Section Modulus calculation and optimization

1. 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.

Note:

• Refer to the grading policy.

• The report is to be made available in the provided Answer Box.

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.

There fore

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)

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

Material Data: Stainless steel AISI 302 – cold rolled

Where yield strength of stainless steel is 520MPa = 520 N/mm^2

here, we are creating the section modulus for the hood assembly for that we are taking the section at one point and taking section from the there model

i have taking the section from the above area, and getting the section as shown

and by using the section inerita analaysis command we are geting the data of inerita

Case 1:

we got the values

max moment of inertia : 5.7094e +06(mm^4)

min moment of inertia : 2.42970e +04(mm^4)

section length : 3.8375e +03mm

section area : 76.75060 mm^2

as persection modulus equation S = I/Y

I = MOI(min) value is considered

Y = total distance/2, total distance= 928.4111mm

Y = 928.4111/2

= 464.20555

S = 24297.0mm^4/464.2055mm

= 52.34104mm^3.

Case 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.

we got the values

max moment of inertia : 5.7094e +06(mm^4)

min moment of inertia : 2.3830e +04(mm^4)

section length : 3.8375e +03mm

section area : 76.751337 mm^2

as persection modulus equation S = I/Y

I = MOI(min) value is considered

Y = total distance/2, total distance= 928.4111mm

Y = 928.4111/2

= 464.20555

S = 23830.0mm^4/464.2055mm

= 51.33354mm^3.

Hence when comparing case 1 & case 2

1. The section modulus value of case 1 is higher than the case 2

2. As we decrease the section area, the moment of inertia & section modulus decrease which means strength and load material also decresase