Section Modulus calculation and optimization

Section Modulus Calculation and Optimization

Objective

To calculate, optimize and improve the section modulus of the hood

Section Modulus

• Section Modulus is a geometric property for a given cross section. It is used in the design of beam or flexural members. The section modulus is directly proportional to the moment of inertia. If the stiffness of the section is high it is more resistance to bending. Section modulus is a measure of strength of a cross sectional area about any reference axis.
• Section modulus is also defined as ratio of total moment resisted by the section to the stress in the extreme fiber
• Section modulus gives as an idea about how much load the given structure can take.
• As the section modulus increases the moment of resistance and the load bearing capacity of the structures also increases.
• The overall strength of the structure/body increases or decreases with movement from the positive or negative side from the neutral axis.

Moment of inertia

• Moment of inertia is the property of the body due to which it resists angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation
• Newton’s First Law also says that higher the mass higher the force is required to move the body, as the body exerts equal resistance force to the applied force

Formula used to calculate the section modulus :-

S=I/Y

I- Area Moment of Inertia

Y- Distance from the neutral axis and the extreme end of the object

Calculation

Section 1

Neutral Fiber =486.886/2 = 243.433 mm

I_max = 4.9136 e⁶ mm⁴

I_{min =} 1.4613 e⁶ mm⁴

Section modulus (max) s = I/Y = 2.018*10⁴ mm³

Section modulus (min) s = I/Y = 0.06001*10³ mm3

Section 2

the section 2 is obtained by offsetting the inner panel curve by 2mm thus increasing the cross sectional area

Neutral Fiber =486.886/2 = 243.433 mm

I_max = 4.8793 e⁶ mm⁴

I_{min =} 1.5324 e⁶ mm⁴

Section modulus (max) s = I/Y = 2.0043*10⁴ mm³

Section modulus (min) s = I/Y = 0.062*10³ mm3

From the above calculations we can see that

• The increase in the cross sectional area resulted in increase in the strength
• The strength of the material is directly proportional to the moment of inertia

Conclusion

Thus the section modulus of the section was calculated using Nx Cad. From the above comparison we can see that Section modulus is directly proportional to area of cross section. If the section modulus increases the strength of the material also increases.