a.) Roller - produces a reaction force perpendicular to the support plane.
b.) Pin (or Hinge) - produces a vertical and horizontal reaction.
R y R x
c.) Fixed - produces a reaction force in any direction and Moment.
R y R x M
2.) Apply laws of equilibrium to find R AX, R AY, R BY
F x = 0 F y = 0 M z = 0
R AX R AY R BY
B.) Internal Shear
1.) Shear - find by cutting a section at the point of interest and F y = 0 on the FBD.
R y R x F.B.D. V
B.) Internal Bending Moment
2.) Moment - find by cutting a section at the point of interest and M = 0 on the FBD.
R y R x F.B.D. V M
If you were to find the internal shear and moment at several locations along the length of a beam, you could plot a graph shear vs. length and a graph of moment vs. length and find where the maximum shear and moment occur.
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B V(k)
4’ 4’ 4’ 8’ 8’ 2k/ft 3k 4k A B M (k-ft)
C.) Shear Diagram - Simpler Way to Draw
1.) Sketch the beam with loads and supports shown (this is the LOAD DIAGRAM).
2.) Compute the reactions at the supports and show them on the sketch.
3.) Draw Shear Diagram baseline
(shear = zero) below the load diagram
a horizontal line.
4.) Draw vertical lines down from the load
diagram to the shear diagram at:
b.) point loads
c.) each end of distributed loads
5.) Working from left to right , calculate
the shear on each side of each
support and point load and at each
end of distributed loads:
a.) For portions of a beam that have
no loading, the shear diagram is
a horizontal line.
b.) Point loads (and reactions) cause
a vertical jump in the shear
- The magnitude of the
jump is equal to the magnitude of the load (or reaction).
- Downward loads cause a negative change in shear.
c.) For portions of a beam under
i.) the slope of the shear diagram is equal to the intensity (magnitude) of the uniformly distributed load (w).
ii.) the change in shear between two
points is equal to the area under the load diagram between those two points.