Lecture 33
Example Problem
Internal Forces
Given:
the left cut section shown below.
Determine:
all forces acting at the cut section using the equations of equilibrium
Solution:
The shearing force applied to the end of the cut section is exactly equal to the internal shearing force before we cut the beam. It is numerically equal (but opposite in sense) to the algebraic sum of the loads and reactions on the FBD to the left of the
cut. Thus it has the value of 300# upwards. We can see then, that this creates a couple which clearly puts the segment out of equilibrium.
By the sum M = 0, we can then find the moment required to put the system back into equilibrium. Remember, the magnitude of a couple is equal to the magnitude of one force multiplied by the distance between the forces.
(300#)(3ft) + M = 0
M = 900 #ft counterclockwise
The moment that has been applied to the cut section of the FBD is the internal or resisting moment. Is it obvious that it is numerically equal to, but opposite in direction, to the moment caused by the loads and reactions?
Copyright © 1995 by Chris Luebkeman and Donald Peting