Vibration of continuous systems

Tidewater's largest product is the VT-12 vertical turbine pump, used in municipal raw-water intakes and cooling-water systems. Its bowl assembly hangs deep in a wet pit on a long column, and on top, at floor level, a cast discharge head carries a 75 kW vertical motor weighing 1,150 kg. A tall, heavy motor on a relatively slender head makes a structure that sways like an inverted pendulum, and its first bending frequency, known in the pump industry as the reed critical frequency, is often uncomfortably close to running speed. For a new intake order, Farah has a revised discharge head with large access windows for the seal, and she asks Owen to check its reed frequency before the drawings are released. The head and motor stand have mass and stiffness spread along their height, so the lumped models used so far need to be reconsidered. This lesson treats continuous systems: strings, bars, shafts and beams.

The vibrating string and the wave equation

A taut string of tension T and mass per unit length ρA, displaced by y(x, t), obeys the wave equation

Separating variables, \(y = Y(x)\cos\omega t\), gives \(Y'' + (\omega/c)^2 Y = 0\) with solutions \(Y = A\sin(\omega x/c) + B\cos(\omega x/c)\). For a string fixed at both ends, the boundary conditions allow only the frequencies \(\omega_n = n\pi c/L\), n = 1, 2, 3, and so on, with mode shapes \(\sin(n\pi x/L)\). A continuous system has an infinite number of natural frequencies, here in the exact ratios 1 : 2 : 3, and each mode shape has n − 1 nodes. The string itself is rarely found in pump work, except as a model of guy wires and long instrument cables, but it shows the pattern for all continuous systems: a partial differential equation, a separation of variables, and boundary conditions that pick out the allowed frequencies.

Axial and torsional vibration of rods

The same equation governs axial vibration of a bar, with \(c = \sqrt{E/\rho}\), and torsional vibration of a circular shaft, with \(c = \sqrt{G/\rho}\). For steel, the axial wave speed is about 5,100 m/s and the torsional wave speed about 3,200 m/s. For a bar fixed at one end and free at the other, the frequencies are \(\omega_n = (2n - 1)\pi c/(2L)\). A 10 m steel line shaft in a deep-well pump therefore has its first axial mode at about 128 Hz and its first torsional mode, if fixed at the motor end, at about 80 Hz, with the impellers' inertia lowering both. These frequencies are found by the lumped methods of Lesson 9 once the impellers are included; the continuous solution checks the shaft's own contribution.

Bending of beams

Transverse vibration of a slender beam, the Euler-Bernoulli model, is governed by a fourth-order equation:

Harmonic solutions give \(Y'''' = \beta^4 Y\) with \(\beta^4 = \rho A\omega^2/EI\), whose general solution combines sin βx, cos βx, sinh βx and cosh βx. Applying two boundary conditions at each end gives a characteristic equation whose roots β_nL fix the natural frequencies:

Unlike strings, the frequencies of beams are not in integer ratios: for a cantilever, the second mode is 6.27 times the first. This matters in diagnosis: a structure that responds at several frequencies in harmonic ratios is more likely being forced by harmonics of running speed than resonating in several of its own modes. The Euler-Bernoulli model neglects shear deformation and rotary inertia, which become important for short, deep beams and for higher modes; the Timoshenko beam model includes them.

Cantilever with a tip mass

The VT-12 head and motor are modeled as a uniform cantilever of height L = 1.6 m to the motor's center of mass, with an effective bending stiffness EI = 4.5×10⁷ N·m² that accounts for the access windows, a distributed mass of 650 kg and a tip mass of 1,150 kg. The tip mass changes the boundary condition at the free end: the shear force there must accelerate the mass. The characteristic equation becomes

which is solved numerically. With a heavy tip mass, the root falls from 1.875 for the bare cantilever to a much smaller value.

# Continuous systems: the VT-12 vertical turbine pump structure and a suction pipe span
import math
f_run = 1780 / 60                       # VT-12 speed, Hz
# 1. Discharge head and motor stand as a uniform cantilever with the motor as a tip mass
L = 1.6                                 # height to the motor center of mass, m
EI = 4.5e7                              # effective bending stiffness of the head with its access windows, N m^2
m_head = 650.0                          # distributed mass of the head and stand, kg
m_motor = 1150.0                        # 75 kW vertical motor, kg
rhoA = m_head / L
def char(bl):                           # frequency equation of a cantilever with a tip mass
    r = m_motor / (rhoA * L)
    return 1 + math.cos(bl) * math.cosh(bl) + r * bl * (math.cos(bl) * math.sinh(bl) - math.sin(bl) * math.cosh(bl))
lo, hi = 0.05, 1.875                    # first root lies below the bare-cantilever value 1.875
for _ in range(100):
    mid = (lo + hi) / 2
    if char(lo) * char(mid) <= 0: hi = mid
    else: lo = mid
bl = (lo + hi) / 2
f_exact = bl**2 / L**2 * math.sqrt(EI / rhoA) / (2 * math.pi)
print(f"Exact (Euler-Bernoulli with tip mass): beta L = {bl:.4f}, f1 = {f_exact:.2f} Hz")
# Rayleigh with the static deflection shape: tip mass plus 33/140 of the beam mass
k = 3 * EI / L**3
f_ray = math.sqrt(k / (m_motor + 33 / 140 * m_head)) / (2 * math.pi)
print(f"Rayleigh estimate: k = {k/1e6:.1f} MN/m, f1 = {f_ray:.2f} Hz (error {(f_ray/f_exact-1)*100:+.2f} %)")
# Rayleigh-Ritz with two shape functions (x/L)^2 and (x/L)^3: solve the 2x2 eigenproblem
n = 400; dx = L / n
phis = [lambda x: (x / L)**2, lambda x: (x / L)**3]
d2 = [lambda x: 2 / L**2, lambda x: 6 * x / L**3]
Kr = [ [sum(EI * d2[i]((j + .5) * dx) * d2[k]((j + .5) * dx) * dx for j in range(n)) for k in range(2)] for i in range(2)]
Mr = [ [sum(rhoA * phis[i]((j + .5) * dx) * phis[k]((j + .5) * dx) * dx for j in range(n)) + m_motor * phis[i](L) * phis[k](L)
       for k in range(2)] for i in range(2)]
a = Mr[0][0] * Mr[1][1] - Mr[0][1]**2
b = -(Kr[0][0] * Mr[1][1] + Kr[1][1] * Mr[0][0] - 2 * Kr[0][1] * Mr[0][1])
c = Kr[0][0] * Kr[1][1] - Kr[0][1]**2
lam = (-b - math.sqrt(b * b - 4 * a * c)) / (2 * a)
print(f"Rayleigh-Ritz, two terms: f1 = {math.sqrt(lam)/(2*math.pi):.2f} Hz")
print(f"Running speed {f_run:.2f} Hz: separation {(f_run - f_exact)/f_run:.1%} below running speed")
# Stiffness needed to put the first mode 25 % above running speed (frequency scales with the square root of EI)
EI_need = EI * (1.25 * f_run / f_exact)**2
print(f"For 25 % above running speed: EI about {EI_need/1e6:.0f} MN m^2 ({EI_need/EI:.2f} times the present head)")
# A flexible foundation (soil-supported slab) adds a rotational spring at the base: model as k_theta
for kth in (2e9, 8e8, 4e8):
    keff = 1 / (1 / k + L**2 / kth)
    f = math.sqrt(keff / (m_motor + 33 / 140 * m_head)) / (2 * math.pi)
    print(f"  base rotational stiffness {kth/1e6:6.0f} MN m/rad: f1 about {f:.1f} Hz")

# 2. Suction pipe span, pinned at both supports, full of water
E = 200e9; do, t = 0.2191, 0.00818     # 8 in schedule 40 steel pipe, m
di = do - 2 * t
I = math.pi * (do**4 - di**4) / 64
m_line = 7850 * math.pi * (do**2 - di**2) / 4 + 1000 * math.pi * di**2 / 4
for span in (4.0, 6.0, 8.0):
    fs = [(k * math.pi / span)**2 * math.sqrt(E * I / m_line) / (2 * math.pi) for k in (1, 2)]
    print(f"Pipe span {span:.0f} m: f1 = {fs[0]:5.1f} Hz, f2 = {fs[1]:5.1f} Hz")

Exact (Euler-Bernoulli with tip mass): beta L = 1.1059, f1 = 25.31 Hz
Rayleigh estimate: k = 33.0 MN/m, f1 = 25.31 Hz (error +0.02 %)
Rayleigh-Ritz, two terms: f1 = 25.31 Hz
Running speed 29.67 Hz: separation 14.7% below running speed
For 25 % above running speed: EI about 97 MN m^2 (2.15 times the present head)
  base rotational stiffness   2000 MN m/rad: f1 about 24.8 Hz
  base rotational stiffness    800 MN m/rad: f1 about 24.1 Hz
  base rotational stiffness    400 MN m/rad: f1 about 23.0 Hz
Pipe span 4 m: f1 =  27.9 Hz, f2 = 111.5 Hz
Pipe span 6 m: f1 =  12.4 Hz, f2 =  49.6 Hz
Pipe span 8 m: f1 =   7.0 Hz, f2 =  27.9 Hz

The exact solution gives βL = 1.106 and a reed frequency of 25.3 Hz. Running speed is 1,780 rpm, 29.7 Hz, so the first mode lies 14.7 percent below running speed: the pump passes through it on every start, and runs only slightly above it. Rayleigh's method, with the cantilever's static deflection as the shape and an effective tip mass of 1,150 + 33/140 × 650 kg, gives the same 25.3 Hz to within 0.02 percent, because the heavy tip mass dominates. The two-term Rayleigh-Ritz model agrees too. When a lumped mass dominates, simple methods are very accurate; when the distributed mass dominates, the shape assumption matters more.

Foundation flexibility

The calculation assumes the head is fixed rigidly at its base. In reality it is bolted to a sole plate on a concrete floor over the wet pit, and both the floor and the bolts add flexibility. A rotational spring k_θ at the base acts in series with the cantilever, so the effective tip stiffness becomes \(1/k_{eff} = 1/k + L^2/k_\theta\). The program shows the reed frequency falling to 24.8, 24.1 and 23.0 Hz as the base rotational stiffness drops from 2,000 to 400 MN·m/rad. Field-measured reed frequencies on vertical pumps are often 10 to 20 percent below calculated fixed-base values for this reason, and pump industry guidance, such as that of the Hydraulic Institute, recommends that the calculation account for the foundation and that the result be checked by an impact test on site.

Design: how much separation?

Guidance for vertical pumps asks for a separation of the reed frequency from running speed of the order of 25 percent, either above or below. Placing it below means the structure is soft, and the pump runs in the mass-controlled region, as an isolated machine does; placing it above means a stiff structure, and the pump runs in the stiffness-controlled region. For a fixed-speed pump, either is acceptable; for a variable-speed pump, the reed frequency must lie outside the whole speed range plus the margin. The program finds that a stiffness 2.15 times the present value, EI of about 97 MN·m², would put the reed frequency 25 percent above running speed. Alternatively, a frequency of 22.3 Hz or below would be 25 percent below; the flexible-foundation case already approaches that, but relying on uncertain foundation flexibility is poor design.

Farah's choice is to stiffen the head: closing two of the four windows with bolted covers that carry load and adding ribs raises the effective EI to about 1.0×10⁸ N·m², putting the reed frequency near 37.7 Hz. The seal can still be reached through the remaining windows. Owen notes that the change also raises the second bending mode, which was nowhere near a forcing frequency, and that the motor manufacturer must confirm that its own frame frequency, which adds in series, is high enough.

Rayleigh-Ritz and orthogonality in continuous systems

The Rayleigh-Ritz method generalizes Rayleigh's method by assuming the deflection as a sum of several shape functions, \(y = \sum a_i\psi_i(x)\), each satisfying at least the geometric boundary conditions. Substituting into the energy expressions gives stiffness and mass matrices, \(K_{ij} = \int EI\,\psi_i''\psi_j''\,dx\) and \(M_{ij} = \int \rho A\,\psi_i\psi_j\,dx\) plus terms for point masses, and the eigenvalue problem of Lesson 9. The result is always an upper bound to the true frequencies, improving as functions are added. The finite element method is a Rayleigh-Ritz method with simple local shape functions on many small elements, which is why MEC-450 builds on this lesson. The modes of continuous systems are orthogonal in the same sense as discrete ones, \(\int \rho A\,Y_iY_j\,dx = 0\), and modal superposition applies in the same way.