Modeling Reinforced Concrete Stiffness
Part 4: Indeterminate Structures & Force Distribution

By Adam Stordahl, PE • Published September 5, 2025

Recap

We've made it to the final post in our series on reinforced concrete stiffness, thank you for following along. We have moved from materials, to cross-sections, to individual elements. In each post we've seen how reinforced concrete stiffness depends heavily on how much it is loaded. We'll now try to find the design forces in an indeterminate concrete structure. Once again, we'll see that the nonlinear coupling of stiffness and load is an important consideration.

Problem Introduction

Consider two cantilever concrete columns of different heights. Both are pinned at the top to a rigid diaphragm. The diaphragm is then subjected to a shear force as shown in the sketch below. Both columns use the same 12 inch by 16 inch cross section that we have used for the previous posts in this series.

Our goal is to find the design moment in each column. Since the columns are cantilevers, this will be the moment at their base. This problem is statically indeterminate since the two columns are tied together, and we'll need to consider the stiffness of each column to determine how the total shear force is shared between them.

Linear Solution

One way to find our design moments would be to model the structure in VisualAnalysis. If we treat the columns as parametric rectangles and pick the 4 ksi concrete from the material database, we'll get the following design moments for the two columns (here is the VA project file).

As expected, the shorter column on the left is stiffer and ends up taking significantly more of the shear force, leading to a larger design moment.

At this point hopefully some alarm bells from the previous posts in this series are ringing. Sure the shorter column is stiffer at first, but we've only modeled its gross-concrete elastic stiffness. As the moment approaches 1320 in-kips, it is going to crack and the reinforcing steel will yield. Will this cause some of the load to move to the longer column?

Nonlinear Section Stiffness

The moment-stiffness plot below is derived from the section stiffness investigation we did in Part 2. I've simplified it a bit by replacing the curved yielding regions with discrete drops in stiffness, basically treating the rebar as points that yield all at once. The plot shows three distinct drops in stiffness. First, when the section cracks; second, when the bottom layer of reinforcement yields, and finally when the middle bars yield and the section is at full capacity.

The plot immediately confirms our suspicion that softening will be important to this problem. The stiffness around 1300 in-kips is a lot less than the initial stiffness of the section.

Model Development

The key to this problem is figuring out how much shear force goes into each column. From there, the moment is just that force times the length of the column. The rigid diaphragm forces both columns to displace the same amount at their tops, so this is really a one degree-of-freedom problem that can be modeled with two springs as shown below.

This model does not "look" like our structure, but it captures the behavior we are interested in, and that is what actually matters.

Section Stiffness to Spring Stiffness

Now we have to figure out how to convert the nonlinear section stiffness (M vs EI) from the plot above, to spring stiffness (F vs K). Recall the equation for a tip-loaded cantilever beam:

By definition, stiffness is:

Combining these two equations gives the stiffness we need:

Applying this transformation to our two columns leads to the force displacement curves shown below. You can download the spreadsheet I used for the calculations here.

The plots include both the cracked and uncracked behavior of the columns. Cracking is a different sort of stiffness loss. When a section yields, it will regain stiffness if you unload it. When it cracks that stiffness is gone forever. It is clear from the plots that the load levels we are interested are well past cracking. So we will consider only the cracked curves going forward.

Nonlinear Solution

With the spring force-displacement relationship defined, we are ready to solve the nonlinear problem. There are a number of ways to do this. We could set up a Newton-Raphson iteration. We could model the problem in VisualAnalysis and step the applied loads, reducing spring stiffness and applying forces to capture the yielded springs. Or we could use a work-energy principal. We'll get the same answer each way (I solved it two different ways in the spreadsheet). This blog is titled "VirtualWork" so I can't resist taking that approach.

The displacement at equilibrium will minimize the following functional (seems like magic, but it’s true):

The plot below shows Π vs u. The minimum occurs when u = 2.9 inches, our displacement when the nonlinear system is in equilibrium.

We can now find the force in each spring (Fs) when displacement is 2.9 inches from our force-displacement curves. The design moments we want are found from M = Fs · L.

Conclusion

The table below compares our first and second order results.

Accounting for the nonlinear stiffness of the section increased the design moment in the right column by 20%. A significant increase that may change how the reinforcing in that column is detailed.

I should mention here that the procedure outlined in this post does not produce the "exact" answer. We have taken the typical linear analysis one step further by considering that concrete softens as you load it and found that it changed the results in a meaningful way. We have not considered confinement, bar slip, and on and on. At the end of the day, reinforced concrete is rocks and sand glued together with steel bars. Pretending we can model that perfectly is foolish.

I do hope this series will cause you to pause and think about load level when analyzing concrete structures. Thanks again for reading.