Section Modulus calculation and optimization

OBJECTIVE:

To Calculate the section modulus for the previously designed hood and optimized hood.

To Come up with a new section that has improved the section modulus of the previous section.

SECTION MODULUS:

Section modulus is the geometric property of a given cross-section used in the design of flexural members. Section modulus depends only on the cross-section shape of the beam or structure. For symmetric sections, the value of section modulus is the same for both above and below of centroid of the cross-section. An asymmetric section has two values for the top and bottom of the centroid, the maximum section modulus (S and minimum section modulus (S ).

S = I/y

Where,

S = Section Modulus

I = Moment of Inertia

y = distance between the neutral axis and the extreme end of the object

There are two types of section modulus:

• Elastic section modulus(S).
• Plastic section modulus(Z).

Significance of Section Modulus:

• Section modulus is an important factor for designing beams and flexural members.
• The higher the section modulus, the higher will be the resistance to bending.
• Easier to calculate strength and stresses in beams.
• If two beams are made of the same material and the section modulus of two beams are being compared and vice versa, the
  beam with a larger section modulus will be tougher and more capable to withstand larger loads.

We know that stress (σ) in a beam or structure is given by

σ=M⋅yIσ=M⋅yI...................(1)

Where I = Moment of Inertia
Y = distance between the neutral axis and the extreme end of the object
M = the moment about the neutral axis

Section Modulus is given by

S=IyS=Iy.......................(2)
From equations (1) and (2),
σ=MSσ=MS........................(3)

Thus from equation (3), it can be said that the higher the section modulus, the higher will be the resistance to bending.

DESIGN 1: SECTION MODULUS OF INITIAL HOOD DESIGN

From the above analysis, following minimum and maximum moment of inertia are obtained

I max = 1.566753196 ×10^7×mm^4

I min = 1.376522594 ×10^5×mm^4

y = 487.4152/2 = 243.7076 mm

S max= I / y = 15667531.96 / 243.7 = 64290.24 mm^3

S min = I / y = 137652.25 / 243.7 = 564.84 mm^3

DESIGN 2: SECTION MODULUS OF OPTIMISED HOOD DESIGN.

By increasing the depth of the inner panel by 2 mm

From the above analysis, following minimum and maximum moment of inertia are obtained

I max = 1.566361139 ×10^7×mm^4

I min = 1.560940841 ×10^5×mm^4

y = 487.4152/2 = 243.7076 mm

S max= I / y = 15663611.39 / 243.7 = 64274.15 mm^3

S min = I / y = 156094.08 / 243.7 = 640.51 mm^3

CONCLUSION:
The strength of the material is directly proportional to the moment of inertia, hence increase in sectional area there is have increase in overall mass of an object ,hence have increase in strength.