Section Modulus calculation and optimization

SECTION MODULES OF HOOD

• Section modules is a direct measure of strenth of any sectional area of an about an reference axis (Neutral axis ).
• Strength is nothing but ,capable of supporting greater loads, so higher the strenth it has a high capable ofsupporting greater loads.
• This strength can measured directly from section modules formula,and defined by below formula.
• section modules represented with a symbol of 'S'.

S=I/Y

Here "I" moment of inersia and "Y' is the distance from the reference line neutral axis to one of extreme end of an object.

• Moment of inertia is nothing but offering resistance against of moment,bending or twisting, i.e if we applied any external load on the body , how much resistance its offering to tend to deform.
• So higher the value of moment of inertia ,have higher resistance to deform under certain load,hence we can say that moment of inertia directly proportional to the strength of the material ,and strength of the material can be calculated with the help of section modulus.
• If we vary the nuetral axis towards positive or negative side the overall strength of the body will decrease ,because when nuetral axis the line lies on the middle of the section only have higher strength at particular point,so here "Y" value was inversely proportional of section modulus.

That's why formula represented as above.

By using the same principle i going to calculate the strength of an hood panel.

Here the moment of inertia was calculated by using the NX-12 software obtained values are

Imax - 8.210815480x10^5mm^4

Imin - 3.9573846x10^3mm^3

Total length of 'Y' value was 482.1073mm

So that neutral axis we need 'Y' value - Y/2=241.05365

1.Section modulus (s) :- I/Y =8.21081540x10^5mm^4/241.05365mm

=3.41x10^3mm^3 ( maximum strenght )

2. Setion modulus (s) :- I/Y =3.95738468x10^3mm^4/241.05365mm

=0.0164x10^3mm^3 ( minimum strength)

So the strength of an hood varies from 0.0164x10^3mm to 3.41x10^3mm

Now, one trail is that here the outer panel curve offsetting inside by decreasing,the results shown below.

I obtained the value of

Imax:- 8.08443x10^5mm^4

Imin:- 3.68660x10^3mm^4

total length of 'Y' value was 482.1073

So that neutral axis we need 'Y'value :- Y/2 =482.1073/2=241.05365

1.Section modulus (s):- I/Y=8.08443x10^5mm^4/241.05365mm

=3.354x10^3mm^3 ( Maximum strength)

2.Section modulus (s):- I/Y=3.68660x10^3mm^4/241.05365mm

=0.0153x10^3mm^3 ( Minimum strength )

So the strength of an hood varies from 0.0153x10^3mm^3 to 3.354x10^3mm^3 .

• 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 decreased inside sectional area,value of minimum and maximum strength was decreased from ( Min:- 0.164x10^3mm^3 to 0.153x10^3mm^3) and ( max:- 3.41x10^3mm^3 to 3.35x10^3mm^3 ) ,Hence with decreased in cross sectional area results in decrease in strength.
• So Newton's first law also expalin that if higher the mass ,the force required to move the object is also high due to offering more resistance in order to move the object .This resisitance we called it as a Moment of Inertia.
• In case of hood also same principle will work .
• The Strenght of the material also depends upon the composition of material,thickness and cross sectional area of an object.

CONCLUSION:-

The strength of the material is directly proportional to the moment of inertia ,hence increase in sectional area there is have increase and decrease in sectional area thera is have decrease in overall mass of an object, Hence have decrease or increase in strength.