WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below members ABC, ADE, and DB are assumed to be solid rigid members. Members ABC and ADE are pinned to the wall at point A. Member ADE is supported by a roller at point E. 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 DB.
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 + A_x = 0
Sum F_y = A_y - 12,000 lbs = 0
Sum T_A = E_x(8 ft) - (12,000 lbs)(12 ft) = 0
Solving for the unknowns:
E_x= 18,000 lbs; A_x= -18,000 lbs; A_y = 12,000 lbs

PART C - Now find internal force in member DB

STEP 1: Draw a free body diagram of a member that DB 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 + DB cos (33.7^o) = 0
Sum F_y = A_cy + DB sin (33.7^o) - 12,000 lbs = 0
Sum T_A = DB sin (33.7^o)(6 ft) - (12,000 lbs)(12 ft) = 0
Solving for the unknowns:

DB = 43,300 lbs; A_cx = -36,000 lbs; A_cy = -12,000 lbs

These are external forces acting on member ABC.

The force in DB is 43,300 lbs (c).