The Ultimate Guide to Your Theory of Structures II Past Year Question Bank

Flexibility Matrix (F): The physical significance of a flexibility coefficient, $f_{ij}$, is the displacement at coordinate i due to a unit force applied at coordinate j. The matrix is an organized collection of these coefficients.

Example: In a 2-DOF system, $f_{21}$ is the vertical deflection at point 2 caused by a 1 kN load at point 1.

Stiffness Matrix (K): The physical significance of a stiffness coefficient, $k_{ij}$, is the force required at coordinate i to produce a unit displacement at coordinate j, while all other coordinates are held at zero displacement.

Example: For a beam, $k_{11}$ could be the force at point 1 needed to cause a 1-unit deflection at point 1 (while point 2 is fixed).

Flexibility Coefficient ($f_{ij}$): It is defined as the displacement at coordinate i due to a unit action (force or moment) applied at coordinate j. It represents the flexibility of the structure at i when loaded at j.

Stiffness Coefficient ($k_{ij}$): It is defined as the action (force or moment) required at coordinate i to cause a unit displacement at coordinate j, while all other coordinates are restrained. It represents the stiffness of the structure.

Static Indeterminacy ($D_s$): This refers to the number of unknown forces (reactions or internal member forces) that are in excess of the available static equilibrium equations. A structure is “statically indeterminate” if $D_s > 0$.

$D_s = $ (Total Unknown Forces) – (Available Equilibrium Equations)

Example (Beam): A fixed beam has 6 unknown reactions (3 at each end) and 3 equilibrium equations ($\Sigma F_x=0, \Sigma F_y=0, \Sigma M=0$). $D_s = 6 – 3 = 3$. It is statically indeterminate to the 3rd degree.

Example (Truss): For a 2D truss, $D_s = (m + r) – 2j$, where m = members, r = reactions, j = joints.

Kinematic Indeterminacy ($D_k$): Also known as the degree of freedom, this refers to the total number of independent joint displacements (translations and rotations) that are possible in a structure. It defines the number of unknowns in the displacement method.

Example (Beam): A propped cantilever has a roller support. This joint is free to rotate but not translate. $D_k = 1$ (rotation at roller). The fixed end has $D_k = 0$. Total $D_k = 1$.

Example (Frame): A 2D rigid-jointed frame joint can have 3 displacements (2 translations, 1 rotation). $D_k$ is the sum of all possible joint displacements minus the number of restrained displacements by supports.

Flexibility: A measure of the displacement caused by a unit load. A flexible structure deforms significantly under load.

Stiffness: A measure of the force required to cause a unit displacement. A stiff structure resists deformation.

Proof of Reciprocal Relation:

Let {P} be the vector of forces and {$\Delta$} be the vector of corresponding displacements.

By definition of the Flexibility Matrix F:

{$\Delta$} = F {P} — (1)

By definition of the Stiffness Matrix K:

{P} = K {$\Delta$} — (2)

Substitute (2) into (1):

{$\Delta$} = F (K {$\Delta$})

{$\Delta$} = (FK) {$\Delta$}

For this to be true, the term (FK) must be an identity matrix I.

FK = I

Similarly, substituting (1) into (2):

{P} = K (F {P})

{P} = (KF) {P}

KF = I

Since FK = KF = I, the flexibility matrix F and the stiffness matrix K are the inverse (reciprocal) of each other.

F = K$^{-1}$ and K = F$^{-1}$

Feature	Flexibility	Stiffness
Definition	Displacement per unit force ($\Delta/P$).	Force per unit displacement ($P/\Delta$).
Concept	Measures the ease of deformation.	Measures the resistance to deformation.
Coefficient ($f_{ij}$)	Displacement at i due to unit force at j.	Force at i due to unit displacement at j.
Used in	Force Method (Flexibility Method).	Displacement Method (Stiffness Method).
Relation	They are reciprocal to each other. F = K$^{-1}$.

Structural Idealization: This is the process of creating a simplified mathematical or computational model of a complex, real-world structure. This is done to make analysis feasible.

Examples:

• Representing a 3D building frame as a 2D plane frame for analysis.
• Modeling a beam or column, which has depth, as a simple 1D line element.
• Assuming connections are either perfectly “pinned” (free to rotate) or “fixed” (no rotation), rather than semi-rigid.
• Assuming supports are unyielding (e.g., a “fixed” support does not move or rotate at all).

Global Coordinate System (GCS): This is a single, common coordinate system (e.g., X, Y, Z) that defines the geometry and loading for the entire structure. All final results (like nodal displacements) are expressed in this system.

Local Coordinate System (LCS): This is a separate coordinate system (e.g., x’, y’, z’) defined for an individual member of the structure. It is typically oriented along the member’s axis. This system is used to simplify the calculation of member-specific properties, like the member’s stiffness matrix, before it is transformed into the GCS.

Basis	Static Indeterminacy ($D_s$)	Kinematic Indeterminacy ($D_k$)
Definition	Excess unknown forces (reactions/internal) beyond available equilibrium equations.	Number of independent displacements (translations/rotations) at joints.
Also Known As	Degree of Redundancy.	Degree of Freedom (DOF).
If $D_s = 0$	Structure is statically determinate.	(No specific conclusion, depends on supports).
If $D_k = 0$	Structure has no possible movement (e.g., all joints are fixed).	(No specific conclusion, depends on members).
Governing Method	Determines the number of unknowns in the Force Method (Flexibility Method).	Determines the number of unknowns in the Displacement Method (Stiffness Method).
Example	A fixed beam has $D_s = 3$.	A fixed beam has $D_k = 0$.

Basis	Force Method (Flexibility Method)	Displacement Method (Stiffness Method)
Primary Unknowns	Redundant forces or moments.	Nodal displacements (translations/rotations).
Governing Equations	Compatibility equations (ensuring displacements fit together).	Equilibrium equations (ensuring forces balance at nodes).
Coefficients	Uses Flexibility coefficients ($f_{ij}$).	Uses Stiffness coefficients ($k_{ij}$).
Indeterminacy	Suitable when Static Indeterminacy ($D_s$) is low.	Suitable when Kinematic Indeterminacy ($D_k$) is low.
Examples	Method of Consistent Deformations, Three-Moment Theorem.	Slope-Deflection Method, Moment Distribution, Finite Element Method.

Statically Determinate Structures	Statically Indeterminate Structures
Can be fully analyzed using only the 3 static equilibrium equations.	Cannot be analyzed by equilibrium equations alone; require compatibility conditions.
Less sensitive to temperature changes, support settlement, or fabrication errors.	Subject to additional stresses due to temperature changes, settlement, etc.
Bending moment and shear force are independent of material and cross-section.	Bending moment and shear force depend on material and cross-section (i.e., on $EI$).
Generally less stiff, resulting in larger deflections.	Generally more stiff, resulting in smaller deflections.
(Example: Simply supported beam)	(Example: Fixed beam, Propped cantilever)

Structural Idealization: This is the process of creating a simplified mathematical or computational model of a complex, real-world structure. (See Q.6 for examples).

Stability of a Truss (2D):

Let $m$ = number of members, $r$ = number of reactions, $j$ = number of joints.

The equilibrium equations available are $2j$.

The total unknowns are $m + r$.

Necessary Condition: $m + r = 2j$.

Sufficient Condition: This condition ($m + r = 2j$) is necessary but not sufficient. The truss must also be geometrically stable, meaning it must not form a mechanism or be improperly supported. For example, if all supports are parallel rollers, the truss is unstable even if $m + r = 2j$.

Degree of Static Indeterminacy ($D_s$): This refers to the number of unknown forces (reactions or internal member forces) that are in excess of the available static equilibrium equations.

Determination for a Plane Truss:

A plane truss is analyzed by considering the equilibrium of its joints. At each joint, there are two equilibrium equations ($\Sigma F_x=0, \Sigma F_y=0$).

Total equilibrium equations available = $2 \times j$ (where $j$ = number of joints)

Total unknown forces = (Number of members, $m$) + (Number of external reactions, $r$)

Degree of Static Indeterminacy ($D_s$) = (Total Unknowns) – (Total Equations)

$D_s = (m + r) – 2j$

External Indeterminacy ($D_{se}$): $D_{se} = r – 3$ (comparing reactions to the 3 equations for the whole body)

Internal Indeterminacy ($D_{si}$): $D_{si} = m – (2j – 3)$

$D_s = D_{se} + D_{si} = (r – 3) + (m – 2j + 3) = m + r – 2j$

Example: A simple triangular truss on a hinge and a roller ($m=3, j=3, r=3$).

$D_s = (3 + 3) – 2(3) = 6 – 6 = 0$. It is statically determinate.

Statement: For a linearly elastic structure, if the total strain energy ($U$) is expressed as a function of the external loads ($P_1, P_2, …, P_n$), the partial derivative of the total strain energy with respect to any single external load ($P_i$) gives the displacement ($\delta_i$) at the point of application of that load, in the direction of that load.

Mathematical Form: $\delta_i = \frac{\partial U}{\partial P_i}$

Proof:

Consider a structure subjected to loads $P_1, P_2, …, P_n$, causing displacements $\delta_1, \delta_2, …, \delta_n$. The total strain energy $U$ is stored.

Assume the strain energy $U$ is expressed as a function of the loads: $U = f(P_1, P_2, …, P_n)$.

Now, apply an incremental load $dP_i$ at the point of application of $P_i$. The total energy of the system increases. The new strain energy $U’$ is:

$U’ = U + dU = U + \frac{\partial U}{\partial P_i} dP_i$ (from the definition of a partial derivative).

The change in strain energy $dU$ from adding $dP_i$ can be found by considering the work done. The total external work $W$ is a function of the loads. The change in $W$ is $dW = \delta_i \cdot dP_i$.

The total strain energy $U’$ after the increment is $U’ = U + dW_{ext}$. The external work done $dW_{ext}$ by adding $dP_i$ to the already deflected structure is $\delta_i \cdot dP_i$.

Therefore, the change in strain energy is $dU = U’ – U = \delta_i \cdot dP_i$.

We now have two expressions for $dU$:

From calculus: $dU = \frac{\partial U}{\partial P_i} dP_i$

From work/energy: $dU = \delta_i \cdot dP_i$

Equating them: $\frac{\partial U}{\partial P_i} dP_i = \delta_i \cdot dP_i$

Therefore, $\delta_i = \frac{\partial U}{\partial P_i}$.

Several theorems, often derived from energy principles, are used to determine the displacement (deflection and rotation) of structures. The key theorems include:

Castigliano’s Second Theorem:

Concept: This is the most common method for finding displacements. It states that for a linearly elastic structure, the displacement at a point in the direction of an applied load ($P$) is equal to the partial derivative of the total strain energy ($U$) with respect to that load.

Formula: $\delta = \frac{\partial U}{\partial P}$

Illustration: To find the vertical deflection ($\delta_C$) at the tip ‘C’ of a cantilever beam, we can apply a real or fictitious (dummy) vertical load $P$ at ‘C’. We then find the total strain energy $U = \int \frac{M_x^2}{2EI} dx$, where the moment $M_x$ is expressed as a function of $P$. The deflection is $\delta_C = (\frac{\partial U}{\partial P})_{P \to \text{actual value}}$. If $P$ was a dummy load, its actual value is 0.

Maxwell’s Reciprocal Theorem:

Concept: This theorem establishes a symmetrical relationship in structural response. It states that the displacement at point A due to a load at point B is equal to the displacement at point B due to the same load at point A (given the loads and displacements are in corresponding directions).

Illustration: Consider a simply supported beam. The vertical deflection at the center (point C) caused by a 10 kN vertical load at the quarter-span (point B) is exactly the same as the vertical deflection at the quarter-span (point B) if the 10 kN vertical load were moved to the center (point C).

Betti’s Law (or Betti-Maxwell Theorem):

Statement: In any structure where the material is linearly elastic (follows Hooke’s law), supports are unyielding, and the temperature is constant, the external virtual work done by a first system of forces ($P_m$) during the deformations caused by a second system of forces ($P_n$) is equal to the external virtual work done by the second system of forces ($P_n$) during the deformations caused by the first system ($P_m$).

Mathematical Form: $\sum P_{m} \cdot \delta_{mn} = \sum P_{n} \cdot \delta_{nm}$ (where $\delta_{mn}$ is the displacement in the $m$-system’s direction caused by the $n$-system).

Use: Betti’s Law is a fundamental principle that is not often used directly for problem-solving. Its primary use is as a theoretical foundation to derive other powerful theorems, most notably Maxwell’s Reciprocal Theorem and the Muller-Breslau Principle for influence lines.

Maxwell’s Reciprocal Theorem:

Statement: As a special case of Betti’s Law, this theorem states that for a linearly elastic structure, the displacement at a point ‘A’ (in direction $i$) due to a unit load applied at point ‘B’ (in direction $j$) is equal to the displacement at point ‘B’ (in direction $j$) due to a unit load applied at point ‘A’ (in direction $i$).

Mathematical Form: $\delta_{AB} = \delta_{BA}$ (for unit loads) or more generally, $P_A \cdot \delta_{AB} = P_B \cdot \delta_{BA}$.

Use: This theorem is extremely useful in structural analysis.

• Influence Lines: It is the theoretical basis for the Muller-Breslau Principle, which provides a simple qualitative (and quantitative) method for constructing influence lines for indeterminate structures.
• Symmetry in Analysis: It proves that the flexibility matrix of a structure is symmetric, which simplifies computational analysis.

Castigliano’s First Theorem (Theorem of Strain Energy)

Statement: If the total strain energy ($U$) of a linearly elastic structure is expressed as a function of its independent displacement components ($\delta_1, \delta_2, …, \delta_n$), the partial derivative of the total strain energy with respect to any single displacement component ($\delta_i$) is equal to the corresponding force ($P_i$) at that point and in the direction of that displacement.

Mathematical Form: $P_i = \frac{\partial U}{\partial \delta_i}$

Proof:

Consider a structure subjected to forces $P_1, P_2, …, P_n$ which cause corresponding displacements $\delta_1, \delta_2, …, \delta_n$. The total strain energy $U$ is stored in the structure and is a function of the displacements: $U = f(\delta_1, \delta_2, …, \delta_n)$.

Now, let one displacement, $\delta_i$, be given a small, arbitrary increment $d\delta_i$, while all other displacements are held constant.

The change in internal strain energy ($dU$) due to this incremental displacement is: $dU = \frac{\partial U}{\partial \delta_i} d\delta_i$.

By the principle of conservation of energy, this change in internal strain energy ($dU$) must be equal to the external work done ($dW$) during this incremental displacement.

Since only displacement $\delta_i$ changed, the external work is done only by the corresponding force $P_i$. This work is $dW = P_i \cdot d\delta_i$.

Equating the external work and the change in internal energy:

$dW = dU$

$P_i \cdot d\delta_i = \frac{\partial U}{\partial \delta_i} d\delta_i$

Dividing by $d\delta_i$, we get: $P_i = \frac{\partial U}{\partial \delta_i}$.

Castigliano’s Second Theorem (Theorem of Deflection)

Statement: For a linearly elastic structure, if the total strain energy ($U$) is expressed as a function of the external loads ($P_1, P_2, …, P_n$), the partial derivative of the total strain energy with respect to any single external load ($P_i$) gives the displacement ($\delta_i$) at the point of application of that load, in the direction of that load.

Mathematical Form: $\delta_i = \frac{\partial U}{\partial P_i}$

Proof:

Consider a structure subjected to loads $P_1, P_2, …, P_n$. The total strain energy $U$ is stored. By the principle of superposition, the total work done (and stored as energy) is $U = \frac{1}{2} \sum P_i \delta_i$.

Assume the strain energy $U$ is expressed as a function of the loads: $U = f(P_1, P_2, …, P_n)$.

Now, apply an incremental load $dP_i$ at the point of application of $P_i$. The change in strain energy $dU$ must equal the complementary work done, which, for a linear system, is $\delta_i \cdot dP_i$.

From calculus: $dU = \frac{\partial U}{\partial P_i} dP_i$

From work/energy: $dU = \delta_i \cdot dP_i$

Equating them: $\frac{\partial U}{\partial P_i} dP_i = \delta_i \cdot dP_i$

Therefore, $\delta_i = \frac{\partial U}{\partial P_i}$.

This question is answered by combining the proofs for the first and second theorems.

Castigliano’s First Theorem (Theorem of Strain Energy)

Statement: If the total strain energy ($U$) of a linearly elastic structure is expressed as a function of its independent displacement components ($\delta_1, \delta_2, …, \delta_n$), the partial derivative of the total strain energy with respect to any single displacement component ($\delta_i$) is equal to the corresponding force ($P_i$) at that point and in the direction of that displacement.

Mathematical Form: $P_i = \frac{\partial U}{\partial \delta_i}$

Proof:

Consider a structure subjected to forces $P_1, P_2, …, P_n$ which cause corresponding displacements $\delta_1, \delta_2, …, \delta_n$. The total strain energy $U$ is stored in the structure and is a function of the displacements: $U = f(\delta_1, \delta_2, …, \delta_n)$.

Now, let one displacement, $\delta_i$, be given a small, arbitrary increment $d\delta_i$, while all other displacements are held constant.

The change in internal strain energy ($dU$) due to this incremental displacement is: $dU = \frac{\partial U}{\partial \delta_i} d\delta_i$.

By the principle of conservation of energy, this change in internal strain energy ($dU$) must be equal to the external work done ($dW$) during this incremental displacement.

Since only displacement $\delta_i$ changed, the external work is done only by the corresponding force $P_i$. This work is $dW = P_i \cdot d\delta_i$.

Equating the external work and the change in internal energy:

$dW = dU$

$P_i \cdot d\delta_i = \frac{\partial U}{\partial \delta_i} d\delta_i$

Dividing by $d\delta_i$, we get: $P_i = \frac{\partial U}{\partial \delta_i}$.

Castigliano’s Second Theorem (Theorem of Deflection)

Statement: For a linearly elastic structure, if the total strain energy ($U$) is expressed as a function of the external loads ($P_1, P_2, …, P_n$), the partial derivative of the total strain energy with respect to any single external load ($P_i$) gives the displacement ($\delta_i$) at the point of application of that load, in the direction of that load.

Mathematical Form: $\delta_i = \frac{\partial U}{\partial P_i}$

Proof:

Consider a structure subjected to loads $P_1, P_2, …, P_n$. The total strain energy $U$ is stored.

Assume the strain energy $U$ is expressed as a function of the loads: $U = f(P_1, P_2, …, P_n)$.

Now, apply an incremental load $dP_i$ at the point of application of $P_i$. The total energy of the system increases. The new strain energy $U’$ is:

$U’ = U + dU = U + \frac{\partial U}{\partial P_i} dP_i$

The change in strain energy $dU$ from adding $dP_i$ can be found by considering the work done. The change in $W$ is $dW = \delta_i \cdot dP_i$.

The total strain energy $U’$ after the increment is $U’ = U + dW_{ext}$. The external work done $dW_{ext}$ by adding $dP_i$ to the already deflected structure is $\delta_i \cdot dP_i$.

Therefore, the change in strain energy is $dU = U’ – U = \delta_i \cdot dP_i$.

We now have two expressions for $dU$:

From calculus: $dU = \frac{\partial U}{\partial P_i} dP_i$

From work/energy: $dU = \delta_i \cdot dP_i$

Equating them: $\frac{\partial U}{\partial P_i} dP_i = \delta_i \cdot dP_i$

Therefore, $\delta_i = \frac{\partial U}{\partial P_i}$.

Statement: If the total strain energy ($U$) of a linearly elastic structure is expressed as a function of its independent displacement components ($\delta_1, \delta_2, …, \delta_n$), the partial derivative of the total strain energy with respect to any single displacement component ($\delta_i$) is equal to the corresponding force ($P_i$) at that point and in the direction of that displacement.

Mathematical Form: $P_i = \frac{\partial U}{\partial \delta_i}$

Proof:

Consider a structure subjected to forces $P_1, P_2, …, P_n$ which cause corresponding displacements $\delta_1, \delta_2, …, \delta_n$. The total strain energy $U$ is stored in the structure and is a function of the displacements: $U = f(\delta_1, \delta_2, …, \delta_n)$.

Now, let one displacement, $\delta_i$, be given a small, arbitrary increment $d\delta_i$, while all other displacements are held constant.

The change in internal strain energy ($dU$) due to this incremental displacement is, by the definition of a partial derivative: $dU = \frac{\partial U}{\partial \delta_i} d\delta_i$.

By the principle of conservation of energy, this change in internal strain energy ($dU$) must be equal to the external work done ($dW$) during this incremental displacement.

Since only displacement $\delta_i$ changed, the external work is done only by the corresponding force $P_i$. This work is $dW = P_i \cdot d\delta_i$.

Equating the external work and the change in internal energy:

$dW = dU$

$P_i \cdot d\delta_i = \frac{\partial U}{\partial \delta_i} d\delta_i$

Dividing by $d\delta_i$, we get: $P_i = \frac{\partial U}{\partial \delta_i}$.

Statement: Maxwell’s Reciprocal Theorem states that for a linearly elastic structure, the displacement at a point ‘A’ in a specific direction, caused by a load $P$ applied at a point ‘B’ in a second specific direction, is equal to the displacement at point ‘B’ in the second direction, caused by the same load $P$ applied at point ‘A’ in the first direction.

Simplified Form: Let $\delta_{AB}$ be the deflection at point A caused by a load $P_B$ at point B. Let $\delta_{BA}$ be the deflection at point B caused by a load $P_A$ at point A. If $P_A = P_B = P$ (and they are in corresponding directions), then $\delta_{AB} = \delta_{BA}$.

Proof (using Betti’s Law):

Betti’s Law states that for two load systems, ‘m’ and ‘n’, the work done by system ‘m’ going through the displacements of system ‘n’ is equal to the work done by system ‘n’ going through the displacements of system ‘m’.

$\sum P_m \cdot \delta_{mn} = \sum P_n \cdot \delta_{nm}$

Let’s define two simple loading systems:

System m: A single load $P_A$ applied at point A.

System n: A single load $P_B$ applied at point B.

Let’s define the displacements:

$\delta_{BA}$ = displacement at B due to load at A (from system m).

$\delta_{AB}$ = displacement at A due to load at B (from system n).

Apply Betti’s Law:

Work of ‘m’ on ‘n’s displacements: $W_{mn} = P_A \cdot \delta_{AB}$

Work of ‘n’ on ‘m’s displacements: $W_{nm} = P_B \cdot \delta_{BA}$

Equating them gives the general form of the theorem:

$P_A \cdot \delta_{AB} = P_B \cdot \delta_{BA}$

If we choose the loads to be of equal magnitude, $P_A = P_B = P$, the equation simplifies to:

$P \cdot \delta_{AB} = P \cdot \delta_{BA}$

$\delta_{AB} = \delta_{BA}$

This proves that the displacement at A due to a load at B is the same as the displacement at B due to the same load at A.

Statement: Maxwell’s Reciprocal Theorem states that for a linearly elastic structure, the displacement at a point ‘A’ in a specific direction, caused by a load $P$ applied at a point ‘B’ in a second specific direction, is equal to the displacement at point ‘B’ in the second direction, caused by the same load $P$ applied at point ‘A’ in the first direction.

Simplified Form: Let $\delta_{AB}$ be the deflection at point A caused by a load $P_B$ at point B. Let $\delta_{BA}$ be the deflection at point B caused by a load $P_A$ at point A. If $P_A = P_B = P$ (and they are in corresponding directions), then $\delta_{AB} = \delta_{BA}$.

Proof (using Betti’s Law):

Betti’s Law states that for two load systems, ‘m’ and ‘n’, the work done by system ‘m’ going through the displacements of system ‘n’ is equal to the work done by system ‘n’ going through the displacements of system ‘m’.

$\sum P_m \cdot \delta_{mn} = \sum P_n \cdot \delta_{nm}$

Let’s define two simple loading systems:

System m: A single load $P_A$ applied at point A.

System n: A single load $P_B$ applied at point B.

Let’s define the displacements:

$\delta_{BA}$ = displacement at B due to load at A (from system m).

$\delta_{AB}$ = displacement at A due to load at B (from system n).

Apply Betti’s Law:

Work of ‘m’ on ‘n’s displacements: $W_{mn} = P_A \cdot \delta_{AB}$

Work of ‘n’ on ‘m’s displacements: $W_{nm} = P_B \cdot \delta_{BA}$

Equating them gives the general form of the theorem:

$P_A \cdot \delta_{AB} = P_B \cdot \delta_{BA}$

If we choose the loads to be of equal magnitude, $P_A = P_B = P$, the equation simplifies to:

$P \cdot \delta_{AB} = P \cdot \delta_{BA}$

$\delta_{AB} = \delta_{BA}$

This proves that the displacement at A due to a load at B is the same as the displacement at B due to the same load at A.

Statement: Castigliano’s Second Theorem states that for a linearly elastic structure, if the total strain energy ($U$) is expressed as a function of the external loads ($P_i$), the partial derivative of the total strain energy with respect to any single external load gives the displacement ($\delta_i$) at the point of application of that load, in the direction of that load.

Mathematical Form: $\delta_i = \frac{\partial U}{\partial P_i}$ (for linear displacement) or $\theta_i = \frac{\partial U}{\partial M_i}$ (for angular displacement/rotation).

Use for finding deflection:

This theorem provides a direct method to calculate the deflection or rotation at any point on a structure (beam, frame, or truss). The procedure is as follows:

To find Deflection ($\delta$):

Step 1: Apply a point load $P$ at the specific point and in the specific direction for which the deflection is required.

Step 2: If a load $P$ does not already exist at that point, apply a fictitious (dummy) load $P$.

Step 3: Express the total internal strain energy ($U$) of the entire structure as a function of all loads, including $P$. For a beam, this is $U = \int \frac{M_x^2}{2EI} dx$, where $M_x$ must be a function of $P$.

Step 4: Take the partial derivative of the strain energy expression with respect to $P$. This is often simplified using the Leibniz integral rule to:

$\delta = \frac{\partial U}{\partial P} = \int \frac{M_x}{EI} \left( \frac{\partial M_x}{\partial P} \right) dx$

Step 5: After differentiation, substitute the actual value for $P$. If $P$ was a dummy load, its actual value is 0. If $P$ was a real load, its numerical value is substituted.

To find Rotation ($\theta$):

The process is identical, but a real or dummy moment $M$ is applied at the point of interest, and the rotation is found using $\theta = \frac{\partial U}{\partial M}$.

Statement: Castigliano’s First Theorem states that if the total strain energy ($U$) of a linearly elastic structure is expressed as a function of its independent displacement components ($\delta_1, \delta_2, …, \delta_n$), the partial derivative of the total strain energy with respect to any single displacement component ($\delta_i$) is equal to the corresponding force ($P_i$) at that point and in the direction of that displacement.

Mathematical Form: $P_i = \frac{\partial U}{\partial \delta_i}$

i) Primary Structure: A primary structure is a structure that is derived from the original indeterminate structure by removing a sufficient number of restraints (supports or internal forces) to make it statically determinate and stable. The choice of the primary structure is not unique, but it must be stable.

ii) Redundant Force (or Redundant): A redundant force (or “redundant”) is an unknown force or moment that is in excess of what is required to maintain static equilibrium. In the force method, these are the forces or moments that are “released” (removed) from the original structure to create the primary structure. The number of redundants is equal to the degree of static indeterminacy.

Example of Selecting Redundants: Consider a propped cantilever beam (fixed at one end, roller at the other). It has 4 unknown reactions and 3 equilibrium equations. Degree of Indeterminacy = 4 – 3 = 1. We need to select one redundant. Option 1: Choose the vertical reaction at the roller support as the redundant ($R_B$). The primary structure becomes a simple cantilever beam. Option 2: Choose the moment reaction at the fixed support as the redundant ($M_A$). The primary structure becomes a simply supported beam. The choice depends on which primary structure is easier to analyze.

iii) Flexibility Coefficient ($f_{ij}$): A flexibility coefficient, $f_{ij}$, is a measure of the displacement at point ‘i’ (in the direction of coordinate ‘i’) caused by a unit load applied at point ‘j’ (in the direction of coordinate ‘j’). The displacement and force can be linear or rotational. For example, $f_{21}$ would be the vertical deflection at point 2 due to a unit vertical load applied at point 1. In the flexibility matrix, the diagonal elements ($f_{ii}$) represent the displacement at a point ‘i’ due to a unit load at that same point ‘i’.

Statement of the Three-Moment Theorem: The Three-Moment Theorem (or Clapeyron’s Theorem) gives a relationship between the internal bending moments at three successive, continuous supports ($A$, $B$, and $C$) in a continuous beam.

For a general case with no support settlement and constant $EI$ within each span, the equation is:

$$M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = -\frac{6A_1 \bar{x}_1}{I_1 L_1} – \frac{6A_2 \bar{x}_2}{I_2 L_2}$$

Terms Involved:

• $M_A, M_B, M_C$: Internal bending moments at supports $A$, $B$, and $C$.
• $L_1, L_2$: Lengths of the two adjacent spans ($AB$ and $BC$).
• $I_1, I_2$: Moments of inertia of the beam in spans $L_1$ and $L_2$.
• $A_1$: Area of the Bending Moment Diagram (BMD) for span $L_1$, considering it as a simply supported beam under its external loads.
• $\bar{x}_1$: Centroid of the $A_1$ area, measured from the left support $A$.
• $A_2$: Area of the BMD for span $L_2$, considering it as a simply supported beam under its external loads.
• $\bar{x}_2$: Centroid of the $A_2$ area, measured from the right support $C$.

Physical Meaning: The theorem is fundamentally a compatibility equation. It enforces that the slope of the elastic curve at the central support ($B$) must be continuous. It states that the slope of the beam at the end $B$ of span $AB$ ($\theta_{BA}$) must be equal and opposite to the slope at the end $B$ of span $BC$ ($\theta_{BC}$), i.e., $\theta_{BA} + \theta_{BC} = 0$. The theorem relates the rotations caused by the support moments ($M_A, M_B, M_C$) and the rotations caused by the external loads (the $A\bar{x}$ terms) and ensures this continuity of slope is satisfied.

Derivation/Proof Concept (Outline):

1. Primary Structure: The continuous beam is split into a series of simply supported beams. The unknown support moments ($M_A, M_B, M_C$, etc.) are the redundants.
2. Superposition: The final slope at any support (e.g., support $B$) is the sum of: The slope caused by the external loads on the simply supported span, and the slopes caused by the redundant moments $M_A$, $M_B$, and $M_C$.
3. Calculate Slopes: The slopes are calculated using the Moment-Area Theorem or the Unit Load Method.
4. Apply Compatibility: The compatibility condition is that the structure is continuous over support $B$, so the final slope just to the left of $B$ must be equal to the final slope just to the right of $B$: $\theta_{BA} + \theta_{BC} = 0$.
5. Assemble Equation: Substituting the expressions for these slopes into the compatibility equation yields the Three-Moment Theorem.

Compatibility Conditions: Compatibility conditions are equations that enforce the geometric continuity and boundary conditions of a structure. In the context of the force method, they state that the final deformation of the original, indeterminate structure must be consistent with its supports.

Physical Meaning: The physical meaning is that the structure must “fit together” without any gaps or overlaps, and it must respect its support conditions.

For example, when we use the method of consistent deformation:

1. We remove a redundant (e.g., a roller support) to get a primary structure (e.g., a cantilever beam).
2. We apply the external loads, and the primary structure deflects at the removed support (e.g., the cantilever tip moves down by $\Delta_B$).
3. We then apply the redundant force (e.g., $R_B$) as an unknown load on the primary structure, causing a deflection in the opposite direction (e.g., the tip moves up by $\Delta_{BB} = R_B \cdot f_{BB}$).
4. The compatibility condition enforces the original structure’s geometry. Since the original structure had a roller support at that point, the total vertical deflection must be zero.
5. Therefore, the compatibility equation is: $\Delta_B + R_B \cdot f_{BB} = 0$. This equation allows us to solve for the unknown redundant $R_B$.

Consistent Deformation Method (Force Method): The Consistent Deformation Method (also known as the Force Method or Flexibility Method) is a procedure for analyzing statically indeterminate structures. Its primary objective is to determine the unknown redundant forces.

The “consistent deformation” name comes from the core principle: the method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

Procedures for Analysis:

1. Determine Degree of Indeterminacy (DSI): Find the number of redundant forces (let’s say ‘n’).
2. Select a Primary Structure: Remove the ‘n’ redundant forces/moments from the original structure to create a statically determinate and stable primary structure.
3. Apply External Loads: Calculate the displacements (or rotations) at the points where redundants were removed, caused only by the original external loads on the primary structure. Let’s call these displacements $\{\Delta_L\}$.
4. Apply Redundants as Loads: One by one, apply a unit value of each redundant force ($X_1, X_2, … X_n$) to the primary structure (with no external loads).
5. Calculate Flexibility Coefficients: For each unit redundant load $X_j=1$, calculate the displacements ($f_{ij}$) at all ‘n’ redundant locations ‘i’. These are the flexibility coefficients.
6. Formulate Compatibility Equations: For each redundant location ‘i’, write an equation stating that the total displacement (from external loads + all redundant forces) must equal the known final displacement of the original structure (which is usually zero, e.g., at a support). This is the principle of superposition. For $n=2$, the equations would be: $\Delta_{1L} + f_{11}X_1 + f_{12}X_2 = 0$ $\Delta_{2L} + f_{21}X_1 + f_{22}X_2 = 0$ In matrix form: $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants: Solve the ‘n’ simultaneous compatibility equations for the ‘n’ unknown redundant forces $\{X\}$.
8. Find Final Forces/Moments: Place the now-known redundant forces (along with the original external loads) back onto the primary structure. Use static equilibrium to find all other reactions, shear forces, and bending moments.

Statement of the Three-Moment Theorem: The Three-Moment Theorem (or Clapeyron’s Theorem) gives a relationship between the internal bending moments at three successive, continuous supports ($A$, $B$, and $C$) in a continuous beam.

For a general case with no support settlement and constant $EI$ within each span, the equation is:

$$M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = -\frac{6A_1 \bar{x}_1}{I_1 L_1} – \frac{6A_2 \bar{x}_2}{I_2 L_2}$$

Terms Involved:

• $M_A, M_B, M_C$: Internal bending moments at supports $A$, $B$, and $C$. These are the redundants.
• $L_1, L_2$: Lengths of the two adjacent spans ($AB$ and $BC$).
• $I_1, I_2$: Moments of inertia of the beam in spans $L_1$ and $L_2$.
• $A_1$: Area of the Bending Moment Diagram (BMD) for span $L_1$, considering it as a simply supported beam under its external loads.
• $\bar{x}_1$: Centroid of the $A_1$ area, measured from the left support $A$.
• $A_2$: Area of the BMD for span $L_2$, considering it as a simply supported beam under its external loads.
• $\bar{x}_2$: Centroid of the $A_2$ area, measured from the right support $C$.

The terms on the right side ($-\frac{6A \bar{x}}{IL}$) are called the “fixed-end moment” terms for a simply supported beam and represent the effect of the external loads.

Temperature Effects in Trusses (Force Method): Temperature changes cause members in a truss to expand or contract. If the truss is statically determinate, this change in length simply causes the truss to change its shape, but it does not induce any stress (assuming uniform temperature change). If the truss is statically indeterminate, the members are restrained from deforming freely, which induces internal stresses and forces (thermal stresses).

In the force method, temperature effects are handled by modifying the compatibility equation:

1. Primary Structure: Select a primary (determinate) structure by removing a redundant member (e.g., member $X$).
2. Calculate Displacements:
   • $\Delta_{TL}$: Calculate the relative displacement (gap or overlap) at the location of the removed redundant $X$, caused by the temperature change ($\Delta T$) on the primary structure. This is found by $\Delta_{TL} = \sum (\alpha \cdot \Delta T \cdot L) \cdot u$, where ‘$u$’ is the force in each member due to a unit load applied at the redundant.
   • $\Delta_{LL}$: Calculate the displacement at the redundant location due to the external loads (if any).
   • $f_{XX}$: Calculate the flexibility coefficient: the displacement at the redundant location due to a unit force ($X=1$) applied to the redundant.
3. Formulate Compatibility Equation: The compatibility equation states that the total final displacement at the redundant location must be zero (to close the gap). Total Displacement = (Disp. from Loads) + (Disp. from Temp) + (Disp. from Redundant) $$\Delta_{LL} + \Delta_{TL} + f_{XX} \cdot X = 0$$
4. Solve for Redundant: The unknown redundant force $X$ (the thermal force) is solved: $$X = – \frac{\Delta_{LL} + \Delta_{TL}}{f_{XX}}$$ If there are no external loads, the equation simplifies to $X = – \frac{\Delta_{TL}}{f_{XX}}$. This $X$ is the internal force in the redundant member caused by the temperature change.

Consistent Deformation Method (Force Method): The Consistent Deformation Method (also known as the Force Method or Flexibility Method) is a procedure for analyzing statically indeterminate structures. The “consistent deformation” name comes from the core principle: the method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

Procedures for Analysis:

1. Determine Degree of Indeterminacy (DSI): Find the number of redundant forces (let’s say ‘n’).
2. Select a Primary Structure: Remove the ‘n’ redundant forces/moments from the original structure to create a statically determinate and stable primary structure.
3. Apply External Loads: Calculate the displacements (or rotations) at the points where redundants were removed, caused only by the original external loads on the primary structure. Let’s call these displacements $\{\Delta_L\}$.
4. Apply Redundants as Loads: One by one, apply a unit value of each redundant force ($X_1, X_2, … X_n$) to the primary structure (with no external loads).
5. Calculate Flexibility Coefficients: For each unit redundant load $X_j=1$, calculate the displacements ($f_{ij}$) at all ‘n’ redundant locations ‘i’. These are the flexibility coefficients.
6. Formulate Compatibility Equations: For each redundant location ‘i’, write an equation stating that the total displacement (from external loads + all redundant forces) must equal the known final displacement of the original structure (which is usually zero, e.g., at a support). This is the principle of superposition. For $n=2$, the equations would be: $\Delta_{1L} + f_{11}X_1 + f_{12}X_2 = 0$ $\Delta_{2L} + f_{21}X_1 + f_{22}X_2 = 0$ In matrix form: $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants: Solve the ‘n’ simultaneous compatibility equations for the ‘n’ unknown redundant forces $\{X\}$.
8. Find Final Forces/Moments: Place the now-known redundant forces (along with the original external loads) back onto the primary structure. Use static equilibrium to find all other reactions, shear forces, and bending moments.

Statement of the Three-Moment Theorem: The Three-Moment Theorem (or Clapeyron’s Theorem) gives a relationship between the internal bending moments at three successive, continuous supports ($A$, $B$, and $C$) in a continuous beam.

For a general case with no support settlement and constant $EI$ within each span, the equation is:

$$M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = -\frac{6A_1 \bar{x}_1}{I_1 L_1} – \frac{6A_2 \bar{x}_2}{I_2 L_2}$$

Derivation/Proof Concept:

1. Primary Structure: The continuous beam is split into a series of simply supported beams. The unknown support moments ($M_A, M_B, M_C$, etc.) are the redundants.
2. Superposition: The final slope at any support (e.g., support $B$) is the sum of: The slope caused by the external loads on the simply supported span, and the slopes caused by the redundant moments $M_A$, $M_B$, and $M_C$.
3. Calculate Slopes: The slopes are calculated using the Moment-Area Theorem or the Unit Load Method.
   • The rotation at $B$ for span $AB$ ($\theta_{BA}$) is found.
   • The rotation at $B$ for span $BC$ ($\theta_{BC}$) is found.
4. Apply Compatibility: The compatibility condition is that the structure is continuous over support $B$, so the final slope just to the left of $B$ must be equal to the final slope just to the right of $B$ (with opposite signs by convention): $\theta_{BA} = -\theta_{BC}$, or $\theta_{BA} + \theta_{BC} = 0$.
5. Assemble Equation: Substituting the expressions for these slopes (in terms of the moments $M_A, M_B, M_C$ and the load terms $A\bar{x}$) into the compatibility equation yields the Three-Moment Theorem.

The theorem is fundamentally a compatibility equation. It enforces that the slope of the elastic curve at the central support ($B$) must be continuous. It states that the slope of the beam at the end $B$ of span $AB$ ($\theta_{BA}$) must be equal and opposite to the slope at the end $B$ of span $BC$ ($\theta_{BC}$), i.e., $\theta_{BA} + \theta_{BC} = 0$. The theorem relates the rotations caused by the support moments ($M_A, M_B, M_C$) and the rotations caused by the external loads (the $A\bar{x}$ terms) and ensures this continuity of slope is satisfied.

Limitations of the Force Method:

1. Choice of Primary Structure: The first step requires choosing a primary structure. This choice is not unique, and a poor choice can lead to a flexibility matrix that is ill-conditioned (difficult to solve accurately) or makes the calculations for displacements (flexibility coefficients) very difficult.
2. Manual Computation: For structures with a high degree of indeterminacy, the force method becomes computationally intensive. It requires setting up and solving a large system of simultaneous equations (size $n \times n$, where $n$ is the DSI).
3. Computer Implementation: The method is more difficult to automate for general-purpose structural analysis software compared to the “Displacement Method” (Stiffness Method). The stiffness method follows a more direct, programmable algorithm that doesn’t require the user to select a primary structure.
4. Flexibility Coefficient Calculation: Calculating the displacements ($\Delta_L$ and $f_{ij}$) can be a tedious process, often requiring methods like the unit load method, virtual work, or Castigliano’s theorem for each coefficient.

Force Method of Analysis: The Force Method (also known as the Flexibility Method or Consistent Deformation Method) is a procedure for analyzing statically indeterminate structures. Its primary objective is to determine the unknown redundant forces. The method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

Limitations of the Force Method:

1. Choice of Primary Structure: The first step requires choosing a primary structure. This choice is not unique, and a poor choice can lead to a flexibility matrix that is ill-conditioned (difficult to solve accurately) or makes the calculations for displacements (flexibility coefficients) very difficult.
2. Manual Computation: For structures with a high degree of indeterminacy, the force method becomes computationally intensive. It requires setting up and solving a large system of simultaneous equations (size $n \times n$, where $n$ is the DSI).
3. Computer Implementation: The method is more difficult to automate for general-purpose structural analysis software compared to the “Displacement Method” (Stiffness Method). The stiffness method follows a more direct, programmable algorithm that doesn’t require the user to select a primary structure.
4. Flexibility Coefficient Calculation: Calculating the displacements ($\Delta_L$ and $f_{ij}$) can be a tedious process, often requiring methods like the unit load method, virtual work, or Castigliano’s theorem for each coefficient.

Primary Structure: A primary structure is a structure that is derived from the original indeterminate structure by removing a sufficient number of restraints (supports or internal forces) to make it statically determinate and stable. The choice of the primary structure is not unique, but it must be stable.

Redundant Force (or Redundant): A redundant force (or “redundant”) is an unknown force or moment that is in excess of what is required to maintain static equilibrium. In the force method, these are the forces or moments that are “released” (removed) from the original structure to create the primary structure. The number of redundants is equal to the degree of static indeterminacy.

Procedures for Force Method:

1. Determine Degree of Indeterminacy (DSI): Find the number of redundant forces (let’s say ‘n’).
2. Select a Primary Structure: Remove the ‘n’ redundant forces/moments from the original structure to create a statically determinate and stable primary structure.
3. Apply External Loads: Calculate the displacements (or rotations) at the points where redundants were removed, caused only by the original external loads on the primary structure. Let’s call these displacements $\{\Delta_L\}$.
4. Apply Redundants as Loads: One by one, apply a unit value of each redundant force ($X_1, X_2, … X_n$) to the primary structure (with no external loads).
5. Calculate Flexibility Coefficients: For each unit redundant load $X_j=1$, calculate the displacements ($f_{ij}$) at all ‘n’ redundant locations ‘i’. These are the flexibility coefficients.
6. Formulate Compatibility Equations: For each redundant location ‘i’, write an equation stating that the total displacement (from external loads + all redundant forces) must equal the known final displacement of the original structure (which is usually zero, e.g., at a support). This is the principle of superposition. For $n=2$, the equations would be: $\Delta_{1L} + f_{11}X_1 + f_{12}X_2 = 0$ $\Delta_{2L} + f_{21}X_1 + f_{22}X_2 = 0$ In matrix form: $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants: Solve the ‘n’ simultaneous compatibility equations for the ‘n’ unknown redundant forces $\{X\}$.
8. Find Final Forces/Moments: Place the now-known redundant forces (along with the original external loads) back onto the primary structure. Use static equilibrium to find all other reactions, shear forces, and bending moments.

Advantages of the Force Method:

1. Fewer Unknowns for Low DSI: For structures with a low degree of static indeterminacy (DSI) but a high degree of kinematic indeterminacy (DOK), the force method is often more efficient. For example, a two-span continuous beam (DSI=1, DOK=2) is easier to solve by hand using the force method.
2. Conceptual Simplicity: The underlying concept of superposition and restoring compatibility is very intuitive and provides a good physical understanding of how indeterminate structures behave.
3. Accuracy: The method is exact (within the assumptions of linear elastic theory).
4. Foundation for Other Methods: It forms the basis for understanding other classical methods, such as the Three-Moment Theorem and Castigliano’s theorems.

Definition of Flexibility: Flexibility is the inverse of stiffness. It represents the displacement caused by a unit force. A flexibility coefficient, $f_{ij}$, is the displacement at point ‘i’ due to a unit load at point ‘j’.

Flexibility Matrix Method (Steps): This is the formal, matrix-based procedure for the Consistent Deformation method, especially useful for structures with a high degree of indeterminacy.

1. Determine DSI (n): Find the number of redundants.
2. Establish Primary Structure: Remove ‘n’ redundants to create a determinate, stable structure.
3. Define Coordinates: Assign ‘n’ coordinates corresponding to the ‘n’ redundant forces $\{X\}$.
4. Calculate Displacement Vector due to Loads ($\{\Delta_L\}$): Find the displacements at each of the ‘n’ coordinates on the primary structure, caused only by the external loads. This results in a vector $\{\Delta_L\}$ of size $(n \times 1)$.
5. Calculate Flexibility Matrix ($[f]$): This is an $(n \times n)$ matrix. The $j$-th column of $[f]$ is found by applying a unit load at coordinate ‘j’ on the primary structure and calculating the displacements at all ‘n’ coordinates ($i=1$ to $n$). $f_{ij} =$ displacement at ‘i’ due to unit load at ‘j’. By Maxwell’s Reciprocal Theorem, $f_{ij} = f_{ji}$, so the matrix is symmetric.
6. Formulate Compatibility Equation: The total displacement at each coordinate is the sum of displacement from external loads and the displacement from all redundant forces. This total displacement must be zero (for rigid supports). $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants ($\{X\}$): Invert the flexibility matrix to solve for the unknown redundant forces. $\{X\} = -[f]^{-1}\{\Delta_L\}$
8. Find Final Response: Use superposition to find the final member forces, moments, or reactions in the original structure. Final Force = (Force in Primary Structure due to Loads) + $\sum$ (Force in Primary Structure due to Unit Redundant $j$) $\times (X_j)$

Three-Moment Equation Derivation:

The Three-Moment Theorem (or Clapeyron’s Theorem) gives a relationship between the internal bending moments at three successive, continuous supports ($A$, $B$, and $C$) in a continuous beam.

For a general case with no support settlement and constant $EI$ within each span, the equation is:

$$M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = -\frac{6A_1 \bar{x}_1}{I_1 L_1} – \frac{6A_2 \bar{x}_2}{I_2 L_2}$$

Derivation/Proof Concept:

1. Primary Structure: The continuous beam is split into a series of simply supported beams. The unknown support moments ($M_A, M_B, M_C$, etc.) are the redundants.
2. Superposition: The final slope at any support (e.g., support $B$) is the sum of: The slope caused by the external loads on the simply supported span, and the slopes caused by the redundant moments $M_A$, $M_B$, and $M_C$.
3. Calculate Slopes: The slopes are calculated using the Moment-Area Theorem or the Unit Load Method.
   • The rotation at $B$ for span $AB$ ($\theta_{BA}$) is found.
   • The rotation at $B$ for span $BC$ ($\theta_{BC}$) is found.
4. Apply Compatibility: The compatibility condition is that the structure is continuous over support $B$, so the final slope just to the left of $B$ must be equal to the final slope just to the right of $B$ (with opposite signs by convention): $\theta_{BA} = -\theta_{BC}$, or $\theta_{BA} + \theta_{BC} = 0$.
5. Assemble Equation: Substituting the expressions for these slopes (in terms of the moments $M_A, M_B, M_C$ and the load terms $A\bar{x}$) into the compatibility equation yields the Three-Moment Theorem.

The theorem is fundamentally a compatibility equation. It enforces that the slope of the elastic curve at the central support ($B$) must be continuous. It states that the slope of the beam at the end $B$ of span $AB$ ($\theta_{BA}$) must be equal and opposite to the slope at the end $B$ of span $BC$ ($\theta_{BC}$), i.e., $\theta_{BA} + \theta_{BC} = 0$. The theorem relates the rotations caused by the support moments ($M_A, M_B, M_C$) and the rotations caused by the external loads (the $A\bar{x}$ terms) and ensures this continuity of slope is satisfied.

Definition of Flexibility: Flexibility is the inverse of stiffness. It represents the displacement caused by a unit force. A flexibility coefficient, $f_{ij}$, is the displacement at point ‘i’ due to a unit load at point ‘j’.

Flexibility Matrix Method (Steps): This is the formal, matrix-based procedure for the Consistent Deformation method, especially useful for structures with a high degree of indeterminacy.

1. Determine DSI (n): Find the number of redundants.
2. Establish Primary Structure: Remove ‘n’ redundants to create a determinate, stable structure.
3. Define Coordinates: Assign ‘n’ coordinates corresponding to the ‘n’ redundant forces $\{X\}$.
4. Calculate Displacement Vector due to Loads ($\{\Delta_L\}$): Find the displacements at each of the ‘n’ coordinates on the primary structure, caused only by the external loads. This results in a vector $\{\Delta_L\}$ of size $(n \times 1)$.
5. Calculate Flexibility Matrix ($[f]$): This is an $(n \times n)$ matrix. The $j$-th column of $[f]$ is found by applying a unit load at coordinate ‘j’ on the primary structure and calculating the displacements at all ‘n’ coordinates ($i=1$ to $n$). $f_{ij} =$ displacement at ‘i’ due to unit load at ‘j’. By Maxwell’s Reciprocal Theorem, $f_{ij} = f_{ji}$, so the matrix is symmetric.
6. Formulate Compatibility Equation: The total displacement at each coordinate is the sum of displacement from external loads and the displacement from all redundant forces. This total displacement must be zero (for rigid supports). $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants ($\{X\}$): Invert the flexibility matrix to solve for the unknown redundant forces. $\{X\} = -[f]^{-1}\{\Delta_L\}$
8. Find Final Response: Use superposition to find the final member forces, moments, or reactions in the original structure. Final Force = (Force in Primary Structure due to Loads) + $\sum$ (Force in Primary Structure due to Unit Redundant $j$) $\times (X_j)$

Statement of the Three-Moment Theorem: The Three-Moment Theorem (or Clapeyron’s Theorem) gives a relationship between the internal bending moments at three successive, continuous supports ($A$, $B$, and $C$) in a continuous beam.

For a general case with no support settlement and constant $EI$ within each span, the equation is:

$$M_A \frac{L_1}{I_1} + 2M_B \left( \frac{L_1}{I_1} + \frac{L_2}{I_2} \right) + M_C \frac{L_2}{I_2} = -\frac{6A_1 \bar{x}_1}{I_1 L_1} – \frac{6A_2 \bar{x}_2}{I_2 L_2}$$

Physical Meaning: The theorem is fundamentally a compatibility equation. It enforces that the slope of the elastic curve at the central support ($B$) must be continuous. It states that the slope of the beam at the end $B$ of span $AB$ ($\theta_{BA}$) must be equal and opposite to the slope at the end $B$ of span $BC$ ($\theta_{BC}$), i.e., $\theta_{BA} + \theta_{BC} = 0$. The theorem relates the rotations caused by the support moments ($M_A, M_B, M_C$) and the rotations caused by the external loads (the $A\bar{x}$ terms) and ensures this continuity of slope is satisfied.

Derivation/Proof Concept:

1. Primary Structure: The continuous beam is split into a series of simply supported beams. The unknown support moments ($M_A, M_B, M_C$, etc.) are the redundants.
2. Superposition: The final slope at any support (e.g., support $B$) is the sum of: The slope caused by the external loads on the simply supported span, and the slopes caused by the redundant moments $M_A$, $M_B$, and $M_C$.
3. Calculate Slopes: The slopes are calculated using the Moment-Area Theorem or the Unit Load Method.
   • The rotation at $B$ for span $AB$ ($\theta_{BA}$) is found.
   • The rotation at $B$ for span $BC$ ($\theta_{BC}$) is found.
4. Apply Compatibility: The compatibility condition is that the structure is continuous over support $B$, so the final slope just to the left of $B$ must be equal to the final slope just to the right of $B$ (with opposite signs by convention): $\theta_{BA} = -\theta_{BC}$, or $\theta_{BA} + \theta_{BC} = 0$.
5. Assemble Equation: Substituting the expressions for these slopes (in terms of the moments $M_A, M_B, M_C$ and the load terms $A\bar{x}$) into the compatibility equation yields the Three-Moment Theorem.

Neutral Point (or Point of Zero Shear/Point of Contraflexure): A “neutral point” in an unloaded span of a continuous beam (typically discussed in the context of influence lines) is the point in that span where the shear force is zero. This term is not standard in the same way as a “point of contraflexure” (zero moment).

A more likely interpretation of this question, especially given its context, might be the “Point of Contraflexure” or “Inflection Point,” which is where the bending moment is zero.

Assuming “Neutral Point” refers to “Point of Contraflexure” (Zero Moment): In an unloaded span of a continuous beam (e.g., span $BC$, with loads only on $AB$ or $CD$), the bending moment diagram is purely a straight line, connecting the support moments $M_B$ and $M_C$. If $M_B$ and $M_C$ have opposite signs, the moment diagram will cross the zero axis. This point of zero moment is the point of contraflexure (or “neutral point” in this context). Its location can be found by simple similar triangles on the moment diagram.

Derivation of Recurrent Formula: If the question is about a “recurrent formula,” it likely refers to an iterative application of the Three-Moment Theorem, which is itself a “recurrent formula” as it is applied to successive pairs of spans.

For an unloaded span $BC$ between loaded spans $AB$ and $CD$:

1. Apply the Three-Moment Theorem to spans $AB$ and $BC$. This gives a relation between $M_A$, $M_B$, and $M_C$, with load terms from $AB$.
2. Apply the Three-Moment Theorem to spans $BC$ and $CD$. This gives a relation between $M_B$, $M_C$, and $M_D$, with load terms from $CD$.
3. Since span $BC$ is unloaded, its load term ($6A\bar{x}/IL$) is zero.
4. Solving this system of equations gives the moments $M_B$ and $M_C$.

The “neutral point” (point of zero moment) $x$ from support $B$ is found from the linear moment equation: $M(x) = M_B + (M_C – M_B) \frac{x}{L_{BC}}$.

Set $M(x) = 0$ and solve for $x$:

$$x = – \frac{M_B \cdot L_{BC}}{M_C – M_B}$$

This $x$ is the location of the neutral point, and the formula to find it is “recurrent” in the sense that it depends on the moments $M_B$ and $M_C$ which are found using the recurrent Three-Moment Theorem.

Compatibility Conditions: Compatibility conditions are equations that enforce the geometric continuity and boundary conditions of a structure. In the context of the force method, they state that the final deformation of the original, indeterminate structure must be consistent with its supports.

Physical Meaning: The physical meaning is that the structure must “fit together” without any gaps or overlaps, and it must respect its support conditions.

For example, when we use the method of consistent deformation:

1. We remove a redundant (e.g., a roller support) to get a primary structure (e.g., a cantilever beam).
2. We apply the external loads, and the primary structure deflects at the removed support (e.g., the cantilever tip moves down by $\Delta_B$).
3. We then apply the redundant force (e.g., $R_B$) as an unknown load on the primary structure, causing a deflection in the opposite direction (e.g., the tip moves up by $\Delta_{BB} = R_B \cdot f_{BB}$).
4. The compatibility condition enforces the original structure’s geometry. Since the original structure had a roller support at that point, the total vertical deflection must be zero.
5. Therefore, the compatibility equation is: $\Delta_B + R_B \cdot f_{BB} = 0$. This equation allows us to solve for the unknown redundant $R_B$.

Redundant Force (or Redundant): A redundant force (or “redundant”) is an unknown force or moment that is in excess of what is required to maintain static equilibrium. In the force method, these are the forces or moments that are “released” (removed) from the original structure to create the primary structure. The number of redundants is equal to the degree of static indeterminacy.

Example of Selecting Redundants: Consider a propped cantilever beam (fixed at one end, roller at the other).

It has 4 unknown reactions (Moment, Vertical, and Horizontal at the fixed end; Vertical at the roller) and 3 equilibrium equations.

Degree of Indeterminacy = 4 – 3 = 1.

We need to select one redundant.

Option 1: Choose the vertical reaction at the roller support as the redundant ($R_B$). The primary structure becomes a simple cantilever beam (which is determinate and stable).

Option 2: Choose the moment reaction at the fixed support as the redundant ($M_A$). The primary structure becomes a simply supported beam (which is also determinate and stable).

The choice depends on which primary structure is easier to analyze.

Consistent Deformation Method (Force Method): The Consistent Deformation Method (also known as the Force Method or Flexibility Method) is a procedure for analyzing statically indeterminate structures. As the PDF snippet for Page 1 states, its primary objective is to determine the unknown redundant forces.

The “consistent deformation” name comes from the core principle: the method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

Procedures for Analysis:

1. Determine Degree of Indeterminacy (DSI): Find the number of redundant forces (let’s say ‘n’).
2. Select a Primary Structure: Remove the ‘n’ redundant forces/moments from the original structure to create a statically determinate and stable primary structure.
3. Apply External Loads: Calculate the displacements (or rotations) at the points where redundants were removed, caused only by the original external loads on the primary structure. Let’s call these displacements $\{\Delta_L\}$.
4. Apply Redundants as Loads: One by one, apply a unit value of each redundant force ($X_1, X_2, … X_n$) to the primary structure (with no external loads).
5. Calculate Flexibility Coefficients: For each unit redundant load $X_j=1$, calculate the displacements ($f_{ij}$) at all ‘n’ redundant locations ‘i’. These are the flexibility coefficients.
6. Formulate Compatibility Equations: For each redundant location ‘i’, write an equation stating that the total displacement (from external loads + all redundant forces) must equal the known final displacement of the original structure (which is usually zero, e.g., at a support). This is the principle of superposition. For $n=2$, the equations would be: $\Delta_{1L} + f_{11}X_1 + f_{12}X_2 = 0$ $\Delta_{2L} + f_{21}X_1 + f_{22}X_2 = 0$ In matrix form: $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants: Solve the ‘n’ simultaneous compatibility equations for the ‘n’ unknown redundant forces $\{X\}$.
8. Find Final Forces/Moments: Place the now-known redundant forces (along with the original external loads) back onto the primary structure. Use static equilibrium to find all other reactions, shear forces, and bending moments.

Consistent Deformation Method (Force Method): The Consistent Deformation Method (also known as the Force Method or Flexibility Method) is a procedure for analyzing statically indeterminate structures. Its primary objective is to determine the unknown redundant forces.

The “consistent deformation” name comes from the core principle: the method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

The method involves:

1. Determining the degree of static indeterminacy
2. Selecting a primary structure by removing redundant forces
3. Calculating displacements caused by external loads
4. Calculating flexibility coefficients
5. Writing compatibility equations
6. Solving for redundant forces
7. Finding final forces and moments through superposition

Consistent Deformation Method (Force Method): The Consistent Deformation Method (also known as the Force Method or Flexibility Method) is a procedure for analyzing statically indeterminate structures. As the PDF snippet for Page 1 states, its primary objective is to determine the unknown redundant forces.

The “consistent deformation” name comes from the core principle: the method involves writing compatibility equations that ensure the deformations of the (simplified) primary structure are consistent with the boundary conditions (supports) and continuity of the original, indeterminate structure.

Procedures for Analysis:

1. Determine Degree of Indeterminacy (DSI): Find the number of redundant forces (let’s say ‘n’).
2. Select a Primary Structure: Remove the ‘n’ redundant forces/moments from the original structure to create a statically determinate and stable primary structure.
3. Apply External Loads: Calculate the displacements (or rotations) at the points where redundants were removed, caused only by the original external loads on the primary structure. Let’s call these displacements $\{\Delta_L\}$.
4. Apply Redundants as Loads: One by one, apply a unit value of each redundant force ($X_1, X_2, … X_n$) to the primary structure (with no external loads).
5. Calculate Flexibility Coefficients: For each unit redundant load $X_j=1$, calculate the displacements ($f_{ij}$) at all ‘n’ redundant locations ‘i’. These are the flexibility coefficients.
6. Formulate Compatibility Equations: For each redundant location ‘i’, write an equation stating that the total displacement (from external loads + all redundant forces) must equal the known final displacement of the original structure (which is usually zero, e.g., at a support). This is the principle of superposition. For $n=2$, the equations would be: $\Delta_{1L} + f_{11}X_1 + f_{12}X_2 = 0$ $\Delta_{2L} + f_{21}X_1 + f_{22}X_2 = 0$ In matrix form: $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants: Solve the ‘n’ simultaneous compatibility equations for the ‘n’ unknown redundant forces $\{X\}$.
8. Find Final Forces/Moments: Place the now-known redundant forces (along with the original external loads) back onto the primary structure. Use static equilibrium to find all other reactions, shear forces, and bending moments.

Definition of Flexibility: Flexibility is the inverse of stiffness. It represents the displacement caused by a unit force. A flexibility coefficient, $f_{ij}$, is the displacement at point ‘i’ due to a unit load at point ‘j’.

Flexibility Matrix Method (Steps): This is the formal, matrix-based procedure for the Consistent Deformation method, especially useful for structures with a high degree of indeterminacy.

1. Determine DSI (n): Find the number of redundants.
2. Establish Primary Structure: Remove ‘n’ redundants to create a determinate, stable structure.
3. Define Coordinates: Assign ‘n’ coordinates corresponding to the ‘n’ redundant forces $\{X\}$.
4. Calculate Displacement Vector due to Loads ($\{\Delta_L\}$): Find the displacements at each of the ‘n’ coordinates on the primary structure, caused only by the external loads. This results in a vector $\{\Delta_L\}$ of size $(n \times 1)$.
5. Calculate Flexibility Matrix ($[f]$): This is an $(n \times n)$ matrix. The $j$-th column of $[f]$ is found by applying a unit load at coordinate ‘j’ on the primary structure and calculating the displacements at all ‘n’ coordinates ($i=1$ to $n$). $f_{ij} =$ displacement at ‘i’ due to unit load at ‘j’. By Maxwell’s Reciprocal Theorem, $f_{ij} = f_{ji}$, so the matrix is symmetric.
6. Formulate Compatibility Equation: The total displacement at each coordinate is the sum of displacement from external loads and the displacement from all redundant forces. This total displacement must be zero (for rigid supports). $\{\Delta_L\} + [f]\{X\} = \{0\}$
7. Solve for Redundants ($\{X\}$): Invert the flexibility matrix to solve for the unknown redundant forces. $\{X\} = -[f]^{-1}\{\Delta_L\}$
8. Find Final Response: Use superposition to find the final member forces, moments, or reactions in the original structure. Final Force = (Force in Primary Structure due to Loads) + $\sum$ (Force in Primary Structure due to Unit Redundant $j$) $\times (X_j)$

The compatibility equation for a two-hinged arch (which is statically indeterminate to the first degree) is derived by setting the net horizontal displacement at one of the supports (say, support B) equal to any known displacement, such as yielding ($\delta_H$).

The total horizontal displacement ($\Delta_B$) is the sum of the displacement on the primary (determinate) structure due to all effects (loads, temperature) and the displacement caused by the redundant horizontal thrust $H$.

Compatibility Equation:

$$\Delta_{B, \text{loads}} + \Delta_{B, \text{temp}} + \Delta_{B, \text{H}} = \Delta_{B, \text{yield}}$$

This can be expanded into the full compatibility equation used to solve for $H$:

$$\left( \int \frac{M_s y ds}{EI} \right) + (L \alpha t) – H \left( \int \frac{y^2 ds}{EI} + \int \frac{\cos^2 \theta ds}{AE} \right) = \delta_H$$

Where:

• External Loads: $\int \frac{M_s y ds}{EI}$ is the horizontal displacement from external loads, where $M_s$ is the simple beam moment on the primary structure.
• Temperature: $L \alpha t$ is the horizontal displacement due to a uniform temperature rise $t$ ($\alpha$ = coefficient of thermal expansion, $L$ = span).
• Redundant Thrust $H$: $- H \left( \int \frac{y^2 ds}{EI} \right)$ is the horizontal displacement (inward) caused by $H$ due to bending.
• Rib Shortening: $- H \left( \int \frac{\cos^2 \theta ds}{AE} \right)$ is the horizontal displacement (inward) caused by $H$ due to axial compression (rib shortening). $A$ is the cross-sectional area and $E$ is the modulus of elasticity.
• Support Yielding: $\delta_H$ is the known horizontal displacement (outward) of the support.

To find the horizontal thrust $H$, we use the compatibility equation, ignoring temperature, rib shortening, and yield effects for this derivation.

The general formula for $H$ is:

$$H = \frac{\int \frac{M_s y ds}{EI}}{\int \frac{y^2 ds}{EI}}$$

This integral is complex. A common and reasonable simplification for arches is to assume that the flexural rigidity varies as $EI = EI_c \sec \theta$, where $EI_c$ is the rigidity at the crown. This simplifies the differential $ds/EI$ to $dx/EI_c$.

Assuming $EI = EI_c \sec \theta$:

$$H = \frac{\int_0^L M_s y dx}{\int_0^L y^2 dx}$$

1. Define Geometry and Loads:

Parabolic Arch: The equation for a parabola with origin at the left hinge (A) is:
$y = \frac{4hx}{L^2}(L-x)$

UDL: For a uniformly distributed load $w$ over the entire span, the simple beam moment $M_s$ at any point $x$ is:
$M_s = \frac{wLx}{2} – \frac{wx^2}{2} = \frac{wx(L-x)}{2}$

2. Calculate the Numerator ($\int_0^L M_s y dx$):

$$\text{Numerator} = \int_0^L \left[ \frac{wx(L-x)}{2} \right] \left[ \frac{4hx(L-x)}{L^2} \right] dx$$

$$= \frac{2wh}{L^2} \int_0^L x^2 (L-x)^2 dx$$

$$= \frac{2wh}{L^2} \int_0^L (L^2 x^2 – 2Lx^3 + x^4) dx$$

$$= \frac{2wh}{L^2} \left[ \frac{L^2 x^3}{3} – \frac{2Lx^4}{4} + \frac{x^5}{5} \right]_0^L$$

$$= \frac{2wh}{L^2} \left( \frac{L^5}{3} – \frac{L^5}{2} + \frac{L^5}{5} \right)$$

$$= \frac{2wh}{L^2} \left( \frac{10 – 15 + 6}{30} \right) L^5 = \frac{2wh}{L^2} \left( \frac{L^5}{30} \right) = \frac{whL^3}{15}$$

3. Calculate the Denominator ($\int_0^L y^2 dx$):

$$\text{Denominator} = \int_0^L \left[ \frac{4hx(L-x)}{L^2} \right]^2 dx$$

$$= \frac{16h^2}{L^4} \int_0^L x^2 (L-x)^2 dx$$

We already solved this integral in the numerator: $\int_0^L x^2(L-x)^2 dx = \frac{L^5}{30}$

$$= \frac{16h^2}{L^4} \left( \frac{L^5}{30} \right) = \frac{16h^2 L}{30} = \frac{8h^2 L}{15}$$

4. Find H:

$$H = \frac{\text{Numerator}}{\text{Denominator}} = \frac{whL^3 / 15}{8h^2 L / 15}$$

$$H = \frac{wL^2}{8h}$$

Cause: An arch, by its nature, carries load primarily through axial compression (thrust, $N_x$). This constant compressive force causes the arch rib material to shorten slightly along its length, just as a column would. This effect is known as rib shortening.

Mechanism: This physical shortening of the arch’s length causes the two ends (hinges) to pull inward toward each other.

Effect on Thrust: The arch’s horizontal thrust $H$ is generated to resist the outward spread caused by vertical loads. Since rib shortening causes an inward pull, it partially counteracts the outward spread. This “relieves” the arch, meaning less horizontal thrust $H$ is needed to maintain compatibility.

Effect on Moment: The bending moment in the arch is given by $M_x = M_s – Hy$. Since rib shortening reduces $H$, the negative (hogging) moment component $Hy$ is smaller. This results in a larger net bending moment $M_x$ in the arch rib.

In summary: Rib shortening reduces the horizontal thrust ($H$) but increases the bending moment in the arch.

The procedure for drawing the Influence Line Diagram (ILD) for the horizontal thrust $H$ can be explained using the Müller-Breslau Principle.

State the Principle: The Müller-Breslau Principle states that the ILD for a redundant reaction (like $H$) is, to some scale, the deflected shape of the structure when that restraint is removed and a unit displacement is applied in the direction of the reaction.

Apply the Principle (Procedure):

Step 1: Start with the two-hinged arch. The horizontal thrust $H$ is the redundant.

Step 2: Remove the restraint corresponding to $H$. This is done by replacing one hinge (e.g., at support B) with a roller hinge, creating a statically determinate primary structure.

Step 3: Apply a unit horizontal displacement ($\Delta_H = 1$) at the roller (support B), pushing it inward toward support A.

Step 4: The resulting deflected shape of the arch rib is the qualitative shape of the ILD for $H$. The ordinates of this shape are proportional to the actual ILD ordinates.

Finding Numerical Values (Practical Procedure):
To plot the ILD numerically, you must find the value of $H$ for a single unit load ($P=1$) moving across the span.

Step 1: Place a unit load $P=1$ at a specific point $k$ along the span.

Step 2: Calculate the simple beam moment diagram $M_{s,k}$ for this unit load.

Step 3: Calculate $H$ using the formula:

$$H(k) = \frac{\int M_{s,k} \cdot y \cdot ds}{ \int y^2 ds }$$

(Assuming the $EI = EI_c \sec \theta$ simplification, $ds$ can be replaced with $dx$).

Step 4: The denominator $\int y^2 dx$ is a constant property of the arch (e.g., $\frac{8h^2 L}{15}$ for a parabola). The numerator $\int M_{s,k} \cdot y \cdot dx$ is the value that changes as the load moves.

Step 5: Repeat this calculation for several points $k$ across the span $L$ and plot the resulting values of $H(k)$. This plotted curve is the ILD for $H$.

Cause: Support yielding is the non-rigid horizontal movement of the arch supports. When the arch pushes outward under load, an imperfect foundation may “yield” or spread apart by a small amount, $\delta_H$.

Mechanism: The horizontal thrust $H$ is the reaction that prevents the arch supports from spreading. If the supports do spread (yield) outward, they are “giving in” to the outward push from the loads.

Effect on Thrust: This yielding “relieves” the arch. Since the support moves, the arch does not need to generate as much internal thrust to maintain compatibility. Therefore, an outward yielding of the supports reduces the horizontal thrust $H$.

Effect on Moment: The bending moment in the arch is given by $M_x = M_s – Hy$. Since support yielding reduces $H$, the negative (hogging) moment component $Hy$ is smaller. This results in a larger net bending moment $M_x$ in the arch rib.

In summary: Support yielding (spreading) reduces the horizontal thrust ($H$) but increases the bending moment in the arch.

The slope-deflection equation for a typical span AB is:

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

Where:

• $M_{AB}$: The final end moment at support A for span AB.
• $M_{FAB}$: The Fixed End Moment at support A for span AB, due to external loads.
• $E$: The Modulus of Elasticity of the beam material.
• $I$: The Moment of Inertia of the beam section.
• $L$: The length of the span AB.
• $\theta_A$: The rotation (slope) of the joint at support A, in radians.
• $\theta_B$: The rotation (slope) of the joint at support B, in radians.
• $\psi$: The chord rotation of the span AB, where $\psi = \frac{\Delta}{L}$.
• $\Delta$: The relative settlement of support B with respect to support A.

The corresponding equation for the other end (B) is:

$M_{BA} = M_{FBA} + \frac{2EI}{L} (2\theta_B + \theta_A – 3\psi)$

Assumptions:

• Deformations are due to bending only. Axial and shear deformations are considered negligible.
• All joints in the structure are rigid, meaning the original angle between members meeting at a joint is maintained after deformation.
• The material is linear, elastic, and follows Hooke’s Law.

Derivation:

The final moment at the end of a member is the algebraic sum of:

1. The Fixed End Moment (FEM) due to external loads (assuming ends are fixed).
2. The moment caused by the rotation $\theta_A$ at the near end (A).
3. The moment caused by the rotation $\theta_B$ at the far end (B).
4. The moment caused by the relative settlement $\Delta$ between supports (chord rotation $\psi = \Delta/L$).

We can derive this using the principle of superposition.

Step 1: Fixed End Moments (FEMs)

Assume the beam is fixed at A and B. The external loads (e.g., UDL, point loads) will induce fixed end moments, $M_{FAB}$ and $M_{FBA}$.

Step 2: Moment due to rotation $\theta_A$

If we apply a rotation $\theta_A$ at end A (while B is fixed and no load), the moments induced are:

At the near end A: $M’_{AB} = \frac{4EI}{L} \theta_A$

At the far end B (carry-over): $M’_{BA} = \frac{2EI}{L} \theta_A$

Step 3: Moment due to rotation $\theta_B$

Similarly, if we apply a rotation $\theta_B$ at end B (while A is fixed and no load), the moments induced are:

At the near end B: $M”_{BA} = \frac{4EI}{L} \theta_B$

At the far end A (carry-over): $M”_{AB} = \frac{2EI}{L} \theta_B$

Step 4: Moment due to chord rotation $\psi$ (Settlement $\Delta$)

If support B settles by $\Delta$ relative to A, it causes a clockwise chord rotation $\psi = \Delta/L$. The moments required to hold the ends fixed (i.e., prevent them from rotating with the chord) are:

At end A: $M”’_{AB} = -\frac{6EI}{L^2} \Delta = -\frac{6EI}{L} \psi$

At end B: $M”’_{BA} = -\frac{6EI}{L^2} \Delta = -\frac{6EI}{L} \psi$

(The negative sign is because a clockwise chord rotation $\psi$ requires counter-clockwise moments to maintain the fixed condition).

Step 5: Superposition

The final end moments are the sum of all components from steps 1-4.

For end A ($M_{AB}$):

$M_{AB} = M_{FAB} + M’_{AB} + M”_{AB} + M”’_{AB}$

$M_{AB} = M_{FAB} + \frac{4EI}{L} \theta_A + \frac{2EI}{L} \theta_B – \frac{6EI}{L} \psi$

Factoring out $\frac{2EI}{L}$:

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

For end B ($M_{BA}$):

$M_{BA} = M_{FBA} + M’_{BA} + M”_{BA} + M”’_{BA}$

$M_{BA} = M_{FBA} + \frac{2EI}{L} \theta_A + \frac{4EI}{L} \theta_B – \frac{6EI}{L} \psi$

Factoring out $\frac{2EI}{L}$:

$M_{BA} = M_{FBA} + \frac{2EI}{L} (2\theta_B + \theta_A – 3\psi)$

Definition of Fixed End Moments (FEMs):

Fixed End Moments are the reaction moments developed at the ends of a structural member (like a beam) when its ends are assumed to be fully fixed (i.e., prevented from rotating) and subjected to external loads. They are a key component in the slope-deflection method and moment distribution method.

Derivation for Fixed-Fixed Beam with UDL ‘w’:

We can use the Moment-Area method or superposition. Let’s use superposition.

Consider a simply supported beam AB of span L with a UDL ‘w’. The slopes at the ends are:

At A: $\theta_{A,load} = \frac{wL^3}{24EI}$ (clockwise, so positive)

At B: $\theta_{B,load} = -\frac{wL^3}{24EI}$ (counter-clockwise, so negative)

Now, we apply end moments $M_{FAB}$ and $M_{FBA}$ to this simply supported beam to make the final slopes zero (to replicate a fixed condition).

The slopes caused by the end moments are:

Slope at A due to $M_{FAB}$ and $M_{FBA}$: $\theta_{A,M} = \frac{M_{FAB} L}{3EI} + \frac{M_{FBA} L}{6EI}$

Slope at B due to $M_{FAB}$ and $M_{FBA}$: $\theta_{B,M} = \frac{M_{FAB} L}{6EI} + \frac{M_{FBA} L}{3EI}$

Note: This assumes $M_{FAB}$ and $M_{FBA}$ are clockwise (positive).

For a fixed beam, the total slope at each end is zero:

$\theta_A = \theta_{A,load} + \theta_{A,M} = 0$

$\frac{wL^3}{24EI} + \frac{M_{FAB} L}{3EI} + \frac{M_{FBA} L}{6EI} = 0$

Multiplying by $6EI/L$: $\frac{wL^2}{4} + 2M_{FAB} + M_{FBA} = 0$ (Eq. i)

$\theta_B = \theta_{B,load} + \theta_{B,M} = 0$

$-\frac{wL^3}{24EI} + \frac{M_{FAB} L}{6EI} + \frac{M_{FBA} L}{3EI} = 0$

Multiplying by $6EI/L$: $-\frac{wL^2}{4} + M_{FAB} + 2M_{FBA} = 0$ (Eq. ii)

By symmetry, the load ‘w’ will cause a counter-clockwise moment at A and a clockwise moment at B, so $M_{FBA} = -M_{FAB}$.

Let $M_{FBA} = M$. Then $M_{FAB} = -M$.

Substitute into (Eq. ii):

$-\frac{wL^2}{4} + (-M) + 2(M) = 0$

$-\frac{wL^2}{4} + M = 0 \implies M = \frac{wL^2}{4}$

This seems incorrect. Let’s re-check the superposition sign convention.

Let’s use the standard formulas and sign convention (Clockwise positive):

$\theta_{A,load} = +\frac{wL^3}{24EI}$

$\theta_{B,load} = -\frac{wL^3}{24EI}$

Slopes from moments (using flexibility coefficients, clockwise positive):

$\theta_A = (\frac{L}{3EI})M_A – (\frac{L}{6EI})M_B$

$\theta_B = -(\frac{L}{6EI})M_A + (\frac{L}{3EI})M_B$

This is for moments applied at A and B.

Let’s use the correct moment-rotation relationships (stiffness method):

$\theta_{A,M} = \frac{M_{FAB}L}{3EI} – \frac{M_{FBA}L}{6EI}$

$\theta_{B,M} = -\frac{M_{FAB}L}{6EI} + \frac{M_{FBA}L}{3EI}$

Total slope = 0:

$\theta_A = \frac{wL^3}{24EI} + \frac{M_{FAB}L}{3EI} – \frac{M_{FBA}L}{6EI} = 0$

$\frac{wL^2}{4} + 2M_{FAB} – M_{FBA} = 0$ (Eq. i)

$\theta_B = -\frac{wL^3}{24EI} – \frac{M_{FAB}L}{6EI} + \frac{M_{FBA}L}{3EI} = 0$

$-\frac{wL^2}{4} – M_{FAB} + 2M_{FBA} = 0$ (Eq. ii)

From symmetry, $M_{FBA} = -M_{FAB}$. Let $M_{FBA} = M$. Then $M_{FAB} = -M$.

Substitute into (Eq. ii):

$-\frac{wL^2}{4} – (-M) + 2(M) = 0$

$-\frac{wL^2}{4} + 3M = 0 \implies M = \frac{wL^2}{12}$

So:

$M_{FBA} = +\frac{wL^2}{12}$ (Clockwise)

$M_{FAB} = -\frac{wL^2}{12}$ (Counter-clockwise)

We use the general slope-deflection equations derived earlier, for a beam AB with no external loads (so $M_{FAB} = M_{FBA} = 0$).

General Equations:

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

$M_{BA} = \frac{2EI}{L} (2\theta_B + \theta_A – 3\psi)$

Case 1: Support Settlement ($\Delta$)

This case is for a fixed beam, so $\theta_A = 0$ and $\theta_B = 0$.

Assume support B settles down by $\Delta$.

This creates a clockwise chord rotation: $\psi = \frac{\Delta}{L}$.

Substitute these conditions into the general equations:

$M_{AB} = \frac{2EI}{L} (2(0) + (0) – 3(\frac{\Delta}{L})) = \frac{2EI}{L} (-\frac{3\Delta}{L})$

$M_{AB} = -\frac{6EI\Delta}{L^2}$ (Counter-clockwise)

$M_{BA} = \frac{2EI}{L} (2(0) + (0) – 3(\frac{\Delta}{L})) = \frac{2EI}{L} (-\frac{3\Delta}{L})$

$M_{BA} = -\frac{6EI\Delta}{L^2}$ (Counter-clockwise)

How settlement is incorporated: Support settlement ($\Delta$) is incorporated into the slope-deflection method as a chord rotation term, $\psi = \Delta/L$. This term generates end moments equal to $-\frac{6EI\Delta}{L^2}$ in the member, which are added to the fixed end moments and moments from joint rotations.

Case 2: Support Rotation ($\theta_A$)

This case is for a fixed beam, so $\theta_B = 0$ (far end is fixed).

There is no settlement, so $\Delta = 0$ and $\psi = 0$.

The left support A rotates clockwise by $\theta_A$.

Substitute these conditions into the general equations:

$M_{AB} = \frac{2EI}{L} (2\theta_A + (0) – 3(0)) = \frac{2EI}{L} (2\theta_A)$

$M_{AB} = \frac{4EI\theta_A}{L}$ (Clockwise)

$M_{BA} = \frac{2EI}{L} (2(0) + \theta_A – 3(0)) = \frac{2EI}{L} (\theta_A)$

$M_{BA} = \frac{2EI\theta_A}{L}$ (Clockwise, this is the “carry-over” moment)

The sign conventions used in the standard slope-deflection method are as follows:

• End Moments ($M$): Moments acting on the ends of the member (or on the joint) are considered positive if they are clockwise and negative if they are counter-clockwise.
• Rotations ($\theta$): Rotations of the joints (slopes) are considered positive if they are clockwise and negative if they are counter-clockwise.
• Chord Rotation ($\psi = \Delta/L$): The rotation of the member’s chord (due to support settlement) is considered positive if the chord rotates clockwise and negative if it rotates counter-clockwise. For a span AB, a settlement $\Delta$ at B relative to A causes a positive (clockwise) chord rotation.

The slope-deflection method is a kinematic method (also known as a displacement method or stiffness method) because its primary unknowns are kinematic quantities—that is, joint displacements and rotations.

In this method:

• The unknowns are the joint rotations ($\theta$) and joint translations (sway, $\Delta$).
• The governing equations (slope-deflection equations) relate the end moments to these unknown displacements.
• The final set of equations is solved by applying equilibrium equations (e.g., sum of moments at each joint is zero, $\Sigma M_{joint} = 0$, and shear equilibrium for sway, $\Sigma F_x = 0$).
• Once the unknown displacements ($\theta$ and $\Delta$) are found, the member end moments and other forces are calculated.

This is in contrast to a force method (or flexibility method), where the primary unknowns are redundant forces, and the governing equations are based on compatibility of displacements.

The slope-deflection method is applied to frames with sway (frames that are free to translate horizontally) by introducing an additional unknown and an additional equilibrium equation.

Application Steps:

1. Identify Unknowns: The unknowns include the joint rotations (e.g., $\theta_B, \theta_C$) just as in a non-sway frame, plus one or more unknown sway displacements ($\Delta$) for each story that can move.
2. Write Slope-Deflection Equations: The standard S-D equations are written for every member (beams and columns). For vertical columns, the chord rotation $\psi = \Delta/L$ (where $\Delta$ is the horizontal sway and L is the column height) will be non-zero. This $\Delta$ term (or $\psi$ term) appears in the S-D equations for all columns in that story.
3. Write Joint Equilibrium Equations: For each of the n unknown joint rotations ($\theta$), a moment equilibrium equation is written (e.g., $\Sigma M_B = 0, \Sigma M_C = 0$). This provides n equations.
4. Write Shear (Sway) Equilibrium Equation(s): Since there is an additional unknown ($\Delta$), an additional equilibrium equation is needed. This is the shear equation. A free-body diagram of the columns in the story is drawn. The horizontal shear force at the end of each column is expressed in terms of its end moments (e.g., $H_{AB} = (M_{AB} + M_{BA})/L$). The sum of all horizontal shear forces in the story’s columns must equal the sum of all external horizontal forces applied to that story ($\Sigma F_x = 0$). This shear equation provides the final (n+1)th equation.
5. Solve the System: The system of (n+1) linear equations is solved simultaneously to find all unknown $\theta$ values and the unknown sway $\Delta$.
6. Find Moments: The calculated $\theta$ and $\Delta$ values are substituted back into the individual S-D equations to find all the final member end moments.

The Moment Distribution Method is an iterative displacement method, developed by Prof. Hardy Cross, used to analyze statically indeterminate beams and frames.

Principle:

The method is based on the concept of successively correcting end moments by “unlocking” and “balancing” joints in a structure.

1. Lock Joints: All joints in the structure are initially assumed to be fixed (locked) against rotation.
2. Calculate FEMs: The Fixed-End Moments (FEMs) are calculated for all members based on the external loads.
3. Balancing: At each joint, the sum of all moments (FEMs and any moments carried over) will generally not be zero. This sum is the “unbalanced moment.” The joint is “unlocked,” and a balancing moment, which is equal in magnitude and opposite in sign to the unbalanced moment, is applied. This balancing moment is distributed among all members connected to the joint based on their relative stiffness (using Distribution Factors).
4. Carry-Over: When a balancing moment is applied to one end of a member, it induces a moment at the far end (if that end is fixed). This is the “carry-over moment.” Typically, half of the balancing moment is carried over to the far end (Carry-Over Factor = 0.5).
5. Iteration: This process of balancing and carrying over is repeated from joint to joint until the unbalanced moments become negligibly small.
6. Final Moments: The final end moment for each member is the algebraic sum of its initial FEM, all distributed balancing moments, and all received carry-over moments.

Example:

Consider a simple two-span continuous beam A-B-C, fixed at A and C, and on a roller support at B.

1. Lock Joint B: Assume joint B is fixed.
2. FEMs: Calculate FEM_AB, FEM_BA, FEM_BC, and FEM_CB due to loads on the spans.
3. Unbalanced Moment: At joint B, the unbalanced moment is M_unbalanced = FEM_BA + FEM_BC.
4. Balance: Apply a balancing moment M_balanced = -M_unbalanced to joint B. This moment is distributed to members BA and BC.
   • M_dist_BA = M_balanced * DF_BA
   • M_dist_BC = M_balanced * DF_BC
   • (Where DF is the Distribution Factor, DF_BA + DF_BC = 1).
5. Carry-Over:
   • Carry over (1/2) * M_dist_BA to joint A (member AB).
   • Carry over (1/2) * M_dist_BC to joint C (member CB).
6. Final Moments: Since A and C are fixed supports, they absorb these carry-over moments and the process stops (as there are no more joints to balance). The final moments are:
   • M_AB = FEM_AB + (1/2) * M_dist_BA
   • M_BA = FEM_BA + M_dist_BA
   • M_BC = FEM_BC + M_dist_BC
   • M_CB = FEM_CB + (1/2) * M_dist_BC

Stiffness Factor (K):

The stiffness factor of a member is the moment required to be applied at its near end to produce a unit rotation (one radian) at that end. The value depends on the support condition at the far end.

• Far End Fixed: For a prismatic member of length L and flexural rigidity EI, the moment M needed to rotate the near end by θ = 1 while the far end is fixed is K = 4EI/L.
• Far End Hinged/Pinned: If the far end is hinged, the moment needed to rotate the near end by θ = 1 is K = 3EI/L. This is often called the modified stiffness.

Distribution Factor (DF):

The distribution factor is a ratio that determines what proportion of an unbalanced moment at a joint is distributed to a specific member connected to that joint. It is the ratio of the stiffness factor (K) of that member to the sum of the stiffness factors of all members meeting at the joint.

Formula: DF_i = K_i / ΣK_j (where i is the member in question and j represents all members at the joint).

The sum of distribution factors at any joint must always equal 1 (ΣDF = 1).

A sketch would show a joint ‘J’ with members JA, JB, and JC. An unbalanced moment M_u at J is shown. Arrows pointing in the opposite direction on each member represent the distributed moments M_JA = -M_u * DF_JA, M_JB = -M_u * DF_JB, etc.

Carry-Over Moment (COM) & Carry-Over Factor (COF):

Carry-Over Moment: When a moment is applied to one end (the near end) of a member, it induces a moment at the far end, provided the far end is fixed. This induced moment at the far end is called the carry-over moment.

Carry-Over Factor: This is the ratio of the moment induced at the far end to the moment applied at the near end.

• Far End Fixed: For a prismatic member, if a moment M_A is applied at end A, it induces a moment M_B = +0.5 * M_A at the fixed end B. The Carry-Over Factor (COF) is +0.5. The positive sign indicates both moments are in the same direction (e.g., both clockwise).
• Far End Hinged: If the far end is hinged, it cannot resist a moment, so the carry-over moment is zero. The COF is 0.

A sketch would show a beam A-B, fixed at B. An applied moment M_A (clockwise) at end A is shown. This results in a carry-over moment M_B = 0.5 * M_A (also clockwise) at the fixed end B.

What is Sway?

Sway (or sidesway) is the lateral (horizontal) displacement of the joints of a frame. In portal frames, sway occurs when the frame is subjected to:

• Asymmetrical Loading: Unbalanced horizontal loads or asymmetrically placed vertical loads.
• Asymmetrical Geometry: The frame itself is not symmetric in terms of column lengths, moments of inertia (EI), or support conditions, even if the loading is symmetric.

Procedure for Sway Analysis:

The moment distribution method, in its basic form, only accounts for joint rotations (assuming no joint translation). To handle sway, the principle of superposition is used, breaking the problem into two parts:

Part 1: Non-Sway Analysis:

1. First, apply an artificial horizontal restraint (a “prop”) to the frame at a level that prevents any lateral movement (e.g., at the beam level).
2. Perform a standard moment distribution analysis on this propped, non-sway frame, considering all the actual external loads.
3. After finding the “non-sway moments” (M_O), calculate the horizontal shear forces in the columns.
4. Determine the magnitude of the propping force, P_R, that was required to prevent the sway. This is found from the horizontal static equilibrium equation for the frame.

Part 2: Sway-Only Analysis (Correction):

1. Now, analyze the frame for the effect of the sway force. This is done by applying a force equal and opposite to the propping force (-P_R) to the frame (with no external loads).
2. To do this, we assume an arbitrary lateral displacement (sway), Δ, for the frame.
3. This assumed sway Δ induces fixed-end moments (sway FEMs) in the columns (e.g., FEM = -6EIΔ/L²).
4. Perform a new moment distribution analysis using only these sway FEMs. This gives a set of moments (let’s call them M_S).
5. From these moments M_S, calculate the total horizontal shear force, P_S, that corresponds to the assumed sway Δ.
6. Since moments are linearly proportional to the sway force, the actual sway correction moments are found by scaling:
   • Correction Factor = (-P_R) / P_S
   • Actual Sway Moments = M_S * ((-P_R) / P_S)

Part 3: Final Moments (Superposition):

The final moments in the frame are the algebraic sum of the moments from Part 1 (non-sway) and Part 2 (sway correction).

M_Final = M_O + (M_S * ((-P_R) / P_S))

Stiffness modification is a technique used to simplify and speed up the moment distribution process. It adjusts the stiffness factor (K) of a member, which often eliminates iteration steps.

1. Member with a Hinged/Pinned End:

• Standard Stiffness: The standard stiffness factor for a member AB (calculating K_AB for joint A) assumes the far end B is fixed: K_AB = 4EI/L.
• Modified Stiffness: If the far end B is a known pin or hinge, its final moment must be zero. We can account for this from the start by using a modified stiffness for the member at joint A:
  • K_modified = 3EI/L
• Benefits:
  • Distribution Factor: This K_modified value is used to calculate the Distribution Factor (DF) at joint A.
  • No Carry-Over: When joint A is balanced, the carry-over factor from A to the hinged end B is zero. This is the main advantage. It prevents moments from being carried over to the pin (where they would have to be balanced back to zero anyway), thus simplifying the calculations and reducing the number of iterations.

2. Due to Symmetry:

Symmetry in a structure and its loading can be used to analyze only half of the structure, treating the centerline as a special support.

• Symmetrical Loading (No Sway):
  • If the structure and loading are perfectly symmetrical, the deformation is also symmetrical.
  • At the axis of symmetry, the slope must be zero (θ = 0).
  • A support that allows no rotation (θ = 0) is, by definition, a fixed support.
  • Procedure: We can analyze only half of the structure, treating the cut at the axis of symmetry as a fixed support. The standard stiffness K = 4EI/L is used for the member connecting to this virtual fixed support.
• Anti-Symmetrical Loading (Sway):
  • If the structure is symmetrical but the loading is anti-symmetrical (e.g., a horizontal force at the top), the deformation is anti-symmetrical.
  • At the axis of symmetry, the moment must be zero (M = 0) and there is lateral displacement (sway).
  • A support that allows rotation and translation but has zero moment (M = 0) is, by definition, a hinged support (or more accurately, a “guided roller” in this context, but for stiffness, it acts as a hinge).
  • Procedure: We can analyze only half of the structure, treating the cut at the axis of symmetry as a hinged support. The modified stiffness K_modified = 3EI/L can be used for the member connecting to this virtual hinge, simplifying the analysis.

Sway condition in portal frames is handled using the principle of superposition, breaking the analysis into two parts: non-sway analysis and sway-only analysis.

Procedure:

1. Non-Sway Analysis:
   • Apply an artificial horizontal restraint to prevent lateral movement.
   • Analyze the frame with actual loads using standard moment distribution.
   • Calculate the restraining force (P_R) required to prevent sway.
2. Sway-Only Analysis:
   • Apply the negative of the restraining force (-P_R) to the frame.
   • Assume an arbitrary sway displacement (Δ).
   • Calculate fixed-end moments due to this sway.
   • Perform moment distribution analysis for these sway moments.
   • Calculate the shear force (P_S) corresponding to the assumed sway.
3. Final Moments:
   • Scale the sway moments by the factor (-P_R/P_S).
   • Add non-sway moments and scaled sway moments to get final moments.

The moment distribution method for analyzing indeterminate frames follows the same principles as for beams, with additional considerations for frame geometry and potential sway.

Procedure:

1. Determine Degrees of Freedom: Identify all joints that can rotate and/or translate.
2. Calculate Stiffness Factors: Determine K = 4EI/L for each member with fixed far end, or 3EI/L for pinned far end.
3. Calculate Distribution Factors: DF = K_member/ΣK_joint for each member at every joint.
4. Calculate Fixed-End Moments: Determine FEMs for all members due to applied loads.
5. Moment Distribution Process:
   • Lock all joints initially
   • Unlock joints one by one, distribute unbalanced moments
   • Carry over moments to far ends
   • Repeat until moments converge
6. Check for Sway: If frame is unsymmetrical or has horizontal loads, perform sway analysis as described in Q.3.
7. Calculate Final Moments: Sum all distributed moments, carry-over moments, and initial FEMs.

Stiffness Factor (K):

The stiffness factor of a member is defined as the moment required to produce a unit rotation at the near end when the far end is fixed. For a prismatic member:

• K = 4EI/L (when far end is fixed)
• K = 3EI/L (when far end is hinged – modified stiffness)

Where E is modulus of elasticity, I is moment of inertia, and L is length of member.

Carry-Over Factor (C):

The carry-over factor is defined as the ratio of the moment induced at the far end to the moment applied at the near end. For a prismatic member:

• C = 0.5 (when far end is fixed)
• C = 0 (when far end is hinged)

The carry-over factor indicates how much of the moment applied at one end is transmitted to the other end.

Stiffness Factor (K):

The moment required to produce a unit rotation at the near end when the far end is fixed. For prismatic members:

• K = 4EI/L (far end fixed)
• K = 3EI/L (far end hinged)

Distribution Factor (DF):

The ratio of stiffness of a particular member to the sum of stiffnesses of all members meeting at a joint. It determines how much of the unbalanced moment at a joint is distributed to each member:

• DF = K_member / ΣK_joint
• ΣDF = 1 at any joint

Carry-Over Factor (C):

The ratio of moment induced at the far end to the moment applied at the near end:

• C = 0.5 (far end fixed)
• C = 0 (far end hinged)

Distribution Factor Concept:

Consider a joint J connected to three members JA, JB, and JC with stiffness factors K_JA, K_JB, and K_JC respectively. When an unbalanced moment M_u occurs at joint J:

• Total stiffness at joint: ΣK = K_JA + K_JB + K_JC
• Distribution factors:
  • DF_JA = K_JA / ΣK
  • DF_JB = K_JB / ΣK
  • DF_JC = K_JC / ΣK
• Distributed moments:
  • M_JA = -M_u × DF_JA
  • M_JB = -M_u × DF_JB
  • M_JC = -M_u × DF_JC

Carry-Over Factor Concept:

Consider a member AB fixed at end B. When a moment M_A is applied at end A:

• Carry-over moment at B: M_B = C × M_A
• For prismatic members with far end fixed: C = 0.5
• Therefore: M_B = 0.5 × M_A

[Note: A sketch would show a joint with three members and arrows indicating distributed moments, and a separate beam showing applied moment and carry-over moment.]

Carry-Over Factor:

The carry-over factor is defined as the ratio of the moment induced at the far end of a member to the moment applied at the near end. It represents how much moment is transmitted from one end of a member to the other end.

• For prismatic members with far end fixed: C = 0.5
• For prismatic members with far end hinged: C = 0

Distribution Factor:

The distribution factor at a joint for a particular member is the ratio of the stiffness of that member to the sum of stiffnesses of all members meeting at that joint. It determines the proportion of unbalanced moment that is distributed to each member.

• DF = K_member / ΣK_joint
• The sum of all distribution factors at a joint equals 1.

A stiffness matrix [K] is a matrix that relates a set of nodal forces (or moments) {P} to a corresponding set of nodal displacements (or rotations) {D} for a structure. The relationship is expressed by the fundamental equation of the stiffness method:

$$\{P\} = [K] \{D\}$$

The properties of the stiffness matrix are:

1. Square Matrix: The stiffness matrix is always a square matrix of size $n \times n$, where ‘n’ is the total number of degrees of freedom (coordinates) considered for the structure.
2. Symmetric: The matrix is symmetric, meaning $K_{ij} = K_{ji}$. This is a consequence of Maxwell’s Reciprocal Theorem.
3. Positive Diagonal Elements: All elements on the main diagonal ($K_{ii}$) are always positive. A diagonal element $K_{ii}$ represents the force required at coordinate ‘i’ to cause a unit displacement at that same coordinate ‘i’ (while all others are restrained). This force must be positive (in the same direction as the displacement) to do work on the structure.
4. Singularity (Unstable Structure): For an unsupported or unstable structure (one that can move as a rigid body), the stiffness matrix is singular, meaning its determinant is zero ($\det[K] = 0$). This is because a rigid-body motion {D} can occur with zero corresponding forces {P}.
5. Non-Singularity (Stable Structure): For a properly supported and stable structure, the stiffness matrix is non-singular (invertible). This allows for the unique solution of displacements: $\{D\} = [K]^{-1} \{P\}$.

A stiffness coefficient, $K_{ij}$, is a single element of the stiffness matrix [K]. It is defined as the force (or moment) required at coordinate ‘i’ to produce a unit displacement (or rotation) at coordinate ‘j’, while all other coordinates ($k \neq j$) are held restrained (i.e., their displacements are kept at zero).

The key differences between the force (flexibility) method and the displacement (stiffness) method are:

Feature	Force (Flexibility) Method	Displacement (Stiffness) Method
Primary Unknowns	Redundant forces.	Nodal displacements (translations and rotations).
Basis of Unknowns	Degree of Static Indeterminacy (DSI).	Degree of Kinematic Indeterminacy (DKI) or Degrees of Freedom.
Governing Equations	Compatibility equations (displacements must be consistent).	Equilibrium equations (forces must balance at nodes).
Core Matrix	Flexibility Matrix [f]. $\{D\} = [f]\{P\}$	Stiffness Matrix [K]. $\{P\} = [K]\{D\}$
Matrix Elements	$f_{ij}$ = displacement at ‘i’ due to a unit force at ‘j’.	$K_{ij}$ = force at ‘i’ due to a unit displacement at ‘j’.
Primary System	Created by removing restraints to make the structure statically determinate.	Created by adding restraints to prevent all nodal displacements (all joints fixed).
Suitability	Efficient for structures with a low DSI.	Efficient for structures with a low DKI.
Automation	Difficult to generalize and automate.	Highly systematic and easily programmed; forms the basis of modern structural analysis software (FEM).

Stiffness: Stiffness is the force or moment required to produce a unit displacement or unit rotation at a specific point (coordinate) on a structure, while all other specified coordinates are held in a restrained condition (zero displacement). It is a measure of a structure’s resistance to deformation.

Derivation of Truss Element Stiffness Matrix:

1. Local Coordinates: First, consider a truss member with cross-sectional area $A$, Young’s modulus $E$, and length $L$, aligned with a local $x’$-axis. The nodes are ‘i’ (at $x’=0$) and ‘j’ (at $x’=L$). The local displacements are $u’_i$ and $u’_j$. The relationship between axial force $\{P’\}$ and axial displacement $\{d’\}$ is:

   $$\begin{Bmatrix} P’_i \\ P’_j \end{Bmatrix} = \frac{AE}{L} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{Bmatrix} u’_i \\ u’_j \end{Bmatrix} \quad \implies \quad \{P’\} = [k’]\{d’\}$$

   This is the $2 \times 2$ local stiffness matrix $[k’]$.

2. Global Coordinates: Now, place this member in a 2D global (x-y) plane. The member is at an angle $\theta$ to the global x-axis. Each node ‘i’ and ‘j’ has two global DOFs: ($u_i, v_i$) and ($u_j, v_j$). The global displacement vector is $\{d\} = \{u_i, v_i, u_j, v_j\}^T$. The global force vector is $\{P\} = \{F_{xi}, F_{yi}, F_{xj}, F_{yj}\}^T$.
3. Transformation Matrix [T]: We must relate the local axial displacements $\{d’\}$ to the global displacements $\{d\}$. The axial displacement $u’_i$ is the projection of the global displacements $u_i$ and $v_i$ onto the member’s $x’$-axis. Let $l = \cos(\theta)$ and $m = \sin(\theta)$.

   $u’_i = u_i \cos(\theta) + v_i \sin(\theta) = l \cdot u_i + m \cdot v_i$

   $u’_j = u_j \cos(\theta) + v_j \sin(\theta) = l \cdot u_j + m \cdot v_j$

   In matrix form, $\{d’\} = [T]\{d\}$:

   $$\begin{Bmatrix} u’_i \\ u’_j \end{Bmatrix} = \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix} \begin{Bmatrix} u_i \\ v_i \\ u_j \\ v_j \end{Bmatrix}$$

   This $2 \times 4$ matrix is the transformation matrix [T].

4. Global Stiffness Matrix [k]: The global stiffness matrix $[k]$ is derived using the principle of virtual work, which gives the standard transformation:

   $$[k] = [T]^T [k’] [T]$$

   Substituting the matrices:

   $$[k] = \begin{bmatrix} l & 0 \\ m & 0 \\ 0 & l \\ 0 & m \end{bmatrix} \left( \frac{AE}{L} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \right) \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix}$$

   First multiplication:

   $$[k] = \frac{AE}{L} \begin{bmatrix} l & -l \\ m & -m \\ -l & l \\ -m & m \end{bmatrix} \begin{bmatrix} l & m & 0 & 0 \\ 0 & 0 & l & m \end{bmatrix}$$

   Final multiplication:

   $$[k] = \frac{AE}{L} \begin{bmatrix} l^2 & lm & -l^2 & -lm \\ lm & m^2 & -lm & -m^2 \\ -l^2 & -lm & l^2 & lm \\ -lm & -m^2 & lm & m^2 \end{bmatrix}$$

   This is the $4 \times 4$ member stiffness matrix for a truss element in global coordinates.

Rotation matrices (more accurately called transformation matrices) are essential in the stiffness method for the following reasons:

• Local vs. Global Coordinates: It is easiest to derive a member’s stiffness matrix ($[k’]$) in its own local coordinate system (e.g., along the member’s axis). However, the overall structure exists in a single, common global coordinate system (e.g., a common X-Y grid).
• Assembly of Members: To build the stiffness matrix for the entire structure (the global stiffness matrix $[K]$), all member stiffness matrices must be “assembled” or added together. This assembly is only valid if all forces and displacements for all members are expressed in the same coordinate system.
• Transformation: A rotation matrix is the tool used to convert a member’s stiffness matrix from its simple local system to the common global system ($[k] = [T]^T [k’] [T]$).
• Compatibility: This transformation ensures that the displacements and forces at the nodes where members connect are compatible and that equilibrium can be enforced in the global X and Y directions. Without this, adding the $k_{11}$ (axial) stiffness of a vertical member to the $k_{11}$ (axial) stiffness of a horizontal member would be meaningless.

Support settlement is incorporated by partitioning the main stiffness equation to separate known, prescribed displacements (the settlements) from unknown displacements.

1. Partition the Matrix: The main equation $\{P\} = [K]\{D\}$ is partitioned into free (‘f’) coordinates and support (‘s’) coordinates.

   $$\begin{Bmatrix} P_f \\ P_s \end{Bmatrix} = \begin{bmatrix} K_{ff} & K_{fs} \\ K_{sf} & K_{ss} \end{bmatrix} \begin{Bmatrix} D_f \\ D_s \end{Bmatrix}$$

   • $\{D_f\}$ = Unknown displacements at free coordinates.
   • $\{D_s\}$ = Known displacements at support coordinates. For a settlement $\Delta$, $D_s = \Delta$. For a rigid support, $D_s = 0$.
   • $\{P_f\}$ = Known applied loads at free coordinates.
   • $\{P_s\}$ = Unknown support reactions.
2. Expand the Equation: We expand the top row of the partitioned equation:

   $$\{P_f\} = [K_{ff}]\{D_f\} + [K_{fs}]\{D_s\}$$

3. Solve for Unknown Displacements: Since $\{P_f\}$ and $\{D_s\}$ are known, we rearrange the equation to solve for the unknown displacements $\{D_f\}$:

   $$[K_{ff}]\{D_f\} = \{P_f\} – [K_{fs}]\{D_s\}$$

   $$\{D_f\} = [K_{ff}]^{-1} ( \{P_f\} – [K_{fs}]\{D_s\} )$$

   The term $\{P_{f,equiv}\} = -[K_{fs}]\{D_s\}$ is treated as an equivalent nodal load at the free coordinates, which is caused by the support settlement.

4. Find Reactions: Once $\{D_f\}$ is found, the unknown support reactions $\{P_s\}$ can be calculated using the bottom row of the partitioned equation:

   $$\{P_s\} = [K_{sf}]\{D_f\} + [K_{ss}]\{D_s\}$$

Member Stiffness Matrix [k]: This matrix relates the nodal forces and nodal displacements for a single structural element (e.g., one beam or one truss bar). It represents the stiffness of just that individual member. It can be expressed in the member’s local coordinate system or (after transformation) in the global coordinate system.

Global Stiffness Matrix [K]: This matrix relates the total external nodal forces and the total nodal displacements for the entire assembled structure. It is constructed by summing the stiffness contributions of all individual member stiffness matrices (which must all be in the global coordinate system) at their corresponding degrees of freedom. It represents the stiffness of the entire structure as a single system.

The primary advantages of the stiffness method over the flexibility method are:

• Systematic and General: The procedure is highly algorithmic and repetitive, regardless of the structure’s complexity. The same steps (form [k], transform, assemble [K], apply boundary conditions, solve) apply to trusses, beams, and frames.
• Ideal for Automation: This systematic nature makes it an ideal method for computer programming. It is the basis for virtually all modern structural analysis software and the Finite Element Method (FEM).
• No Choice of Redundants: The method does not require the analyst to choose a primary structure or identify redundants, which is a key (and sometimes complex) step in the flexibility method. The degrees of freedom are uniquely defined by the nodes.
• Local Effects: Member properties are defined locally and then assembled, which is often a more direct and conceptually simpler process than calculating flexibility coefficients, which can depend on integrals over the entire structure.

Derivation of Beam Element Stiffness Matrix (Flexural Only):

1. Coordinates: Consider a beam element of length $L$, modulus $E$, and inertia $I$. We only consider flexural (bending) deformation, so we ignore axial effects. The element has two nodes, ‘i’ (left) and ‘j’ (right). The degrees of freedom (DOFs) are the transverse displacement $v$ and rotation $\theta$ at each node.

   Node i: $v_i$ (Coord 1), $\theta_i$ (Coord 2)

   Node j: $v_j$ (Coord 3), $\theta_j$ (Coord 4)

   The displacement vector is $\{d\} = \{v_i, \theta_i, v_j, \theta_j\}^T$.

   The force vector is $\{P\} = \{F_{yi}, M_i, F_{yj}, M_j\}^T$.

2. Method: We find the $4 \times 4$ stiffness matrix $[k]$ by applying a unit displacement at each DOF (coordinate) while holding all others at zero, and then calculating the required restraining forces. These forces form the corresponding column of $[k]$. We use the standard fixed-end beam formulas.
3. Derivations (Columns of [k]):
   • Column 1: Apply $v_i = 1$ (all others = 0). This is a unit settlement at node ‘i’.

     $F_{yi} = 12EI/L^3$

     $M_i = 6EI/L^2$

     $F_{yj} = -12EI/L^3$

     $M_j = 6EI/L^2$

     Col 1: $\{12EI/L^3, \ 6EI/L^2, \ -12EI/L^3, \ 6EI/L^2\}^T$

   • Column 2: Apply $\theta_i = 1$ (all others = 0). This is a unit rotation at node ‘i’.

     $F_{yi} = 6EI/L^2$

     $M_i = 4EI/L$ (near end)

     $F_{yj} = -6EI/L^2$

     $M_j = 2EI/L$ (far end)

     Col 2: $\{6EI/L^2, \ 4EI/L, \ -6EI/L^2, \ 2EI/L\}^T$

   • Column 3: Apply $v_j = 1$ (all others = 0). This is a unit settlement at node ‘j’.

     $F_{yi} = -12EI/L^3$

     $M_i = -6EI/L^2$

     $F_{yj} = 12EI/L^3$

     $M_j = -6EI/L^2$

     Col 3: $\{-12EI/L^3, \ -6EI/L^2, \ 12EI/L^3, \ -6EI/L^2\}^T$

   • Column 4: Apply $\theta_j = 1$ (all others = 0). This is a unit rotation at node ‘j’.

     $F_{yi} = 6EI/L^2$

     $M_i = 2EI/L$ (far end)

     $F_{yj} = -6EI/L^2$

     $M_j = 4EI/L$ (near end)

     Col 4: $\{6EI/L^2, \ 2EI/L, \ -6EI/L^2, \ 4EI/L\}^T$

4. Assemble the Matrix [k]:

   $$[k] = \begin{bmatrix} 12EI/L^3 & 6EI/L^2 & -12EI/L^3 & 6EI/L^2 \\ 6EI/L^2 & 4EI/L & -6EI/L^2 & 2EI/L \\ -12EI/L^3 & -6EI/L^2 & 12EI/L^3 & -6EI/L^2 \\ 6EI/L^2 & 2EI/L & -6EI/L^2 & 4EI/L \end{bmatrix}$$

   This can also be factored as:

   $$[k] = \frac{EI}{L^3} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix}$$

Definition of Influence Line Diagram (ILD): An Influence Line Diagram (ILD) is a graph that represents the variation of a specific function at a fixed point in a structure (such as support reaction, shear force at a section, or bending moment at a section) as a single unit load moves across the span of the structure.

Statement of Muller-Breslau Principle: The influence line for a force function (like a reaction, shear, or moment) is to the same scale as the deflected shape of the structure when the structure is given a unit displacement (or rotation) at the point and in the direction of the force function.

Proof of Muller-Breslau Principle:

We can prove the principle using the principle of virtual work for the vertical reaction $R_A$ at support A.

Consider two independent states:

Real System (Force): The beam is in its original state with a unit load ($P=1$) applied at point $x$. The reaction at A is $R_A(x)$.

Virtual System (Displacement): The constraint at A is removed, and a unit vertical displacement ($\delta_A^* = 1$) is applied upwards at A, causing deflection $y(x)$.

Applying virtual work: $W_{ext} = \sum (F_{real} \times \delta_{virtual}) = 0$

$(P=1 \text{, down}) \times (y(x) \text{, up}) + (R_A(x) \text{, up}) \times (\delta_A^*=1 \text{, up}) = 0$

$-1 \times y(x) + R_A(x) \times 1 = 0$

$R_A(x) = y(x)$

This proves that the ILD ordinate $R_A(x)$ equals the deflection $y(x)$ when unit displacement is applied at A.

Statement of Muller-Breslau Principle: The influence line for a force function (reaction, shear, or moment) is to the same scale as the deflected shape of the structure when the structure is given a unit displacement (or rotation) at the point and in the direction of the force function.

Proof:

Using the principle of virtual work for a reaction $R_A$:

Real System: Beam with unit load at $x$, reaction $R_A(x)$ at A.

Virtual System: Remove constraint at A, apply unit displacement $\delta_A^* = 1$, causing deflection $y(x)$.

Virtual work equation: $(P=1) \times y(x) + R_A(x) \times 1 = 0$

Since $P=1$ acts downward and $y(x)$ is upward, we get: $-y(x) + R_A(x) = 0$

Therefore: $R_A(x) = y(x)$

This proves that the influence line ordinate equals the deflection curve when unit displacement is applied.

To draw the ILD for bending moment ($M_B$) at an intermediate support B of a continuous beam using Muller-Breslau principle:

1. Release the Moment Constraint: Introduce a hinge at support B to release the bending moment constraint.
2. Apply Unit Rotation: Apply a unit relative rotation ($\theta_B = 1$) at the hinge in the positive direction of the moment.
3. Draw the Deflected Shape: The resulting deflected shape of the entire continuous beam is the ILD for $M_B$.

Characteristics:

• Deflection at support B and all other supports is zero
• Sharp “kink” at support B with discontinuous slope
• Curved deflection in adjacent spans returning to zero at other supports
• Typically shows negative ordinates in adjacent spans

For a propped cantilever beam fixed at A and propped at B, to draw ILD for vertical reaction $R_A$:

Using Muller-Breslau Principle:

1. Release Constraint: Remove vertical restraint at A while maintaining moment restraint
2. Apply Unit Displacement: Apply unit vertical displacement ($\delta_A = 1$) upwards at A
3. Draw Deflected Shape: The elastic curve under conditions: $\delta_A = 1$, $\theta_A = 0$, $\delta_B = 0$

ILD Characteristics:

• At fixed support A: ordinate = 1
• Horizontal tangent at A (zero slope)
• Curve slopes downward, crossing x-axis
• At propped support B: ordinate = 0

Alternatively, using statics: $R_A(x) = 1 – \text{ILD for } R_B(x)$

Importance of Influence Line Diagrams (ILDs) in structural analysis:

1. Analysis of Moving Loads: Primary tool for analyzing structures subjected to variable or moving loads like bridges and crane girders
2. Determining Worst-Case Load Position: Shows critical positions to place loads for maximum reaction, shear, or moment
3. Calculating Maximum Effects:
   • For concentrated loads: $\sum (P_i \times y_i)$
   • For UDL: $w \times \text{area under ILD}$
4. Qualitative Design Guidance: Provides understanding of structural behavior under moving loads
5. Efficient Design: Helps in determining most unfavorable loading conditions for design

Muller-Breslau Principle: The influence line for a force function is to the same scale as the deflected shape of the structure when given a unit displacement at the point and in the direction of the force function.

Application to Indeterminate Beams:

1. Identify Function: Determine reaction, shear, or moment for ILD
2. Release Constraint: Remove corresponding structural restraint
   • Reaction: Remove support
   • Shear: Introduce vertical “slide”
   • Moment: Insert hinge
3. Apply Unit Displacement: Apply unit displacement/rotation in positive direction
4. Draw Shape: Resulting deflected shape is the ILD

For indeterminate beams, ILDs are curved elastic deflection curves.

Muller-Breslau Principle: The influence line for a force function equals the deflected shape when unit displacement is applied at that function’s location.

Example Application:

For Reaction $R_B$ in Propped Cantilever:

1. Remove prop (roller) at B
2. Apply unit vertical displacement ($\delta_B = 1$) upwards
3. The deflected shape of cantilever (fixed at A, free at B) when end B is pushed up by 1 unit is the ILD for $R_B$

Result: Curve starts at A (zero deflection, zero slope) and rises to deflection of 1 at B. This curved shape is the ILD for $R_B$.

Muller-Breslau Principle: The influence line for a force function is to the same scale as the deflected shape when unit displacement is applied at that function’s point.

Application to Redundant Reaction $R_B$ in Propped Cantilever:

Beam: Cantilever fixed at A, propped with roller at B

Procedure:

1. Release: Remove the prop (roller support) at B. Structure becomes determinate cantilever.
2. Displace: Apply unit vertical displacement ($\delta_B = 1$) upwards at point B.
3. Shape: The deflected shape of the cantilever when its end B is pushed up by 1 unit is the ILD for $R_B$.

Characteristics:

• Starts at A with zero deflection and zero slope
• Rises to deflection of 1 at B
• This curved shape is the ILD for $R_B$
• Exact numerical ordinates require analysis of deflected shape (e.g., conjugate beam method)

Muller-Breslau Principle: The influence line for any force function (reaction, shear, or moment) is to the same scale as the deflected shape of the structure when the structure is given a unit displacement (or rotation) at the point and in the direction of the force function.

Explanation:

The principle provides a direct graphical method for constructing influence lines. Instead of calculating ordinates point by point, we can visualize the entire ILD by considering the structure’s deformation behavior.

Key Points:

• For reactions: Apply unit translation in direction of reaction
• For shear: Apply unit relative translation (slide)
• For moment: Apply unit relative rotation (hinge)
• The resulting elastic curve represents the ILD
• For determinate structures: ILDs are straight lines
• For indeterminate structures: ILDs are curved lines

The principle is based on the principle of virtual work and provides both qualitative and quantitative information about influence lines.

Muller-Breslau Principle: The influence line for a force function (reaction, shear, or moment) is to the same scale as the deflected shape of the structure when the structure is given a unit displacement (or rotation) at the point and in the direction of the force function.

Muller-Breslau Principle: The influence line for a force function equals the deflected shape of the structure when unit displacement is applied at that function’s location.

Explanation:

The principle states that if we want to draw the influence line for any force quantity (reaction, shear, moment), we should:

1. Remove the constraint that provides that force
2. Apply a unit displacement in the direction of the force
3. The resulting deflected shape is the influence line

This works because the deflected shape represents how the structure would deform under the application of that particular force, which corresponds exactly to the influence line definition.

Muller-Breslau Principle: The influence line for a force function is to the same scale as the deflected shape when unit displacement is applied at that function’s point.

Application to Indeterminate Structures:

For indeterminate structures, the Muller-Breslau principle is particularly valuable because:

1. Qualitative ILDs: Allows quick sketching of ILD shape without complex calculations
2. Curved ILDs: For indeterminate structures, ILDs are curved lines representing elastic deflection curves
3. Procedure:
   • Release constraint corresponding to desired function
   • Apply unit displacement/rotation
   • The elastic curve is the ILD
4. Accuracy: While exact ordinates require analysis, the shape helps identify critical load positions

This method transforms the complex problem of calculating ILD ordinates into the more manageable problem of analyzing a deflected shape.

Plastic Hinge: A plastic hinge is a yielded zone that forms in a ductile flexural member (like a steel beam) when the bending moment reaches the plastic moment capacity ($M_p$) of the section. At this point, the entire cross-section has yielded (both in tension and compression). The section can no longer resist additional moment but can undergo large rotations at this constant plastic moment, behaving like a “hinge.”

Difference between Plastic Hinge and Mechanical Hinge:

Feature	Mechanical Hinge (e.g., a Pin Joint)	Plastic Hinge
Formation	Physically built into the structure.	Forms in a ductile material when the section yields.
Moment Resistance	Resists zero bending moment ($M=0$).	Resists a constant, maximum moment (the plastic moment, $M_p$).
Rotation	Allows free rotation from the start.	Allows rotation only after the moment $M_p$ has been reached.
Nature	A real, physical connection.	A concept representing a localized yielded region of the member.
Permanence	Permanent.	Temporary; it exists only under the collapse load. If the load is removed, the section becomes rigid again (though with residual stresses).

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$). It is a property of the cross-section’s geometry.

$$S = \frac{M_p}{M_y} = \frac{Z_p \cdot \sigma_y}{Z_e \cdot \sigma_y} = \frac{Z_p}{Z_e}$$

Where $Z_p$ is the plastic section modulus and $Z_e$ is the elastic section modulus.

Load Factor (F): The ratio of the ultimate (collapse) load that a structure can sustain to the working (service) load it is designed to carry.

$$F = \frac{\text{Collapse Load}}{\text{Working Load}}$$

The two basic theorems of plastic analysis are the Lower Bound (Static) Theorem and the Upper Bound (Kinematic) Theorem. They are used to find the true collapse load ($W_c$).

1. Lower Bound (Static) Theorem:

Statement: The collapse load ($W_c$) is the largest load that can be applied to a structure for which a “safe” or “statically admissible” bending moment distribution can be found. A bending moment distribution is “safe” if it satisfies:

• Equilibrium: The moments are in equilibrium with the external loads.
• Yield Condition: The bending moment at any point in the structure is not greater than the plastic moment capacity ($|M| \le M_p$).

Explanation: This theorem states that any load calculated from a “safe” bending moment distribution will be less than or equal to the true collapse load ($W \le W_c$). It provides a lower-limit or “safe” value for the collapse load.

2. Upper Bound (Kinematic) Theorem:

Statement: The collapse load ($W_c$) is the smallest load that can be found for which a kinematically admissible collapse mechanism exists. A mechanism is “kinematically admissible” if it satisfies:

• Mechanism: The number of plastic hinges is sufficient to create a mechanism.
• Work-Energy Equation: The internal work done by the plastic hinges rotating ($W_i$) is equal to the external work done by the loads moving through the mechanism’s displacements ($W_e$).

Explanation: This theorem states that any load calculated from an assumed collapse mechanism (which may not be the true one) will be greater than or equal to the true collapse load ($W \ge W_c$). It provides an upper-limit or “unsafe” value for the collapse load.

Uniqueness Theorem: The true collapse load is the one where both the static and kinematic conditions are satisfied simultaneously.

Feature	Elastic Analysis	Plastic Analysis
Material Behavior	Assumes a linear stress-strain relationship (Hooke’s Law).	Assumes an idealized elasto-plastic stress-strain relationship.
Stress Limit	Design is based on the first yield stress ($\sigma_y$). The maximum stress must not exceed this limit.	Design is based on the plastic moment capacity ($M_p$). It allows yielding and stress redistribution.
Governing Factor	Strength is governed by the first yield at a single point.	Strength is governed by the formation of a collapse mechanism.
Applicability	Used for all materials and for serviceability checks (e.g., deflection).	Applicable only to ductile materials (like steel) and for ultimate strength checks.
Load Factor	A factor of safety is applied to the stress.	A load factor is applied to the working loads to determine the ultimate (collapse) load.

A collapse mechanism is the fundamental concept used in plastic analysis to determine the ultimate (collapse) load of a structure. It describes the state at which the structure becomes unstable and can undergo infinitely large deformations at a constant load.

The concept is as follows:

• As load is applied to a statically indeterminate structure, the most highly stressed section eventually reaches its plastic moment capacity ($M_p$) and forms a plastic hinge.
• This does not mean the structure has failed. The structure redistributes the load to other sections, and the plastic hinge rotates at a constant moment $M_p$.
• As the load continues to increase, other sections reach $M_p$ and form additional plastic hinges.
• This process continues until a sufficient number of plastic hinges have formed to transform the structure (or a part of it) into an unstable “mechanism,” similar to a chain of links.
• Once this mechanism forms, the structure cannot resist any additional load and will “collapse” under large deformations.

For a structure with a degree of indeterminacy ‘r’, the number of plastic hinges ‘n’ required to form a complete collapse mechanism is generally $n = r + 1$.

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$). It is a property of the cross-section’s geometry.

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

Where $Z_p$ is the plastic section modulus and $Z_e$ is the elastic section modulus.

1. Lower Bound (Static) Theorem

Statement: If a bending moment distribution can be found that satisfies equilibrium with the external loads ($W_s$) and does not exceed the plastic moment capacity at any point ($|M| \le M_p$), then the load $W_s$ is less than or equal to the true collapse load ($W_c$).

$$W_s \le W_c$$

Proof: Let $W_c$ be the true collapse load with its corresponding mechanism and moment distribution $M_c$. Let $W_s$ be any other load with a “safe” (statically admissible) moment distribution $M_s$, where $|M_s| \le M_p$ everywhere.

By the principle of virtual work, for the true collapse mechanism:

External Work ($W_e$) by $W_c$ = Internal Work ($W_i$) by $M_c$.

If we apply the same mechanism to the “safe” load $W_s$ and its moment distribution $M_s$:

$W_e(\text{by } W_s) = W_i(\text{by } M_s)$.

The ratio of external works is equal to the ratio of loads: $\frac{W_e(\text{by } W_s)}{W_e(\text{by } W_c)} = \frac{W_s}{W_c}$.

This gives: $W_e(\text{by } W_s) = \frac{W_s}{W_c} \cdot W_e(\text{by } W_c) = \frac{W_s}{W_c} \cdot W_i(\text{by } M_c)$.

So, $W_i(\text{by } M_s) = \frac{W_s}{W_c} \cdot W_i(\text{by } M_c)$.

At the plastic hinges, $|M_s| \le M_p = |M_c|$. Therefore, the internal work done by $M_s$ must be less than or equal to the internal work done by $M_c$.

$W_i(\text{by } M_s) \le W_i(\text{by } M_c)$.

Substituting, we get: $\frac{W_s}{W_c} \cdot W_i(\text{by } M_c) \le W_i(\text{by } M_c)$.

This implies: $W_s \le W_c$. (Hence, proved).

2. Upper Bound (Kinematic) Theorem

Statement: If a load ($W_u$) is calculated from any assumed (kinematically admissible) collapse mechanism by equating internal and external work, this load $W_u$ will be greater than or equal to the true collapse load ($W_c$).

$$W_u \ge W_c$$

Proof: Let $W_u$ be the load calculated from an arbitrary, assumed mechanism. Let $W_c$ be the true collapse load with its correct (and “safe”) moment distribution $M_c$, where $|M_c| \le M_p$ everywhere.

Let the work equation for the assumed mechanism be:

$W_e(\text{by } W_u) = W_i(\text{by } M_p \text{ at assumed hinges})$

By virtual work, applying the assumed mechanism to the true load $W_c$:

$W_e(\text{by } W_c) = W_i(\text{by } M_c \text{ at assumed hinge locations})$

At the hinge locations of the assumed mechanism, the true moment $M_c$ must be less than or equal to $M_p$. Therefore, the internal work done by $M_c$ is less than or equal to the internal work done by $M_p$.

$W_i(\text{by } M_c) \le W_i(\text{by } M_p \text{ at assumed hinges})$.

This gives: $W_e(\text{by } W_c) \le W_e(\text{by } W_u)$.

Since the external work is proportional to the load ($W_e \propto W$), it follows that:

$W_c \le W_u$. (Hence, proved).

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$). It indicates the reserve strength of a section beyond its first yield.

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

Plastic Modulus ($Z_p$): Also known as the plastic section modulus, it is a geometric property of a cross-section used to calculate the plastic moment capacity ($M_p = Z_p \cdot \sigma_y$). It is defined as the first moment of area about the plastic neutral axis (the axis that divides the cross-section into two equal areas).

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Plastic Moment ($M_p$): The maximum bending moment that a fully yielded (plastic) cross-section can resist. It is calculated as the product of the yield stress ($\sigma_y$) and the plastic section modulus ($Z_p$). $M_p = Z_p \cdot \sigma_y$.

Load Factor (F): The ratio of the ultimate (collapse) load that a structure can sustain to the working (service) load it is designed to carry.

$$F = \frac{\text{Collapse Load}}{\text{Working Load}}$$

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$).

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

Plastic Modulus ($Z_p$): Also known as the plastic section modulus, it is a geometric property of a cross-section used to calculate the plastic moment capacity ($M_p = Z_p \cdot \sigma_y$). It is defined as the first moment of area about the plastic neutral axis (the axis that divides the cross-section into two equal areas).

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$).

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Shape Factor (S): The ratio of the plastic moment capacity ($M_p$) of a section to its elastic yield moment ($M_y$).

$$S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e}$$

Plastic Section Modulus ($Z_p$): A geometric property of a cross-section used to calculate the plastic moment capacity ($M_p = Z_p \cdot \sigma_y$). It is defined as the first moment of area about the plastic neutral axis (the axis that divides the cross-section into two equal areas).

Collapse Mechanism: The state reached when a sufficient number of plastic hinges have formed in a structure (or a part of it) to make it unstable, allowing it to undergo large, uncontrolled deformations at a constant load (the collapse load).

Plastic Hinge: A yielded zone in a ductile member where the section has fully yielded due to bending, allowing it to rotate at a constant plastic moment ($M_p$).

Plastic Moment Capacity ($M_p$): The maximum bending moment that a fully yielded (plastic) cross-section can resist. It is the moment required to form a plastic hinge and is calculated as the product of the yield stress ($\sigma_y$) and the plastic section modulus ($Z_p$). $M_p = Z_p \cdot \sigma_y$.

The fundamental assumptions made in plastic theory are:

• Plane Sections: Plane sections of the member normal to the axis before bending remain plane and normal after bending.
• Material Behavior: The stress-strain relationship is idealized as perfectly elasto-plastic (follows Hooke’s law up to the yield stress and then deforms plastically at a constant stress).
• Small Deformations: The deformations are small, so the equations of equilibrium can be applied to the undeformed structure.
• Ductility: The material is sufficiently ductile to allow for large rotations at plastic hinges to occur without fracture.
• No Axial Load: The analysis typically considers only bending, assuming axial loads are negligible.
• No Instability: The effects of instability (like local or lateral-torsional buckling) are prevented or not considered.

Shape Factor (S): The shape factor is a property of a cross-section’s geometry. It is defined as the ratio of the plastic moment capacity ($M_p$) of the section to its elastic yield moment ($M_y$). It represents the reserve capacity of the section beyond the point of first yield.

$$S = \frac{M_p}{M_y} = \frac{Z_p \cdot \sigma_y}{Z_e \cdot \sigma_y} = \frac{Z_p}{Z_e}$$

Where:

• $Z_p$ = Plastic Section Modulus
• $Z_e$ = Elastic Section Modulus