I just need a little help on this problem. The support has two reactions and a m
ID: 1822653 • Letter: I
Question
I just need a little help on this problem. The support has two reactions and a moment. I have found the equation of the parabola using point-vertex form to be y=(-100/3)*x^2+400x-300 . I integrated this from 0 to 6 to find what I thought would be the magnitude of the point load, which I found to be 3000 N.
I just need help figuring out where to go from here, and if what I started with at this point is correct.
I just need a little help on this problem. The support has two reactions and a moment. I have found the equation of the parabola using point-vertex form to be y=(-100/3)*x^2+400x-300 . I integrated this from 0 to 6 to find what I thought would be the magnitude of the point load, which I found to be 3000 N. I just need help figuring out where to go from here, and if what I started with at this point is correct.Explanation / Answer
You can calculate the centroid and the area of the two portions of the distributed loads. However, it is easier to replace the given loading with the eqivalent loading shown below. NOte the lower portion is pointing up.
Now, from the figure given in the textbook, we have
RI = (6)(300) = 1800 N and RII =(2/3) (6)(1200) = 4800 N
Since the beam is in equilibrium, we have
Fx =0: Ax =0 <= ANs.
Fy =0: Ay +RI - RII=0
Ay = 3000 N <= ANs.
MA =0: MA +RI(3) - RII (15/4) =0
MA = 12600 N.m <= ANs.