CELE Structural Theory & Analysis — Indeterminate Structures: Displacement MethodsStudy Notes

Displacement methods solve indeterminate structures by treating unknown joint displacements and rotations as the primary unknowns, then using member stiffness relationships to express member-end moments in terms of these displacements. The fundamental premise is that at any joint in a rigidly connected frame, the sum of all member-end moments must equal zero (joint equilibrium). This contrasts with force methods, where redundant reactions are the unknowns.

The key advantage for engineers: displacement methods scale naturally to large structures. Once you understand the stiffness relationships for one member, you apply them to all members. This linear scaling made displacement methods the basis for finite element analysis (FEA) and all modern structural software used in Philippine practice.

In practical terms, when you design a multi-story building to NSCP 2015 lateral load requirements, your analysis software uses displacement methods to distribute forces to beams and columns proportionally to their stiffness. Understanding the underlying theory helps you recognize when a software result seems unreasonable and debug the model.

The two primary displacement methods for hand calculation are:

1. Slope-deflection method—direct, systematic, writes equations for every member end, solves simultaneously.
2. Moment distribution method—iterative, practical for hand calculation, converges to the same answer through successive balancing and carry-over operations.

Both methods require the same starting point: fixed-end moments (FEM) tables.

Fixed-end moments are the moments that develop at the ends of a member when both ends are held completely fixed (zero rotation and zero displacement at both ends) while subject to applied loads. These are reference moments—they represent the internal moment distribution within the member under fully restrained conditions. When you release the restraints (allow rotation), the actual member-end moments differ from these FEM values by an amount that depends on the member stiffness and the end rotations.

Sign Convention: In Philippine practice and most textbooks using the slope-deflection method, the convention is clockwise-moment-on-the-member is positive. This means a moment vector pointing into the page (right-hand rule, thumb into page) is positive. When you draw a free body diagram of a member, a positive moment at the left end rotates the member clockwise; a positive moment at the right end also rotates clockwise.

Memorizing the FEM table is non-negotiable for board examination success. You will not have time to derive these during the exam. Recognizing load patterns (UDL, point load, triangular, etc.) and applying the correct formula is a baseline skill.

For a member span of length L:

Uniformly Distributed Load (UDL) of intensity w over full span:
FEM_AB = −wL²/12 (left end, negative)
FEM_BA = +wL²/12 (right end, positive)
This is the most common case in continuous beams.

Central Point Load P at midspan:
FEM_AB = −PL/8
FEM_BA = +PL/8
Use this for symmetric loading; for unsymmetric loads use the general point load formula.

Point Load P at distance 'a' from left end A, distance 'b' from right end B (where a + b = L):
FEM_AB = −Pab²/L² (left end)
FEM_BA = +Pa²b/L² (right end)
This is the general formula; substitute a = b = L/2 to recover the midspan case.

Triangular Load: zero at one end, maximum intensity w at the other end:
FEM (zero-end) = −wL²/30 (moment at the end where load is zero)
FEM (peak-end) = +wL²/20 (moment at the end where load is maximum)
This case appears less frequently but is worth remembering for complex loading scenarios.

Practical Note for Philippine Practice: When analyzing bridges, building frames, or floor systems under gravity loads, the UDL formula is used most frequently. When analyzing industrial structures with concentrated equipment loads, the point-load formula dominates. Triangular loads appear in soil pressure diagrams and distributed loads that taper along the span.

The slope-deflection method is the most fundamental displacement method because it directly relates member-end moments to end rotations through linear equations. Once you write the equation for every member end, you impose joint equilibrium conditions and solve the system—no iteration required.

The fundamental slope-deflection equation expresses the moment at end 1 of a member (connecting joints 1 and 2) as:

M₁₂ = (2EI/L)[2θ₁ + θ₂ − 3ψ] + FEM₁₂

where:

• M₁₂ = moment at end 1 acting on the member (clockwise-positive)
• E = Young's modulus (usually cancels in relative stiffness ratios)
• I = second moment of inertia of the member
• L = span length
• θ₁ = rotation at joint 1 (clockwise-positive)
• θ₂ = rotation at joint 2 (clockwise-positive)
• ψ = chord rotation = Δ/L, where Δ is the relative horizontal displacement of joint 2 minus joint 1 (positive when joint 2 moves down or right relative to joint 1, meaning the member chord rotates clockwise)
• FEM₁₂ = fixed-end moment at end 1 from applied loads

Breaking down the equation:

• The term (2EI/L) represents member stiffness; larger EI or shorter spans create stiffer members
• The coefficients 2 and 1 on θ₁ and θ₂ reflect that the near-end rotation has twice the effect on the near-end moment compared to the far-end rotation
• The term −3ψ captures the effect of chord rotation (support settlement, frame sidesway): a clockwise chord rotation decreases the moments at both ends (it relieves the moment demands)
• FEM₁₂ is added as the base moment from loads under fixed conditions

For the opposite end (moment at end 2):

M₂₁ = (2EI/L)[2θ₂ + θ₁ − 3ψ] + FEM₂₁

Note that the moment at end 2 also depends on ψ with the same coefficient. This ensures carry-over consistency.

Step-by-step procedure:

1. Identify all unknown rotations. For a continuous beam on n supports, there are (n−2) unknown rotations at interior supports. For a rigid frame, count the number of rigid joints not at fixed supports.
2. Compute fixed-end moments using the FEM table for each member and load combination.
3. Write slope-deflection equations for every member end. If a member connects joints i and j, write equations M_ij and M_ji.

4. Apply boundary conditions:

   • At a fixed support: rotation θ = 0
   • At a simple (pinned) support: moment M = 0 (write the equation and set it to zero to solve for the required rotation)
   • At a free (cantilever) end: moment M = 0
   • For no support settlement and no frame sidesway: ψ = 0
5. Apply joint equilibrium. At each rigid joint, sum all member-end moments on that joint and set equal to zero. For example, at an interior joint where three members meet: M₁₂ + M₃₂ + M₄₂ = 0.
6. Solve the system of linear equations simultaneously for all unknown rotations.
7. Back-substitute into the original equations to find all member-end moments.
8. Use member-end moments and shear equilibrium to find member shear forces and reactions.

Practical Advantages:

• Systematic and straightforward once you set up the equations
• No iteration—you get the exact answer in one pass
• Naturally handles complex loading and boundary conditions
• Provides a clear audit trail for checking your work

Practical Disadvantages:

• For hand calculation, the number of equations grows quickly; even a four-span continuous beam requires solving 2×2 or 3×3 systems
• Matrix operations by hand are tedious and error-prone
• Moment distribution is usually faster for hand problems with fewer unknowns

Moment distribution is an iterative procedure that converges to the same final answer as slope-deflection but is typically much faster for hand calculation on beams and simple frames. The method is based on the physical principle that at each joint, unbalanced moments are distributed to adjacent members in proportion to their stiffness.

Key Concepts:

Member Stiffness (k): The stiffness of a member for moment distribution purposes is the relative ability of the member to resist rotation at a joint. The stiffness factor is:

k = 4EI/L for a member with both ends fixed or both ends continuous (interior member)

k = 3EI/L for a member with one end pinned and the other fixed (or cantilevered at one end). The pinned/free end does not carry moment, so that end is effectively removed from the stiffness calculation.

At a joint where multiple members meet, the total stiffness is Σk (sum of all member stiffnesses meeting at that joint).

Distribution Factor (DF): At a joint, the distribution factor for each member is the ratio of that member's stiffness to the total stiffness at the joint:

DF_i = k_i / Σk

The sum of all distribution factors at a joint equals 1.0. Distribution factors represent the fraction of an unbalanced moment that is distributed to each member.

Carry-Over Factor: When you apply a balancing moment to a member at one end (say, joint B), that moment causes the member to rotate at that end, which in turn induces a moment at the opposite end (the far end). For a member with both ends initially in the same state (both continuous or both fixed), the carry-over factor is 1/2. Thus:

Carry-over moment at far end = (1/2) × (moment distributed at near end)

For a member with one end pinned, no carry-over occurs to the pinned end (it cannot develop a moment), so you set that stiffness to 3EI/L and assign it zero distribution factor or ignore it in the distribution process.

Step-by-step Moment Distribution Procedure:

1. Compute Fixed-End Moments (FEM)
   Write FEM values for every member under the applied loads, using the FEM table. Include the sign: negative for the near end relative to the load, positive for the far end.

2. Determine Stiffness Factors (k)
   For each member, decide if both ends are continuous (k = 4EI/L) or one end is pinned/free (k = 3EI/L). Typically, in a continuous beam, all interior members have k = 4EI/L, and if the structure is part of a frame, check the boundary conditions.

   For ease of computation, you often assume EI is constant and work with relative stiffness values. For example, if all members have the same I and E, you can use k = 4/L or k = 3/L without the EI term. Alternatively, if spans are unequal but EI is constant, express k as 4/L. The key is that relative stiffness matters, not absolute values.

3. Compute Distribution Factors (DF)
   At each joint, sum the stiffness factors of all members meeting at that joint (DF_i = k_i / Σk). Express as decimals (e.g., 0.5 for equal members) or fractions. Note: A fixed support or a free end has DF = 0 (zero unbalanced moment is distributed there because the rotation is already zero or the moment must be zero).
4. Cycle Through Distribution and Carry-Over
   Iteration cycle:
   a) Identify the joint with the largest unbalanced moment.
   b) Distribute the negative of the unbalanced moment to each member at that joint, using the distribution factors. That is, if the unbalanced moment is M_unbal = FEM_1 + FEM_2 + ... (sum of all FEM values at the joint from all members), then distribute −M_unbal in proportion to DF: ΔM_i = −DF_i × M_unbal.
   c) Carry over half of each distributed moment to the far end of the corresponding member.
   d) Repeat for the next joint with an unbalanced moment (usually the adjacent joint, since you just added carry-overs there).
   e) Continue cycling through joints until the unbalanced moments at each joint are negligible (typically when they drop below 0.1 kN·m or about 0.1% of the largest FEM).

5. Tabulation
   Create a table with:

   • One column for each member-end
   • Rows for: FEM, first distribution, first carry-over, second distribution, second carry-over, etc.
   • At the bottom, sum each column to get the final moment.
6. Sign Convention
   Moments are recorded in the table with their sign (negative or positive). When you distribute, you compute −(unbalanced moment) and distribute that using DF. The result might be positive or negative depending on the sign of the unbalanced moment.

Practical Notes:

• Use a spreadsheet or table format for clarity; each member-end is a column.
• For a two-span beam, you typically need only 2–3 cycles of distribution before convergence.
• For a three-span beam, expect 3–4 cycles.
• For frames, if there is sidesway, the method becomes more complex (you must account for chord rotation, which requires a separate analysis). Most hand problems do not include sidesway, so assume ψ = 0.
• When a member has a pinned far end, treat its stiffness as 3EI/L, and do not carry over a moment to that end. Instead, recognize that the pinned end develops zero moment, and the distributed moment at the pinned end is consumed entirely in creating a reaction moment at that support.

Comparison with Slope-Deflection:

Both methods arrive at identical final member-end moments. Slope-deflection is direct but requires solving a system of equations. Moment distribution is iterative but each iteration is simple (multiply by a distribution factor, divide by 2). For a two-span beam by hand, moment distribution is faster. For a complex frame with multiple unknown rotations, both methods are tedious, and you would use software.

Sign Convention Consistency: Throughout both slope-deflection and moment distribution, you must maintain a consistent sign convention. The convention used in this chapter (and most Philippine civil engineering textbooks aligned with NSCP 2015) is:

• Moments are positive when they act clockwise on the member (using the right-hand rule: thumb into the page).
• At a joint, the moment exerted by a member on the joint is the negative of the moment exerted on the member (Newton's third law).
• When writing equilibrium equations at a joint, sum moments exerted on the joint (not on the members).

For a horizontal beam oriented left-to-right, a positive (clockwise) moment at the left end causes the left end to rotate clockwise, which curves the beam in a manner that puts the bottom fiber in tension and the top fiber in compression (sagging). A positive moment at the right end also causes clockwise rotation at that end, consistent with a sagging moment diagram.

Application to Continuous Beams:

Continuous beams are the most common application in building floors (every floor system over supports is a continuous beam) and bridge decks. The analysis steps are:

1. Identify all load cases (dead load, live load, wind, seismic per NSCP 2015).
2. For each load case, compute FEM for each span.
3. Use moment distribution or slope-deflection to find moments at each support.
4. From moments and shear equilibrium, find reactions.
5. Check that the sum of vertical reactions equals the total applied load.
6. Design each section based on the governing moment and shear envelope.

Application to Portal Frames:

A portal frame (typically a single-story rectangular frame) under lateral load (wind or seismic per NSCP 2015) has unknown joint rotations and possibly sidesway (chord rotation ψ ≠ 0). The analysis steps are:

1. Compute FEM for members under applied loads (e.g., UDL on the beam due to gravity, or assume the wind load is applied at the joint for simplified analysis).
2. Write slope-deflection equations including ψ for members that experience chord rotation.
3. Impose equilibrium at each joint and the sidesway constraint (e.g., if the frame is braced, ψ = 0; if free to sway, apply the shear equilibrium condition to solve for ψ).
4. Solve for all unknowns (rotations and possibly ψ).
5. Back-substitute to find moments and then reactions.

Propped Cantilevers:

A propped cantilever (fixed at one end, supported by a pin at the other end, cantilever extends beyond the pin) has a single unknown: the rotation at the pinned support. This is one of the simplest indeterminate structures and is often used in examination problems to test understanding of the method without the complexity of multiple unknowns.

Sections with Variable Cross-Sections:

If a member has a variable I (e.g., a tapered beam in a high-rise), the stiffness varies along the length. Slope-deflection formulas must be modified, or you use integration methods. Most hand problems assume constant I within each member, so we do not detail this here, but it is important to recognize the limitation when analyzing real structures.

Effect of Support Settlement:

If a support settles by an amount Δ (downward), the chord rotation ψ = Δ/L affects the member-end moments. In slope-deflection, you include −3ψ(2EI/L) in each end-moment equation. In moment distribution, you must account for settlement through an auxiliary analysis or by adjusting the FEM values (complex for hand calculation). NSCP 2015 and ACI 318 do not explicitly address support settlement in their standard design procedures, but it can be important for structures on soft soil or with differential foundation movement. If encountered in an examination problem, use slope-deflection to set up the equations and include the ψ term.

Convergence of Moment Distribution:

The moment distribution method converges to the exact answer, and the rate of convergence depends on the structure. For a two-span continuous beam, typically 2–3 cycles are sufficient (unbalanced moments drop to near-zero). For a three-span beam, 3–4 cycles. For a frame with multiple joints, you may need 4–5 cycles. The convergence is faster when the stiffnesses of adjacent members are similar and slower when they differ greatly (e.g., a very stiff central beam between flexible corner beams).

In practice, you can stop distributing when the unbalanced moments at each joint are less than about 0.1 kN·m (for engineering structures) or 0.01 kN·m (for high precision). On an examination, aim for accuracy to about 1 kN·m; achieving this usually requires 3–4 cycles.

Accuracy Considerations:

1. Rounding: Round intermediate values to 0.1 kN·m to avoid accumulating rounding errors, but keep one extra digit until the final answer.
2. FEM Computation: Ensure you compute FEM correctly using the table formula. A sign error here propagates through all subsequent calculations.
3. Stiffness Factors: If members have equal EI and equal L, you can set k = 1 for all and work with relative stiffness. Avoid computing 4EI/L with large numbers; instead, use ratios.
4. Carry-Over Arithmetic: When carrying over half a moment, round to a consistent number of decimal places. For example, if you distribute +18 kN·m, the carry-over is 9 kN·m; if you distribute −18.25 kN·m, the carry-over is −9.125 kN·m, which you can round to −9.1 or keep as is depending on the required precision.
5. Tabulation: Organize your moment distribution table clearly. Rows represent cycles; columns represent member-ends. Sum each column at the end. A mistake in tabulation (e.g., forgetting to carry over to a particular end) is easy to make and hard to spot, so double-check your table.

Comparison: Slope-Deflection vs. Moment Distribution

Slope-Deflection:

• Pros: Direct and systematic; you solve the problem once, no iteration; easier to check intermediate steps; naturally handles support settlement and sidesway.
• Cons: Requires setting up and solving a system of linear equations; for hand calculation, solving a 3×3 or larger system is tedious and error-prone.
• When to use: For structures with few unknowns (one or two interior support rotations); for problems with unusual boundary conditions (support settlement, prescribed rotation); for verifying a moment distribution result.

Moment Distribution:

• Pros: Iterative and intuitive; each step is simple (multiply by DF, add, divide by 2); easier to spot errors in individual cycles; converges quickly for most practical beams.
• Cons: Iteration can be tedious if convergence is slow; sidesway requires auxiliary analysis (not easily done by hand); it's easy to forget a carry-over or make a tabulation error.
• When to use: For continuous beams with multiple spans; for problems without support settlement or sidesway; for practicing the procedure quickly.

Examination Strategy:

1. Read the problem carefully. Identify the structure type (continuous beam, frame, cantilever), boundary conditions, and loads.
2. Check if displacement method is the intended approach. (If the problem asks for "moments at all supports" or "moments at each section," displacement methods are indicated. If it asks for a redundant reaction, force method may be faster.)
3. Compute FEM. Use the table; double-check signs.
4. For a two-span beam or simple frame: Use moment distribution if no sidesway or settlement; use slope-deflection if sidesway or settlement is present.
5. For a single span with one or two interior supports (e.g., propped cantilever): Slope-deflection is often faster because there's only one unknown.
6. Organize your work. Use a clear table for moment distribution; write slope-deflection equations in order (one per member-end).
7. Check your final answer. Verify that reactions sum correctly and that the moment diagram is reasonable (sagging where you expect it, zero at free ends, etc.).
8. If time permits, re-check one critical calculation (e.g., the FEM formula or the largest moment value).

Common Mistakes on Examinations:

1. Sign Error in FEM: Forgetting that FEM_near-end is negative and FEM_far-end is positive (or vice versa depending on convention).
   Catch: Always verify the sign using physical reasoning. A UDL on a fixed beam causes the near end to rotate toward the load, so the moment there should be negative (counter to the load's rotational effect).
2. Distribution Factor Arithmetic: Computing DF incorrectly, e.g., summing the wrong stiffnesses or forgetting to normalize.
   Catch: Check that all DF values at a joint sum to 1.0.
3. Forgetting Carry-Over: In moment distribution, carrying over to some member-ends but not others, or forgetting the 1/2 factor.
   Catch: Use a consistent table format; every distributed moment should have a corresponding carry-over.
4. Sign Error in Joint Equilibrium: Distributing the unbalanced moment instead of its negative, or summing moments incorrectly at the joint.
   Catch: The unbalanced moment is the sum of all moments at a joint (including FEM and carry-overs); you distribute the negative of this sum to balance the joint.
5. Mishandling Boundary Conditions: Treating a pinned end as fixed, or vice versa; forgetting that a free end has M = 0.
   Catch: Re-read the problem statement and clearly identify each support type on your diagram.
6. Inconsistent Sign Convention: Switching between clockwise-positive and counter-clockwise-positive partway through.
   Catch: Write your sign convention clearly at the top of your work and stick to it throughout.