A Problem on Forces

1. Aug 24, 2013

smutangama

1. The problem statement, all variables and given/known data

I'm supposed to find the forces in the members as functions of P ans θ.
Same Young's Modulus and area for all of them which is E and A

and then calculate the vertical deflection.

2. Relevant equations

none

3. The attempt at a solution

I named the members from theft to right A, B and C and the forces TA, TB and TC respectively.

I started with equilibrium equations and got

TA=TC

2 TAcosθ + TB= P

Now I tried to use the modulus equation σ/ε or FL/Aex

and I cant figure out a way forward from here. Any suggestions?

Plus my first gut feeling tells me the force in B is P and the rest is 0 but its a past paper and the question was 8 marks which means there's more to it obviously.

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2. Aug 24, 2013

haruspex

Since you are told the modulus, there is some stretching involved, so the geometry is altered by the load. Are the angle theta and length L for the loaded or unloaded geometry?

3. Aug 24, 2013

Staff: Mentor

In terms of L and θ, what are the original lengths of the sides of the triangles? If the point P displaces downward by a distance δ, what is the new length of the vertical wire? Since the side of the triangle at the wall doesn't increase in length when the point P is moved downward, what does the pythagorean theorem tell you about the new lengths of the slanted wires? What are the strains in the vertical wire and the slanted wires? In terms of δ, what are the tensions in the wires? Assuming that, for purposes of the force balance, the angle θ changes negligibly, what is the downward displacement δ?

4. Aug 24, 2013

smutangama

@harupex, it doesn't really say

original lengths of sides would be L/cosθ

new length vertical wire δ+L

new length when p displaced is (L+δ)/cosθ

strains in vertical δ/L

For side bars

strain in side beam(((L+δ)/cosθ)-(L/cosθ))/(L/cosθ) or δ/L

so δ/L is strain and using young's modulus P = (δ/L) x E x A

For Vertical bar

P= E x (δ/L) x A

I still don't get it in a force in both function P and θ, plus should the δ still be in the equation?

5. Aug 25, 2013

haruspex

I have to disagree with Chet about ignoring the change in theta. It is important that the proportional increase in length of the angled wires is not the same as that in the vertical wire.
The distances which remain constant are between the attachments at the tops of the wires. So I suggest forgetting about L and using that distance as the base for expressing all lengths. Call it a or x or whatever.
It won't matter much whether we take theta to be the angle before or after stretching as long as we're consistent. Suppose it's the after angle, and let θ+Δθ be the relaxed angle.
In terms of a and θ+Δθ, what are the three lengths unstretched? What are the three stretched lengths? What does that give you for the three tensions?

6. Aug 25, 2013

Staff: Mentor

Hi haruspex.

I didn't really say that the change in theta should be ignored. I only said it could be ignored in the force balance. In quantifying the kinematics of the deformation, it is essential to take it into account, as you indicated.

The original three sides of the triangle are L, Lsecθ, and Ltanθ. As you indicated, the Ltanθ side remains constant during the deformation. The new lengths of the sides should be L+δ, Ltanθ, and $\sqrt{(L+δ)^2+(L\tanθ)^2}$. If we linearize the latter with respect to δ, we find that $\sqrt{(L+δ)^2+(L\tanθ)^2}≈L\secθ+δ\cosθ$. So the strain in the slanted wires is $\frac{δ}{L}\cos^2θ$.
So the tension in the vertical wire is $$T_B=EA\frac{δ}{L}$$
and the tensions in the slanted wires are $$T_A=EA\frac{δ}{L}\cos^2θ$$
These can be substituted into the vertical force balance to determine δ.

EDIT: There was a small error in the previous version of this post that I corrected. It did not affect the final results.

Chet

Last edited: Aug 25, 2013
7. Aug 25, 2013

smutangama

how did you get the cos2θ? I can't seem to follow the math

8. Aug 25, 2013

Staff: Mentor

If the deformed length of the slanted wires is Lsecθ+δcosθ, then the strain is:

$$ε=\frac{(L\secθ+δ\cosθ)-L\secθ}{L\secθ}=\frac{δ\cosθ}{L\secθ}=\frac{δ\cos^2θ}{L}$$

Does that help?

Chet

9. Aug 25, 2013

haruspex

OK.
Since we're posting solutions, here's mine.
Using a = distance between adjacent attachment points, stretched length L = a cot(θ):
$T_B = \lambda(a \cot(θ) - a \cot(θ+δθ) ) / (a \cot(θ)) = \lambda δθ\sec(θ)/\sin(θ)$
$T_A = \lambda(a .cosec(θ) - a .cosec(θ+δθ) ) / (a .cosec(θ)) = \lambda δθ\cot(θ) = T_B \cos^2(θ)$
Which is the same as you got.

10. Aug 25, 2013

Staff: Mentor

Yes. This works just as well. Still, it is more to my taste to work directly with the downward displacement δ than with the intermediate parameter dθ. But certainly what you did is correct.

Chet