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6.2 Structural Analysis Methods

Key Takeaways

  • A 2D structure's static determinacy and stability are classified using member, reaction, joint, and release equations, such as 3m + r = 3j + ci for frames and b + r = 2j for trusses.

  • Geometric instability occurs when reactions are concurrent or parallel, or when internal mechanisms are created, rendering the structure unstable regardless of equation results.

  • Zero-force members are truss elements carrying no load under a specific configuration; they can be identified by inspection at unloaded joints with two or three members.

  • Conjugate beam theorems state that slope equals conjugate shear (V_c) and deflection equals conjugate moment (M_c), with support conditions transformed accordingly.

  • The Virtual Work Method uses virtual unit loads to calculate deflections via the sum of (n * N * L) / (A * E) for trusses and the integral of (m * M) / (E * I) dx for beams.

Last updated: July 2026

Structural Analysis Methods

Structural analysis is the determination of the effects of loads on physical structures and their components. This section covers structural determinacy, truss analysis, shear and moment diagrams, and deflection calculations, which are central topics on the PE Civil exam.


Structural Determinacy and Stability

A structure is categorized based on the relationship between the number of available equations of static equilibrium and the number of unknown support reactions and internal member forces.

1. Classification of Structures

  • Statically Determinate: A structure is determinate if all reactions and internal forces can be solved using the equations of static equilibrium alone: ∑Fx=0,∑Fy=0,∑Mz=0\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_z = 0
  • Statically Indeterminate: A structure is indeterminate if it has more unknown forces than available equilibrium equations. The excess unknowns are called redundants, and the number of redundants defines the degree of static indeterminacy.
  • Statically Unstable: A structure is unstable if it cannot resist loads from any arbitrary direction without undergoing rigid-body motion.

2. Determinacy Equations for 2D Systems

To evaluate determinacy, we compare the unknowns (reactions rr, members/bars mm or bb) against the equations provided by joints jj and internal releases cic_i.

Beams and Rigid Frames

For a 2D beam or frame consisting of mm members, rr support reaction components, jj joints, and cic_i internal release equations:

  • 3m+r<3j+ci  ⟹  3m + r < 3j + c_i \implies Statically Unstable
  • 3m+r=3j+ci  ⟹  3m + r = 3j + c_i \implies Statically Determinate (if geometrically stable)
  • 3m+r>3j+ci  ⟹  3m + r > 3j + c_i \implies Statically Indeterminate (degree of indeterminacy: id=3m+r−3j−cii_d = 3m + r - 3j - c_i)

Note: For an internal hinge connecting nn members, the number of release equations is ci=n−1c_i = n - 1. For a hinge connecting 2 members, ci=1c_i = 1.

Plane Trusses

For a 2D truss with bb bar members, rr support reaction components, and jj joints:

  • b+r<2j  ⟹  b + r < 2j \implies Statically Unstable
  • b+r=2j  ⟹  b + r = 2j \implies Statically Determinate (if geometrically stable)
  • b+r>2j  ⟹  b + r > 2j \implies Statically Indeterminate (degree of indeterminacy: id=b+r−2ji_d = b + r - 2j)

3. Geometric Instability

A structure may satisfy the determinacy equations (3m+r≥3j+ci3m+r \ge 3j+c_i or b+r≥2jb+r \ge 2j) but still be unstable if it is geometrically unstable. Instability occurs if:

  • Concurrent Reactions: The lines of action of all support reactions intersect at a single point, allowing the structure to rotate about that point.
  • Parallel Reactions: All support reactions are parallel, allowing the structure to translate freely in the perpendicular direction.
  • Internal Mechanisms: The arrangement of internal members or releases creates a mechanism (e.g., three collinear hinges in a beam).

Analysis of Plane Trusses

Trusses are structures composed of straight members connected at joints, which are assumed to act as frictionless pins. Truss members carry only axial forces (tension or compression); bending moments and shear forces are assumed to be zero.

1. Method of Joints

The Method of Joints involves isolating each joint and applying the equations of equilibrium:

∑Fx=0,∑Fy=0\sum F_x = 0, \quad \sum F_y = 0
  • Procedure:
    1. Calculate support reactions for the entire truss.
    2. Select a joint with no more than two unknown member forces.
    3. Draw the free-body diagram of the joint, assuming unknown member forces are in tension (pulling away from the joint). A negative solved value indicates compression (pushing toward the joint).
    4. Solve the two equilibrium equations.
    5. Move to an adjacent joint and repeat.

2. Method of Sections

The Method of Sections is used when the forces in only a few specific members are required.

  • Procedure:
    1. Calculate support reactions.
    2. Pass a cutting plane (section cut) through the truss, cutting through no more than three members with unknown forces (one of which is the target member).
    3. Draw the free-body diagram of one of the cut parts.
    4. Apply the three equations of equilibrium: ∑Fx=0,∑Fy=0,∑MO=0\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0
    5. Choose the moment center OO at the intersection of the lines of action of two unknown forces. This eliminates them from the moment equation, allowing the third force to be solved directly.

3. Zero-Force Members

Zero-force members carry no load under a specific loading condition. Identifying them simplifies truss analysis.

  • Rule 1: If only two non-collinear members meet at a joint with no external load or support reaction, both are zero-force members.
  • Rule 2: If three members meet at a joint where two are collinear and the third is non-collinear, the non-collinear member is a zero-force member, provided there is no external load or support reaction at the joint.

Shear and Moment Diagrams

Shear and bending moment diagrams plot the internal shear force (VV) and bending moment (MM) along the length of a structural member.

1. Mathematical Relationships

The relationships between the distributed load w(x)w(x), shear V(x)V(x), and moment M(x)M(x) are governed by:

dVdx=−w(x)  ⟹  V(x2)−V(x1)=−∫x1x2w(x)dx\frac{dV}{dx} = -w(x) \implies V(x_2) - V(x_1) = -\int_{x_1}^{x_2} w(x) dx dMdx=V(x)  ⟹  M(x2)−M(x1)=∫x1x2V(x)dx\frac{dM}{dx} = V(x) \implies M(x_2) - M(x_1) = \int_{x_1}^{x_2} V(x) dx

2. Graphical Rules

  • Load to Shear:
    • A concentrated point load causes a vertical step change in the shear diagram equal to the load magnitude.
    • A uniform downward load (ww) results in a downward-sloping linear shear diagram with slope equal to −w-w.
  • Shear to Moment:
    • The slope of the moment diagram at any point is equal to the value of the shear force at that point.
    • The change in moment between two points is equal to the area under the shear diagram between those points.
    • A point where the shear diagram crosses zero (or has a discontinuity crossing zero) corresponds to a local maximum or minimum bending moment.
    • A concentrated moment causes a vertical step change in the moment diagram. (A clockwise external moment causes an upward step, and a counter-clockwise external moment causes a downward step in the moment diagram, depending on the sign convention used).

Deflection Calculations

Deflection calculations are essential to verify serviceability limits (e.g., L/360L/360 for live load deflection).

1. Moment-Area Method

The Moment-Area Method is a semi-graphical method based on two theorems relating the geometry of the elastic curve to the area of the M/EIM/EI diagram.

  • First Theorem: The angle θB/A\theta_{B/A} between the tangents to the elastic curve at points AA and BB is equal to the area under the M/EIM/EI diagram between those two points: θB−θA=∫ABMEIdx=Area[A−B](MEI)\theta_B - \theta_A = \int_A^B \frac{M}{EI} dx = \text{Area}_{[A-B]} \left( \frac{M}{EI} \right)
  • Second Theorem: The tangential deviation tB/At_{B/A} of point BB relative to a tangent line drawn at point AA is equal to the first moment of the area under the M/EIM/EI diagram between AA and BB about point BB: tB/A=∫ABxMEIdx=Area[A−B](MEI)×xˉBt_{B/A} = \int_A^B x \frac{M}{EI} dx = \text{Area}_{[A-B]} \left( \frac{M}{EI} \right) \times \bar{x}_B Where xˉB\bar{x}_B is the distance from the centroid of the area to point BB.

Note: Tangential deviation is NOT deflection; it is the vertical distance from point B on the elastic curve to the tangent line projected from point A. Deflection must be calculated by relating tangential deviations using geometry.

2. Conjugate Beam Method

The Conjugate Beam Method transforms the physical beam deflection problem into an equivalent beam loading and shear/moment analysis.

  • Principles:
    • The slope θ\theta at any point in the real beam is equal to the shear force VcV_c at the corresponding point in the conjugate beam: θ=Vc\theta = V_c.
    • The deflection Δ\Delta at any point in the real beam is equal to the bending moment McM_c at the corresponding point in the conjugate beam: Δ=Mc\Delta = M_c.
  • Boundary Condition Conversions:
    Real Beam SupportConjugate Beam SupportReason
    Fixed End (θ=0,Δ=0\theta = 0, \Delta = 0)Free End (Vc=0,Mc=0V_c = 0, M_c = 0)Matches zero slope/deflection.
    Free End (θ≠0,Δ≠0\theta \ne 0, \Delta \ne 0)Fixed End (Vc≠0,Mc≠0V_c \ne 0, M_c \ne 0)Matches non-zero slope/deflection.
    Pin/Roller (End) (θ≠0,Δ=0\theta \ne 0, \Delta = 0)Pin/Roller (End) (Vc≠0,Mc=0V_c \ne 0, M_c = 0)Matches non-zero slope, zero deflection.
    Roller (Internal) (θcont,Δ=0\theta_{\text{cont}}, \Delta = 0)Internal Hinge (Vc disc.,Mc=0V_c \text{ disc.}, M_c = 0)Hinge allows slope discontinuity (shear step) but zero moment.
    Internal Hinge (θ disc.,Δcont\theta \text{ disc.}, \Delta_{\text{cont}})Internal Roller/Support (Vc cont.,Mc≠0V_c \text{ cont.}, M_c \ne 0)Roller allows moment (deflection) but maintains shear.

3. Virtual Work Method

The Virtual Work Method (or Unit Load Method) is a powerful, general method for calculating deflections in both determinate and indeterminate structures.

  • Truss Deflection: 1⋅Δ=∑nNLAE1 \cdot \Delta = \sum \frac{n N L}{A E} Where:
    • NN = axial force in a truss member due to actual external loads.
    • nn = axial force in a truss member due to a virtual unit load applied at the joint and in the direction of the desired deflection.
    • LL = member length.
    • AA = member cross-sectional area.
    • EE = modulus of elasticity.
  • Beam and Frame Deflection: 1⋅Δ=∫0LmMEIdx1 \cdot \Delta = \int_0^L \frac{m M}{EI} dx Where:
    • MM = internal bending moment in the beam due to actual external loads.
    • mm = internal bending moment due to a virtual unit load applied at the point and in the direction of the desired deflection.
    • EE = modulus of elasticity.
    • II = moment of inertia.

These integrals are typically evaluated using product integration tables or standard formulas for common shapes, which saves significant time on the exam.

Test Your Knowledge

Determine the static determinacy and stability of a 2D truss with 11 members, 6 joints, and 3 support reactions.

A

Statically unstable

B

Statically determinate and stable

C

Statically indeterminate to the 1st degree

D

Statically indeterminate to the 2nd degree

Test Your Knowledge

Which real beam support condition converts to an internal hinge in the conjugate beam method?

A

Fixed support

B

Internal support (roller or pin)

C

Free end

D

End roller or pin

Test Your Knowledge

Using the virtual work method, what is the correct formula to calculate the deflection of a statically determinate truss due to external loads?

A

Sum of (n * N * L) / (A * E)

B

Sum of (n * N * L^2) / (A * E)

C

Integral of (m * M) / (E * I) dx

D

Sum of (n * N) / (A * E)

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