Jump to

ADVERTISEMENT

I/H Section Calculator

- By Dr. Minas E. Lemonis, PhD - Updated: July 8, 2020

This tool calculates the properties of an I/H cross-section, also known as double-tee, I-beam or I-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 =
Properties related to x-x axis:
Ix =
Sx =
Zx =
Rgx =

ADVERTISEMENT

Properties related to y-y axis:
Iy =
Sy =
Zy =
Rgy =
Other properties:
Iz =
I section geometry
Share this
Table of Contents
See also
Properties of unequal I/H section
Properties of a Rectangular Tube
Properties of a Circular section
Moment of Inertia of a I/H section
All Cross Section tools

ADVERTISEMENT

ADVERTISEMENT

Definitions

Table of contents

Geometry

The area A and the perimeter P, of an I/H cross-section, can be found with the next formulas:

\begin{split} & A & = 2b t_f + (h-2t_f)t_w \\ & P & = 4b + 2h - 2t_w \end{split}

Moment of Inertia

The moment of inertia of an I/H section can be found if the total area is divided into three, smaller subareas, A, B, C, as shown in figure below. The final area, may be considered as the additive combination of A+B+C. However, since the flanges are equal, a more straightforward combination can be (A+B+C+2V)-2V. This way, the moment of inertia I_x of the I/H section, relative to centroidal x-x axis, is determined like this:

I_x = \frac{b h^3}{12} - \frac{(b-t_w) (h-2t_f)^3}{12}

where h the section height, b the width of the flanges, t_f the thickness of the flanges and t_w the thickness of the web.

The moment of inertia I_y of the I/H section, around centroidal y-y axis, is found by combining the individual moments of inertia, from the two flanges (subareas B and C) and the web (subarea A). The centroids of these subareas are aligned with the cross-section y axis, and therefore adding them together is acceptable (otherwise we should have employed the parallel axis theorem). The moment of inertia I_y is:

I_y = 2\frac{t_f b^3}{12} + \frac{(h-2t_f) t_w^3}{12}

which is simplified to:

I_y =\frac{t_f b^3}{6} + \frac{h_w t_w^3}{12}

where h_w=h-2t_f , is the clear web height.

I-section subareas for Ixx and Iyy calculation

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 .

ADVERTISEMENT

Elastic modulus of I section

The elastic section modulus S_x of any cross section, for a given axis x-x (centroidal), describes the response of the section under elastic flexural bending, around this axis. It is defined as:

S_x = \frac{I_x}{Y}

where I_x the moment of inertia of the section around x-x axis and Y the offset of a given section fiber (parallel to the same x-x axis) from centroid. Typically the elastic modulus related to the most distant fiber is of interest. If a cross-section is symmetric (the I-section is) around an axis (e.g. x-x) and its dimension perpendicular to this axis is h, then Y=h/2 and the above formula becomes:

S_x = \frac{2 I_x}{h}

Similarly, for the section modulus S_y , around y-y axis, which, for the I/H section, happens to be axis of symmetry too, the above definitions are written as:

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 the neutral axis the stresses are by definition 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 of I section

The plastic section modulus is similar to the elastic one, but defined with the assumption of full plastic yielding, of the cross section, as a result of 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 is a result of equilibrium of internal forces in the cross-section, under plastic bending. Indeed, the compressive force would be A_cf_y , assuming the yield stress is equal to f_y , in compression, and that the material over the entire compressive area has yielded (thus the stresses are equal to f_y everywhere). Similarly, the tensile force would be A_t f_y , using the same assumptions. Enforcing equilibrium:

A_cf_y = A_t f_y\Rightarrow

A_c= A_t

The 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 I-section however, is indeed a symmetric cross-section, both around strong and weak axis. As a result the I-section plastic neutral axis coincides with the elastic one. 

plastic bending
icon

The plastic neutral axis divides the cross section in two equal areas, provided the material features equal tensile and compressive yield strength.

Around x axis

For an I-section, the x-x axis, that is parallel to flanges, is typically the strong axis of the cross-section. The plastic modulus of a cross-section is given by the general formula (assuming bending around x axis):

Z_x = 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 .

For the case of an I-beam, the plastic neutral axis passes through centroid, dividing the whole section into two equal parts. Taking advantage of symmetry it is: Y_c=Y_t . Finding these centroids, is straightforward. We will consider the part above the neutral axis (assumed in compression). The centroid of this part is located at a distance Y_c from the plastic neutral axis. It is convenient to assume that the whole part, is equivalent to the difference between an outer rectangle, having dimensions, b and h/2 minus two smaller rectangles, as shown in the next figure. The centroid of the whole part is then found, by making the static moment of inertia (first moment of area) equal between the whole part and the difference of its outer and the smaller rectangular areas. Remember that the examined part features half the total section area:

{A\over2} Y_c = {b h \over 2} {h\over 4} - 2{b-t_w\over2}\left({h\over2}-t_f\right) { h/2-t_f \over 2}\Rightarrow

Y_c ={1\over A} \left({b h^2 \over 4} - (b-t_w){(h-2t_f)^2\over4}\right)\Rightarrow

Y_c ={1\over A} \left({b h^2 \over 4} -{ (b-t_w) h_w^2 \over 4}\right)

where h_w=h-2t_f , the clear web height.

Half I-section: finding plastic modulus around strong axis
Upper half of an I-section

Taking into consideration that Y_c=Y_t (due to symmetry) and given that A_c=A_t , the plastic modulus, of an I-section, around x axis, is calculated like this:

Z_x = A_c Y_c + A_t Y_t \Rightarrow

Z_x = 2 A_c Y_c \Rightarrow

Z_x = 2 {A\over2} {1\over A} \left({b h^2 \over 4} -{ (b-t_w) h_w^2 \over 4}\right) \Rightarrow

Z_x = \frac{bh^2}{4} - { (b-t_w) h_w^2 \over 4}

One may wonder why we didn't find the plastic moduli of the flanges and the web around their respective local centroidal axes, and then use a parallel axis theorem, to get the respective moduli around the section centroid, similarly to the procedure we follow for the moment of inertia. Unfortunately, there is no parallel axes theorem for the calculation of plastic modulus. It is not possible to find the plastic modulus around a parallel axis, without recalculating it from start.

icon

There is no parallel axes theorem for plastic modulus calculations!

Around y axis

For an I-section, the y-y axis, that is perpendicular to flanges, is typically the weak axis of the cross-section. Instead of finding the centroid of the compressive and tensile areas, these time we will follow a different approach, in order to find the plastic modulus. It is not difficult to prove that the plastic modulus of a rectangle, with dimensions b and h, around a centroidal axis, parallel to b is:

Z={bh^2\over4}

The whole I-section is now considered a combination of three rectangles, two for the flanges and one for the web, as illustrated in the following figure.

I-section subareas for weak axis plastic modulus finding

All three rectangles share the same centroidal axis, which is also the y-axis, of the entire cross section. It would be not possible to proceed otherwise. Applying the above formula, for each one of these rectangles, we get the plastic modulus, of the entire I-section, around y axis:

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

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

Radius of gyration

Radius of gyration R_g 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. The I-section, would have considerably higher radius of gyration, particularly around its x-x axis, because much of its cross-sectional area is located far from the centroid, at the two flanges.

I/H section formulas

The following table, lists the main formulas, related to the mechanical properties of the I/H section (also called double-tee section).

I/H section properties
I/H cross-section

QuantityFormula
Area: A = 2b t_f + h_w t_w
Perimeter: P = 4b + 2h - 2t_w
Moments of inertia

I_x =\frac{b h^3}{12} - \frac{(b-t_w)h_w^3}{12}

I_y =\frac{t_f b^3}{6} + \frac{h_w t_w^3}{12}

Elastic modulus:

S_x = \frac{2I_x}{h}

S_y = \frac{2 I_y}{b}

Plastic modulus:

Z_x = \frac{bh^2}{4} - \frac{(b-t_w) h_w^2}{4}

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

where:

h_w=h-2t_f

Related pages

Properties of unequal I/H section
Properties of a Rectangular Tube
Properties of a Circular section
Moment of Inertia of a I/H section
All Cross Section tools

Liked this page? Share it with friends!

ADVERTISEMENT

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