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Industrial safety and maintenance

Mechanical Engineering (B.E./B.Tech) · Industrial engineering · free study notes for exams, placements and interviews

In short

Industrial safety is the systematic prevention of injury, illness, fire and property loss in a workplace; maintenance is the set of actions that keeps plant and equipment in a condition where it performs its function safely and economically. The two are inseparable — most industrial accidents happen on neglected equipment or during repair work, and most unplanned downtime is a maintenance failure. Examination of this topic covers accident causation and costing, hazard identification techniques (HAZOP, FMEA, FTA), statutory duties under the Occupational Safety, Health and Working Conditions Code, 2020 (in force from 21 November 2025; it repealed the Factories Act 1948, whose section numbers older papers still quote as the former law), maintenance classification, TPM and OEE. It is examined in industrial engineering papers, GATE (ME/PI) and the Diploma in Industrial Safety, and asked in interviews for safety officer, maintenance engineer, plant engineer and EHS roles at Tata Steel, L&T, Reliance, BPCL, NTPC, Maruti Suzuki and the chemical and pharma sector.

Explained

Basic terms and accident causation

Hazard identification and risk assessment techniques

Indian statutes, codes and institutions

Specific hazards and their control

Maintenance: types and organisation

TPM, OEE and the safety–maintenance link

Key points

Formulas and examples

Frequency Rate FR = (Number of lost-time injuries × 1,000,000) / Man-hours worked
Severity Rate SR = (Total days lost or charged × 1,000,000) / Man-hours worked
Incidence Rate = (Number of injuries × 1,000) / Average number of persons employed
Average days lost per injury = SR / FR
Frequency–Severity Index FSI = √(FR × SR / 1,000)
Safe-T-Score STS = (FR_present − FR_past) / √(FR_past ÷ (present man-hours ÷ 1,000,000)); |STS| ≤ 2 means no significant change, > +2 worse, < −2 better
Risk = Likelihood × Severity; RPN = S × O × D (FMEA, each 1–10)
Total accident cost = Direct cost + Indirect cost (Heinrich's rule of thumb: indirect ≈ 4 × direct)
OEE = Availability × Performance × Quality
Availability = (Loading time − Downtime) / Loading time
Performance = (Ideal cycle time × Total parts produced) / Operating time
Quality = Good parts / Total parts produced
Total maintenance cost per period = (PM cost per service × number of services) + (Expected breakdowns × cost per breakdown); optimum PM interval is where this is minimum
Maintenance cost ratio = Annual maintenance cost / Replacement asset value (typically 2–5%)

Units: man-hours in hours actually worked (not paid); FR and SR are per million man-hours; days lost in calendar days charged; costs in rupees.

Example 1 — Safety statistics for a plant. A factory employs 500 workers who each work 48 hours a week for 50 weeks. In the year there were 9 lost-time injuries, one of them fatal. Days lost on the eight non-fatal cases total 240; the fatality is charged 6,000 days.

Example 2 — Has safety improved? Last year FR was 10.0; this year the plant worked 1,200,000 man-hours and FR fell to 7.5.

Example 3 — Cost of one accident. A machine-guarding accident costs the firm ₹1,20,000 in direct costs (compensation, medical, insurance excess). Using Heinrich's 1:4 ratio, indirect costs (lost production, overtime, investigation, retraining, damaged tooling, legal and reputation) ≈ 4 × 1,20,000 = ₹4,80,000; total ≈ ₹6,00,000. If the firm's profit margin is 4%, sales of 6,00,000 / 0.04 = ₹1.5 crore are needed to recover it — while the interlocked guard that would have prevented it costs about ₹25,000.

Example 4 — OEE of a CNC machine. Shift of 8 hours (480 min) with a 30-minute planned lunch break. Breakdown 30 min, setup 15 min. Ideal cycle time 0.5 min/part; 720 parts produced, 18 rejected.

Example 5 — Optimum preventive-maintenance interval. A PM service costs ₹20,000; a breakdown costs ₹50,000 including lost output. Expected breakdowns per year rise as the interval lengthens:

PM intervalServices/yearPM cost (₹)Breakdowns/yearBreakdown cost (₹)Total (₹)
1 month122,40,0001.050,0002,90,000
2 months61,20,0002.51,25,0002,45,000
3 months480,0004.02,00,0002,80,000
4 months360,0006.03,00,0003,60,000
Breakdown only009.04,50,0004,50,000

The U-shaped total cost is lowest at a 2-month interval (₹2,45,000 per year) — a saving of ₹2,05,000 against run-to-failure. Note that if the failure mode is safety-critical, the interval is set by the risk, not by this cost minimum.

BasisBreakdownPreventivePredictive (CBM)
TriggerAfter failureFixed time or usageMeasured condition
DowntimeUnplanned, longPlanned, shortPlanned, shortest
Spares heldHigh, unpredictableModerate, knownLow, ordered on evidence
InvestmentNilLowInstruments and skilled analysts
SuitsCheap, redundant, non-critical itemsWear-out items with known lifeCostly rotating and electrical plant
RiskHigh safety and output riskOver-maintenance, infant mortality after workMisreading of data, sensor failure

Where it is used

Interview questions

Common mistakes

Related topics

Maintenance and reliability engineering · Statistical quality control · Lean manufacturing and Six Sigma · Work study and ergonomics · Production planning and control · Engineering economics and costing

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