Solving Beam Bending Qs: Find Beam Dimensions with 25 MNm^2 Stiffness

[HorseRidingTic]

Bonjouro!
Firstly, as this is my first post here, I would like to say thank you to everyone who is a part of the site and Hello!
I've used this forum a number of times, but this is the first time I've posted here as this user!

If the flexural stiffness of a beam is 25 MNm^2, and the beam is made of steel, what is the breadth and depth of the beam (or at least beam dimensions that would work)

I did this by saying
I = bd^3/12

EI = 25,000,000,000 Nmm^2
E = 200,000 MPa
I = 1,500,000mm^4
Using I = b*d^3 / 12
Assuming breadth (b) is 50mm, I get depth (d) as 31.072mm, but apparently this is wrong and I cannot figure out why :'(

My question comes from this thing I found on the internet, which gives me these FEA results, but they don't match up!
View post on imgur.com

Please help PhysicsForum guys! You're my only hope!

 

[SteamKing]

Bonjouro!
Firstly, as this is my first post here, I would like to say thank you to everyone who is a part of the site and Hello!
I've used this forum a number of times, but this is the first time I've posted here as this user!

If the flexural stiffness of a beam is 25 MNm^2, and the beam is made of steel, what is the breadth and depth of the beam (or at least beam dimensions that would work)

I did this by saying
I = bd^3/12

EI = 25,000,000,000 Nmm^2
E = 200,000 MPa
I = 1,500,000mm^4
Using I = b*d^3 / 12
Assuming breadth (b) is 50mm, I get depth (d) as 31.072mm, but apparently this is wrong and I cannot figure out why :'(

My question comes from this thing I found on the internet, which gives me these FEA results, but they don't match up!
View post on imgur.com

Please help PhysicsForum guys! You're my only hope!

I think your problem comes down to a mistake in handling the units of EI and E somewhere.

Let's take the original data:

EI = 25 MN-m², where E = 200 GPa = 200 × 10⁹ N/m²

By dividing EI by E, we get:

I = EI / E = 25 × 10⁶ N-m² / (200 × 10⁹ N/m²) = 1.25 × 10^-4 m⁴

Since there are 10³ mm / m, there are 10¹² mm⁴/m⁴

So I = 1.25 × 10^-4 m⁴ = 1.25 × 10^-4 m⁴ ⋅ 10¹² mm⁴/m⁴ = 1.25 × 10⁸ mm⁴

taking b = 50 mm and assuming a rectangular cross-section for I:

I = bh³ / 12 = 50h³ / 12 = 1.25 × 10⁸ mm⁴

h³ = 3 × 10⁷ mm³

h ≈ 310.7 mm

 

[HorseRidingTic]

Brother! You have saved me. iT was the units, that were the issue. Now I know better. Thank you!