Section Modulus calculation and optimization

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. The higher the section modulus of a structure, the more 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 the design include an area for tension and shear, the radius of gyration for compression, and the 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. There are two types of section moduli, the elastic section modulus, and the plastic section modulus.

S=I/Y

Here,

S = Section modulus

I= Moment of Inertia (Unit mm 4)

Y= Distance from the Neutral axis to the extreme end (unit mm)

The Unit of a “section modulus” is (mm3)

Elastic section modulus

For general design, the elastic section modulus 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 a moment of inertia) and y is the distance from the neutral axis to any given fiber. It is often reported using y = c, where c is the distance from the neutral axis to the most extreme fiber, as seen in the table below. It is also often used to determine the yield moment (My) such that My = S × σy, where σy is the yield strength of the material.

Plastic section modulus

The plastic section modulus is used for materials where elastic yielding is acceptable and plastic behavior 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 neutral axis (PNA). The PNA is defined as the axis that splits the cross-section such that the compression force from the area in compression equals the tension force 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 plastic 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 denoted by 'k', used for an indication of capacity beyond the elastic limit of the material. This could be shown mathematically with the formula:-

shape factor for a rectangular section is 1.5.

So here we are Calculating Elastic Section Modulus.

Calculation:

Case 1: Before Optimization

The hood which was already created is imported and the section inertia analysis is done.

We have to make sure that the curve obtained is a single/closed curve.

We get,

I (max)= 8.308735380e+05 (mm^4)

I (min)= 4.893856163e+03 (mm^4)

As per section modulus equation:

S= I/Y

I= MOI (min) is required to consider

Y=total distance /2 = 479.5mm/2 =239.75mm

Therefore, S= 4893.856163 mm^4 / 239.75 mm

S= 20.4123 mm^3

Case 2: After Optimization

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

We get,

I (max)= 8.335048775e+05 (mm^4)

I (min)= 5.267904135e+03 (mm^4)

As per section modulus equation:

S= I/Y

I= MOI (min) is required to consider

Y=total distance /2 = 479.5mm/2 =239.75mm

Therefore, S= 5267.904135 mm^4 / 239.75 mm

S= 21.972 mm^3

Conclusion:

Resulting by comparing case 1 & case 2,

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