Section Modulus calculation and optimization

AIM :

To calculate the section modules of the hood design respectively.

PROCEDURE:

y = distance between the neutral axis and the extreme end of the object
Area moment of Inertia:
The area moment of inertia of a beams cross sectional 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 = Youngs 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 fibres of the beams are subjected to maximum tension


Y Distance = 439.35 mm

case 1


MOI (max) = 4.837720e+06 (mm^4)
MOI (min) = 1.5950780e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =878.7mm
Y = 878.7 / 2 = 439.35 mm
Therefore, S =1.5950780e+04 mm^4/439.35 mm

S = 36.30540571mm^3

Case 2


MOI (max) = 4.817020e+06 (mm^4)
MOI (min) = 1.6015960e+04 (mm^4)
As per the section modulus equation,
S = I/Y
I = MOI (min) value is considered
Y = Total distance /2, Total distance =878.7mm
Y = 878.7 / 2
= 439.35 mm

Therefore, S =1.6015960e+04 mm^4/439.35 mm
S = 36.453761mm^3

Conclusion:

Hence when comparing case 1 & case 2,

1. The section modulus value of case 2 is higher than the case 1