WORLD FOR CIVIL

Topic 3.3b: Statically Indeterminate Structures - Example 2

The structure shown in Diagram 1 consists of horizontal member BCE, which is supported by two steel cables, AB and EF, and a Brass cable, BD pinned to the ceiling as shown. A downward external load of 50,000 lb. is applied at point C. For this structure we would like to determine the stress in the steel and brass cables, and the movement of point C due to the load. As we analyze this structure, we ignore the fact that member BCE will experience some bending (which in a sense is a deformation, and which we will deal with when the topic of beams is discussed). Ignoring the bending in this case effects the result only to a small degree.
Our first step is to apply our static equilibrium procedure to our structure

Part I - Static Equilibrium Analysis
1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. In Diagram 2, a FBD of the structure is shown. In this structure the load simply puts all the cables in tension, and the ceiling acts on the structure as shown in Diagram 2. Notice that at this point we do not assume the force the ceiling exerts on each steel cable is the same (although the symmetry of problem indicates that should be true), but rather we indicate two different forces on the steel cables, F_st1 and F_st2. Our static equilibrium equations will make clear the relationship between the forces in the two steel cables.

2: Resolve forces into x and y components. (All forces are either in x or y direction.)
3: Apply the equilibrium conditions.
: (no external x forces.)
: F_St1 + F_St2 + F_Br - 50,000 lb. = 0
: - F_St1 (4 ft.) + F_St2 (4 ft.) = 0
From the torque equation, we see that F_st1 = F_st2, and so we will now call the force in the steel cables just F_St. And we can rewrite the sum of y- forces equation as:
F_St + F_St + F_Br - 50,000 lb. = 0, or 2 F_St + F_Br -50,000 lb. = 0
However, we still have too many unknowns. We have two unknowns at this point, and only one independent equations left - and we cannot solve. Static equilibrium conditions alone are not enough to solve this problem - it is statically indeterminate. Another way to state the problem is that we need another independent equation to solve for the unknowns. The deformations of the brass and steel members will give us this additional equation..

Part II. - Deformation Equation
Step 1 is to find some general relationship between the deformations of the members of the structure. We may get this from the way the problem is stated, or often from the geometry of the structure - as in this case. The effect of the 50,000 lb. load, because of the symmetry of the problem, will be to elongate both the steel and the brass members equally, and cause horizontal member BCE to move downward as shown in Diagram 3.

We now write a general relationship between the deformations of the members which we can get from the geometry of the problem. For our structure:
Deformation of Steel = Deformation of Brass
or symbolically:
This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure. At first it may not be clear how we can use this equation, since it involves deformations, not forces as in the equilibrium equations. However, if we recall our stress/strain/deformation relationships, we see we can write the deformation of a member as:

deformation = [ force in member * length of member] / [ young's modulus of member * area of member], or def = FL/EA. Substituting this relationship into our deformation equation we have: [FL/EA]_St = [FL/EA]_Br We now have our additional relationship between the forces. After substituting in the values for the members we have:
( F_St *72")/(30 x 10⁶ lb./in² * .5 in²) = ( F_Br * 48")/(15 x 10⁶ lb./in² * .75 in² )
If we simplify this equation we obtain: F_St = .89 F_Br We now substitute this into our sum of y-forces equation from static equilibrium equations.
: (no external x forces.)
: 2 F_St + F_Br -50,000 lb. = 0
: - F_St1 (4 ft.) + F_St2 (4 ft.) = 0
and obtain: 2 (.89 F_Br) + F_Br - 50,000 lb. = 0; and solving we have:
F_Br = 18,000 lb., and F_St = 16,000 lb.
We find the stress from: Stress Brass = 18,000 lb./ .75 in² = 24,000 lb./in², Stress Steel = 16,000 lb./.5 in² = 32,000 lb./in².
And finally we can find the movement of point C simply by finding the elongation of member CD, the Brass member, from Def. = FL/EA = (18,000 lb. * 48 in.)/(15 x 10⁶ lb.in² * .75 in²) = .0768 in.

We have now solved our statically indeterminate problem and determined the values of the external support reactions acting on the structure, the stresses, and the movement of point C.