WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below members AD, DC, and ABC are assumed to be solid rigid members. Member ED is a cable. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the values of all the support forces acting on the structure.
C. Determine the force (tension or compression) in member DC.
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 cos (37^o) + A_x = 0
Sum F_y = A_y + E sin (37^o) - 10,000 lbs - 8,000 lbs = 0
Sum T_A = E cos (37^o)(12ft) - (10,000 lbs)(4 ft) - (8,000 lbs)(12ft) = 0
Solving for the unknowns: E = 14,200 lbs; A_y = 9,480 lbs; A_x = 11,400 lbs

PART C: Now find internal force in member DC
STEP 1: Draw a free body diagram of a member that DC acts on - member ABC.
STEP 2: Resolve all forces into x and y components (see diagram).
STEP 3: Apply the equilibrium conditions.
Sum F_x = A_cx - DC cos (56.3^o) = 0
Sum F_y = A_cy - 10,000 lbs - 8,000 lbs + DC sin (56.3^o) = 0
Sum T_A = (-10,000 lbs)(4 ft) - (8,000 lbs)(12 ft) +DC sin (56.3^o)(12 ft) = 0
Solving for the unknowns: DC = 13,600 lbs; A_cx = 7,560 lbs; A_cy = 6,670 lbs
These are external forces acting on member ABC. The force in DC is 13,600 lbs (c).