Section Modulus calculation and optimization
REPORT
ON
SECTION MODULUS
AIM:- To calculate the section modulus of hood panel and then optimizing the hood panel for better section modulus.
INTRODUCTION:-
Section modulus is geometry for a given cross- section which is used to design of beam or flexural members.
Formula of section modulus:-
it may be defines as the ratio of total moment resisted(inertia) by the section to the stress in the extreme fiber which is equal to yield stress.
there,
S=I/Y
Where
S=section modulus(mm^3)
I=Moment of inertia of the section about the bending axis(mm^4)
Y=Distance between the neutral axis and extrem end of the object(mm)
where as,
M= Bending moment
I= Area moment of inertia
E=Radius of gyration .
NEUTRAL AXIS:-when beam is subjected to pure bending the top most fiber of the beam are subjected to maximum compression , while the bottom most fiber of the beam are subjected to maximum tension .
Material of hood:- stainless steel AISI 302-cold rolled
where yield strength steel is 520MPa=520N/mm^2
calculation :-
CASE 1:- Before Optimization
The Hood was already created is imported and the section inertia analysis is done.
MOI(max)=4.851962186e+06(mm^4)
MOI(MIN)=1.566679550e+04(mm^4)
As per the section modulus equation .
S1=I/Y
Y=Total distance /2
=880.826/2
=440.413mm
At minimum value of MOI.
S1=I(MIN)/Y
S1=(15667.9550mm^4)/440.413mm
=35.5755960882172 mm^3.
At Maximum value MOI.
S=I(MAX)/Y
S=4851962.186mm^4/440.413mm
=1.1016845x10^6mm^3
CASE 2: After optimization
the hood inner panel's curve is offset by 0.5mm outwards and section inertia analysis is done same case 1 for closed curve.
MOI(max)=4.940889463e+06[mm^4]
MOI(MIN)=2.101772197e+04[mm^4]
as per section modulus equation
S2=I/Y
I=MOI(min) value is considerd
Y=Total distance /2
=880.826/2
=440.413mm
At minmum value
S2= 21017.72197/440.413
=47.72275mm^3
At maximum value
S2=4940889.46[mm^4]/440.413mm
=1.12187638x10^4mm^3
case-3
Again the hood inner panel's curve is offset by 0.5mm outwards and section inertia analysis is done same case 2 for closed curve.
MOI(max)=5.050434992e+06[mm^4]
MOI(MIN)=2.482791660e+04[mm^4]
as per section modulus equation
S3=I/Y
I=MOI(min) value is considerd
Y=Total distance /2
=880.826/2
=440.413mm
At minmum value
S3= 2.482791660e+04[mm^4]/440.413
=56.37416mm^3
At maximum value
S3= 5050434.992[mm^4]/440.413mm
=1.14674975x10^4mm^3
CONCLUSION:-
- As we knew larger section modulus than it will be stronger and capable to supporting greater load. so from the above we can see that 3rd case more section modulus as by increasing area we can get more stronger graeter load supporter.
- As we notice that if we increase the cross section the MOI increase due to strength also increase .
- Hood should we design in such a way that it should absorb all maximum loads when a crash happens.so while design we have in such a way that it should not be more stiffer and not more weeken.