Section Modulus calculation and optimization

Question :

Use the Section from your Hood design and calculate the Section Modulus using the formula

S = I/y

Where,

S = Section Modulus

I = Moment of Inertia

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

Come up with a new section that has an improved Section Modulus compared to the previous one and mention what changes you made which resulted in the increased strength.

Solution :

SECTION MODULUS OF THE HOOD

1. Section Modulus:

• The section modulus (S) is geometric property of the cross section used for designing beams and flexural members. It does not represent anything physically.
• Formula of Section Modulus: It may be defined as the ration of total moment resisted by the section to the stress in the extreme fibre which is equal to yield stress.

Therefore,

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)

2. Area moment of Inertia:

• The area moment of inertia of a beams cross sectional area measures the beam ability to resist bending. The larger the moment of inertia less will be the bend.

Where,

• M = Bending moment
• I = Area moment of Inertia
• E = Youngs modulus
• R = Radius of gyration

3. Neutral axis:

• When beam is subjected to pure bending the top most fibres of the beam are subjected to maximum compression, while the bottom most fibres of the beams are subjected to maximum tension.
• Material Data: Stainless steel AISI 302 – cold rolled.
• Where yield strength of stainless steel is 520MPa = 520 N/mm^2.

Calculation:

• CASE 1: Before Optimization ( Section -1 )

• The Hood which was already created is imported and the section inertia analysis is done.
• We have to make sure that the curve obtained is a single/closed curve.

MOI (max) = 1.558404417x10^7 (mm^4)

MOI (min) = 1.447665485x10^5 (mm^4)

As per the section modulus equation,

S = I/Y

I = MOI (min) value is considered

Y = Total distance /2,

Total distance =486.8955mm

Y = 486.8955 / 2

Y = 243.44775 mm

Therefore,

S = 1.447665485x10^5 mm^4/243.44775 mm

S = 594.6514129 mm^3.

• CASE 2: After Optimization ( Section -2 )

The hood inner panel’s curve is offset by 0.5mm outwards and section inertia analysis is done same as case 1 for a closed curve.

MOI (max) = 1.575026162x10^7 mm^4

MOI (min) = 1.487543369x10^5 mm^4

As per section modulus equation,

S = I /Y

I = MOI (min) value is considered

Y = Total distance /2,

Total distance =486.8955mm,

Y = 486.8955 / 2,

Y = 243.44775 mm.

Therefore,

S = 1.487543369x10^5 mm^4 / 243.44775 mm

S = 611.0325097 mm^3.

Conclusion:

Hence when comparing case 1 & case 2,

• The section modulus value of case 2 is higher than the case 1.
• As we increase the section area, the moment of inertia & section modulus increases which means strength and load of material also increases.