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Angle (L) Section Calculator

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

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

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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 =
shape details
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Table of Contents
See also
Properties of a Rectangular Tube
Properties of I/H section
Properties of unequal I/H section
Moment of Inertia of an Angle
All Cross Section tools

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Definitions

Table of contents

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 x_c , and from the bottom edge y_c , 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}

We have a special article, about the centroid of compound areas, and how to calculate it. Should you need more details, you can find it here.

L section geometry

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 the 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 better done around the non-centroidal x0,y0 axes, followed by application of the the Parallel Axes Theorem.

First, the moments of inertia Ix0, Iy0 and Ix0y0 of the angle section, around the x0, and y0 axes, are found like this:

\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 around 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}

where, x_c the distance of the centroid from the y0 axis and x_c the distance of the centroid from x0 axis. Expressions for these distances are given in the previous section.

Take in mind, that x, y axes are not the natural ones, the L-section would prefer to bend around, if left unrestrained. These would be the principal axes, that are inclined in respect to the geometric x, y axes, as described in the next section.

Principal axes

Principal axes are those, for which the product of inertia Ixy, of the cross-section becomes zero. Typically, the principal axes are symbolized with I and II and are perpendicular, one with the other. The moments of inertia, when defined around the principal axes, are called principal moments of inertia and are the maximum and minimum ones. Specifically, the moment of inertia, around principal axis I, is the maximum one, while the moment of inertia around principal axis II, is the minimum one, compared to any other axis of the cross-section. For symmetric cross-sections, the principal axes match the axes of symmetry. However, there is no axis of symmetry in an L section (unless for the special case of an angle with equal legs), and as a result the principal axes are not apparent, by inspection alone. They must be calculated, and in particular, their inclination, relative to some convenient geometric axis (e.g. x, y), should be determined. 

Knowing the moments of inertia I_x , I_y and the product of inertia I_{xy} , of the L-section, around centroidal x, y axes, it is possible to find the principal moments of inertia I_I, I_{II} , around principal axes I and II, respectively, and the inclination angle \theta , of the principal axes from the x, y ones, with the following 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_I is considered the major principal moment (maximum one) and I_{II} the minor principal moment (minimum one). It follows that: I_I>I_{II} .

Principal axes of L-section

Moment of inertia and bending

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.

Polar moment of inertia of L-section

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 , I_y , the moments of inertia around axes x-x and y-y, respectively, which are mutually perpendicular to z-z and meet at a common origin.

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

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Elastic section modulus

The elastic section modulus S_x of any cross section around centroidal axis x-x, describes the response of the section under elastic flexural bending. 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 distance from centroid of a given section fiber (that is parallel to the axis). For the angle section, due to its unsymmetry, the S_x is different for a top fiber (at the tip of the vertical leg) or a bottom fiber (at the base of the horizontal leg). Normally, the more distant fiber (from centroid) is considered when finding the elastic modulus. This happens to be at the tip of the vertical leg (for bending around x-x). Using the possibly bigger Y , we get the smaller S_x , which results in higher stress calculations, as will be shown shortly after. This is usually preferable for the design of the section. Therefore:

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

S_{x,\textit{max}} = \frac{I_x}{y_c}

where the “min” or “max” designations are based on the assumption that y_c \lt h-y_c , which is valid for any angle section.

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

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

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

elastic bending

If a bending moment M_x 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). Over the neutral axis the stresses are by definition zero. Absolute maximum stress 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 .

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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, 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 an 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} \ge t \\ y_{pna} b = \frac{A}{2} & \text{ , if } y_{pna} \lt t \\ \end{array} \right.

which becomes:

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

where y_\textit{pna} , the distance of the plastic neutral axis from the bottom end 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.

plastic neutral axis in L section for bending around a horizontal axis

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 vertical leg, the plastic modulus can be found like this:

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

which becomes:

Z_x =t\ { h_1^2-b^2 + 2bh\over 4} \quad , t \le {A\over2 b}

where h_1=h-t .

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

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

which can be simplified to:

Z_x ={b t^2\over 4} + {h t h_1\over 2} -{ t^2 h_1^2 \over4b} \quad , t \gt {A\over2 b}

where h_1=h-t .

Around y axis

The plastic section modulus around y axis can be found in a similar way. If we orient the L-section, so that the vertical leg becomes horizontal, then the resulting shape is similar in form with the originally oriented one. Thus, the derived equations should have the same form, as found for the x-axis. We only have to swap h for b and vice-versa. This way, the exact position of the plastic neutral axis is given by the following formula:

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

where x_\textit{pna} , the distance of the plastic neutral axis from the left end 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 (see figure below).

For the first case, that is when the y-axis crosses the horizontal leg, the plastic modulus is found by the formula:

Z_y =t\ { b_1^2-h^2 + 2h b\over 4} \quad , t \le {A\over2 h}

where b_1=b-t .

For the second case, that is when the y-axis crosses the vertical leg, the plastic modulus is found by the formula:

Z_y ={h t^2\over 4} + {b t b_1\over 2} -{ t^2 b_1^2 \over4h} \quad , t \gt {A\over2 h}

where b_1=b-t .

plastic neutral axis in L section for bending around a vertical axis

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.

Angle (L) section formulas

The following table, lists the main formulas for the mechanical properties of the angle (L) cross section.

Angle (L) section formulas

QuantityFormula
Area: A = (h+b-t)t
Perimeter: P = 2b + 2h
Centroid:

x_c = \frac{t}{2A}\left( b^2 + h t - t^2\right)

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

Moments of inertia around centroid \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}
Principal axes and moments of inertia: \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}
Elastic modulus:

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

S_y = \frac{I_y}{b-x_c}

Plastic modulus:

Z_x = \left\{\begin{array}{ll} t\ { h_1^2-b^2 + 2bh\over 4} &\quad , t \le {A\over2 b} \\ {b t^2\over 4} + {h t h_1\over 2} -{ t^2 h_1^2 \over4b} & \quad , t \gt {A\over2 b} \end{array}\right.

Z_y = \left\{\begin{array}{ll} t\ { b_1^2-h^2 + 2h b\over 4} &\quad , t \le {A\over2 h} \\ {h t^2\over 4} + {b t b_1\over 2} -{ t^2 b_1^2 \over4h} & \quad , t \gt {A\over2 h} \end{array}\right.

Plastic neutral axis:

(distances from bottom or left)

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

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

where:

b_1=b-t

h_1=h-t

\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}

Related pages

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

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