Skip to main content

Chapter 1 · 4 hours

Introduction

IOE past exam questions

Past questions and answers

11 questions set from this chapter, 5 of them more than once; 3 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 7 of 24 exams
  • Asked 7 times
  • 2080 Chaitra · 4 marks
  • 2078 Chaitra · 4 marks
  • 2076 Baisakh · 4 marks
  • 2076 Bhadra · 4 marks
  • 2075 Baisakh · 4 marks
  • 2072 Magh · 4 marks
  • 2072 Asoj · 4 marks

Explain the principle of superposition with suitable examples, stating the conditions for its application and its limitations.

Answer

The principle of superposition states that for a linearly elastic structure, the total effect (reaction, internal force, stress or displacement) caused by several loads acting together is equal to the algebraic sum of the effects caused by each load acting separately.

Conditions for application

  1. The material obeys Hooke's law (stress is proportional to strain), i.e. the material is linearly elastic.
  2. Displacements are small, so the geometry of the structure does not change appreciably under load (equilibrium equations are written on the undeformed geometry).
  3. The support conditions do not change when loads are applied (no settlement-dependent or yielding supports).
  4. Loads are applied statically and do not change the stiffness of the structure.

Examples

1. Beam deflection. A simply supported beam carries a central point load PP and a UDL ww over the full span. The mid-span deflection is the sum of the two separate deflections:

δ=PL348EI+5wL4384EI\delta = \frac{PL^3}{48EI} + \frac{5wL^4}{384EI}

2. Reactions and bending moment. For a beam with several point loads, the reaction at a support is the sum of the reactions due to each load alone. The bending moment at a section is the sum of the moments due to each load alone.

3. Truss member force. The force in a member under combined loading is the sum of the forces due to each joint load applied separately.

Uses

  • Complicated loading is split into simple standard cases whose results are tabulated.
  • Basis of influence lines, the force method (compatibility) and the method of consistent deformations for indeterminate structures.

Limitations

  • It is not valid if the material is non-linear (stress beyond the proportional limit, cracked concrete, plastic hinges).
  • It is not valid if deformations are large, so the geometry changes (cables with large sag, slender columns, PP-Δ\Delta effects).
  • It fails when the effect of one load changes the effect of another, e.g. a beam-column where axial load changes the bending stiffness: the deflection under P1+P2P_1+P_2 is not the sum of the separate deflections.
  • It cannot be used when supports change their nature under load (e.g. a gap closing or a support lifting off).
  • It does not apply to quantities that are not linear in load, such as strain energy: U≠U1+U2U \ne U_1 + U_2, since UU depends on the square of the load.
  • Most repeated · 3 of 24 exams
  • Asked 3 times
  • 2073 Magh · 4 marks
  • 2069 Poush · 6 marks
  • 2075 Baisakh · 4 marks

Differentiate between linear and non-linear behaviour of structures with suitable force-displacement diagrams and explain their uses in theory of structures.

Answer

A structure shows linear behaviour when displacements are directly proportional to the applied loads, and non-linear behaviour when this proportionality does not hold.

 Force                       Force
  |        /                  |       ___---
  |      /                    |   _--
  |    /  slope = k           | /   slope changes
  |  /                        |/
  +-------------- Disp.       +-------------- Disp.
   (a) Linear                  (b) Non-linear
PointLinear behaviourNon-linear behaviour
Force-displacementStraight line through originCurve, slope (stiffness) changes with load
Stiffness kkConstantVaries with load or displacement
Principle of superpositionValidNot valid
Governing equationsLinear algebraic equationsNon-linear equations, solved iteratively
Result with double loadResponse doublesResponse is not doubled
CausesHooke's law, small displacementMaterial yielding, large displacement, changing support conditions
Analysis effortSimple, one-stepLoad applied in increments, Newton-Raphson iteration
UnloadingReturns along the same lineResidual deformation may remain (if inelastic)

Types of non-linearity

  • Material non-linearity: stress-strain curve is not straight (steel after yield, concrete).
  • Geometric non-linearity: large displacements change the geometry, e.g. cables, slender columns.

Uses in theory of structures

  • Linear analysis is the basis of most classical methods (moment area, virtual work, influence lines, slope-deflection), used for service-load design.
  • Superposition allows complex loadings to be solved as combinations of simple cases.
  • Non-linear analysis is used to find collapse load and ductility, to design slender and cable structures, and to study post-yield behaviour in earthquake design.
  • Most repeated · 3 of 24 exams
  • Asked 3 times
  • 2079 Chaitra · 2 marks
  • 2077 Chaitra · 2 marks
  • 2070 Bhadra · 3 marks

What are the two basic approaches of structural analysis? Explain them briefly.

Answer

Structural analysis is carried out by one of two basic approaches, depending on which quantity is taken as the unknown. Both satisfy equilibrium, compatibility and force-displacement relations.

1. Force method (flexibility method)

  • The unknowns are redundant forces (reactions or internal forces).
  • The structure is made statically determinate by removing redundants, and compatibility conditions at the removed points are written in terms of flexibility coefficients.
  • Used mainly for structures with a low degree of static indeterminacy.
  • Examples: consistent deformation, three-moment equation, Castigliano's second theorem.

2. Displacement method (stiffness method)

  • The unknowns are joint displacements (rotations and translations).
  • Equilibrium equations of joints are written in terms of stiffness coefficients and solved for displacements, then member forces are found.
  • Used when the degree of kinematic indeterminacy is small, and it is easy to programme.
  • Examples: slope-deflection method, moment distribution, matrix stiffness method.
PointForce methodDisplacement method
UnknownsRedundant forcesJoint displacements
Conditions writtenCompatibilityEquilibrium
CoefficientsFlexibilityStiffness
Suitable forFew redundantsFew degrees of freedom
  • Asked 2 times
  • 2077 Chaitra · 2 marks
  • 2068 Bhadra · 4 marks

Explain briefly the use of computer based methods in structural analysis.

Answer

Computer based methods use matrix formulation of the force or displacement method (matrix stiffness method, finite element method) so that large structures can be analysed quickly by software such as SAP2000, ETABS, STAAD.Pro and ANSYS.

Steps in computer analysis

  1. Model the structure: joints (nodes), members (elements), supports, material and section properties.
  2. Apply loads and load combinations.
  3. The program forms member stiffness matrices, assembles the global matrix [K][K], and solves [K]{Δ}={F}[K]\{\Delta\} = \{F\} for joint displacements.
  4. Member forces, reactions and stresses are back-calculated and presented as diagrams and tables.

Uses and advantages

  • Handles highly indeterminate structures (multi-storey frames, 3D frames, shells, plates) that are impractical by hand.
  • Very fast and accurate; many load combinations (dead, live, wind, earthquake) can be run quickly.
  • Allows non-linear, dynamic, buckling and soil-structure interaction analysis.
  • Easy to repeat the analysis after changing member sizes in design iterations.
  • Gives graphical output (deformed shape, bending moment diagrams).

Cautions

  • Results depend on the correctness of the model, so the engineer must check them, e.g. equilibrium of reactions and approximate hand calculations.
  • "Garbage in, garbage out": wrong supports or loads give wrong results.
  • Asked 2 times
  • 2073 Bhadra · 4 marks
  • 2069 Bhadra · 4 marks

Describe the types of structures based on the material used and the methods of their analysis.

Answer

Structures can be classified by the principal material used. The material decides the strength, stiffness, behaviour (linear or not) and hence the method of analysis.

TypeFeaturesMethod of analysis
Steel structuresHigh strength, ductile, nearly linear elastic until yield; trusses, frames, bridges, towersElastic analysis with the usual methods; plastic analysis for collapse load; second-order analysis for slender members
Reinforced concrete (RCC)Concrete is weak in tension, so steel carries tension; composite and non-linear after cracking; buildings, bridges, damsElastic analysis for member forces (frame analysis, moment distribution, matrix methods); limit state design for sections; redistribution of moments
Prestressed concreteInitial compression offsets tension; long spansElastic analysis for stresses, plus prestress effects; checks at transfer and service
Timber structuresLight, anisotropic, strength depends on grain direction; trusses, roofsElastic analysis with allowable stresses; special attention to joints
Masonry structuresStrong in compression, weak in tension; walls, arches, domesThrust line analysis for arches; stress check with no-tension assumption
Composite structuresSteel and concrete act together (steel girder with concrete slab)Transformed section (modular ratio) method
Aluminium and othersLight, lower stiffness; special structuresSimilar to steel but allowing for the lower modulus

All of them are analysed first as linearly elastic with the principle of superposition. Non-linear methods are used when the load nears failure.

  • 2075 Bhadra · 1+3 marks

What is structure analysis? Explain the basic approaches of structural analysis.

Answer

Structural analysis is the process of determining the reactions, internal forces (axial force, shear force, bending moment, torsion), stresses and displacements of a structure under given loads, so that it can be designed safely and economically.

Basic approaches

Every method of analysis must satisfy three conditions: equilibrium, compatibility (continuity of displacements) and the force-displacement (constitutive) relation. Depending on the unknown chosen, two approaches exist.

1. Force (flexibility) method

  • Unknowns: redundant forces.
  • Remove redundants to get a statically determinate (released) structure, then apply compatibility equations.
  • Examples: method of consistent deformation, three-moment theorem.

2. Displacement (stiffness) method

  • Unknowns: joint displacements (rotations and translations).
  • Write joint equilibrium equations using stiffness coefficients.
  • Examples: slope-deflection, moment distribution, matrix stiffness method.

The force method suits structures with few redundants; the displacement method suits structures with few degrees of freedom and is used in computer programs.

  • 2071 Bhadra · 6 marks

Explain the characteristics of structural mechanics and describe, with suitable examples, the two basic approaches of structural analysis.

Answer

Structural mechanics is the branch of mechanics that studies how structures respond to loads, i.e. the forces, stresses, strains and displacements produced in them.

Characteristics of structural mechanics

  1. It is based on three fundamental conditions: equilibrium of forces, compatibility of displacements, and the material's force-displacement relation (Hooke's law).
  2. It idealises real structures into simple models: members as lines (beams, columns, bars), supports as hinge, roller or fixed, loads as point, line or distributed loads.
  3. It assumes materials are homogeneous, isotropic and linearly elastic, and displacements are small.
  4. It uses the principle of superposition for linear structures.
  5. It separates structures into statically determinate (solved by equilibrium alone) and indeterminate (need compatibility as well).
  6. It supplies the data (forces and displacements) needed for design of members and for checking serviceability.

Two basic approaches of structural analysis

Force (flexibility) method: The unknowns are redundant forces. Example: a propped cantilever carrying a UDL is made determinate by removing the prop. The deflection of the cantilever at the prop point is equated to the prop's deflection, so the prop reaction is found:

RB=3wL8R_B = \frac{3wL}{8}

Displacement (stiffness) method: The unknowns are joint displacements. Example: in a continuous beam the support rotations θB\theta_B are taken as unknown and found from the joint moment equilibrium ∑MB=0\sum M_B = 0 (slope-deflection method); member end moments are found afterwards.

PointForce methodDisplacement method
UnknownRedundant forcesJoint displacements
Condition solvedCompatibilityEquilibrium
Best forFew redundantsFew displacement unknowns, computers
  • 2071 Magh · 4 marks

Explain with suitable force-displacement diagrams the elastic, inelastic, linear and non-linear behaviour of structure.

Answer

Behaviour of a structure is described by its force-displacement (load-deflection) curve during loading and unloading.

 F |   /            F |    ___--      
   |  /               |  _-          
   | /  loading &     | /  loading    
   |/   unloading     |/  _-- unload  
   +---------- D      +--/--------- D
  (a) Linear elastic   (b) Non-linear elastic
 F |     ___---       F |    _--
   |    /             |  _/ 
   |   /  unload      | /   \ unload
   |  /  parallel     |/     \ (residual)
   +--/-/-------- D   +-------\---- D
  (c) Inelastic        (d) Non-linear

1. Linear elastic: The load-deflection curve is a straight line through the origin. The structure returns to its original shape on unloading along the same line. Example: steel beam under working loads.

2. Non-linear elastic: The curve is not straight, but unloading follows the same curve back to the origin; no permanent deformation. Example: a rubber band, or a cable with sag change.

3. Inelastic: Beyond the elastic limit the loading and unloading paths differ. Unloading is along a line parallel to the initial elastic slope, leaving residual (permanent) deformation. Example: steel stressed beyond yield.

4. Non-linear: The stiffness (slope of the curve) changes with load. It arises from material non-linearity (yielding, cracking) or geometric non-linearity (large displacement). The principle of superposition is not valid.

Linear and non-linear describe the shape of the curve; elastic and inelastic describe whether unloading recovers the deformation.

  • 2074 Bhadra · 4 marks

Explain material non-linearity and geometrical non-linearity with neat sketches.

Answer

Non-linearity arises when displacement is not proportional to load. The two main sources are the material and the geometry.

Material non-linearity

The stress-strain relation of the material is not linear, so the stiffness changes with stress even for small displacements.

  • Examples: steel beyond the yield point, concrete in compression, cracked concrete, soil.
  • Equilibrium is still written on the undeformed geometry.
 stress           ___________ yield
   |            /
   |          /
   |        /
   |      /
   +------------------ strain
   elastic-plastic curve

Geometric non-linearity

Displacements are large enough that the geometry of the structure changes and equilibrium must be written on the deformed shape, even if the material is linear.

  • Examples: slender columns (P-delta effect), cables and suspension bridges, thin shells, arches near buckling.
 Load          P  |
   |          ___ v  deflected
   |       _--   -----------> axial load P
   |     /           acting on deflection
   |   /             adds extra moment P*delta
   +------------------ deflection

Here the extra moment PΔP\Delta makes the load-deflection curve non-linear: stiffness decreases as axial load increases.

PointMaterialGeometric
CauseStress-strain curve non-linearLarge change of geometry
Equilibrium onOriginal shapeDeformed shape
ExampleYielding steelCable, slender column
  • 2070 Bhadra · 3 marks

Explain briefly the non-linearity in structural analysis.

Answer

Non-linearity in structural analysis means that the response (displacement, strain) is not directly proportional to the applied load, so the stiffness of the structure changes during loading and superposition cannot be used.

Sources

  1. Material non-linearity: non-linear stress-strain behaviour such as yielding of steel, cracking and crushing of concrete.
  2. Geometric non-linearity: large displacements that change the geometry, e.g. cables, slender columns.
  3. Boundary (contact) non-linearity: supports that change with load, such as a gap closing or a support lifting off.

Consequences

  • The force-displacement curve is curved, not straight.
  • Principle of superposition is not valid; results depend on load history.
  • Equations are non-linear and are solved by incremental-iterative methods (e.g. Newton-Raphson), usually by computer.

Linear analysis is adequate for service loads. Non-linear analysis is needed to find the true ultimate load, collapse behaviour and the performance of slender or cable structures.

  • 2070 Magh · 4+2 marks

Explain with an example how you would use the method of superposition in determining deflections. Also explain why it is necessary to determine deflections in the design of a structure.

Answer

Use of superposition to find deflection

For a linear elastic beam, the deflection under several loads is the sum of the deflections due to each load acting alone. The loading is split into simple cases for which standard deflection formulas are known, and the results are added algebraically.

Example: A cantilever of length LL carries a point load PP at the free end and a UDL ww over the whole length. Find the free end deflection.

  1. Case 1, point load alone: δ1=PL33EI\delta_1 = \dfrac{PL^3}{3EI}
  2. Case 2, UDL alone: δ2=wL48EI\delta_2 = \dfrac{wL^4}{8EI}
  3. Total deflection:
δ=δ1+δ2=PL33EI+wL48EI\delta = \delta_1 + \delta_2 = \frac{PL^3}{3EI} + \frac{wL^4}{8EI}

For example, with P=10P = 10 kN, w=5w = 5 kN/m, L=4L = 4 m, EI=20000EI = 20000 kNm2^2: δ=10×643×20000+5×2568×20000=0.01067+0.008=0.0187\delta = \dfrac{10\times 64}{3\times 20000} + \dfrac{5\times 256}{8\times 20000} = 0.01067 + 0.008 = 0.0187 m (18.7 mm).

This works only if the material is linearly elastic and deflections are small.

Why deflection must be determined in design

  • Serviceability: Excessive deflection causes cracking of plaster and partitions, damage to finishes and fixtures, and discomfort to users. Codes limit it, e.g. L/250L/250 for total deflection and L/350L/350 for live load.
  • Appearance and function: Sagging beams look unsafe and may pond water on roofs or disturb machinery alignment.
  • Indeterminate analysis: Deflections (compatibility) are needed to find the redundant forces.
  • Vibration control: Stiffness (related to deflection) controls vibration and natural frequency.
  • Check of analysis: Computed deflection can be compared with field measurement.

Questions from Old Question Collection (CE 551) (IOE exam papers from 2065 to 2081 (24 papers)). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗