WORLD FOR CIVIL

Topic 2.2 Trusses - Example 2

The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and supported by a roller at point D. For this structure we wish to determine the value of all the support forces acting on the structure, and to determine the force in member FC by method of joints.

For the first part of the problem we proceed using our normal static's procedure.

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. (Note in Diagram 2, we have replaced the pinned support by an unknown x and y force (A_x , A_y), and replaced the roller support by the vertical unknown force D_y.

STEP 2: Break any forces not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions

: A_x = 0
: A_y + D_y - 12,000 lbs - 20,000 lbs = 0
: (-12,000 lbs)(4 ft) - (20,000 lbs)(12 ft) + D_y(24 ft) = 0
Solving for the unknowns: D_y = 12,000 lbs; A_y = 20,000 lbs. These are the external support reactions acting on the structure.

PART 2: Determine the internal force in member FC by method of joints. We begin at a joint with only two unknowns acting, joint D.
JOINT D:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and load, and including any needed angles.(Diagram 3) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.

STEP 2: Resolve all forces into x and y components. (Diagram 3).
STEP 3: Apply equilibrium conditions:
: -CD + ED cos (66.4^o) = 0
: 12,000 lbs - ED sin (66.4^o)= 0
Solving for the unknowns: ED = 13,100 lbs (C); CD = 5,240 lbs (T)

Now that we have calculated the values for ED and CD we can move to joint E. We could not solve joint E initially as it had too many unknowns forces acting on it.

JOINT E:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any needed angles. (Diagram 4) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.
STEP 2: Resolve all forces into x and y components. (Diagram 4).

STEP 3: Apply equilibrium conditions:
FE -(13,100 lbs) cos (66.4^o) - CE cos (66.4^o) = 0
:(13,100 lbs) sin (66.4^o) - CE sin (66.4^o) = 0
Solving for the unknowns: FE = 10,500 lbs (C); CE = 13,100 lbs (T) (Since force values were positive, the initial direction chosen for the forces was correct.)
Now that we have calculated the values for FE and CE we move to joint C. We could not solve joint C initially as it had too many unknowns forces acting on it.

JOINT C:
STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any need angles. (Diagram 5) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.
STEP 2: Resolve all forces into x and y components. (Diagram 5).

STEP 3: Apply equilibrium conditions:
: 5,450 + (13,100 lbs) cos (66.4^o) + FC cos (66.4^o) - BC = 0
: 13,100 lbs sin (66.4^o) - FC sin (66.4^o) = 0
Solving for the unknowns: FC = 13,100 lbs (c); BC = 15,950 lbs (t) Thus, member FC is in compression with a force of 13, 100 lbs.