Section Modulus calculation and optimization

AIM

Use the Section from your Hood design and calculate the section modulus,Come up with a new section that has improved the section modulus of the previous section and mention what change you made that has increased the strength.

SECTION OF MODULUS

Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members. Other geometric properties used in design include area for tension and shear, radius of gyration for compression, and moment of inertia and polar moment of inertia for stiffness.

There are two types of section moduli, the elastic section modulus and the plastic section modulus.

It is termed as the ratio of second moment of area and distance from N.A (Neutral axis) to the extreme fiber. Also it is the measure of strength of given member. The stress in the outermost section of beam is computed with the help of section modulus. It is indicated by S.

Section modulus is indicated as follows:

S = Section Modulus

I = Moment of Inertia

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

Here, I is “moment of inertia” and y is distance from “neutral axis” to top or bottom of fiber.

Unit of a “section modulus” is in (mm3).

Section modulus depends only on the cross section shape of the beam. Cross section shapes like rectangular, square, circular, I section and T, composite section etc.

For symmetric sections, the value of “section modulus” is for both above and below of centroid of the fiber.

An asymmetric section has two values for top and bottom of the fiber due to different distance from centroid to the top and bottom of the material, the maximum “section modulus” is S m ax and the minimum section modulus is S min.

Importance:

• It is significant for beam and flexural member design.

• More the section modulus, more the member is resistant to bending of the beam.

• It is easier to calculate stresses in beam and an exact measure of strength of steel.

• If two beams are made of same material and comparing the “section modulus” of two beams, the beam with bigger section modulus will be tougher and more capable to withstand larger loads.

SIGNIFICANCE OF SECTION MODULUS

The section modulus of the cross-sectional shape is of significant importance in designing beams. It is a direct measure of the strength of the beam. A beam that has a larger section modulus than another will be stronger and capable of supporting greater loads.

1. Section modulus is the important factor for design of beam and flexural member
2. Higher the value of section modulus, higher will be the resistance of member to bending
3. It is required to calculate stresses in beams.
4. It is used to calculate strength of the steel structure
5. More the section modulus, it can withstand more load and it is also considered to be more tougher.

By using the Hood design I am going to calculate the Hood Section modulus.

CASE 1

Below pic referes to the 2D sketch of the Hood part.By this we can calculate the hood section modulus.

In NX OR CATIA V5 software we have a command called Section Inertia,By this we can automatically get the Moment of Interia values.

Finally I got my Moment of Inertia (max. & min.)values of Hood part.

MOI (max) :(1.7714e-06 * 10^05) [mm^4]

MOI (min) : (6.345e-004 * 10^03) [mm^4]

Below picture referes to the distance between hood from one end to extreme end.

The total distance of the hood Y = (881.926 mm.)

Calculation for the Hood Section of Modulus

Let,

Moment of Inertia (max. & min.)values of Hood part.

MOI (max) : (1.7714e-06 * 10^05) [mm^4]

MOI (min) : (6.345e-004 * 10^03) [mm^4]

The total distance of the hood Y = (881.926 mm.)

Distance between the neutral axis to the extreme end = (881.926/2) = 440.963 mm.

For Asymmetric has two values for the top and bottom of the centroid ie., S(max) and S(min).

For S(max) and S(min)

Section modulus is the ratio of second moment of area and distance from N.A (Neutral axis) to the extremeend is given by S = I/Y

S(max) = I(max)/Y = (1.7714e-06 * 10^05) / (440.963 ) = (401.698) [mm^3]

S(min) = I(min)/Y = (6.345e-004 * 10^03) / ( 440.963 ) = (14.3798) [mm^3]

CASE 2

In the Case 2 we are offsetting the hood inner panel about 0.5 mm downwards.Why because Higher the value of section modulus, higher will be the resistance of member to bending so thats why i increased the hood inner section.

Below picture referes that 2D sketch of the inner panel is offset by 0.5 mm downwards as shown in the picture.

After Offsetting the inner panel about 0.5 mm then i got MOI of this particular section,Lets see whether there is increasing in value or not.When comparing to CASE 1

Finally I got my Moment of Inertia (max. & min.)values of Hood part.

MOI (max) : (1.86144e-06 * 10^05) [mm^4]

MOI (min) : (6.636e-004 * 10^03) [mm^4]

Below picture referes to the distance between hood from one end to extreme end.

The total distance of the hood Y = (881.926 mm.)

Calculation for the Hood Section of Modulus

Let,

Moment of Inertia (max. & min.)values of Hood part.

MOI (max) : (1.86144e-06 * 10^05) [mm^4]

MOI (min) : (6.636e-004 * 10^03) [mm^4]

The total distance of the hood Y =(881.926 mm.)

Distance between the neutral axis to the extreme end = (881.926 /2) = (440.963) mm.

For Asymmetric has two values for the top and bottom of the centroid ie., S(max) and S(min).

For S(max) and S(min)

Section modulus is the ratio of second moment of area and distance from N.A (Neutral axis) to the extremeend is given by S = I/Y

S(max) = I(max)/Y = ( 1.86144e-06 * 10^05) / (440.963 ) =(422.1170) [mm^3]

S(min) = I(min)/Y = ( 6.636e-004 * 10^03) / (440.963 ) =(15.0398) [mm^3]

Comparing Case 1 and Case 2

We considering the Section Modulus minimum value only

After Calculating we got the values of Case 1 and Case 2 are

CASE 1

S(min) = (14.3798) [mm^3]

CASE 2

S(min) = (15.0398) [mm^3]

• When Comparing this two Hood values.As you can see there is increasing in the values in both S max and S min.

• When i increased Sectional area by 0.5 mm therefore the S(min) value in case 2 increased and also increased in strength From (14.3798) [mm^3] to (15.0398) [mm^3]

• When we see the values of S(max) there is slightly increases from (401.698) [mm^3] to (422.1170) [mm^3]

• Hence comparing these two cases,we can say that When there is increase in cross sectional area similarlly there will be increase in strength.

• Increase in cross sectional area by 0.5 mm Increased the minimum load carrying capacity or strength of the Hood.

CONCLUSION

Thus I calculated the section modulus of Hood design and also there is improved in section modulus of the previous section Finally i concluded that there is Increase in Sectional area it increases the Overall mass of an object and also strength of an object.