force method structural analysis examples

So, after you have analyzed all the forces using some Structural Analysis Method, all you have to do is be a good bookkeeper to draw the shear and moment diagrams.

Given that there is an evenly distributed force along the length of the span we can draw the approximate parabolic moment diagrams for each beam.

The above example is a beam in which we know all the intermediate span end forces.

A hinge is an internal reaction that doesn't allow moments and can only transfer shear force (summing moments about a hinge = 0, hinges also allow for one extra equation). Who cares what it looks like.Īnother example worked out in metric with the addition of a hinge.

From 5L/8 to L, we know the moment must decrease non-linearly (2nd power).

To find the Maximum moment, add the shear force at L/2 to the area of the shear triangle from L/2 to the location of zero shear (5L/8) discovered from the third bullet.

The point of inflection of the the moment diagram (slope = 0 = flat) is where the shear force = 0.

it increases linearly from 0 to L/2 with a maximum value equivalent to the area of the shear diagram. Just like the evenly distributed load affected the right side of the shear/moment diagram, we can use the same concept on the moment diagram.

The first half of the moment diagram is a slam dunk.

We find the slope of the shear by subtracting Ra from Rb and dividing by the acting length (L/2) At L/2 we know the shear force must change linearly as the evenly distributed load acts form L/2 to L. Now that we have Ra, we know the shear force from support A to L/2.To find the reactions, we place a statically equivalent load in the middle and sum the forces and moments about on of the support locations and calculate Ra and Rb.Both ends are pinned so there can be no moment at either end. The simply supported beam above is subjected to a evenly distributed load from L/2 to L.

Force method structural analysis examples