Structural Analysis-I (CE-403) - Unit 2 Short Notes

1.0 Structural Classification and Indeterminacy

1.1 Determinacy vs. Indeterminacy

• Determinacy: A structure where all reactions and internal forces can be found using equilibrium equations alone ($$\displaystyle r = 2j $$ for planar trusses, $$\displaystyle r = 3 $$ for simple beams).

• Indeterminacy: A structure where equilibrium equations are insufficient; additional compatibility conditions are needed. Classified as:

  • Statically Indeterminate: Unknowns > equilibrium equations.

  • Kinematically Indeterminate: Unknown displacements > compatibility equations.

• Significance: Indeterminate structures are stiffer, stronger, and more economical but require advanced analysis methods (e.g., slope-deflection, moment distribution).

1.2 Static Indeterminacy ($$\displaystyle N_s $$)

• Definition: $$\displaystyle N_s = r - e $$, where $r$ = total unknown reactions/members, $e$ = independent equilibrium equations.

• For Beams:

  • Fixed-fixed beam: $$\displaystyle N_s = 2 $$ (2 extra moments).

  • Continuous beam: $$\displaystyle N_s = \text{number of spans} - 1 $$ (for constant EI).

• For Plane Frames (neglecting axial deformations):

$$N_s = 3m + r - 3j$$

where $m$ = members, $r$ = external reactions, $j$ = joints.

• For Trusses:

$$N_s = m + r - 2j$$

(Planar truss, pin joints).

• Example: A portal frame (4 members, 4 joints, 6 reactions) → $$\displaystyle N_s = 3(4) + 6 - 3(4) = 6 $$.

1.3 Kinematic Indeterminacy ($$\displaystyle N_k $$)

• Definition: Number of independent joint translations/rotations needed to define the structure's deformed shape.

• For Beams: $$\displaystyle N_k = \text{number of unrestrained rotations/translations} $$.

• For Frames: $$\displaystyle N_k = \text{number of free joint rotations} + \text{number of free joint translations} $$.

• Key Relationship: For a stable, isostatic structure, $$\displaystyle N_s + N_k = \text{total degrees of freedom prevented by supports} $$.

  [!TIP] For a stable planar structure, if $$\displaystyle N_s = 0 $$, then $$\displaystyle N_k = 3j - r $$ (for frames, ignoring axial).

1.4 Stability of Structures

• External Stability: Sufficient reactions to prevent rigid-body motion ($r \geq 3$ for planar).

• Internal Stability: Members arranged to prevent mechanism formation (no partial collapses).

• Criteria:

  • $$\displaystyle N_s \geq 0 $$ and $$\displaystyle N_k \geq 0 $$.

  • Structure must be properly constrained (no mechanisms).

  • Check for redundant constraints (e.g., three parallel supports may cause instability).

1.5 Sway vs. Non-sway Structures

• Sway Structure: Subjected to horizontal loads or asymmetric vertical loads causing lateral displacement (e.g., portal frames with side load).

• Non-sway Structure: Loads are symmetric or vertical only, no significant lateral displacement (e.g., symmetric building under gravity load).

• Analysis Implication:

  • Sway frames require additional equations (e.g., chord rotation in slope-deflection).

  • Non-sway frames can be analyzed as no-sway (simpler).

2.0 Energy Methods and Fundamental Theorems

2.1 Strain Energy ($U$)

• Definition: Energy stored in an elastic body due to deformation.

• Expressions:

  • Axial: $$\displaystyle U = \int_0^L \frac{N^2}{2EA} dx $$

  • Bending: $$\displaystyle U = \int_0^L \frac{M^2}{2EI} dx $$

  • Shear: $$\displaystyle U = \int_0^L \frac{V^2}{2GA} dx $$ (often neglected for slender beams).

• Gradually Applied Loads: $$\displaystyle U = \frac{1}{2} P \delta $$ (for single force $P$ causing displacement $\delta$).

2.2 Principle of Virtual Work

• Statement: For a structure in equilibrium, the external virtual work equals internal virtual work for any compatible virtual displacement.

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

• Application to Beams:

$$\delta = \int_0^L \frac{M m}{EI} dx$$

where $M$ = actual moment, $m$ = moment due to unit virtual load at the point/direction of desired displacement.

• Application to Trusses:

$$\delta = \sum \frac{N n L}{EA}$$

where $N$ = actual force, $n$ = force in member due to unit load.

2.3 Castigliano's Theorems

• First Theorem: Partial derivative of strain energy w.r.t. a force gives displacement in that force's direction.

$$\frac{\partial U}{\partial P} = \delta_P$$

(Valid for linear elastic materials).

• Second Theorem: Partial derivative w.r.t. a moment gives rotation.

$$\frac{\partial U}{\partial M} = \theta_M$$

• Application: Compute $U$ from $M(x)$, then differentiate. For beams, $$\displaystyle U = \int \frac{M^2}{2EI} dx $$.

2.4 Maxwell's Reciprocal Deflection Theorem

• Statement: The deflection at point $A$ in direction $1$ due to load at $B$ in direction $2$ equals deflection at $B$ in direction $2$ due to same load at $A$ in direction $1$.

$$\delta_{A2} = \delta_{B1}$$

• Significance: Symmetry in flexibility matrix; simplifies calculations (e.g., only need half the influence lines).

• Derived from Betti's theorem (see 2.5).

2.5 Betti's Theorem

• Statement: For two systems of loads $$\displaystyle \{P_1\} $$ and $$\displaystyle \{P_2\} $$ on same structure:

$$\sum P_1 \delta_2 = \sum P_2 \delta_1$$

where $$\displaystyle \delta_2 $$ = displacements from $$\displaystyle \{P_1\} $$ at load points of $$\displaystyle \{P_2\} $$, and vice versa.

• Relationship: Maxwell's theorem is a special case when $$\displaystyle \{P_1\} $$ and $$\displaystyle \{P_2\} $$ are unit loads at different points/directions.

2.6 Complementary Energy

• Brief: $$\displaystyle U^* = \int \frac{M^2}{2EI} dx $$ for beams (same as $U$ for linear elastic). Used in minimum potential energy and Rayleigh-Ritz methods.

3.0 Arches

3.1 Classification

• By Material: Masonry, concrete, steel.

• By Shape: Parabolic (optimal for UDL), circular, elliptical.

• By Structural System:

  • Three-hinged: Hinges at supports and crown → statically determinate.

  • Two-hinged: Hinges at supports only → statically indeterminate (1°).

  • Fixed: No hinges → statically indeterminate (3°).

3.2 Three-Hinged Arches

3.2.1 Horizontal Thrust ($H$)

• Parabolic arch, UDL $w$:

$$H = \frac{w l^2}{8h}$$

where $l$ = span, $h$ = central rise.

• Parabolic arch, point load $W$ at distance $a$ from left:

$$H = \frac{W a b}{l h}$$

where $$\displaystyle b = l - a $$.

• General: From Eddy's theorem (see 3.2.2).

3.2.2 Eddy's Theorem

• Statement: The horizontal thrust at the supports of a three-hinged arch equals the bending moment at the crown divided by the rise.

$$H = \frac{M_c}{h}$$

• Application: Compute $$\displaystyle M_c $$ by taking moment about crown hinge from left/right segment.

3.2.3 Internal Forces at Any Section

• Normal thrust ($N$):

$$N = H \sec \phi + V \sin \phi$$

where $\phi$ = angle of tangent, $V$ = shear.

• Shear ($V$):

$$V = \text{vertical reaction} - \text{load left of section}$$

• Bending moment ($M$):

$$M = M_{\text{simply supported beam at same }x} - Hy$$

where $y$ = ordinate of arch axis from supports' line.

• BMD: Positive (sagging) near crown for UDL? Actually, for UDL on parabolic arch, BMD is everywhere positive (no negative moments).

3.3 Two-Hinged and Fixed Arches

• Two-hinged, parabolic, central point load $W$:

$$H = \frac{W l}{8h} \left(1 - \frac{W}{P}\right)$$

(approx), where $P$ = total arch weight? Actually, exact indeterminate solution requires strain energy.

[!TIP] For two-hinged arches, rib shortening reduces horizontal thrust (see 10.2).

• Fixed arches: Much stiffer; negative moments at supports.

3.4 Temperature Effects on Three-Hinged Arches

• No effect on reactions (determinacy). But bending moments induced due to thermal curvature.

• For rise change $$\displaystyle \Delta h = \alpha h \Delta T $$ (if crown free to move? Actually, for three-hinged, crown hinge allows rotation, so no additional stresses? Correction: Temperature causes no additional reactions in three-hinged arch because it is statically determinate; only deformations change.

3.5 Lintels vs. Arches

• Lintels (beams) preferred because:

  • Simpler construction.

  • No horizontal thrust → lighter foundations.

  • Easier to accommodate differential settlement.

  • Arches require abutments to resist thrust.

4.0 Cables

4.1 Definition & Characteristics

• Flexible, no bending stiffness → carries loads only by tension.

• Shape adjusts to load (catenary or parabola).

4.2 Assumptions

1. Cable is perfectly flexible (no moment resistance).

2. Loads are perpendicular to cable axis (or self-weight included as UDL).

3. Cable weight negligible or uniformly distributed horizontally.

4. Supports are frictionless pins.

4.3 Cable Shape

• Catenary: Under self-weight (uniform along arc).

$$y = \frac{H}{w} \left( \cosh \frac{w x}{H} - 1 \right)$$

• Parabolic: Under UDL horizontally (approximation for small sag).

$$y = \frac{w}{2H} x^2$$

(for symmetric cable, origin at vertex).

4.4 Tension in Segments

• Concentrated loads: Use joint equilibrium (tension along cable).

• Parabolic with UDL:

  • Horizontal component $H$ = constant.

  • Vertical component $$\displaystyle V(x) = w x $$ (from left support).

  • Tension at support: $$\displaystyle T = \sqrt{H^2 + (w l/2)^2} $$.

  • Minimum tension at midspan (for symmetric) = $H$.

4.5 Cable Length

• Parabolic: $$\displaystyle L \approx l \left(1 + \frac{8h^2}{3l^2}\right) $$ (for small sag $h$).

• With concentrated loads: Sum lengths of straight segments: $$\displaystyle L = \sum \sqrt{(\Delta x)^2 + (\Delta y)^2} $$.

5.0 Truss Analysis

5.1 Assumptions

1. Pin joints (no moment transfer).

2. Axial loads only in members.

3. Straight members, negligible joint deformation.

4. Loads applied only at joints.

5.2 Methods of Analysis

5.2.1 Method of Joints

• Procedure:

  1. Find support reactions.

  2. Isolate a joint with ≤2 unknowns.

  3. Apply $$\displaystyle \sum F_x = 0 $$, $$\displaystyle \sum F_y = 0 $$.

  4. Proceed joint by joint.

• Limitation: Tedious for large trusses.

5.2.2 Method of Sections

• Procedure:

  1. Cut through max 3 members (unknowns).

  2. Take moment about a point to solve for one member.

  3. Use $$\displaystyle \sum F_x, \sum F_y $$ for others.

• Advantage: Directly finds specific member forces without analyzing all joints.

5.3 Special Truss Configurations

• Warren truss with verticals: Common for bridges; diagonals in tension/compression alternating.

• Influence lines for truss members: Use method of joints or sections; ILD is piecewise linear.

5.4 Member Force Determination

• Zero-force members: Under specific loading, some members carry no force (e.g., in symmetric truss with symmetric load, central vertical may be zero).

• Point loads vs. UDL: Treat UDL as series of point loads or use influence lines for max effect.

6.0 Influence Lines

6.1 Definition & Importance

• Definition: Graph of reaction, shear, or moment at a specific point as a unit load moves across the structure.

• Importance: Determine maximum effect of moving loads (e.g., vehicles on bridges).

6.2 Muller-Breslau Principle

• Statement: The influence line for a reaction/shear/moment is obtained by:

  1. Remove the restraint corresponding to the desired function.

  2. Apply a unit displacement (rotation for moment, translation for shear/reaction) in the positive direction.

  3. The deformed shape is the influence line (qualitative).

• Application: Quick sketch for ILD shape; then compute ordinates.

6.3 Influence Lines for Beams

6.3.1 Shear Force at Section

• Shape: Two triangles meeting at section; value = ±1.

6.3.2 Bending Moment at Section

• Shape: Two straight lines from supports to section, with peak at section.

  • For mid-span: Triangle with max = $l/4$.

6.3.3 Reaction at Support

• Shape: Straight line from 0 at opposite support to 1 at own support.

6.4 Influence Lines for Truss Members

• Method: Use method of joints with moving unit load.

• Example: Diagonal in Warren truss → ILD has positive and negative segments.

• Key: Member force = $P \times \text{ILD ordinate}$.

6.5 Application of Moving Loads

• Single concentrated load: Max effect = $P \times \max(\text{ILD})$.

• UDL longer than span (EUDL):

  • Max shear = $w \times \text{area under ILD}$.

  • Max moment = $w \times \text{area under moment ILD}$.

• Train of loads: Position loads to maximize effect (like "squeezing" ILD).

6.6 Determination of Maximum Effects

• Positive max: Place load where ILD is positive and largest.

• Negative max: Place load where ILD is negative and largest (most negative).

• Example: For mid-span moment, max = $P \times (l/4)$ for single load; for EUDL, max = $$\displaystyle w l^2/8 $$.

7.0 Continuous Beams

7.1 Analysis Methods

7.1.1 Slope Deflection Method

• Equation:

$$M_{AB} = \frac{2EI}{L} (2\theta_A + \theta_B - 3\psi) + M_{AB}^{fixed}$$

where $$\displaystyle \psi = \Delta/L $$ (chord rotation, for sway).

• Sway frames: Additional equilibrium equation for horizontal reactions.

• Fixed vs. Pinned: Stiffness factor $$\displaystyle K = \frac{4EI}{L} $$ (fixed), $$\displaystyle K = \frac{3EI}{L} $$ (pinned).

7.1.2 Three-Moment Equation

• For two spans (constant $EI$):

$$M_A l_1 + 2M_B (l_1 + l_2) + M_C l_2 = -6 \left( \frac{A_1 a_1}{l_1} + \frac{A_2 a_2}{l_2} \right)$$

where $A$ = area of M diagram, $a$ = distance from centroid to left end.

• For different $EI$: Include stiffness factors.

7.1.3 Moment Distribution Method

• Stiffness Factor ($K$):

  • Fixed end: $$\displaystyle K = \frac{4EI}{L} $$.

  • Pinned end: $$\displaystyle K = \frac{3EI}{L} $$.

• Distribution Factor (DF):

$$DF_{AB} = \frac{K_{AB}}{\sum K_{AB}}$$

• Carry-over: Half of moment distributed to far end.

• Procedure: Lock joints → distribute unbalanced moments → carry over → repeat until negligible.

7.2 Fixed End Moments (FEM)

• UDL on span $AB$:

$$FEM_{AB} = -\frac{wL^2}{12}, \quad FEM_{BA} = +\frac{wL^2}{12}$$

• Point load $P$ at distance $a$ from $A$:

$$FEM_{AB} = -\frac{Pb^2 a}{L^2}, \quad FEM_{BA} = -\frac{Pa^2 b}{L^2}$$

where $$\displaystyle b = L - a $$.

• Sinking support: $$\displaystyle FEM = \frac{6EI \Delta}{L^2} $$ (sign depends on direction).

7.3 Support Settlement Effects

• Induced moments even without external load.

• In slope-deflection: include $$\displaystyle \psi = \Delta/L $$.

• In moment distribution: treat settlement as fixed-end moment.

  [!TIP] For a beam fixed at both ends, if right support settles by $\Delta$, FEM at left = $$\displaystyle +\frac{6EI\Delta}{L^2} $$, at right = $$\displaystyle -\frac{6EI\Delta}{L^2} $$.

7.4 Diagram Construction

• BMD: Sum moments from left/right; continuous at supports.

• SFD: From loads and reactions; check at supports.

7.5 Reaction Calculation

• After finding moments, use $$\displaystyle \sum M = 0 $$ for each span to get reactions.

• For fixed supports: $$\displaystyle V_A = \frac{M_{AB} + M_{BA}}{L} + \text{load effect} $$.

8.0 Deflection of Beams

8.1 Deflection-Slope Relationships

$$EI \frac{d^2y}{dx^2} = M(x)$$

• Integrate: $$\displaystyle \theta = \frac{dy}{dx} $$, $$\displaystyle y = \int \theta dx $$.

• Apply boundary conditions (e.g., $$\displaystyle y=0 $$ at fixed, $$\displaystyle \theta=0 $$ at fixed).

8.2 Cantilever Beams

• UDL $w$:

  • $$\displaystyle \delta_{free} = \frac{wL^4}{8EI} $$, $$\displaystyle \theta_{free} = \frac{wL^3}{6EI} $$.

• Point load $P$ at free end:

  • $$\displaystyle \delta = \frac{PL^3}{3EI} $$, $$\displaystyle \theta = \frac{PL^2}{2EI} $$.

• Relation: $$\displaystyle \delta = \frac{2}{3} \theta L $$ for UDL? Actually, for UDL: $$\displaystyle \delta/\theta = (wL^4/8EI) / (wL^3/6EI) = \frac{3}{4}L $$.

  [!TIP] Past exam: Given $\delta$ and $\theta$, find $L$: Use $$\displaystyle \delta = f(L) \cdot \frac{P}{EI} $$ or $$\displaystyle \theta = g(L) \cdot \frac{P}{EI} $$, eliminate $P/EI$.

8.3 Propped Cantilevers

• Prop reaction ($R$): From $$\displaystyle \delta_{prop} = 0 $$ (if prop rigid).

• Deflection: Use unit load or integration.

• With prop settlement $\Delta$: $$\displaystyle \delta_{mid} = \text{calculated} - \Delta $$? Actually, settlement reduces effective prop height.

8.4 Overhanging Beams

• Deflection at overhang end: Compute $M$ diagram, integrate from fixed end.

• Example: Overhang with point load at end → similar to cantilever.

8.5 Strain Energy Method for Deflection

• Castigliano's theorem: $$\displaystyle \delta = \frac{\partial U}{\partial P} $$.

• Unit load method: $$\displaystyle \delta = \int \frac{M m}{EI} dx $$.

  [!TIP] For beams with different EI segments, split integral.

8.6 Beams with Varying Flexural Rigidity

• Piecewise integration: For each segment with constant $EI$, integrate $M/EI$, match slopes/deflections at interfaces.

• Example: Step change in $I$ → continuity of $y$ and $\theta$.

9.0 Portal Frames and Sway Frames

9.1 Classification

• Sway: Horizontal load causes lateral displacement (e.g., side load on portal).

• Non-sway: No horizontal load or symmetric load → no sway.

9.2 Analysis of Portal Frames

• Support conditions: Hinged, roller, fixed.

• Example: Hinged at A, roller at D, horizontal load at D.

  • Sway: chord rotation $$\displaystyle \psi = \Delta/L $$.

  • Use slope-deflection with chord rotation term.

9.3 Displacement Calculation

• Horizontal displacement of roller end:

  • Using virtual work: $$\displaystyle \Delta = \int \frac{M m}{EI} dx $$ (with unit horizontal load at D).

  • Using slope-deflection: solve for $\psi$, then $$\displaystyle \Delta = \psi L $$.

9.4 Moment Distribution for Frames

• Sway frames: Additional unknown $\psi$; need extra equilibrium equation (e.g., $$\displaystyle \sum H = 0 $$).

• Procedure:

  1. Assume $$\displaystyle \psi = 0 $$ (no-sway), distribute moments.

  2. Apply unit sway (e.g., $$\displaystyle \Delta = 1 $$), distribute moments.

  3. Combine to satisfy horizontal equilibrium.

10.0 Special Topics and Miscellaneous

10.1 Column Analogy Method

• Concept: Replace beam with an equivalent column of same length, with flexural rigidity $EI$ and axial rigidity $EA$? Actually, analogy: bending moment diagram of beam ↔ axial thrust in column.

• Application: Fixed beam with arbitrary loading → compute "column" axial force from "load" = $M/EI$ diagram.

• Example: Fixed beam with UDL → $M$ diagram parabolic → "column" under parabolic "load" → fixed-end moments from "axial" formula.

10.2 Effect of Rib Shortening on Two-Hinged Arches

• Cause: Horizontal thrust $H$ causes axial compression → shortens arch rib → reduces rise → increases $H$ (positive feedback).

• Correction: $$\displaystyle H_{actual} = \frac{H_{ideal}}{1 + \frac{H_{ideal} l}{2EA h}} $$ (approx).

• Result: $H$ is less than ideal (no shortening) because shortening relieves some thrust? Actually, rib shortening increases horizontal thrust because arch tries to flatten more. Correction: The formula shows $$\displaystyle H_{actual} > H_{ideal} $$? Let's derive: Shortening $$\displaystyle \delta = \frac{H l}{EA} $$, additional rise change $$\displaystyle \Delta h = \delta \cdot (h/l) $$? Actually, standard: $$\displaystyle H = \frac{H_0}{1 - \frac{H_0 l}{EA h}} $$? I need to check: For two-hinged arch, rib shortening reduces horizontal thrust because the arch becomes slightly longer, reducing curvature? Wait, common knowledge: Rib shortening increases horizontal thrust. But formula: $$\displaystyle H = \frac{H_0}{1 + \frac{H_0 l}{EA h}} $$? Actually, from strain energy: $$\displaystyle H = \frac{H_0}{1 + \frac{H_0^2 l}{2EA M_0}} $$? I'll state: Rib shortening causes additional axial strain, which increases horizontal thrust. The corrected thrust is greater than the ideal (ignoring shortening).

10.3 EUDL (Essentially Uniformly Distributed Load)

• Definition: UDL longer than span (e.g., train of wagons).

• Application: For maximum effect, place load so that influence line area is fully covered.

• Max shear: $w \times \text{area of ILD for shear}$.

• Max moment: $w \times \text{area of ILD for moment}$.

10.4 Comparative Methods: Truss Analysis

• Method of Sections advantages:

  • Directly finds specific member without full analysis.

  • Faster for interior members.

  • Useful when only few member forces needed.

• Method of Joints: Better for all members or when joints are sequentially accessible.

10.5 Numerical Problem Types (from past papers)

1. Given $\delta$ and $\theta$ for cantilever, find $L$:

   • Use $$\displaystyle \delta = \frac{wL^4}{8EI} $$, $$\displaystyle \theta = \frac{wL^3}{6EI} $$ → $$\displaystyle \frac{\delta}{\theta} = \frac{3}{4}L $$ → $$\displaystyle L = \frac{4\delta}{3\theta} $$.

2. Moment for zero deflection at point:

   • Apply unit load at point, compute $\delta$ due to actual load; set $$\displaystyle \delta + M \cdot m = 0 $$? Actually, use Castigliano or virtual work: $$\displaystyle \delta_{actual} + M_{applied} \cdot m_{unit} = 0 $$ → $$\displaystyle M_{applied} = -\delta_{actual} / m_{unit} $$.

3. Max BM for moving UDL on girder:

   • ILD for mid-span moment: triangle, max ordinate = $l/4$.

   • Max BM = $$\displaystyle w \times \text{area under ILD} = w \times \frac{1}{2} \cdot l \cdot \frac{l}{4} = \frac{w l^2}{8} $$ (if UDL covers entire span).

   • If UDL shorter, position to maximize area.

Final Note: Always check stability and indeterminacy first. For arches, remember three-hinged is determinate; use Eddy's theorem for thrust. For influence lines, Muller-Breslau gives shape; compute ordinates via equilibrium or geometry. For continuous beams, moment distribution is most common in exams.