The structure shown below is a truss which is
pinned to the wall at point E, and supported by a roller at point
A. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the value of all the support forces acting on the
structure.
C. Determine the force in member GC by method of sections.
Unless otherwise indicated, all joints and support
points are assumed to be pinned or hinged joints.
Solution:
PARTS A & B:
STEP 1: Draw a free body diagram showing and
labeling all load forces and support (reaction) forces, as well
as any needed angles and dimensions.
STEP 2: Break any forces not already in x and y
direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum Fx = Ex = 0
Sum Fy = Ay + Ey - 12,000 lbs -
6,000 lbs = 0
Sum TE = (12,000 lbs)(4 ft) - Ay(12 ft) = 0
Solving for the unknowns:
Ey = 14,000 lbs; Ay = 4,000 lbs
PART C: - Now find the internal force in member GC by method of
section
STEP 1: 
Cut the structure into "2 sections" with a
vertical line which cuts through members GE, GC and BC, and is
just to the right of point G (see dia-
gram). The internal forces in members GE, GC and BC now become
external forces acting on the left hand section as shown. (We
chose directions for these forces which may or may not be
correct, but which will become clear when we solve for their
values.)
STEP 2: Now treat the section shown as a new
structure and apply statics procedure
- Draw a free body diagram of the left hand section.
- Resolve all forces into x an y components
(see diagram).
- Apply equilibrium conditions:
Sum Fx = - GE cos (37o) + GC cos (37o)
+ BC = 0
Sum Fy = 4,000 lbs - GE sin (37o) - GC sin
(37o) = 0
Sum TG = (-4,000 lbs)(4 ft) + BC(3 ft) = 0
Solving for the unknowns:
BC = 5,330 lbs; GE = 6,670 lbs; GC = 0 lbs