UNIT 3: Structural Analysis-I - Short Notes

1. Fundamental Concepts of Indeterminacy & Stability

Static Indeterminacy (Ns): Degree to which the equilibrium equations ($$\displaystyle \Sigma F_x=0, \Sigma F_y=0, \Sigma M=0 $$) are insufficient to determine all reactions and internal forces.

• External Indeterminacy (reactions): $$\displaystyle N_s = r - e $$, where $r$ = total reaction components, $e$ = total equilibrium equations (3 for plane).

• Internal Indeterminacy (members): For trusses (axial only): $$\displaystyle N_s = m - 2j - r + 3 $$, where $m$ = members, $j$ = joints. For frames (bending + axial): $$\displaystyle N_s = 3m + r - 3j $$.

• Assumption: Axial deformation often neglected in frames → reduces $$\displaystyle N_s $$ by number of axially loaded members.

Kinematic Indeterminacy (Nk): Number of independent joint displacements (rotations $\theta$ & translations $\Delta$) possible.

• For a plane frame: $$\displaystyle N_k = 2j - r - m' $$, where $m'$ = members with both ends fixed (prevent rotation/translation).

• Key Relationship: For a stable, determinate structure: $$\displaystyle N_s + N_k = 0 $$ (or total constraints = total releases).

Stability: A structure is stable if it maintains its shape under load. Criteria:

1. External Stability: Support system must prevent all rigid body motion (3 in plane).

2. Internal Stability: Members must be arranged to prevent mechanism formation.

[!TIP] A structure with $$\displaystyle N_s < 0 $$ is unstable (mechanism). $$\displaystyle N_s > 0 $$ is statically indeterminate. $$\displaystyle N_s = 0 $$ is statically determinate and stable (provided no internal mechanisms).

Sway vs. Non-Sway:

• Non-Sway: No significant horizontal displacement of joints due to unsymmetrical loading or support settlement. e.g., Symmetrically loaded symmetric frame.

• Sway: Significant horizontal displacement occurs. e.g., Portal frame with side load.

• Importance: Sway frames require separate consideration of joint translations in methods like Slope Deflection & Moment Distribution.

2. Energy Methods & Theorems

Strain Energy (U): Energy stored in a deformed elastic body.

• Axial load: $$\displaystyle U = \frac{P^2 L}{2AE} $$

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

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

Principle of Virtual Work (PVW): For a structure in equilibrium, the external virtual work equals internal virtual work.

Application for Deflection:

1. Apply a unit virtual load at the point & direction where deflection $\delta$ is sought.

2. Calculate internal virtual forces (bending moments $m$, shears $v$, axial $n$) due to unit load.

3. Calculate real internal forces ($M, V, N$) due to actual loads.

4. $$\displaystyle \delta = \int_0^L \frac{M m}{EI} dx + \int_0^L \frac{V v}{GA} dx + \int_0^L \frac{N n}{AE} dx $$. For beams, axial/shear terms often negligible: $$\displaystyle \boxed{\delta = \int_0^L \frac{M m}{EI} dx} $$.

Castigliano's Theorems:

• First Theorem: The partial derivative of total strain energy $U$ with respect to an applied force gives the displacement in the direction of that force.

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

• Second Theorem: The partial derivative of total strain energy $U$ with respect to a redundant force gives the displacement in the direction of that redundant (used in force method).

  $$\displaystyle \boxed{\delta = \frac{\partial U}{\partial X}} $$, where $X$ is redundant.

[!TIP] Use Second Theorem for finding deflections in statically indeterminate structures by releasing a redundant and enforcing compatibility.

Betti's Theorem (Reciprocal Work Theorem): For two different load systems (1 & 2) on the same structure: $$\displaystyle \boxed{\sum (P_{1i} \delta_{2i}) = \sum (P_{2i} \delta_{1i})} $$

Where $P$ = force, $\delta$ = displacement at point/direction of force.

Maxwell's Reciprocal Deflection Theorem: Direct consequence of Betti's theorem. $$\displaystyle \boxed{\delta_{12} = \delta_{21}} $$

Deflection at point 1 in direction of force $$\displaystyle P_2 $$ equals deflection at point 2 in direction of force $$\displaystyle P_1 $$.

[!TIP] Enables calculation of deflection at point A due to load at B by applying unit load at B and finding deflection at A.

3. Analysis of Trusses

Assumptions:

1. Members are pin-connected at joints (no moment transfer).

2. Loads & reactions act only at joints.

3. Members are two-force members (axial force only).

4. Self-weight neglected or distributed to joints.

5. Joints are frictionless hinges.

Method of Joints:

1. Solve for reactions.

2. Isolate a joint with ≤2 unknown forces.

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

4. Proceed joint by joint.

Zero-Force Members: If only 2 non-collinear members at a joint with no external load, both have zero force. If 3 members, 2 collinear, third is zero-force.

Method of Sections (Method of Moments):

1. Calculate reactions.

2. Pass a section through ≤3 members (or ≤2 if 2 unknowns).

3. Isolate one part; take moments about a point to solve for desired member forces directly.

Advantage: Directly finds force in specific interior members without solving entire truss.

Influence Lines for Truss Members (ILD):

• Concept: Variation of axial force in a specific member as a unit load moves across the structure.

• Construction: Use Muller-Breslau Principle (remove member, apply tension/compression as "release") or method of joints/sections for each position of unit load.

• Application: Find maximum member force under moving loads (UDL, point loads) by placing load where ILD ordinate is max/min.

For UDL longer than span: $$\displaystyle \text{Max Force} = \text{Area under positive ILD} \times \text{UDL intensity} $$.

4. Analysis of Beams & Frames: Classical Methods

Slope-Deflection Method:

• Fundamental Equation for member $AB$ (end $A$ fixed, $B$ may rotate/translate):

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

  where $$\displaystyle \psi = \frac{\Delta}{L} $$ (chord rotation due to joint translation), $$\displaystyle M_{AB}^{f} $$ = fixed-end moment.

• Stiffness Factor (K): $$\displaystyle K = \frac{EI}{L} $$ for far end pinned; $$\displaystyle K = \frac{4EI}{L} $$ for far end fixed.

• Carry-Over Factor (COF): Moment at far end due to moment at near end. COF = 0.5 for prismatic member with far end pinned or fixed.

• Distribution Factor (DF): For joint with multiple members:

  $$\displaystyle \boxed{DF_{AB} = \frac{K_{AB}}{\sum K_{\text{all members at joint}}}} $$

  Sum of DFs at a joint = 1.

• Procedure: Write slope-deflection equations for all members → solve simultaneous equations for unknown rotations/translations → compute moments → draw BMD.

Sway Frames: Include chord rotation term $\psi$. Often separate into "sway" and "non-sway" analyses or use modified slope-deflection equations with shear equation.

Three-Moment Equation (Clausen's Theorem):

• For continuous beam with 2 spans (AB, BC) at support B:

  $$\displaystyle \boxed{M_A L_{AB} + 2M_B (L_{AB} + L_{BC}) + M_C L_{BC} = -6 \left( \frac{A_1 \bar{x}_1}{L_{AB}} + \frac{A_2 \bar{x}_2}{L_{BC}} \right)} $$

  where $A\bar{x}$ = moment of area of M/EI diagram about the nearer end.

• With Support Settlement: Add term $$\displaystyle -6EI \left( \frac{\Delta_a}{L_{ab}} + \frac{\Delta_c}{L_{bc}} \right) $$ on RHS, where $\Delta$ = settlement (positive downward).

• Application: Write equation for each internal support → solve for unknown moments → compute reactions from moment equilibrium of spans.

Moment Distribution Method:

1. Calculate Fixed-End Moments (FEM) for each span under given loads (tables standard).

2. Calculate Distribution Factors (DF) at each joint (except fixed/roller supports).

3. Release each joint in turn:

   • Unbalanced moment = sum of FEM + carried moments.

   • Distribute moment to each member = Unbalanced moment × (-DF).

   • Carry-over half of distributed moment to far end.

4. Repeat cycles until moments are negligible.

5. Sum moments at each member end.

Sway Frames: Analyze in two parts: (a) Non-sway (all joints fixed against translation) → get "non-sway" moments. (b) Apply arbitrary horizontal force to cause sway → get "sway" moments. Combine using ratio from shear equation or column analogy.

Column Analogy Method:

• Analogy: A beam of length $L$ with flexural rigidity $EI$ is analogous to a column of length $L$, cross-sectional area $$\displaystyle A = 1 $$, and flexural rigidity $$\displaystyle EI_f = EI $$, on an elastic foundation with modulus $$\displaystyle k = 1 $$.

• Procedure:

  1. Replace beam loads with equivalent column loads.

  2. Find "column" reactions (analogous to beam fixed-end moments).

  3. Find "column" deflections (analogous to beam slopes).

  4. Use beam-column relations to get beam moments/slopes.

Used primarily for finding fixed-end moments of beams with various loadings and support conditions.

Effect of Support Settlements:

• Settlements induce additional moments and reactions in statically indeterminate beams.

• Analysis: Treat settlements as "loads" causing chord rotations $$\displaystyle \psi = \frac{\Delta}{L} $$ in slope-deflection equations, or add terms in three-moment equation.

• Fixed Beam with one support settled by $\delta$: Fixed-end moments induced: $$\displaystyle M_{AB} = -\frac{6EI\delta}{L^2} $$, $$\displaystyle M_{BA} = \frac{6EI\delta}{L^2} $$ (A settled, B fixed).

5. Arches

Classification:

• By Hinges: Three-hinged (determinate), Two-hinged (indeterminate), Fixed (indeterminate).

• By Shape: Parabolic (uniformly distributed load), Circular, etc.

• By Material: Masonry (thrust), Steel/Concrete (thrust + moment).

Three-Hinged Parabolic Arch (UDL):

• Horizontal Thrust: $$\displaystyle \boxed{H = \frac{wL^2}{8h}} $$

  where $w$ = load/length, $L$ = span, $h$ = central rise.

• Reactions: Use $$\displaystyle \Sigma M_{hinge}=0 $$ to find $H$, then $$\displaystyle \Sigma F_y=0 $$, $$\displaystyle \Sigma F_x=0 $$ for vertical reactions.

• Bending Moment at Section: $$\displaystyle M_x = V_x x - H y $$, where $x,y$ from section to origin.

• Normal Thrust & Shear: $$\displaystyle N_x = H \cos\theta + V_x \sin\theta $$, $$\displaystyle Q_x = V_x \cos\theta - H \sin\theta $$, where $\theta$ = slope of tangent.

Eddy's Theorem:

"The bending moment at any section of an arch is proportional to the vertical intercept between the actual arch axis and the linear arch (funicular polygon) for the given loads."

Mathematically: $$\displaystyle M_x = H \cdot y_{\text{linear arch}} $$, where $H$ = horizontal thrust (constant for three-hinged arch). Useful for quick BM determination.

Effects on Arches:

• Rib Shortening: Axial compression in arch rib causes shortening → increases horizontal thrust in two-hinged/fixed arches. Negligible in three-hinged.

• Temperature Effects: Expansion/contraction induces additional thrust/moments. For two-hinged: $$\displaystyle \Delta H = \frac{\alpha \Delta T EA}{L} $$ (if expansion prevented).

• Moving Loads: Position loads to maximize thrust/BM. For three-hinged parabolic arch, maximum positive BM under load occurs when load is at section of maximum ordinate to linear arch.

6. Cables

Assumptions:

1. Flexible (no bending stiffness).

2. Inextensible (constant length).

3. Loads perpendicular to cable axis (often horizontal for suspension bridges).

4. Cable weight negligible compared to applied loads.

Analysis (Uniformly Distributed Horizontal Load $w$):

• Shape: Parabolic (approx. for small sag): $$\displaystyle \boxed{y = \frac{w x^2}{2H}} $$, where $H$ = horizontal tension (constant).

• Equilibrium: At any point, $H$ constant, $$\displaystyle V = w x $$, $$\displaystyle T = \sqrt{H^2 + V^2} $$.

• Support Tensions: $$\displaystyle T_A = T_B = \sqrt{H^2 + \left(\frac{wL}{2}\right)^2} $$.

• Sag ($f$): At midspan ($$\displaystyle x=L/2 $$): $$\displaystyle f = \frac{w (L/2)^2}{2H} = \frac{wL^2}{8H} $$ → $$\displaystyle \boxed{H = \frac{wL^2}{8f}} $$.

• Relation: $$\displaystyle T_{\text{max}} - T_{\text{min}} = wL $$ (difference between support and midspan tension).

Cables with Concentrated Loads:

• Shape is piecewise linear (straight between loads).

• Tension in each segment constant.

• Use equilibrium at each load point (force polygon method) to find tensions.

• Geometry from catenary equations or parabolic segments.

7. Influence Line Diagrams (ILD)

Definition: Graph showing variation of a response function (reaction, shear, moment, member force) at a specific point as a unit load moves across the structure.

Muller-Breslau Principle (MBP):

"The ILD for any response function is of the same shape as the deflected shape of the structure when that function is removed (i.e., the corresponding constraint is released) and a unit value of that function is applied."

• Steps:

  1. Remove the constraint corresponding to the response function (e.g., for reaction at A, remove support at A).

  2. Impose a unit displacement/rotation in the direction of the response (e.g., unit vertical displacement at A for reaction).

  3. The resulting deflected shape (within elastic limit) is the ILD (with correct sign convention).

ILD for Beams:

• Reaction at Support: Triangular shape. Ordinate at load position = reaction due to unit load at that position.

• Shear Force at Section: Two triangles meeting at section. Sign changes at section.

• Bending Moment at Section: Two triangles meeting at section. Value at load position = moment due to unit load at that position.

Application of ILD for Moving Loads:

1. Single Concentrated Load $P$: Max effect = $P \times$ (max ordinate of ILD).

2. Train of Loads: Place loads on ILD to maximize positive/negative effect (usually heaviest load on max positive ordinate).

3. UDL of Finite Length $w$: Max effect = $w \times$ (area under ILD covered by UDL).

4. Equivalent UDL (EUDL): For UDL longer than span, max effect = $w \times$ (area under entire ILD).

[!TIP] For UDL longer than span, the maximum effect is simply $w \times$ (area under entire ILD), as the load can always cover the entire diagram.

8. Deflection of Beams

Elastic Curve Equation: $$\displaystyle \boxed{EI \frac{d^2 y}{d x^2} = M(x)} $$

Integrate twice: $$\displaystyle \frac{dy}{dx} = \theta(x) $$, $y(x)$. Apply boundary conditions (deflection/slope at supports).

Standard Formulas (Cantilever, Simply Supported, Fixed):

• Cantilever, UDL $w$:

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

• Simply Supported, Central Point Load $P$:

  $$\displaystyle \delta_{\text{mid}} = \frac{PL^3}{48EI} $$

• Fixed-Fixed, UDL $w$:

  $$\displaystyle \delta_{\text{mid}} = \frac{wL^4}{384EI} $$, $$\displaystyle M_{\text{ends}} = \frac{wL^2}{12} $$ (hogging).

Deflection in Composite/Non-Prismatic Beams:

• Beam split into segments with different $EI$.

• Integrate $M/EI$ piecewise.

• Continuity Conditions: At interfaces, slope $\theta$ and deflection $y$ must be continuous.

• Solve integration constants from boundary conditions and continuity.

Propped Cantilevers:

• Analysis Methods:

  1. Consistent Deformation (Force Method): Prop reaction $R$ is redundant. Compatibility: deflection at prop = 0 (or given settlement).

  2. ILD: ILD for reaction at prop. Max $R$ = area under ILD for moving load.

  3. Integration: Apply boundary conditions.

• Effect of Prop Sinking: If prop sinks by $\delta$, compatibility: $$\displaystyle y_{\text{prop}} = \delta $$. This induces additional reaction and alters internal moments.

[!TIP] For propped cantilever with UDL, prop reaction $$\displaystyle R = \frac{3}{8}wL $$ (if prop at free end? No, typically at some point). Always check boundary conditions carefully.