13.3 Statically indeterminate beams

Statically Indeterminate Beams

Statically indeterminate beams have more unknown reactions than equilibrium equations can handle, so standard statics alone won't get you to a solution. These beams show up constantly in real structures (think continuous bridge spans or fixed-end beams in building frames), and solving them requires combining equilibrium with additional conditions based on how the beam actually deforms.

Identifying Statically Indeterminate Beams

A beam is statically determinate when you can solve for every unknown reaction using just the three equilibrium equations: $\Sigma F_x = 0$, $\Sigma F_y = 0$, and $\Sigma M = 0$. When the number of unknown reactions exceeds three, you've got a statically indeterminate beam.

Common configurations that produce indeterminacy:

• Propped cantilever: A fixed support on one end and a roller or pin on the other gives you four unknowns (three from the fixed support, one from the roller) but only three equilibrium equations.
• Fixed-fixed beam: Both ends are fixed, producing six reaction unknowns total.
• Continuous beams: Multi-span beams over three or more supports. Each interior support adds an unknown vertical reaction.

The degree of static indeterminacy is simply the number of unknowns minus the number of independent equilibrium equations. A propped cantilever is indeterminate to the 1st degree. A fixed-fixed beam is indeterminate to the 3rd degree. The higher the degree, the more additional equations you'll need to solve the problem.

Solving for Reactions in Indeterminate Beams

Compatibility Equations

Since equilibrium alone isn't enough, you need compatibility equations to close the gap. These equations enforce the physical requirement that the beam's deflections and rotations remain continuous and consistent with the support conditions.

For example, if a propped cantilever has a roller at one end, you know the deflection at that roller is zero. That's your compatibility condition. You can then express the deflection at that point in terms of the unknown redundant reaction, set it equal to zero, and solve.

The general approach:

1. Identify the degree of indeterminacy (how many extra unknowns you have).
2. Choose that many reactions as redundants and temporarily remove them, creating a "released" or "primary" structure that is statically determinate.
3. Write compatibility equations that enforce the known displacement or rotation at each removed support.
4. Solve the compatibility equations (using superposition of deflections from the applied loads and the redundant reactions) to find the redundants.
5. Back-substitute into equilibrium equations to find all remaining reactions.

Slope-Deflection Method

The slope-deflection method writes the end moments of each beam segment in terms of the unknown joint rotations and displacements. The general slope-deflection equation for a member spanning from node A to node B is:

$M_{AB} = \frac{2EI}{L}(2\theta_A + \theta_B - 3\psi) + FEM_{AB}$

where $\theta_A$ and $\theta_B$ are the rotations at ends A and B, $\psi$ is the chord rotation (relative settlement divided by span length), $L$ is the span, $EI$ is the flexural rigidity, and $FEM_{AB}$ is the fixed-end moment due to the applied loads.

Steps to apply the method:

1. Calculate fixed-end moments (FEMs) for each loaded span, treating every joint as fixed.
2. Write slope-deflection equations for both ends of every span.
3. Apply equilibrium at each joint: the sum of moments from all members meeting at a joint equals zero.
4. Apply boundary conditions (e.g., $\theta = 0$ at a fixed support, $M = 0$ at a pin or roller).
5. Solve the resulting system of equations for the unknown rotations and displacements.
6. Substitute back into the slope-deflection equations to get the final end moments.

This method works well for continuous beams and frames, especially when you're comfortable solving simultaneous equations.

Moment-Distribution Method

The moment-distribution method is an iterative technique developed by Hardy Cross that avoids solving simultaneous equations directly. It's particularly efficient for hand calculations.

Key concepts:

• Stiffness factor: For a prismatic member of length $L$ and flexural rigidity $EI$, the stiffness is $\frac{4EI}{L}$ if the far end is fixed, or $\frac{3EI}{L}$ if the far end is pinned.
• Distribution factor (DF): At any joint, each member's DF equals its stiffness divided by the total stiffness of all members at that joint. The DFs at a joint always sum to 1.
• Carry-over factor: When a moment is distributed to one end of a member, half of that moment carries over to the far end (for a fixed far end).

Steps:

1. Calculate fixed-end moments for all loaded spans.
2. Compute distribution factors at each joint.
3. Identify the unbalanced moment at each joint (the algebraic sum of FEMs).
4. Distribute the unbalanced moment to each member at the joint according to the DFs (with opposite sign to balance).
5. Carry over half of each distributed moment to the far end of each member.
6. Repeat steps 3–5 until the carry-over moments become negligibly small.
7. Sum all moments at each member end to get the final end moments.

Convergence is usually fast, often requiring only 3–4 cycles for reasonable accuracy.

Deflection and Slope of Indeterminate Beams

Once you've found the reactions and internal moments, you still need to determine deflections and slopes. Several methods work for indeterminate beams.

Moment-Area Method

This method uses two theorems that relate the $M/EI$ diagram to slopes and deflections:

• First theorem: The change in slope between two points A and B equals the area under the $M/EI$ diagram between those points.

$\theta_B - \theta_A = \int_A^B \frac{M}{EI}\, dx$

• Second theorem: The vertical distance of point B from the tangent drawn at point A equals the first moment of the $M/EI$ diagram area (between A and B) taken about point B.

$t_{B/A} = \int_A^B \frac{M}{EI} \cdot \bar{x}_B \, dx$

where $\bar{x}_B$ is the horizontal distance from each differential area element to point B.

To use this method, you sketch the $M/EI$ diagram (often breaking it into simple shapes like triangles and parabolas), compute areas and centroids, and apply the theorems. It works best when the $M/EI$ diagram has a clean geometric shape.

Conjugate-Beam Method

The conjugate-beam method converts the deflection problem into a familiar shear-and-moment problem on an imaginary beam.

The rules:

• Load the conjugate beam with the $M/EI$ diagram of the original beam as a distributed load.
• The shear in the conjugate beam at any point equals the slope of the original beam at that point.
• The moment in the conjugate beam at any point equals the deflection of the original beam at that point.

The support conditions of the conjugate beam are transformed from the original:

This transformation ensures the math stays consistent. Once you set up the conjugate beam, you just analyze it like any other loaded beam using equilibrium.

Direct Integration Method

This is the most fundamental approach. You start from the moment-curvature relationship:

$EI \frac{d^2y}{dx^2} = M(x)$

1. Express $M(x)$ as a function of position along the beam (you'll need the reactions first).
2. Integrate once to get the slope: $EI \frac{dy}{dx} = \int M(x)\, dx + C_1$
3. Integrate again to get the deflection: $EI \cdot y = \iint M(x)\, dx\, dx + C_1 x + C_2$
4. Apply boundary conditions (known deflections and slopes at supports) to solve for the constants $C_1$ and $C_2$.
5. For indeterminate beams with multiple segments, enforce continuity of slope and deflection at the junction between segments.

Direct integration is straightforward when the moment equation is simple, but it gets tedious for beams with multiple load regions.

Principle of Virtual Work

Virtual work (also called the unit-load method) lets you find the deflection at any specific point on the beam.

1. Remove all real loads and apply a virtual unit load ($P = 1$) at the point and in the direction of the desired deflection.
2. Solve for the internal virtual bending moment $m(x)$ throughout the beam due to this unit load.
3. The real deflection $\Delta$ at that point is:

$\Delta = \int_0^L \frac{M(x) \cdot m(x)}{EI}\, dx$

where $M(x)$ is the real bending moment from the actual loads.

For slope at a point, apply a virtual unit moment instead of a unit force. This method handles complex loading and variable cross-sections well, making it one of the most versatile tools for indeterminate beam analysis.

Support Settlements in Indeterminate Beams

Effects of Support Settlements

In a determinate beam, if a support settles (moves downward), the beam simply follows it without developing any new internal forces. Indeterminate beams behave very differently. Because the beam is constrained at multiple points, a settlement at one support forces the beam to deform in a way that generates additional bending moments, shear forces, and reaction changes throughout the structure.

The severity of these induced forces depends on:

• The magnitude of the settlement
• The location of the settling support relative to other supports
• The beam's flexural rigidity $EI$ (stiffer beams develop larger forces from the same settlement)
• The span lengths and overall configuration

Analysis Techniques for Support Settlements

The same methods used for loading analysis can be adapted for settlements. The key difference is that the compatibility equations now include known nonzero displacements instead of zeros.

Using the slope-deflection method: The chord rotation term $\psi = \Delta / L$ (where $\Delta$ is the relative settlement between two adjacent supports) gets plugged directly into the slope-deflection equations. The rest of the procedure stays the same.

Using the moment-distribution method: Settlements modify the fixed-end moments. You calculate the FEMs caused by the settlement (using $FEM = \pm \frac{6EI\Delta}{L^2}$ for a prismatic member with relative settlement $\Delta$), then distribute and carry over as usual.

Using superposition: You can analyze the beam under applied loads (ignoring settlements) and under settlements alone (ignoring applied loads), then add the results together. This separation often simplifies the work considerably.

Using the conjugate-beam method: The settlements appear as additional boundary conditions on the conjugate beam, modifying the reactions and therefore the slope and deflection results.

Accounting for support settlements is critical in practice because differential settlement between supports is one of the most common causes of distress in continuous beam and frame structures.