Section Modulus calculation and optimization
Aim : To do section modulus calculation and optimization
Section modulus :
- Section modulus is the ratio of bending to the farthest fiber from the neutral axis , or it is the measue of the strength of the beam .
- Ratio of area moment of inertia to the distance from the neutral axis to the point where the bending normal stress is determined .
- It is the geometric property of the given cross- section of any shaft or beam structures .
- It is direct measurement of the strength of shaft or beam section .
- Higher the section modulus of a structure , the more the resistive it becomes to bending .
Section modulud (S) = I/Y
where ,
I = Moment of inertia of structure .
Y = Distance between the neutral axis and the extreme end of the objects .
Section modulus of different shapes :
classification of section modulus :
1 ) Elastic section modulus:
The elastic section modulus is applicable upto the yield point of material.
It is used in most of engineering application .
S = I/Y
where I = moment of inertia ,
Y = Distance between neutral axis and outermost fibber.
2 ) Plastic section modulus :
It is necessary to calculate the plastic momemt capacity of the section .
It is addition of the moment of area of area under compression about PNA (plastic neutral axis ) and the moment of area of area under tensile stress .
Calulation of section modulus for different cases :
case 1 :
- Initially we have taken the hood we have made .
- Now we have taken intersection of hood to find 2D curves .
- we have closed all the loop for making it closed loop for section modulus analysis .
Moment of inertia Max = 7.52 x 10^6 mm^4
Moment of inertia min = 2.30 x 10^4 mm^4
y = total distance / 2
y = 897.67/2
y = 448.835 mm
s=i/y
s = 2.3 x 10^4 / 448.835
= 51.24 mm^3
Case 2 :
- Here we have done offset of inner panel by 1 mm .
- Here area will increase .
Moment of inertia Max = 9.9 x 10^6 mm^4
Moment of inertia min = 3.36 x 10^4 mm^4
y = total distance / 2
y = 897.67/2
y = 448.835 mm
s=i/y
s = 3.36 x 10^4 / 448.835
= 74.86 mm^3
Conclusion :
The strength of materia is directly proportional to area and moment of inertia .So as we have increase area in 2nd case its strength also increases .