cmmcnamara said:
I think that is why I am having a bit of trouble. To me the beam acts as a spring in the vertical direction but acts differently depending on the force configuration, but once a location is fixed its stiffness should be a fixed rate. What I think I should do is essentially replace the beam connected to the spring with another spring that has the spring rate of the beam's stiffness. Since the springs are connected end to end the springs are in series and the equivalent stiffness should be k=(1/k_beam+1/k_spring)^-1. However the solutions manual for this problem claims that the springs are in parallel which I don't understand at all.
I can see how you might conclude that the "springs" are in series, but they are not. When 2 springs of different stiffness constants 'k' are in series, they each carry the same load but each deflects by a
different amount. As an example, consider 2 springs in series, subject to a hanging weight force of 10 N. One spring has a k value of 200 N/m and the other has a k value of 400 N/m. They each experience 10 N of force. Since per Hookes Law F=kx, spring one deflects 0.05 m and spring 2 deflects 0.025 m. The total deflection is 0.075 m, and the equivalent stiffness per your formula is k=133, which is consistent with total deflection = 10/133 = 0.075 m, but the important thing to note is that the deflections of each springs are
different.
Now look at the problem at hand with a spring support at midspan. When the beam is now loaded, both the beam and the spring will each deflect the
same amount at midspan. They are effectively connected together at their "top" in parallel, not 'end to top' if they were in series.
Anyway, the problem as you likely know is statically indeterminate to the first degree, and you must use the deformation compatability method to calculate the force delivered by the spring.