Jump to content
Engineer Boards
 photo CHPE_AnimatedWebBanner_650x1202_zps5704d467.gif
ezzieyguywuf

What is effective eccentricity?

Recommended Posts

I'm working on problem 3 in chapter 53 of the MERM companion. In the solution on page 53-5, an equation in given for "effective eccentricity". I have checked in the MERM, in Shigley, and done some googling. I can't figure out where this equation came from. It appears to be a way of determining an effective force to used for a combined axial and eccentric load on a column.

Share this post


Link to post
Share on other sites

ezzieyguywuf,

Here's whats happening: You have two forces applied, the concentric one of 100kip and the eccentric one of 150kip. The eccentric one is applied at a distance of 3.33 inches from the neutral axis. In order to use the secant formula, you need a single equivalent force, not two.

The magnitude of this equivalent force is of course, 250kip. But where is it applied? It will be applied at a distance e from the neutral axis so that the moment about the neutral axis due to this equivalent force matches the moment about the neutral axis due to the original system of forces. The moment due to the original force system is (150kip x 3.33in). The moment due to the equivalent force is (250kip x e). Set those two equal and you get the value of e.

Does that help? If not, let me know. I can create a sketch that will illustrate it better. 

Edited by Slay the P.E.
correct minor typo

Share this post


Link to post
Share on other sites
On 12/9/2017 at 1:02 PM, Slay the P.E. said:

ezzieyguywuf,

Here's whats happening: You have two forces applied, the concentric one of 100kip and the eccentric one of 150kip. The eccentric one is applied at a distance of 3.33 inches from the neutral axis. In order to use the secant formula, you need a single equivalent force, not two.

The magnitude of this equivalent force is of course, 250kip. But where is it applied? It will be applied at a distance e from the neutral axis so that the moment about the neutral axis due to this equivalent force matches the moment about the neutral axis due to the original system of forces. The moment due to the original force system is (150kip x 3.33in). The moment due to the equivalent force is (250kip x e). Set those two equal and you get the value of e.

Does that help? If not, let me know. I can create a sketch that will illustrate it better. 

Yes that helps! This is a simple statics problem it makes perfect sense now, thank you.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now


×