Section Modulus calculation and optimization

Aim:

To calculate the section modulus for given hood master section with our design.

Formula:

The formula of section modulus is given by,

S=I/Y

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

Steps to calculate elastic section modulus:-

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

How it depends on strength of the structure:

In simple terms, the section modulus is the ratio of bending moment to bending stress for steel. If your steel has a high section modulus it will be harder to bend and can withstand a high moment without having high bending stress. You need to divide the maximum bending moment by the section modulus to get the bending stress and then compare the bending stress to the allowable tensile stress to see if the steel can take that much moment. All bending equipment have section modulus ratings.

A three-roll section bender can be designed to bend steel with section modulus between 0.4 to 500 in3. So, to determine if the equipment can bend your steel section, you need to calculate the section modulus mathematically or by referring to engineering tables. Then, if that amount is equal or less than that of the calculated strength of the bending machine, your machine should be able to curve the section.

Steps to find the section modulus in catia

In catia we could find moment of inertia, from that we apply the formula and find the section modulus.

1. intersect the cross-section and get the line of the surface

join the gaps in the lines

just extract only the outer line of the surface

and join with require points

join every lines as a single closed curve

then fill it and analyse the moment of inertia

where, M1 is minimum and M2 is maximum moment of inertia

I will go with minimum

after to find the distance between the cross section

and y = D/2

here we check the section modulus for the design given in master section.

y=484.983/2 =242.5mm

M1 = 6.761*10^-7m^4 (min) = 6.761*10^5mm^4

so section modulus is S= M/y

S = 6.761*10^5mm^4/242.5mm = 2788mm^3

lets check with another surface area for offset of 0.5mm :

y=484.983/2 =242.5mm

M1 = 7.137*10^-7m^4 (min) = 7.137*10^5mm^4

so section modulus is S= M/y

S =7.137*10^5mm^4/242.5mm = 2943mm^3

conclusion:

From this challenge i have learned about section modulus, moment of inertia and how it supports for strength of the surface or cross-section.

So from the above calculation by increasing the area of the surface i found that we get more section modulus and more strength.