Section Modulus calculation and optimization

AIM

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

SECTION MODULOUS

Section Modulus is a geometry property for a given cross-section which is used in the design of part

• Other geometric properties used in design include tension, radius of gyration for compression and moment of inertia for stiffness.
• In this challenge, we will be focusing on the Moment of inertia to identify the strength and stiffness of the part.
• If the section modulus increases the moment of inertia increases and the strength and load bearing capacity also increases.
• Section modulus can be improved by varying cross sectional geometry of the part.
• It is that simple to increase the section modulus or the strength of the part.

FORMULA FOR SECTION MODULUS

Section modulus is the ratio of Moment of inertia to the Distance between the neutral axis and extreme end of the object (mm) and the formula is given as follows.

S = I/Y,

Where, S= Section Modulus (mm^3)

I= Moment of Inertia of the section about the bending axis (mm^4)

Y= Distance between the neutral axis and extreme end of the object (mm)

PROCEDURE AND CALCULATION OF SECTION MODULUS:

* In this challenge, I have done the section modulus on the car hood and calculated accordingly.

PROCEUDRE

STEP: 1

Section is taken along the Y0 region with help of the plane using INTERSECTION CURVE .

STEP 2:

• If we have any gap in the intersection curve, it can be filled using the sketch command.
• After filling the gaps, Moment of inertia is identified using the SECTION INERTIA in the ANALYSIS drop down list.

Go to Menu < Analysis < Section Inertia.

CALCULATION

CASE 1 or SECTION 1

• The above image shows the value of the Distance between the neutral axis and extreme end of the object (Y) (mm).

Moment of Inertia (I) (Max) = 8.28329.3851 e+05 (mm^4)

Moment of Inertia (I) (Min) = 4.121153109 e+03 (mm^4)

The Formula to calculate is

S = I/Y

I(min) value should be taken for calculation.

Y= 487.1899/2 (Half of the value to taken from the neutral axis)

Y = 243.59495

Therefore, S = 412115.3109/ 243.54945

= 1692.121 mm^3

CASE 2 OR SECTION 2 (OPTIMIZATION):

In this case, I Have optimized the section modulus by increasing the distance of the inner panel from the outer panel by offsetting the inner panel by 0.5mm.

Moment of Inertia (I) (Max) = 8.239976617 e+05 (mm^4)

Moment of Inertia (I) (Min) = 4.504467490 e+03 (mm^4)

The Formula to calculate is

S = I/Y

I(min) value should be taken for calculation.

Y= 487.1899/2 (Half of the value to taken from the neutral axis)

Y = 243.59495

Therefore, S = 450446.7490/ 243.54945

= 1849.508 mm^3

RESULTS & CONCLUSION:

In this challenge, I have calculated 2 section modulus as shown above to identify and to achieve a higher moment of inertia.

In the First case, I just did a section to identify the section modulus of the design.

In the Second Case, I have tried optimzing the Section modulus of the previous section to achieve the Higher Moment of inertia and it is done by offsetting the Inner Hood panel for 0.5mm outward.

From the Case 1 or Section 1, I have got the section modulus value of 1692.121 mm^3.

From the Case 2 or Section 2, Section Modulus is optimized and the output values is 1849.508 mm^3.

Hence, it is clearly visible that increasing the Moment of inertia we can improve the section modulus and the section modulus is directly dependent on the Moment of inertia and the distance from the neutral axis.