Section Modulus calculation and optimization

SECTION MODULUS CALCULATION OF HOOD

AIM:

• To calculate the section modulus of the previously designed hood for analyzing its strength and also optimizing the design to see the difference in the bending strength of the hood.Higher the section modulus of a structure, the more the resistive it becomes to bending.

SECTION MODULUS:

• Section modulus is a geometric property for a given cross-section used in the design of beams or flexural members, it is also referred to as the polar modulus or the Torsional sectional modulus. 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.
• Any relationship between these properties is highly dependent on the shape in question.Equations for the section moduli of common shapes are given below. These are two types of section moduli, the elastic modulus and the plastic section modulus.
• S=I/Y

Here,

• S= Section modulus
• I= Moment of Inertia( unit mm 4)
• Y= Distance from Netural axis to the extreme end (unit mm)

The Unit of a "section of modulus" is (mm3)

Changing the position can show the difference in bending which is because of the centroid distance from extreme fiber end.

In Layman's terms, in this position there is more material from axis to the specific direction, to resist the load which makes less bend amd improves the strength.

ELASTIC SECTION MODULUS:

• For general design, the elstic section moduls is used, applying up to the yield point for most metals and other common materials. The elastic section modulus is defined as S= I/Y, Where I is the second moment of area (or area moment of inertia, not to be confused with moment of inertia) and Y is the distance from the netural axis to any given fibre.
• It is often reported using Y=c, where c is the distance from the netural axis to the most extreme fibre, as seen in the table below.It is also often used to determine the yield moment (My) Such that my yield strength.

PLASTIC SECTION MODULUS:

• The plastic section modulus is used for materials where elastic yielding is acceptable and plastic behaviour is assumed to be an acceptable limit.Designs generally strive to ultimately remain below the plastic limit to avoid permanent deformations, often comparing the plastic capacity against amplified forces or stresses.
• The plastic section modulus depends on the location of the plastic netural axis (PNA). The PNA is defined as the axis that splits the cross section such that the compression forces from the area in compression equals the tension forces from the area in tension. So, for sections with constant yielding stress, the area above and below the PNA will be equal,but for composite sections,this is not necessarily the case.
• The platic section modulus is the sum of the areas of the cross section on each side of the PNA(which may or may not be equal) multiplied by the distance from the local centroids of the two areas to the PNA: the plastic section modulus can also be called the 'First moment of area' .
• The plastic section modulus is used to calculate the plastic moment, Mp, or full capacity of a cross-section.The two terms are related by the yield strength of the material in question,Fy by MP=Fy*Z. Plastic section modulus and elastic section modulus are related by a shape factor which can be denotted by 'k', used for an indication of capacity beyond elastic limit of material. This could be shown mathematically with the formula:- shape factor for rectangular section is 1.5. So here we are calculating Elastic section modulus

FIRST DESIGN STUDY:

• The intersection of the design was taken and then section inertia Analysis was done-

Moment of inertia (max) = 4.74259x10^6xmm^4

Moment of inertia (min) = 1.409157x10^4xmm^4

Y=872.1503/2 = 436.07515 mm

S=I/Y

S=1.409157x10^4/436.07515=32.31 mm^3

HOOD DESIGN OPTIMIZATION:

By increasing the depth of the inner panel by 0.5mm

Here,

Moment of inertia(max) = 4.734219x10^6xmm^4

Moment of interia(min) = 1.431526x10^4xmm^4

Y= 872.1503/2 = 436.07515 mm

S= I/Y = 1.431526x10^4/ 436.07515 = 32.82 mm^3

Conclusion:

• So,here the difference can be seen in the moment of inertia and the changes in the section modulus can be seen by change in the design (depth).