Maintenance programmes

Why this matters

UAT 321 taught how to build a maintenance programme from the manufacturer’s due items. This course asks the next question: what should those intervals be, and should parts be replaced on schedule at all? The course case is a fleet of four survey drones at one agency, with two years of repair records. The drone knowledge base’s unit on maintenance programmes covers scheduled inspections, flight hours, component life and logbooks, and ASTM F2909-19 addresses the continued airworthiness of lightweight UAS, covering the aircraft, control station, C2 link and launch and recovery equipment. The numbers in this module are hypothetical.

Reliability-centred maintenance

The Reliability-Centered Maintenance report by Nowlan and Heap (1978), which changed how United Airlines maintained its aircraft, found that most components do not fail because of age. NASA’s RCM guide (2008) summarises several studies: random failures make up about 77–92% and age-related failures only 8–23%. The key conclusion is that scheduled replacement only works for components that wear out with age; components that fail randomly should be monitored for condition instead. SAE JA1011 sets the criteria a process must meet to be called RCM.

Failure patterns and Weibull

The Weibull distribution describes many component lives through a shape parameter $\beta$ and a characteristic life $\eta$. Following the NIST statistics handbook, $\beta <1$ means a decreasing failure rate (early failures), $\beta =1$ a constant rate (random failures), and $\beta >1$ an increasing rate (wear-out). Estimating $\beta$ from real data therefore tells you which strategy to use.

import math
import numpy as np

hours = np.array([310, 420, 480, 530, 600, 640, 700, 760, 830, 950.])   # flight hours at failure, sorted
n = len(hours)
F = (np.arange(1, n + 1) - 0.3) / (n + 0.4)      # median rank (Bernard)
x, y = np.log(hours), np.log(-np.log(1 - F))
beta, a = np.polyfit(x, y, 1)
eta = math.exp(-a / beta)
r2 = np.corrcoef(x, y)[0, 1] ** 2
mean_life = eta * math.gamma(1 + 1 / beta)
print(f"beta {beta:.2f}, eta {eta:.0f} h, R^2 {r2:.3f}, mean life {mean_life:.0f} h")
print("wear-out: planned replacement can help" if beta > 1 else "not wear-out: use condition monitoring")

beta 3.30, eta 694 h, R^2 0.999, mean life 623 h
wear-out: planned replacement can help

The most economical interval

If a motor is replaced as planned at age $T$, or on failure if it fails before $T$, each cycle of use ends with cost $C_p$ (planned replacement) with probability $R(T)$, or $C_f$ (in-flight failure including the damage that follows) with probability $1-R(T)$. The mean cycle length is ${\int }_0^{T}R(t)dt$. The cost per hour is therefore the expected cost per cycle divided by the expected cycle length.

import math
import numpy as np

BETA, ETA = 3.30, 694.4                          # from Example 1
CP, CF = 1500, 30000                             # THB

def R(t):
    return math.exp(-(t / ETA) ** BETA)

def cost_rate(T):
    ts = np.linspace(0, T, 2001)
    return (CP * R(T) + CF * (1 - R(T))) / np.trapezoid([R(u) for u in ts], ts)

best = min(range(50, 901, 10), key=cost_rate)
mean_life = ETA * math.gamma(1 + 1 / BETA)
print(f"best interval {best} h: {cost_rate(best):.2f} THB/h, R({best}) = {R(best):.3f}")
for T in (150, 300, 400):
    print(f"interval {T} h: {cost_rate(T):.2f} THB/h")
print(f"run to failure: {CF / mean_life:.2f} THB/h")

best interval 220 h: 9.75 THB/h, R(220) = 0.978
interval 150 h: 11.22 THB/h
interval 300 h: 10.93 THB/h
interval 400 h: 14.94 THB/h
run to failure: 48.16 THB/h

Choosing a strategy

• Scheduled replacement when a component clearly wears out ($\beta >1$) and failure has serious consequences
• Condition monitoring when there is a measurable warning before failure, such as vibration, battery internal resistance or motor temperature
• Run to failure when failure does not affect safety and replacement is cheap, such as status lights
• If failure affects safety and no strategy works, redesign, for example by adding redundancy

Module lab

Common mistakes

Summary

• RCM found that most components fail randomly; scheduled replacement only works for parts that wear out with age
• Weibull $\beta$ shows the failure pattern: below 1 early failures, 1 random, above 1 wear-out
• The most economical interval minimises the expected cost per cycle divided by the expected cycle length
• Choose a strategy from the failure pattern and its consequences

Check your understanding

1. What does $\beta =0.7$ tell you about a component?
2. $\eta =500$ h and $\beta =2$. What is the reliability at 250 h?
3. With 5 data points, what is the median rank of the first?
4. Why is the cost of an in-flight failure $C_f$ usually much higher than the part price?
5. Which strategy suits a component that fails randomly?