WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below member ABD is a solid rigid member pinned to the wall at A, supported by steel cable BC, and connected to member EFG by steel cable DE.
(Cables BC and DE each have a cross sectional area of .5 square inches.)
Member EFG is supported by a roller at F and is loaded with 12000 lbs at G.
For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in cable BC.
C. Determine the movement of point G due to the applied load.
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Part A:
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: If we were to go about this problem in the normal fashion, we would have four unknowns and only three equations. There would be no way to solve this problem using that approach. Therefore, we must take a different perspective. By taking the member apart and analyzing member EFG there will only be two unknowns.

Apply the equilibrium conditions:
Sum F_y = F_y - E_y - 12,000 lbs = 0
Sum T_E = (-12,000 lbs)(6 ft) + F_y (2 ft) = 0

Solving for the unknowns:
F_y = 36,000 lbs; E_y = 24,000 lbs

Now we can analyze member ABD in a similar fashion.

Apply the equilibrium conditions:
Sum F_x = A_x = 0
Sum F_y = A_y - C_y + 24,000 lbs = 0
Sum T_B = -A_y (3 ft) + (24,000 lbs)(3 ft) = 0

Solving for the unknowns:
A_y = 24,000 lbs; C_y = 48,000 lbs
Part B. Stress_BC = F/A = 48,000 lbs/ .5 in² = 96,000 psi

Part C. Def = Deformation
Point G moves due to the deformation of cable BC and Cable DE.
STEP 1: Mov. B = Deformation cable BC
Mov. B = (FL / EA) = (48,000 lbs)(24 in) / (30*106 lbs/in²)(.5 in²) = .0768 in

STEP 2: Movement of point D is proportional to movement of point B, and we can write:
Mov. D / 6 ft = Mov. B / 3 ft
Mov. D / 6 ft = .0768 in / 3 ft
Mov. D = .1536 in

STEP 3: Movement of point E is equal to movement of point D plus the elongation of cable DE.
Mov. E = Mov. D + (FL / EA)_DE = .1536 in + (24,000 lbs)(24 in) / (30*10⁶ lbs/in²)(.5 in²)
Mov. E = .1536 in + .0384 in = .192 in

STEP 4: Finally the movement of point G is proportional to the movement of point E and we may write:
Mov. G / 4 ft = Mov. E / 2 ft
Mov. G / 4 ft = .192 in / 2 ft
Mov. G = .384 in