Section Modulus calculation and optimization

REPORT:

SECTION MODULUS: It is the total moment resisited by the section to the stress in the extreme fibre. it gives us an idea about how much of load the given structure can take

formula :

S=I/Y

where,

S - section modulus

I - moment of inertia

y - distance between the neutral axis and the extreme end of the object

CASE 1

shown below is the picture in 2D of hood part. by this i calculated the hood section modulus. for that i took thw intersection between the hood inner panal and the outer panel. after taking the intersection i joined both intersection into one single intersection. after the joined intersection i made a closed surface using fill comand

CALCULATIONS:

formula:

S=I/Y

I (MAX)= 1.436e-004*10^8 mm^4

I(MIN)= 9.064e-007*10^5 mm^4

for caculating y that is the distanve between the neutrale axis to the extreme end of the object. for finding y i drawn a scketch which showing the diemntions between the extreme ends of the obeject and the value is divided inte half for the calulation . the sketch is shown below

here ,

Y= 478/2 =239

section modulus calculations

S(MAX)=(1.436e-004*10^8 mm^4)/ 239

S(MAX)= 600836.82mm^4

S(MIN)=(9.064e-007*10^5 mm^4)/239

S(MIN)=3792.46mm^4

CASE 2:

shown below is the picture in 2D of hood part. i took the intersection between the hood inner panal and the outer panel.after that i increased the value by 0.5 by using offset command. after taking the intersection i joined both intersection into one single intersection. after the joined intersection i made a closed surface using fill comand

CALCULATIONS:

formula:

S=I/Y

I (MAX)= 1.485e-004*10^8 mm^4

I(MIN)= 9.495e-007*10^5 mm^4

for caculating y that is the distanve between the neutrale axis to the extreme end of the object. for finding y i drawn a scketch which showing the diemntions between the extreme ends of the obeject and the value is divided inte half for the calulation . the sketch is shown below

here ,

Y= 478/2 =239

section modulus calculations

S(MAX)=(1.485e-004*10^8 mm^4)/ 239

S(MAX)= 621338.912mm^4

S(MIN)=(9.495e-007*10^5 mm^4)/239

S(MIN)=3972.8mm^4

we are concidering the minimum values of section modulus.while comapring both case 1 and case 2 we have

CASE 1

S(MIN)=3792.46mm^4

CASE 2

S(MIN)=3972.8mm^4

CONCLUSION:

As we can see while i increased the value by 0.5 in case 2 we got the better section modulus. if the section modulus increases the moment of resisitance increases and hence the total load bearing capacity also increases