TABULATION OF CASES

2a t ; displacement (2) Increasing force ₁

Note: After u_m is reached in (3), a reverse cycle of motion begins.

C

ASE

3.2 B

ASIC

O

SCILLATOR

, T

RANSLATIONALOR

R

OTATIONAL (a) Symmetric triangular. Refer to text.

(b) Triangular increasing.

Maximum response takes place for t > t₀. The response is then the same as for (c) below.

(c) Triangular decreasing.

C

ASE

3.4 O

SCILLATOR

R

ESPONSETO

O

THER

C

OMPACT

P

ULSES

1. Solution limit for t_x → ∞: step loading with W0; u_m = 2u_st (Max. response) approximation for u_m (accuracy per tabulation below):

Note: The fi rst tabulated u_m/u_st is the true maximum value from the equation for u(t).

2

Note: For a further application of the concept of limiting load types see Chapter 13.

C

ASE

3.6 O

SCILLATOR

R

ESPONSETO

S

LOPE

-

AND

-S

TEP

P

ULSE

Initial velocity v₀ or l0 prescribed Translational:

0 0

P − M (force as a function of displacement) Rotational:

t C (Rotation as function of velocity) T_d(t) = C′λ(t) = 0 ²

(C) ( )t

T J

− ′ α (Torque as function of rotation)

Note: t₀₅ is the time needed for the initial velocity to drop by half.

C

ASE

3.8 D

AMPER

-O

NLY

O

SCILLATOR

. I

NITIAL

B

ASE

V

ELOCITYV₀

P

RESCRIBED

Note: Refer to Case 3.7 to get analogous expressions for rotational case.

v₀ M

C

u_b

u

C

ASE

3.9 D

AMPED

O

SCILLATOR

, T

RANSLATIONALOR

R

OTATIONAL Moderate damping, ζ < 1.0

= 2

1 χ ζ

− ζ ; damping parameter

st d d d Small damping, ζ << 1.0

u_m = [1 + exp(−πζ)]ust; at t₁ = π/ω = τ/2

(b) Exponential decay loading P = P₀ exp(−bt):

When the spring relaxes during reverse motion, mass M separates with a rebound velocity V.

Upper bound of spring force: P_sm = v₀ kM (when

(Note: This is accurate to within a few percent over a large range of k, M, and C.) Rebound velocity V and contact duration t₀:

Method A: 2 _{e 0}

=π ; contact duration for a marked rebound

where

Method B: Refer to Case 12.9. (A more general approach.)

Notes: Method A gives good results for a fi rm damper, when v_e << v₀. Otherwise, Method B may give better results. In those latter cases the rebound may be slower and, in effect, only the duration of the loading phase, lasting t₀/2, may be reasonably predicted by the above. Rebound velocities calculations are valid for a single-impact event only.

k M u₂ u₁

v₀ C

C

ASE

3.12 O

SCILLATORWITH

I

NITIAL

V

ELOCITYV₀

(

OR

l

₀

) H

AS

I

TS

B

ASE

S

UDDENLY

S

TOPPED

Equivalent to Giving the Fixed Oscillator in Case 3.2 a Short Impulse S = Mv₀ or S = Jl₀ Translational:

u_m = v₀/ω; at t = τ/4; vm = v₀; Rm=v0 kM Rotational:

α_m = λ₀/ω; at t = τ/4; λ_m = λ₀; Tm= λ0 KJ

C

ASE

3.13 D

AMPED

O

SCILLATORWITH

I

NITIAL

V

ELOCITYV₀

(

OR

l

₀

)

The base is suddenly stopped and held.

Moderate damping, ζ < 1.0

1 2

χ = ζ

− ζ ; damping parameter

0

Note: In general, the peak reaction may be larger than either the damper force (at the beginning) or the spring force (at maximum defl ection).

Intermediate damping, ζ ≤ 0.2 ^m = ⁰⎛⎜⎝ − ζ²⎞⎟⎠ ⎡⎢⎣⎛⎜⎝ ζ −π⎞⎟⎠ζ⎤⎥⎦

1/2

v exp 2

1 2

R Mk ; peak base reaction.

Small damping, ζ << 1.0

⎛−πζ⎞

C

ASE

3.14 S

IMPLE

O

SCILLATOR

, P

RESCRIBED

B

ASE

M

OVEMENT

(b) Short acceleration pulse or velocity step

0 st

u = ωv at t = τ/4 (peak relative displacement)

P_m = v0 kM (peak reaction) Note: Absolute values are shown.

C

ASE

3.15 D

AMPED

O

SCILLATOR

, P

RESCRIBED

B

ASE

M

OVEMENT

u

_b

(–

u = u − u_b

)

Small damping, ζ < < 1.0 Note: Absolute values are shown.

C

ASE

3.16 C

ANTILEVER

B

EAMASA

R

IGID

S

EGMENTWITHAN

A

NGULAR

S

PRING

M L (maximum spring moment) (b) Rectangular impulse S = W₀t₀.

(a) Short impulse S = W₀t₀, equivalent to initial velocity v₀ = S/M_e. u_m = v₀/ω; at t = τ/4; vm = v₀; b= ^{b m}

2K u

M L (maximum base spring moment)

= ^{c m}

c

4K u

M L (maximum center spring moment) (b) Rectangular impulse S = W₀t₀.

Equivalent center mass: M_e = M/3;

u_st = W0/k

(a) Short impulse S = W₀t₀, equivalent to initial velocity v₀ = S/M_e. u_m = v₀/ω; at t = τ/4; v_m = v₀; m= ^{c m}

4K u

M L (maximum spring moment) (b) Rectangular impulse S = W₀t₀.

C

ASE

3.19 S

UPPORTEDAND

C

LAMPED

B

EAM AS

T

WO

R

IGID

S

EGMENTS

The expressions for the response are the same as in Case 3.17.

C

ASE

3.20 C

LAMPED

–C

LAMPED

B

EAMWITH

O

NE

E

ND

G

UIDED

P

RESENTEDASA

M L (maximum spring moment) (b) Rectangular impulse S = W₀t₀.

(c) Step load W = W₀H(t)

Stiffness, peak displacement, peak reaction, maximum bending, respectively:

Stiffness, peak displacement, peak reaction, maximum bending, respectively:

C

ASE

3.24 M

ASSLESS

B

EAM

S

TRUCKBY

M

ASS

M

WITH

N

O

R

EBOUND

Stiffness, peak displacement, peak reaction, maximum bending, respectively, are

Stiffness, peak reaction, peak displacement, respectively:

Edges supported: 16₂ 1 3

C

ASE

3.28 M

ASSLESS

S

QUARE

P

LATE

S

TRUCKAT

M

IDPOINTBY

M

ASS

M

WITH

N

O

R

EBOUND

. S

TIFFNESS

, P

EAK

R

EACTION

, P

EAK

D

ISPLACEMENT

, R

ESPECTIVELY

a v₀

E, h M

Edges supported: 86.18D₂

k= a ; Rm =v0 kM; u_m = R_m/k; (at t = τ/4)

Edges clamped: 178.7D₂

k= a ; Rm =v0 kM; u_m = R_m/k; (at t = τ/4)

C

ASE

3.29 C

IRCULAR

, M

ODERATELY

T

HICK

R

ING

, P

RESSURIZED

E, A, m h

w(t)

R a

u

Static stress, strain, radial defl ection (at R), and stretching work due to static load w₀ (N/m):

σ =st w a⁰ ;

A ε =st w a⁰ ;

EA st= ⁰ u w aR

EA ; = πR( 0 )² EA w a L

(a) Short impulse S = w₀t₀:

⎛ ⎞

= ω = ⎜⎝ ⎟⎠

1/2 0

m 0

v v m

u R

EA at t = τ/4

(b) Step load, w = w₀H(t)

u_m = 2u_st at t = τ/2

C

ASE

3.30 C

YLINDRICAL

, M

ODERATELY

T

HICK

S

HELL

, S

UDDENLY

P

RESSURIZED

(Refer to the illustration in Case 3.29)

Static stress, strain, radial defl ection (at R), and stretching work due to static pressure p₀:

σ = ε = = ″ = = π

Note: The equations pertain to a unit-long segment of shell.

C

ASE

3.31 S

PHERICAL

S

HELL

, M

ODERATELY

T

HICK

, P

RESSURIZED

(Refer to the illustration in Case 3.29)

Static stress, strain, radial defl ection (at R), and stretching work due to static pressure p₀:

σ = ε = = ′ = = π

= + ν ; radial displacement at distance r

u pa1

Note: Refer to Appendix D.

C

ASE

3.33 S

PHERICAL

C

AVITY

, I

NFINITE

E

LASTIC

B

ODY

(Refer to the illustration in Case 3.32)

3

= + ν ; radial displacement at distance r

= + ν

p r r radial and hoop stress, respectively, positive under tension Note: Refer to Appendix D.

C

ASE

3.34 R

OTATIONAL

O

SCILLATOR

, S

UBJECTEDTO

P

ERIODIC

F

ORCING

τζ ; peak rotation after many cycles, in presence of damping.

Note: This is equally valid for a translational oscillator. See text for details and also Ref. [44].

T

K J α

C

ASE

3.35 R

OTATIONAL

O

SCILLATOR

, S

UBJECTEDTO

C

YCLIC

F

ORCING WITH

R

EPEAT

P

ERIODOF

t

_P

S S₀

t₁ t_p t_p

S₀ S₀ S₀

t

Short angular impulse S₀.

0 0

S

λ = J ; initial velocity induced by S₀

α1 = λ0/ω; peak rotation after fi rst application of S0

t₁ is a multiple of a τ + 3τ/4; tp is a multiple of τ

α_m = nα₁; peak rotation after n cycles (no damping) Ref. [44]

Note: This is equally valid for a translational oscillator. Time points t₁ and t_p are selected to maximize the response.

C

ASE

3.36 B

ASIC

D

AMPED

O

SCILLATOR

, S

UBJECTEDTO

S

INUSOIDAL

F

ORCING AT

N

ATURAL

F

REQUENCY

, t

_P

= t

W

W₀ t

τ τ

u_st = W₀/k; static defl ection under W₀

st

m 2

u =u

ζ; upper defl ection limit after large number of half-sine pulses

u_t ≈ u_m(1 − exp(−ζωt)); amplitude after p-pulses, or after t = p(τ/2) or ωt = πp; p is odd number

≈ ω st= m

2

u tu u ; approximation for a small number of applied cycles

= π τ

0 0

S 2W ; total impulse in one full sine cycle for this harmonic forcing S₁; impulse during one period applied by a periodic, nonsinusoidal pulse

1

t m

0

(1 exp( ))

u S u t

≈ S − −ζω ; amplitude after p-pulses for the above impulse

Note: This is equally valid for a rotational oscillator. The above harmonic forcing pattern is regarded as a continuous sequence of sine impulses. The direction of force as well as of movement reverses after each half-period.

C

ASE

3.37 R

OTATIONAL

D

AMPED

O

SCILLATOR

, S

UBJECTEDTO

P

ERIODIC

, R

ECTANGULAR

P

ULSES

The period of forcing t_p = nτ is a multiple of the natural

t₀ = m(τ/2); load duration, an odd multiple of half-period.

t₁ = p(τ/2) = tp − t₀; no load, an odd multiple of half-period.

m st

⎝ ⎠; close approximation

Note: This is equally valid for a translational oscillator [49].

C

ASE

3.38 B

ASIC

O

SCILLATOR

, I

NITIAL

S

TATIC

L

OAD

W

_st

F

OLLOWEDBYA

u₁ is the maximum displacement from this fi rst impulse alone u₀ = W₀/k; static defl ection from sustained load W₀ alone (a) Step load W₀(t₀ > τ/2)

u(t) = u₀(1 − cos ωt) + v¹ sin tω

ω ; resultant defl ection

(b) Initial impulse followed by a rectangular pulse

um ≈

(

up²+u1^{2 1/2}

)

; where u_p is the effect of the rectangular pulse acting alone.

Note: The second equality in the fi rst subcase is useful for more general systems acting in a SDOF mode. The term v₁/ω is the maximum displacement attained as a result of the initial impulse acting alone.

EXAMPLES

E

XAMPLE

3.1 M

ASS

M S

UPPORTEDBY

T

WO

C

OLUMNSAND

I

MPACTED

L

ATERALLY

Treating the beams as massless, fi nd the peak defl ection and bending stress for W = 20 kN applied in the form of a rectangular pulse lasting longer than the natural period.

The “universal beam” or the I-section is used, with the stronger direction opposing defl ection. The depth of beam is 612 mm while I = 986 × 10⁶ mm⁴, H = 3,000 mm, and E = 200,000 MPa.

---Case 3.2 shows that maximum defl ection can be u_m = 2u_st for an impulse lasting that long.

To fi nd stiffness, refer to Case 2.16:

× × ×

Static defl ection: u_st = W/k = 20,000/175,290 = 0.1141 mm Peak dynamic defl ection: u_m = 2u_st = 0.2282 mm

Both ends of the beam are fi xed. It may be found in any reference text, that in this case the maximum moment is M = QL/2, where Q is the maximum lateral force per beam. The peak dynamic internal loading may be recovered from the peak defl ection: Q_m = ku_m/2 = 175,290 × 0.2288/2 ≈ 20,000 N (per beam)

(It is quite apparent here that the dynamic loading per beam would be twice the static load-ing. It is not always so clear in practical applications.)

M_m = 20,000 × 3000/2 = 30 × 10⁶ N-mm.

Peak bending stress: σ = M_mc/I = 30 × 10⁶ (612/2)/986 × 10⁶ = 9.31 MPa.

A shear check is desirable as well.

E

XAMPLE

3.2 A T

ORQUE

S

UDDENLY

A

PPLIEDTOA

D

ISK

The shaft-disk arrangement is as shown in Case 3.2. The torque is a triangular, decreasing pulse, as in Case 3.2d with the maximum of T₀. The torque duration is only 0.1τ. Estimate the peak torque experienced by the shaft, as a multiple of T₀.

---Due to a short application time the pulse may be treated as rectangular with an impulse of S = T₀t₀/2, equivalent to T₀/2 applied for t₀. From Case 3.2c, the maximum angle of

where u_st is the angle of rotation induced by the static torque T₀/2. The peak shaft torque can therefore be estimated as T_m = 0.618T₀/2 = 0.309T₀. The exact solution presented in Harris and Crede [30, pp. 8–34], clearly shows that for this ratio of t₀/τ the approximation is suf-fi ciently accurate.

E

XAMPLE

3.3 S

UDDEN

R

EMOVAL OF

S

UPPORT

Weight W = 1000 N is suspended on a cable so that the spring below the weight is unstressed. If the cable suddenly breaks, what will be the maximum downward defl ection and maximum velocity attained?

How high will the weight go after the rebound? Use k = 100 N/mm.

---The mass is W/g = 1000/9.81 = 101.94 kg. Static displacement due to W is u_st = 1000/100 = 10 mm. The cable breaking is equivalent to a sudden application of the force of gravity.

From Case 3.2d with W as step load one gets u_m = 2u_st = 20 mm

m st st

v 100 10 0.3132 mm/ms

101,940

u k u

= ω = M = ⋅ = ; attained at midpoint of downward travel.

After reaching peak downward defl ection the weight returns to its original position, as may be inferred from the energy balance.

E

XAMPLE

3.4 S

UDDEN

C

HANGEOF

D

IRECTIONOF

M

OVEMENT

The fi gure shows a single-axis trailer moving along the road, about to run over a discon-tinuity in the form of slope change. Assuming the vertical axis of the assembly to remain vertical after entering the slope, while forward velocity v₀ remains parallel to the surface, fi nd the peak spring force and the peak damper force applied to the mass M:

v₀ = 12 m/s, M = 700 kg, k = 50 N/mm, ζ = 0.3.

M k C v₀

15 3

(a)

α vα₀ Δ

v₀ (b)

---The velocity diagram in (b) shows that the change in the vertical direction is Δv ≈ v0 sin α = 12 × 0.1961 = 2.353 m/s

k u g

Cable M

This velocity increase is acquired instantly and sustained, which means Case 3.15b, mod-erate damping, is applicable. (Except that v₀ in 3.15 is the same as Δv here). The dynamic parameters are

ω = (50/700,000)^1/2 = 8.452 × 10⁻³ kHz C_c = 2Mω = 11,833; critical damping per Equation 3.3

C = ζC_c = 3550 N-ms/mm; ωd = ω(1 − ζ²)^1/2 = 8.063 Hz

sin exp( ) sin 72.54 exp( 0.3982)

8.063 10

A weightless block separating the damper (C = 47.54 N-s/mm) and the spring (k = 100 N/mm) can slide without friction. The spring is preloaded and the system is motionless. At t = 0 both preloading forces P₀ = 1164 N are released and mass M = 2.1 g is pushed to the right.

When the force compressing the spring drops to nil, the mass separates from the spring.

Determine peak velocity v_f with which M fl ies off by assuming that one-half of the stored energy is lost due to damper action in the expansion phase.

---An important thing to realize here is that after the release, the spring and the damper have the same force, as the connecting block is massless. Suppose the damper is rigid, i.e., the spring is grounded. Equating the strain energy in the spring, P₀² /(2k), with the kinetic energy of the moving mass, one has the fl y-off velocity as

2 2

However, only one-half of the stored energy is recovered and converted to the kinetic energy because of the energy loss in the damper during expansion. In the equation above, the right side representing the strain energy has to be reduced by 1/2. Therefore, the fl y-off velocity may be estimated as

2 separate such events are recorded, each giving a pulse that can be simplifi ed to a decreas-ing triangular shape. (Refer to Case 3.3c). The fi rst event gave p₀ = 3.64 MPa and impulse S = 72.41 kPa-ms, while for the second the quantities are 3.22 and 54.51, respectively.

Determine the expected peak strain. The shell material data: E = 82,760 MPa; v = 0.3 while density is uncertain. It should be recovered from the experimental observation that the natural frequency of radial motion is, on average, 2225 Hz. (The test results were reported by Johnson [40].)

Natural frequency f = ω/(2π) from Case 2.46:

ω ′ ×

ρ = 0.01481 g/mm³. (Note that the thickness term in f was negligible). The density being so high (14.81 g/cm³) is indicative of some additional, nonstructural material lining the shell.

The period is τ = 1/f = 0.4494 ms. To estimate impulse duration, assume it has a triangular decreasing shape (Chapter 13). Then S = p₀t₀/2 or t₀ = 2S/p₀.

Here t₀ = 2 × 0.07241/3.64 = 0.0398 ms. (Impulse must be expressed in MPa-ms here, to have consistent units.) Noting that t₀ < 0.1τ, the impulse is decisively short, the shape of the pulse does not matter much and Case 3.2c may be invoked to fi nd peak dynamic displacement:

0

m st

0.0398

2 sin 2 0.0968sin 0.05318 mm

0.4494

u = u πt = × π =

τ

(This is smaller than static displacement under the peak load.) The peak hoop strain is εh = u_m/R

εh = 0.05318/285.8 = 186 × 10⁻⁶.

Two test results are quoted in Ref. [40]; 168 × 10⁻⁶ and 187 × 10⁻⁶, but it seems uncertain which one belongs to which event.

E

XAMPLE

3.7 A

MPLITUDE

G

ROWTH DURING

R

ESONANCE

The magnitude of force W₀ applied to an oscillator is such that it induces the static displace-ment u_st = 10 mm. If this force is applied with the natural frequency of the oscillator, fi nd the amplitude of motion reached after two cycles and then after 20 cycles. Damping coef-fi cient is ζ = 0.04.

---Apply Equation 3.35a, without the oscillating part: ^m≈ ζ^st(1 exp− (−ωζ)) 2

u u t after two cycles t = 2τ = 4π/ω, therefore ωt = 4π and after 20 cycles ωt = 40π.

Peak displacement after two cycles: ^m = ×^st (1 exp 4 0.04))− (− π =4.94 ^st 2 0.04

u u u = 49.4 mm

Similarly, for 20 cycles: u_st = 124.2 mm.

Note that the upper limit of the amplitude is ^m = ζ^st = ×10

2 2 0.04

u u = 125 mm

To test the accuracy of a further approximation expressed by Equation 3.37, namely

stcos 2

u≈ωtu ωt

use the above for one half-cycle only, namely for a half-sine pulse. Then ωt = π and

m st 1.571 st

u ≈2πu = u

According to Table 3.1, DF(u) for a half-sine pulse is 1.77, therefore the accuracy of the above approach for a one-half cycle is moderate.

E

XAMPLE

3.8 R

ESPONSETO A

G

ROUND

S

HOCK

In a velocity record of an earthquake with a strong vertical component, the bell-shaped curve described by

0 b

v v (1 cos )

2 t

= − Ω ; with Ωt0 = 2π when the pulse ends

was identifi ed in vertical ground movement. If a building is idealized to an oscillator, fi nd the peak relative (ground-to-mass) acceleration that can take place.

---v₀

t t₀

Equation 3.40 tells us that if the relative motion only is of interest, then the ground accel-eration is the same as the relative mass-ground accelaccel-eration, except for the sign. When the expression for v_b is differentiated, the acceleration is obtained as

0 b

v sin

a =Ω2 Ωt; which has the peak of Ωv₀/2.

This represents a full sine wave during Ωt₀ = 2π with the force amplitude:

b b 0

1 v

W =Ma =2MΩ

The relative static displacement, which this force can induce, is simply W_b/k. To maximize response, we set Ω = ω:

Because of small number of cycles involved (one full cycle or two half-sine impacts) one can use Equation 3.37, with ωt = 2π, to obtain peak amplitude:

0 0

m st

2 v v

2 2 2 2

u =ωtu = π ω = π ω

For the sake of comparison note that in Case 3.12, when a sudden application of velocity v₀ takes place, one obtains u_m = v₀/ω. In this problem velocity is applied smoothly, but the pulse shape is such that it results in a larger response.

E

XAMPLE

3.9 A S

TEAM

T

URBINEWITHA

P

ARTIAL

A

DMISSION

Such a turbine has its blades subjected to the driving steam force only on a portion of its circumference. The fact that during each rotation the blades become loaded and unloaded again may set up a resonant condition. Let a blade of such turbine have 40 vibratory cycles in one revolution. The steam is acting during just a little more than a half of the revolution.

For the fundamental period, the damping level is ζ = 0.001. The steam force is of such a magnitude, that when statically applied it induces a 0.1 mm defl ection. Find the peak defl ec-tion under the described condiec-tions.

---The above tells us that the steam forces act during over 20τ, say 20.5τ, while there are no forces during the remainder of one revolution, the latter being the period of forcing.

Referring to Case 3.37, which, while formulated for angles of rotation, applies to transla-tions as well. It tells us that the relevant multiples are: n = 40, m = 41, and p = 39. We have u_st = 0.1 mm and

πζp = π0.001 × 39 = 0.1225, πζm = 0.1288, and 2πζn = 0.2513 Substitution gives

0.1225

st 2 st 0.2513 st st

1 1

8.482 ; 8.457

1 1

p n

e e

b u u u d u

e e

−πζ −

− πζ −

+ +

= = = =

− −

= + + =

m 1 (1 8.482 8.457) st 8.97 st

u 2 u u ; average maximum amplitude

u_m ≈ u_st ⎛⎜⎝ +πζ ⎞⎟⎠ = ^st⎛⎜⎝ +π × ⎞⎟⎠

1 1

1 1

0.001 40

n u = 8.958u_st; approximation of the above The approximate expression gives a very close result.

95

In document Formulas for Mechanical and Structural Shock and Impact BBS (Page 88-116)