WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

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 F_x = E_x = 0
Sum F_y = A_y + E_y - 12,000 lbs - 6,000 lbs = 0
Sum T_E = (12,000 lbs)(4 ft) - A_y(12 ft) = 0
Solving for the unknowns:
E_y = 14,000 lbs; A_y = 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 F_x = - GE cos (37^o) + GC cos (37^o) + BC = 0
Sum F_y = 4,000 lbs - GE sin (37^o) - GC sin (37^o) = 0
Sum T_G = (-4,000 lbs)(4 ft) + BC(3 ft) = 0
Solving for the unknowns:
BC = 5,330 lbs; GE = 6,670 lbs; GC = 0 lbs