Second moment of area I and deflecton relationship

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SUMMARY

The relationship between the second moment of area (I) and beam deflection is mathematically defined, confirming that doubling I will halve the maximum deflection (y max) when I appears in the denominator of the deflection equation. The relevant equations include I = bd³ / 12 and y max = -5wL⁴ / 384EI. Additionally, increasing Young's modulus (E) or the beam's depth by approximately 26% also achieves the same effect of halving deflection. However, there is no general rule of thumb for reducing beam length to halve deflection, as deflection is proportional to L⁴.

PREREQUISITES
  • Understanding of beam theory and deflection mechanics
  • Familiarity with the second moment of area (I) and its calculation
  • Knowledge of Young's modulus (E) and its impact on material properties
  • Basic grasp of structural engineering principles
NEXT STEPS
  • Study the derivation and implications of the equation y max = -5wL⁴ / 384EI
  • Explore the effects of varying the second moment of area (I) on beam performance
  • Investigate the relationship between Young's modulus and material selection in structural design
  • Learn about advanced beam deflection analysis techniques for complex loading scenarios
USEFUL FOR

Structural engineers, civil engineering students, and professionals involved in beam design and analysis will benefit from this discussion, particularly those focused on optimizing beam performance and understanding deflection mechanics.

LDC1972
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Second moment of area "I" and deflecton relationship

Homework Statement



What I value will halve the beam deflection

Homework Equations



I = bd^3 / 12

and

y max = - 5wL^4 / 384EI

The Attempt at a Solution



Transpose y max = - 5wL^4 / 384EI to:

I = - 5wL^4 / 384Ey max

This makes the value of I (2nd moment of area) double to achieve a halved deflection. Is this just a commonly known relationship?

Obviously I'm not inferring that the beam CSA be doubled, only 'I', and I'm also aware that when thinking of CSA, it takes approximately 26% of a rectangular beam depth increase to halve deflection.
 
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LDC1972: Both of your ymax equations in post 1 are the same. Did you make a typographic mistake?
 
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I don't think so?

I transposed:

y max = - 5wL^4 / 384EI to: I = - 5wL^4 / 384Ey max

To get a new value for I. I already have all values, then the question asks what new value of I is required to halve max deflection. So after the transposition, I used y max with a new value (half of original y max) and the end result was a perfectly doubled I.

This lead me to wonder; is this true all the time? Does doubling I always halve deflection?
 
LDC1972: OK, sounds good. That is correct.

LDC1972 said:
Does doubling I always halve deflection?
Yes, but only when deflection is a function with I in each denominator.
 
Last edited:
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Wow, that is surprising! Thank you very much for the confirmation.
 
So, to take this philosophy a stage further...

If you doubled "E" (youngs modulus) i.e. the material being used, then this would also halve deflection.

So to halve deflection you have 3 options:

Double I
Double E
Increase original depth by approximately 26%

Or complicate things by a mixture of all 3!

These facts are making sense of what is currently theory on my course and also my understanding of their interaction.
 
I forgot the 4th option: decrease the beams length.

Does this also have a "rule of thumb", such as halving the length will halve the deflection?
 
For this loading, deflection is proportional to L^4. Do you think that halving the length will halve the deflection?

BTW, these are not 'rules of thumb'. As you have found, it is a definite mathematical relationship.

The math worked for you with I. Why not try it again for L?

Math class is more than just providing jobs for math teachers, you know.
 
LDC1972: Yes, that is correct. And a fifth option is to double b, if the beam has a prismatic, rectangular cross section.

No, there is no general rule of thumb for length reduction. You would just solve for L, when ymax is one half.
 
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  • #10
nvn said:
LDC1972: Yes, that is correct. And a fifth option is to double b, if the beam has a prismatic, rectangular cross section.

No, there is no general rule of thumb for length reduction. You would just solve for L, when ymax is one half.

Thanks NVN! I have "thanked you".

To the previous poster:

I am philosophising, of course math is definitive - did you think I though otherwise??

When you are examining the safety of a 200 m tower crane, you rarely take formulas and a calculator (read NEVER). That is the responsibility of the manufacturer. As a surveyor, I put the onus on the manufacturer to provide the materials properties / safe working loads etc, then using 25 years experience, decide whether I need further proof beyond testing deflection and strain gauge operation (and if I make such a decision, this puts a £1000 a day tower crane out of action, not to mention crippling site production for the day).

So "rule of thumb" is very useful in the real world, after the math class has done the initial and very accurate formulae!
 
  • #11
You have been playing around with formulas for simple beams loaded in simple ways. If you apply your 'rules of thumb' without proper consideration to a more complex structure loaded in a more complex manner, you could find yourself in a spot of trouble.

It's better to do a lot of analysis and calculation before you get to the field. If I were in charge of examining the safety of a crane, I would insist on making certain that all was OK, even if I had to seek outside technical help.
 

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