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13.1 Indeterminate Structures: Approximate and Classical Methods
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13.1 Indeterminate Structures: Approximate and Classical Methods 13.2 Moment Distribution Method and Slope Deflection 13.3 Influence Lines for Determinate and Indeterminate Structures 13.4 NSCP Minimum Design Loads: Dead, Live, Wind, and Earthquake
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13.1 Indeterminate Structures: Approximate and Classical Methods
Key Takeaways Explain This Topic AI
The degree of static indeterminacy (DSI) defines the number of redundant forces required to solve a system, calculated as DSI = 3m + r - (3j + c) for 2D rigid frames and DSI = b + r - 2j for 2D pin-jointed trusses.
The Portal Method models low-rise frames under lateral loads by placing inflection points at column mid-heights and beam mid-spans, distributing story shear such that interior columns carry twice the shear of exterior columns.
The Cantilever Method models tall, slender frames under lateral loads by assuming column axial stresses vary linearly with distance from the column group centroid, reflecting global flexural cantilever action.
The Method of Consistent Deformations (Force Method) releases redundants to form a stable determinate primary structure, solving compatibility equations where flexibility coefficients satisfy Maxwell-Betti reciprocity (δ_ij = δ_ji).
Castigliano's Second Theorem calculates structural deflections and redundant reactions by taking the partial derivative of internal strain energy with respect to the applied load or redundant force (∂U/∂R_i = Δ_i).
Last updated: October 2026
13.1 Indeterminate Structures: Approximate and Classical Methods
Statically indeterminate structures form the backbone of modern civil engineering infrastructure. Unlike determinate structures—where internal shears, axial forces, and bending moments are obtained purely from equations of static equilibrium—indeterminate structures require consideration of material deformations and geometric compatibility. In Philippine civil engineering practice and licensure examinations, structural engineers must readily determine the degree of static indeterminacy ( D S I DSI D S I ) , execute rapid approximate analyses for preliminary sizing, and apply rigorous classical force methods for exact solutions.
1. Degree of Static Indeterminacy ( D S I DSI D S I )
A structure is statically determinate when the number of independent equilibrium equations equals the total number of unknown reaction and internal force components. When unknowns exceed equilibrium equations, the structure is statically indeterminate ( D S I > 0 DSI > 0 D S I > 0 ). If unknowns are fewer than available equations, or if constraints are configured parallel or concurrent, the structure is geometrically unstable ( D S I < 0 DSI < 0 D S I < 0 ).
Planar Rigid Frames and Continuous Beams
For two-dimensional planar frames consisting of rigid joints, each member carries three internal actions (axial force, shear force, bending moment), and each joint provides three equations of static equilibrium ( ∑ F x = 0 \sum F_x = 0 ∑ F x = 0 , ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 , ∑ M = 0 \sum M = 0 ∑ M = 0 ).
D S I = 3 m + r − ( 3 j + c ) DSI = 3m + r - (3j + c) D S I = 3 m + r − ( 3 j + c )
Where:
m m m = total number of structural members (beam and column segments between joints).
r r r = total number of independent support reaction components (roller = 1 = 1 = 1 , pinned hinge = 2 = 2 = 2 , fixed = 3 = 3 = 3 ).
j j j = total number of joints, including support foundations and beam-column intersections.
c c c = total number of internal condition equations (releases). An internal hinge connecting k k k members introduces c = k − 1 c = k - 1 c = k − 1 moment releases ( M = 0 M = 0 M = 0 ). An internal shear slide release introduces c = 1 c = 1 c = 1 shear equation ( V = 0 V = 0 V = 0 ).
An alternative formulation based on closed loops is:
D S I = 3 ( loops ) − c − r rel DSI = 3(\text{loops}) - c - r_{\text{rel}} D S I = 3 ( loops ) − c − r rel
where each fully closed rigid cell is indeterminate to the 3rd degree.
Planar Pin-Jointed Trusses
In a planar pin-connected truss, each member carries only one unknown axial force ( b b b or m m m ), each support provides r r r reaction components, and each pin joint yields two equilibrium equations ( ∑ F x = 0 \sum F_x = 0 ∑ F x = 0 , ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 ):
D S I = b + r − 2 j DSI = b + r - 2j D S I = b + r − 2 j
External Indeterminacy ( D S I e x t DSI_{ext} D S I e x t ): D S I e x t = r − 3 DSI_{ext} = r - 3 D S I e x t = r − 3 . If r > 3 r > 3 r > 3 , support reactions cannot be found from overall statics alone.
Internal Indeterminacy ( D S I i n t DSI_{int} D S I in t ): D S I i n t = b − ( 2 j − 3 ) DSI_{int} = b - (2j - 3) D S I in t = b − ( 2 j − 3 ) . If extra diagonal bracing bars exist beyond the minimum required for a stable triangulation network, the truss is internally redundant.
Total static indeterminacy is the sum: D S I = D S I e x t + D S I i n t DSI = DSI_{ext} + DSI_{int} D S I = D S I e x t + D S I in t .
Structure Type Determinacy Formula Determinacy Condition Instability Condition
Planar Truss D S I = b + r − 2 j DSI = b + r - 2j D S I = b + r − 2 j D S I = 0 DSI = 0 D S I = 0 (and stable) b + r < 2 j b + r < 2j b + r < 2 j or geometric mechanism
Planar Frame D S I = 3 m + r − 3 j − c DSI = 3m + r - 3j - c D S I = 3 m + r − 3 j − c D S I = 0 DSI = 0 D S I = 0 (and stable) 3 m + r < 3 j + c 3m + r < 3j + c 3 m + r < 3 j + c or unstable arrangement
Space Frame (3D) D S I = 6 m + r − 6 j − c DSI = 6m + r - 6j - c D S I = 6 m + r − 6 j − c D S I = 0 DSI = 0 D S I = 0 (and stable) 6 m + r < 6 j + c 6m + r < 6j + c 6 m + r < 6 j + c
2. Approximate Methods for Building Frames under Lateral Loads
Before finite-element software is deployed, approximate manual methods provide an immediate check on computer output and serve as primary questions in board examinations. Lateral loads (wind or earthquake base shear) induce horizontal racking on building frames. To render an n n n -bay, m m m -story indeterminate frame statically determinate, engineers introduce physical assumptions regarding internal inflection points (points of contraflexure where bending moment M = 0 M = 0 M = 0 ) and column shear or axial distributions.
The Portal Method
The Portal Method was developed by Albert Smith in 1915. It is best suited for low-rise, broad building frames (height-to-width ratio H / B < 4 H/B < 4 H / B < 4 ) where shear deformation ("racking") dominates over global overturning flexure.
Governing Assumptions:
Inflection Points in Columns: An inflection point ( M = 0 M = 0 M = 0 ) occurs at the mid-height of each column ( h / 2 h/2 h /2 ) in every story.
Inflection Points in Girders: An inflection point ( M = 0 M = 0 M = 0 ) occurs at the mid-span of each girder ( L / 2 L/2 L /2 ) in every bay.
Shear Distribution Among Columns: The total horizontal story shear V story V_{\text{story}} V story at any level is distributed among the columns such that each interior column carries twice the shear of each exterior column (for frames with equal bay widths):
V ext = V story 2 n , V int = 2 V ext = V story n V_{\text{ext}} = \frac{V_{\text{story}}}{2n}, \quad V_{\text{int}} = 2 V_{\text{ext}} = \frac{V_{\text{story}}}{n} V ext = 2 n V story , V int = 2 V ext = n V story
where n n n is the total number of bays across the story. If bay widths are unequal, column shears are assigned proportionally to their tributary span widths.
Step-by-Step Procedure:
Column Shears: Compute V ext V_{\text{ext}} V ext and V int V_{\text{int}} V int at each story level using the shear distribution assumption.
Column Moments: Column end moments at the joint and base are determined directly from the distance to the inflection point: M col = V col × ( h / 2 ) M_{\text{col}} = V_{\text{col}} \times (h/2) M col = V col × ( h /2 ) .
Girder Moments: Isolate each beam-column joint. Enforce moment equilibrium ( ∑ M = 0 \sum M = 0 ∑ M = 0 ) to calculate girder end moments. At an exterior joint with one column above and below: M girder = M col,top + M col,bot M_{\text{girder}} = M_{\text{col,top}} + M_{\text{col,bot}} M girder = M col,top + M col,bot .
Girder Shears: With the girder inflection point at mid-span ( L / 2 L/2 L /2 ), girder shear is: V girder = 2 M girder L V_{\text{girder}} = \frac{2 M_{\text{girder}}}{L} V girder = L 2 M girder .
Column Axial Forces: Sum vertical forces ( ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 ) at each joint from the roof downward. Exterior columns resist the full girder shear as axial tension (windward side) or compression (leeward side). At interior columns, opposing girder shears cancel each other out when bays and spans are identical, resulting in zero net axial load from lateral forces.
The Cantilever Method
The Cantilever Method was introduced by A. C. Wilson in 1908. It is specifically formulated for tall, slender multi-story frames ( H / B ≥ 4 H/B \ge 4 H / B ≥ 4 ) where the building deflects as a vertical flexural cantilever beam and overall overturning moment dominates.
Governing Assumptions:
Inflection Points in Columns: Hinges occur at the mid-height of all columns.
Inflection Points in Girders: Hinges occur at the mid-span of all girders.
Column Axial Stress Distribution: The direct axial stress ( σ i \sigma_i σ i ) in each column is directly proportional to its horizontal distance ( d i d_i d i ) from the centroid of the cross-sectional areas of all columns in that story:
x ˉ = ∑ A i x i ∑ A i , σ i = C ⋅ d i , P i = A i σ i = A i C d i \bar{x} = \frac{\sum A_i x_i}{\sum A_i}, \quad \sigma_i = C \cdot d_i, \quad P_i = A_i \sigma_i = A_i C d_i x ˉ = ∑ A i ∑ A i x i , σ i = C ⋅ d i , P i = A i σ i = A i C d i
For columns with equal cross-sectional areas ( A i = A A_i = A A i = A ):
P i = C ′ ⋅ d i P_i = C' \cdot d_i P i = C ′ ⋅ d i
Taking the moment of the column axial forces about the column group centroid at the level of the column hinges, and equating it to the total overturning moment M overturning M_{\text{overturning}} M overturning caused by lateral loads above that level:
M overturning = ∑ i = 1 k P i d i = C ′ ∑ i = 1 k d i 2 ⟹ C ′ = M overturning ∑ d i 2 M_{\text{overturning}} = \sum_{i=1}^k P_i d_i = C' \sum_{i=1}^k d_i^2 \implies C' = \frac{M_{\text{overturning}}}{\sum d_i^2} M overturning = ∑ i = 1 k P i d i = C ′ ∑ i = 1 k d i 2 ⟹ C ′ = ∑ d i 2 M overturning
Once column axial forces P i P_i P i are determined:
Girder shears are calculated from joint vertical equilibrium ( ∑ F y = 0 \sum F_y = 0 ∑ F y = 0 ).
Girder end moments are obtained from girder shear: M girder = V girder × ( L / 2 ) M_{\text{girder}} = V_{\text{girder}} \times (L/2) M girder = V girder × ( L /2 ) .
Column shears and column moments follow from joint moment equilibrium ( ∑ M = 0 \sum M = 0 ∑ M = 0 ).
3. Approximate Gravity Load Analysis of Continuous Frames
Under uniformly distributed vertical gravity loads ( w w w ), continuous girders develop negative moments at the support faces and positive moments near mid-span. Because rigid joints rotate very little under balanced gravity spans:
Inflection points in girders are assumed at 0.10 L 0.10 L 0.10 L from each column face/joint support.
The central girder segment between inflection points has an effective span of 0.80 L 0.80 L 0.80 L and behaves as a simply supported beam:
M midspan + ≈ w ( 0.80 L ) 2 8 = 0.08 w L 2 = w L 2 12.5 M^+_{\text{midspan}} \approx \frac{w (0.80 L)^2}{8} = 0.08 w L^2 = \frac{w L^2}{12.5} M midspan + ≈ 8 w ( 0.80 L ) 2 = 0.08 w L 2 = 12.5 w L 2
The vertical shear transferred from the central span to each cantilever end is V = w ( 0.80 L ) 2 = 0.40 w L V = \frac{w (0.80 L)}{2} = 0.40 w L V = 2 w ( 0.80 L ) = 0.40 w L .
The negative moment at the face of the column is computed from the 0.10 L 0.10 L 0.10 L overhang:
M support − ≈ w ( 0.10 L ) 2 2 + ( 0.40 w L ) ( 0.10 L ) = 0.005 w L 2 + 0.040 w L 2 = 0.045 w L 2 ≈ w L 2 22.2 M^-_{\text{support}} \approx \frac{w (0.10 L)^2}{2} + (0.40 w L)(0.10 L) = 0.005 w L^2 + 0.040 w L^2 = 0.045 w L^2 \approx \frac{w L^2}{22.2} M support − ≈ 2 w ( 0.10 L ) 2 + ( 0.40 w L ) ( 0.10 L ) = 0.005 w L 2 + 0.040 w L 2 = 0.045 w L 2 ≈ 22.2 w L 2
Columns carry axial gravity forces equal to tributary floor areas; column moments under balanced gravity loads are assumed zero or distributed between upper and lower columns based on stiffness ratios.
4. Classical Force Method (Consistent Deformations)
The Method of Consistent Deformations (also known as the Force Method) solves statically indeterminate structures by releasing redundant constraints to establish a stable, determinate primary structure .
Formulation and Compatibility Equations
Identify the degree of indeterminacy ( D S I = n DSI = n D S I = n ). Select n n n redundant forces or moments ( R 1 , R 2 , … , R n R_1, R_2, \dots, R_n R 1 , R 2 , … , R n ).
Remove these redundants to create the primary determinate structure.
Apply the actual external loads to the primary structure and calculate deflections at the released coordinates: Δ 10 , Δ 20 , … , Δ n 0 \Delta_{10}, \Delta_{20}, \dots, \Delta_{n0} Δ 10 , Δ 20 , … , Δ n 0 .
Apply unit virtual loads ( R i = 1 R_i = 1 R i = 1 ) at each released coordinate to determine flexibility coefficients δ i j \delta_{ij} δ ij (deflection at coordinate i i i due to a unit force at coordinate j j j ):
Δ i 0 = ∫ M 0 m i E I d x , δ i j = ∫ m i m j E I d x \Delta_{i0} = \int \frac{M_0 m_i}{EI} dx, \quad \delta_{ij} = \int \frac{m_i m_j}{EI} dx Δ i 0 = ∫ E I M 0 m i d x , δ ij = ∫ E I m i m j d x
where M 0 M_0 M 0 is the bending moment due to applied service loads, and m i , m j m_i, m_j m i , m j are moments caused by unit loads at coordinates i i i and j j j .
Formulate the compatibility equations enforcing displacement boundary conditions:
[ δ 11 δ 12 … δ 1 n δ 21 δ 22 … δ 2 n ⋮ ⋮ ⋱ ⋮ δ n 1 δ n 2 … δ n n ] [ R 1 R 2 ⋮ R n ] + [ Δ 10 Δ 20 ⋮ Δ n 0 ] = [ Δ 1 Δ 2 ⋮ Δ n ] \begin{bmatrix} \delta_{11} & \delta_{12} & \dots & \delta_{1n} \\ \delta_{21} & \delta_{22} & \dots & \delta_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \delta_{n1} & \delta_{n2} & \dots & \delta_{nn} \end{bmatrix} \begin{bmatrix} R_1 \\ R_2 \\ \vdots \\ R_n \end{bmatrix} + \begin{bmatrix} \Delta_{10} \\ \Delta_{20} \\ \vdots \\ \Delta_{n0} \end{bmatrix} = \begin{bmatrix} \Delta_1 \\ \Delta_2 \\ \vdots \\ \Delta_n \end{bmatrix} δ 11 δ 21 ⋮ δ n 1 δ 12 δ 22 ⋮ δ n 2 … … ⋱ … δ 1 n δ 2 n ⋮ δ nn R 1 R 2 ⋮ R n + Δ 10 Δ 20 ⋮ Δ n 0 = Δ 1 Δ 2 ⋮ Δ n
If supports are unyielding, the right-hand vector equals zero. If a support settles by Δ settlement \Delta_{\text{settlement}} Δ settlement , the corresponding compatibility displacement equals − Δ settlement -\Delta_{\text{settlement}} − Δ settlement .
Maxwell-Betti Reciprocal Theorem
The Maxwell-Betti Reciprocal Theorem proves that for any linear elastic structure, the deflection at coordinate j j j caused by a unit load at coordinate k k k equals the deflection at coordinate k k k caused by a unit load at coordinate j j j :
δ j k = δ k j \delta_{jk} = \delta_{kj} δ jk = δ kj
This guarantees that the structural flexibility matrix [ δ ] [\delta] [ δ ] is always symmetric , reducing the number of flexibility coefficient integrations required.
5. Castigliano's Second Theorem
Carlo Alberto Castigliano stated in 1879 that for any linearly elastic structure subjected to external loads, the first partial derivative of total internal strain energy ( U U U ) with respect to any applied force or redundant reaction ( R i R_i R i ) equals the displacement ( Δ i \Delta_i Δ i ) of the point of application in the direction of that force:
∂ U ∂ R i = Δ i \frac{\partial U}{\partial R_i} = \Delta_i ∂ R i ∂ U = Δ i
For beams and frames governed by flexure, the total strain energy is U = ∫ M 2 2 E I d x U = \int \frac{M^2}{2EI} dx U = ∫ 2 E I M 2 d x . Differentiating inside the integral yields:
Δ i = ∫ M E I ( ∂ M ∂ R i ) d x \Delta_i = \int \frac{M}{EI} \left(\frac{\partial M}{\partial R_i}\right) dx Δ i = ∫ E I M ( ∂ R i ∂ M ) d x
For an unyielding redundant support, the boundary displacement is zero, yielding the condition of minimum strain energy (Least Work principle):
- ∂ U ∂ R i = 0 \frac{\partial U}{\partial R_i} = 0 ∂ R i ∂ U = 0
- 6. Comprehensive Worked Examples
- Worked Example 1: Portal Method Analysis of a Building Frame
Problem: A single-story, two-bay building frame has two equal spans of L = 6.0 m L = 6.0\text{ m} L = 6.0 m and a story height of h = 4.0 m h = 4.0\text{ m} h = 4.0 m . A lateral wind shear force of H = 60 kN H = 60\text{ kN} H = 60 kN is applied at the roof girder level. Using the Portal Method, calculate: (a) column shears, (b) base bending moments, (c) girder shears, and (d) exterior column axial forces.
Solution:
Step 1: Column Shears: With n = 2 n = 2 n = 2 bays and 3 columns (2 exterior, 1 interior): V ext = H 2 n = 60 kN 2 ( 2 ) = 15.0 kN V_{\text{ext}} = \frac{H}{2n} = \frac{60\text{ kN}}{2(2)} = 15.0\text{ kN} V ext = 2 n H = 2 ( 2 ) 60 kN = 15.0 kN V int = 2 V ext = 30.0 kN V_{\text{int}} = 2 V_{\text{ext}} = 30.0\text{ kN} V int = 2 V ext = 30.0 kN Check: 15.0 + 30.0 + 15.0 = 60.0 kN 15.0 + 30.0 + 15.0 = 60.0\text{ kN} 15.0 + 30.0 + 15.0 = 60.0 kN . (Satisfies horizontal equilibrium).
Step 2: Column Bending Moments: With hinges at column mid-height ( h / 2 = 2.0 m h/2 = 2.0\text{ m} h /2 = 2.0 m ): M col,ext = V ext × 2.0 m = 15.0 × 2.0 = 30.0 kN ⋅ m M_{\text{col,ext}} = V_{\text{ext}} \times 2.0\text{ m} = 15.0 \times 2.0 = 30.0\text{ kN}\cdot\text{m} M col,ext = V ext × 2.0 m = 15.0 × 2.0 = 30.0 kN ⋅ m M col,int = V int × 2.0 m = 30.0 × 2.0 = 60.0 kN ⋅ m M_{\text{col,int}} = V_{\text{int}} \times 2.0\text{ m} = 30.0 \times 2.0 = 60.0\text{ kN}\cdot\text{m} M col,int = V int × 2.0 m = 30.0 × 2.0 = 60.0 kN ⋅ m
Step 3: Girder Moments and Shears: At the windward exterior joint: ∑ M = 0 ⟹ M girder = M col,ext = 30.0 kN ⋅ m \sum M = 0 \implies M_{\text{girder}} = M_{\text{col,ext}} = 30.0\text{ kN}\cdot\text{m} ∑ M = 0 ⟹ M girder = M col,ext = 30.0 kN ⋅ m .
Girder inflection point is at mid-span ( L / 2 = 3.0 m L/2 = 3.0\text{ m} L /2 = 3.0 m ): V girder = M girder L / 2 = 30.0 3.0 = 10.0 kN V_{\text{girder}} = \frac{M_{\text{girder}}}{L/2} = \frac{30.0}{3.0} = 10.0\text{ kN} V girder = L /2 M girder = 3.0 30.0 = 10.0 kN
Step 4: Column Axial Loads: At the windward exterior column: upward girder shear pulls the column ⟹ P windward = 10.0 kN \implies P_{\text{windward}} = 10.0\text{ kN} ⟹ P windward = 10.0 kN (Tension).
At the leeward exterior column: downward girder shear pushes the column ⟹ P leeward = 10.0 kN \implies P_{\text{leeward}} = 10.0\text{ kN} ⟹ P leeward = 10.0 kN (Compression).
At the interior column: girder shear from bay 1 acts downward while girder shear from bay 2 acts upward ⟹ P int = 10.0 − 10.0 = 0 kN \implies P_{\text{int}} = 10.0 - 10.0 = 0\text{ kN} ⟹ P int = 10.0 − 10.0 = 0 kN .
Worked Example 2: Propped Cantilever via Consistent Deformations
Problem: A propped cantilever beam A B AB A B of span L L L carries a uniform load w w w . Support A A A is fixed; support B B B is an unyielding roller. Find the prop reaction R B R_B R B .
Solution:
Release the redundant reaction R B R_B R B upward at B B B . The primary structure is a cantilever beam fixed at A A A .
Downward deflection at B B B due to load w w w : Δ B 0 = − w L 4 8 E I \Delta_{B0} = -\frac{w L^4}{8EI} Δ B 0 = − 8 E I w L 4
Upward deflection at B B B due to unit redundant load R B = 1 R_B = 1 R B = 1 : δ B B = + L 3 3 E I \delta_{BB} = +\frac{L^3}{3EI} δ BB = + 3 E I L 3
Compatibility equation at support B B B (zero settlement): Δ B 0 + R B δ B B = 0 ⟹ − w L 4 8 E I + R B ( L 3 3 E I ) = 0 \Delta_{B0} + R_B \delta_{BB} = 0 \implies -\frac{w L^4}{8EI} + R_B \left(\frac{L^3}{3EI}\right) = 0 Δ B 0 + R B δ BB = 0 ⟹ − 8 E I w L 4 + R B ( 3 E I L 3 ) = 0 R B = w L 4 / 8 E I L 3 / 3 E I = 3 8 w L = 0.375 w L R_B = \frac{w L^4 / 8EI}{L^3 / 3EI} = \frac{3}{8} w L = 0.375 w L R B = L 3 /3 E I w L 4 /8 E I = 8 3 w L = 0.375 w L
Bending moment at fixed support A A A : M A = w L 2 2 − R B L = w L 2 2 − 3 8 w L 2 = 1 8 w L 2 = 0.125 w L 2 M_A = \frac{w L^2}{2} - R_B L = \frac{w L^2}{2} - \frac{3}{8} w L^2 = \frac{1}{8} w L^2 = 0.125 w L^2 M A = 2 w L 2 − R B L = 2 w L 2 − 8 3 w L 2 = 8 1 w L 2 = 0.125 w L 2
7. Licensure Exam Pitfalls & Review Notes
Warning
Pitfall 1: Internal Hinge Releases vs. Member Multiplicity When an internal hinge connects more than two members, the condition equation is c = k − 1 c = k - 1 c = k − 1 , where k k k is the number of connected members. If four members frame into a single hinge pin, it provides 4 − 1 = 3 4 - 1 = 3 4 − 1 = 3 equations of condition, not 1.
Caution
Pitfall 2: Portal Method Column Shear Ratios with Unequal Spans The 1 : 2 : 1 1 : 2 : 1 1 : 2 : 1 column shear ratio applies exclusively to equal bay widths. If Bay 1 is 4 m 4\text{ m} 4 m and Bay 2 is 8 m 8\text{ m} 8 m , shears are distributed based on tributary bay widths: V 1 ∝ 4 / 2 = 2 V_1 \propto 4/2 = 2 V 1 ∝ 4/2 = 2 , V 2 ∝ ( 4 + 8 ) / 2 = 6 V_2 \propto (4+8)/2 = 6 V 2 ∝ ( 4 + 8 ) /2 = 6 , V 3 ∝ 8 / 2 = 4 V_3 \propto 8/2 = 4 V 3 ∝ 8/2 = 4 . Interior column shear is not simply double the exterior shear in non-uniform frames.
Tip
Pitfall 3: Maxwell-Betti Coordinate Definitions Flexibility coefficients δ i j \delta_{ij} δ ij must maintain strict directional conventions. If coordinate 1 is downward deflection and coordinate 2 is counterclockwise rotation, δ 12 \delta_{12} δ 12 represents the linear deflection at 1 caused by a unit counterclockwise couple at 2.