Section Modulus calculation and optimization

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 fiber 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 fibers of the beam are subjected to maximum compression, while the bottom most fiber of the beams are subjected to maximum tension.

This pictures contains the following scene, A beam on its narrow edge supported at both ends. A man is walking on a straight beam. 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 :

Case 1 :

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

I (MAX) = 9.001583509e+07 (mm^4)

I (MIN) = 2.853179612e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = Total distance / 2

Y=879.6091/2

Y=439.80455

Therefore,

S = 285317.9612 mm^3 /439.80455 mm

S1 = 648.73808mm^3

Case 2 : Trying to increase section modulus by shifting inner panel curves 0.5 mm far.

By offsetting inner panel wall we get below result.

I (MAX) = 8.98479968e+07 (mm^4)

I (MIN) = 2.88885203e+05 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) Value is considered

Y = 439.80455 mm (same as above)

Therefore,

S = 288885.203/439.80455 mm

S2= 656.84905 mm^3

Results:

After the calculations of the above 2 cases the section modulus of an S1 and S2 the difference is 8.11097mm^3

The section modulus value of case 2 is increased with 8.11097mm^3