Horizontal cable with one vertical point force

[Slimjimi]

Homework Statement

Hang a weight (F) from the middle of a horizontal wire of length L, attached on both sides. The wire has an effective modulus (E) and effective area (A). Find an equation for the new length of the wire L_F. (NOTE: Do not try to solve for a form L_F =, this is too hard).

Homework Equations

To see the equations I thought were relevant, see my attempt at the solution.

The Attempt at a Solution

Force Balance Equation: T = $\frac{Fcosθ}{2}$

Compatibility Equation: ∂ = ∂₁ + ∂₂. ∂₁ = ∂₂.

Force-Displacement Equation: ∂₁ = $\frac{TL}{2EA}$ = $\frac{FLcosθ}{4EA}$.

This gives: ∂=2∂₁ = $\frac{FLcosθ}{2EA}$

Which gives a final answer of: L_F=L+∂ = L + $\frac{FLcosθ}{2EA}$.

But I'm not confident in my work, because professor said it would be too difficult to solve for the solution in this form.

 

[Spinnor]

From where do you measure θ? What is θ when F goes to zero?

 

[Slimjimi]

Theta is the angle made at the support ends. Theta is the same on the left and right side. When F=0, Theta=0, theoretically.

 

[Spinnor]

Shouldn't you have then,

2Tsinθ = F

 

[ansrox]

Okay I've got the same problem as part of a homework.
I agree that you need 2Tsin(A)=F

 

[Slimjimi]

So then, with the new force balance equation, we get:

∂₁ = $\frac{TL}{2EA}$ = $\frac{FL}{4EAsinθ}$

Which gives us:

∂=∂₁ + ∂₂ = 2∂₁ = $\frac{FL}{2EAsinθ}$

So the final answer then would be:

L_F=L+$\frac{LF}{2EAsinθ}$