The structure shown below is a truss which is
pinned to the floor at point A, and supported by a roller at
point D. 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 FC by method of joints.
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 = Ax = 0
Sum Fy = Ay + Dy - 12,000 lbs -
20,000 lbs = 0
Sum TA = (-12,000 lbs)(4 ft) - (20,000 lbs)(12 ft) + Dy(24
ft) = 0
Solving for the unknowns: Dy = 12,000 lbs; Ay
= 20,000 lbs
PART C - Now find internal force in member FC by method of
joints.
JOINT D:
STEP 1: Draw a free body diagram of the joint.
STEP 2: Resolve all forces into x and y
components (see diagram).
STEP 3: Apply equilibrium
conditions:
Sum Fx = -CD + ED cos (66.4o) = 0
Sum Fy = 12,000 lbs - ED sin (66.4o)= 0
Solving for the unknowns:
ED = 13,100 lbs (c); CD = 5,450 lbs (t)
JOINT E:
STEP 1: Draw a free body diagram of the joint.
STEP 2: Resolve all forces into x and y
components (see diagram).
STEP 3: Apply equilibrium conditions:
Sum Fx = (-13,100 lbs) cos (66.4o) + CE cos
(66.4o) = 0
Sum Fy = (13,100 lbs) sin (66.4o) - CE sin
(66.4o) = 0
Solving for the unknowns: FE = 10,500 lbs (c); CE = 13,100 lbs
(t)
JOINT C:
STEP 1: Draw a free body diagram of the joint.
STEP 2: Resolve all forces into x and y
components (see diagram).
STEP 3: Apply equilibrium conditions:
Sum Fx = 5,450 + (13,100 lbs) cos (66.4o) +
FC cos (66.4o) - BC = 0
Sum Fy = (13,100 lbs) sin (66.4o) - FC sin
(66.4o) = 0
Solving for the unknowns: FC = 13,100 lbs (c); BC = 15,950 lbs
(t)