# Second moment of area I and deflecton relationship

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.

nvn
Homework Helper
LDC1972: Both of your ymax equations in post 1 are the same. Did you make a typographic mistake?

• 1 person
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?

nvn
Homework Helper
LDC1972: OK, sounds good. That is correct.

Does doubling I always halve deflection?
Yes, but only when deflection is a function with I in each denominator.

Last edited:
• 1 person
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?

SteamKing
Staff Emeritus
Homework Helper
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.

nvn
Homework Helper
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.

• 1 person
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!

SteamKing
Staff Emeritus