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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 9. Radius of Gyration of an area about an Axis

**9.1 PARALLEL AXIS THEOREM**

Statement – Moment of inertia of an area about any reference axis is equal to the sum of moment of inertia of the same area about its centroidal axis parallel to reference axis and the product of area and the square of the distance between the reference and centroidal axes.

* I _{ref}* =

*I*+

_{cent}*Ah*

^{2}(9.1)

Fig.1 Parallel Axis Theorem

An axis passing through the centroid C of an area is called centroidal axis. The theorem can be proved as:

The moment of inertia of an area about a given reference axis is written using Eq.(8.3a) as

* I _{ref}* = ∫(

*h+y*)

^{2 }

*dA*

Here (*h+y*)^{2} = *h*^{2} + 2*hy* +*y*^{2}

Therefore,

*I _{ref}* = ∫

*h*

^{2 }

*dA*+ ∫2

*hydA*+ ∫

*y*

^{2}

*dA*

*I _{ref}* =

*h*

^{2}∫

*dA*+ 2

*h*∫

*ydA*+ ∫

*y*

^{2}

*dA*

In the above expression ∫*y*^{2}*dA* = I* _{cent}* as per Eq.(8.3a), ∫

*dA*=

*A*and ∫

*ydA*= first moment of an area about centroidal axis which will be equal to zero.

Hence, *I _{ref}* =

*h*

^{2}

*A*+

*I*

_{cent}or* I _{ref}* =

*I*+

_{cent}*Ah*

^{2}

**Example:** The area in Fig… is symmetrical with respect to the *x* and *y* axis. The area is 150 m^{2} and the moment of inertia with respect to the ‘*b*’ axis is 4200 m^{4}. Determine the moment of inertia of the area with respect to the ‘*a*’ axis.

**Solution: **According to the parallel axis theorem,** **

*I _{b}*=

*I*+

_{cent}*Ah*

^{2}

4200 = *I _{cent }*+ 150(4)

^{2}

*I _{cent }*= 2400 – 4200

*I _{cent }*= 1800 m

^{4}

*I _{a}*

_{ }=

*I*+

_{cent}*Ah*

^{2}

= 1800 + 150(6)^{2}

*I _{a}*

_{ }= 7200 m

^{4}

**9.2 PERPENDICULAR AXIS THEOREM**

Statement: Moment of inertia of a plane area about an axis perpendicular to the plane of the figure (*z*-axis) is equal to the sum of moment of inertia of the same area about two rectangular axes in the plane of the area (*x* and *y* axes).

Fig.2 Perpendicular axis theorem

*I _{z}* =

*I*+

_{x}*I*(9.2)

_{y}According to the definition of moment of inertia [Eq.8.3a] about *z*-axis is given by

*I _{z}*

_{ }= ∫

*r*

^{2}

*dA*

From Fig. 2, *r*^{2} = *x*^{2} + *y*^{2} can be used.

Therefore,

*I _{z}* = ∫(

*x*

^{2}+

*y*

^{2})

*dA*

*Iz* = ∫*x*^{2}*dA* + ∫*y*^{2}*dA*

*I _{z}* =

*I*+

_{y}*I*

_{x}Or *Iz *= *I _{x}* +

*I*

_{y} **9.3 MOMENT OF INERTIA OF GEOMETRICAL FIGURES ABOUT CENTROIDAL AXES**

Using parallel Axis Theorem and moment of inertia of geometrical figures about one of their edges, the moment of inertia about their centroidal axis can be determined. Centroidal moment of inertia for a rectangle, triangle, semicircle and quarter circle are obtained as follows:

**9.3.1 Rectangular Area**

Moment of inertia of a rectangle about x-axis as shown in Fig.3 is given as [Eq.(8.5)]

*I _{x}*

_{ }= \[{{b{d^3}} \over {12}}\]

Fig.3 Moment of inertia of a rectangle about its centroidal axis

From parallel axis theorem,

*I _{cent}*

_{ }=

*I*–

_{ref}*Ah*

^{2}

*Ic _{x}* =

*I*

_{x}_{ }–

*Ah*

^{2}= \[{{b{d^3}} \over 3} - bd\] \[{\left( {{d \over 2}} \right)^2}\]

*I*= \[{{b{d^3}} \over {12}}\] (11.12a)

_{cx}Similarly

* I _{cx}* = \[{{b{d^3}} \over {12}}\] (11.12b)