Follow the steps below to replace the forcing system shown by a statically equiv
ID: 3278103 • Letter: F
Question
Follow the steps below to replace the forcing system shown by a statically equivalent system of a single force and a moment at A.
Follow the steps below to replace the forcing system shown by a statically equivalent system of statically equivalent system of a single force and a moment at A. Replace the uniformly distributed loading in the sense of static equivalence by a single force. What is its magnitude and direction and where it applies? Replace each of the two triangularly distributed loadings in the sense of static equivalence by a single force, respectively. What are their magnitudes and directions and where they apply? Replace the original forcing system in the figure by a statically equivalent forcing system of a single force and single moment applied at A.Explanation / Answer
Going according to the steps in the question
a. the uniform loading is 600 lb/ft, acting for 2 ft, so
net force applied by the uniform distributed load, F1 = 600*2 = 1200 lb
this force acts at a distance of 2ft from point A
b. comsider the two triangularly distributed loads. these net forces can be found out by finding the area of the triasngles, with the forces acting through the centroid.
so, consider ythe upper triangular force
force, F2 = Area of triangle = 0.5*2*250 = 250 lb, acting at, 4/3 ft from the point B
similiarly, the bottom half force F3 = 0.5*2*250 = 250 lb, acting at 4/3 ft from the point B
c. now, let the net force onthe system be F, forces along y axis be have unit vector j and those along x axis may have unit vector i
so, from force balance we get
F = -300i - F2 + F3 -400j + F1
F = -300i - 250i + 250i - 400j + 1200j = -300i + 1600j
so, F = -300i + 1600j