WORLD FOR CIVIL

STATICS / STRENGTH OF MATERIALS - Example

In the structure shown below members ABC and CDE are assumed to be solid rigid members. Member ABC is pinned to the wall at A and is supported by a roller at point C. Member CDE is pinned to the wall at point E, and is supported by steel cable DF. Cable DF has a diameter of .75 inch. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress in cable DF.
C. Determine the movement of point B 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 five 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 ABC there will only be three unknowns.

Apply the equilibrium conditions:
Sum F_x = A_x = 0
Sum F_y = A_y + C_y - 12,000 lbs = 0
Sum T_A = (-12,000 lbs)(6 ft) + C_y (8 ft) = 0
Solving for the unknowns:
C_y = 9,000 lbs; A_y = 3,000 lbs

Now we can analyze member CDE in a similar fashion.
Apply the equilibrium conditions:

Sum F_x = E_x = 0
Sum F_y = F_y + E_y - 9,000 lbs = 0
Sum T_E = -F_y (2 ft) + (9,000 lbs)(4 ft) = 0

Solving for the unknowns:
F_y = 18,000 lbs; E_y = -9,000 lbs
Part B. Stress_DF = F/A = 18,000 lbs/ .4418 in² = 40,740 psi

Part C. Def = Deformation

STEP 1: Point C moves due to the deformation of cable FD. So we first determine the deformation of FD.
Def_FD = (FL / EA) = (18,000 lbs)(6 ft)(12 in/ft) / (30 *10⁶)(.4417 in²) = .0978 in

STEP 2: Point C moves in proportion to how much point D moves (see diagram).

Mov. C / 4 ft = Mov. D / 2 ft
Mov. C / 4 ft = .0978 in/ 2 ft
Mov. C = .1956 in

STEP 3: Member ABC is resting on member CDE therefore, at point C the movement is the same for members ABC and CDE (see diagram).

Mov. B / 6 ft = Mov. C / 8 ft
Mov. B / 6 ft = .1956 in/ 8 ft
Mov. B = .1467 in