 2 ) = 5, 26M 0 (17a) Thus, at this frequency the amplitude is largely reduced by the damping. 13
Tekniska Högskolan i inköping, IKP Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, 070316 kl 14-18 DE 2 - (Problemdel med hjälpmedel) 7. En stor plåt, belastad med en en-axlig spänning, spänning/stress (MPa) utsätts för en belastningssekvens enligt figur. 300 Denna sekvens upprepas. Materialet har en Wöhlerkurva som ges av sambandet 200 σ a = 50 logn + 400 (MPa) 100 där σ a är spänningsamplituden. Bestäm förväntat antal sekvenser till utmattningsbrott. 0 Använd Palmgren-Miners delskadehypotes. tid/time Inverkan av spänningens medelvärde får försummas. English: 7. A large plate, loaded in uni-axial tension, is subjected to a load sequence according to the figure. This sequence is repeated. The material has a Wöhler curve given by the equation σ a = 50 logn + 400 (MPa) where σ a is the stress amplitude. Determine the expected number of sequences to fatigue failure. Use the Palmgren-Miner damage accumulation rule. The influence of the stress mean value can be neglected. ösning: Rain-flow count gives 1 cycle from 0 to 300 MPa, giving σ a = 150 MPa, 1 cycle from 50 to 300 MPa, giving σ a = 125 MPa, 2 cycles between 50 and 250 MPa, giving σ a = 100 MPa, and 1 cycle between 50 and 200 MPa, giving σ a = 75 MPa. These stress amplitudes give (from the Wöhler curve) N = 100 000, 316 228, 1000000, and 3 162 278 cycles, respectively The Palmgren-Miner damage accumulation rule gives 1 D = 100 000 + 1 316 228 + 2 1 000 000 + 1 3 162 278 = 1 64 600 Thus, failure is expected after approximately 64 000 sequences (giving 320 000 cycles). 14
Tekniska Högskolan i inköping, IEI/IKP Tore Dahlberg TENTAMEN i Mekaniska svängningar och utmattning, 040312 kl 8-12 DE 2 - (Problemdel med hjälpmedel) spänning/stress (MPa) 250 100-0,0025-250 - 330 töjning/ strain 0,003 8. Diagrammet visar tre stabiliserade hysteresslingor för ett material utsatt för en lastsekvens bestående av tre cykler. Använd Morrows ekvation (med hänsyn tagen till spänningens medelvärde) för att beräkna förväntat antal lastsekvenser till brott. Materialdata: E = 200 GPa, ν = 0,3, σ U = σ B = 700 MPa, Ψ = 0,65, σ f = 900 MPa, ε f = 0,26, b = 0,095, och c = 0,47. English: 8. The diagram shows three stabilized hysteresis loops for a material subjected to a load sequence composed of three cycles. Using the Morrow relationship, and taking the mean stress into account, determine the expected number of load sequences to fatigue failure. Material properties: E = 200 GPa, ν = 0.3, σ U = 700 MPa, Ψ = 0.65, σ f = 900 MPa, ε f = 0.26, b = 0.095, and c = 0.47. ösning/solution: The diagram gives three cycles: - One cycle with strain range ε = 0,0055, giving strain amplitude ε a = 0,00275. The stress mean value is σ m = (250 + ( 330)) /2 = 40 MPa. Thus, according to Morrow, one obtains ε a = σ f σ m E (2N)b +ε f (2N) c giving 0, 00275 = 900 ( 40) 200 000 (2N) 0,095 + 0, 26 (2N) 0,47 Solving for N gives N = 49 000 cycles (2N is load reversals to failure). - One cycle with strain range ε = 0,003, giving ε a = 0,0015. The stress mean value is σ m = 0. Then Morrow gives 0, 0015 = 900 0 200 000 (2N) 0,095 + 0, 26 (2N) 0,47 Solving for N gives N = 750 000 cycles. - One cycle with strain range ε = 0,0025, giving ε a = 0,00125. The stress mean value is σ m = 115 MPa. Morrow gives 0, 00125 = 900 + 115 200 000 (2N) 0,095 + 0, 26 (2N) 0,47 15
Solving for N gives N = 4 500 000 cycles Palmgren-Miner now gives 1 D = 49 000 + 1 750 000 + 1 4500 000 = 1 45 500 Expected number of sequences to failure is 45 500. 16