SS - More on Stiffness

The Stiffness Method of Analysis is the basis of all commercial structural analysis programs. Unlike before, where we described stiffness as the unit displacement over a net force, here the focus of this blog will be the development of equations that only take into account bending deformations, i.e., ignore axial member deformations. Within the assumptions, the stiffness method for beam and frame structures is “exact”.

In the stiffness method of analysis, we write equilibrium equations in terms of unknown joint (node) displacements. The number of unknowns in the stiffness method of analysis is known as the degree of kinematic indeterminacy, which refers to the number of node/joint displace-ments that are unknown and are needed to describe the displaced shape of the structure.

One major advantage of the stiffness method of analysis is that the kinematic degrees of freedom are well-defined.

Definitions and Terminology

Positive Sign Convention: Counterclockwise moments and rotations along with transverse forces and displacements in the positive y-axis direction.

Fixed-End Forces: Forces at the “fixed” supports of the kinematically restrained structure.

Member-End Forces: Calculated forces at the end of each element/member resulting from the applied loading and deformation of the structure.

Stiffness Analysis Procedure

The steps to be followed in performing a stiffness analysis can be summarized as:

1. Below is a portal frame with column height "h" and beam length "2h".  The beam has a gross moment of inertia half that of the column.  Determine the needed displacement unknowns at the nodes/joints and label them u1, u2, …, un in sequence where n = the number of displacement unknowns or degrees of freedom.

2. Modify the structure such that only one displacement (u) or rotational displacement is applied.  Let the other translations/rotations equal zero.  This is virtual work that must be analyzed, in other words.

3. Indicate the locations of individual stiffness acting in the frame.  Note: this happens at each degree of freedom.  There are no degrees of freedom at the base, since the base is fixed, but there would be two more degrees of freedom at the base were the columns pinned instead of fixed.



4. Indicate the locations of individual stiffness acting in the frame. Note: this happens at each degree of freedom. There are no degrees of freedom at the base, since the base is fixed, but there would be two more degrees of freedom at the base were the columns pinned instead of fixed.  Below is a diagram which shows what stiffness coefficient is used for either a rotational displacement of 1 radian or a translational displacement of 1.  Note how the end conditions don't exactly match the end conditions in the portal frame as drawn above.  The diagram is just for acquiring stiffnesses, not for replicating end conditions in a structure.



5. Next, after having found all of the stiffness coefficients, place them into a 3x3 matrix and multiply them by the elastic modulus and the moment of inertia of the column (recall the inertia of the beam is half that of the column).  The following matrix should look as follows:

5. Once the matrix is set, divide the matrix into fourths, which will respond to (A) the stiffness caused by translation (k11) and (B) the stiffness caused by rotation at the joints (k12, k21 and k22). 

6.  Lastly, solve for the effective stiffness by putting each structural stiffness component together in the following equation:

So what is the point of all this?  The dynamic movement of a building is a relation of the mass, the damping (to be posted later on the blog) and the effective stiffness of the structure.  The first two are generally investigated or already known.  The latter has to be analyzed.  This blog was to show how that can be done. 

Finally, by taking all three components, a universal equation governing displacement and force can be derived as follows:

This equation is essentially the most important governing equation in all of structural analysis.  Dynamically, it helps predict stiffness, displacement and force all in one unified equation.  By understanding why this equation works the way it does, an architect not only begins to speak the language of structures - rather, he becomes a master of structures entirely. 

SS - Determinacy

In statics, when the static equilibrium equations are sufficient to determine the internal forces and reactions of the structure, that structure is said to be statically determinate.

In solving for determinacy, three summations of forces (which are obtained by Newton's laws of motion) are used to determine whether a structure is determinate. If for three equations, three variables can be solved, then that structure is called statically determinate.

The three equations for equilibrium state the following:

Σ N = 0: the sum of the horizontal components of the forces equals zero;

Σ V = 0: the sum of the vertical components of forces equals zero;

Σ M = 0: the sum of the horizontal components of the forces equals zero;

Where N is considered the normal force, V the shear force and M the moment force. The units for N and V are either pounds (lb), kips (thousands of pounds) (k) or Newtons (N) / kiloNewtons (kN).

The Free Body Diagram

In beams, columns or other composite structural members, it is best to break the key components of bar forces into separate reactions acting against or along those members. This is called a free body diagram, and it is the fundamental building block of all statics in determining force vectors.

The classical scenario typically used for physics is the block on a slope. The block has some weight, w, which acts downward toward the surface of the earth. Against the slope, however, a partial fraction of that weight, called the normal force (-N) acts perpendicularly against the slope. An equal, but opposite reaction (N) is the result of the sum of vertical forces equaling zero.

In structural beams, N is a horizontal force, acting linearly throughout the member. A change of connotation in how one determines force is all structural analysis suggests. Instead of negative forces acting downward and positive forces acting upward, structural analysis is more interested in what members are in tension, and which are in compression. If a member is in tension, that force is outward and its value is positive. Conversely, if it is in compression, the force is inward and its value is negative.

Or put it more simply, when your muscles are in tension, you build bigger biceps, and that typically is a positive outcome. Structurally, you're better with bigger, stronger muscles. Alternatively, when your muscles are compressed, there is a force acting against you, a negative force (which may include a bench press, dumbbells or some other voluntary load). You must prove elastic against that force in order to withstand it (which is the definition of compression).

A standard free body diagram may look something like this:

Where there may be a point load (or concentrated load), distributed load (in kN/m), reaction forces (at A in this diagram), distances for the loading forces and reaction forces, and subsequent shear vectors (V) and moment vectors (M) at cut cross-sections in the member. As a result of where the cut is taken, V and M with vary either linearly or parabolically. Or not at all. It all depends on the degree of loading (whether it is one point, continuously linear or curvilinear). The way it affects V and M, however, is another blog, however, and will not be included here.

All in all, though, there are two main considerations when analyzing a structure. One is stability, which is also another post. And the other one (which tells you something about stability) is determinacy. Without determinacy, it is very hard to predict what happens for a structure. Therefore, it is the preeminent consideration for all of structural analysis.

Now take a look at the top diagram. Why are the members and trusses either determinate or indeterminate (not solvable by equilibrium equations)? Recall that roller reactions have only one force they exhibit (vertical to loading), pins have two (vertical and horizontal) and fixed connections have three (vertical, horizontal and rotational to loading). Consider this a bit of a quiz. But feel free to take your time and look closely.

SS - Free Body Diagram and Determinacy Problems