Section Modulus calculation and optimization

AIM:Section modulus calculation and optimization

INTRODUCTION: It 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. Any relationship between these properties is highly dependent on the shape.

There are two types of section modulus,

1. The elastic section modulus
2. The plastic section modulus

Section Modulus can be calculated by using the Formula:

S=I/y

Where,

S-Section Modulus I- Moment of inertia

y- Distance from the neutral axis to the most extreme end of the object

1.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 is used to calculate the plastic moment, M_p, or full capacity of a cross-section. The two terms are related by the yield strength of the material in question, F_y, by M_p=F_y*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 elastic limit of material. This could be shown mathematically with the formula

2.Elastic section modulus

The elastic section modulus is defined as S = I / y, where I is the Second moment area (or area moment of inertia, not to be confused with moment of inertia) and y is the distance from the neutral axis to any given fibre. It is often reported using y = c, where c is the distance from the neutral axis to the most extreme fibre, as seen in the table below. It is also often used to determine the yield moment (M_y) such that M_y = S × σ_y, where σ_y is the Yield strength of the material.

Calulation of section modulus for different cases :

case 1 :

• Initially we have taken the hood we have made .
• Now we have taken intersection of hood to find 2D curves .
• we have closed all the loop for making it closed loop for section modulus analysis .

Moment of inertia Max = 7.52 x 10^6 mm^4

Moment of inertia min = 2.30 x 10^4 mm^4

y = total distance / 2

y = 897.67/2

y = 448.835 mm

s=i/y

s = 2.3 x 10^4 / 448.835

= 51.24 mm^3

Case 2 :

• Here we have done offset of inner panel by 1 mm .
• Here area will increase .

Moment of inertia Max = 9.9 x 10^6 mm^4

Moment of inertia min = 3.36 x 10^4 mm^4

y = total distance / 2

y = 897.67/2

y = 448.835 mm

s=i/y

s = 3.36 x 10^4 / 448.835

= 74.86 mm^3

Conclusion :

The strength of materia is directly proportional to area and moment of inertia .So as we have increase area in 2nd case its strength also increases .