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Tee (T) Section Calculator

- By Dr. Minas E. Lemonis, PhD - Updated: July 1, 2020
Home > Cross Sections > Tee (T)

This tool calculates the properties of a T 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 =
icon

Geometric properties:
Area =
Perimeter =
yc =
Properties related to x-x axis:
Ix =
Sx =
Zx =
ypna =
Rgx =

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Properties related to y-y axis:
Iy =
Sy =
Zy =
Rgy =
Other properties:
Iz =
Tee (T) cross section geometry
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Table of Contents
-  Calculator
-  Definitions
-  Geometry
-  Moment of Inertia
-  Elastic modulus
-  Plastic modulus
-  Around x axis
-  Around y axis
-  Radius of gyration
-  Tee section formulas
-  Related pages
See also
Properties of I/H section
Properties of a Channel (U) section
Properties of unequal I/H section
Moment of Inertia of a Tee
All Cross Section tools

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Definitions

Table of contents

-  Geometry
-  Moment of Inertia
-  Elastic modulus
-  Plastic modulus
-  Around x axis
-  Around y axis
-  Radius of gyration
-  Tee section formulas
-  Related pages

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, from the top edge, can be calculated if we consider that the first moment of area (also called static moment) of the entire T-section should be equal to the combined static moments of the web and the flange:

Ay_c = (t_w h) {h\over2} +\Big(\left(b-t_w\right)t_f \Big) {t_f\over2}\Rightarrow

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

We have a special article about finding the centroid of a compound area. If you need more details about the procedure, get it here.

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 I_{x0} of the tee section, relative to non-centroidal 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, t_f , the thickness of the flange (parallel to x-x) and t_w , the thickness of the web (perpendicular to x-x).

Knowing I_{x0} and y_{c} , the moment of inertia I_{x} around the centroidal x axis, can 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 I_y of the tee section, around centroidal 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}

Tee (T) cross section: separate areas for Ixx Iyy finding

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 \kappa , 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 I_z around an axis z-z (perpendicular to the section), can be done with the Perpendicular Axes Theorem:

I_z = I_x + I_y

where the I_x and I_y are the moments of inertia around 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 modulus

The elastic section modulus S_x , of any cross section, around an axis x-x that is centroidal, describes the response of the section under elastic flexural bending, around the same axis. It is defined as:

S_x = \frac{I_x}{Y}

where I_x the moment of inertia of the section around x axis and Y the offset from centroid, of a section fiber, parallel to x-x axis. Typically the most distant fiber is of interest. For the T-section, the elastic section modulus S_x , around the x-x axis, is not the same for a top-end and a bottom-end fiber. Since the T-section is not symmetric around x-axis, the distances of the two end fibers (top and bottom), from this axis, are different. The bigger Y results in the smaller S_x , 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 S_y , around y-y axis, which, for the T section, happens to be an axis of symmetry, the section modulus is found by the following formula:

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

elastic bending

If a bending moment M_x is applied around 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). Over neutral axis the stresses are zero. Absolute maximum \sigma , 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 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, A_t in tension and A_c in compression, separated by the neutral axis. This axis is called plastic neutral axis, and for non-symmetric sections, is not 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 Y_c the distance of the centroid of the compressive area A_c from the plastic neutral axis and Y_t the respective distance of the centroid of the tensile area A_t .

plastic bending

Around x axis

For the case of a T section, under x-x bending, the position of the plastic neutral axis 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} \ge t_f \\ y_{pna} b = \frac{A}{2} & \text{ , if } y_{pna} \lt t_f \\ \end{array} \right.

which becomes:

y_{pna} =\left \{ \begin{array}{ll} h- \frac{A}{2t_w} & \text{ , if: } t_f \le {A\over2 b} \\ \frac{A}{2b} & \text{ , if: } t_f \gt {A\over2 b} \\ \end{array} \right.

where y_\textit{pna} 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 becomes valid when the axis passes through the flange. Generally, it can't be known which equation is relevant beforehand.

plastic neutral axis in a tee cross-section

Once the plastic neutral axis is determined, the calculation of the centroids of the compressive and tensile areas becomes straightforward. For the first case, that is when the axis crosses the web, the plastic modulus can be found like this:

\begin{split} Z_x =&{t_w(h-y_\textit{pna})^2\over 2} + { by_\textit{pna}^2\over 2} \\&- {(b-t_w) (y_\textit{pna}-t_f)^2\over2 } \quad , t_f \le {A\over2 b}\end{split}

which can be simplified to:

Z_x ={t_w (h-t_f)^2 \over4} + {b h t_f\over 2} - {b^2 t_f^2\over 4t_w} \quad , t_f \le {A\over2 b}

For the second case, that is when the axis passes through the flange, the plastic modulus is found with equation:

\begin{split} Z_x =&{b y_\textit{pna}^2\over2 } + {t_w (h-y_\textit{pna})^2\over 2} \\&+ { (b-t_w)(t_f-y_\textit{pna})^2\over 2} \quad , t_f \gt {A\over2 b} \end{split}

which becomes:

Z_x = {t_w h^2 \over2} +{b t_f^2\over4} -{h t_f t_w\over2} - {(h-t_f)^2 t_w^2\over 4b}\quad , t_f \gt {A\over2 b}

Around y axis

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[\frac{1}{Ac} \left(\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}

Radius of gyration

Radius of gyration Rg of a cross-section, relative to an axis, is given by the formula:

R_g = \sqrt{\frac{I}{A}}

where I the moment of inertia of the cross-section around 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.

Tee section formulas

The following table, lists the formulas, for the calculation the main mechanical properties of a T section.

Tee (T) section formulas
Tee (T) cross section

QuantityFormula
Area: A = b t_f + h_w t_w
Perimeter: P = 2b + 2h
Centroid: y_c = \frac{1}{2A}\left( t_w h^2 + b_1{t_f}^2 \right)
Moments of inertia

I_{x} = I_{x0} - A y_c^2

I_y = \frac{h_w t_w^3}{12} + \frac{t_f b^3}{12}

Elastic modulus:

S_{x} = \frac{I_x}{h-y_c}

S_y = \frac{2 I_y}{b}

Plastic modulus:

Z_x = \left\{\begin{array}{ll} {t_w h_w^2 \over4} + {b h t_f\over 2} - {b^2 t_f^2\over 4t_w} &\quad , t_f \le {A\over2 b} \\ {t_w h^2 \over2} +{b t_f^2\over4} -{h t_f t_w\over2} - {h_w^2 t_w^2\over 4b} &\quad , t_f \gt {A\over2 b} \end{array}\right.

Z_y = \frac{t_f b^2}{4} + \frac{h_w t_w^2}{4}

Plastic neutral axis:

(distance from top)

 y_{pna} =\left \{ \begin{array}{ll} h- \frac{A}{2t_w} & \text{ , if: } t_f \le {A\over2 b} \\ \frac{A}{2b} & \text{ , if: } t_f \gt {A\over2 b} \\ \end{array} \right.

where:

b_1=b-t_w

h_w=h-t_f

I_{x0} = \frac{t_w h^3}{3} +\frac{b_1 t_{f}^3}{3}

Related pages

Properties of I/H section
Properties of a Channel (U) section
Properties of unequal I/H section
Moment of Inertia of a Tee
All Cross Section tools

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See also
Properties of I/H section
Properties of a Channel (U) section
Properties of unequal I/H section
Moment of Inertia of a Tee
All Cross Section tools