Section Modulus calculation and optimization

AIM - To find the section modulus of the hood and optimization to improve the overall section modulus.

Objective: Use the section from your Hood design and calculate the section modulus using the formula S= I/y.

Section Modulus:

• Section modulus is the ratio of total moment resisted by the section to the stress in the extreme fibre.
• Section modulus give us idea about how much of load the given structure can take.
• If the section modulus increases the moment of resistance increases and hence its load bearing capacity.
• • S=I/Y ,

  Where S= section modulus

  I=moment of inertia of a section

  Y=distance from neutral axis to any given fibre

  Area Moment of Inertia:

  The area moment of inertia of a beam cross section 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 = Young modulus

  R = Radius of gyration

  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 fibre of the beams are subjected to maximum tension.

  

  
• This picture contains the following scene, Diagram of a beam on its narrow edge supported at both ends. A man is walking on a straight beam. Diagram of a beam resting on its flat side supported at both ends. A man is walking on the straight beam and its bending in the middle.

• Question-using the section from previous hood design calculate the section modulus and also come up with a new section that has improved the section modulus of the previous section and mention what changes made that increased the length

CASE 1

Hood design

Moment of inertia (max)=4.78578`*10^4mm^4

Moment of inertia(min)=1.88739⋅104mm41.88739⋅104mm4

value of y=440mm

section modulus=moi (min)/y

1.88739⋅1041.88739⋅104/ 440

=42.89mm^3

CASE 2

hood optimised design

by adding +0.6 mm in the depth of hood

MOI (max) = 7.46052`*10^4 mm^4

MOI (MIN)= 2.94221*10^4 mm^4

s=66.89 mm^3

comparing case 1 and 2

percentage increase in optimised one= (66.89-42.89)/42.89

=5.5%

Conclusion:

Resulting by comparing case 1 & case 2,

1. The section modulus value of case 2 is higher than the case 1
2. As we increase the section area, the moment of inertia & section modulus increase which means strength and load material also increase. Which the moment of inertia directly proportional to the strength of the material.
3. By comparing the values we can clearly say that the section modulus of the optimized hood increases by 5.5%
4. moment of inertia increases by increasing depth of hood.
5. it seems if we increase the gap between inner and outer panel moi also increases.