Structural Analysis-I (CE-403) - Unit 5 Short Notes

UNIT 5: Structural Analysis-I - Short Notes

I. Structural Indeterminacy and Stability

Indeterminacy

A structure is indeterminate if the number of independent equilibrium equations is insufficient to determine all unknown reactions and internal forces solely from equilibrium conditions. Indeterminacy provides redundancy, enhancing safety and serviceability but requiring advanced analysis methods.

Static Indeterminacy ($$\displaystyle N_s $$)

• Definition: Number of unknown reactions/internal forces minus number of independent equilibrium equations.

• Calculation for Plane Structures:

  • Beams/Frames: $$\displaystyle N_s = R - 3 $$ (where $R$ = total external reactions). For frames with rigid joints, also consider internal moments at joints.

  • Trusses (Pin-jointed): $$\displaystyle N_s = m + r - 2j $$ (where $m$ = members, $r$ = external reactions, $j$ = joints).

  • General Formula (Plane Frame): $$\displaystyle N_s = 3m + r - 3j - s' $$ (where $m$ = members, $r$ = external reactions, $j$ = joints, $s'$ = number of internal hinges).

Kinematic Indeterminacy ($$\displaystyle N_k $$)

• Definition: Number of independent joint displacements (translations/rotations) possible in a structure.

• Calculation for Beams/Frames (neglecting axial deformations):

  • Count all unknown joint rotations and independent joint translations (sway).

  • For a fixed support: Restrains 2 translations + 1 rotation → contributes 3 to $$\displaystyle N_k $$.

  • For a hinged/pinned support: Restrains 2 translations → contributes 2 to $$\displaystyle N_k $$.

  • For a roller support: Restrains 1 translation (perpendicular to roller) → contributes 1 to $$\displaystyle N_k $$.

  • At a rigid joint: If joint has $n$ members, it can have 1 rotation and 2 translations (if free to sway). If no sway, only 1 rotation is possible.

  [!TIP] Exam Tip: For a plane frame, Degree of Indeterminacy = $$\displaystyle N_s - N_k $$. If $$\displaystyle N_s > N_k $$, structure is statically indeterminate. If $$\displaystyle N_s = N_k $$, structure is statically determinate and stable. If $$\displaystyle N_s < N_k $$, structure is unstable.

Stability of Structures

• External Stability: The structure as a whole must be externally stable. The support system must prevent all 3 rigid body motions (2 translations, 1 rotation) in a plane. Check: $R \geq 3$ and supports must be properly arranged (non-parallel, non-concurrent reaction lines).

• Internal Stability: The structural geometry must not deform into a mechanism. This requires that $$\displaystyle N_s \geq N_k $$ and that there are no mechanisms formed by member alignment (e.g., three members meeting at a joint without a fixed support can form a mechanism).

Sway and Non-sway Frames

• Non-sway Frame: No possibility of lateral displacement (sway) of joints under vertical loads. Analysis can be done without considering joint translations.

• Sway Frame: Subjected to horizontal loads or asymmetric vertical loading causing lateral displacement. Requires consideration of sway displacements in analysis (e.g., in slope deflection, additional equations from joint equilibrium are needed).

Pin-jointed vs. Rigidly Jointed Structures

II. Energy Methods

Principle of Virtual Work

• Statement: For a structure in equilibrium, the external virtual work done by a system of virtual forces in equilibrium is equal to the internal virtual work done by the corresponding virtual stresses/strains.

• Application to Flexural Members: For a beam, if a virtual unit load is applied at the point/direction where deflection $\delta$ is desired, the deflection is:

$$\delta = \int \frac{M m}{EI} dx$$

where $M$ = actual bending moment due to real loads, $m$ = bending moment due to virtual unit load.

Strain Energy ($U$)

• Definition: Energy stored in a deformable body due to applied loads, as it strains elastically.

• Expressions:

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

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

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

Complementary Energy

• Concept: Energy expressed in terms of stresses. For linear elastic structures, Strain Energy = Complementary Energy.

Castigliano's Theorems

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

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

> **Condition:** Load $$\displaystyle P_i $$ must be the only load acting in its direction during differentiation; other loads are kept constant.

2. Second Theorem: The partial derivative of the total strain energy $U$ with respect to a displacement $$\displaystyle \delta_i $$ gives the force $$\displaystyle P_i $$ required to produce that displacement.

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

> **Application:** Used to find reactions (forces) when a support settlement (displacement) is known.

Betti's Theorem

• Statement: For a linearly elastic structure, the work done by a first system of loads acting through the displacements caused by a second system of loads is equal to the work done by the second system acting through the displacements caused by the first.

$$\sum P_i \delta_i' = \sum P_i' \delta_i$$

where primed quantities correspond to the second load system.

Maxwell's Reciprocal Deflection Theorem

• Statement: A direct consequence of Betti's Theorem. The deflection at point $A$ in the direction of force $P$ due to load $P$ at $B$ is equal to the deflection at point $B$ in the direction of force $P$ due to the same load $P$ applied at $A$.

$$\delta_{A,B} = \delta_{B,A}$$

• Derivation: Apply Betti's Theorem to two load systems: (1) Unit load at A, (2) Unit load at B.

III. Analysis of Indeterminate Beams

Slope Deflection Method

• Slope Deflection Equation (for member AB):

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

where $$\displaystyle \psi = \frac{\Delta}{L} $$ (chord rotation due to support settlement or sway), $$\displaystyle M_{AB}^{fixed} $$ = fixed-end moment due to external loads on member AB (with ends fixed).

• Fixed-End Moments (FEM) for Standard Loads (Span L):

  • Point Load $W$ at distance $a$ from A, $b$ from B:

    $$\displaystyle M_{AB}^f = -\frac{Wb^2a}{L^2} $$, $$\displaystyle M_{BA}^f = -\frac{Wa^2b}{L^2} $$

  • UDL $w$ over entire span:

    $$\displaystyle M_{AB}^f = M_{BA}^f = -\frac{wL^2}{12} $$

  • Couple $M$ at A (clockwise):

    $$\displaystyle M_{AB}^f = M $$, $$\displaystyle M_{BA}^f = -M $$

• Application: Write equations for each member. For sway frames, an additional equation from joint equilibrium (∑H=0 or ∑V=0) is needed to solve for the sway displacement $\Delta$.

Moment Distribution Method

• Distribution Factor (DF): Fraction of unbalanced moment at a joint distributed to a connecting member.

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

where $k$ = stiffness factor.

• Stiffness Factor ($k$): For a member AB:

  • Far end fixed: $$\displaystyle k = \frac{4EI}{L} $$

  • Far end pinned/roller: $$\displaystyle k = \frac{3EI}{L} $$

  • Far end free (cantilever): $$\displaystyle k = \frac{EI}{L} $$ (or 0 for distribution)

• Carry-over Factor (CO): When a moment is applied at one end of a member, half of it is "carried over" to the other end. CO = 0.5 (for prismatic member with far end fixed or pinned).

• Procedure: 1) Calculate FEM. 2) Calculate DFs. 3) Balance & distribute moments at each joint iteratively until moments are negligible. 4) Sum FEM + distributed moments + carried-over moments for final end moments.

Three-Moment Equation

• For Continuous Beams with Constant I:

$$M_{A-1}L_{A-1,A} + 2M_A(L_{A-1,A} + L_{A,A+1}) + M_{A+1}L_{A,A+1} = -6 \left( \frac{A_{A-1}\bar{x}_{A-1}}{L_{A-1,A}} + \frac{A_{A}\bar{x}_{A}}{L_{A,A+1}} \right)$$

where $A$ = area of BMD for simple span between two supports, $\bar{x}$ = centroid distance from left support.

• For Variable I: The equation includes terms for $$\displaystyle \frac{1}{I} $$:

$$\frac{M_{A-1}L_{A-1,A}}{I_{A-1,A}} + 2M_A \left( \frac{L_{A-1,A}}{I_{A-1,A}} + \frac{L_{A,A+1}}{I_{A,A+1}} \right) + \frac{M_{A+1}L_{A,A+1}}{I_{A,A+1}} = -6 \left( \frac{A_{A-1}\bar{x}_{A-1}}{I_{A-1,A}L_{A-1,A}} + \frac{A_{A}\bar{x}_{A}}{I_{A,A+1}L_{A,A+1}} \right)$$

Column Analogy Method

• Analogy: A beam of length $L$ and flexural rigidity $EI$ under end moments and transverse loads is analogous to a column of length $L$ and axial rigidity $EA$ under axial loads and lateral forces.

• Application: The bending moment diagram for the beam is analogous to the axial load diagram for the column. The shear force diagram is analogous to the lateral force diagram. Fixed-end moments and reactions can be found by solving the analogous column problem.

Effects of Support Settlement

• Induces additional moments and reactions in statically indeterminate beams.

• Analysis: In Slope Deflection, chord rotation $$\displaystyle \psi = \frac{\Delta}{L} $$ (where $\Delta$ = relative settlement) is included in the equation. In Moment Distribution, equivalent fixed-end moments due to settlement are applied first.

  • For a beam AB with support B settling by $\Delta$ (downwards):

    $$\displaystyle M_{AB}^f = \frac{6EI\Delta}{L^2} $$, $$\displaystyle M_{BA}^f = -\frac{6EI\Delta}{L^2} $$

Temperature Effects on Beams

• Cause: Differential thermal expansion if top and bottom fibers are at different temperatures.

• Induced Moment: For a beam with uniform temperature change $\Delta T$, no stress. If $\Delta T$ varies linearly through depth (e.g., top at $$\displaystyle T_1 $$, bottom at $$\displaystyle T_2 $$), curvature $$\displaystyle \kappa = \frac{\alpha (T_2 - T_1)}{h} $$.

• Fixed-end moments due to temperature gradient (for prismatic beam):

$$M_{AB}^f = M_{BA}^f = \frac{EI \alpha \Delta T}{h} \quad \text{(for linear gradient)}$$

where $$\displaystyle \Delta T = T_{bottom} - T_{top} $$, $h$ = depth.

IV. Arches

Classification

Three-Hinged Arches

• Horizontal Thrust ($H$): Determined using Eddy's Theorem or equilibrium.

• Eddy's Theorem: The bending moment at any section of a three-hinged arch is equal to the bending moment at that section of a simply supported beam of the same span carrying the same loads, minus the product of the horizontal thrust $H$ and the vertical intercept from the arch's linear arch (funicular polygon) to the actual arch at that section.

$$M_x = M_x^{simply\,supported} - H \cdot y_x$$

• At any section: Normal thrust $N$, shear force $V$, and bending moment $M$ can be found from equilibrium of a segment.

Two-Hinged Arches

• Horizontal Thrust: Indeterminate. Depends on rib shortening and temperature.

• Rib Shortening: Axial compression in the arch rib causes shortening, which reduces the horizontal thrust $H$.

• Temperature Effects: Temperature change causes expansion/contraction. For a two-hinged arch, this induces an additional horizontal thrust.

Fixed Arches

• Fully restrained against rotation and translation at supports.

• No horizontal thrust at supports (if symmetric and symmetric loading). Moments at supports are large.

• Temperature & Rib Shortening cause significant secondary stresses.

Special Effects

• Rib Shortening: Axial force $N$ causes shortening $$\displaystyle \delta = \frac{NL}{AE} $$. This shortening tends to close the arch, reducing the horizontal thrust $H$ in two-hinged arches.

• Temperature Effects:

  • Three-hinged: $\Delta T$ causes no additional stress (hinges allow free movement).

  • Two-hinged/Fixed: $\Delta T$ induces additional axial force and thus additional horizontal thrust.

$$\Delta H_T = \frac{EA \alpha \Delta T}{L} \cdot \frac{h}{L} \quad \text{(approx for parabolic arch)}$$

Practical Considerations

• Lintels are preferred over arches in modern construction due to: ease of construction, smaller foundation loads, larger headroom, and compatibility with modern masonry/concrete which is weak in tension.

V. Cables

Assumptions

1. Cables are perfectly flexible (no bending stiffness).

2. Cables are inextensible (length constant).

3. Loads are perpendicular to the cable axis (or weight is negligible compared to applied loads).

4. Cable is homogeneous and frictionless at supports.

Analysis under Concentrated Loads

• Cable shape is polygonal.

• Tension in segments: Found by equilibrium of joints (method of joints). Tension is constant along a straight segment between loads.

• Geometry: Use geometry to relate vertical and horizontal components. Horizontal component $H$ is constant throughout the cable (from ∑M=0 about any point).

Analysis under Uniformly Distributed Load (Horizontal)

• Cable Shape: Parabolic ($$\displaystyle y = \frac{w}{2H}x^2 $$ for symmetric cable with vertex at origin).

• Horizontal Thrust ($H$):

$$H = \frac{wL^2}{8h} \quad \text{(for symmetric parabolic cable, span L, sag h)}$$

• Tension:

  • At lowest point (vertex): $$\displaystyle T_{min} = H $$ (horizontal only).

  • At supports: $$\displaystyle T_{max} = \sqrt{H^2 + V^2} $$, where $$\displaystyle V = \frac{wL}{2} $$ (vertical reaction).

$$T_{max} = H \sqrt{1 + \left( \frac{L}{2h} \right)^2}$$

• Cable Length (Parabolic):

$$L_{cable} \approx L \left[ 1 + \frac{8}{3} \left( \frac{h}{L} \right)^2 \right] \quad \text{(for small sag)}$$

Difference between Greatest and Least Tensions

• $$\displaystyle \Delta T = T_{max} - T_{min} = H \left( \sqrt{1 + \left( \frac{L}{2h} \right)^2} - 1 \right) $$

VI. Influence Lines

Definition and Importance

• Influence Line Diagram (ILD): A graph showing the variation of a response function (reaction, shear, moment, truss force) at a specific point in a structure as a unit load moves across the structure.

• Importance: Used to determine the maximum effect (positive/negative) of a moving load system (vehicles, trains) on a structural member.

Muller-Breslau Principle

• Statement: The influence line for a response function (reaction, shear, moment) is proportional to the deflected shape of the structure when the constraint providing that response function is removed and a unit displacement (rotation or translation) is imposed in the positive direction of the function.

• Application: 1) Remove the restraint corresponding to the desired function. 2) Impose a unit displacement (1 for reaction, 1 rotation for moment) in the positive sense. 3) The resulting elastic curve is the ILD.

  • For shear at a section: Cut the beam at the section, impose a vertical displacement of 1 unit.

  • For moment at a section: Release the moment restraint (insert a hinge), impose a relative rotation of 1 radian.

Influence Lines for Beams

• Shear Force at Section: ILD is a step function (±1 on either side of the section).

• Bending Moment at Section: ILD is triangular (for simply supported) or composed of lines meeting at the section.

• For Continuous Beams: Use Muller-Breslau. The ILD for moment at an interior support will have negative ordinates on adjacent spans.

Influence Lines for Truss Members (Panel Method)

1. Cut the truss through the member whose force ILD is required.

2. Apply a unit force in the direction of the member (tension positive) at the cut.

3. Solve the resulting statically determinate truss (using method of joints/sections) for the vertical reactions at the supports due to this unit force.

4. The vertical reaction diagram (as the load moves) is the ILD for the member force. The value at a load position is the algebraic sum of vertical reactions at the two ends of the panel where the load is applied.

Moving Loads on Beams

• Single Point Load ($P$): Maximum effect = $P \times$ (maximum ordinate of ILD at load position).

• UDL Longer than Span: Maximum effect = UDL intensity $\times$ (Area under ILD over the span).

• Equivalent Uniformly Distributed Load (EUDL): For a train of point loads, the maximum effect is found by placing the load system such that the leading wheel is at the maximum positive ordinate of the ILD. The effect is the sum of each load multiplied by its corresponding ILD ordinate.

• Train of Point Loads: Maximum effect = $$\displaystyle \sum (P_i \times y_i) $$, where $$\displaystyle y_i $$ is ILD ordinate under load $$\displaystyle P_i $$.

VII. Trusses

Assumptions in Simple Truss Analysis

1. All members are pin-connected at joints.

2. All external loads are applied at the joints only.

3. Members are weightless (or their weight is distributed to joints).

4. The structure is plane and loads are in the plane.

5. Axial forces only in members (no bending/shear at joints).

Method of Joints

• Procedure: Isolate each joint, apply $$\displaystyle \sum F_x = 0 $$, $$\displaystyle \sum F_y = 0 $$. Start at a joint with ≤2 unknowns.

• Sequence: Solve joints systematically, using known forces from previous joints.

Method of Sections

• Procedure: Pass a section through no more than 3 members (whose forces are unknown). Isolate one part, apply $$\displaystyle \sum F_x = 0 $$, $$\displaystyle \sum F_y = 0 $$, $$\displaystyle \sum M = 0 $$.

• Advantage over Method of Joints: Directly finds forces in specific members without solving the entire truss. Particularly useful for members in the middle of a large truss.

Determination of Member Forces

• Tension: Member is pulled; force points away from joint.

• Compression: Member is pushed; force points towards joint.

• Zero Force Members: Identified by inspection:

  • Joint with only 2 non-collinear members and no external load → both members are zero force.

  • Joint with 3 members, two collinear, no external load → third member is zero force.

VIII. Deflection of Beams

Cantilever Beams (Fixed at A, Free at B)

• Under Point Load $W$ at Free End:

  • Deflection at B: $$\displaystyle \delta_B = \frac{WL^3}{3EI} $$ (downwards)

  • Slope at B: $$\displaystyle \theta_B = \frac{WL^2}{2EI} $$ (clockwise)

• Under UDL $w$ over entire span:

  • Deflection at B: $$\displaystyle \delta_B = \frac{wL^4}{8EI} $$

  • Slope at B: $$\displaystyle \theta_B = \frac{wL^3}{6EI} $$

• Under Couple $M$ at Free End:

  • Deflection at B: $$\displaystyle \delta_B = \frac{ML^2}{2EI} $$

  • Slope at B: $$\displaystyle \theta_B = \frac{ML}{EI} $$

Propped Cantilevers

• A cantilever with an additional support (prop) at some point.

• Prop Reaction ($R$): Found by enforcing zero deflection at prop location (using moment-area or integration).

• Deflection at Various Points: Calculated considering the effect of both fixed-end moment and prop reaction.

Deflection Calculation Methods

• Moment Area Method:

  • Theorem 1: Change in slope between two points = $$\displaystyle \frac{Area\,of\,M/EI\,diagram\,between\,points}{L} $$.

  • Theorem 2: Deflection of point B relative to tangent at A = $$\displaystyle \frac{Moment\,of\,Area\,of\,M/EI\,diagram\,between\,A\,and\,B\,about\,B}{EI} $$.

• Strain Energy Method (Castigliano's Theorem):

  • Compute total strain energy $$\displaystyle U = \int \frac{M^2}{2EI} dx $$.

  • Desired deflection $$\displaystyle \delta = \frac{\partial U}{\partial P} $$, where $P$ is the load causing the deflection.

Relationship between Deflection, Slope, and Beam Length

• For a cantilever with UDL, $$\displaystyle \frac{\delta_B}{\theta_B} = \frac{L}{3} $$.

• For a cantilever with point load at end, $$\displaystyle \frac{\delta_B}{\theta_B} = \frac{L}{1.5} $$.

• These relationships are used in problems where deflection and slope are given to find $L$ or other parameters.

IX. Special Definitions and Concepts (Integrated Above)

• Eddy's Theorem: (See IV.2.3) $$\displaystyle M_x = M_x^{simple} - H \cdot y_x $$

• EUDL (Equivalent Uniformly Distributed Load): (See VI.5.3) For moving point loads, the maximum effect is equivalent to a UDL of length equal to the structure's span and intensity such that its total load equals the sum of point loads, but placed to maximize effect.

• Distribution Factor & Stiffness Factor: (See III.2) DF = $$\displaystyle k_{member}/\sum k $$, $$\displaystyle k = 4EI/L $$ (far fixed), $3EI/L$ (far pinned).

• Strain Energy Method: (See II.2 & VIII.4) $$\displaystyle U = \int \frac{M^2}{2EI}dx $$, $$\displaystyle \delta = \frac{\partial U}{\partial P} $$.

• Influence Line Diagram (ILD): (See VI.1) Graph of response function vs. position of unit moving load.

• Muller-Breslau Principle: (See VI.2) ILD ∝ displaced shape after removing restraint and applying unit displacement.

• Castigliano's Theorems: (See II.4) $$\displaystyle \delta_i = \frac{\partial U}{\partial P_i} $$, $$\displaystyle P_i = \frac{\partial U}{\partial \delta_i} $$.

• Betti's Theorem: (See II.6) $$\displaystyle \sum P_i \delta_i' = \sum P_i' \delta_i $$.

• Principle of Virtual Work: (See II.1) $$\displaystyle \delta = \int \frac{M m}{EI} dx $$ for beams.

• Temperature Effect on Arches: (See IV.5.2) Causes additional thrust in two-hinged/fixed arches; no effect on three-hinged.

• Rib Shortening: (See IV.5.1) Axial compression shortens arch, reducing $H$ in two-hinged arches.