Mechanics of Materials - Deflection of a rigid bar

[ginrummy]

Homework Statement

http://img.waffleimages.com/5b5524d885644e9f1c76a800c4b0701a0606268a/problem6.jpg

(The image will not work unless it is copied and pasted into the address bar)

I am given this drawing and stress v strain diagram, and am told to find the deflection, ∂, at 4 (the end of the beam), if P = 50 kN.

Homework Equations

When solving by geometry:
E = stress/strain
∂ = PL/AE
F = k∂
k = AE/L

The Attempt at a Solution

Using the stress v strain diagram provided, I solve for the elastic modulus (Young's modulus):

E = (700x10^6 N/m^2)/(.35) = 2x10^9 N/m^2

Take the moment about point A to solve for tension in the cable, T2:

(clockwise is positive) ∑MA= 0= (1.4 m)(-T2 N)+(2.2 m)(50,000 N) => T2= 78,571.43 N

From there, I do a compatibility sketch. Unfortunately I do not have software on my computer to draw up a diagram to post, but it is a right triangle, with the length of the bar as the "height," and ∂4 and ∂2 serving as the bases in a similar triangle relationship.

Therefore:

(∂4/2.2)=(∂2/1.4) leading to ∂4=(2.2/1.4)∂2

Therefore, using the Hooke's Law equation:

T2= k2∂2 => ∂2= T2/k2
k2= AE/L= [(1.25x10^-4 m^2)(2x10^9 N/m^2)]/3m => k2= 83,333.3 N/m

Therefore:
∂2= 0.943 m
∂4= 1.48 m


These deflections seem WAY too high to be correct. I feel confident that I'm using the right process, but there has to be some mistake I'm making somewhere.

 

[pongo38]

the website http://img.waffleimages.com/5b5524d8...a/problem6.jpg
didn't work for me.