Section Modulus calculation and optimization

Section Modulus calculation and optimization

Aim:

Section modulus calculation and optimization of the Hood design using NX CAD

Section Modulus:

• Section modulus is the geometric property of any component or shaft or beam structural cross-sections.It determines a object's strength and ability to resist external forces which flex the object.The moment carrying capacity of an any object is directly dependent on geometrical property and material property.
• Geometry of an object plays an important role in load carrying capacity of an object.A component having larger section modulus will be stronger, stiffness and capable of supporting greater loads.Therefore section modulus is the predominant factor which evidences the strength of an object.
• Section modulus is the ratio of the momemt of inertia of area and the distance from the neutral axis to any given (or extreme) fiber.

Section modulus, S=I/y (unit is mm³)

where,

• I-Area moment of inertia(mm⁴)
• y-Distance between neutral axis and extreme end of fiber(mm).

Calculation of Section modulus of the hood assembly:

1.Using NX, in the view tab option click the edit section and select save copy of section curves

2. Project the curve in to a plane and inorder to analyse the section, close the section using sketch command

3. Measure the distance from neutral axis to end of the section using refernece lines

4. Use section inertia analysis and select the existing curve to obtain the parameters

CASE 1

4. Calculate the section modulus from the obtained parameters as follows:

Moment of inertia of area(I) minimum = 1.308791659e+04[mm⁴]

= 1.308791659 X 10⁴mm⁴

Distance, y= 426.2mm

Therefore,

Section modulus, S=Moment of inertia of area(I)/Distance between neutral axis and extreme end of fiber

S= 1.308791659 X 10⁴mm⁴/426.2mm

S= 30.70 mm³

5. According to the definition of section modulus, if the area of cross section of object changes,the area moment of inertia will also changes and result in changing the section modulus because of direct proportionality, so change the area by increasing distance between outer and ineer panel by 0.5 mm and join the curves as followed in previous steps

CASE-2

6. Calculate the section modulus for the new optimised section

Moment of inertia of area(I) minimum= 1.321117820e+04[mm⁴]

= 1.321117820X 10⁴mm⁴

Distance, y= 426.2mm

Therefore,

Section modulus, S=Moment of inertia of area(I)/Distance between neutral axis and extreme end of fiber

S= 1.321117820X 10⁴mm⁴/426.2mm

S= 31.97mm3

Conclusion

Here a increased value of section modulus can be seen, as already known, higher the section modulus of a structure, the more the strength and resistance against bending.so as in the comparison of above cases, section after optimisation is having more section modulus than the previous section because of increasing area of section due to increased distance between hood outer and inner panel