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CE-503 (A) · Structural analysis-II/Quick Revision Short Notes

Structural analysis-II (CE-503 (A)) - Unit 4 Short Notes

UNIT 4: Advanced Structural Analysis

I. Fundamental Concepts and Definitions

Term Definition Key Formula/Concept
Static Indeterminacy Degree of redundancy; number of extra reactions/forces beyond equilibrium equations. For beams/frames: r - e (r = reactions, e = 3 for 2D frames)
Kinematic Indeterminacy Number of independent joint displacements (rotations & translations). For beams: 2j - r - c (j = joints, r = reactions, c = constraints)
Load Factor (γ) Factor by which working loads are multiplied to obtain ultimate/plastic collapse load. P_u = γ * P_w
Plastic Modulus (Z_p) Section property for plastic moment capacity. Sum of areas on each side of PNA × distance from their centroids to PNA. M_p = f_y * Z_p
Plastic Neutral Axis (PNA) Axis dividing the cross-section into two equal areas under compression and tension at yield. A_c = A_t (areas above & below PNA)
Relative Stiffness (K) Stiffness of a member relative to others at a joint. K = (I/L) for far end fixed; K' = (4I/L) for far end pinned
Distribution Factor (DF) Portion of unbalanced moment at a joint distributed to a connecting member. DF_AB = (K_AB) / (ΣK at joint A)
Carry-Over Factor (COF) Fraction of moment carried over to the far end when a moment is applied at the near end. COF = 0.5 for prismatic members with far end fixed.
Factor of Safety (FoS) Ratio of material strength to allowable/working stress. FoS = f_y / f_allow
Shape Factor (S) Ratio of plastic moment capacity to yield moment capacity. S = M_p / M_y = Z_p / Z
Plastic Hinge A zone in a beam where plastic moment M_p is reached, allowing rotation like a mechanical hinge. Formation of 3 hinges in a beam causes collapse mechanism.

[!TIP] Exam Focus: Be prepared to define all key terms (as in Nov 2023 paper). Understand the difference between plastic and elastic section modulus and the significance of shape factor (S > 1 for I-sections).


II. Methods of Analysis for Indeterminate Structures

A. Moment Distribution Method (MDM)

  • Principle: Iterative method to solve for joint rotations by distributing unbalanced fixed-end moments.

  • Assumptions: Joints are rigid (no translation), members are prismatic.

  • Key Steps:

    1. Calculate Fixed-End Moments (FEM) for all members under given loads.

    2. Calculate Distribution Factors (DF) at each joint (ΣDF = 1).

    3. Lock all joints (apply COF of 0.5 to far ends).

    4. Release joints one by one: distribute unbalanced moment, carry over half to far ends.

    5. Repeat until moments are negligible.

    6. Sum FEM + distributed moments for final member end moments.

  • Fixed-End Moments (Common Cases - for member AB, A left, B right):

    • UDL of intensity w over entire span L: FEM_AB = -wL²/12, FEM_BA = +wL²/12

    • Point load P at mid-span: FEM_AB = -PL/8, FEM_BA = +PL/8

    • Point load P at distance a from A, b from B: FEM_AB = -Pb²L / L², FEM_BA = -Pa²L / L²

  • Kani’s Method: A variation where distribution factors are calculated for the next cycle based on current moments, often faster for frames with sidesway.

[!TIP] Common Pitfall: Forgetting to carry over distributed moments. For frames, sway correction must be applied separately if lateral displacement exists.

B. Flexibility Method (Force Method)

  • Basic Idea: Release enough redundants to make the structure statically determinate (primary structure). Apply compatibility (displacement) conditions to solve for redundants.

  • Procedure:

    1. Determine degree of indeterminacy n.

    2. Select n redundants (reactions or internal forces). Release them → primary determinate structure.

    3. Calculate flexibility coefficients δ_ij (displacement at i due to unit load at j).

    4. Write compatibility equations: δ_i0 + Σ(δ_ij * X_j) = 0 (where X_j are redundants, δ_i0 is displacement due to original loads).

    5. Solve for X_j. Find all other reactions/moments by superposition.

  • Flexibility Coefficient: δ_ij = ∫(m_i m_j / EI) dx (using Moment-Area or Conjugate Beam).

C. Stiffness Method (Displacement Method)

  • Basic Idea: Assume unknown joint displacements (rotations & translations). Write equilibrium equations for joints in terms of displacements and stiffness coefficients.

  • Procedure:

    1. Determine kinematic indeterminacy k.

    2. Assign unknown displacement degrees of freedom (DOF).

    3. Apply fixed-end forces (equivalent joint loads) for each DOF.

    4. Form stiffness matrix [K] where K_ij = force at i due to unit displacement at j.

    5. Solve [K]{D} = {F} for displacements {D}.

    6. Compute member end moments from displacements.

  • Stiffness Coefficient: K_ij for rotation at end A due to unit rotation at same end (far fixed) = 4EI/L.

[!TIP] Comparison: Flexibility Method is easier for structures with fewer redundants (e.g., continuous beams). Stiffness Method is better for frames with many joints and is the basis for computer analysis.


III. Influence Lines for Indeterminate Structures

A. Müller-Breslau Principle (MBP)

  • Statement: The influence line for any reaction (or internal force) in a structure is obtained by removing the restraint corresponding to that quantity and introducing a corresponding displacement (or rotation) of unit magnitude in the direction of the positive influence quantity, while keeping the structure statically determinate.

  • Key: The deformed shape of the released determinate structure is the influence line.

  • Procedure:

    1. Identify the quantity (reaction, shear, moment) for which IL is needed.

    2. Release the corresponding restraint (e.g., for vertical reaction R_A, remove support at A).

    3. Apply a unit displacement in the positive direction of the quantity (e.g., +1 unit upward for R_A).

    4. Determine the deformed shape (using static equilibrium or qualitative reasoning). This shape is the influence line.

    5. For shear or moment at a section, make a cut at the section and apply the unit displacement/rotation that would be caused by the internal force.

[!TIP] Crucial: MBP gives the correct influence line shape but not the scale for indeterminate structures. To get ordinates, you must use the actual indeterminate structure (e.g., via method of consistent deformations) or use the principle of virtual work.

B. Influence Lines for Beams (Indeterminate)

  • Example: Double Overhanging Beam (Continuous)

    • Reaction at B: Release support B, apply +1 upward. The IL is a continuous curve with positive and negative ordinates.

    • Shear at section n: Cut at n, apply +1 shear (relative displacement). IL consists of two lines meeting at the cut.

    • Moment at section n: Cut at n, apply +1 rotation (relative rotation). IL is a continuous curve with a peak/valley at the section.

    • Ordinate Calculation: Use virtual work: Q = Σ (m_q * M / EI) over all members, where m_q is IL ordinate from MBP shape.

C. Influence Lines for Frames

  • Apply MBP by releasing the specific support or making a cut at the section of interest.

  • For sway frames, the influence line for a reaction will have contributions from both joint rotations and translations.

D. Continuous Beams

  • Use MBP by releasing the support/reaction. The resulting shape is a piecewise linear or curved diagram.

  • Ordinates are best found using Clough's Theorem (for equal spans) or method of moments.


IV. Plastic Analysis

A. Plastic Moment and Hinge Formation

  • Plastic Moment Capacity (M_p): Maximum moment a section can resist when entire cross-section yields. M_p = f_y * Z_p.

  • Collapse Mechanism: Structure becomes a mechanism when sufficient plastic hinges form. For a beam, 3 hinges are needed (2 at supports/points of zero moment, 1 in span).

  • Collapse Load: Load at which a mechanism forms. Found using static theorem (equilibrium of mechanism) or kinematic theorem (virtual work).

  • Load Factor (γ): γ = M_p / M_max (from elastic analysis). Lower γ means more economical design.

B. Plastic Analysis of Beams

  • Propped Cantilever (UDL):

    • Hinges form at fixed end (M = M_p), prop (R = reaction, but moment zero), and somewhere in span (M = M_p).

    • Use virtual work: wL * Δ = M_p * θ1 + M_p * θ2. Solve for w.

  • Simply Supported Beam (UDL):

    • Hinges form at both supports and mid-span.

    • Collapse load: w_u L² / 8 = M_p → w_u = 8M_p / L².

  • Fixed-Fixed Beam (UDL):

    • Hinges form at both ends and one or two interior points. Usually 4 hinges total (2 per span if continuous).

    • For single span: hinges at ends and at L/4 from each support? Actually, for UDL, hinges form at supports and at points of contraflexure (near L/4 from ends). Collapse load w_u = 16M_p / L².

C. Plastic Section Properties

  • Plastic Modulus (Z_p):

    • Rectangular (b×h): Z_p = (b h²) / 4

    • I-Section: Sum of (A_i * y_i) for areas above & below PNA. PNA is at centroid for symmetric sections.

    • Circular (radius R): Z_p = (4R³) / 3

  • Shape Factor (S):

    • S = M_p / M_y = Z_p / Z (Z = elastic section modulus).

    • For I-section: S ≈ 1.10 - 1.20 (depends on flange/web ratio).

    • For rectangular: S = 1.5.

    • Significance: Indicates reserve strength beyond yield. Higher S means more economical plastic design.

  • Design: Select section where M_p ≥ M_applied. Use Z_p = M_p / f_y.

[!TIP] Exam Problem: You may be asked to locate plastic hinges and compute collapse load for a given beam (like propped cantilever). Always draw the collapse mechanism first.


V. Lateral Load Analysis for Tall Buildings

A. Wind and Earthquake Loads

  • Wind Loads (IS 875 Part 3):

    • Basic wind speed (V_b) based on region.

    • Design wind pressure: p_z = K_1 * K_2 * K_3 * V_b² (risk, terrain, height factors).

    • Applied as: Lateral point loads at floor levels (equivalent static) or distributed pressure.

  • Earthquake Loads (IS 1893):

    • Base shear method: V_b = A_h * W (seismic coefficient A_h = (Z * I * S_a) / (R / g)).

    • Vertical distribution: F_i = (w_i * h_i) / Σ(w_j * h_j) * V_b.

    • Dynamic analysis (response spectrum) for irregular/tall buildings.

B. Analysis Methods for Frames

  • Portal Method:

    • Assumptions: Points of contraflexure at mid-height of columns and mid-span of beams.

    • Shear distribution: Equal shear in columns of a bay. Odd bays take odd stories' shear.

    • Used for low-rise frames.

  • Cantilever Method:

    • Assumptions: Points of contraflexure at mid-height of columns. Axial forces in columns proportional to distance from C.G.

    • Shear distribution: Proportional to column stiffness and distance from C.G.

    • Better for high-rise buildings.

  • Factor Method (Kani’s for Sway):

    • Distribution factors are modified to account for joint translations.

    • More accurate but complex.

C. Stability and Drift Criteria

  • Story Drift (Δ): Lateral displacement difference between two consecutive floors.

  • Limits (IS 1893/IS 456):

    • For wind/earthquake: Δ / h ≤ 0.004 (h = story height) for moment-resisting frames.

    • For serviceability (wind): Δ / h ≤ 0.002 (comfort).

  • Overall Stability (P-Δ effects): Check if secondary moments from lateral displacement are significant. Usually neglected if Δ / h < 0.01.

  • Torsion: Check eccentricity between center of mass and center of stiffness.

[!TIP] Key Point: Portal method assumes equal column shear in a bay. Cantilever method assumes column forces proportional to distance from C.G. Know when to use which.


DiagramCANVAS: Sketch showing plastic hinge formation in a propped cantilever under UDL, with hinges at fixed end, prop, and mid-span, and the collapse mechanism shape.
DiagramCANVAS: Influence line for reaction at B in a continuous beam using Müller-Breslau, showing a continuous curve with positive and negative ordinates.
DiagramCANVAS: Portal method assumptions for a 3-bay 3-story frame, showing inflection points at mid-heights and mid-spans, and shear distribution in columns.
DiagramCANVAS: Comparison of plastic and elastic stress distributions in an I-section, showing PNA at centroid for plastic case and linear elastic distribution.
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