## Properties of Tee (T) Cross-Section

This tool calculates the properties of a tee cross-section. Enter the shape dimensions h, b, tf and tw below. The calculated results will have the same units as your input. Please use consistent units for any input.

 h = b = tf = tw = Geometric properties: Area = Perimeter = yc = Properties related to x-x axis: Ix = Sx = Zx = ypna = Rgx = Properties related to y-y axis: Iy = Sy = Zy = Rgy = Other properties: Iz =

## Definitions

### Geometry

The area A and the perimeter P of a tee cross-section, can be found with the next formulas:

$\begin{split} & A & = b t_f + (h-t_f)t_w \\ & P & = 2b + 2h \end{split}$

The distance of the centroid along y axis, from the top edge can be found using the first moments of area of the web and the flange:

$y_c = \frac{1}{A}\left( \frac{t_w h^2}{2} + \frac{(b-t_w) {t_f}^2}{2} \right)$

### Moment of Inertia

The moment of inertia of a tee section can be found if the total area is divided into two, smaller ones, A, B, as shown in figure below. The final area, may be considered as the additive combination of A+B. Therefore, the moment of inertia Ix of the tee section, relative to non-centroidal x0-x0 axis, is determined like this:

$I_{x0} = \frac{t_w h^3}{3} +\frac{(b-t_w) t_{f}^3}{3}$

where h the tee height, b the width of the flange, tf the thickness of the flange (parallel to x-x) and tw the thickness of the web (perpendicular to x-x).

Knowing Ix0 and yc , the moment of inertia Ix relative to centroidal x-x axis, may be determined, using the Parallel Axes Theorem:

$\begin{split} & I_{x0} & = I_{x} + A y_c^2 \Rightarrow \\ & I_{x} & = I_{x0} - A y_c^2 \end{split}$

The moment of inertia Iy of the tee section, relative to centroidal y-y axis, can be found directly, by the additive combination of C+D sub-areas:

$I_y = \frac{(h-t_f) t_w^3}{12} + \frac{t_f b^3}{12}$

The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation:

$M = E\times I \times \kappa$

where E is the Young's modulus, a property of the material, and κ the curvature of the beam due to the applied load. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the developed curvature is reversely proportional to the moment of inertia I.

The polar moment of inertia, describes the rigidity of a cross-section against torsional moment, likewise the planar moments of inertia described above, are related to flexural bending. The calculation of the polar moment of inertia Iz about an axis z-z (perpendicular to the section), can be done with the Perpendicular Axes Theorem:

$I_z = I_x + I_y$

where the Ix and Iy are the moments of inertia about axes x-x and y-y that are mutually perpendicular with z-z and meet at a common origin.

The dimensions of moment of inertia are $$[Length]^4$$.

### Elastic section modulus

The elastic section modulus Sx of any cross section about axis x-x (centroidal), describes the response of the section under elastic flexural bending. It is defined as:

$S_x = \frac{I_x}{Y}$

where Ix the moment of inertia of the section about x-x axis and Y the distance from centroid of a section point (aka fiber, typically the most distant one), measured perpendicularly to x-x axis. For the tee section, the elastic section modulus Sx , relative to the x-x axis, due to the unsymmetry, is different for Y measured from the top or the bottom fiber. The bigger Y results in the smaller Sx , which is usually preferable for the design of the section. Therefore:

$S_{x,min} = \frac{I_x}{h-y_c}$

where the min designation is based on the assumption that $$y_c < h-y_c$$, which is valid for any tee section.

For the section modulus Sy , relative to y-y axis, which, for this section, happens to be a symmetry axis, the section modulus is found by:

$S_y = \frac{I_y}{X} \Rightarrow S_y = \frac{2 I_y}{b}$

If a bending moment Mx is applied on axis x-x, the section will respond with normal stresses, varying linearly with the distance from the neutral axis (which under elastic regime coincides to the centroidal x-x axis). Along neutral axis the stresses are zero. Absolute maximum σ will occur at the most distant fiber, with magnitude given by the formula:

$\sigma = \frac{M_x}{S_x}$

From the last equation, the section modulus can be considered for flexural bending, a property analogous to cross-sectional A, for axial loading. For the latter, the normal stress is F/A.

The dimensions of section modulus are $$[Length]^3$$.

### Plastic section modulus

The plastic section modulus is similar to the elastic one, but defined with the assumption of full plastic yielding of the cross section due to flexural bending. In that case the whole section is divided in two parts, one in tension and one in compression, each under uniform stress field. For materials with equal tensile and compressive yield stresses, this leads to the division of the section into two equal areas, At in tension and Ac in compression, separated by the neutral axis. The axis is called plastic neutral axis, and for non-symmetric sections, isn't the same with the elastic neutral axis (which again is the centroidal one). The plastic section modulus is given by the general formula:

$Z = A_c Y_c + A_t Y_t$

where Yc the distance of the centroid of the compressive area Ac from the plastic neutral axis and Yt the respective distance of the centroid of the tensile area At .

For the case of a tee cross-section, the plastic neutral axis for x-x bending, can be found by either one of the following two equations:

$\left \{ \begin{array}{ll} (h-y_{pna})t_w = \frac{A}{2} & \text{ , if } y_{pna} \gt t_f \\ y_{pna} b = \frac{A}{2} & \text{ , if } y_{pna} \le t_f \\ \end{array} \right.$

where ypna the distance of the plastic neutral axis from the top edge of the flange. The first equation is valid when the plastic neutral axis passes through the web, while the second one when it passes through the flange. Generally, it can't be known which equation is relevant beforehand. Once the plastic neutral axis is determined, the calculation of the centroids of the compressive and tensile areas becomes straightforward. Expressions for these are not included here.

For y-y bending, the plastic neutral axis passes through centroid (due to the symmetry). The calculation of the plastic modulus can be easily formulated:

$\begin{split} & Z_y & = 2 A_c Y_c \Rightarrow\\ & Z_y & = 2 A_c \left[\left(\frac{1}{Ac} (\frac{b t_f}{2}\frac{b}{4} +\frac{(h-t_f) t_w}{2}\frac{t_w}{4} \right) \right] \Rightarrow\\ & Z_y & = \frac{t_f b^2}{4} + \frac{(h-t_f) t_w^2}{4} \end{split}$

$R_g = \sqrt{\frac{I}{A}}$
where I the moment of inertia of the cross-section about the same axis and A its area. The dimensions of radius of gyration are $$[Length]$$. It describes how far from centroid the area is distributed. Small radius indicates a more compact cross-section. Circle is the shape with minimum radius of gyration, compared to any other section with the same area A.