WORLD FOR CIVIL

Topic 3.2a: Statically Determinate - Example 1

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In the structure shown in Diagram 1, member ABCD is 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 we would like to determine the axial stress in cable CE, the deformation of member CE, the strain in member CE, and finally to determine the movement of point D due to the applied loads. (Young's Modulus for steel = 30 x 10⁶ lb/in².)

Step I: As the first step in our solution, we apply static equilibrium conditions to determine the value of the external support reactions (the process we studied in Topic 1 - Statics). Our, hopefully familiar, procedure is as follows:

1: Draw a free body diagram showing and labeling all load forces and support forces, as well as any needed angles and dimensions (Diagram 2).

2: Resolve all forces into their x and y components.
3: Apply the static 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) = 0
Solving for the unknowns: E = 25,600 lb.; A_x = 22,170 lb.; A_y = 17,200 lb.
Step II. Now that we have the values of the external support forces, we determine the value of the force in the internal member in which we would like to find the stress. In this particular problem, this is quite easy once we recognize that at point E, there is only one axial member (CE) attached to the wall. Therefore the force of the wall acting on member CE is equal to the internal force in the member itself: Force in CE = 25,600 lb.

Once we have the force in member CE, we can apply the appropriate stress/strain relationships and solve for the quantities of interest.

A.) Stress in CE = F/A = 25,600 lbs/(3.14*(.5 in)²= 32,600 psi.
B.) Deformation of CE = (FL/ EA)_CE = (25,600 lbs)(16 ft * 12 in/ft) / (30*10⁶ lbs/in² )(3.14*(.5 in)² ) = .209 in
C.) Strain in CE = (Deformation of CE)/(Length of CE) = (.209"/192") = .00109
D.) Movement of point D. This part of the problem requires a bit of reflection on the geometry of the problem. As is shown in diagram 3, when member CE elongates, member ACD - which is pinned at point A - rotates downward. The amount that point D moves is related to the amount point C moves, and point C moves the same amount that cable CE deforms. We can relate the movement of the two points from geometry by:
[Movement of C / 12 ft = Movement of D / 20 ft] or [.209 in / 12 ft = Movement of D / 20 ft], and solving gives us Movement of D = .348 in.