WORLD FOR CIVIL

Topic 3.2b: Statically Determinate - Example 2

In the structure shown in Diagram 1, members ABC and BDE are assumed to be solid rigid members. Member BDE is supported by a roller at point E, and is pinned to member ABC at point B. Member ABC is pinned to the wall at point A. Member ABC is an aluminum rod with a diameter of 1 inch. (Young's Modulus for Aluminum is 10 x 10⁶ lb/in²)
For this structure we would like to determine the axial stress in member ABC both in section AB and section BC, and to determine the movement of point C due to the applied loads.

Part I. To solve the problem we first need to determine the external support forces acting on the structure. We proceed using our static equilibrium procedure (from Topic 1 - Statics)

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

2: Resolve forces into x and y components.
3: Apply the equilibrium conditions.
Sum F_x = A_x = 0
Sum F_y = A_y + E_y -16,000 lbs - 12,000 lbs = 0
Sum T_E = (16,000 lbs)(12 ft) + (12,000 lbs)(8 ft) - A_y(12 ft) = 0
Solving for the unknowns: A_y = 24,000 lbs; E_y = 4000 lbs

Part II. An interesting aspect to this problem is that member ABC is not an axial member, and so it is not in simple uniform tension or compression. However, we are fortunate in that it is not a complex non-axial member. It is not in shear, but rather simply is in different amounts of tension above and below point B. Therefore to determine the amount of stress in each part of ABC, we first make a free body diagram of member ABC and apply static equilibrium principles.

1: FBD of member ABC. (Diagram 3)

2. Resolve all forces into x and y components.
3. Apply equilibrium conditions.
Sum F_x = B_x = 0
Sum F_y = 24,000 lb. - B_y - 16,000 lbs = 0
Solving for the unknowns:B_y = 8,000 lb.

Now to find the force in section AB of member ABC. Cut the member between points A and B,and analyze the top section. We can do this since if a member is in static equilibrium, then any portion of the member is also in static equilibrium.

Looking at Diagram 4 (which is the free body diagram of the upper section of member ABC), we see that for the section of AB shown to be in equilibrium, the internal force (which becomes external when we cut the member) must be equal and opposite to the 24,000 lb force of the wall on the member at point A.

Once we know the tension in section AB of member ABD, we find the stress from our relationship Stress = F/A = 24,000 lb/(3.14 x .5²) = 30,600 psi.

We then use the same approach with section BC of member ABC. We cut member ABC between point B and point C, and apply static equilibrium principles to the top section. Diagram 5 is the free body diagram of that section, and by simply summing forces in the y-direction, we see that the internal force BC (which becomes an external force when we cut the member) must be 16,000 lb. for equilibrium.

The stress in section BC is then given by Stress = F/A = 16,000 lb/(3.14 x .5²)
Stress (BC) = 20,400 psi.

Part III. To determine the movement of point C is a relatively simple problem when we realize that the movement of point C will be equal to the deformation (elongation) of section AB plus the deformation (elongation) of section BC.
That is, the Movement of C = Def_AB + Def_BC
Movement of. C = [ (FL / EA)_AB + (FL / EA)_BC ]
Movement of C = [ (24,000 lbs)(72 in) / (10*10⁶ psi)(3.14*(.5 in)²)]_AB + [ (16,000 lbs)(48 in) /
(10*10⁶ psi)(3.14*(.5 in)²)]_BC
Movement of C = (.220 in) + (.0978 in) = .318 in