WORLD FOR CIVIL

Topic 2.2 Trusses -Example 3

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 H. 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 DG by method of sections.

We begin by determining the external support reactions acting on the structure.

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 1, 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 H_y
STEP 2: Break any forces not already in x and y direction into their x and y components. (All forces in x/y directions.)
STEP 3: Apply the equilibrium conditions.
: A_x = 0
: A_y + H_y -12,000 lbs - 20,000 lbs - 10,000 lbs = 0
: (-12,000 lbs)(20 ft) - (20,000 lbs)(40 ft) - (10,000 lbs)(60 ft) + H_y(80 ft) = 0
Solving for the unknowns: A_y = 21,500 lbs; H_y = 20,500 lbs. These are the external support reactions acting on the structure.

Part 2: Now we will find internal force in member DG by method of sections. Cut the truss vertically with a line passing through members DF, DG, and EG. We have shown the section of the truss to the right of the cut. We now treat this section of the truss as if it were a completely new structure. The internal forces in members DF, DG, and EG now become external forces with respect to this section. We have represented these forces with the arrows shown. The forces must act along the direction of the cut member (since all members in a truss are axial members), and we have selected an initial direction either into or away from the section for each of the forces. If we have selected an incorrect initial direction for a force, when we solve for the value of the force, the value will be negative indicating the force acts in the opposite direction of the one chosen initially. We may now proceed with the analysis of this structure using standard Static's techniques.

I. Draw a Free Body Diagram of the structure (section), showing and labeling all external forces, and indicating needed dimensions and angles. (Diagram 2)
II. Resolve (break) all forces into their x and y-components. (Diagram 2)
III. Apply the Equilibrium Equations ()
:EG + DG cos (51.3^o) + DF cos (22.6^o) = 0
: -10,000 lbs + 20,500 lbs - DG sin (51.3^o) - DF sin (26.6^o) = 0
: -DF cos (26.6^o)(15 ft) + (20,500 lbs)(20 ft) = 0
Solving for the unknowns: DF = 30,600 lbs (C); DG = -4,090 (opposite direction)= 4,090 lbs (T); EG = -24,800(opposite direction)= 24,800 lbs (T)