Linköpings Universitet, Hållfasthetslära, IEI/IKP Tore Dahlberg ...

Linköpings Universitet, Hållfasthetslära, IEI/IKP
Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, TMMI09
EXAMINATION in Mechanical Vibrations and Fatigue
2007-10-16 kl 14-18
L Ö S N I N G A R ---- SOLUTIONS
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.
Lösning:
ω=2πf and f = 1 T
2.
Förklara vad en egenfrekvens är.
English: Explain what en eigenfrequency is.
Lösning/Solution:
En frekvens ett system svänger med om systemet lämnas att svänga fritt.
A frequency a system will vibrate with if it is left to vibrate freely.
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Tekniska Högskolan i Linköping, IEI/IKP
Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, 071016 kl 14-18
EXAMINATION in Mechanical Vibrations and Fatigue
DEL 1 - (Teoridel utan hjälpmedel)
3.
spänning/stress (MPa) Hur många cykler erhålls ur vidstående
svängningssekvens? Ange medelvärde och
300
amplitud för varje cykel. Använd rain-flow
200
count-metoden
English: How many cycles is obtained from the
100
sequence given? Give mean value and
amplitude of each cycle. Use the rain-flow
0
tid/time count method.
Lösning/Solution:
There will be four cycles:
two cycles with stress σ=150 ± 150 MPa,
one cycle with stress σ=150 ± 50 MPa, and
one cycle with σ=100 ± 100 MPa.
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.
Lösning/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 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.
σ ∞
9

Tekniska Högskolan i Linköping, IEI/IKP
Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, 071016 kl 14-18
EXAMINATION in Mechanical Vibrations and Fatigue
DEL 2 - (Problemdel med hjälpmedel)
5. En massa M hängs upp i tre fjädrar (styvhet
k
k 1 = k, k 2 =2k och k 3 =3k).
1 k 2 k1
k (a) Bestäm egenvinkelfrekvensen för systemet
M
3
k2
om massan och fjädrarna monteras enligt figur
k3
M
(a).
(a)
(b) (b) Vad blir egenvinkelfrekvensen om massan
och fjädrarna monteras enligt figur (b)?
English: 5. A mass M is mounted with three springs (stiffness k 1 = k, k 2 =2k,
and och k 3 =3k).
(a) Determine the (angular) eigenfrequency of the system if the mass and the
springs are mounted as shown in figure (a).
(b) What will the eigenfrequency be if mass and springs are mounted as in
figure (b).
Lösning/Solution:
(a) The equation of motion of the mass is Mẍ =−F 1
− F 2
+ F 3
(a)
where
This gives
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
F 1
= k 1
x F 2
= k 2
x and F 3
=−k 3
x
Mẍ +(k 1
+ k 2
+ k 3
)x = 0
ω e
= √⎯⎯⎯⎯
(k 1 + k 2 + k 3 )
= √⎯6 √⎯ k M
M = 2.45 √⎯ 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 1 k 2
k 1 + k 2
+ k 3
⎞ ⎟⎠ x = 0
ω e
= √⎯⎯⎯⎯⎯
⎛ k 1 k 2
⎜
⎝ (k 1 + k 2 ) + k 3
Thus, the eigenfrequency in case (b) goes down, but not that much, because
the stiffest spring k 3 will dominate the behaviour of the mass in both cases.
⎞
⎟
⎠
(b)
1
M = √⎯ 11
3 √⎯ k M = 1.91 √⎯ k M
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Tekniska Högskolan i Linköping, IEI/IKP/Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, 071012 kl 14-18
EXAMINATION in Mechanical Vibrations and Fatigue
DEL 2 - (Problemdel med hjälpmedel)
6.
Härled rörelseekvationen för en dämpad pendel
l
l
enligt figur (massa M, längd L, gravitation g,
dämpkonstant c). Pendeln gör små svängningar
c
runt sitt vertikala jämviktsläge.
M
Mg
English: Deduce the equation of motion of a damped pendulum according to
figure (mass M, length L, gravitation g, damping constant c). The pendulum
performs small oscillations around its vertical equilibrium position.
Lösning/Solution:
The pendulum is subjected to two forces: the gravity force Mg and the
damping force S d = c ⋅ l ˙φ . The equation of motion, J ¨φ=∑moment, becomes
J ¨φ=−Mg sinφ−c ⋅ l ˙φ
Using J = Ml 2 and sin φ = φ give
Ml 2 ¨φ+c ⋅ l ˙φ+Mg φ=0
which is the equation asked for.
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Tekniska Högskolan i Linköping, IEI/IKP
Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, 071016 kl 14-18
EXAMINATION in Mechanical Vibrations and Fatigue
DEL 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.
Lösning/Solution:
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).
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Tekniska Högskolan i Linköping, IEI/IKP Tore Dahlberg
TENTAMEN i Mekaniska svängningar och utmattning, 071012 kl 14-18
EXAMINATION in Mechanical Vibrations and Fatigue
DEL 2 - (Problemdel med hjälpmedel)
8. Ett material utsätts för en lastsekvens som
töjning/ strain
ger töjningar enligt figur. Använd Morrows
0,003
ekvation för att beräkna förväntat antal
0,001
lastsekvenser till brott. Inverkan av spänningens
tid/time medelvärde får försummas.
Materialdata: E = 200 GPa, ν = 0,3, σ U = σ B =
- 0,0025
700 MPa, Ψ = 0,65, σ’ f = 900 MPa, ε’ f = 0,26,
b = −0,095, och c = −0,47.
English: 8. A material is subjected to a load sequence giving the strain shown
in the figure. Use the Morrow relationship to determine the expected number
of load sequences to fatigue failure. The influence of the mean stress can be
neglected. 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.
Lösning/Solution: The diagram gives four cycles:
- One cycle with strain range Δε = 0,0055, giving strain amplitude ε a =
0,00275. According to Morrow one obtains (neglecting the mean stress)
ε a
= ⎛ σ’ f ⎞
⎜ ⎟ ⎝ E ⎠ ⋅(2N)b +ε’ f
⋅(2N) c
giving
0, 00275 = ⎛ 900 ⎞
⎜ ⎟ ⎝ 200 000⎠ ⋅(2N)− 0,095 + 0, 26 ⋅(2N) − 0,47
Solving for N gives N = 44 550 cycles (2N is load reversals to failure).
- One cycle with strain range Δε = 0,0025, giving ε a = 0,00125. Morrow gives
0, 00125 = ⎛ 900 ⎞
⎜ ⎟ ⎝ 200 000⎠ ⋅(2N)− 0,095 + 0, 26 ⋅(2N) − 0,47
Solving for N gives N = 2 180 000 cycles
- Two cycles with strain range Δε = 0,003, giving ε a = 0,0015. Morrow gives
0, 0015 = ⎛ 900 ⎞
⎜ ⎟ ⎝ 200 000⎠ ⋅(2N)− 0,095 + 0, 26 ⋅(2N) − 0,47
Solving for N gives N = 720 000 cycles.
Palmgren-Miner now gives
1
D =
44 550 + 2
720 000 + 1
2180 000 = 1
39 000
Expected number of sequences to failure is 39 000.
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