Section Modulus calculation and optimization

Answer:

Section modulus:

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. Any relationship between these properties is highly dependent on the shape in question.

There are two types of section modulus,

1) Elastic section modulus

2) Plastic section modulus

Elastic section modulus:

It is defined as the ratio of the moment of inertia (I) of a section about its centroidal axis and to the distance of the extreme layer from the neutral axis (y).

S=I/y

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.

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.

Calculation of section modulus:

• I am taking my hood assembly challenge for my section modulus calculation.
• For calculating section modulus we need to find the value of moment of inertia of my hood design, for taking moment of inertia we taken inner and outer panel intersection with respect to ZY plane.
• Then make a skeck with respect to intersection by using project curve option.
• Also there is no open loops for checking inertia, the sketch shoud be closed.
• Then find moment of inertia by using section inertia analysis comment.
• After that we need to check the distance (y) of the object. Hood section is irregular so we take the total distance from one end to another end and taken the half value of that distance.
• For calculating section modulus we using S=I/y formula.

Fig 1 (My hood assembly)

From above image,

Moment of inertia I = 6.713128253 e+03 mm^4 = 6713.128253 mm^4

Distance y = 485.8852/2 mm (ie, we need distance from neutral axis to extreme end so i took half of the distance)

= 242.94 mm

Hence, S = 27.63 mm^3

Fig 2 (After increasing the distance of the inner panel by 1 mm)

From above image,

Moment of inertia I = 7.138374748 e+03 mm^4 = 7138.374748 mm^4

Distance y = 242.94 mm

Hence, S = 29.38 mm^3

Fig 2 (After increasing the distance of the inner panel by 2 mm)

From above image,

Moment of inertia I = 7.588393332 e+03 mm^4 = 7588.393332 mm^4

Distance y = 242.94 mm

Hence, S = 31.23 mm^3

Conclusion:

From the above result if we increasing the distance between inner and outer panel of the Hood assembly the inertia value incresing therefore Section modulus value also increased, hence the strength also increased.