To do your math modeling, sketch a bow of the type you want to build at full draw. Fill in the major forces. The sum of all forces should equal zero unless the bow is being thrown. The force always acts on the bow in a direction. There are basically two forces on the bow, the bow hand force pushing the bow away from the archer at 40lbs and the string hand force pushing the string towards the archer. Now draw a free body diagram of the point where the string is being drawn. You have a 40 lb force pulling at zero degrees. You know that strings can only pull in the direction of the string so you have two other forces at other angles. For the sake of illustration let's say the angle is 60 degrees from horizontal top and bottom. So, the horizontal component of these forces has to be equal to the 40 lbs going the opposite way. The string leaves the draw point in two directions symmetrically which makes it mathematically easy. 2*20-40=0. However that is not the total force on the string. That is the force on the string in the horizontal direction. To find the force on the string draw a triangle with the horizontal leg being 20, the vertical leg being unknown. The hypotenuse is unknown in length as well however, the angles are 30,60,90 with the 20 lb leg going from the 60 to 90. From this we can use trig functions on the calculator to find out that the magnitude of the tension in the string is 40 lbs. now draw a new free body diagram at the tip of the bow to calculate the forces on the tip. Once you have these you can draw a free body diagram of the segment and work your way around the bow. The technique for doing this is a full quarter in engineering school. If you want to do this kind of research pick up an old book on statics at a university library and dig in. You'll want to have at least trig under your belt to get started. Also, the numbers I used came out even for illustration. A real bow would be far more difficult. I have the math to do this. I but never do. Archery is fun not school.