11.3 Structural Trusses, Rigid Frames & Energy Methods

PE Mechanical: Machine Design and Materials Exam

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11.3 Structural Trusses, Rigid Frames & Energy Methods

Key Takeaways Explain This Topic AI

Planar trusses achieve static determinacy when 2 j = m + r 2j = m + r 2 j = m + r ; if m + r > 2 j m + r > 2j m + r > 2 j , the truss is statically indeterminate to degree ( m + r − 2 j ) (m + r - 2j) ( m + r − 2 j ) , requiring deformation compatibility or energy formulations.

Zero-force members can be identified by inspection: if two non-collinear members meet at an unloaded joint, both carry zero force; if three members meet at an unloaded joint where two are collinear, the third transverse member carries zero force.

Total elastic strain energy includes axial ( U = ∫ P 2 2 A E d x U = \int \frac{P^2}{2AE} dx U = ∫ 2 A E P 2 ​ d x ), flexural bending ( U = ∫ M 2 2 E I d x U = \int \frac{M^2}{2EI} dx U = ∫ 2 E I M 2 ​ d x ), and torsional ( U = ∫ T 2 2 G J d x U = \int \frac{T^2}{2GJ} dx U = ∫ 2 G J T 2 ​ d x ) components; flexural bending energy dominates by orders of magnitude in slender structural frames ( L / h > 10 L/h > 10 L / h > 10 ).

Castigliano's Second Theorem establishes that the partial derivative of total elastic strain energy with respect to an applied point load P i P_i P i ​ equals the displacement δ i \delta_i δ i ​ along that load's line of action ( δ i = ∂ U / ∂ P i \delta_i = \partial U / \partial P_i δ i ​ = ∂ U / ∂ P i ​ ).

Fictitious dummy loads ( Q → 0 Q \to 0 Q → 0 ) permit calculation of deflections or slopes at points lacking active external forces, while enforcing zero displacement at redundant supports ( ∂ U / ∂ R = 0 \partial U / \partial R = 0 ∂ U / ∂ R = 0 ) resolves statically indeterminate reactions.

Last updated: October 2026

11.3 Structural Trusses, Rigid Frames & Energy Methods

Structural frameworks in mechanical engineering—including overhead gantry crane girders, robotic arm links, engine test stands, and heavy equipment chassis—must support severe spatial load distributions without excessive deflection or instability. Analyzing these assemblies requires mastery of static determinacy, internal force equilibrium, and energy methods such as Castigliano's Second Theorem for both determinate and statically indeterminate systems.

Structural Trusses: Classification & Static Determinacy

A truss is an assemblage of straight structural members interconnected at pin joints (nodes) forming triangular sub-units. Ideal truss analysis assumes:

Members are joined at their ends by frictionless, spherical/rotational pins.

Loads and support reactions are applied strictly at the joints.

Centroidal axes of all concurrent members at any joint intersect at a single common point.

Members carry pure axial tension or compression; bending moments and transverse shears are zero. Planar Truss Determinacy Relationships

m = Number of Members, j = Number of Joints, r = Number of Reaction Components

• +--------------------+------------------------------------------------------------+
• | m + r < 2j | Statically Unstable (Mechanism: structure collapses) |
• +--------------------+------------------------------------------------------------+
• | m + r = 2j | Statically Determinate (Solvable via statics if stable) |
• +--------------------+------------------------------------------------------------+
• | m + r > 2j | Statically Indeterminate (Degree of Indeterminacy = m+r-2j)|
• +--------------------+------------------------------------------------------------+

For a three-dimensional space truss, each joint has three translational degrees of freedom, yielding the determinacy criterion: m + r = 3 j m + r = 3j m + r = 3 j

Note

Satisfying m + r = 2 j m + r = 2j m + r = 2 j is a necessary but not sufficient condition for stability. A truss can have sufficient members overall but possess a localized unstable rectangular panel or collinear support reactions that render it a mechanism.

Truss Analysis: Method of Joints vs. Method of Sections Method of Joints (Joint Equilibrium) Method of Sections (Cutting Plane)

• F_AB
• ^ Cut Plane
• | |
• Joint A *-----> F_AC =====*==|==*===== Top Chord
• / \ / \ | / \
• / \ / \|/ \
• v v *=====*=====*== Bottom Chord
• R_Ax R_Ay |
• 2 equations per joint (Fx = 0, Fy = 0) - Cut through max 3 unknown members
• Marches across every node sequentially - Solves target internal member directly
• 1. Method of Joints

Isolates individual joints as concurrent force particles in equilibrium:

∑ F x = 0 , ∑ F y = 0 \sum F_x = 0, \quad \sum F_y = 0 ∑ F x ​ = 0 , ∑ F y ​ = 0

Strategy : Begin at an exterior support joint having at least one known load/reaction and no more than two unknown member forces .

Sign Convention : Always assume unknown member forces act in tension (pulling away from the joint). A positive calculated result confirms tension ( T T T ); a negative result indicates compression ( C C C ).

2. Method of Sections

Passes an imaginary cutting plane completely through the truss, slicing through no more than three non-concurrent, non-parallel members whose internal forces are desired. The isolated section is treated as a rigid body governed by coplanar statics:

∑ F x = 0 , ∑ F y = 0 , ∑ M O = 0 \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0 ∑ F x ​ = 0 , ∑ F y ​ = 0 , ∑ M O ​ = 0

Advantage : Determines internal forces in deep web or chord members directly in a single step without calculating forces across intervening joints.

Moment Pivot Selection : Sum moments about the intersection node of two unknown members to solve for the third unknown force directly in one equation.

Rapid Identification of Zero-Force Members

Zero-force members carry no axial force under a specific static loading condition. Identifying them by visual inspection dramatically reduces the algebraic effort required on the PE exam. The Three Zero-Force Member Inspection Rules

• Rule 1: Two Non-Collinear Members Rule 2: Three Members (Two Collinear)
• Unloaded Joint Unloaded Joint
• * *
• / \ /|\
• F_1 = 0 / \ F_2 = 0 F_1 = F_2/ | \ F_3 = 0 (Transverse)
• / \ / | \
• Rule 1: Two Non-Collinear Members at an Unloaded Joint

If only two non-collinear members meet at a joint that has no external load and no support reaction , both members are zero-force members:

∑ F ⊥ = 0 ⟹ F 1 = 0 , F 2 = 0 \sum F_{\perp} = 0 \implies F_1 = 0, \quad F_2 = 0 ∑ F ⊥ ​ = 0 ⟹ F 1 ​ = 0 , F 2 ​ = 0

Rule 2: Three Members at an Unloaded Joint (Two Collinear)

If three members meet at an unloaded joint where two of the members are collinear:

The third, non-collinear (transverse) member is a zero-force member ( F 3 = 0 F_3 = 0 F 3 ​ = 0 ).

The two collinear members carry equal and opposite forces ( F 1 = F 2 F_1 = F_2 F 1 ​ = F 2 ​ ).

Rule 3: Two Members at a Loaded Joint Collinear with One Member

If two members meet at a joint where an external load is applied parallel to one member, the non-collinear member is a zero-force member, and the collinear member balances the external load.

Tip

Why include zero-force members in a design if they carry zero load? They provide essential lateral bracing to prevent compressive buckling of slender primary chords, accommodate secondary load cases (such as wind or moving cranes), and support the structure during fabrication and transport.

Rigid Frames & Moment-Resisting Connections

Unlike pin-jointed trusses, a rigid frame features moment-resisting connections (welded flanges, gusset plates) that preserve the angular orientation between intersecting members under load. Planar Portal Frame Mechanics

• Rigid Joint B Rigid Joint C
• +-----------------------------+
• | Transverse Bending M(x) |
• | Axial Force P(x) |
• Column AB | Shear Force V(x) | Column CD
• | |
• | |
• + +
• Fixed Base A Fixed Base D
• Key Internal Actions in Frames

Every cross section of a rigid frame member resists three internal force resultants:

Axial Force ( P P P ) : Tension or compression acting normal to the cross section.

Transverse Shear Force ( V V V ) : Direct shear acting tangential to the cross section.

Bending Moment ( M M M ) : Internal flexural couple that bends the member and creates normal stresses σ = − M y / I \sigma = -M y / I σ = − M y / I .

In portal frames resisting lateral wind or equipment crane loads, columns experience massive bending moments that transfer continuously around the 90° rigid corners into the horizontal cross beams.

Elastic Strain Energy Formulations

Elastic strain energy ( U U U ) represents the internal potential energy stored within an elastic body during deformation. For linear-elastic materials obeying Hooke's Law, the general strain energy expressions for individual structural actions are:

1. Axial Loading (Bars & Truss Members) U axial = ∫ 0 L P ( x ) 2 2 A E d x U_{\text{axial}} = \int_0^L \frac{P(x)^2}{2 A E}\,dx U axial ​ = ∫ 0 L ​ 2 A E P ( x ) 2 ​ d x

For a truss composed of n n n uniform members with constant axial force P i P_i P i ​ : U truss = ∑ i = 1 n P i 2 L i 2 A i E i U_{\text{truss}} = \sum_{i=1}^n \frac{P_i^2 L_i}{2 A_i E_i} U truss ​ = i = 1 ∑ n ​ 2 A i ​ E i ​ P i 2 ​ L i ​ ​

2. Flexural Bending (Beams & Frames) U bending = ∫ 0 L M ( x ) 2 2 E I d x U_{\text{bending}} = \int_0^L \frac{M(x)^2}{2 E I}\,dx U bending ​ = ∫ 0 L ​ 2 E I M ( x ) 2 ​ d x

3. Torsional Loading (Circular Shafts) U torsion = ∫ 0 L T ( x ) 2 2 G J d x = T 2 L 2 G J U_{\text{torsion}} = \int_0^L \frac{T(x)^2}{2 G J}\,dx = \frac{T^2 L}{2 G J} U torsion ​ = ∫ 0 L ​ 2 G J T ( x ) 2 ​ d x = 2 G J T 2 L ​

4. Transverse Shear (Timoshenko Energy Term) U shear = ∫ 0 L f s V ( x ) 2 2 G A d x U_{\text{shear}} = \int_0^L \frac{f_s V(x)^2}{2 G A}\,dx U shear ​ = ∫ 0 L ​ 2 G A f s ​ V ( x ) 2 ​ d x

where f s f_s f s ​ is the transverse shear form factor ( f s = 6 / 5 = 1.2 f_s = 6/5 = 1.2 f s ​ = 6/5 = 1.2 for rectangular cross sections; f s = 10 / 9 ≈ 1.11 f_s = 10/9 \approx 1.11 f s ​ = 10/9 ≈ 1.11 for solid circular sections).

Neglecting Shear & Axial Compliance in Slender Frames

For slender flexural members having length-to-depth ratios L / h > 10 L / h > 10 L / h > 10 , the ratio of shear strain energy to flexural bending strain energy scales as: U shear U bending ∝ ( h L ) 2 < 0.01 ( 1 % ) \frac{U_{\text{shear}}}{U_{\text{bending}}} \propto \left( \frac{h}{L} \right)^2 < 0.01 \quad (1\%) U bending ​ U shear ​ ​ ∝ ( L h ​ ) 2 < 0.01 ( 1% )

Similarly, axial deformation contributes negligible compliance compared to bending. Therefore, in slender frame analysis, flexural bending energy U bending U_{\text{bending}} U bending ​ accounts for > 98 % > 98\% > 98% of total structural deformation and is analyzed exclusively.

Castigliano's Second Theorem & The Dummy Load Method

Published by Alberto Castigliano in 1879, Castigliano's Second Theorem is the most powerful energy tool for calculating structural deflections and rotations.

Theorem Statement

In any linear-elastic structure supported such that rigid-body motion is prevented, the partial derivative of the total elastic strain energy ( U U U ) with respect to any applied external point force ( P i P_i P i ​ ) equals the deflection ( δ i \delta_i δ i ​ ) of the point of application along the line of action of that force: δ i = ∂ U ∂ P i \delta_i = \frac{\partial U}{\partial P_i} δ i ​ = ∂ P i ​ ∂ U ​

Similarly, the partial derivative with respect to an applied moment ( M i M_i M i ​ ) yields the rotational angle ( θ i \theta_i θ i ​ ) in radians: θ i = ∂ U ∂ M i \theta_i = \frac{\partial U}{\partial M_i} θ i ​ = ∂ M i ​ ∂ U ​

Leibniz Integral Rule Formulation (Differentiating Under the Integral)

Differentiating the squared terms before integrating eliminates squaring complex polynomials and drastically simplifies algebraic execution: δ i = ∫ 0 L M ( x ) E I ( ∂ M ( x ) ∂ P i ) d x + ∑ j = 1 n P j L j A j E j ( ∂ P j ∂ P i ) + ∫ 0 L T ( x ) G J ( ∂ T ( x ) ∂ P i ) d x \delta_i = \int_0^L \frac{M(x)}{E I} \left( \frac{\partial M(x)}{\partial P_i} \right)\,dx + \sum_{j=1}^n \frac{P_j L_j}{A_j E_j} \left( \frac{\partial P_j}{\partial P_i} \right) + \int_0^L \frac{T(x)}{G J} \left( \frac{\partial T(x)}{\partial P_i} \right)\,dx δ i ​ = ∫ 0 L ​ E I M ( x ) ​ ( ∂ P i ​ ∂ M ( x ) ​ ) d x + j = 1 ∑ n ​ A j ​ E j ​ P j ​ L j ​ ​ ( ∂ P i ​ ∂ P j ​ ​ ) + ∫ 0 L ​ G J T ( x ) ​ ( ∂ P i ​ ∂ T ( x ) ​ ) d x

The Fictitious Dummy Load Method

To calculate deflection or slope at a point or direction where no actual load acts :

Place an imaginary, fictitious external force Q Q Q (or dummy moment C C C ) at the target point in the desired direction.

Express the internal bending moment M ( x , Q ) M(x, Q) M ( x , Q ) as a function of physical loads and Q Q Q .

Compute the partial derivative ∂ M ( x , Q ) ∂ Q \frac{\partial M(x, Q)}{\partial Q} ∂ Q ∂ M ( x , Q ) ​ .

Evaluate the Castigliano integral and immediately set Q = 0 Q = 0 Q = 0 : δ = ∫ 0 L M ( x , Q ) E I ( ∂ M ( x , Q ) ∂ Q ) d x ∣ Q = 0 \delta = \left. \int_0^L \frac{M(x, Q)}{E I} \left( \frac{\partial M(x, Q)}{\partial Q} \right)\,dx \,\right|_{Q = 0} δ = ∫ 0 L ​ E I M ( x , Q ) ​ ( ∂ Q ∂ M ( x , Q ) ​ ) d x ​ Q = 0 ​

Analysis of Statically Indeterminate Structures

When a structure possesses more supports or internal constraints than available equations of static equilibrium, it is statically indeterminate . Castigliano Indeterminate Solution Procedure

• 1. Identify Redundant Reaction(s) R ---> Express Degree of Indeterminacy D_I = r - equations
• 2. Formulate Internal Moment M(x, R) ---> In terms of applied loads and redundant R

3. Apply Compatibility Condition ---> Displacement at rigid support is zero:

delta_R = partial U / partial R = 0

4. Solve Algebraic Equation for R ---> Directly yields reaction force or moment!

By enforcing the geometric boundary condition at an unyielding redundant support: ∂ U ∂ R = 0 \frac{\partial U}{\partial R} = 0 ∂ R ∂ U ​ = 0

Castigliano's theorem reduces complex indeterminate boundary-value differential equations into straightforward single-variable linear algebraic equations.

Realistic Worked Engineering Scenario: Deflection & Redundant Reaction in an L-Shaped Crane Frame L-Shaped Gantry Support Frame

• B (Rigid Joint) C (Tip)
• +-----------------------------+
• | |
• | Horizontal Boom BC | ---> Fictitious Load Q
• | Length L = 96 in |
• | v Concentrated Load P = 5,000 lbf
• Column AB | (or Redundant Roller R_C)
• Height |
• H = 120 in |
• |
• |
• + A (Fixed Base: Fully Clamped)
• Problem Statement

An industrial jib frame A B C ABC A BC consists of a vertical column A B AB A B of height H = 120 in H = 120\text{ in} H = 120 in and a horizontal cantilever boom B C BC BC of length L = 96 in L = 96\text{ in} L = 96 in rigidly joined at B B B ( 90 ∘ 90^\circ 9 0 ∘ ). The frame is fully clamped (fixed) at base A A A and fabricated from a structural steel wide-flange section with uniform flexural rigidity:

• E = 29.0 × 10 6 psi E = 29.0 \times 10^6\text{ psi} E = 29.0 × 1 0 6 psi
• I = 110.0 in 4 I = 110.0\text{ in}^4 I = 110.0 in 4

Flexural rigidity: E I = 29.0 × 10 6 × 110.0 = 3.19 × 10 9 lbf ⋅ in 2 EI = 29.0 \times 10^6 \times 110.0 = 3.19 \times 10^9\text{ lbf}\cdot\text{in}^2 E I = 29.0 × 1 0 6 × 110.0 = 3.19 × 1 0 9 lbf ⋅ in 2

Calculate:

The vertical deflection δ C , v \delta_{C,v} δ C , v ​ at tip C C C under a vertical downward load P = 5 , 000 lbf P = 5,000\text{ lbf} P = 5 , 000 lbf .

The horizontal deflection δ C , h \delta_{C,h} δ C , h ​ at tip C C C using the fictitious dummy load method.

Statically Indeterminate Extension : A vertical roller support is installed at tip C C C preventing vertical deflection ( δ C , v = 0 \delta_{C,v} = 0 δ C , v ​ = 0 ). A vertical equipment load W = 6 , 000 lbf W = 6,000\text{ lbf} W = 6 , 000 lbf is applied at the midpoint D D D of boom B C BC BC ( x = 48 in x = 48\text{ in} x = 48 in from B B B ). Determine the upward redundant reaction force R C R_C R C ​ provided by the roller.

• Step-by-Step Mathematical Solution
• Part 1: Vertical Deflection δ C , v \delta_{C,v} δ C , v ​ under Point Load P P P

Establish coordinate frames:

For horizontal boom B C BC BC : let x x x run from 0 0 0 at tip C C C to L = 96 in L = 96\text{ in} L = 96 in at corner B B B . M 1 ( x ) = P x , ∂ M 1 ∂ P = x M_1(x) = P x, \quad \frac{\partial M_1}{\partial P} = x M 1 ​ ( x ) = P x , ∂ P ∂ M 1 ​ ​ = x

For vertical column A B AB A B : let y y y run from 0 0 0 at corner B B B downward to H = 120 in H = 120\text{ in} H = 120 in at base A A A .

The vertical load P P P at distance L L L from the column creates a constant overturning moment everywhere along column A B AB A B : M 2 ( y ) = P L , ∂ M 2 ∂ P = L M_2(y) = P L, \quad \frac{\partial M_2}{\partial P} = L M 2 ​ ( y ) = P L , ∂ P ∂ M 2 ​ ​ = L

Apply Castigliano's Second Theorem: δ C , v = ∂ U ∂ P = ∫ 0 L M 1 E I ( ∂ M 1 ∂ P ) d x + ∫ 0 H M 2 E I ( ∂ M 2 ∂ P ) d y \delta_{C,v} = \frac{\partial U}{\partial P} = \int_0^L \frac{M_1}{EI}\left(\frac{\partial M_1}{\partial P}\right)dx + \int_0^H \frac{M_2}{EI}\left(\frac{\partial M_2}{\partial P}\right)dy δ C , v ​ = ∂ P ∂ U ​ = ∫ 0 L ​ E I M 1 ​ ​ ( ∂ P ∂ M 1 ​ ​ ) d x + ∫ 0 H ​ E I M 2 ​ ​ ( ∂ P ∂ M 2 ​ ​ ) d y δ C , v = 1 E I ∫ 0 L ( P x ) ( x ) d x + 1 E I ∫ 0 H ( P L ) ( L ) d y \delta_{C,v} = \frac{1}{EI} \int_0^L (P x)(x)\,dx + \frac{1}{EI} \int_0^H (P L)(L)\,dy δ C , v ​ = E I 1 ​ ∫ 0 L ​ ( P x ) ( x ) d x + E I 1 ​ ∫ 0 H ​ ( P L ) ( L ) d y δ C , v = P L 3 3 E I + P L 2 H E I = P L 2 E I ( L 3 + H ) \delta_{C,v} = \frac{P L^3}{3 EI} + \frac{P L^2 H}{EI} = \frac{P L^2}{EI} \left( \frac{L}{3} + H \right) δ C , v ​ = 3 E I P L 3 ​ + E I P L 2 H ​ = E I P L 2 ​ ( 3 L ​ + H )

Substitute the numerical parameters:

L 2 = 96 2 = 9 , 216 in 2 L^2 = 96^2 = 9,216\text{ in}^2 L 2 = 9 6 2 = 9 , 216 in 2

L / 3 + H = 96 / 3 + 120 = 32 + 120 = 152 in L/3 + H = 96/3 + 120 = 32 + 120 = 152\text{ in} L /3 + H = 96/3 + 120 = 32 + 120 = 152 in δ C , v = 5 , 000 lbf × 9 , 216 in 2 × 152 in 3.19 × 10 9 lbf ⋅ in 2 = 7.00416 × 10 9 3.19 × 10 9 = 2.1957 in ≈ 2.20 in \delta_{C,v} = \frac{5,000\text{ lbf} \times 9,216\text{ in}^2 \times 152\text{ in}}{3.19 \times 10^9\text{ lbf}\cdot\text{in}^2} = \frac{7.00416 \times 10^9}{3.19 \times 10^9} = 2.1957\text{ in} \approx 2.20\text{ in} δ C , v ​ = 3.19 × 1 0 9 lbf ⋅ in 2 5 , 000 lbf × 9 , 216 in 2 × 152 in ​ = 3.19 × 1 0 9 7.00416 × 1 0 9 ​ = 2.1957 in ≈ 2.20 in

Tip C C C deflects vertically downward by 2.20 in 2.20\text{ in} 2.20 in .

Part 2: Horizontal Deflection δ C , h \delta_{C,h} δ C , h ​ via Fictitious Dummy Load Method

Apply a fictitious horizontal dummy force Q Q Q directed rightward at tip C C C .

Formulate bending moments including Q Q Q :

For boom B C BC BC : the horizontal force Q Q Q produces axial compression but zero bending moment : M 1 ( x ) = P x , ∂ M 1 ∂ Q = 0 M_1(x) = P x, \quad \frac{\partial M_1}{\partial Q} = 0 M 1 ​ ( x ) = P x , ∂ Q ∂ M 1 ​ ​ = 0

For column A B AB A B : horizontal force Q Q Q acts with moment arm y y y measured from corner B B B : M 2 ( y ) = P L + Q y , ∂ M 2 ∂ Q = y M_2(y) = P L + Q y, \quad \frac{\partial M_2}{\partial Q} = y M 2 ​ ( y ) = P L + Q y , ∂ Q ∂ M 2 ​ ​ = y

Apply Castigliano's theorem and evaluate at Q = 0 Q = 0 Q = 0 : δ C , h = ∫ 0 H M 2 ( y ) E I ( ∂ M 2 ∂ Q ) d y ∣ Q = 0 = 1 E I ∫ 0 H ( P L ) ( y ) d y = P L H 2 2 E I \delta_{C,h} = \left. \int_0^H \frac{M_2(y)}{EI} \left( \frac{\partial M_2}{\partial Q} \right)\,dy \,\right|_{Q=0} = \frac{1}{EI} \int_0^H (P L)(y)\,dy = \frac{P L H^2}{2 EI} δ C , h ​ = ∫ 0 H ​ E I M 2 ​ ( y ) ​ ( ∂ Q ∂ M 2 ​ ​ ) d y ​ Q = 0 ​ = E I 1 ​ ∫ 0 H ​ ( P L ) ( y ) d y = 2 E I P L H 2 ​

Substitute the numerical parameters:

P L = 5 , 000 × 96 = 480 , 000 lbf ⋅ in P L = 5,000 \times 96 = 480,000\text{ lbf}\cdot\text{in} P L = 5 , 000 × 96 = 480 , 000 lbf ⋅ in

H 2 = 120 2 = 14 , 400 in 2 H^2 = 120^2 = 14,400\text{ in}^2 H 2 = 12 0 2 = 14 , 400 in 2

Numerator = 480 , 000 × 14 , 400 = 6.912 × 10 9 lbf ⋅ in 3 = 480,000 \times 14,400 = 6.912 \times 10^9\text{ lbf}\cdot\text{in}^3 = 480 , 000 × 14 , 400 = 6.912 × 1 0 9 lbf ⋅ in 3

Denominator = 2 × 3.19 × 10 9 = 6.38 × 10 9 lbf ⋅ in 2 = 2 \times 3.19 \times 10^9 = 6.38 \times 10^9\text{ lbf}\cdot\text{in}^2 = 2 × 3.19 × 1 0 9 = 6.38 × 1 0 9 lbf ⋅ in 2 δ C , h = 6.912 × 10 9 6.38 × 10 9 = 1.0834 in ≈ 1.08 in \delta_{C,h} = \frac{6.912 \times 10^9}{6.38 \times 10^9} = 1.0834\text{ in} \approx 1.08\text{ in} δ C , h ​ = 6.38 × 1 0 9 6.912 × 1 0 9 ​ = 1.0834 in ≈ 1.08 in

Tip C C C sways horizontally to the right by 1.08 in 1.08\text{ in} 1.08 in .

Part 3: Statically Indeterminate Reaction R C R_C R C ​

With a roller installed at tip C C C , the redundant upward force is R C R_C R C ​ . An equipment load W = 6 , 000 lbf W = 6,000\text{ lbf} W = 6 , 000 lbf acts downward at midpoint D D D ( x = L / 2 = 48 in x = L/2 = 48\text{ in} x = L /2 = 48 in from C C C ).

Establish moment expressions in terms of redundant R C R_C R C ​ :

On boom segment C D CD C D ( 0 ≤ x ≤ L / 2 0 \le x \le L/2 0 ≤ x ≤ L /2 ): M 1 ( x ) = R C x , ∂ M 1 ∂ R C = x M_1(x) = R_C x, \quad \frac{\partial M_1}{\partial R_C} = x M 1 ​ ( x ) = R C ​ x , ∂ R C ​ ∂ M 1 ​ ​ = x

On boom segment D B DB D B ( L / 2 ≤ x ≤ L L/2 \le x \le L L /2 ≤ x ≤ L ): M 2 ( x ) = R C x − W ( x − L 2 ) , ∂ M 2 ∂ R C = x M_2(x) = R_C x - W\left(x - \frac{L}{2}\right), \quad \frac{\partial M_2}{\partial R_C} = x M 2 ​ ( x ) = R C ​ x − W ( x − 2 L ​ ) , ∂ R C ​ ∂ M 2 ​ ​ = x

On column A B AB A B ( 0 ≤ y ≤ H 0 \le y \le H 0 ≤ y ≤ H ):

The moment transferred to the top of column B B B is constant along its height: M 3 ( y ) = R C L − W ( L 2 ) , ∂ M 3 ∂ R C = L M_3(y) = R_C L - W\left(\frac{L}{2}\right), \quad \frac{\partial M_3}{\partial R_C} = L M 3 ​ ( y ) = R C ​ L − W ( 2 L ​ ) , ∂ R C ​ ∂ M 3 ​ ​ = L

Apply the geometric compatibility condition at the rigid roller: ∂ U ∂ R C = 0 \frac{\partial U}{\partial R_C} = 0 ∂ R C ​ ∂ U ​ = 0 . 1 E I [ ∫ 0 L / 2 ( R C x ) ( x ) d x + ∫ L / 2 L [ R C x − W ( x − L 2 ) ] ( x ) d x + ∫ 0 H ( R C L − W L 2 ) ( L ) d y ] = 0 \frac{1}{EI} \left[ \int_0^{L/2} (R_C x)(x)\,dx + \int_{L/2}^L \left[R_C x - W\left(x - \frac{L}{2}\right)\right](x)\,dx + \int_0^H \left(R_C L - \frac{WL}{2}\right)(L)\,dy \right] = 0 E I 1 ​ [ ∫ 0 L /2 ​ ( R C ​ x ) ( x ) d x + ∫ L /2 L ​ [ R C ​ x − W ( x − 2 L ​ ) ] ( x ) d x + ∫ 0 H ​ ( R C ​ L − 2 W L ​ ) ( L ) d y ] = 0

Combine the R C R_C R C ​ terms: ∫ 0 L R C x 2 d x + ∫ 0 H R C L 2 d y = R C ( L 3 3 + L 2 H ) \int_0^L R_C x^2\,dx + \int_0^H R_C L^2\,dy = R_C \left( \frac{L^3}{3} + L^2 H \right) ∫ 0 L ​ R C ​ x 2 d x + ∫ 0 H ​ R C ​ L 2 d y = R C ​ ( 3 L 3 ​ + L 2 H )

Evaluate the load W W W integrals: ∫ L / 2 L W ( x 2 − L 2 x ) d x = W [ x 3 3 − L x 2 4 ] L / 2 L = W [ ( L 3 3 − L 3 4 ) − ( L 3 24 − L 3 16 ) ] = W ( L 3 12 + L 3 48 ) = 5 48 W L 3 \int_{L/2}^L W \left(x^2 - \frac{L}{2}x\right)dx = W \left[ \frac{x^3}{3} - \frac{L x^2}{4} \right]_{L/2}^L = W \left[ \left( \frac{L^3}{3} - \frac{L^3}{4} \right) - \left( \frac{L^3}{24} - \frac{L^3}{16} \right) \right] = W \left( \frac{L^3}{12} + \frac{L^3}{48} \right) = \frac{5}{48} W L^3 ∫ L /2 L ​ W ( x 2 − 2 L ​ x ) d x = W [ 3 x 3 ​ − 4 L x 2 ​ ] L /2 L ​ = W [ ( 3 L 3 ​ − 4 L 3 ​ ) − ( 24 L 3 ​ − 16 L 3 ​ ) ] = W ( 12 L 3 ​ + 48 L 3 ​ ) = 48 5 ​ W L 3 ∫ 0 H ( W L 2 ) L d y = W L 2 H 2 \int_0^H \left(\frac{WL}{2}\right) L\,dy = \frac{W L^2 H}{2} ∫ 0 H ​ ( 2 W L ​ ) L d y = 2 W L 2 H ​

Equating to zero yields the closed-form expression for R C R_C R C ​ : R C ( L 3 3 + L 2 H ) = W ( 5 48 L 3 + 1 2 L 2 H ) R_C \left( \frac{L^3}{3} + L^2 H \right) = W \left( \frac{5}{48} L^3 + \frac{1}{2} L^2 H \right) R C ​ ( 3 L 3 ​ + L 2 H ) = W ( 48 5 ​ L 3 + 2 1 ​ L 2 H )

Dividing both sides by L 2 L^2 L 2 : R C = W ( 5 48 L + 1 2 H 1 3 L + H ) R_C = W \left( \frac{\frac{5}{48} L + \frac{1}{2} H}{\frac{1}{3} L + H} \right) R C ​ = W ( 3 1 ​ L + H 48 5 ​ L + 2 1 ​ H ​ )

Substitute the numerical values ( L = 96 in L = 96\text{ in} L = 96 in , H = 120 in H = 120\text{ in} H = 120 in , W = 6 , 000 lbf W = 6,000\text{ lbf} W = 6 , 000 lbf ):

Numerator factor: 5 48 ( 96 ) + 1 2 ( 120 ) = 10 + 60 = 70 in \frac{5}{48}(96) + \frac{1}{2}(120) = 10 + 60 = 70\text{ in} 48 5 ​ ( 96 ) + 2 1 ​ ( 120 ) = 10 + 60 = 70 in

Denominator factor: 1 3 ( 96 ) + 120 = 32 + 120 = 152 in \frac{1}{3}(96) + 120 = 32 + 120 = 152\text{ in} 3 1 ​ ( 96 ) + 120 = 32 + 120 = 152 in R C = 6 , 000 lbf × ( 70 152 ) = 6 , 000 × 0.46053 = 2 , 763.16 lbf ≈ 2 , 763 lbf R_C = 6,000\text{ lbf} \times \left( \frac{70}{152} \right) = 6,000 \times 0.46053 = 2,763.16\text{ lbf} \approx 2,763\text{ lbf} R C ​ = 6 , 000 lbf × ( 152 70 ​ ) = 6 , 000 × 0.46053 = 2 , 763.16 lbf ≈ 2 , 763 lbf

The redundant roller support provides an upward reaction force of 2 , 763 lbf 2,763\text{ lbf} 2 , 763 lbf , completing the indeterminate solution with exact mathematical elegance.