Static Equivalent of Miles’ Equation

Abstract

The replacement of dynamic random response analysis by a simple approximate quasi-static analysis in combination with Miles’ equation is discussed in this chapter. Equivalent static acceleration and force fields are considered. A procedure to perform an equivalent finite element (stress) analysis is presented. Examples are given to illustrate this equivalent quasi-static approach. Problems with solutions are provided to gain more insight in using Miles’ equation in quasi-static applications.

Problems

3.1

Prove with the aid of Fig. 3.10 that the natural frequency $f_n$ (Hz) of the SDOF system is given by

$$\begin{aligned} f_n=\frac{1}{2 \pi } \sqrt{\frac{1}{\delta _{F=1} m}}. \end{aligned}$$

(3.24)

Fig. 3.10

Static displacement of SDOF system

3.2

Calculate the first and second moment of area $W_{xx}$ and $I_{xx}$, respectively, around xx axis and center of gravity of the T-section in Fig. 3.3, neglecting higher-order terms of the thickness t.

Answers: $0.3333b^2t, ~0.1389b^3t$

3.3

Calculate the overall SPL of the sound pressures given in Table 3.5.

Answer: OASPL $=$ 146 dB

3.4

The fixed-pinned beam is shown in Fig. 3.3, but in addition the beam is simply supported at point B. The material and geometrical properties are given in Table 3.4, and the PSD function $W_p(f)$ is presented in Table 3.5. The modal damping ratio is $\zeta =0.05$.

• 1. Calculate for the fixed-pinned beam the reaction shear forces and bending moment in point A and B; $D_A$, $M_A$ and $D_B$, respectively.

• 2. Calculate with theoretical formula first natural frequency $f_n$ of fixed-pinned beam and associated PSD value $W_p(f_n)$.

• 3. Calculate with the equivalent force field method $D_{A,rms}$, $M_{A,rms}$, $D_{B,rms}$.

• 4. Calculate with the FEA² the first two natural frequencies $fn_i,~i=1,2$ and the r.m.s. of the reaction forces and bending moment $D_{A,rms}$, $D_{B,rms}$, and $M_{A,rms}$.

Answers:

• 1. $D_A=5/8 pb L_b$, $M_A=1/8 pb L_b^2$, and $D_B=3/8 pb L_b$.

• 2. $f_n=610.74$ Hz, $W_p(f_n)=71.38$ Pa$^2$/Hz.

• 3. $D_{A,rms}=361.42$ N, $M_{A,rms}=90.35$ Nm, $D_{B,rms}=216.85$ N.

• 4. $f_n=610.75, ~1,979.5$ Hz, $D_{A,rms}=261.53$ N, $D_{B,rms}=179.27$ N, $M_{A,rms}=82.36$ Nm.

3.5

A fixed-free (cantilevered) beam with tip mass is illustrated in Fig. 3.11. The bending stiffness of the massless beam is EI (Nm$^2$), and the length is L (m). The tip mass is m (kg), and the area of the massless plate welded to the mass is $A_p$ (m$^2$). The amplification factor is Q. The PSD of the pressure is $W_p$ and is constant in the frequency interval of interest. Calculate the natural frequency $f_n$ (Hz) and the r.m.s. values of the displacement X, the bending moment $M_A$, and the shear force $D_A$ in accordance with the flowchart given in Fig. 3.7.

Answers: $f_n=\sqrt{3EI/mL^3}$, $x_{rms}=A_pL^3/3EI \times \sqrt{(\pi /2) f_nQW_p}$, $M_{A,rms}=A_pL\times \sqrt{(\pi /2) f_nQW_p}$, $D_{A,rms}=A_p \times \sqrt{(\pi /2) f_nQW_p}$

Fig. 3.11

Fixed-free beam with tip mass

3.6

For the circular simply supported plate, the radial and tangential bending moments $M_r$ and $M_t$ are calculated when the plate is enforced random accelerated along the rim with a constant single-sided PSD $W_{\ddot{U}}$ (g$^2$/Hz). The theory of plates and shell and nomenclature are taken from the famous book of Timoshenko and Krieger [11].

The bending stiffness of the plate is given by D, r is the running radius, the outer radius is a, the mass per unit of area is $m_p$, and Poisson’s ratio is $\nu$.

The random dynamic response analyses is performed applying the assumed modes method [12]. There are two different assumed modes considered:

• The static deflection mode $\varPhi _1(r)=A(a^2-r^2)(\frac{5+\nu }{1+\nu } a^2-r^2)$ taken from [11].

• The assumed mode taken from [13]; $\varPhi _2(r)=B(4+\nu )a^3-3(2+\nu )ar^2+2(1+\nu )r^3$.

Both assumed modes fulfill the boundary condition $\varPhi _1(a)=\varPhi _2(a)=0$, and the shapes are similar to the first vibration mode of the plate [4].

Evaluate the natural frequency corresponding to both assumed modes with the aid of Rayleigh quotient.

Answers: $\omega _{n,1}^2=24.92 D/(m a^4)$, $\omega _{n,2}^2=24.89 D/(m a^4)$.

The assumed modes shall fulfill the orthogonally relation $\int _A \varPhi ^2 dA=1$. Define the scaling constants A and B.

Answers: $A=0.2120/(\pi a^{10}\,m)$, $B=0.1950/(\pi a^{8}\,m)$

Calculate the radial bending moment $M_r(r=0)$ corresponding to the assumed modes and with unit acceleration $\ddot{u}=1$ g or unit pressure $\text {p}=1$ Pa.

Answers: Enforced acceleration: $M_r(r=0)=-13.32m\ddot{u}\big |_{=1} a^2 A D$, $M_r(r=0)=-18.59m\ddot{u}\big |_{=1} a B D$, pressure field: $M_r(r=0)=-13.32p\big |_{=1} a^2 A D$, $M_r(r=0)=-18.59p \big |_{=1} a B D$

The bending stiffness of the circular sandwich panel is given by $0.5 E h^2 t$, where Young’s modulus of the face sheets is $E=70$ GPa, and the thickness of the face sheets is $t=0.5$ mm. The lowest natural frequency of the circular plate is $f_n=75$ Hz. The diameter of the circular plate is $d=1$ m. The PSD of the enforced acceleration is $W_{\ddot{U}}=0.04$ g$^2$/Hz in between 20–200 Hz. The PSD of the pressure field is $W_p=100$ Pa$^2$/Hz in between 20–200 Hz. Define the core height h considering both assumed modes. Calculate for both the enforced random acceleration and random pressure field and both assumed modes the r.m.s. radial bending stress in the face sheets at $r=0$. The bending resistance is given by $W_b=I/0.5h=ht$.

Fig. 3.12

Simply (pinned)-spring supported beam [14]

3.7

A simply spring supported beam is illustrated in Fig. 3.12. The beam is loaded by a random force F with a PSD $W_F(f)$ (N$^2$/Hz). Calculate the r.m.s. displacement $\delta _{rms}$ at location B.

The solution can be achieved by performing the following steps:

• Calculate the static displacement $\delta$ for $F=1$ N using the Myosotis equations [15].

• Calculate with Rayleigh’s quotient the approximation of the natural frequency $f_n$, when the assumed mode is $y(x)=\sin \left( \frac{\pi x}{2L}\right)$.

• Calculate $\delta _{rms}$ using Miles’ equation Eq. (3.14).

Answers: $\delta |_{F=1}=-\frac{L^3 \alpha }{3EI}\left[ \alpha ^3-2\alpha ^2 +\alpha (1+\gamma ) \right]$, $\alpha =\frac{x_0}{L}$, $\gamma =\frac{3EI}{kL^3}$, $f_n=\frac{\sqrt{32 k{{L}^{3}}+{{\pi }^{4}}EI}}{8 \pi \sqrt{m}{{L}^{2}}}$

The second part of this problem is a numerical evaluation:

• The length of the Al-alloy beam is $L=1$ m, Young’s modulus $E=70$ GPa, the density $\rho =2,700$ kg/m$^3$, and the spring stiffness $k=\frac{3EI}{L^3}$ (N/m) ($\gamma =1$). Define a preliminary cross section (h, t) of the beam such that the lowest natural frequency is $100 \le f_n \le 120$ Hz. Use the approximate equation for $f_n$.

• Repeat the modal analysis with your favorite FEA package and generate a FEM consisting of ten simple (Hermitian) beam elements. Define a more accurate cross section (h, t) to achieve the lowest natural frequency objective.

• Check with your FEM the displacement $\delta |_{F=1}$ for $\alpha =0.6$.

Answers: $A= \frac{9}{4}ht$, $I\approx \frac{1}{3}h^3t$, $h=100$ mm, $t=2$ mm, $f_n=108.4461$ Hz, $h=120$ mm, $t=2.4$ mm, $f_n=109.1148$ Hz, $\delta |_{F=1}=1.4385 \times 10^{-6}$ versus $1.4385 \times 10^{-6}$ m (FEM)

3.8

Evaluate the strong and weak points of the flowchart of the equivalent static approach in conjunction with Miles’ equation as presented in Fig. 3.7.