---
title: "Derive economical depth of girder"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:economical-depth-girder"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/economical-depth-girder
site: BCT Engineering
---

# Derive economical depth of girder

Design of Steel and Timber Structure · Chapter 7: Flexure Members

- Asked in: 2068 Magh (old course), 6 marks
- Video: [Economic Depth Of Plate Girder || Derivation || VVI || Plate Girder](https://www.youtube.com/watch?v=AGZEZO3DgM8) — Sabin Dhungel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/economical-depth-girder
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

Derive a relation for economical depth of a girder.

## Short answer

For minimum weight, total area A = 2M/(f d) + d²/k with k = d/t_w. Differentiating yields d³ = M k/f, so optimal depth d = (M k/f)^(1/3). At optimum, each flange area equals the web area.

## Answer

**Aim.** Find the depth $d$ of an I-girder (plate girder) for which the total weight (area of steel per unit length) is minimum for a given bending moment $M$.

#### Assumptions
- The flanges carry the entire bending moment (the web's contribution to bending is neglected), so $A_f = \dfrac{M}{f\,d}$, where $f$ is the permissible (design) stress in the flange, $f = f_y/\gamma_{m0}$.
- The web has depth $d$ and thickness $t_w = d/k$, where $k = d/t_w$ is a fixed web slenderness chosen from the buckling limits. Web area $A_w = d\,t_w = \dfrac{d^2}{k}$.
- Stiffeners and weld metal are neglected.

#### Derivation
Total cross-sectional area (two flanges plus web):
$$
A = 2A_f + A_w = \frac{2M}{f\,d} + \frac{d^2}{k}
$$
For minimum weight, differentiate with respect to $d$ and set the result to zero:
$$
\frac{dA}{dd} = -\frac{2M}{f\,d^2} + \frac{2d}{k} = 0
$$
$$
\frac{M}{f\,d^2} = \frac{d}{k} \quad \Rightarrow \quad d^3 = \frac{M\,k}{f}
$$
$$
\boxed{d = \left(\frac{M\,k}{f}\right)^{1/3}}
$$
The second derivative $d^2A/dd^2 = 4M/(fd^3) + 2/k > 0$, so this is a minimum.

#### Result and use
- At the economical depth, $A_f = \dfrac{M}{fd} = \dfrac{d^2}{k} = A_w$: each flange has the same area as the web, so the two flanges together have **twice** the web area.
- If the web is allowed to carry bending (an area $A_w/6$ is added to each flange), the same working gives $d = (1.5\,M\,k/f)^{1/3}$ approximately; in practice $d \approx L/10$ to $L/12$, and $k = 100$ to $200$ for webs with stiffeners.
- Then $t_w = d/k$, $A_f = M/(fd)$, and the flange width is chosen at about $d/3$ to $d/5$.

## Check yourself

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

1. In the derivation, the flanges carry the whole bending moment M at design stress f. What is the area of one flange?
   A. A_f = M d / f
   B. A_f = M / (f d)
   C. A_f = M f / d
   D. A_f = f d / M
   Answer: B. Moment M = f A_f d, so A_f = M/(f d). (Key point: Flange area from moment)

2. The total area is A = 2M/(f d) + d^2/k. What condition gives the economical depth?
   A. Set A = 0
   B. Set dA/dd = 0, giving d^3 = M k / f
   C. Set d^2A/dd^2 = 0
   D. Set the web area equal to one flange area
   Answer: B. Minimum weight occurs where dA/dd = -2M/(f d^2) + 2d/k = 0, leading to d = (Mk/f)^(1/3). (Key point: Minimisation condition)

3. At the economical depth, how does the area of the two flanges compare with the web area?
   A. Equal
   B. Flanges are half the web
   C. Flanges are twice the web
   D. Flanges are four times the web
   Answer: C. 2A_f = 2M/(f d) = 2d^2/k = 2A_w, so the two flanges together have twice the web area. (Key point: Flange to web area at optimum)

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