Structural Analysis-I (CE-403) Unit 4 Notes

UNIT 4: Structural Analysis-I Short Notes

I. FUNDAMENTAL CONCEPTS & INDETERMINACY

Static Indeterminacy

Definition: The state where the number of unknown reaction and/or internal member forces exceeds the number of independent equilibrium equations available.
Significance: Requires compatibility conditions (deformations) in addition to equilibrium for solution.
Calculation for Plane Frames (Neglecting Axial Deformation):
- Formula: $$\displaystyle N_s = 3j + r_e - 3m - r_i $$
  - $j$ = number of joints (including supports)
  - $$\displaystyle r_e $$ = number of external reaction components
  - $m$ = number of members
  - $$\displaystyle r_i $$ = number of internal reaction components (like fixed end moments, which are 2 per fixed joint)
Calculation for Trusses:
- Formula: $$\displaystyle N_s = m + r_e - 2j $$
  - $m$ = number of members
  - $$\displaystyle r_e $$ = number of external reaction components
  - $j$ = number of joints

Kinematic Indeterminacy (Degree of Freedom)

Definition: The number of independent joint displacements (translations/rotations) possible in a structure.
Significance: Equals the number of unknown displacement/rotation parameters needed to describe the deformed shape. For a stable, determinate structure: $$\displaystyle N_k = N_s $$.
Calculation: Count independent joint translations (perpendicular to members for frames) and rotations. For a plane frame, each rigid joint has 3 potential DOF (2 trans, 1 rot), but constraints reduce this.

Stability of Structures

Concept: Ability of a structure to maintain its geometry under load without undergoing mechanism (collapse).
Criteria:
1. External Stability: The structure as a whole must be a stable system against rigid body motion. $$\displaystyle r_e \geq 3 $$ for 2D frames.
2. Internal Stability: The structure must not contain internal mechanisms. This is checked by ensuring $$\displaystyle N_s \geq 0 $$ and the structure is properly connected.
Relationship: A structure is unstable if $$\displaystyle N_s < 0 $$ (excess constraints can also cause instability if improperly arranged).

Sway vs. Non-Sway Structures

Sway Structure: A structure that can undergo a lateral (horizontal) displacement of joints under vertical loads due to lack of symmetric support or loading. Requires consideration of sway effects in analysis (e.g., in slope deflection, include chord rotations $\psi$).
Non-Sway Structure: Symmetric in geometry, loading, and support conditions such that no net horizontal reaction or joint translation occurs under vertical loads. Analysis simplifies as chord rotations $$\displaystyle \psi = 0 $$.

Feature	Sway Frame	Non-Sway Frame
Horizontal Displacement	Possible under vertical load	Not possible
Analysis	Requires sway equations (include $\psi$)	Simpler, $$\displaystyle \psi = 0 $$
Example	Un-symmetric portal frame	Symmetric multi-span beam

[!TIP] Exam Pitfall: Always check for symmetry in geometry, loading, and support conditions to identify sway. A fixed-ended beam is non-sway; a portal frame with one roller support is typically sway.

II. ENERGY THEOREMS & PRINCIPLES

Strain Energy (U)

Definition: Energy stored in a deformed elastic body due to applied loads.
Expressions:
- Axial Load: $$\displaystyle U = \int_0^L \frac{N^2}{2EA} dx $$
- Bending: $$\displaystyle U = \int_0^L \frac{M^2}{2EI} dx $$ (Most Common)
- Shear: $$\displaystyle U = \int_0^L \frac{V^2}{2GA} dx $$ (Often neglected for slender beams)

Castigliano's Theorems

First Theorem (For Forces): The partial derivative of total strain energy with respect to an applied force gives the displacement in the direction of that force.

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

*   Use for finding displacement at a point where a load is applied.

Second Theorem (For Displacements): The partial derivative of total strain energy with respect to a load gives the displacement at the point of application in the direction of the load. If the load is not present, apply a dummy load $P$, find $U$ in terms of $P$, then set $$\displaystyle P=0 $$ after differentiation.

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

*   Use for finding unknown forces (reactions, moments) causing known displacements (e.g., support settlement).

> **Key:** For beam bending, $$\displaystyle U = \int \frac{M^2}{2EI} dx $$. Express $M$ as function of real + dummy load.

Principle of Virtual Work

Statement: For a structure in equilibrium, the total work done by the external forces during any virtual displacement compatible with constraints is zero.

$$ \delta W_{ext} = 0 $$

Application to Deformable Bodies (Trusses/Beams):
- Apply a virtual (unit) load in the direction of desired displacement at the point of interest.
- Calculate internal forces ($m, n$ for real; $\bar{m}, \bar{n}$ for virtual).
- Displacement Formula:
  - Truss: $$\displaystyle \delta = \sum \frac{n \bar{n} L}{EA} $$
  - Beam: $$\displaystyle \delta = \int \frac{m \bar{m}}{EI} dx $$ or $$\displaystyle \theta = \int \frac{m \bar{m}}{EI} dx $$ (for slope, $\bar{m}$ from unit rotation)

Maxwell's Reciprocal Deflection Theorem

Statement: The deflection at point A in the direction of force $$\displaystyle P_1 $$ due to $$\displaystyle P_1 $$ is equal to the deflection at point B in the direction of force $$\displaystyle P_2 $$ due to $$\displaystyle P_2 $$.

$$ \delta_{12} = \delta_{21} $$

*   Where $$\displaystyle \delta_{12} $$ = deflection at 1 due to load at 2; $$\displaystyle \delta_{21} $$ = deflection at 2 due to load at 1.

Derivation: Direct consequence of Betti's Theorem:

$$ P_1 \delta_{12} + P_2 \delta_{21} = P_2 \delta_{21} + P_1 \delta_{12} $$

(trivially true), but more generally for two systems of forces:

$$ \sum_{i=1}^{n} P_i \delta'_{i} = \sum_{i=1}^{n} P'_i \delta_{i} $$

Setting one system as a unit load at A, the other as a unit load at B yields Maxwell's theorem.

Complementary Energy (Brief)

Concept: Energy expressed in terms of stresses. For linear elastic systems, it equals strain energy. Used in the Principle of Minimum Complementary Energy (the true stress state minimizes complementary energy among all statically admissible stress fields).

III. ANALYSIS OF TRUSSES

Assumptions

All members are pin-jointed (two-force members).
Loads and reactions act only at joints.
Members are weightless or self-weight neglected.
Joints are frictionless hinges.

Methods of Analysis

Method of Joints:
1. Find support reactions.
2. Isolate a joint with ≤2 unknown forces.
3. Apply $$\displaystyle \sum F_x=0, \sum F_y=0 $$.
4. Proceed joint by joint.
Method of Sections:
1. Calculate support reactions.
2. Pass a section through max 3 members (with unknowns).
3. Consider equilibrium of one part (LHS/RHS).
4. Use $$\displaystyle \sum M=0 $$ for direct solution of desired member.
Advantage: Directly finds force in specific interior members without solving entire truss.

Influence Lines for Truss Members (Muller-Breslau)

Procedure:
1. Remove the member and apply a tensile force of +1 at its ends (or shear/vertical reaction for chords).
2. The resulting displaced shape (with hinges at other joints) is the Influence Line Diagram (ILD) for that member's force.
Warren Truss with Verticals:
- Top Chord (in compression): ILD is similar to ILD for support reaction at that end.
- Bottom Chord (in tension): ILD is similar to ILD for shear at that section.
- Diagonals: ILD consists of straight lines between panel points. Sign (+ve for tension, -ve for compression) determined by the assumed +1 tension.
UDL Longer Than Span: Maximum member force = (Intensity of UDL) × (Area under ILD for that member).

IV. ANALYSIS OF BEAMS & CONTINUOUS BEAMS

Fixed Beams & Cantilevers

Fixed End Moments (FEM) for Standard Loads (Span = L, EI constant):

| Load Type | FEM at A | FEM at B | | :--- | :--- | :--- | | UDL, w | $$\displaystyle wL^2/12 $$ | $$\displaystyle wL^2/12 $$ | | Point Load, P at mid | $PL/8$ | $PL/8$ | | Point Load, P at distance a from A | $$\displaystyle Pb^2(2a+b)/L^2 $$ | $$\displaystyle Pa^2(a+2b)/L^2 $$ | | Couple, M | $M/2$ | $M/2$ |
Effect of Support Settlement:
- If support B settles by $\Delta$ (downwards positive), it induces a sinking force.
- Fixed End Moments due to settlement $$\displaystyle \Delta_B $$:

$$ FEM_{AB} = - \frac{6EI \Delta_B}{L^2}, \quad FEM_{BA} = \frac{6EI \Delta_B}{L^2} $$

*   Treat as additional loading and superimpose.

Continuous Beams - Analysis Methods

Slope Deflection Method

Fundamental Equation:

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

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

*   $$\displaystyle m_{AB} $$ = moment at A end of member AB (clockwise on AB).

*   $$\displaystyle \theta_A, \theta_B $$ = rotations of joints A, B (clockwise positive).

*   $$\displaystyle \psi = \frac{\Delta}{L} $$ = chord rotation (sagging positive).

*   $$\displaystyle FEM_{AB} $$ = fixed end moment for member AB due to external load only.

Procedure:
1. Calculate FEMs for each span.
2. Write slope-deflection equations for each member end.
3. Apply Joint Equilibrium: $$\displaystyle \sum m_{joint} = 0 $$ at each joint (except supports with known rotation).
4. Solve for unknown rotations $\theta$.
5. Back-substitute to find all end moments.
6. Draw BMD by balancing moments at each section.

Three Moment Equation (Clapeyron's)

General Form (for spans i-1, i, i+1 with constant EI):

$$ M_{i-1}L_{i-1} + 2M_i(L_{i-1}+L_i) + M_{i+1}L_i = -6 \left( \frac{A_{i-1}\bar{x}_{i-1}}{L_{i-1}} + \frac{A_i\bar{x}_i}{L_i} \right) + 6EI \left( \psi_{i-1} + \psi_i \right) $$

*   $A$ = area of M diagram (from simple supports) for the span.

*   $\bar{x}$ = distance from left support to centroid of $A$.

*   $\psi$ = chord rotation (settlement effect: $$\displaystyle \psi = \frac{\Delta}{L} $$).

For Varying I: Replace $EI$ terms with $$\displaystyle H_i = \frac{6EI}{L} $$ for each span. Equation becomes:

$$ H_{i-1}M_{i-1} + 2(H_{i-1}+H_i)M_i + H_iM_{i+1} = -6 \left( \frac{A_{i-1}\bar{x}_{i-1}}{L_{i-1}} \cdot \frac{EI}{L} \text{ terms} \right) $$

Moment Distribution Method

Definitions:
- Stiffness Factor (k): For a member AB, $$\displaystyle k_{AB} = \frac{4EI}{L} $$ if far end fixed; $$\displaystyle k_{AB} = \frac{3EI}{L} $$ if far end pinned/hinged.
- Distribution Factor (DF): Fraction of unbalanced moment distributed to a member at a joint.

$$ DF_{AB} = \frac{k_{AB}}{\sum k_{all\,members\,at\,joint}} $$

*   **Carry-Over Factor (CO):** Always **0.5** for prismatic members with far end fixed. Moment carried to far end = 0.5 × distributed moment.

Procedure:
1. Calculate fixed end moments (FEM) for all members (consider sign: sagging +).
2. Calculate stiffness factors (k) and distribution factors (DF) for each joint.
3. Release joints one by one (in cycles):
  - Find Unbalanced Moment = $$\displaystyle \sum FEM $$ at joint.
  - Distribute this moment to each connected member using DF: $$\displaystyle M_{dist} = - (Unbalanced) \times DF $$.
  - Carry-over half of each distributed moment to the far end of the member.
4. Repeat cycles until moments are negligible.
5. Sum FEM + all distributed + carried-over moments for each end to get final moments.
6. Draw BMD and find reactions.

Column Analogy Method

Concept: Analogy between bending of a beam on fixed supports and buckling of a column with equivalent length.
Procedure for Fixed Beam:
1. Replace beam by a column of length = beam span, with flexural rigidity = $EI$.
2. The load on column is the bending moment diagram of the beam for the given loading, but with sagging +ve.
3. The deflection curve of column under this load gives the slope diagram of the beam.
4. The horizontal thrust in the column analogy is proportional to the fixed end moment.
5. For a fixed-fixed beam, the column is fixed at both ends. Use standard column formulas or conjugate beam method on the column to find thrust (which gives FEM).
Application: Particularly useful for beams with varying moment of inertia or with coupled loads.

V. INFLUENCE LINES

Definition & Importance

Definition: A graph showing the variation of a response function (reaction, shear, moment) at a specific point in a structure as a unit load moves across it.
Importance: Determines maximum effect of moving loads (vehicles, trains) for design.

Muller-Breslau Principle

Statement: The influence line for a reaction (or shear, moment) has the same shape as the deformed elastic curve of the structure when the corresponding constraint is removed and a unit displacement (rotation for moment, translation for reaction/shear) is imposed in the positive direction of the function.
Procedure:
1. Remove the constraint for the desired function (e.g., remove support for reaction, insert hinge for shear, insert hinge and release moment for moment).
2. Impose a unit displacement in the positive direction of the function.
3. The resulting displaced shape (with hinges at other original rigid joints) is the ILD.

ILD for Beams (Simply Supported)

Reaction at Support A: Straight line from 1 at A to 0 at B.
Shear Force at Section C:
- Left of C: ordinate = 1 (constant)
- Right of C: ordinate = -1 (constant)
- Jump of -2 at C.
Bending Moment at Section C:
- Triangle with peak at C: ordinate = $x(L-x)/L$, where $x$ = distance from left support.

Application of ILD

Concentrated Loads: Max effect = $$\displaystyle \sum (Load \times ILD\,ordinate\,at\,load\,position) $$.
Uniformly Distributed Load (UDL):
- Max effect = (Intensity) × (Area under ILD for the loaded length).
- Position for Max Effect: Place UDL such that it covers the positive areas (for max +ve) or negative areas (for max -ve) of ILD.
Equivalent Uniformly Distributed Load (EUDL): For a load longer than span, the maximum effect equals the intensity times the total area under the ILD (since ILD is linear, area is trapezoidal/triangular).

Key: For a simply supported beam, max +ve BM at mid-span under UDL = $$\displaystyle wL^2/8 $$ (area of triangle = 1/2 × L × 1 = L/2, so $w \times L/2$? Correction: ILD ordinate at mid = 0.5, area = 0.5×L/2? Standard Result: Max BM = $$\displaystyle wL^2/8 $$. ILD area for full span = 1/2 × L × 1 = L/2? Wait: ILD for mid-span moment is triangle with max 1 at mid? No: For simply supported beam, ILD for mid-span moment has max ordinate = L/4? Recall: For point at distance a from left, ILD ordinate = a(L-a)/L. At mid (a=L/2), ordinate = (L/2)(L/2)/L = L/4. Area under full ILD (triangle base L, height L/4) = 1/2 × L × L/4 = L²/8. Then max BM = w × (L²/8) = wL²/8. Correct.

VI. ARCHES

Classification

By Material: Masonry, Concrete, Steel, Timber.
By Shape: Parabolic (for UDL), Circular, Elliptical, Catenary (for self-weight).
By Structural System:
- Three-Hinged: Hinges at two supports + crown. Statically determinate.
- Two-Hinged: Hinges at supports only. Statically indeterminate to degree 1.
- Fixed (No Hinges): Statically indeterminate to degree 3.

Three-Hinged Arches

Horizontal Thrust (H):
- General: Found from equilibrium of entire arch or using condition that moment at hinge (crown) is zero.
- Parabolic Arch, UDL (w): $$\displaystyle \boxed{H = \frac{wL^2}{8h}} $$, where L = span, h = rise.
- Parabolic Arch, Point Load W at distance a from left:

$$ H = \frac{W}{h} \left( a - \frac{L}{2} \right) \left( \text{sign depends on position} \right) $$

(Derived from $$\displaystyle \sum M_{crown}=0 $$).

Reactions: Found from global equilibrium ($$\displaystyle \sum F_x, \sum F_y, \sum M $$) using known H.
Bending Moment, Shear, Normal Thrust at any section:
- Consider left/right segment as a free body.
- $$\displaystyle M_x = V_x \cdot x - H \cdot y $$ (where x, y are coordinates of section from left support).
- Shear $$\displaystyle V_x $$ and normal thrust $$\displaystyle N_x $$ from resolving forces parallel/perpendicular to tangent.
Eddy's Theorem:
- Statement: In a three-hinged arch, the bending moment at any section is proportional to the vertical intercept between the actual arch axis and the linear arch (the funicular polygon for the given loads).
- Application: Quick determination of BMD. $$\displaystyle M_x = H \cdot y_{deviation} $$.

Two-Hinged & Fixed Arches

Effect of Temperature Change ($\Delta T$):
- Causes additional horizontal thrust in two-hinged arches.
- For two-hinged parabolic arch: $$\displaystyle \Delta H = \frac{\alpha \Delta T E A}{1 + \frac{12EI}{H L^2}} $$ (approx). More significant for long-span arches.
Effect of Rib Shortening:
- Axial compression in arch causes shortening, inducing additional horizontal thrust.
- Similar in form to temperature effect: $$\displaystyle \Delta H = \frac{EA \delta}{L} $$ where $\delta$ is shortening.

Parabolic Arch under UDL

Horizontal Thrust: $$\displaystyle H = \frac{wL^2}{8h} $$ (as above).
Shape of Bending Moment Diagram:
- Since $$\displaystyle M_x = V_x x - H y $$ and for parabolic arch $$\displaystyle y = \frac{4h}{L^2} x(L-x) $$, and $$\displaystyle V_x $$ varies linearly.
- Result: BMD is also parabolic but with opposite curvature to the arch axis. Max BM occurs at quarter points.

VII. CABLES

Assumptions in Force Analysis

Perfectly flexible: No bending stiffness, force is always tangential (cable force = tension).
Loads are perpendicular to the cable axis (or horizontal for suspension cables).
Self-weight may be negligible or considered as UDL horizontally.

Cable with Uniformly Distributed Load (Horizontal)

Shape: Parabolic ($$\displaystyle y = \frac{w}{2H} x^2 $$ for symmetric cable with lowest point at origin).
Equation: For span L, sag (dip) d at mid-span:

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

where H = horizontal component of tension (constant throughout).

Tensions:
- At lowest point: $$\displaystyle T_{min} = H $$.
- At support (A or B): $$\displaystyle T_{max} = \sqrt{H^2 + V_A^2} $$, where $$\displaystyle V_A = \frac{wL}{2} $$.
- $$\displaystyle T_{max} = H \sqrt{1 + \left( \frac{wL}{2H} \right)^2 } = H \sqrt{1 + \left( \frac{4d}{L} \right)^2 } $$.
Relationship: $$\displaystyle H \propto \frac{1}{d} $$ (for given w, L). Larger sag reduces horizontal thrust.

Cable with Concentrated Loads

Shape: Composed of straight line segments between load points.
Equilibrium: For each segment, apply $$\displaystyle \sum F_x=0, \sum F_y=0 $$ at each joint (load point).
Procedure:
1. Find support reactions (global equilibrium).
2. Consider equilibrium of each cable segment (free body diagram).
3. Tension in segment = resultant of horizontal and vertical components at its ends.
4. Total Length: Sum lengths of straight segments: $$\displaystyle L_{total} = \sum \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} $$.

[!TIP] Exam Focus: Be able to derive parabolic shape equation from equilibrium of a cable element: $$\displaystyle dT = w dx $$, and $$\displaystyle T \sin\theta = V $$, $$\displaystyle T \cos\theta = H = const $$. Then $$\displaystyle \frac{dy}{dx} = \tan\theta = \frac{V}{H} = \frac{wx}{H} $$, integrate to get $$\displaystyle y = \frac{w}{2H}x^2 $$.

Account

Structural Analysis-I (CE-403) - Unit 4 Short Notes