Section Modulus calculation and optimization

SECTION MODULUS

Section modulus is the ratio of area moment of inertia about the neutral axis and to the distance of outermost fiber from the neutral axis.

The section modulus (S) is geometry property of the cross section used for designing beams and flexural members. It does not represent anything physically.

S= I/Y

S - section modulus

I - Area moment of Inertia

Y - Distance from neutral axis to the outermost fiber

Area moment of inertia = (bd^(3))/3

If b is less it is more prone for bending without changing any propotion we want to add strength for that rotate the bar to 90 degree then the b will be increased then the force that will be exerted can be withstanded.

Practical Demonstration

Relation between Area moment of Inertia and Section modulus

M/I = σ/Y = E/R (bending equation)

σ - Bending stress

Y - distance from neutral axis

E - Young's Modulus

R - curvature

M= I/Y * σ

(I/Y = S)

ie; M= Sσ

SECTION MODULUS STUDY

Steps

• Make a section of the hood assembly and connect those two to make a closed surface.
• Connect all other discontinuities present in the section using studio spline
• Search for Section Inertia analysis
• Select Section inertia analysis and select the closed surface then we will get the information about that model.
• With that information we want to calculate the section modulus
• Then offset the bottom line of the section and attach those two line inside a sketch
• then do the section inertia analysis again
• then calculate the value and compare it with the previous data and check whether the Section modulus has increased
• if increased the strength will also be increased

EXISTING SECTION

Distance from the neutral axis to the outermost fiber (Y) = 879.6306/2 = 439.8153 mm

this value will be constant for the two sections

Section Inertia Analysis

MOI (min) - 2.85x10^5 mm^4

MOI(max) - 8.89x10^7 mm^4

we will consider MOI (min) as I

then the equation for Section Modulus is S = I/Y

S = 2.85x10^5 mm^4 / 439.8153 =647.999 mm^3

OPTIMISED DESIGN

Y is constant = 439.8153 mm

Section Inertia Analysis

MOI (min) - 2.935x10^5 mm^4

MOI (max) - 8.866x10^7 mm^4

S = I/Y

= 2.935x10^5 / 439.8153

= 667.32 mm^3

There is a visible increase in Section Modulus of 19.32 mm^3

CONCLUSION

Strength of material is directly proportional to the moment of interia, thus increasing the sectional area increase the increases strength of component.

Thus sectional modulus for hood design is calculated and optimized design is also done to increase the section modulus of the hood.