Section Modulus calculation and optimization
Aim:
To find the section modulus of the hood and optimization to improve the overall section modulus.
Section Modulus:
Section modulus is defined as the ratio of moment of inertia of a beam about its C.G to the maximum distance of extreme x-section of the beam (Ymax).
The section modulus is classified into two types:-
Elastic section modulus
The elastic section modulus is applicable up to the yield point of material.
It is used in most engineering applications
Plastic section modulus
The plastic section modulus is generally used for material in which plastic behavior is observed.
Elastic section modulus
The elastic section modulus is applicable up to the yield point of material.
The elastic section modulus is defined as S = I / y, where I is the second moment of area (or area moment of inertia, not to be confused with moment of inertia) and y is the distance from the neutral axis to any given fibre.
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).
To Find Moment Of Inertia Of A Rectangle
the formula for finding the moment of inertia of a rectangle is given by I=bd³ ⁄ 3, when the axis of rotation is at the base of the rectangle.
I=bd³ ⁄ 3
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 fibre of the beams are subjected to maximum tension.
This picture contains the following scene, 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 the straight beam and its bending in the middle
Calculation:
Case :01
Offset the outer sketch to inner side and the offset distance is 0.75mm
Steps to be follow
- Start the intersect section of outer and inner surface of the hood and join the inter section
- Join the both inner and outer panel intersection sketch
- And offset sketch to the inner side for thickness of 0.75 mm
- And check the inertia using measure inertia
- We got the value of MIN MOI & MAX MOI
- Then find the distance form starting to end point then distance /2 and we got the y
- And to find the section modules of the surface using the formula
Sketch 01
Offset sketch 01
Join the two sketch and make as fill
The Value Of Min MOI & Max MOI
Distance of two corner (D)
To find the y = d/2
Here we are considered the inner pannel section of the to analysis the section inertia analysis after doing the we got the desired results shown in the above image
I (MAX) = 8.915e^-8 m^4 = 8.915*10^4 mm^4
I (MIN) = 1.533e^-5m^4 = 1.533*10^7 mm^4
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2
Y= 484.225/2
Y=242.11
Therefore,
S = 8.915*10^4 mm^4/242.11 mm
S1 = 368.221mm^3
Case :02
Offset the outer sketch to outside side and the offset distance is 0.75mm
Steps to be follow
- Start the intersect section of outer and inner surface of the hood and join the inter section
- Join the both inner and outer panel intersection sketch
- And offset sketch to the outer side for thickness of 0.75 mm
- And check the inertia using measure inertia
- We got the value of MIN MOI & MAX MOI
- Then find the distance form starting to end point then distance /2 and we got the y
- And to find the section modules of the surface using the formula
Sketch 01
Offset sketch 01
Join the two sketch and make as fill
The Value Of Min MOI & Max MOI
Distance of two corner (D)
To find the y = d/2
Here we are considered the inner pannel section of the to analysis the section inertia analysis after doing the we got the desired results shown in the above image
I (MAX) = 9.88e-008m4 =9.88*10^4 mm^4
I (MIN) = 1.554e-005m4 = 1.554*10^7 mm^4
As per the section modulus equation,
S = I/Y
I = MOI (min) Value is considered
Y = Total distance / 2
Y= 484.281/2
Y=242.1405
Therefore,
S = 9.88*10^4 mm^4/242.1405 mm
S2 = 408.027mm^3
Conclusion
After the calculations of the above 2 cases the section modulus of an S1 and S2 the difference is 39.806 mm
The section modulus value of case 2 is higher than the case 1
As we increase the section area, the moment of inertia & section modulus increase its means strength of the material also increase. Which the moment of inertia directly proportional to the strength of the material.