Statically indeterminate axially loaded problem

[Matthew Heywood]

Homework Statement

stat indet.jpg

Homework Equations

F_AB = F_BC etc...

The Attempt at a Solution

F_AB = F_BC + 300kN = F_CD = F_DE + 600kN

Sorry for another post. I'm really not sure where to go with this one...

 

[SteamKing]

You've got a situation involving axial loads.

How would you calculate the total stretch in the member, assuming the member could stretch under load without coming into contact with the ground?

 

[Matthew Heywood]

Using ΔL = FL/AE ?

 

[SteamKing]

Yes.

 

[Matthew Heywood]

So ∂ = 4.5mm = F_AB(150mm)/125mm²E_AB + (F_BC + 300)(150mm)/125mm²E_BC + F_CD(150mm)/200mm²E_CD + (F_DE + 600)(150mm)/200mm²E_DE ?

 

[SteamKing]

So ∂ = 4.5mm = F_AB(150mm)/125mm²E_AB + (F_BC + 300)(150mm)/125mm²E_BC + F_CD(150mm)/200mm²E_CD + (F_DE + 600)(150mm)/200mm²E_DE ?

Not necessarily.

Read the problem statement carefully. There is a gap of 4.5 mm between the bottom of the member and the ground, when the member is unloaded. Once the loads are applied to the member, it stretches until contact with the ground prevents further extension, which suggests that a reaction force with the ground develops.

You need to write equations of statics here along with calculating the extensions in the member under load, because the member becomes statically indeterminate once contact is made with the ground and the reaction develops. It also means you're probably going to need to use a value for Young's Modulus for the material in the member in order to solve this problem.

 

[Matthew Heywood]

Ah okay. So you'd use ∂ = 0, F_AB = F_BC + 300kN = F_CD = F_DE + 600kN and E = 200GPa as it is steel?

 

[SteamKing]

Ah okay. So you'd use ∂ = 0, F_AB = F_BC + 300kN = F_CD = F_DE + 600kN

No, the deflection δ ≠ 0, since the member is loaded and is going to stretch.

What you need to do first is find out by how much the member would stretch if the ground wasn't in the way, and then determine what reactions would develop so that the member only stretches 4.5 mm when it reaches equilibrium with the ground. (Hint: a free body diagram would be helpful here.)

and E = 200GPa as it is steel?

This is OK.