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MODULE 1. BASIC CONCEPTS

MODULE 2. SYSTEM OF FORCES

MODULE 3.

MODULE 4. FRICTION AND FRICTIONAL FORCES

MODULE 5.

MODULE 6.

MODULE 7.

MODULE 8.

## LESSON 11. RADIUS OF GYRATION OF AN AREA ABOUT AN AXIS

**11.1 RADIUS OF GYRATION OF AN AREA ABOUT AN AXIS **

Consider an area *A* whose moment of inertia about *x* and *y* axes is *I _{x}* and

*I*. Suppose the area concentrated into a thin strip parallel of to

_{y}*x*-axis as shown in Fig.11.1. such that its moment of inertia about

*x*-axis is same as I

_{x}. Then the distance at which this strip is to be placed from

*x*-axis is called

*radius of gyration*of the area about

*x*-axis, denoted by

*R*.

_{x}**Fig.11.1 Radius of gyration**

According to the parallel axis theorem, the moment of inertia of the strip about *x*-axis is

* I _{xx}* =

*I*+

_{cent}*AR*

_{x}^{2}

*(11.19)*

But *I _{cent}* = \[{{b{t^3}} \over {12}}\] is negligible as

*t*is small,

*t*

^{3}is negligibly smaller.

Therefore,

*I _{xx}* =

*AR*

_{x}^{2}

Or *R _{x} *= \[\sqrt {{{{I_{xx}}} \over A}}\] (11.20)

It is expressed in mm or cm or m.

Hence, the radius of gyration of an area about an axis is defined as the distance from the axis where thin strip (having an area equal to the given area) is to be placed so that the moment of inertia of the strip about the axis is equal to the moment of inertia of the given area about that axis.

Similar to Eq. (11.20), the radius of gyration of the area about *y*-axis [Figure 11.13(c)] and pole or polar radius of gyration [Figure 11.13(d)] are obtained as given in Eqs. (11.21) and (11.22).

* R _{y}* = \[\sqrt {{{{I_{yy}}} \over A}} \] (11.21)

*R _{0}* = \[\sqrt {{J \over A}}\] (11.22)

Where *J *= polar moment of inertia using perpendicular axis theorem

*J* = *I _{xx}* +

*I*

_{yy}*AR*_{0}^{2} = *AR _{x}*

^{2}+

*AR*

_{y}^{2}

Hence

* R _{0}^{2 }= R_{x}^{2} + R_{y}^{2}* (11.23)

**Example 1**: Find the moment of inertia of the rectangular section as shown in Figure------- about the faces PQ and RS.

**Fig.11.2**

**Sol**: From the Figure, width b = 50 mm and depth d = 70 mm

The moment of inertia of the entire rectangular area about PQ is given by

*I _{PQ} = I_{xx} + Ah^{2}*

= \[{{b{d^3}} \over {12}} + \left( {b \times d} \right){h^2}\]

= \[{{50 \times {{70}^3}} \over {12}} + \left( {50 \times 70} \right) \times {\left( {35} \right)^2}\]

=1429166.67 + 4287500

= 5716666.67 mm^{4}

*I _{BC} = I_{yy} + Ah^{2}*

= \[{{b{d^3}} \over {12}} + \left( {b \times d} \right){h^2}\]

= \[{{70 \times {{50}^3}} \over {12}} + \left( {70 \times 50} \right) \times {\left( {25} \right)^2}\]

= 729166.67+2187500

= 2916666.67 mm^{4}

**Example 2**: Find the moment of inertia of a hollow rectangular section about two mutually perpendicular axes in its plane and passing through its centroid . The dimensions of the outer rectangle are 80 mm×100 mm and that of the inner rectangle is 60 mm×80 mm.

**Fig.11.3**

**Sol:** Outer rectangle: B = 80 mm and D = 100 mm

Inner rectangle: b = 60 mm and d = 80 mm

The formula is *I _{Gx}* = \[{1 \over {12}}\left( {B{D^3} - b{d^3}} \right)\]

= \[{1 \over {12}}\left( {80 \times {{100}^3} - 60 \times {{80}^3}} \right)\]

= 4106666.67 mm

^{4}

*I _{Gy}* = \[{1 \over {12}}\left( {D{B^3} - d{b^3}} \right)\]

= \[{1 \over {12}}\left( {100 \times {{80}^3} - 80 \times {{60}^3}} \right)\]

= 2826666.67 mm^{4}

**Example 3**: Determine the moment of inertia and radius of gyration about the horizontal axis passing through the centroid of the section as shown in Figure.

**Fig.11.4**

**Sol:** The given I-section is symmetrical about the y-y axis, therefore,

\[\bar x = {{100} \over 2}\] = 50 mm

To find , we have to divide the whole Figure into standard areas.