Mechanics of materials/statics question.?

Question

Here is the question, now included is a link to the diagram, good luck and thank you for the help:

http://i434.photobucket.com/albums/qq65/chris15012/diagram.jpg

Knowing that the central portion of link BD has a uniform cross-sectional area of 800 mm2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 MPa.

Answer from back of book is: 62.7 kN.

Answer 1

First we find the force in the link member BD, by multiplying the allowable stress by the cross-sectional area:

Force in BD = (50 N/mm^2)  * ( 800 mm^2) = 40,000 N = 40 kN.

note 1 MPa = 1 N/mm^2

Now, summing the moments about the point C  and equalizing to zero (static frame) gives :

P * 135mm = BD (sin angle) * 450 mm  = BD (240/510) *450 mm

then  P = 62.745 kN

note :

the angle between BD and CB, has sin =240/510 = opposit/hypotenus