WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

The structure shown below is a truss which is pinned to the wall at point F, 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 (tension or compression) in member EB 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 F_x = A_x + F_x = 0
Sum F_y = F_y - 8,000 lbs - 6,000 lbs = 0
Sum T_A = (-8,000 lbs)(8 ft) - (6,000 lbs)(16 ft) - F_x(10 ft) = 0
Solving for the unknowns:
A_x = 16,000 lbs; F_x = -16,000 lbs; F_y = 14,000 lbs

PART C - Now find internal force in member EB by method of joints.
JOINT F:
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 F_x = FE - 16,000 lbs = 0
Sum F_y = 14,000 lbs - FA = 0
Solving for the unknowns:
FE = 16,000 lbs (t); FA = 14,000 lbs (t)

JOINT A:
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 F_x = 16,000 lbs - AB -AE cos (68.2^o) = 0
Sum F_y = 14,000 lbs - AE sin (68.2^o) = 0
Solving for the unknowns: AE = 15,080 lbs (c); AB = 10,400 lbs (c)

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 F_x = -16,000 lbs + ED + 5,600 lbs + EB cos (68.2^o) = 0
Sum F_y = 14,000 lbs - EB sin (68.2^o) = 0
Solving for the unknowns: ED = 4,800 lbs (t); EB = 15,080 lbs (t)