Sponsored By:

Copyright © 2011
Total Construction Help, LLC
TotalConstructionHelp.com
All rights reserved

If you Find Our Site
Helpful to You,
We would Appreciate
Any Donations to
Help Us Continue to
Provide You with Free
UpToDate
Information


See Useful Software to Help You in Constructing Your Project

Please Visit Our Sponsors
 A Great Source for Tools and Hardware
 Best Selection of Building Construction Books
 Wide Selection of Engineering Books
 Books for Small Homes and Cottages

 Expert Advice Books
 Books on Architecture
 Find the Hardware you need
 Building Code Books


Section Modulus (Homogeneous Sections  Symmetrical)
When Designing a Beam another property used is Section Modulus (Z) also known as Sx.
The section modulus of the crosssectional shape is of significant importance in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.
It includes the idea that most of the work in bending is being done by the extreme fibres of the beam, ie the top and bottom fibres of the section. The distance of the fibres from top to bottom is therefore built into the calculation.
The elastic modulus is denoted by Z or Sx. To calculate Z, the distance (y) to the extreme fibres from the centroid (or neutral axis) must be found as that is where the maximum stress could cause failure.
In structural engineering, the section modulus of a beam is the ratio of a cross section's second moment of area to the distance of the extreme compressive fibre from the neutral axis.
For symmetrical sections this will mean the Zx max and Zx min are equal. For unsymmetrical sections (Tbeam for example) the section modulus used will differ depending on whether the compression occurs in the web or flange of the section.
The section modulus is directly related to the strength of a corresponding beam. It is expressed in units of volume, such as, inches^{3}, m^{3}, mm^{3}.
For general design, the Elastic section modulus is used, applying up to the Yield point for most metals and other common materials.
The Beams Shown are a Wood and Steel Beams

The following are some typical section and equations for some of the most common members:
Assume that the Square Shape is a 6" x 6", therefore, to find the Section
Modulus (Sx):
Sx = d
^{
3
}
/ 6 = 6
^{
3
}
/ 6 =
36 inch
^{
3
}
, for xx through the Center.
Once we know the Section Modulus, the Moment of Inertia can be computer using
the Equation:
Ix = Sx x c = 36 x 3 =
108 inch
^{
4
}
or the Equation Ix = d
^{
4
}
/ 12 may be used.
or
Sx = d
^{
3
}
/ 3 = 6
^{
3
}
/ 3 =
72 inch
^{
3
}
, for xx through the Bottom Edge.
Assume that the Square Shape is a 4" wide x 6" high, therefore, to find the
Section Modulus (Sx):
Sx = bd
^{
2
}
/ 6 = 4 x 6
^{
2
}
/ 6 =
24 inch
^{
3
}
, for xx through the Center.
Once we know the Section Modulus, the Moment of Inertia can be computer using
the Equation:
Ix = Sx x c = 24 x 3 =
72 inch
^{
4
}
or the Equation Ix = bd
^{
3
}
/ 12 may be used.
or
Sx = bd
^{
2
}
/ 3 = 4 x 6
^{
2
}
/ 3 =
48 inch
^{
3
}
, for xx through the Bottom Edge.
Assume that the Hollow Retangular Shape is a 4" (b) wide x 6" (d) high, with a
2" (b1) x 4" (d1) Hollow Center,
therefore, to find the Section Modulus (Sx):
Sx = (bd
^{
3
}
)(b1d1
^{
3
}
) / 6d =
(4 x 6
^{
3
}
)  (2 x 4
^{
3
}
) / 6 x 6 =
20.44 inch
^{
3
}
, for xx through the Center.
Application of Section Modulus (Sx)
The following are Pratical Examples of Loads and Calculations being applied to
Retangular Members
Example of a Simple Supported Beam with a Uniformly Distributed Load with
Equations and Solutions:
Uniformly Distributed Load (100 lb per ft) on a 2 x 10
Example of a Simple Supported Beam with a Point Concentrated Load with
Equations and Solutions:
Point Concentrated Load (600 lb) on a 2 x 12
Example of a Simple Supported Beam with a Point Concentrated Load with
Equations and Solutions:
Point Concentrated Load (1200 lb) on a 4 x 12

