How can I solve this structural mechanics problem accurately?

[notist]

I've been trying to solve a structural mechanics problem from the book Mechanics of Materials, but the results I acquired weren't very satisfying.

Here it is

Homework Statement

[PLAIN]http://img42.imageshack.us/img42/4375/p17n.jpg

Homework Equations

Average Stress: \sigma= F/A

Torque: \vec{M}=\vec{r} x \vec{F}

Equations of static

The Attempt at a Solution

First of all, I defined a load applied at point B and direction DB, resulting from the application of equation of the average stress:

|\vec{F(B)}|=40 KN and \vec{F(B)}= -18.82 ex - 35.29 ey.

Now, let the reactions at points C and D be called Cx/Cy and Dx/Dy respectively, I first assumed the values of Dx and Dy were symetrical to those in F(B), so as to obey the equations of static.
From here I simply calculated the sum of every torque regarding to point C, in order to eliminate Cx and Cy. The equation is as follows

-18.82*0.45 + 0.135P+35.29*0.24=0, resulting in P=0, which is obviously false.

From here I thought I was mistaken about the values of Dx and Dy and decided to treat them as variables. Writing the equations of static, I got

Cx+Dx=18.82
Cy+Dy=P+35.29

And from the total Torque relative to point C,

-18.82*0.45 + 0.135*P+0.24*Dy=0

Obviously, the system is indeterminate, but it's impossible to get more linearly independente equations of static for the system, so I decided to analyse the free body diagram of the two bars separately.

For the bar BD, I calculated the total moment relative do point B, resulting in the equation
-0.45*Dx+0.24*Dy=0

For the bar AC, I calculated the total moment relative to point A:
-0.135*Cx+0.570*Cy=0

The system is now solvable, but it the result is still incorrect. My main question is, if there is any flaw in my logic, where is it? Acording to the solutions, the result is 62.7KN.

 

[PhanthomJay]

I'll have to check the numbers later (or someone else can)..But although the problem is statically indeterminate, the fact that the force in BD must be axial, gives you the extra bit of info you need.

I would not suggest trying to break up the applied load P into components along rotated axes. Instead, apply it at point B with the appropriate moment (couple) also applied at that point, and solve. Free Body Diagrams and sketches are always very helpful.

 

[pongo38]

It IS statically determinate. It is like a 3-hinged arch. Moments about C for the whole thing gives you DY in terms of P. Can you get it from there without more help?

 

[nvn]

-18.82*0.45 + 0.135*P + 35.29*0.24 = 0

notist: Dx is not located at 0.45 m below point C. Try this summation again. And isn't your sign on Dy wrong here?

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 62.7 kN, not 62.7kN. See the international standard for writing units[/color] (ISO 31-0[/color]). Also, always use correct capitalization of unit symbols. E.g., kN, not KN. KN means kelvin Newton; kN means kiloNewton. See NIST[/color] for the correct spelling of any unit symbol.

 

[Filip Larsen]

You are making it much too complicated. First chapter of my edition of that book gives an introduction to stress, so the strain of CB can be ignored and the setup can be considered completely static.

The moment around C from P give rise to a horizontal force at B which is balanced by the projected force through BD. Write up those equations and you get P = 62.7 kN.

 

[pongo38]

Also, you could say that the perpendicular distance from C to BD is (by inspection, or drawing similar triangles) 450*240/510 Hence take moments about C and get Force in BD directly.