Derive economical depth of girder

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 $d/5$.