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Properties of a Rectangular hollow section
Properties of a I/H section
Properties of a non-symmetric I/H section
Moment of Inertia of an Angle
All Cross Section Property tools

Properties of Angle Cross-Section (L)

- By Dr. Minas E. Lemonis, PhD - Updated: March 27, 2019

This tool calculates the properties of an angle cross-section (also called L section). Enter the shape dimensions h, b and t below. The calculated results will have the same units as your input. Please use consistent units for any input.

h =
b =
t =
icon

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

Properties related to y-y axis:
Iy =
Ixy =
Sy =
Zy =
xpna =
Rgy =
Properties related to major principal axis I:
II =
θI (°) =
SI =
RgI =
Properties related to minor principal axis II:
III =
θII (°) =
SII =
RgII =
Other properties:
Iz =
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Definitions

Geometry

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

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

The distance of the centroid from the left edge of the section xc, and from the bottom edge yc, can be found using the first moments of area, of the two legs:

\begin{split} & x_c & = \frac{1}{A}\left( \frac{t}{2}\left( b^2 + h t - t^2\right) \right) \\ & y_c & = \frac{1}{A}\left( \frac{t}{2}\left( h^2 + b t - t^2\right) \right) \end{split}

Moment of Inertia

The moment of inertia of an angle cross section can be found if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. The final area, may be considered as the additive combination of A+B+C. However, a more straightforward calculation can be achieved by the combination (A+C)+(B+C)-C. Also, the calculation is done relative to the non-centroidal x0,y0 axes, followed by application of the the Parallel Axes Theorem. So, the moments of inertia Ix0, Iy0and Ix0y0of the angle section, are given by the formulas:

\begin{split} & I_{x0} & = \frac{t}{3} \left(b t^2 + h^3 -t^3 \right) \\ & I_{y0} & = \frac{t}{3} \left(h t^2 + b^3 -t^3 \right) \\ & I_{x0y0} & = \frac{t^2}{4} \left(b^2 + h^2 -t^2 \right) \end{split}

shape I finding

Application of the Parallel Axes Theorem makes possible to find the moments of inertia in respect with the centroidal axes x,y:

\begin{split} & I_{x} & = I_{x0} - A y_c^2 \\ & I_{y} & = I_{y0} - A x_c^2 \\ & I_{xy} & = I_{x0y0} - A x_c y_c \end{split}

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 Izabout an axis z-z (perpendicular to the section), can be done with the Perpendicular Axes Theorem:

I_z = I_x + I_y

where the Ixand Iyare 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 Sxof 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 Ixthe 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 angle section, due to the unsymmetry, Sxis 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 assumptions that y_c lt h-y_c, which is valid for any angle section.

Similarly, for the elastic section modulus Sy, relative to the y-y axis, the minimum elastic section modulus is found with:

S_{y,min} = \frac{I_y}{b-x_c}

where the min designation is based on the assumptions that x_c lt b-x_c, which is valid for any angle section.

elastic bending

If a bending moment Mxis 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, Atin tension and Acin 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 Ycthe distance of the centroid of the compressive area Acfrom the plastic neutral axis and Ytthe respective distance of the centroid of the tensile area At.

For the case of angle 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 = \frac{A}{2} & \text{ , if } y_{pna} \gt t \\ y_{pna} b = \frac{A}{2} & \text{ , if } y_{pna} \le t \\ \end{array} \right.

where ypnathe distance of the plastic neutral axis from the bottom edge of the section. The first equation is valid when the plastic neutral axis passes through the vertical leg, while the second one when it passes through the horizontal leg. 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.

plastic bending

Similarly, for y-y bending, the plastic neutral axis passes can be found by either one of the following two equations:

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

where xpnathe distance of the plastic neutral axis from the left edge of the section. The first equation is valid when the plastic neutral axis passes through the horizontal leg, while the second one when it passes through the vertical leg.

Principal axes

The principal moments of inertia I_I, I_{II}about principal axes I and II, and the rotation angle \thetaof the principal axes from centroidal x, y ones can be found, using the formulas:

\begin{split} & I_{I,II} & = \frac{I_x+I_y}{2} \pm \sqrt{\left(\frac{I_x-I_y}{2}\right)^2 + I_{xy}^2} \\ & \tan 2\theta & = -\frac{2I_{xy}}{I_x-I_y} \end{split}

By definition, I_Iis considered the major principal moment (maximum one) and I_{II}the minor principal moment (minimum one).

Radius of gyration

Radius of gyration Rgof 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 Rectangular hollow section
Properties of a I/H section
Properties of a non-symmetric I/H section
Moment of Inertia of an Angle
All Cross Section Property tools