---
title: "Gumbel design flood for highway bridge, 4% risk"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-bridge-risk4pct"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-bridge-risk4pct
site: BCT Engineering
---

# Gumbel design flood for highway bridge, 4% risk

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2068 Chaitra, 9 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-bridge-risk4pct
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

A highway bridge has to be designed with an expected life of 50 years and an allowable flood risk of 4%. The flood data of bridge site were well fitted to Gumbel EV distribution and discharges for 50 and 300 years return period are found to be 150 and 650 m$^3$/sec respectively. Estimate the frequency and magnitude of design flood for the bridge.

## Short answer

Design return period from 4% risk in 50 years: T=1225 years. Using Gumbel linearity in reduced variate with known floods (150 m³/s at 50 yr, 650 m³/s at 300 yr), design flood ≈ 1041 m³/s.

## Answer

#### Frequency (return period) of the design flood
Risk $R=0.04$ in a life $n=50$ years:

$$
R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.96)^{1/50}\ \Rightarrow\ T=1225.3 \text{ years}
$$

The design flood is the **1225-year flood**, with annual exceedance probability $1/T=0.00082$.

#### Magnitude of the design flood
Gumbel's equation is linear in the reduced variate $y_T=-\ln[\ln(T/(T-1))]$, so the two known floods define the line:

| T (yr) | $y_T$ | $x_T$ (m³/s) |
|---|---|---|
| 50 | 3.9019 | 150 |
| 300 | 5.7021 | 650 |
| 1225 | 7.1106 | ? |

$$
\frac{\sigma}{S_n}=\frac{650-150}{5.7021-3.9019}=277.75 \text{ m}^3/\text{s}
$$

$$
x_T=150+277.75\,(7.1106-3.9019)=1041 \text{ m}^3/\text{s}
$$

**Answer: design return period ≈ 1225 years; design flood ≈ 1041 m³/s.**

## Check yourself

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

1. For R = 0.04 and n = 50, which gives T?
   A. T = 1/0.04 = 25 years
   B. 1/T = 1 - 0.96^(1/50), T about 1225 years
   C. T = 50/0.04 = 1250 years exactly
   D. 1/T = 0.04^(1/50)
   Answer: B. Solving R = 1 - (1 - 1/T)^n gives T = 1225 years. (Key point: Design return period)

2. How is the Gumbel line fixed from the 50- and 300-year floods?
   A. It cannot be done without the mean
   B. Slope = (650 - 150)/(300 - 50)
   C. Slope = 650/300
   D. Slope = (650 - 150)/(y300 - y50), since x is linear in y_T
   Answer: D. x is linear in y_T, so slope = 500/(5.7021 - 3.9019) = 277.75. (Key point: Using two known floods)

3. With y_1225 = 7.1106, what is the design flood?
   A. 150 + 277.75 x (7.1106 - 3.9019) = 1041 m^3/s
   B. 650 + 277.75 x 7.1106
   C. 150 x 7.1106 = 1066
   D. 277.75 x 7.1106 = 1975
   Answer: A. Start from a known point on the line and add slope times the difference in y_T. (Key point: Extrapolating)

4. What is the annual exceedance probability of the design flood?
   A. 0.04
   B. 1/1225 = 0.00082
   C. 0.02
   D. 1/50
   Answer: B. Annual exceedance probability is 1/T, here 0.00082. The 4% is risk over the whole 50-year life. (Key point: Meaning of the result)

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