Example Problems

Solution: A spreadsheet is used for this problem so that it can be used to solve the problem for a wide array of variables. You will need to down load the spreadsheet for this problem to understand this discussion.

We are also making a big simplifying assumption in this problem. That is that the frame will not deflect sideways due to unsymmetrical stiffness and loading. This is only true if the frame can be assumed to be braced.

You will see, in the "given" section of the spreadsheet that input cells (the grey cells) have been provided for all the variables defined in Figure CB5.1.1. These values can be varied to study the sensitivity of the analysis to the various variables. Feel free to experiment with the values!

Determine Relative Rotational Stiffness Values and Distribution Factors

For this step, we need to know the rotational stiffness of the various members framing into each joint. In this problem, we assume that all members are made of the same material. This means that the modulus of Elasticity term is constant, so all we need are the member lengths and moment of inertias.

The moment of inertias are the moments of inertia resisting bending in the plane of the frame. For the beams, this is normally I_x. For the columns, you need to know how they are oriented before inputting values for the moment of inertia. If the columns are bent about their weak axis in the plane, the enter I_y else enter I_x.

The relative stiffness term for each structural element then is computed as I/L.

The distribution factors are computed by weighting the relative rotational stiffness of the members connected at a joint. For example, at joint A we compute only two distribution factors since there are only two members (column 1 and beam 1) at the joint:

DF_C1 = (I/L)_C1 / [(I/L)_C1 + (I/L)_B1]
DF_B1 = (I/L)_B1 / [(I/L)_C1 + (I/L)_B1]

Similarly, at an internal joint, there are three members at each joint. Taking joint B, the three members are beams 1 and 2 and column 2. The distribution factors for the ends of the members at joint B are computed as:

DF_C2 = (I/L)_C2 / [(I/L)_C2 + (I/L)_B1+ (I/L)_B2]
DF_B1 = (I/L)_B1 / [(I/L)_C2 + (I/L)_B1 + (I/L)_B2]
DF_B2 = (I/L)_B2 / [(I/L)_C2 + (I/L)_B1 + (I/L)_B2]

A similar approach is taken at the other two joints. The results are on a line labeled "DF".

Determining the Fixed End Moments

To start the problem, we find the end moments as if each span has rigid fixed ends. In this case we have two loadings on each span: a uniform distributed load and a point load.

The equations for fixed end moments are readily derived or can be found in a variety of texts and handbooks. The equations used here are:

	Left End	Right End
Distributed Load:	+ wL²/12	-wL²/12
Point Load:	+Pa(L-a)²/L²	+Pa²(L-a)/L²

Note that the signs on the moments are consistent with the right hand rule. These are reaction moments.

The starting fixed end moments at each end of each span are the sum of the fixed end moments due to the two loads.

Balancing the Moments at each Joint

Adding the FEMs at each joint results in a non-zero moment on the joint. This indicates that there is an imbalance. This imbalance is divided among the members rigidly connected to the joint in proportion to the relative stiffenesses of the members. In other words, each attached member is assigned an additional moment equal to the negative of the imbalance multiplied by the member's distribution factor at that joint. The result is that the sum of the FEMs and balancing moments results in a zero value. The joints are then balanced.

The Carryover Step

Once the joints are balanced, the balancing moments cause an additional moment at the opposite end of the member. The additional moment is equal to half the balancing moment.

We only apply the carry over to the beam segments and not the column segments in this case. We assume that the far ends of the columns are fixed and do not "reflect" moments back to the joint. As we are not solving for the whole moments on the columns we are not concerned with the carryover on the columns.

Note that on the carryover ("CO") line in the spread sheet that the moments are equal to half the moments on the balancing line above and at the opposite end of the beam span.

The Summation Line

The summation line, in each case, includes the previous pre-balance moment (FEM or Sum) plus the current balancing and carryover moments.

Subsequent Iterations

The balance, carryover, and sum operations are repeated until the imbalance gets small enough for your purposes. In this case, we took the number of iterations further than we probably needed to.

In the end, the moments are not changing in the third significant figure so, it was deemed sufficient for our needs.

Finding the End Shears

The end shears are found by applying equations of equilibrium to a free body of each span. The general free body diagram is shown in Figure CB5.1.2

Figure CB5.1.2
Single Span FBD

The only two unknowns at this point are V₁ and V₂, so two equilibrium equations are needed to solve for the two unknowns.

V₁ = P(L-a)/L + wL/2 + (M₁ + M₂)/ L
V₂ = Pa/L + wL/2 - (M₁ + M₂)/ L

Finding the Internal Shear and Moment

To do find these forces, use the free body diagram shown in Figure CB5.1.3.

Figure CB5.1.3
Partial Free Body Diagram

Solving for equilibrium equations at "x" yields:

V_X = V₁ -wx - if( a < x then P else 0)
M_X = -M₁ + V₁x -wx²/2 - P*min[0, x-a]

Note that both equations account for the relative positions of "x" and the location of the point load. If x is less than a, then the point load term does not appear in the equation.

In the spreadsheet solution, a table has been created that computes the internal shear and moment at 10th points along the span. This increment is generally close enough to find the maximum moments and shears, however it does not give a great diagram where point load cause discontinuities. The result is sufficient for our current needs.

The results are graphed in the spreadsheet.