WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below member ABCD a solid rigid member pinned to the wall at A, and supported by steel cable CE. Cable CE is pinned to the wall at E and has a diameter of 1 inch. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in cable CE.
C. Determine the movement of point D due to the applied load.
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Part A. External support reaction - Statics:
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 - E cos (30^o) = 0
Sum F_y = A_y + E sin (30^o) - 10,000 lbs - 20,000 lbs = 0
Sum T_A = (20,000 lbs)(4.8 ft) + (10,000 lbs)(16 ft) + E cos (30^o)(2 ft) - E sin (30^o)(23.46 ft)

Solving for the unknowns:
E = 25,600 lbs; A_x = 22,170 lbs; A_y = 17,200 lbs
Part B. Stress_CE = F/A = 25,600 lbs/.785 in² = 32,600 psi
Part C. Def = Deformation
Movement of D: First find deformation of cable CE
STEP 1: Def_CE = (FL / EA)_CE = (25,600 lbs)(16 ft * 12 in/ft) / (30*10⁶ lbs/in² )(3.14*(.5 in)² ) = .209 in
STEP 2: Point C moves the same amount that cable CE deforms, and point D moves a proportional amount compared to point C (see diagram).

move. C / 12 ft = move. D / 16 ft
.209 in / 12 ft = move. D / 16 ft
Solving for movement of D = .348 in