---
title: "Return period for bridge design at 5% risk"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:bridge-return-period-5pct-risk"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/bridge-return-period-5pct-risk
site: BCT Engineering
---

# Return period for bridge design at 5% risk

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2081 Bhadra, 4 marks
- Video: [Frequency Of Rainfall and Probability - Risk, Return Period, Reliability, Flood Occurring- Hydrology](https://www.youtube.com/watch?v=yZOCh-0L4cQ) — Engineers' Fate
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/bridge-return-period-5pct-risk
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

What return period will you adopt in the design of a bridge on a river if you are allowed to accept only 5% risk of flooding in the 25 years of expected life of the bridge?

## Short answer

Risk 0.05 over 25 years yields (1-1/T)^25=0.95, so 1-1/T=0.99795, giving T=488 years.

## Answer

The probability that a flood of return period T occurs at least once in n years of the structure's life is the risk

$$
\bar R=1-\left(1-\frac1T\right)^{n}
$$

Given $\bar R=0.05$ and $n=25$ years:

$$
0.05=1-\left(1-\frac1T\right)^{25}\ \Rightarrow\ \left(1-\frac1T\right)^{25}=0.95
$$

$$
1-\frac1T=(0.95)^{1/25}=0.997950\ \Rightarrow\ \frac1T=0.002050
$$

$$
T=487.9 \text{ years}
$$

**Answer: adopt a return period of about 488 years (design for the ≈ 500-year flood).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which equation relates risk R, return period T and life n?
   A. R = (1/T)^n
   B. R = 1 - (1 - 1/T)^n
   C. R = 1 - T/n
   D. R = n/T
   Answer: B. It is the probability of at least one exceedance in n years. (Key point: Risk equation)

2. With R = 0.05, n = 25, what do we solve?
   A. (1 - 1/T)^25 = 0.05
   B. (1 - 1/T)^25 = 0.95
   C. T = 25/0.05
   D. 1/T = 0.05/25
   Answer: B. 1 - R = 0.95 equals the chance of no flood in 25 years. (Key point: Solving for T)

3. What return period is adopted?
   A. About 100 years
   B. About 20 years
   C. About 488 years
   D. About 25 years
   Answer: C. 1/T = 1 - 0.95^(1/25) = 0.00205, so T = 488 years. (Key point: Result)

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