Section Modulus calculation and optimization

Aim

To calculate section modulus of hood and optimize the design to have improved section modulus.

Section modulus:-

Section modulus is a direct measure of a strength of any cross sectional area of an object about an reference axis.

The formula of section modulus is given by,

S=I / Ymax

Where,
I = Moment of inertia (mm^4)
Y = Distance between neutral axes and outermost fiber (mm)

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. Or in more simple terms, it can be described as a quantity that decides the amount of torque needed for a specific angular acceleration in a rotational axis. Moment of Inertia is also known as the angular mass or rotational inertia. The SI unit of moment of inertia is kg m².

Moment of inertia is usually specified with respect to a chosen axis of rotation. It mainly depends on the distribution of mass around an axis of rotation. MOI varies depending on the axis that is chosen.

In General form,

Moment of Inertia is expressed as I = m × r²

where,

m = Sum of the product of the mass.

r = Distance from the axis of the rotation.

Area Moment of Inertia:

The area moment of inertia of a beam cross section area measures the beam ability to resist

Steps to calculate section modulus:-

• Step 1:- Find location of centroid of given cross section.
• Step 2:- Find moment of inertia about neutral axis (INAINA)
• Step 3:- Find YmaxYmax from neutral axis.
• Step 4:- Find elastic section modulus using the formula provided above.

Moment of inertia is nothing but offering resistance against the moment, bending or twisting. So higher the moment of ineria higher will be its resistance to deform is higher.

Here the sectional modulus of hood design before optimization

Momemt of inertia is calculated by NX-CAD software

I_max – 1.552875789 x 10^7 (mm^4)

I_min – 1.241069295 x 10^5 (mm^4)

Y = Total length / 2

= 486.8738/2

=243.4369 mm

section modulus for max. = 1.552875789 x 10^7 / 243.4369

= 6.37896633 x 10^4

section modulus for min. = 1.241069295 x 10^5 / 243.4369

= 0.050981149 x 10^4

So for this hood design section modulus range from 6.37896633 x 10^4 mm^3 to 0.050981149 x 10^4 mm^3

Optimized design

Inorder to increase the sectional modulus, cross sectional area has to be increased. So here the optmized design:-

I_max – 1.548221723 x 10^7 (mm^4)

I_min – 1.279672082 x 10^5 (mm^4)

Y = Total length / 2

= 486.8738/2

=243.4369 mm

section modulus for max. = 1.548221723 x 10^7 / 243.4369

= 6.35984817 x 10^4

section modulus for min. = 1.279672082 x 10^5 / 243.4369

= 0.05256689 x 10^4

So for this hood design section modulus range from 6.35984817 x 10^4 mm^3 to 0.05256689 x 10^4 mm^3

It is evident that if the section modulus is increased from 6.37896633 x 10^4 mm^3 to 6.35984817 x 10^4 mm^3 by increasing the sectional area.

Conclusion

Strength of material is directly proportional to the moment of interia, thus increasing the sectional area increase the increases strength of component.

Thus sectional modulus for hood design is calculated and optimized design is also done to increase the section modulus of the hood.