Jump to
Table of Contents
Share this
See also
Properties of a I/H section
Properties of a Rectangular section
Moment of Inertia of a I/H section
All Cross Section Property tools

Properties of a Double-Tee Cross-Section with Unequal Flanges

- By Dr. Minas E. Lemonis, PhD - Updated: March 3, 2019

This tool calculates the properties of an I/H shaped cross-section, with unequal flanges. Enter the shape dimensions below. The calculated results will have the same units as your input. Please use consistent units for any input.

h =
tw =
buf =
tuf =
bdf =
tdf =
Geometric properties:
Area =
Perimeter =
hw =
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 =
shape details
Table of Contents
Share this

Theoretical background

Geometry

The area A and the perimeter P of a double-tee, with unequal flanges, can be found from the next two formulas:

\[ \begin{split} & A & = b_{uf} t_{uf} + b_{df} t_{df} +h_w t_w \\ & P & = 2b_{uf}+2b_{df} + 2h - 2t_w \end{split} \]

The clear height of the web, \(h_w\) that appears in the above formula, is the clear distance between the two flanges:

\[ h_{w}=h-t_{uf}-t _{df} \]

Due to symmetry, about the y-y axis, the centroid of the cross-section must lye on the y-y axis. Therefore, \(x_c=0\). However, the same cannot be said for the other axis (x-x) since no symmetry exists about it, due to the unequal flanges. The exact location of the centroid  should be therefore calculated. To find its distance, yc , from a convenient axis of reference, say the bottom edge of the cross-section,  the first moments of area, of the web and the two flanges, relative to the same edge are employed (note: the first moment of area is defined as the area times the distance of the area centroid from the axis of reference). That is:

\[A \cdot y_{c} = A_{df} \cdot y_{c,df}  + A_{uf} \cdot y_{c,uf} +A_w \cdot y_{c,w}\]

where 

  • \(A_{df}=b_{df} t_{df} \), is the area of the lower flange
  • \(A_{uf}=b_{uf} t_{uf} \), is the area of the upper flange,
  • \(A_{w}=h_{w} t_{w} \), is the area of the web,
  • \(y_{df}=\frac{ t_{df}}{2} \), is the distance of the lower flange centroid from the bottom edge,
  • \(y_{uf}=h-\frac{ t_{uf}}{2} \), is the distance of the upper flange centroid from the bottom edge,
  • \(y_{w}=t_{df}+\frac{ h_{w}}{2} \), is the distance of the web centroid from the bottom edge.

Moment of Inertia

Major axis

Similarly, in order to find the moment of inertia of an unsymmetrical double-tee, the total cross-section is divided into three, smaller ones, one for the lower flange, one for the upper one and one for the web. Therefore, the moment of inertia Ix ,relative to centroidal x-x axis, is determined like this:

\[ I_x = I_{x,df} +I_{x,uf} +I_{x,w}  \]

where,

  • \( I_{x,df} = \frac{b_{df} t_{df}^3}{12}+A_{df}\left(y_c-\frac{t_{df}}{2}\right)^2\), the moment of inertia of the lower flange relative to the centroidal x-x axis,
  • \( I_{x,uf} = \frac{b_{uf} t_{uf}^3}{12}+A_{uf}\left(h-\frac{t_{uf}}{2}-y_c\right)^2\), the moment of inertia of the upper flange relative to the centroidal x-x axis,
  • \( I_{x,w} = \frac{t_w h_w^3}{12}+A_w\left(t_{df}+\frac{t_w}{2}-y_c\right)^2\), the moment of inertia of the web relative to the centroidal x-x axis.

For each of the above expressions the so called 'parallel axis theorem' has been employed for the estimation of the respective moments of inertia of  the three parts. For more information about this technique seehere .

Minor axis

Again, the total cross area is divided into three parts, one for the lower flange, one for the upper one and one for the web. This time however, the axis of bending (y-y) is also an axis of symmetry (for all three parts), and therefore the calculations are a bit easier (because the use of the parallel axis theorem is unnecessary). The total moment of inertia \(I_y\) is found as:

\[ I_y = I_{y,df} +I_{y,uf} +I_{y,w}  \]

where,

  • \( I_{y,df} = \frac{t_{df} b_{df}^3}{12}\), the moment of inertia of the lower flange relative to the centroidal y-y axis,
  • \( I_{y,uf} = \frac{t_{uf} b_{uf}^3}{12}\), the moment of inertia of the upper flange relative to the centroidal y-y axis,
  • \( I_{y,w} = \frac{h_w t_w^3}{12}\), the moment of inertia of the web relative to the centroidal y-y axis,

 

 

Polar

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.

Applications

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 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. Because the cross-section is unsymmetrical around x-x axis, the distance Y is different for the upper and bottom fibers, and as a consequence there are two different section moduli. Typically, the smaller one is of interest (the one corresponding to bigger X), because it is related with the more stressed fiber, as explained later in this section.

elastic bending

Similarly, the section modulus Sy , relative to y-y axis, which, for this section, happens to be symmetry axis too, is written as:

\[ S_y = \frac{I_y}{X} \]

where Iy the moment of inertia of the section about y-y axis and X the distance of edge fiber from centroid, measured perpendicularly to y-y axis. This time however, the distance X is the same for the left and right parts of the cross-section and equal to \(\max(0.5b_{uf}; 0.5b_{df})\) 

Elastic bending stress

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 .

plastic bending

Since the plastic neutral axis divides the cross-section to equal areas, it must be:

\[ A_c=A_t  = \frac{A}{2} \]

The above expression can be utilized for the calculation of the position of the neutral axis. For the case of a double-tee with unequal flanges, let's assume that the plastic neutral axis, for major bending, has a distance \(y_{pna}\) from the bottom edge and that it is located somewhere between the two flanges. The area of the lower half part is:

\(\frac{A}{2}=b_{df} t_{df}+ t_w \left(y_{pna}-t_{df}\right) \)

The only unknown is \(y_{pna}\). Rearranging the equation we get:

\[ y_{pna}=t_{df}+\frac{ \frac{A}{2}-b_{df} t_{df} }{t_w} \]

If the value of \( y_{pna}\) doesn't satisfy the assumption we made that the plastic neutral axis is lying between the two flanges (or in mathematical terms: \( t_{df} \le y_{pna} \le h- t_{uf}\)), then an alternative assumption should be made, either that it is located in the upper or in the lower flange. Following the same principle, a valid value of \(y_{pna}\), that satisfies its assumptions can be derived. These cases are rather uncommon though.

Following, the calculation of \( y_{pna}\), the centroids of the two areas at either sides of the neutral axis can be found and the evaluation of the plastic modulus becomes straightforward. 

Plastic bending

The plastic modulus is related to the stresses of a cross-section under bending in the plastic regime, in an analogous manner the elastic modulus does for the elastic regime. If a bending moment Mx is applied on axis x-x, and the entire section becomes fully plastified, so that every fiber has reached yield stress of the material, \(\sigma_y\), then  the following formula is valid:

\[ \sigma_y = \frac{M_x}{Z_x} \]

From the last equation, the bending moment that causes full yield of the cross-section, typically named plastic moment, can be determined .

The dimensions of plastic modulus are \( [Length]^3 \).

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 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.

See also
Properties of a I/H section
Properties of a Rectangular section
Moment of Inertia of a I/H section
All Cross Section Property tools