WORLD FOR CIVIL

Topic 3.7: Stress, Strain & Hooke's Law - Solution Exam Question 1

The structure shown 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.
E_st = 30 x 10 6 psi; E_br = 15 x 10 6 psi; E_al = 10 x 10 6 psi
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.
Solution:
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, drawing a free body diagram of the entire structure, we would have up to 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. If we first take the structure apart and analyze member ABC there will only be three unknowns.(See Diagram) Then we 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
We next analyze member CDE in a similar fashion. Drawing a free body diagram and applying the equilibrium conditions: (See Diagram)

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
Next we find the remaining quantities
B. Stress_DF = F/A = 18,000 lbs/ .4418 in² = 40,740 psi
C. Point B moves due to the deformation of member DF.
Deformation of DF = FL/EA = (18,000 lb * 72")/(30 x 10⁶ lb/in² * (3.14 *.375"²)) = .098". Point C moves down a proportional amount: Move. C/4' = .098"/2 ft, solving Move.C = .196". Finally, Point B moves down a proportional amount to the movement of point C. Move.B/6' = .196"/8'. Solving Move. B =.147"