Question 8.1: A helicopter manufacturer is developing a code of practice f......

Part (a)
From Table 8.1, for the first normal bending mode of a simply supported uniform beam:

where ω_{1} is the natural frequency in rad/s, and E,  I,  L,  and  μ are given above.

Also from Table 8.1, the shape of the first mode, with i = 1, is given by y_{1} = \sin \beta_{1}x.

Since \beta_{1} = \pi/L,   \mathrm{then}  y_{1} = \sin (\pi x / L).   \mathrm{Defining}   y_{1} = y / y_{c} where y is the actual displacement at distance x along the pipe and y_{c} is the maximum displacement at the half-span position, then:

which is sketched in Fig. 8.1 at (b). The slope, dy / dx, and the curvature, d²y / dx², are given by differentiating Eq. (B) twice with respect to x:

These are sketched in Fig. 8.1 at (c) and (d).

We can now derive an equivalent system to describe the vibration of the pipe, lumping the mass, stiffness and damping at the center of the pipe, where the displacement relative to the base is y_{c}. With the base fixed, the kinetic energy, T, in the pipe is

T = \frac{1}{2} \mu \int_{0}^{L}{\dot{y}^{2}}dx                   (E)

Now differentiating Eq. (B) with respect to time:

\dot{y} = \dot{y}_{c} \sin \frac{\pi x}{L}                   (F)

which gives the vertical velocity at every point on the pipe.

Therefore:

T = \frac{1}{2} \mu \dot{y}_{c}^2 \int_{0}^{L}{\sin^{2} \frac{\pi x}{L}}dx = \frac{1}{4} \mu L \dot{y}_{c}^{2}                   (G)

This must be equal to the kinetic energy of the equivalent mass, \underline{m}, which is

The equivalent stiffness, \underline{k}, could be found in a similar way by equating the potential energy U in the pipe to that of the equivalent spring stiffness, \underline{k}, but a short cut is to use the fact that we already know the first natural frequency, ω_{1}, from Eq. (A):

The pipe with damping and base motion can now be represented by the lumped model shown at Fig. 8.1(e). This is seen to be the same as that of the system shown in Fig. 2.3, and the equation of motion is given by Eq. (2.4). In the present notation this is

{m} \ddot{y} +{c} \dot{y} + {k} y = – {m} \ddot{x}                             (2.4)

\underline{m} \ddot{y}_{c} + \underline{c} \dot{y}_{c} + \underline{k} y_{c} = – \underline{m} \ddot{x}_{B}                             (K)

where \ddot{x}_{B} is the base acceleration input, in this case the acceleration at both pipe attachments.
The modulus of the frequency response function \left|y_{c}\right| / \left|x_{B}\right| of the equivalent system represented by Eq. (K) was given as Eq. (4.26):

and

ω is the excitation frequency in rad/s, ω_{1} the undamped natural frequency in rad/s, f the excitation frequency in Hz, f_{1} the undamped natural frequency in Hz and γ the viscous damping coefficient.
Also, from Eq. (4.32), but using the present notation:

\frac{\left|y\right| }{\left|\dot{x} \right| } = \frac{1}{\omega } \cdot \left(\frac{\left|y\right| }{\left|x\right| } \right) , \frac{\left|y\right| }{\left|\ddot{x} \right| } = \frac{1}{\omega^{2}} \cdot \left(\frac{\left|y\right| }{\left|x\right| } \right) , \frac{\left|\dot{y} \right| }{\left|x \right| } =\omega \cdot \left(\frac{\left|y\right| }{\left|x\right| } \right) , \frac{\left|\ddot{y} \right| }{\left|x \right| } =\omega^{2} \cdot \left(\frac{\left|y\right| }{\left|x\right| } \right)                          (4.32)

\frac{\left|y_{c}\right|}{\left|\ddot{x} _{B}\right|} = \frac{1}{\omega ^{2}} \cdot \left(\frac{\left|y_{c}\right|}{\left|x_{B}\right|}\right)                          (M)

From Eqs (L) and (M):

\frac{\left|y_{c}\right|}{\left|\ddot{x} _{B}\right|} = \frac{\Omega ^{2}}{\omega ^{2}\sqrt{(1 – \Omega^{2})^{2} + (2\gamma \Omega )^{2}} } = \frac{1}{\omega_{1}^{2}\sqrt{(1 – \Omega^{2})^{2} + (2\gamma \Omega )^{2}}}                 (N)

Equation (N) now gives the magnitude of the displacement, \left|y_{c}\right| at the center of the pipe in terms of the base acceleration magnitude, \left|\ddot{x} _{B}\right|. The maximum value of \left|y_{c}\right| will occur, for practical purposes, when the excitation frequency coincides with the natural frequency, i.e. when Ω = 1 . Therefore:

\left|y_{c}\right| _{MAX} = \frac{\left|\ddot{x} _{B}\right|}{2\gamma \omega^{2}_{1}}                   (O)

The natural frequency ω_{1}, in rad/s, is given by Eq. (A). Inserting numerical values:

Referring to the vibration specification, the required input at this frequency is ± 5g, so \left|\ddot{x}_{B}\right| = (5 × 386) in. / s² = 1930 in./s² . Substituting these values and γ = 0.02 into Eq. (O), the required single-peak displacement amplitude at the center of the pipe is

Part (b)
From Eq. (D), the curvature of the pipe is given by:

The curvature at the mid point, (d^{2}y / dx^{2})_{c} , where x = L/2, is

\left(\frac{d^{2}y}{dx^{2}}\right)_{c} = y_{c} \frac{\pi^{2}}{L^{2}} \left(- \sin \frac{\pi}{2} \right) = – y_{c} \frac{\pi^{2}}{L^{2}}                        (P)

The corresponding bending moment is M_{c} = EI (d^{2}y / dx^{2})_{c} and the corresponding stress, s_{c}, at the center of the span and distance D/2 from the bending neutral axis, is

s_{c} = M_{c} \frac{D}{2I} = \frac{DE}{2} \left(\frac{d^{2}y}{dx^{2}}\right)_{c} = – \frac{\pi^{2}DE}{2L^{2}} y_{c}                        (Q)

The negative sign in Eq. (Q) is of no importance, since it is arbitrary whether tensile or compressive stress is taken as positive, so we can simply say that

\left|s_{c}\right|_{MAX} = \frac{\pi^{2}DE}{2L^{2}} \left|y_{c}\right|_{MAX}                        (R)

Inserting numerical values into Eq. (R), D = 0.3125 in.; E = 30 × 10^{6}   lbf/in.², L = 17 in. and \left|y_{c}\right|_{MAX} = 0.126   in., gives