Section Modulus calculation and optimization
AIM:
Section modulus calculation and optimization
OBJECTIVE:
Use section of the Hood and calculate the section modulus.
Vary the section in such a way that new section is an improved version of previous section.
THEORY:
It is a way of quantifying a cross-section of a given geometry. It is directly proportional to the area moment of inertia of the cross-section. Higher the section modulus, the more the moment of inertia, more stiffness and thus more resistance to deformation and bending. Section modulus can be improved by varying the geometry of the cross-section. In other words, stiffness can be improved by just improving the geometry without the need for adding extra material.
Calculate the section modulus using the formula S=I/Y
S= Section modulus
I= Moment of Inertia
Y= Distance between neutral axis and the extreme end of the object
the section modulus we need the cross section of hood. For that I created a cross section using intersection curve.
study 1:
FIG: Cross-section of hood
Find the inertia using section inertia tool in NX:
results:
Section#: 1
Type: Hollow
CG: (0.112552359, 744.484845258, 0.000000000) [mm]
MOI (Max):4.798574140e+06[mm^4]
MOI (Min): 1.560195958e+04[mm^4]
Principal Directions:
Max: (-0.000040338, 0.999999999, 0.000000000) Min: (0.999999999, 0.000040338, 0.000000000)
Section Length:3.683900075e+03[mm]
Section Area:73.668275708[mm^2]
1. Here the minimum moment of inertia, I=1.560195958e+04[mm^4]
The distance between neutral axis and extreme end of the object,Y =440.35mm
Therefore the Section Modulus, S = I/Y
s= 1.560195958e+04[mm^4]/440.35mm
=35.44[mm^3](minimum strength).
From the section#: 1, using Minimum Moment of Inertia and extreme distance of layer, we got the section modulus(S) value is 35.44[mm^3]
2.Here the maximum moment of inertia, I =4.798574140e+06[mm^4]
Section modulus(S):-I/Y=4.798574140e+06[mm^4]/440.35mm
=10.898x103[mm^3](Maximum strength).
From the section#: 2, using maximum Moment of Inertia and extreme distance of layer, we got the section modulus(S) value is 10.898x103[mm^3].
Section#:2
In this section modulus to increase of inner panel distance from the previous section#:1. Offset of the inner panel 1.5mm
CG: (-0.005575477, 744.342533614, -931.612102839) [mm]
MOI (Max):4.839266788e+06 mm^4]
MOI (Min):1.691339344+04[mm^4] Principal Directions: Max: (-0.000001942, 0.985084513, 0.172071214) Min: (1.000000000, 0.000000611, 0.000007785)
Section Length:3.693341641e+03[mm]
Section Area:73.866858879 mm^2]
1. Here the minimum moment of inertia, I =1.619339344e04[mm^4]
The distance between neutral axis and extreme end of the object,Y =440.35mm
Therefore the Section Modulus, S = I/Y
S =1.69339344e04[mm^4]/440.35mm
=38.45 [mm^3](minimum strength).
From the section#: 1, using Minimum Moment of Inertia and extreme distance of layer, we got the section modulus(S) value is 38.45[mm^3]
2.Here the maximum moment of inertia, I =4.839266788e+06 mm^4]
Section modulus(S):-I/Y=4.839266788e+06 mm^4]/440.35mm
=10.989x103[mm^3](Maximum strength).
From the section#: 2, using maximum Moment of Inertia and extreme distance of layer, we got the section modulus(S) value is 10.989x103[mm^3].
- So, here when comparing both the hoods there is changes in both MIN and MAX strength values from35.44[mm^3] to 10.898x103[mm^3],Hence with increase in cross sectional area results in increase in strength.
- But, So, here when comparing both the hoods there is changes in both MAX and MIN strength values,what i came to know that when i increased sectional area by 1mm the value of minimum strength was increased from 38.45 [mm^3] to 10.989x103[mm^3],i think this due to design of hood, here what we want to know that when increase in cross sectional area there will be increase in minimum strength or maximum strength or increase in both.
CONCLUSION:-
The strength of the material is directly proportional to the moment of inertia, hence increase in sectional area there is have increase in overall mass of an object ,hence have increase in strength.