Section Modulus calculation and optimization

Aim:

To find the section modulus of the hood and optimization to improve the overall section modulus.

Objective:

Use the section from your Hood design and calculate the section modulus using the formula S= I/y.

Section Modulus:

The section modulus (S) is geometry property of the cross section used for designing beams and flexural members. It does not represent anything physically.

Formula of section Modulus: It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Which is equal to yield stress.

Therefore

S=I/y, Where

S= Section modulus (mm^3)

I = Moment of inertia of the section to the stress in the extreme end of the object (mm^4)

Y= Distance between the neutral axis and extreme end of the object (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 bending. The larger the moment of inertia less will be the bend.

Where,

M = Bending Moment

I = Area moment of Inertia

E = Young modulus

R = Radius of gyration

 

Neutral Axis: When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.

                                          

• This picture contains the following scene, Diagram of a beam on its narrow edge supported at both ends. A man is walking on a straight beam. Diagram of a beam resting on its flat side supported at both ends. A man is walking on the straight beam and its bending in the middle.

Calculation:

Section Modulus :

>Here we are considered the inner Pannel section for analysis the section inertia analysis. After that we got the desired results which is shown in the above image

I (MAX) = 1.5824 x 10⁷ mm⁴

I (MIN) = 1.9984 x 10⁵ mm⁴

As per the section modulus equation,

S = I/Y

I (Minimum MOI) = 1.9984 x 10⁵ mm⁴

Y = Total distance / 2 

= 487.39 mm/2 =243.695 mm

S= I/Y =1.9984 x 10⁵ mm⁴ / 243.695 mm

            =820.0414 mm³ =0.82 x 10³ mm³

>Here we are considered the inner Pannel section for analysis the section inertia analysis. After that we got the desired results which is shown in the above image

I (MAX) = 1.5820 x 10⁷ mm⁴

I (MIN) = 2.2154 x 10⁵ mm⁴

As per the section modulus equation,

S = I/Y

I (Minimum MOI) = 2.2154 x 10⁵ mm⁴

Y = Total distance / 2 

= 487.39/2 =243.695 mm

S= I/Y =2.2154 x 10⁵ mm⁴ / 243.695 mm

            =909.0871 mm³ =0.909 x 10³ mm³

>Here we are considered the inner Pannel section for analysis the section inertia analysis. After that we got the desired results which is shown in the above image

I (MAX) = 8.5983 x 10⁷ mm⁴

I (MIN) = 2.0902 x 10⁵ mm⁴

As per the section modulus equation,

S = I/Y

I (Minimum MOI) = 2.0902 x 10⁵ mm⁴

Y = Total distance / 2 

= 865.2449/2 =432.6224 mm

S= I/Y = 2.0902 x 10⁵ mm⁴/ 432.6224 mm

            = 483.1465 mm³ =0.4831 x 10³ mm³

^.

>Here we are considered the inner Pannel section for analysis the section inertia analysis. After that we got the desired results which is shown in the above image

I (MAX) = 8.5908 x 10⁷ mm⁴

I (MIN) = 2.2601 x 10⁵ mm⁴

As per the section modulus equation,

S = I/Y

I (Minimum MOI) = 2.2601 x 10⁵ mm⁴

Y = Total distance / 2 

= 865.2449/2 =432.6224 mm

S= I/Y = 2.2601 x 10⁵ mm⁴ / 432.6224 mm

            = 522.4186 mm³ =0.5224 x 10³ mm³