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, Fst1 and Fst2. 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.)
: FSt1 + FSt2
+ FBr - 50,000 lb. = 0
: - FSt1 (4 ft.) + FSt2
(4 ft.) = 0
From the torque equation, we see that Fst1 = Fst2, and so we will
now call the force in the steel cables just FSt. And we can rewrite the sum of
y- forces equation as:
FSt + FSt + FBr - 50,000 lb. = 0, or 2 FSt
+ FBr -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:
( FSt *72")/(30 x 106 lb./in2
* .5 in2) = ( FBr * 48")/(15 x 106 lb./in2
* .75 in2 )
If we simplify this equation we obtain: FSt = .89 FBr
We now substitute this into our sum of y-forces equation from static
equilibrium equations.
: (no external x forces.)
: 2 FSt + FBr -50,000 lb. = 0
: - FSt1 (4 ft.) + FSt2 (4 ft.) = 0
and obtain: 2 (.89 FBr) + FBr - 50,000 lb. = 0; and
solving we have:
FBr = 18,000 lb., and FSt = 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.
Return to:
Topic 3.3: Statically Indeterminate Structures
or Select:
Topic 3: Stress, Strain & Hooke's Law - Table of Contents
Strength of Materials Home Page