A triangular load and the force that replaces it
A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.
3 essays call
load-resultant. The drawing above is what it returns with no arguments at all; every
call below passes it something, because a placement that passes nothing draws whichever
member of the family the generator happens to default to rather than the one its essay
argues about.
Where it is called
Changing this generator changes every one of these figures.
The load that is spread out, and the force that replaces it
A distributed load can be swapped for a single force at its centroid. The reactions come out identical and the bending moment does not, and knowing which side of the cut the swap is legitimate on is most of the skill.
Weight is the only thing resisting it
A structure that is strong enough everywhere can still be blown over, and nothing in its material properties has any part in whether it is. The whole answer is a weight and a width — and the failure begins long before anything tips, at the moment one edge stops pressing down.
Moving a force, and what it costs
Every free body on this site begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.