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Aim To calculate the section modulus of the previously designed hood for analyzing its strength and also optimizing the design to see the difference in the bending strength of the hood. Higher the section modulus of a structure, the more the resistive it becomes to bending.Section modulusSection modulus is a…
Shrivardhan Magdum
updated on 15 May 2023
Aim
To calculate the section modulus of the previously designed hood for analyzing its strength and also optimizing the design to see the difference in the bending strength of the hood. Higher the section modulus of a structure, the more the resistive it becomes to bending.
Section modulus
Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members, it is also referred to as the Polar Modulus or the Torsional Sectional Modulus. Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness. Any relationship between these properties is highly dependent on the shape in question. Equations for the section moduli of common shapes are given below. There are two types of section moduli, the elastic section modulus and the plastic section modulus.
S=I/Y
Here,
S = Section modulus
I= Moment of Inertia (Unit mm 4)
Y= Distance from Neutral axis to the extreme end (unit mm)
The Unit of a “section modulus” is (mm3)
Changing the position can show the difference in bending which is because of the centroid distance from extreme fiber end.
In Layman's terms, in this position there is more mterial from axis to the specific direction, to resist the load which makes less bend amd improves the strength.
First Design Study:-
The Intersection of the design was taken and then section Inertia Analysis was done-
Moment of inertia(max) = 8.19398×10^5×mm^4
Moment of inertia(min) = 5.34031×10^3×mm^4
Y=485.7599/2=242.8799 mm
S=I/Y
S=5.34031×10^3/242.87= 21.98 mm^3
Hood Design Optimization:-
By increasing the depth of the inner panel by 0.5 mm
Here,
Moment of inertia(max) = 8.16698×10^5×mm^4
Moment of inertia(min) = 5.51216×10^3×mm^4
Y=485.7599/2=242.8799 mm
S=I/Y
S=5.51216×10^3/242.87=22.69 mm^3
Conclusion
So, here the difference can be seen in the moment of inertia and the change in the section modulus can be seen by the change in the design(depth).
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