WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point F. 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 CD by any method.
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 = 0
Sum F_y = -4,000 lbs - 3,000 lbs + A_y + F_y = 0
Sum T_A = (-4,000 lbs)(12 ft) - (3,000 lbs)(24 ft) + F_y(33 ft) = 0
Solving for the unknowns:
F_y = 3,640 lbs; A_y = 3,360 lbs

PART C - Now find internal force in member CD by sections. Cut vertically through members BD, CD, and CE near to points B and C. Analyze left section as shown in diagram.
STEP 1: Draw a free body diagram of the left section, showing and labeling all external loads and forces.
STEP 2: Resolve all forces into x and y components.
STEP 3: Apply the equilibrium conditions:
Sum F_x = BD cos (14^o) - CD cos (45^o) - CE = 0
Sum F_y = -4,000 lbs + 3,360 lbs - BD sin (14^o) - CD sin (45^o) = 0
Sum T_C = (-3,360 lbs)(12 ft) - BD cos (14^o)(15 ft) = 0
Solving for the unknowns:
BD = -2,770 lbs (opposite direction), 2,770 lbs (c); CD = 43 lbs (c); CE =2,720 lbs (t)