WORLD FOR CIVIL

Topic 3.3: Statically Indeterminate Structures

The fact that real members deform under external forces and loads seems to add a complication when analyzing structures, but for certain types of structures, known as Statically Indeterminate Structures, it is the effect of the deformation that allows use to solve for the external forces on the structure and in the members of the structure. Perhaps the best way to illustrate this is to examine a relatively simple statically indeterminate structure.

Example 1:
The structure shown in Diagram 1 is formed by member ABDF (which is pinned to the wall at point A), steel member BC, and aluminum member DE, both of which are pinned to and support member ABDF, and both of which are pinned to the ceiling as shown. An external load of 10,000 lb. is applied at point F. For this structure some of the things we might wish to know might be: after the 10,000 lb. load is applied, what are the external support forces acting on the structure, what is the stress in the steel and aluminum members, what is the movement of point F due to the load. As we analyze this structure, we are going to ignore the fact that member ABDF 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. As is also the case when the weight of the structural member is ignored in analyzing the structure. This often effects the result only slightly, especially when the external force and loads are much larger than the weight of the members.

To understand what we mean by a statically indeterminate structure, let us first try to analyze the structure using our standard static equilibrium procedure.

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. At point A, where member ABDF is hinged to the wall, we replace the hinge by support forces A_x and A_y. The steel and aluminum members are pinned to the ceiling. Normally we would replace the pins by horizontal and vertical support forces, however in this case we can do better (meaning less unknowns). Since both the steel and aluminum members have forces acting on them at only two points (each end), they are axial members and are in simple tension or compression - in this problem, simple tension. Therefore, the ceiling simply pulls vertically upward on each member with force, F_St and F_Al, as shown in Diagram 2.

2: Resolve forces into x and y components. (All forces are either in x or y direction.)
3: Apply the equilibrium conditions.
: A_x = 0
: -A_y + F_St + F_Al - 10,000 lb. = 0
: + F_St (6 ft.) + F_Al (12 ft.) - 10,000 lb. (18 ft.) = 0
At this point in a statically determinate problem, we would, in most cases, be able to solve for the external support reactions. However, in this case, we observe that we have three unknowns and only two independent equations - and can not solve. (We do see that A_x must be zero, from the first equation, but that is no help with the other two equations, in finding A_y, F_St, F_Al.) We might try to take the structure apart in some way, or redraw the FBD, but none of this will help. Static equilibrium conditions alone are not enough to solve this problem - it is statically indeterminate. Another way to state this difficulty is that we need another independent equation to solve for the unknowns. The deformations of the steel and aluminum 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 10,000 lb. load will be to elongate both the steel and aluminum members which will cause member ABDF to rotate downward about hinged point A. We diagram this, showing and labeling the deformations involved. This is somewhat like a Free Body Diagram, but with deformation effects shown. See Diagram 3. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem. Since member ABDF is pinned at point A, as it rotates downward we see we can write (from similar triangles):

Deformation of Steel / 6 ft = Deformation of Aluminum / 12 ft
or we can rewrite as: 2 * Deformation of Steel = Deformation of Aluminum
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:
2 * [FL/EA]_St = [FL/EA]_Al
We now have our additional relationship between the forces. After substituting in the values for the members we have:
(2 * F_St *72")/(30 x 10⁶ lb/in² * .5 in²) = ( F_Al * 72")/(10 x 10⁶ lb/in² * 1 in² )

If we simplify this equation we obtain: F_St = .75 F_Al We now substitute this into our torque equation from static equilibrium equations (shown below)
: A_x = 0
: -A_y + F_St + F_Al - 10,000 lb. = 0
: + F_St (6 ft.) + F_Al (12 ft.) - 10,000 lb. (18 ft.) = 0

and obtain: (.75 F_Al)(6 ft.) + F_Al (12) - 10,000 lb. (18 ft.) = 0; and solving we have:
F_Al = 10,900 lb., and F_St = 8175 lb. We can now also solve for Ay, finding Ay = 10,075 lb
We find the stress from: Stress Steel = 8175 lb/ .5 in² = 16, 350 lb/in², Stess Aluminum = 10,900 lb/1 in² = 10,900 lb/in².

And finally we can find the movement of point F by first finding the elongation of member DE, the aluminum member, from Def. = FL/EA = (10,900 lb * 72 in.)/(10 x 10⁶ lb.in² * 1 in²) = .0785 in. Point F moves in proportion to the elongation of member DE, and we may write: .0785 in./12 ft = Move. F/18 ft, solving Move. F = .118 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 F.

Summary: As a brief review, our method for solving statically indeterminate problems consists of two basic steps: Step 1 is to apply static equilibrium conditions and write the static equilibrium equations. Step 2 is to find a general relationship between the deformations of the members in the structure, and then to rewrite this relationship using deformation = FL/EA to obtain an additional equation between the forces acting on the structure. Finally using the equations from Steps 1 and 2, we should be able to solve for all the external support forces acting on the structure. To see applications of these relationships, we now will look at several examples.

Continue to:
Example 1 ; Example 2