Understanding the Value of the Second Moment Of Area

In summary, the equation for the Second Moment of Area taken at the base is y=1/2 d while the equation for the Second Moment of Area taken at the centriod is bh^3/3.
  • #1
tomtomtom1
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TL;DR Summary
Understanding the Value of the Second Moment Of Area
Hi all

I was hoping someone could help shed some light clearing some doubt on 2nd Moment of Area.

I know that if i had a beam that was loaded then the top of the beam would experience compressive forces.

As i moved down towards the neutral axis these compressive forces would become zero.

And i as i moved towards the bottom of the beam then the beam experiences tensile forces.

I know that the 2nd moment of area is how the area of a cross section is spread from the neutral axis i. e the more area away from the neutral
axis the less bending the beam will experience.

This is where i get confused.

Given a a rectangle (Breath 3m, Height= 7m), the equation for 2nd moment of area is given by bh^3/12 so my value for the second moment of area is 85.75m^4 as shown below:-

nuteral.JPG


I am trying to visualize what this number means?

Does 85.75 mean that i have 87.75m^4 of area above AND below the neutral axis as shown in green above?

Or

Does 87.75 mean that my total area furthest away from the neutral axis is 87.75?

Finally does the equation Ixx = bh^3/12 only apply to a single point which is at the centriod or along the neutral axis in the cross section?

Can anyone help?

Thank you.
 
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  • #2
tomtomtom1 said:
Summary:: Understanding the Value of the Second Moment Of Area

Hi all

I was hoping someone could help shed some light clearing some doubt on 2nd Moment of Area.

I know that if i had a beam that was loaded then the top of the beam would experience compressive forces.

As i moved down towards the neutral axis these compressive forces would become zero.

And i as i moved towards the bottom of the beam then the beam experiences tensile forces.
Opposite if you drew beam cross-section. Upper half of bent beam will be in tension and lower in compression.
Try to draw such diagrams in 3D. Cross-sections drawings may be misleading.

tomtomtom1 said:
Finally does the equation Ixx = bh^3/12 only apply to a single point which is at the centriod or along the neutral axis in the cross section?
Yes, it is the correct statement a.f.a.i.k.
 
  • #3
Hi Trurle

Thank you for your response.

Can i ask given my example what does 87.75 actually mean?

Does it mean that i have 87.75 of area above the centriod and 87.75 of area below the centriod?

I am trying to actually understand what the value means other than its a value the describes the resistance to bending.

Thanks
 
  • #4
tomtomtom1 said:
Hi Trurle

Thank you for your response.

Can i ask given my example what does 87.75 actually mean?

Does it mean that i have 87.75 of area above the centriod and 87.75 of area below the centriod?

I am trying to actually understand what the value means other than its a value the describes the resistance to bending.

Thanks
1) You have a typo. Correct value is 85.75
2) 85.75 m4 is total rigidity (summing both above and below neutral line). If you take second moment of area integral from -h/2 to h/2 instead of using formula directly, the confusion will not happen.
 
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  • #5
trurle said:
1) You have a typo. Correct value is 85.75
2) 85.75 m4 is total rigidity (summing both above and below neutral line). If you take second moment of area integral from -h/2 to h/2 instead of using formula directly, the confusion will not happen.

Trurle

I did make a typo, apologies.

Your comment about the value 85.75 m4 being the total rigidity made a lot of sense to me until i took the second moment of area at the base thinking that the value for the total rigidity would be the same - but it was not.

I found out that the equation for the second moment of area taken from the base is bh^3/3.
When i applied this equation to my example i got a total rigidity value of 343m4 but the value should be the same as the cross section has not changed.

Below is a sketch of my thought process:-

thought process.JPG


Where am i going wrong?

Thanks
 
  • #6
You are using the wrong formula. The one you used is for the case where y=d. The correct formula for a neutral axis at y = 1/2 d, as shown in your figure, is: bd^3/12 = 85.75.
 
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  • #7
JBA said:
You are using the wrong formula. The one you used is for the case where y=d. The correct formula for a neutral axis at y = 1/2 d, as shown in your figure, is: bd^3/12 = 85.75.

Hi JBA

Thank you for your response.

I have to be honest i don't think i understand.

Are you are you saying that the equation for Second Moment of Area taken at the base is y=1/2 d? i thought it was bh^3/3??

I am trying to compare the Second Moment Of Area taken at the centriod with the Second Moment of Area taken at the base of the shape.

This is what i am trying to do?
 
  • #8
tomtomtom1 said:
Are you are you saying that the equation for Second Moment of Area taken at the base is y=1/2 d? i thought it was bh^3/3??
I apologize, I misunderstood the issue of your confusion.

tomtomtom1 said:
I am trying to compare the Second Moment Of Area taken at the centroid with the Second Moment of Area taken at the base of the shape.
Your diagram and equation results are fully correct for the two cases. The Ix' for the y = h case is a derivative of the central axis Ix case done by applying the "parallel axis theorem": Ix' = Ix+Ad^2, where d is distance of Ix' from Ix.

See the below reference:
https://en.wikipedia.org/wiki/Second_moment_of_area
 

What is the second moment of area?

The second moment of area, also known as the moment of inertia, is a measure of an object's resistance to bending. It is calculated by integrating the square of the distance from an axis of rotation to each point on the cross-sectional area of the object.

Why is the second moment of area important?

The second moment of area is important because it helps engineers and scientists understand the structural integrity of an object. It is a key factor in determining how much an object will deform or bend under a given load.

How is the second moment of area calculated?

The second moment of area is calculated by taking the integral of the square of the distance from an axis of rotation to each point on the cross-sectional area of the object. This calculation can be done using mathematical formulas or with computer software.

What factors affect the second moment of area?

The second moment of area is affected by the shape, size, and distribution of an object's cross-sectional area. Objects with a larger cross-sectional area and a greater distribution of mass away from the axis of rotation will have a higher second moment of area.

How is the second moment of area used in real-world applications?

The second moment of area is used in a variety of real-world applications, including designing and analyzing structures such as bridges, buildings, and aircraft. It is also used in the design and optimization of machine components and in understanding the behavior of materials under stress.

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