Section Modulus calculation and optimization

                                                              Section Modulus calculation and optimization

AIM:-Use the section from your Hood design and calculate the section modulus using the formula S= I/y.

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

 Formula of section Modulus:- It may be defined as the ratio of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Which is equal to yield stress.

Therefore

S=I/y, Where

S= Section modulus (mm^3)

I = Moment of inertia of the section to the stress in the extreme end of the object (mm^4)

Y= Distance between the neutral axis and extreme end of the object (mm).

Area Moment of Inertia:-

 The area moment of inertia of a beam cross section area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.

 Where,

M = Bending Moment

I = Area moment of Inertia

E = Young modulus

R = Radius of gyration

 

Neutral Axis:- When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibre of the beams are subjected to maximum tension.

This picture contains the following scene, Diagram of a beam on its narrow edge supported at both ends. A man is walking on a straight beam. Diagram of a beam resting on its flat side supported at both ends. A man is walking on the straight beam and its bending in the middle.

Use the formula:  where b = breadth (width) and d = depth (height) to calculate the moment of inertia about the x-x axis for beams A and B using information from the diagram.

Section modulus is defined as the ratio of polar moment of inertia to the radius of the shaft or the distance from the neutral axis to the outer fibres.

For a beam which has been bent, the maximum stress σ_max will occur at the maximum distance y_max from the neutral axis.

Thus:

The quantity I/y_max is a purely geometric function and is termed the section modulus Z. Thus we can write:

Therefore :

Z = Section modulus.

I = Moment of Inertia.

Ymax = distance between the neutral axis and the extreme end of the object.

PROCEDURE:- 

we can find the section modulus with two way one is that dantum plane which we use to make a mirror image of body and second make the dantum plane on the middle of the body.

so,i go with the that dantum plane which i used in mirror image of body.

CASE:-01

BEFORE OPTIMIZATION:-

CALCULATIONS:-

MOI(MAX)= 9.171752544e+07[mm^4]

MOI(MIN)= 2.687432493e+05[mm^4]

TOTAL DISTANCE= 877.0288 mm

I = MOI (min) Value is considered

Y = Total distance / 2  ,    Y = 877.0288 mm/2 ,  Y =438.514mm

SO,

I= 2.6874 mm^4

Y= 438.514 mm

S=I/Y

S = 2.6874 mm^4/438.514 mm

S=0.006128 mm^3

CASE:-02

AFTER OPTIMIZATION:-

The hood inner panel’s curve is offset by 0.5 mm outwards and section inertia analysis is done same as case 1 for a closed curve.

MOI(MAX)= 9.156027319e+07[mm^4]

MOI(MIN)= 2.713510271e+05[mm^4]

CALCULATION:-

TOTAL DISTANCE= 877.0288 mm

I = MOI (min) Value is considered

Y = Total distance / 2  ,    Y = 877.0288 mm/2 ,  Y =438.514mm

SO,

I= 2.7135 mm^4

Y= 438.514 mm

S=I/Y

S=2.7135mm^4/438.514mm

S=0.006187 mm^3

Conclusion:-

Resulting by comparing case 1 & case 2,

• The section modulus value of case 2 is higher than the case 1
• As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increase. Which the moment of inertia directly proportional to the strength of the material.