Structural analysis-II (CE-503 (A)) - Unit 3 Short Notes

UNIT 3: Advanced Structural Analysis (CE-503 A)

I. Fundamental Definitions and Principles

[!TIP] Exam Focus: Distinguish clearly between Plastic Modulus (Z_p) and Elastic Section Modulus (Z). Z_p ≥ Z. Their ratio is the Shape Factor.

II. Displacement Methods of Structural Analysis

A. Moment Distribution Method (Hardy Cross)

Core Idea: Iterative distribution of fixed-end moments (FEM) until joints are in equilibrium (unbalanced moment ≈ 0).

Procedure & Key Rules:

1. Calculate Fixed-End Moments (FEM): For each member under given loads (consider sign convention: sagging +, hogging -).

2. Compute Distribution Factors (DF): At each joint (except fixed supports), $$\displaystyle DF = \frac{K}{\sum K} $$.

3. Release Joints Sequentially:

   • Distribute the unbalanced moment at a joint to connected members using their DFs.

   • Carry-over: Distribute half of the distributed moment to the far end of each member (Carry-over Factor = 0.5 for prismatic members).

4. Repeat: Cycle through all joints until moments are negligible (typically < 1% of largest FEM).

5. Sum Moments: Final moment at each end = FEM + Distributed + Carry-overs.

Analysis of Continuous Beams:

• Apply procedure above.

• Draw Bending Moment Diagram (BMD) by moment summation at each section.

• Draw Shear Force Diagram (SFD) from BMD and applied loads.

Analysis of Rigid Frames:

• Treat columns and beams as members.

• Far-end conditions matter: For columns, far end may be fixed (base) or pinned (if footing allows rotation).

• Sway Frames: If sidesway is possible, a sway force must be applied to prevent displacement, and analysis is done in two parts: (i) Non-sway (joint rotations only), (ii) Sway (with artificial horizontal force). Moments from both parts are superimposed.

[!TIP] Common Pitfall: Forgetting to include carry-over moments from previous cycles. Always carry over from distributed moments, not the net moment after carry-over.

B. Stiffness Methods

Kani's Method (Improved Moment Distribution):

• Key Difference: Uses rotation stiffness (K_r) and distribution factors (DF) based on K_r.

• K_r Calculation: $$\displaystyle K_r = \frac{4EI}{L} $$ for far-end fixed, $$\displaystyle K_r = \frac{3EI}{L} $$ for far-end pinned. Same as relative stiffness (K).

• Procedure: Similar to moment distribution but carry-over is automatic in the stiffness formulation. Often faster for frames with many joints.

• Output: Directly gives joint rotations and final end moments.

General Stiffness Method for Beams:

• Based on displacement (slope-deflection) equations.

• Slope-Deflection Equation (for member AB, no settlement):

$$M_{AB} = \frac{2EI}{L} \left( 2\theta_A + \theta_B - 3\psi \right) + FEM_{AB}$$

where $$\displaystyle \psi = \frac{\Delta}{L} $$ (chord rotation due to translation).

• Write equations for all members, solve simultaneous equations for unknown rotations (θ) and/or sway displacements (Δ).

III. Force Method (Flexibility Method)

Core Idea: Choose a basic determinate structure by releasing enough redundants (forces/displacements) to make it statically determinate. Apply compatibility conditions (known displacements) to solve for redundants.

Flexibility Matrix Approach:

1. Select Redundants: Number = degree of indeterminacy (e.g., for a 2-span continuous beam, 1 redundant reaction).

2. Form Basic Structure: Remove redundants.

3. Compute Flexibility Coefficients (f_ij):

   • $$\displaystyle f_{ij} $$ = displacement at the location/direction of redundant i due to a unit load applied at location/direction of redundant j on the basic structure.

   • Use Mohan's Theorem or Conjugate Beam method.

4. Compatibility Equation:

$$\delta_{i0} + \sum_{j=1}^{n} f_{ij} \cdot X_j = 0$$

where $$\displaystyle \delta_{i0} $$ = displacement at *i* due to **original loads** on basic structure, $$\displaystyle X_j $$ = unknown redundants.

5. Solve for $$\displaystyle X_j $$, then find final reactions and moments by superposition.

Application to Continuous Beams:

• Common redundant: One support reaction (e.g., interior support of a 2-span beam).

• Basic structure: Propped cantilever or simply supported spans.

• $$\displaystyle \delta_{i0} $$: Deflection at the released support due to applied loads (must be zero in original structure).

• $$\displaystyle f_{ii} $$: Deflection at that support due to unit redundant.

[!TIP] Exam Tip: For continuous beams, using Mohan's Theorem (area of M/EI diagram from real load / (area of M/EI diagram from unit load) * L) is efficient for calculating $$\displaystyle f_{ij} $$ and $$\displaystyle \delta_{i0} $$.

IV. Plastic Analysis and Design

A. Plastic Hinge Concept & Collapse Mechanisms

• Plastic Hinge: A finite length of a beam where yielding is complete, and rotation can occur at constant moment = $$\displaystyle M_p $$. It behaves like a mechanical hinge but can transmit moment.

• Collapse Mechanism: A kinematically admissible mechanism formed by sufficient plastic hinges (typically number of hinges = degree of static indeterminacy + 1) that causes unlimited displacements.

• Collapse Load (W_u): Found by virtual work:

$$\text{External Work} = \text{Internal Plastic Work}$$

$$W_u \cdot \delta = \sum M_p \cdot \theta$$

where $\delta$ is the **virtual displacement** at the point of load application, and $\theta$ are **plastic rotations** at hinges.

Example: Propped Cantilever under UDL (Span L)

• Hinge Formation: At fixed end (A), prop (B), and mid-span (C).

• Mechanism: Beam becomes a simply supported beam with a hinge at mid-span.

• Virtual Work:

  • External Work: $$\displaystyle w_u L \cdot (\delta/2) $$ (load moves by $\delta/2$ at midspan).

  • Internal Work: $$\displaystyle M_p(\theta_A + \theta_B) + 2M_p \theta_C $$.

  • Geometry gives: $$\displaystyle \theta_A = \theta_B = \theta_C = \delta/L $$.

  • $$\displaystyle \therefore w_u L \cdot (\delta/2) = M_p(2 \cdot \delta/L) + 2M_p(\delta/L) = 4M_p \delta / L $$

  • Collapse Load: $$\displaystyle \boxed{w_u = \frac{8M_p}{L^2}} $$

B. Plastic Design of Steel Sections (I-Sections)

• Design Condition: $$\displaystyle M_{max} \leq M_p = f_y \cdot Z_p $$.

• Procedure:

  1. Calculate ultimate load (factored load): $$\displaystyle W_u = \gamma \cdot W_{working} $$ (γ = 1.5 for UDL in IS 800).

  2. Find maximum design moment $$\displaystyle M_u $$ for the span (e.g., for simply supported beam with UDL: $$\displaystyle M_u = \frac{W_u L}{8} $$).

  3. Select a rolled I-section from tables such that its plastic moment capacity $$\displaystyle M_p \geq M_u $$.

  4. Check for shear and local buckling as per IS 800.

C. Plastic Section Properties

• Section Modulus (Z): Elastic property, $$\displaystyle Z = \frac{I}{y_{max}} $$.

• Plastic Modulus (Z_p): $$\displaystyle Z_p = \sum \frac{A_i \cdot \bar{y}_i}{2} $$ (sum over compression/tension zones).

• Shape Factor (SF): $$\displaystyle SF = \frac{Z_p}{Z} \geq 1 $$. Indicates reserve strength beyond elastic limit.

  • For rectangular section: $$\displaystyle SF = 1.5 $$

  • For I-section: $SF \approx 1.1 - 1.2$

Calculation for Circular Section (Radius R):

• Elastic Z: $$\displaystyle Z = \frac{\pi R^3}{4} $$

• Plastic Z: PNA is diameter. Compression area = half circle.

  $$\displaystyle Z_p = 2 \times \left( \int_0^R y \cdot (2\sqrt{R^2-y^2}) dy \right) = \frac{4R^3}{3} $$

• Shape Factor: $$\displaystyle SF = \frac{4R^3/3}{\pi R^3/4} = \frac{16}{3\pi} \approx 1.697 $$

[!TIP] Remember: For plastic design, use Z_p. For elastic design, use Z. Shape factor shows the extra moment capacity available in plastic range.

V. Influence Lines for Indeterminate Structures

A. Muller-Breslau Principle

Statement: The influence line for any reaction, shear, or moment in an indeterminate structure has the same shape as the deflected shape of the structure when the corresponding constraint is removed and a unit displacement (rotation for moment, vertical for reaction) is imposed in the positive direction.

• Steps:

  1. Remove the constraint corresponding to the desired function (e.g., remove support for reaction, cut section for shear/moment).

  2. Impose a unit displacement (1 for reaction, 1 rotation for moment) in the positive direction.

  3. The resulting elastic curve is the influence line.

  4. Determine ordinates using geometry or method of joints.

B. Influence Line Construction

For Continuous Beams:

• Reaction (e.g., at support B): Remove support B, impose unit upward displacement. The beam becomes a simply supported beam on remaining supports. ILD is piecewise linear.

• Shear Force (at section X): Cut section X, impose unit relative vertical displacement (left side up, right side down). The beam splits into two independent simply supported spans. ILD is piecewise linear.

• Bending Moment (at section X): Insert a hinge at X, impose unit relative rotation. The beam becomes a mechanism. ILD is piecewise linear but with a kink at X.

For Overhanging Beams:

• Similar procedure. The ILD may extend beyond supports into overhangs.

For Propped Cantilevers:

• Reaction at prop: Remove prop, impose unit upward displacement. The cantilever becomes a simply supported beam on the fixed end and the prop location.

• Moment at fixed end: Insert hinge at fixed end, impose unit rotation. The beam becomes a simply supported beam on the prop.

C. Numerical Ordinate Calculation & Diagram Drawing

• Use method of joints or geometry on the deflected shape from Muller-Breslau.

• For continuous beams, calculate ordinates at key points: supports, load points, points of contraflexure.

• Example: Influence Line for Reaction at B (2-span beam AB, BC):

  • Remove B, impose unit ↑ at B.

  • Span AB: Simply supported on A and B. Ordinate at any point x from A: $$\displaystyle y = \frac{x}{L_{AB}} $$ (linear from 0 at A to 1 at B).

  • Span BC: Simply supported on B and C. Ordinate at any point x' from B: $$\displaystyle y = 1 - \frac{x'}{L_{BC}} $$ (linear from 1 at B to 0 at C).

[!TIP] Key Insight: The area under the influence line for a moment or reaction over a span gives the value of that function due to a unit load moving across that span.

VI. Special Topics in Structural Analysis

A. Lateral Load Effects on High-Rise Structures

Wind Load Considerations (IS 875 Part 3):

• Dynamic in nature: Causes sway, torsion, and vortex shedding.

• Analysis Methods:

  • Static Equivalent: Apply wind pressure as lateral loads at each floor (simplified).

  • Dynamic Analysis: For tall/slender buildings, use mode superposition or response spectrum (for earthquake) but adapted for wind.

• Effects: Increases story shear, overturning moment, and drift. Must check P-Δ effects (secondary moments due to lateral displacement).

Earthquake Load Considerations (IS 1893):

• Inertial forces: $$\displaystyle F = m \cdot a_g $$ (mass × ground acceleration).

• Base Shear Method (for low-rise): $$\displaystyle V_b = A_h \cdot W $$, where $$\displaystyle A_h $$ = horizontal seismic coefficient.

• Dynamic Analysis (for high-rise): Response Spectrum Method is standard. Obtain modal participation factors, mode shapes, and combine modal responses (SRSS or CQC).

• Effects: Similar to wind but often larger magnitudes and directional. Requires ductile detailing (IS 13920).

B. Codal Provisions (BIS / IS Codes)

• IS 875 (Parts 1, 2, 3): Provides dead, live, and wind loads.

• IS 1893 (Part 1): Criteria for earthquake resistant design. Gives seismic zone map, importance factor, response spectrum, and lateral force procedure.

• IS 800 (2007): General steel design, including plastic analysis provisions (Clause 6.2).

• IS 13920: Ductile detailing for earthquake resistance.

• For Tall Buildings (IS 16700): Specific guidelines for analysis and design of buildings > 50m. Emphasizes performance-based design, wind tunnel testing for very tall buildings, and advanced analysis methods.

[!TIP] Exam Focus: Know the difference between static and dynamic lateral load analysis. Be familiar with IS 1893's base shear formula and the response spectrum method concept. Always mention relevant IS codes when discussing design.