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 the geometry property of the cross-section used for designing beams and flexural members. It does not represent anything physically.

The formula of section Modulus:

It may be defined as the ratio of the 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 at the end of the object (mm^4)

Y= Distance between the neutral axis and the end of the object (mm).

Moment of inertia:

It 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. The moment of Inertia is also known as the angular mass or rotational inertia.

The SI unit of moment of inertia is kg m².

The moment of inertia is usually specified concerning 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,

The 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's 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 the beam is subjected to pure bending the topmost fibres of the beam are subjected to maximum compression, while the bottommost fibre of the beam is subjected to maximum tension.

• This picture contains the following scene, a 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 a straight beam and it's bending in the middle.

Calculation:

Section Modulus :

Case 1:

Here we are considered the inner panel section of the analysis section inertia analysis after doing this we got the desired results shown in the above image

I (MAX) = 4.590567193e+04(mm^4)

I (MIN) = 4.332774364 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2

Y=479.5191/2

Y=239.75

Therefore,

S = 43327.74364 mm^3 /239.75 mm

S1 = 180.74313mm^3

Case 2:

Here we are considering an section of the inner pannel at the curved region and by doing the analysis we got the desired results shown below

I (MAX) = 7.054060364e+03 (mm^4)

I (MIN) = 5.018879483 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2,

Total distance, Y= 258.095 mm

Therefore,

S = 50188.79483/258.095 mm

S2= 194.458 mm^3

Results:

>After the calculations of the above 2 cases the section model of an S1 and S2 difference is 14.458mm

>The section modulus value of case 2 is higher than case 1

>As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increases. Which is the moment of inertia directly proportional to the strength of the material.