We have already said that the second moment of area or mass moment of inertia, is a measure of the efficiency of the beam to resist bending loads.
Consider the Universal ‘I’-section Beam shown in Figure 3.1.34. The beam is shown with its principal axes X–X and Y–Y. These axes intersect at the centroid.
Normally when considering the second moment of area, it is neces-sary to find this, about the X–X and Y–Y axes. This is shown in a later example, where on substitution of values into the formulae for the sec-ond moment of area, it is found that when the secsec-ond moment of area is taken about the X–X axis, with the greater amount of material further away from the reference axis. Then the higher will be the value of the second moment of area and the higher will be the resistance to bending of the beam.
In general then, the resistance to bending of a beam, will be greater, as its area and depth increase.
This is easily demonstrated by considering a steel rule (Figure 3.1.35), where it is easy bent around cross-section (A–A) but it is difficult to bend around cross-section (B–B). The second moment of area about the prin-cipal axes for a rectangular cross-section beam is shown next, in Example 3.1.11.
37 mm
12 mm
x
158 mm 75 mm
50 mm 100 mm 24 mm
25 mm
8 mm
8 mm
24mm50mm
1
2 3
y
Figure 3.1.33 Cross-section of beam for Example 3.1.10
Similarly, taking moments about A–A, we get:
6388y (3700 37) (2400 12) (288 8) which gives:
Then centroid is 49.94 mm to the right of the B–B datum and 26.3 mmabove the A–A datum.
y 168004 mm.
6388 26 3.
Y
Y
x X
Centroid
Figure 3.1.34 Universal beam showing principle axes
A Easily bent or flexed across the flat
Difficult to bend edge on Figure 3.1.35 Resistance to
bending of a steel rule
Example 3.1.11
Derive the formulae for the second moment of area Ixxabout the X–X axis for the rectangular cross-section, shown (Figure 3.1.36a).
We are required to find the second moment of area about the X–X principal axis. Figure 3.1.36b shows the same rect-angular section, with an elemental strip and its breadth and depth, marked on.
The first stage in determining the second moment of area of a beam cross-section is to find its centroid. This is easily found for a rectangular cross-section and is shown in Figure 3.1.36(a), as being at the intersection of the principal axis.
Then considering this elemental strip having breadth b and depth dy, at a distance y from the X–X axis.
Then the second moment of area of this strip is its area multiplied by distance y (b)(dy)(y2) or by2dy. So the second moment of half of the rectangular area, is the sum of all these quantities.
Then, using the integral calculus to sum these quantities.
We have:
Ixxfor half the rectangular area .
Then for complete rectangular area .
By symmetry the second moment of area about the Y–Y principal axis is given as:
I db
An example for determining the second moment of area of a rectan-gular cross-section, when moments are taken about its base will be found in the calculus section of Chapter 6 on analytical methods, Example 6.3.26, as mentioned before.
The table given below summarises the second moment of area for rectangular and circular shapes about their various axes.
Section detail Second moment of area Rectangular section about
Circular section about principal axes (polar axes) (Figure 3.1.40)
I I d
Example 3.1.12
Consider the I section universal beam shown in Example 3.1.9. Determine the second moment of area about the prin-cipal axes IXXand IYY, for this beam.
The beam is reproduced here in Figure 3.1.41, showing the I-section in its normal vertical position, with the principal axes marked on in their correct positions about the centroid of area.
We know from Example 3.1.9, that the centroid of area for the whole beam acts through the center of the web, at 170.7 mm from the A–A axis.
We can verify the centroid position by taking moments about the base B–B. So that:
14 000x (4500 15) (7500 145) (2000 290) and 14 000x 1 810 000 or,
as expected!
To find IXX, we first divide the I-beam into three rectangular sections, as we did, when finding its centroid. Then using the parallel axis theorem, that is IXX INN AN(SN)2. We find the individual values of IXXfor each of the rectangular areas, then sum them to find the total value of IXXfor the whole beam section.
So, using the parallel axes theorem and taking moments about X–X. We get for the top flange (Area 1):
Similarly, for the web, we get:
And for the bottom flange, we get:
Therefore total second moment of area about X–X for the beam section is,
To determine the second moment of area of any shape about an axis B–B, parallel to another axis A–A, at a perpendicular distance S, then, the area of the shape multiplied by the distance squared (AS2) must be added to the second moment of area about A–A, that is:
IBBⴝ IAAⴙ AS2.
Now to find IYY, we note that all the individual rectangular second moments of area, lie on the Y–Y axis, therefore, S 0 and in all cases, the total second moment of area about Y–Y reduces to the sum of the individual rectangular second moments, that is:
IYYⴝ 10.67 ⴛ 106mm4.
Make sure that you are able to follow the whole argument and obtain, the same numerical values.
I I I I Figure 3.1.41 I – Section
universal beam for example 3.1.12