Virtual work 0910
Upcoming SlideShare
Loading in...5
×
 

Virtual work 0910

on

  • 6,066 views

 

Statistics

Views

Total Views
6,066
Views on SlideShare
6,066
Embed Views
0

Actions

Likes
0
Downloads
378
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Virtual work 0910 Virtual work 0910 Document Transcript

  • Structural Analysis III Virtual Work 3rd Year Structural Engineering 2009/10 Dr. Colin Caprani 1 Dr. C. Caprani
  • Structural Analysis IIIContents1. Introduction ......................................................................................................... 4 1.1 General............................................................................................................. 4 1.2 Background...................................................................................................... 52. The Principle of Virtual Work......................................................................... 14 2.1 Definition....................................................................................................... 14 2.2 Virtual Displacements ................................................................................... 16 2.3 Virtual Forces ................................................................................................ 17 2.4 Simple Proof using Virtual Displacements ................................................... 18 2.5 Internal and External Virtual Work............................................................... 193. Application of Virtual Displacements ............................................................. 21 3.1 Rigid Bodies .................................................................................................. 21 3.2 Deformable Bodies ........................................................................................ 28 3.3 Problems ........................................................................................................ 394. Application of Virtual Forces........................................................................... 41 4.1 Basis............................................................................................................... 41 4.2 Deflection of Trusses..................................................................................... 42 4.3 Deflection of Beams and Frames .................................................................. 49 4.4 Integration of Bending Moments................................................................... 55 4.5 Problems ........................................................................................................ 585. Virtual Work for Indeterminate Structures................................................... 66 5.1 General Approach.......................................................................................... 66 5.2 Using Virtual Work to Find the Multiplier ................................................... 68 5.3 Indeterminate Trusses.................................................................................... 71 5.4 Indeterminate Frames .................................................................................... 75 5.5 Continuous Beams ......................................................................................... 82 5.6 Problems ........................................................................................................ 916. Virtual Work for Self-Stressed Structures ..................................................... 94 2 Dr. C. Caprani
  • Structural Analysis III 6.1 Background.................................................................................................... 94 6.2 Trusses ......................................................................................................... 100 6.3 Beams .......................................................................................................... 106 6.4 Frames.......................................................................................................... 109 6.5 Problems ...................................................................................................... 1147. Past Exam Questions....................................................................................... 116 7.1 Summer 1997............................................................................................... 116 7.2 Summer 1998............................................................................................... 117 7.3 Summer 1999............................................................................................... 118 7.4 Summer 2000............................................................................................... 119 7.5 Summer 2001............................................................................................... 120 7.6 Summer 2002............................................................................................... 121 7.7 Summer 2004............................................................................................... 122 7.8 Summer 2007............................................................................................... 123 7.9 Semester 2, 2008.......................................................................................... 124 7.10 Semester 2, 2009 ...................................................................................... 125 7.11 Semester 2, 2009 ...................................................................................... 1268. References ........................................................................................................ 1279. Appendix – Volume Integrals......................................................................... 128 3 Dr. C. Caprani
  • Structural Analysis III1. Introduction1.1 GeneralVirtual Work is a fundamental theory in the mechanics of bodies. So fundamental infact, that Newton’s 3 equations of equilibrium can be derived from it. Virtual Workprovides a basis upon which vectorial mechanics (i.e. Newton’s laws) can be linkedto the energy methods (i.e. Lagrangian methods) which are the basis for finiteelement analysis and advanced mechanics of materials.Virtual Work allows us to solve determinate and indeterminate structures and tocalculate their deflections. That is, it can achieve everything that all the othermethods together can achieve.Before starting into Virtual Work there are some background concepts and theoriesthat need to be covered. 4 Dr. C. Caprani
  • Structural Analysis III1.2 BackgroundStrain Energy and Work DoneStrain energy is the amount of energy stored in a structural member due todeformation caused by an external load. For example, consider this simple spring:We can see that as it is loaded by a gradually increasing force, F, it elongates. We cangraph this as:The line OA does not have to be straight, that is, the constitutive law of the spring’smaterial does not have to be linear. 5 Dr. C. Caprani
  • Structural Analysis IIIAn increase in the force of a small amount, δ F results in a small increase indeflection, δ y . The work done during this movement (force × displacement) is theaverage force during the course of the movement, times the displacement undergone.This is the same as the hatched trapezoidal area above. Thus, the increase in workassociated with this movement is: F + (F +δ F ) δU = ⋅δ y 2 δ F ⋅δ y = F ⋅δ y + (1.1) 2 ≈ F ⋅δ ywhere we can neglect second-order quantities. As δ y → 0 , we get: dU = F ⋅ dyThe total work done when a load is gradually applied from 0 up to a force of F is thesummation of all such small increases in work, i.e.: y U = ∫ F dy (1.2) 0This is the dotted area underneath the load-deflection curve of earlier and representsthe work done during the elongation of the spring. This work (or energy, as they arethe same thing) is stored in the spring and is called strain energy and denoted U.If the load-displacement curve is that of a linearly-elastic material then F = ky wherek is the constant of proportionality (or the spring stiffness). In this case, the dottedarea under the load-deflection curve is a triangle. 6 Dr. C. Caprani
  • Structural Analysis IIIDraw this: 7 Dr. C. Caprani
  • Structural Analysis IIIAs we know that the work done is the area under this curve, then the work done bythe load in moving through the displacement – the External Work Done, We - is givenby: 1 We = Fy (1.3) 2We can also calculate the strain energy, or Internal Work Done, WI , by: y U = ∫ F dy 0 y = ∫ ky dy 0 1 = ky 2 2Also, since F = ky , we then have: 1 1 WI = U = ( ky ) y = Fy 2 2But this is the external work done, We . Hence we have: We = WI (1.4)Which we may have expected from the Law of Conservation of Energy. Thus:The external work done by external forces moving through external displacements isequal to the strain energy stored in the material. 8 Dr. C. Caprani
  • Structural Analysis IIILaw of Conservation of EnergyFor structural analysis this can be stated as: Consider a structural system that is isolated such it neither gives nor receives energy; the total energy of this system remains constant.The isolation of the structure is key: we can consider a structure isolated once wehave identified and accounted for all sources of restraint and loading. For example, toneglect the self-weight of a beam is problematic as the beam receives gravitationalenergy not accounted for, possibly leading to collapse.In the spring and force example, we have accounted for all restraints and loading (forexample we have ignored gravity by having no mass). Thus the total potential energyof the system, Π , is constant both before and after the deformation.In structural analysis the relevant forms of energy are the potential energy of the loadand the strain energy of the material. We usually ignore heat and other energies.Potential Energy of the LoadSince after the deformation the spring has gained strain energy, the load must havelost potential energy, V. Hence, after deformation we have for the total potentialenergy: Π =U +V 1 (1.5) = ky 2 − Fy 2In which the negative sign indicates a loss of potential energy for the load. 9 Dr. C. Caprani
  • Structural Analysis IIIPrinciple of Minimum Total Potential EnergyIf we plot the total potential energy against y, equation (1.5), we get a quadratic curvesimilar to:Consider a numerical example, with the following parameters, k = 10 kN/m andF = 10 kN giving the equilibrium deflection as y = F k = 1 m . We can plot thefollowing quantities: 1 1 • Internal Strain Energy, or Internal Work: U = WI = ky 2 = 10 y 2 = 5 y 2 2 2 • Potential Energy: V = − Fy = −10 y • Total Potential Energy: Π = U + V = 5 y 2 − 10 y 1 1 • External Work: We = Py = 10 y = 5 y 2 2and we get the following plots (split into two for clarity): 10 Dr. C. Caprani
  • Structural Analysis III Total Potential Energy 35 External Work 30 Internal Work Energy 25 Equilibrium 20 15 10 5 0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 -5 -10 Deflection, y Total Potential Energy 40 Strain Energy 30 Potential Energy Equilibrium Energy 20 10 0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 -10 -20 -30 -40 Deflection, yFrom these graphs we can see that because WI increases quadratically with y, whilethe We increases only linearly, WI always catches up with We , and there will alwaysbe a non-zero equilibrium point where We = WI . 11 Dr. C. Caprani
  • Structural Analysis IIIAdmittedly, these plots are mathematical: the deflection of the spring will not take upany value; it takes that value which achieves equilibrium. At this point we consider asmall variation in the total potential energy of the system. Considering F and k to beconstant, we can only alter y. The effect of this small variation in y is: 1 1 Π ( y + δ y ) − Π ( y ) = k ( y + δ y ) − F ( y + δ y ) − ky 2 + Fy 2 2 2 1 1 = k ( 2 y ⋅ δ y ) − F ⋅ δ y + k (δ y ) 2 (1.6) 2 2 1 = ( ky − F ) δ y + k (δ y ) 2 2Similarly to a first derivate, for Π to be an extreme (either maximum or minimum),the first variation must vanish: δ (1)Π = ( ky − F ) δ y = 0 (1.7)Therefore: ky − F = 0 (1.8)Which we recognize to be the ∑F x = 0 . Thus equilibrium occurs when Π is anextreme.Before introducing more complicating maths, an example of the above variation inequilibrium position is the following. Think of a shopkeeper testing an old type ofscales for balance – she slightly lifts one side, and if it returns to position, and nolarge rotations occur, she concludes the scales is in balance. She has imposed a 12 Dr. C. Caprani
  • Structural Analysis IIIvariation in displacement, and finds that since no further displacement occurs, the‘structure’ was originally in equilibrium.Examining the second variation (similar to a second derivate): 1 δ ( 2 )Π = k (δ y ) ≥ 0 2 (1.9) 2We can see it is always positive and hence the extreme found was a minimum. This isa particular proof of the general principle that structures take up deformations thatminimize the total potential energy to achieve equilibrium. In short, nature is lazy!To summarize our findings: • Every isolated structure has a total potential energy; • Equilibrium occurs when structures can minimise this energy; • A small variation of the total potential energy vanishes when the structure is in equilibrium.These concepts are brought together in the Principle of Virtual Work. 13 Dr. C. Caprani
  • Structural Analysis III2. The Principle of Virtual Work2.1 DefinitionBased upon the Principle of Minimum Total Potential Energy, we can see that anysmall variation about equilibrium must do no work. Thus, the Principle of VirtualWork states that: A body is in equilibrium if, and only if, the virtual work of all forces acting on the body is zero.In this context, the word ‘virtual’ means ‘having the effect of, but not the actual formof, what is specified’. Thus we can imagine ways in which to impose virtual work,without worrying about how it might be achieved in the physical world.Virtual WorkThere are two ways to define virtual work, as follows. 1. Principle of Virtual Displacements: Virtual work is the work done by the actual forces acting on the body moving through a virtual displacement. This means we solve an equilibrium problem through geometry. 2. Principle of Virtual Forces: Virtual work is the work done by a virtual force acting on the body moving through the actual displacements. This means we solve a geometry problem through equilibrium. 14 Dr. C. Caprani
  • Structural Analysis IIIIllustrate the Principle of Virtual Displacements:Illustrate the Principle of Virtual Forces: 15 Dr. C. Caprani
  • Structural Analysis III2.2 Virtual DisplacementsA virtual displacement is a displacement that is only imagined to occur. This isexactly what we did when we considered the vanishing of the first variation of Π ; wefound equilibrium. Thus the application of a virtual displacement is a means to findthis first variation of Π .So given any real force, F, acting on a body to which we apply a virtualdisplacement. If the virtual displacement at the location of and in the direction of F isδ y , then the force F does virtual work δ W = F ⋅ δ y .There are requirements on what is permissible as a virtual displacement. Forexample, in the simple proof of the Principle of Virtual Work (to follow) it can beseen that it is assumed that the directions of the forces applied to P remainunchanged. Thus: • virtual displacements must be small enough such that the force directions are maintained.The other very important requirement is that of compatibility: • virtual displacements within a body must be geometrically compatible with the original structure. That is, geometrical constraints (i.e. supports) and member continuity must be maintained.In summary, virtual displacements are not real, they can be physically impossible butthey must be compatible with the geometry of the original structure and they must besmall enough so that the original geometry is not significantly altered.As the deflections usually encountered in structures do not change the overallgeometry of the structure, this requirement does not cause problems. 16 Dr. C. Caprani
  • Structural Analysis III2.3 Virtual ForcesSo far we have only considered small virtual displacements and real forces. Thevirtual displacements are arbitrary: they have no relation to the forces in the system,or its actual deformations. Therefore virtual work applies to any set of forces inequilibrium and to any set of compatible displacements and we are not restricted toconsidering only real force systems and virtual displacements. Hence, we can use avirtual force system and real displacements. Obviously, in structural design it is thesereal displacements that are of interest and so virtual forces are used often.A virtual force is a force imagined to be applied and is then moved through the actualdeformations of the body, thus causing virtual work.So if at a particular location of a structure, we have a deflection, y, and impose avirtual force at the same location and in the same direction of δ F we then have thevirtual work δ W = y ⋅ δ F .Virtual forces must form an equilibrium set of their own. For example, if a virtualforce is applied to the end of a spring there will be virtual stresses in the spring aswell as a virtual reaction. 17 Dr. C. Caprani
  • Structural Analysis III2.4 Simple Proof using Virtual DisplacementsWe can prove the Principle of Virtual Work quite simply, as follows. Consider aparticle P under the influence of a number of forces F1 ,K, Fn which have a resultantforce, FR . Apply a virtual displacement of δ y to P, moving it to P’, as shown:The virtual work done by each of the forces is: δ W = F1 ⋅ δ y1 + K + Fn ⋅ δ yn = FR ⋅ δ yRWhere δ y1 is the virtual displacement along the line of action of F1 and so on. Nowif the particle P is in equilibrium, then the forces F1 ,K, Fn have no resultant. That is,there is no net force. Hence we have: δ W = 0 ⋅ δ yR = F1 ⋅ δ y1 + K + Fn ⋅ δ yn = 0 (2.1)Proving that when a particle is in equilibrium the virtual work of all the forces actingon it sum to zero. Conversely, a particle is only in equilibrium if the virtual workdone during a virtual displacement is zero. 18 Dr. C. Caprani
  • Structural Analysis III2.5 Internal and External Virtual WorkConsider the spring we started with, as shown again below. Firstly it is unloaded.Then a load, F, is applied, causing a deflection y. It is now in its equilibrium positionand we apply a virtual displacement, δ y to the end of the spring, as shown:A free body diagram of the end of the spring is:Thus the virtual work done by the two forces acting on the end of the spring is: δ W = F ⋅ δ y − ky ⋅ δ yIf the spring is to be in equilibrium we must then have: δW = 0 F ⋅ δ y − ky ⋅ δ y = 0 F ⋅ δ y = ky ⋅ δ y F = ky 19 Dr. C. Caprani
  • Structural Analysis IIIThat is, the force in the spring must equal the applied force, as we already know.However, if we define the following: • External virtual work, δ WE = F ⋅ δ y ; • Internal virtual work, δ WI = ky ⋅ δ y ;We then have: δW = 0 δ WI − δ WE = 0Thus: δ WE = δ WI (2.2)which states that the external virtual work must equal the internal virtual work for astructure to be in equilibrium.It is in this form that the Principle of Virtual Work finds most use.Of significance also in the equating of internal and external virtual work, is that thereare no requirements for the material to have any particular behaviour. That is, virtualwork applies to all bodies, whether linearly-elastic, elastic, elasto-plastic, plastic etc.Thus the principle has more general application than most other methods of analysis.Internal and external virtual work can arise from either virtual displacements orvirtual forces. 20 Dr. C. Caprani
  • Structural Analysis III3. Application of Virtual Displacements3.1 Rigid BodiesBasisRigid bodies do not deform and so there is no internal virtual work done. Thus: δW = 0 δ WE = δ WI (3.1) ∑ F ⋅δ y i i =0A simple application is to find the reactions of statically determinate structures.However, to do so, we need to make use of the following principle:Principle of Substitution of ConstraintsHaving to keep the constraints in place is a limitation of virtual work. However, wecan substitute the restraining force (i.e. the reaction) in place of the restraint itself.That is, we are turning a geometric constraint into a force constraint. This is thePrinciple of Substitution of Constraints. We can use this principle to calculateunknown reactions: 1. Replace the reaction with its appropriate force in the same direction (or sense); 2. Impose a virtual displacement on the structure; 3. Calculate the reaction, knowing δ W = 0 . 21 Dr. C. Caprani
  • Structural Analysis IIIReactions of Determinate and Indeterminate StructuresFor statically determinate structures, removing a restraint will always render amechanism (or rigid body) and so the reactions of statically determinate structures areeasily obtained using virtual work. For indeterminate structures, removing a restraintdoes not leave a mechanism and hence the virtual displacements are harder toestablish since the body is not rigid. 22 Dr. C. Caprani
  • Structural Analysis IIIExample 1Determine the reactions for the following beam:Following the Principle of Substitution of Constraints, we replace the geometricconstraints (i.e. supports), by their force counterparts (i.e. reactions) to get the free-body-diagram of the whole beam:Next, we impose a virtual displacement on the beam. Note that the displacement iscompletely arbitrary, and the beam remains rigid throughout: 23 Dr. C. Caprani
  • Structural Analysis IIIIn the above figure, we have imposed a virtual displacement of δ y A at A and thenimposed a virtual rotation of δθ A about A. The equation of virtual work is: δW = 0 δ WE = δ WI ∑ F ⋅δ y i i =0Hence: VA ⋅ δ y A − P ⋅ δ yC + VB ⋅ δ yB = 0Relating the virtual movements of points B and C to the virtual displacements gives: VA ⋅ δ y A − P (δ y A + a ⋅ δθ A ) + VB (δ y A + L ⋅ δθ A ) = 0And rearranging gives: (VA + VB − P )δ y A + (VB L − Pa ) δθ A = 0 24 Dr. C. Caprani
  • Structural Analysis IIIAnd here is the power of virtual work: since we are free to choose any value we wantfor the virtual displacements (i.e. they are completely arbitrary), we can chooseδθ A = 0 and δ y A = 0 , which gives the following two equations: (VA + VB − P ) δ y A = 0 (VB L − Pa )δθ A = 0 VA + VB − P = 0 VB L − Pa = 0But the first equation is just ∑F y = 0 whilst the second is the same as∑ M about A = 0 . Thus equilibrium equations occur naturally within the virtualwork framework. Note also that the two choices made for the virtual displacementscorrespond to the following virtual displaced configurations: Draw them 25 Dr. C. Caprani
  • Structural Analysis IIIExample 2For the following truss, calculate the reaction VC :Firstly, set up the free-body-diagram of the whole truss:Next we release the constraint corresponding to reaction VC and replace it by theunknown force VC and we apply a virtual displacement to the truss to get: 26 Dr. C. Caprani
  • Structural Analysis IIIHence the virtual work done is: δW = 0 δ WE = δ WI δy −10 ⋅ + VC ⋅ δ y = 0 2 VC = 5 kNNote that the reaction is an external force to the structure, and that no internal virtualwork is done since the members do not undergo virtual deformation. The truss rotatesas a rigid body about the support A.Problem:Find the horizontal reactions of the truss. 27 Dr. C. Caprani
  • Structural Analysis III3.2 Deformable BodiesBasisFor a virtual displacement we have: δW = 0 δ WE = δ WI ∑ F ⋅δ y = ∫ P ⋅δ e i i i iIn which, for the external virtual work, Fi represents an externally applied force (ormoment) and δ yi its virtual displacement. And for the internal virtual work, Pirepresents the internal force in member i and δ ei its virtual deformation. Differentstress resultants have different forms of internal work, and we will examine these.Lastly, the summations reflect the fact that all work done must be accounted for.Remember in the above, each the displacements must be compatible and the forcesmust be in equilibrium, summarized as: Set of forces in equilibrium ∑ F ⋅δ y = ∑ P ⋅δ e i i i i Set of compatible displacements 28 Dr. C. Caprani
  • Structural Analysis IIIThese displacements are completely arbitrary (i.e. we can choose them to suit ourpurpose) and bear no relation to the forces above. For example, the simple trussshown has a set of forces in equilibrium along with its actual deformed shape. Alsoshown is an arbitrary set of permissible compatible displacements: Loading and dimensions Equilibrium forces and actual deformation Compatible set of displacements 29 Dr. C. Caprani
  • Structural Analysis IIIInternal Virtual Work by Axial ForceMembers subject to axial force may have the following: • real force by a virtual displacement: δ WI = P ⋅ δ e • virtual force by a real displacement: δ WI = e ⋅ δ PWe have avoided integrals over the length of the member since we will only considerprismatic members. 30 Dr. C. Caprani
  • Structural Analysis IIIInternal Virtual Work in BendingThe internal virtual work done in bending is one of: • real moment by a virtual curvature: δ WI = M ⋅ δκ • virtual moment by a real curvature: δ WI = κ ⋅ δ MThe above expressions are valid at a single position in a beam.When virtual rotations are required along the length of the beam, the easiest way todo this is by applying virtual loads that in turn cause virtual moments and hencevirtual curvatures. We must sum all of the real moments by virtual curvatures alongthe length of the beam, hence we have: L L δ WI = ∫ M x ⋅ δκ x dx = ∫ κ x ⋅ δ M x dx 0 0 L δMx = ∫Mx ⋅ dx 0 EI x 31 Dr. C. Caprani
  • Structural Analysis IIIInternal Virtual Work in ShearAt a single point in a beam, the shear strain, γ , is given by: V γ= GAvwhere V is the applied shear force, G is the shear modulus and Av is the cross-sectionarea effective in shear which is explained below. The internal virtual work done inshear is thus: • real shear force by a virtual shear strain: δV δ WI = V ⋅ δγ = V ⋅ GAv • virtual shear force by a real shear strain: V δ WI = γ ⋅ δ V = ⋅ δV GAvIn these expressions, the area of cross section effective in shear comes about becausethe shear stress (and hence strain) is not constant over a cross section. The stressV Av is an average stress, equivalent in work terms to the actual uneven stress overthe full cross section, A. We say Av = A k where k is the shear factor for the crosssection. Some values of k are 1.2 for rectangular sections, 1.1 for circular sections,and 2.0 for thin-walled circular sections.The above expressions are valid at a single position in a beam and must be integratedalong the length of the member as was done for moments and rotations. 32 Dr. C. Caprani
  • Structural Analysis IIIInternal Virtual Work in TorsionAt a single point in a member, the twist, φ , is given by: T φ= GJwhere T is the applied torque, G is the shear modulus, J is the polar moment ofinertia. The internal virtual work done in torsion is thus: • real torque by a virtual twist: δT δ WI = T ⋅ δφ = T ⋅ GJ • virtual torque by a real twist: T δ WI = δ T ⋅ φ = δ T ⋅ GJOnce again, the above expressions are valid at a single position in a beam and mustbe integrated along the length of the member as was done for moments and rotations.Note the similarity between the expressions for the four internal virtual works. 33 Dr. C. Caprani
  • Structural Analysis IIIExample 3For the beam of Example 1 (shown again), find the bending moment at C.To solve this, we want to impose a virtual displacement configuration that onlyallows the unknown of interest, i.e. M C , to do any work. Thus choose the following:Since portions AC and CB remain straight (or unbent) no internal virtual work is donein these sections. Thus the only internal work is done at C by the beam movingthrough the rotation at C. Thus: 34 Dr. C. Caprani
  • Structural Analysis III δW = 0 δ WE = δ WI P ⋅ δ yC = M C ⋅ δθCBu the rotation at C is made up as: δθC = δθCA + δθCB δ yC δ yC = + a b ⎛a+b⎞ =⎜ ⎟ δ yC ⎝ ab ⎠But a + b = L , hence: ⎛ L⎞ P ⋅ δ yC = M C ⋅ ⎜ ⎟ δ yC ⎝ ab ⎠ Pab MC = LWe can verify this from the reactions found previously: M C = VB ⋅ b = ( Pa L ) b . 35 Dr. C. Caprani
  • Structural Analysis IIISign Convention for RotationsWhen imposing a virtual rotation, if the side that is already in tension due to the realforce system elongates, then positive virtual work is done.Sketch this: 36 Dr. C. Caprani
  • Structural Analysis IIIExample 4Calculate the force F1 in the truss shown:To do this, we introduce a virtual displacement along the length of member 1. We dothis so that member 2 does not change length during this virtual displacement and sodoes no virtual work. Note also that compatibility of the truss is maintained. Forexample, the members still meet at the same joint, and the support conditions are met. 37 Dr. C. Caprani
  • Structural Analysis IIIThe virtual work done is then: δW = 0 δ WE = δ WI δy −10 ⋅ = − F1 ⋅ δ y 2 10 F1 = = 5 2 kN 2Note some points on the signs used: 1. Negative external work is done because the 10 kN load moves upwards; i.e. the reverse direction to its action. 2. We assumed member 1 to be in compression but then applied a virtual elongation to the member thus reducing its internal virtual work. Hence negative internal work is done. 3. We initially assumed F1 to be in compression and we obtained a positive answer confirming our assumption.Problem:Investigate the vertical and horizontal equilibrium of the loaded joint by consideringvertical and horizontal virtual displacements separately. 38 Dr. C. Caprani
  • Structural Analysis III3.3 Problems 1. For the truss shown, calculate the vertical reaction at C and the forces in the members, using virtual work. 2. For the truss shown, find the forces in the members, using virtual work: 39 Dr. C. Caprani
  • Structural Analysis III 3. Using virtual work, calculate the reactions for the beams shown, and the bending moments at salient points. Beam 1 Beam 2 Beam 3 40 Dr. C. Caprani
  • Structural Analysis III4. Application of Virtual Forces4.1 BasisWhen virtual forces are applied, we have: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i iAnd again note that we have an equilibrium set of forces and a compatible set ofdisplacements: Set of compatible displacements ∑ y ⋅δ F = ∑e ⋅δ P i i i i Set of forces in equilibriumIn this case the displacements are the real displacements that occur when the structureis in equilibrium and the virtual forces are any set of arbitrary forces that are inequilibrium. 41 Dr. C. Caprani
  • Structural Analysis III4.2 Deflection of TrussesIllustrative ExampleFor the truss shown below, we have the actual displacements shown with twopossible sets of virtual forces. Actual Loading and (Compatible) Displacements 42 Dr. C. Caprani
  • Structural Analysis III Virtual Force (Equilibrium) SystemsIn this truss, we know we know: 1. The forces in the members (got from virtual displacements or statics); 2. Thus we can calculate the member extensions, ei as: ⎛ PL ⎞ ei = ⎜ ⎟ ⎝ EA ⎠i 3. Also, as we can choose what our virtual force δ F is (usually unity), we have: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ PL ⎞ y ⋅δ F = ∑⎜ ⎟ ⋅ δ Pi ⎝ EA ⎠i 4. Since in this equation, y is the only unknown, we can calculate the deflection of the truss. 43 Dr. C. Caprani
  • Structural Analysis IIIExample 5Given that E = 200 kN/mm2 and A = 100 mm2 for each member, for the truss shownbelow, calculate the vertical and horizontal deflection of joint B.In these problems we will always choose δ F = 1 . Hence we apply a unit virtual forceto joint B. We apply the force in the direction of the deflection required. In this wayno work is done along deflections that are not required. Hence we have: 44 Dr. C. Caprani
  • Structural Analysis IIIThe forces and elongations of the truss members are: • Member AB: PAB = +50 kN ⎛ +50 ⋅ 5000 ⎞ eAB = ⎜ ⎟ ⎝ 200 ⋅ 100 ⎠ = +12.5 mm • Member BC: PBC = −30 kN ⎛ −30 ⋅ 3000 ⎞ eBC = ⎜ ⎟ ⎝ 200 ⋅ 100 ⎠ = −4.5 mmNote that by taking tension to be positive, elongations are positive and contractionsare negative.Horizontal Deflection: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i y ⋅ 1 = (12.5 ⋅ 0 ) AB + ( −4.5 ⋅ +1) BC y = −4.5 mmVertical Deflection: ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ 5⎞ ⎛ 3⎞ y ⋅ 1 = ⎜12.5 ⋅ ⎟ + ⎜ −4.5 ⋅ − ⎟ ⎝ 4 ⎠ AB ⎝ 4 ⎠ BC y = +18.4 mmNote that the sign of the result indicates whether the deflection occurs in the samedirection as the applied force. Hence, joint B moves 4.5 mm to the left. 45 Dr. C. Caprani
  • Structural Analysis IIIExample 6Determine the vertical and deflection of joint D of the truss shown. Take E = 200kN/mm2 and member areas, A = 1000 mm2 for all members except AE and BD whereA = 1000√2 mm2.The elements of the virtual work equation are: • Compatible deformations: The actual displacements that the truss undergoes; • Equilibrium set: The external virtual force applied at the location of the required deflection and the resulting internal member virtual forces.Firstly we analyse the truss to determine the member forces in order to calculate theactual deformations of each member: 46 Dr. C. Caprani
  • Structural Analysis IIINext, we apply a unit virtual force in the vertical direction at joint D. However, bylinear superposition, we know that the internal forces due to a unit load will be 1/150times those of the 150 kN load.For the horizontal deflection at D, we apply a unit horizontal virtual force as shown:Equations of Virtual Work δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ P0 L ⎞ yDV ⋅ 1 = ∑ ⎜ ⎟ ⋅ δ Pi 1 ⎝ EA ⎠i ⎛ P0 L ⎞ yDH ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi 2 ⎝ EA ⎠iIn which: • P 0 are the forces due to the 150 kN load; • δ P1 are the virtual forces due to the unit virtual force applied in the vertical direction at D: P0 δ P1 = 150 • δ P 2 are the virtual forces due to the unit virtual force in the horizontal direction at D. 47 Dr. C. Caprani
  • Structural Analysis IIIUsing a table is easiest because of the larger number of members: P0 ⎛ P0 L ⎞ ⎛ P0 L ⎞ ⎟ ⋅δ P ⎟ ⋅δ P 1 2 L A P 0 δP = 1 δ P2 ⎜ ⎜ Member 150 ⎝ A ⎠ ⎝ A ⎠ (mm) (mm2) (kN) (kN) (kN) (kN/mm)×kN (kN/mm)×kN AB 2000 1000 +150 +1 0 +300 0 AE 2000√2 1000√2 +150√2 +1√2 0 +600 0 AF 2000 1000 0 0 0 0 0 BC 2000 1000 0 0 0 0 0 BD 2000√2 1000√2 +150√2 +1√2 0 +600 0 BE 2000 1000 -150 -1 0 +300 0 CD 2000 1000 0 0 0 0 0 DE 2000 1000 -150 -1 +1 +300 -300 EF 2000 1000 -300 -2 +1 +1200 -600 ∑= 3300 -900E is left out because it is common. Returning to the equations, we now have: 1 ⎛ P0 L ⎞ yDV ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi 1 E ⎝ A ⎠i +3300 yDV = = +16.5 mm 200Which indicates a downwards deflection and for the horizontal deflection: 1 ⎛ P0 L ⎞ yDH ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi 2 E ⎝ A ⎠i −900 yDH = = −4.5 mm 200The sign indicates that it is deflecting to the left. 48 Dr. C. Caprani
  • Structural Analysis III4.3 Deflection of Beams and FramesExample 7Using virtual work, calculate the deflection at the centre of the beam shown, giventhat EI is constant.To calculate the deflection at C, we will be using virtual forces. Therefore the tworelevant sets are: • Compatibility set: the actual deflection at C and the rotations that occur along the length of the beam; • Equilibrium set: a unit virtual force applied at C which is in equilibrium with the internal virtual moments it causes.Compatibility Set:The external deflection at C is what is of interest to us. To calculate the rotationsalong the length of the beam we have: L Mx κ =∫ dx 0 EI xHence we need to establish the bending moments along the beam: 49 Dr. C. Caprani
  • Structural Analysis IIIFor AC the bending moment is given by (and similarly for B to C): P Mx = x 2Equilibrium Set:As we choose the value for δ F = 1 , we are only left to calculate the virtual moments:For AC the internal virtual moments are given by: 1 δMx = x 2 50 Dr. C. Caprani
  • Structural Analysis IIIVirtual Work Equation δW = 0 δ WE = δ WI ∑ y ⋅ δ F = ∑κ i i i ⋅δ MiSubstitute in the values we have for the real rotations and the virtual moments, anduse the fact that the bending moment diagrams are symmetrical: L 2 ⎡M ⎤ y ⋅ 1 = 2 ∫ ⎢ x ⎥ ⋅ δ M x dx 0 ⎣ EI ⎦ L 2 2 ⎛P ⎞ ⎛1 ⎞ y= EI ∫ ⎜ 2 x ⎟ ⋅ ⎜ 2 x ⎟ dx 0 ⎝ ⎠ ⎝ ⎠ L2 2P = ∫x 2 dx 4 EI 0 L 2 P ⎡ x3 ⎤ = 2 EI ⎢ 3 ⎥ 0 ⎣ ⎦ P L3 = ⋅ 6 EI 8 PL3 = 48 EIWhich is a result we expected. 51 Dr. C. Caprani
  • Structural Analysis IIIExample 8Find the vertical and horizontal displacement of the prismatic curved bar cantilevershown:Even though it is curved, from the statics, the bending moment at any point is stillforce by distance. Hence, at any angle φ , we have: M φ = P ( R − R cos φ ) = PR (1 − cos φ )To find the displacements, we follow our usual procedure and place a unit load at thelocation of, and in the direction of, the required displacement. 52 Dr. C. Caprani
  • Structural Analysis IIIVertical DisplacementThe virtual bending moment induced by the vertical unit load shown, is related to thatfor P and is thus: δ M φ = R (1 − cos φ )Thus our virtual work equations are: δW = 0 δ WE = δ WI 1 ⋅ δ BV = ∫ κ ⋅ δ M ⋅ ds π 2 Mφ = ∫0 EI ⋅ δ M φ ⋅ RdφIn which we have used the relation ds = R dφ to change the integration from alongthe length of the member to around the angle φ . Next we introduce our equations forthe real and virtual bending moments: 53 Dr. C. Caprani
  • Structural Analysis III π 2 PR (1 − cos φ ) 1 ⋅ δ BV = ∫ ⋅ R (1 − cos φ ) ⋅ Rdφ 0 EI π 2 PR 3 ∫ (1 − cosφ ) 2 = ⋅ dφ EI 0 3 PR PR 3 = ( 3π − 4 ) ≈ 5.42 EI EIHorizontal DisplacementIn this case, the virtual bending moment is: δ M φ = R sin φThus the virtual work equations give: π 2 PR (1 − cos φ ) 1 ⋅ δ BH = ∫0 EI ⋅ R sin φ ⋅ Rdφ π 2 PR 3 = EI ∫ ( sin φ − sin φ cosφ ) ⋅ dφ 0 3π 2 PR ⎛ sin 2φ ⎞ = EI ∫ ⎜ sin φ − 0⎝ 2 ⎠ ⎟ ⋅ dφ PR ⎛ 1 ⎞ PR 3 3 = ⎜ ⎟= EI ⎝ 2 ⎠ 2 EI 54 Dr. C. Caprani
  • Structural Analysis III4.4 Integration of Bending MomentsWe are often faced with the integration of being moment diagrams when using virtualwork to calculate the deflections of bending members. And as bending momentdiagrams only have a limited number of shapes, a table of ‘volume’ integrals is used:This table is at the back page of these notes for ease of reference.Example 9Using the table of volume integrals, verify the answer to Example 7.In this case, the virtual work equation becomes: L 2 ⎡M ⎤ y ⋅ 1 = 2 ∫ ⎢ x ⎥ ⋅ δ M x dx 0 ⎣ EI ⎦ 2 y= [ M x shape] × [δ M x shape] EI 2 ⎡ 1 PL L L ⎤ = ⋅ ⋅ ⋅ EI ⎢ 3 4 4 2 ⎥ ⎣ ⎦ 3 PL = 48 EI 1In which the formula jkl is used from the table. 3 55 Dr. C. Caprani
  • Structural Analysis IIIExample 10 – Summer ’07 Part (a)For the frame shown, determine the horizontal deflection of joint C. Neglect axialeffects in the members and take EI = 36 × 103 kNm 2 . 100 kN B C 20 kN/m 4m A D 4m 4mFirstly we establish the real bending moment diagram:Next, as usual, we place a unit load at the location of, and in the direction of, therequired displacement: 56 Dr. C. Caprani
  • Structural Analysis IIINow we have the following for the virtual work equation: δW = 0 δ WE = δ WI 1 ⋅ δ CH = ∫ θ ⋅ δ M ⋅ ds M =∫ ⋅ δ M ⋅ ds EINext, using the table of volume integrals, we have: 1 ⎧⎡ 1 ⎤ ⎫ M ∫ EI ⋅ δ M ⋅ ds = (⎤ ⎡1 ) ⎨ ⎢ ( 440 )( 2 ) 4 2 ⎥ + ⎢ ( 2 )(160 + 2 ⋅ 440 )( 4 ) ⎥ ⎬ EI ⎩ ⎣ 3 ⎦ AB ⎣ 6 ⎦ BC ⎭ 1659.3 1386.7 = + EI EI 3046 = EIHence: 3046 3046 1 ⋅ δ CH = = × 103 = 84.6 mm EI 36 × 10 3 57 Dr. C. Caprani
  • Structural Analysis III4.5 Problems1. Determine the vertical deflection of joint E of the truss shown. Take E = 200 kN/mm2 and member areas, A = 1000 mm2 for all members except AC where A = 1000√2 mm2. (Ans. 12.2 mm). B C 3m D A E 60 kN 3m 4m2. Determine the vertical and horizontal deflection of joint C of the truss shown. Take E = 10 kN/mm2 and member areas, A = 1000 mm2 for all members except AC where A = 1000√2 mm2 and CE where A = 2500 mm2. (Ans. 32/7 mm horizontal, 74/7 mm vertical). 50 kN 3 4 B C D 3m A E 3m 4m 58 Dr. C. Caprani
  • Structural Analysis III3. Determine the horizontal deflection of joint A and the vertical deflection of joint B of the truss shown. Take E = 200 kN/mm2 and member areas, A = 1000 mm2 for all members except BD where A = 1000√2 mm2 and AB where A = 2500 mm2. (Ans. 15.3 mm; 0 mm) C D 3m B E 4m 90 kN 3m A4. Determine the vertical and horizontal deflection of joint C of the truss shown. Take E = 200 kN/mm2 and member areas, A = 150 mm2 for all members except AC where A = 150√2 mm2 and CD where A = 250 mm2. (Ans. 8.78 mm; 0.2 mm) 100 kN A B 3m C 4m D 3m 59 Dr. C. Caprani
  • Structural Analysis III5. Determine the vertical deflection of joint D of the truss shown. Take E = 200 kN/mm2 and member areas, A = 1000 mm2 for all members except BC where A = 1000√2 mm2. (Ans. 24 mm) C 3m B A E D 90 kN 3m 2m 2m6. Verify that the deflection at the centre of a simply-supported beam under a uniformly distributed load is given by: 5wL4 δC = 384 EI7. Show that the flexural deflection at the tip of a cantilever, when it is subjected to a point load on the end is: PL3 δB = 3EI8. Show that the rotation that occurs at the supports of a simply supported beam with a point load in the middle is: PL2 θC = 16 EI 60 Dr. C. Caprani
  • Structural Analysis III9. Show that the vertical and horizontal deflections at B of the following frame are: PR 3 PR 3 δ By = ; δ Bx = 2 EI 4 EI10. For the frame of Example 10, using virtual work, verify the following displacements in which the following directions are positive: vertical upwards; horizontal to the right, and; clockwise: 1176.3 • Rotation at A is − ; EI 3046 • Vertical deflection at B is − ; EI 8758.7 • Horizontal deflection at D is ; EI 1481.5 • Rotation at D is . EI 61 Dr. C. Caprani
  • Structural Analysis III11. For the cantilever beam below, show that the vertical deflection of B, allowing PL3 PL for both flexural and shear deformation, is + . EI GAv 62 Dr. C. Caprani
  • Structural Analysis IIIProblems 12 to 16 are considered Genius Level!12. Show that the vertical deflection at the centre of a simply supported beam, 5wL4 wL2 allowing for both flexural and shear deformation is + . 384 EI 8GAv 459413. For the frame shown, show that the vertical deflection of point D is ↓. EI Neglect axial deformation and take EI = 120 × 103 kNm 2 . 63 Dr. C. Caprani
  • Structural Analysis III14. For the frame shown, determine the vertical deflection of C and the rotation at D about the global x axis. Take the section and material properties as follows: A = 6600 mm 2 E = 205 kN/mm 2 G = 100 kN/mm 2 I = 36 × 106 mm 4 J = 74 × 106 mm 4 .(Ans. 12.23 mm ↑ ; 0.0176 rads ACW)15. Determine the horizontal and vertical deflections at B in the following frame. Neglect shear and axial effects. 2 wR 4 3π wR 4(Ans. δ Bx = → ; δ By = ↑) EI 2 EI 64 Dr. C. Caprani
  • Structural Analysis III16. For the frame shown, neglecting shear effects, determine the vertical deflection of points B and A.(Ans. δ By = ( P + wa ) b3 ↓ ; δ = 1 ⎡ Pa 3 wa 4 ( P + wa ) b3 ⎤ ⎛ + + wa 2 ⎞ ab Ay ⎢ ⎥ + ⎜ Pa + ⎟ ↓) 3EI EI ⎣ 3 8 3 ⎦ ⎝ 2 ⎠ GJ 65 Dr. C. Caprani
  • Structural Analysis III5. Virtual Work for Indeterminate Structures5.1 General ApproachUsing the concept of compatibility of displacement, any indeterminate structure canbe split up into a primary and reactant structures. In the case of a one-degreeindeterminate structure, we have: Final = Primary + ReactantAt this point the primary structure can be analysed. However, we further break up thereactant structure, using linear superposition: Reactant = Multiplier × Unit ReactantWe summarize this process as: M = M 0 +αM1 • M is the force system in the original structure (in this case moments); • M 0 is the primary structure force system; • M 1 is the unit reactant structure force system. 66 Dr. C. Caprani
  • Structural Analysis IIIFor a truss, the procedure is the same: Final = Primary + Reactant Reactant = Multiplier × Unit ReactantThe final system forces are: P = P 0 + α P1The primary structure can be analysed, as can the unit reactant structure. Therefore,the only unknown is the multiplier, α .We use virtual work to calculate the multiplier α . 67 Dr. C. Caprani
  • Structural Analysis III5.2 Using Virtual Work to Find the MultiplierWe must identify the two sets for use: • Displacement set: We use the actual displacements that occur in the real structure; • Equilibrium set: We use the unit reactant structure’s set of forces as the equilibrium set. We do this, as the unit reactant is always a determinate structure and has a configuration similar to that of the displacement set.Show this process graphically: 68 Dr. C. Caprani
  • Structural Analysis IIIThe virtual work equation (written for trusses) gives: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ PL ⎞ 0 ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi 1 ⎝ EA ⎠iThere is zero external virtual work. This is because there is no external virtual forceapplied. Also note that the real deformations that occur in the members are in termsof P, the unknown final forces. Hence, substituting P = P 0 + α ⋅ δ P1 (where δ is nowused to indicate virtual nature): ⎛ ( P 0 + α ⋅ δ P1 ) L ⎞ 0 = ∑⎜ ⎟ ⋅ δ Pi1 ⎜ EA ⎟ ⎝ ⎠i ⎛ P0 L ⎞ ⎛ δ P1L ⎞ = ∑⎜ ⎟ ⋅ δ Pi + α ⋅ ∑ ⎜ 1 ⎟ ⋅ δ Pi 1 ⎝ EA ⎠i ⎝ EA ⎠i 0=∑ P ⋅ δ Pi ⋅ Li 0 1 +α ⋅∑ (δ P ) i 1 2 Li EAi EAiFor beams and frames, the same development is: δW = 0 δ WE = δ WI ∑ y ⋅ δ F = ∑κ i i i ⋅δ Mi Mi 0 ⋅1 = ∑ ⋅δ Mi EIWhere again there is no external displacement of the virtual force. 69 Dr. C. Caprani
  • Structural Analysis IIIAlso, substitute M = M 0 + α ⋅ δ M 1 to get: 0 = ∑∫ L (M 0 + α ⋅δ M 1) δ M i1dx 0 EI i 0 = ∑∫ M 0 ⋅ δ M i1 L dx + α ⋅ ∑ ∫ (δ M i ) dx L 1 2 0 EI i 0 EI iThus in both bases we have a single equation with only one unknown, α . We canestablish values for the other two terms and then solve for α and the structure as awhole. 70 Dr. C. Caprani
  • Structural Analysis III5.3 Indeterminate TrussesExample 11Calculate the forces in the truss shown and find the horizontal deflection at C. TakeEA to be 10×104 kN for all members.Solve for the Truss Member ForcesChoose member AC as the redundant:Next analyse for the P 0 and P1 force systems. 71 Dr. C. Caprani
  • Structural Analysis III P0 System P1 SystemUsing a table: L P0 δ P1 P 0 ⋅ δ P1 ⋅ L (δ P )1 2 ⋅L Member (mm) (kN) (kN) × 104 × 104 AB 4000 +40 - 4/5 -12.8 0.265 BC 3000 0 - 3/5 0 0.108 CD 4000 0 - 4/5 0 0.256 AC 5000 0 1 0 0.5 BD 5000 -50 1 -25 0.5 ∑= -37.8 1.62Hence: 0=∑ P ⋅ δ Pi ⋅ Li 0 1 +α ⋅∑ (δ P ) i 1 2 Li EAi EAi −37.8 × 104 1.62 × 104 = +α ⋅ EA EA 72 Dr. C. Caprani
  • Structural Analysis IIIAnd so −37.8 α =− = 23.33 1.62The remaining forces are obtained from the compatibility equation: P0 δ P1 P = P 0 + α ⋅ δ P1 Member (kN) (kN) (kN) AB +40 - 4/5 21.36 BC 0 - 3/5 -14 CD 0 - 4/5 -18.67 AC 0 1 23.33 BD -50 1 -26.67Note that the redundant always has a force the same as the multiplier.Calculate the Horizontal Deflection at CTo calculate the horizontal deflection at C, using virtual work, the two relevant setsare: • Compatibility set: the actual deflection at C and the real deformations that occur in the actual structure; • Equilibrium set: a horizontal unit virtual force applied at C to a portion of the actual structure, yet ensuring equilibrium.We do not have to apply the virtual force to the full structure. Remembering that theonly requirement on the virtual force system is that it is in equilibrium; choose theforce systems as follows: 73 Dr. C. Caprani
  • Structural Analysis IIIThus we have: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ PL ⎞ yCH ⋅ 1 = ∑ ⎜ ⎟ ⋅ δ Pi ⎝ EA ⎠i ⎛ 23.33 × 5000 ⎞ 5 ⎛ −18.67 × 4000 ⎞ −4 =⎜ ⎟⋅ +⎜ ⎟⋅ ⎝ 10 × 10 ⎠ 3 ⎝ 10 × 10 4 4 ⎠ 3 yCH = 2.94 mmBecause we have chosen only two members for our virtual force system, only thesemembers do work and the calculation is greatly simplified. 74 Dr. C. Caprani
  • Structural Analysis III5.4 Indeterminate FramesExample 12For the frame shown, calculate the reactions and draw the bending moment diagram.Determine the horizontal deflection at joint C. Take EI = 40×103 kNm2, constant forall members.Break the frame up into its reactant and primary structures: M System = M0 System + α × M1 SystemEstablish the M 0 and M 1 force systems: 75 Dr. C. Caprani
  • Structural Analysis III M0 System M1 SystemApply the virtual work equation: 0 = ∑∫ M 0 ⋅ δ M i1 L dx + α ⋅ ∑ ∫ (δ M i ) dx L 1 2 0 EI i 0 EI iWe will be using the table of volume integrals to quicken calculations. Therefore wecan only consider lengths of members for which the correct shape of bending momentdiagram is available. Also, we must choose sign convention: we consider tension onthe outside of the frame to be positive.As each term has several components we consider them separately: M 0 ⋅ δ M i1 LTerm 1 - ∑ ∫ EI dx : 0 i • AD: we have graphically, and from the volume integral table: 1 1 j ( k1 + k2 ) l = ( −6 )( 600 + 360 ) 4 = −11520 2 2 76 Dr. C. Caprani
  • Structural Analysis III • DB: Similarly: jkl = ( −6 )( 360 ) 4 = −8640 • BC: 1 1 jkl = ( −6 )( 360 ) 6 = −3240 4 4Hence: M 0 ⋅ δ M i1 −23400 L ∑ ∫ EI dx = EI 0 iTerm 2: L (δ M ) 1 2 1 ⎧ ⎡1 ⎤ ⎫ ∑∫ dx = ⎨[ jkl ] AB + ⎢ jkl ⎥ ⎬ i 0 EI i EI ⎩ ⎣ 3 ⎦ BC ⎭ 1 ⎧ ⎡1 ⎤ ⎫ = ⎨ ⎣( −6 )( −6 ) 8⎦ AB + ⎢ ( −6 )( −6 ) 6 ⎥ ⎬ ⎡ ⎤ EI ⎩ ⎣3 ⎦ BC ⎭ 360 = EINote that Term 2 is always easier to calculate as it is only ever made up of straightline bending moment diagrams.Thus we have: 77 Dr. C. Caprani
  • Structural Analysis III 0 = ∑∫ M 0 ⋅ δ M i1 L L dx + α ⋅ ∑ ∫ (δ M i ) dx 1 2 0 EI i 0 EI i −23400 360 = +α ⋅ EI EIAnd so 23400 α= = 65.0 360Thus the vertical reaction at C is 65.0 kN upwards. With this information we cansolve for the moments using M = M 0 + α M 1 (or by just using statics) and theremaining reactions to get: 78 Dr. C. Caprani
  • Structural Analysis IIITo calculate the horizontal deflection at C using virtual work, the two relevant setsare: • Compatibility set: the actual deflection at C and the real deformations (rotations) that occur in the actual structure; • Equilibrium set: a horizontal unit virtual force applied at C to a determinate portion of the actual structure.Choose the following force system as it is easily solved:Thus we have: δW = 0 δ WE = δ WI ∑ y ⋅ δ F = ∑θ ⋅ δ M i i i i ⎡M ⎤ L yCH ⋅ 1 = ∑ ∫ ⎢ x ⎥ ⋅ δ M x dx 0⎣ EI ⎦ 79 Dr. C. Caprani
  • Structural Analysis IIIThe real bending moment diagram, M, is awkward to use with the integral table.Remembering that M = M 0 + α M 1 simplifies the calculation by using:To give: 80 Dr. C. Caprani
  • Structural Analysis IIIAnd so using the table formulae, we have: ⎡1 ( −4 )( −6 ⋅ α ) 4 + ( 2 ⋅ 360 + 600 )( −4 ) 4⎤ + 1 ⎢2 ⎣ 6 ⎥ ⎦ AD ⎡1 1 ⎤ ⎢ 2 ( −4 − 8 )( −6 ⋅ α ) 4 + 2 ( −4 − 8 )( 360 ) 4 ⎥ + ⎣ ⎦ DB ⎡1 ( −8)( −6 ⋅ α ) 6 + ( −8)( 360 ) 6⎤ 1 ⎢3 ⎣ 4 ⎥ ⎦ BCWhich gives us: ⎡Mx ⎤ L 1 ∑ ∫ ⎢ EI ⎥ ⋅ δ M ⎣ ⎦ x dx = ⎡ 288α + ( −16480 ) ⎤ EI ⎣ ⎦ 0 2240 = EI = 0.056 = 56 mmThe answer is positive, indicating that the structure moves to the right at C: the samedirection in which the unit virtual force was applied. 81 Dr. C. Caprani
  • Structural Analysis III5.5 Continuous BeamsBasisIn this section we will introduce structures that are more than 1 degree staticallyindeterminate. We do so to show that virtual work is easily extensible to multiply-indeterminate structures, and also to give a method for such beams that is easilyworked out, and put into a spreadsheet.Consider the example 3-span beam. It is 2 degrees indeterminate, and so we introduce2 hinges (i.e. moment releases) at the support locations and unit reactant moments intheir place, as shown: 82 Dr. C. Caprani
  • Structural Analysis IIIAlongside these systems, we have their bending moment diagrams:Using the idea of the multiplier and superposition again, we can see that: M = M 0 + α1 ⋅ δ M 1 + α 2 ⋅ δ M 2The virtual work equation is: δW = 0 δ WE = δ WI 0 ⋅1 = ∑ ∫θ ⋅ δ M 83 Dr. C. Caprani
  • Structural Analysis IIIThere is no external virtual work done since the unit moments are applied internally.Since we have two virtual force systems in equilibrium and one real compatiblesystem, we have two equations: M M 0 = ∑∫ ⋅ δ M 1 and 0 = ∑∫ ⋅δ M 2 EI EIFor the first equation, expanding the expression for the real moment system, M: (M 0 + α1 ⋅ δ M 1 + α 2 ⋅ δ M 2 ) ∑∫ EI ⋅δ M 1 = 0 M 0 ⋅δ M 1 δ M 1 ⋅δ M 1 δ M 2 ⋅δ M 1 ∫ EI + α1 ∫ EI + α2 ∫ EI =0In which we’ve dropped the summation over all members – it being understood thatwe sum for all members. Similarly for the second virtual moments, we have: M 0 ⋅δ M 2 δ M 1 ⋅δ M 2 δ M 2 ⋅δ M 2 ∫ EI + α1 ∫ EI + α 2 ∫ EI = 0Thus we have two equations and so we can solve for α1 and α 2 . Usually we writethis as a matrix equation: ⎧ M 0 ⋅δ M 1 ⎫ ⎡ δ M 1 ⋅δ M 1 δ M 2 ⋅δ M 1 ⎤ ⎪∫ ⎪ EI ⎪ ⎢∫ ⎪ EI ∫ EI ⎥ ⎧α1 ⎫ ⎧0 ⎫ ⎨ 2⎬ +⎢ ⎥⎨ ⎬ = ⎨ ⎬ ⎪ M ⋅δ M ⎪ ⎢ δ M ⋅δ M 0 1 2 δ M 2 ⋅ δ M 2 ⎥ ⎩α 2 ⎭ ⎩0 ⎭ ⎪∫ ⎩ EI ⎪ ⎣∫ ⎭ ⎢ EI ∫ EI ⎦ ⎥Each of the integral terms is easily found using the integral tables, and the equationsolved. 84 Dr. C. Caprani
  • Structural Analysis IIISimilarly the (simpler) equation for a 2-span beam is: M 0 ⋅δ M 1 δ M 1 ⋅δ M 1 ∫ EI + α1 ∫ EI = 0And the equation for a 4-span beam is: ⎧ M 0 ⋅δ M 1 ⎫ ⎡ δ M 1 ⋅δ M 1 δ M 2 ⋅δ M 1 ⎤ ⎪∫ EI ⎪ ⎢∫ EI ∫ EI 0 ⎥ ⎪ ⎪ ⎢ ⎥ ⎧α1 ⎫ ⎧0 ⎫ ⎪ M ⋅δ M ⎪ ⎢ δ M ⋅δ M 2 0 2 1 δ M ⋅δ M 2 2 δ M 3 ⋅δ M 2 ⎥ ⎪ ⎪ ⎪ ⎪ ⎨∫ ⎬ + ⎢∫ ∫ EI ∫ EI ⎥ ⎨α 2 ⎬ = ⎨0⎬ ⎪ EI ⎪ ⎢ EI ⎥ ⎪ ⎪ ⎪0 ⎪ 3 ⎩α 3 ⎭ ⎩ ⎭ ⎪ M ⋅δ M ⎪ ⎢ 0 3 δ M 2 ⋅δ M 3 δ M ⋅δ M ⎥ 3 ⎪∫ ⎪ ⎢ 0 ∫ EI ∫ EI ⎦ ⎥ ⎩ EI ⎭ ⎣Sketch out why there are no terms involving δ M 3 ⋅ δ M 1 : 85 Dr. C. Caprani
  • Structural Analysis IIIExample 13Using virtual work, find the bending moment diagram for the following beam:Proceeding as described above, we introduce releases (hinges) at the support points,apply the unit virtual moments, and find the corresponding bending momentdiagrams:Next we need to evaluate each term in the matrix virtual work equation. We’ll takethe two ‘hard’ ones first: 86 Dr. C. Caprani
  • Structural Analysis III M 0 ⋅δ M 1 1 ⎡1 ( 50 )( −1)( 4 + 2 )⎤ + ⎡1 ( 45)( −1)( 6 )⎤ 1 ∫ EI = EI ⎢ 6 ⎣ ⎥ ⎦ AB 2 EI ⎢3 ⎣ ⎥ ⎦ BC • 1 95 = ( −50 − 45) = − EI EINote that since δ M 1 = 0 for span CD, there is no term for it above. Similarly, for thefollowing evaluation, there will be no term for span AB: M 0 ⋅δ M 2 1 ⎡1 ( 45)( −1)( 6 ) ⎤ + ⎡ ( 93.75)( −1)( 5)⎤ 1 1 ∫ EI = 2 EI ⎢ 3 ⎣ ⎥ ⎦ BC EI ⎢ 3 ⎣ ⎥ ⎦ CD • 1 201.25 = ( −45 − 156.25) = − EI EIThe following integrals are more straightforward since they are all triangles: δ M 1 ⋅δ M 1 1 ⎡1 ⎤ 1 ⎡1 ⎤ 2.333 • ∫ EI = ⎢ 3 ( −1)( −1)( 4 ) ⎥ + 2 EI ⎢ 3 ( −1)( −1)( 6 ) ⎥ = EI EI ⎣ ⎦ AB ⎣ ⎦ BC δ M 2 ⋅δ M 1 1 ⎡1 ( −1)( −1)( 6 ) ⎤ = 0.5 • ∫ EI = 2 EI ⎢ 6 ⎣ ⎥ ⎦ BC EI δ M 1 ⋅δ M 2 0.5 δ M 2 ⋅δ M 1 • ∫ EI = EI , since it is equal to ∫ EI by the commutative property of multiplication. δ M 2 ⋅δ M 2 1 ⎡1 ⎤ 1 ⎡1 ⎤ 2.667 • ∫ EI = 2 EI ⎢ 3 ( −1)( −1)( 6 ) ⎥ + EI ⎢ 3 ( −1)( −1)( 5 ) ⎥ = EI ⎣ ⎦ BC ⎣ ⎦ CDWith all the terms evaluated, enter them into the matrix equation: 87 Dr. C. Caprani
  • Structural Analysis III ⎧ M 0 ⋅δ M 1 ⎫ ⎡ δ M 1 ⋅δ M 1 δ M 2 ⋅δ M 1 ⎤ ⎪∫ ⎪ EI ⎪ ⎢∫ ⎪ EI ∫ EI ⎥ ⎧α1 ⎫ ⎧0 ⎫ ⎨ ⎬+ ⎢ ⎥⎨ ⎬ = ⎨ ⎬ ⎪ M ⋅δ M ⎪ ⎢ δ M ⋅δ M 0 2 1 2 δ M 2 ⋅ δ M 2 ⎥ ⎩α 2 ⎭ ⎩0 ⎭ ⎪∫ ⎩ EI ⎪ ⎣∫ ⎭ ⎢ EI ∫ EI ⎦ ⎥To give: 1 ⎧ −95 ⎫ 1 ⎡ 2.333 0.5 ⎤ ⎧α1 ⎫ ⎧0 ⎫ ⎨ ⎬+ ⎢ ⎥⎨ ⎬ = ⎨ ⎬ EI ⎩−201.25⎭ EI ⎣ 0.5 2.667 ⎦ ⎩α 2 ⎭ ⎩0 ⎭And solve, as follows: ⎡ 2.333 0.5 ⎤ ⎧α1 ⎫ ⎧ 95 ⎫ ⎢ 0.5 2.667 ⎥ ⎨α ⎬ = ⎨201.25⎬ ⎣ ⎦⎩ 2⎭ ⎩ ⎭ ⎧α1 ⎫ 1 ⎡ 2.667 −0.5 ⎤ ⎧ 95 ⎫ ⎨ ⎬= ⎢ ⎥⎨ ⎬ ⎩α 2 ⎭ ( 2.333 ⋅ 2.667 − 0.5 ⋅ 0.5 ) ⎣ −0.5 2.333⎦ ⎩201.25⎭ ⎧25.57 ⎫ =⎨ ⎬ ⎩70.67 ⎭Now using our superposition equation for moments, M = M 0 + α1 ⋅ δ M 1 + α 2 ⋅ δ M 2 ,we can show that the multipliers are just the hogging support moments: M B = 0 + 25.57 ⋅ 1 + 70.67 ⋅ 0 = 25.57 kNm M C = 0 + 25.57 ⋅ 0 + 70.67 ⋅ 1 = 70.67 kNmFrom these we get the final BMD: 88 Dr. C. Caprani
  • Structural Analysis III 89 Dr. C. Caprani
  • Structural Analysis IIISpreadsheet SolutionA simple spreadsheet for a 3-span beam with centre-span point load and UDL capabilities, showing Example 13, is: Span 1 Span 2 Span 3 VW EquationsLength m 4 6 5EI x 1 2 1 -95.000 + 2.333 0.500 * alpha1 = 0PL kN 50 0 0 -201.250 0.500 2.667 alpha2 0UDL kN/m 0 10 30BMDs PL 50 0 0 2.333 0.500 * alpha1 = 95.000 UDL 0 45 93.75 0.500 2.667 alpha2 201.250 alpha1 = 0.446512 -0.083721 * 95.000 M0 x M1 Span 1 Span 2 Span 3 Totals alpha2 -0.083721 0.390698 201.250 PL -50.000 0.000 0.000 UDL 0.000 -90.000 0.000 alpha1 = 25.57 Total -50.000 -45.000 0.000 -95.000 alpha2 70.67 M1 x M1 Span 1 Span 2 Span 3 Total 1.333 1.000 0.000 2.333 Mid-Span and Support Moments M2 x M1 Span 1 Span 2 Span 3 Mab 37.2 kNm Total 0.000 0.500 0.000 0.500 Mb -25.6 kNm Mbc -3.1 kNm M0 x M2 Span 1 Span 2 Span 3 Mc -70.7 kNm PL 0.000 0.000 0.000 Mcd 58.4 kNm UDL 0.000 -90.000 -156.250 Total 0.000 -45.000 -156.250 -201.250 M1 x M2 Span 1 Span 2 Span 3 Total 0.000 0.500 0.000 0.500 M2 x M2 Span 1 Span 2 Span 3 Total 0.000 1.000 1.667 2.667 90 Dr. C. Caprani
  • Structural Analysis III5.6 Problems 1. For the truss shown, calculate the vertical deflection at C when member EB is not present and when it is present. Members AB, DC and DE have EA = 60 × 103 kN; members EB, BC and AD have EA = 100 × 103 kN, and; member BD has EA = 80 × 103 kN. (Ans. 15 mm and 11.28 mm) 2. Autumn 2007 For the truss shown, determine the force in each member and determine the horizontal deflection of joint B. Take EA = 200 × 103 kN for all members. (Ans. Choosing BD: α = 50 2 ; 2 mm) B 100 kN 2m A D C 2m 2m 91 Dr. C. Caprani
  • Structural Analysis III 3. For the frame shown, calculate the bending moment diagram, and verify the 856.2 327.6 following deflections: δ Bx = → and δ Cy = ↓. (Ans. EI EI M B = 30.1 kNm ) 4. Summer 2007, Part (b): For the frame shown, draw the bending moment diagram and determine the horizontal deflection of joint C. Neglect axial effects in the members and take EI = 36 × 103 kNm 2 . (Ans. H D = 75.8 kN ; 19.0 mm) 100 kN B C 20 kN/m 4m A D 4m 4m 92 Dr. C. Caprani
  • Structural Analysis III 5. For the beam of Example 13, show that the deflections at the centre of each span, taking downwards as positive, are: 41.1 • Span AB: ; EI 23.9 • Span BC: − ; EI 133.72 • Span CD: . EI 6. Using virtual work, analyse the following prismatic beam to show that the support moments are M B = 51.3 kNm and M C = 31.1 kNm , and draw the bending moment diagram: 93 Dr. C. Caprani
  • Structural Analysis III6. Virtual Work for Self-Stressed Structures6.1 BackgroundIntroductionSelf-stressed structures are structures that have stresses induced, not only by externalloading, but also by any of the following: • Temperature change of some of the members (e.g. solar gain); • Lack of fit of members from fabrication: o Error in the length of the member; o Ends not square and so a rotational lack of fit; • Incorrect support location from imperfect construction; • Non-rigid (i.e. spring) supports due to imperfect construction.Since any form of fabrication or construction is never perfect, it is very important forus to know the effect (in terms of bending moment, shear forces etc.) that such errors,even when small, can have on the structure.Here we introduce these sources, and examine their effect on the virtual workequation. Note that many of these sources of error can exist concurrently. In suchcases we add together the effects from each source. 94 Dr. C. Caprani
  • Structural Analysis IIITemperature ChangeThe source of self-stressing in this case is that the temperature change causes amember to elongate: ∆LT = α L ( ∆T )where α is the coefficient of linear thermal expansion (change in length, per unitlength, per degree Celsius), L is the original member length and ∆T is thetemperature change.Since temperature changes change the length of a member, the internal virtual workis affected. Assuming a truss member is being analysed, we now have changes inlength due to force and temperature, so the total change in length of the member is: PL e= + α L ( ∆T ) EAHence the internal virtual work for this member is: δ WI = e ⋅ δ P ⎛ PL ⎞ =⎜ + α L ( ∆T ) ⎟ ⋅ δ P ⎝ EA ⎠ 95 Dr. C. Caprani
  • Structural Analysis IIILinear Lack of FitFor a linear lack of fit, the member needs to be artificially elongated or shortened tofit it into place, thus introducing additional stresses. This is denoted: λLConsidering a truss member subject to external loading, the total change in lengthwill be the deformation due to loading and the linear lack of fit: PL e= + λL EAHence the internal virtual work for this member is: δ WI = e ⋅ δ P ⎛ PL ⎞ =⎜ + λL ⎟ ⋅ δ P ⎝ EA ⎠ 96 Dr. C. Caprani
  • Structural Analysis IIIRotational Lack of FitA rotational lack of fit, which applies to frames only, occurs when the end of amember is not square. Thus the member needs to be artificially rotated to get it intoplace, as shown below. This is denoted as: λθConsidering a frame member which has a lack of fit, λθ and a virtual moment δ M atthe same point, then the internal virtual work done at this point is: δ WI = λθ ⋅ δ MThis must be added to the other forms of internal virtual work. Not also that the signsmust be carefully chosen so that the virtual moment closes the gap – we will see thismore clearly in an example. 97 Dr. C. Caprani
  • Structural Analysis IIIErrors in Support LocationThe support can be misplaced horizontally and/or vertically. It is denoted: λSA misplaced support affects the external movements of a structure, and so contributesto the external virtual work. Denoting the virtual reaction at the support, in thedirection of the misplacement as δ R , then we have: δ We = λS ⋅ δ R 98 Dr. C. Caprani
  • Structural Analysis IIISpring SupportsFor spring supports we will know the spring constant for the support, denoted: kSSince movements of a support are external, spring support movements affect theexternal virtual work. The real displacement ∆ S that occurs is: ∆ S = R kSIn which R is the real support reaction in the direction of the spring. Further, since Rwill be known in terms of the multiplier and virtual reaction, δ R , we have: R = R0 + α ⋅ δ RHence: ∆ S = ( R 0 + α ⋅ δ R ) kSAnd so the external work done is: δ We = ∆ S ⋅ δ R (δ R ) 2 R0 = ⋅δ R + α ⋅ kS kSThe only unknown here is α which is solved for from the virtual work equation. 99 Dr. C. Caprani
  • Structural Analysis III6.2 TrussesExample 14Here we take the truss of Example 6 and examine the effects of: • Member ED was found to be 5 mm too long upon arrival at site; • Member AB is subject to a temperature increase of +100 ˚C.For this truss, E = 200 kN/mm2; member areas, A = 1000 mm2 for all members exceptAE and BD where A = 1000√2 mm2, and; coefficient of expansion isα = 10 × 10−6 °C-1 .The change in length due to the temperature change is: ∆LT = α L ( ∆T ) = (10 × 10−6 ) ( 2000 )( +100 ) = 2 mmVertical Displacement of Joint DThe virtual work equation is now: 100 Dr. C. Caprani
  • Structural Analysis III δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ P0 L ⎞ yDV ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi + 5 ⋅ δ PED + 2 ⋅ δ PAB 1 1 1 ⎝ EA ⎠iIn Example 6 we established various values in this equation: ⎛ P0 L ⎞ • ∑⎜ ⎟ ⋅ δ Pi = 16.5 1 ⎝ EA ⎠i • δ PED = −1 1 • δ PAB = +1 1Hence we have: yDV ⋅ 1 = 16.5 + 5 ( −1) + 2 ( +1) yDV = 13.5 mmHorizontal Displacement of Joint D:Similarly, copying values from Example 6, we have: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ P0 L ⎞ yDH ⋅1 = ∑ ⎜ ⎟ ⋅ δ Pi + 5 ⋅ δ PED + 2 ⋅ δ PAB 2 2 2 ⎝ EA ⎠i = −4.5 + 5 ( +1) + 2 ( 0 ) yDH = +0.5 mm to the right 101 Dr. C. Caprani
  • Structural Analysis IIIExample 15Here we use the truss of Example 11 and examine, separately, the effects of: • Member AC was found to be 3.6 mm too long upon arrival on site; • Member BC is subject to a temperature reduction of -50 ˚C; • Support D is surveyed and found to sit 5 mm too far to the right.Take EA to be 10×104 kN for all members and the coefficient of expansion to beα = 10 × 10−6 °C-1 .Error in Length:To find the new multiplier, we include this effect in the virtual work equation: δW = 0 δ WE = δ WI ∑ y ⋅δ F = ∑e ⋅δ P i i i i ⎛ PL ⎞ 0 = ∑⎜ ⎟ ⋅ δ Pi + λL ⋅ δ PAC 1 1 ⎝ EA ⎠iBut P = P 0 + α ⋅ δ P1 , hence: 102 Dr. C. Caprani
  • Structural Analysis III ⎛ ( P 0 + α ⋅ δ P1 ) L ⎞ 0 = ∑⎜ ⎟ ⋅ δ Pi1 + λL ⋅ δ PAC 1 ⎜ EA ⎟ ⎝ ⎠i ⎛ P0 L ⎞ ⎛ δ P1L ⎞ = ∑⎜ ⎟ ⋅ δ Pi + α ⋅ ∑ ⎜ 1 ⎟ ⋅ δ Pi + λL ⋅ δ PAC 1 1 ⎝ EA ⎠i ⎝ EA ⎠i 0=∑ P ⋅ δ Pi ⋅ Li 0 1 +α ⋅∑ (δ P ) i 1 2 Li + λL ⋅ δ PAC 1 EAi EAiAs can be seen, the λL ⋅ δ PAC term is simply added to the usual VW equation. From 1before, we have the various values of the summations, and so have: −37.8 × 104 1.62 × 104 0= +α ⋅ + ( 3.6 )( +1) EA EAHence: 37.8 × 104 − 3.6 EA α= = +1.11 1.62 × 104Thus member AC is 1.1 kN in tension. Note the change: without the error in fit it was23.3 kN in tension and so the error in length has reduced the tension by 22.2 kN.Temperature Change:The same derivation from the VW equation gives us: (δ Pi1 ) Li + ∆L ⋅ δ P1 2 P 0 ⋅ δ Pi1 ⋅ Li 0=∑ +α ⋅∑ T BC EAi EAiThe change in length due to the temperature change is: 103 Dr. C. Caprani
  • Structural Analysis III ∆LT = α L ( ∆T ) = (10 × 10−6 ) ( 3000 )( −50 ) = −1.5 mmNoting that the virtual force in member BC is δ PBC = 0.6 , we have: 1 −37.8 × 104 1.62 × 104 0= +α ⋅ + ( −1.5 )( 0.6 ) EA EAGiving: 37.8 × 104 + 0.9 EA α= = +28.9 1.62 × 104Thus member AC is 28.9 kN in tension, an increase of 5.4 kN.Error in Support Location:Modifying the VW equation gives: (δ Pi1 ) Li 2 P 0 ⋅ δ Pi1 ⋅ Li λS ⋅ δ H D = ∑ +α ⋅∑ EAi EAiThe value of the virtual horizontal reaction is found from the virtual force system tobe 0.6 kN to the right. Hence: −37.8 × 104 1.62 × 104 + ( 5 )( 0.6 ) = +α ⋅ EA EA 104 Dr. C. Caprani
  • Structural Analysis IIINote the sign on the support displacement: since the real movement is along the samedirection as the virtual force, it does positive virtual work. Solving: −3EA + 37.8 × 104 α= = +4.8 1.62 × 104Thus member AC is 4.8 kN in tension; a reduction of 18.5 kN.All Effects Together:In this case, the virtual work equation is: λS ⋅ δ H D = ∑ P ⋅ δ Pi ⋅ Li 0 1 +α ⋅∑ (δ P ) i 1 2 Li + λL ⋅ δ PAC + ∆LT ⋅ δ PBC 1 1 EAi EAiSubstituting the various values in gives: −37.8 × 104 1.62 × 104 + ( 0.5 )( 0.6 ) = +α ⋅ + ( 3.6 )( +1) + ( −1.5 )( 0.6 ) EA EAAnd solving: α= ( −3.6 + 0.9 − 3) EA + 37.8 × 104 = −11.9 1.62 × 104And so member AC is 11.9 kN in compression. Thus should be the same as theoriginal force of 23.3 kN plus the changes induced by the errors: 23.3 − 22.2 + 5.4 − 18.5 = −12 kN 105 Dr. C. Caprani
  • Structural Analysis III6.3 BeamsExample 16Consider the following 2-span beam with central spring support. Determine anexpression for the central support reaction in terms of its spring stiffness.To do this, we will consider the central support as the redundant:The virtual work equation, accounting for the spring, is: 106 Dr. C. Caprani
  • Structural Analysis III 1⋅ ∆S = ∑ ∫ M 0 ⋅ δ M i1 L dx + α ⋅ ∑ ∫ L (δ M i ) dx 1 2 0 EI i 0 EI iThe deflection at the central support will be: VB α ∆S = = k kSince the reaction at B is 1 ⋅ α . Noting that the deflection of the spring will beopposite to the unit load, and using the volume integrals, we have: α⎡ 5 ⎛ l ⎞ ⎛ wl 2 ⎞ ⎤ 2 2 ⎡ 1 ⎛ l ⎞⎛ l ⎞ ⎤ − = ⎢ ⎜ − ⎟⎜ ⎟ ( l )⎥ + α ⋅ ⎢ 3 ⎜ − 2 ⎟⎜ − 2 ⎟ ( l ) ⎥ k ⎣12 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ EI EI ⎣ ⎝ ⎠⎝ ⎠ ⎦ α −5wl 4 l3 − = +α ⋅ k 24 EI 6 EI ⎛ EI l 3 ⎞ −5wl 4 α ⎜− − ⎟ = ⎝ k 6⎠ 24And finally: 5wl 4 α= 24 EI + 4l 3 k 5wl = EI 24 3 + 4 klIt is important we understand the implications of this result: 107 Dr. C. Caprani
  • Structural Analysis III• For no support present, k = 0 and so 24 EI k → ∞ meaning α → 0 and there is no support reaction (as we might expect);• For k = ∞ we have the perfectly rigid (usual) roller support and so 24 EI k → 0 5 giving us α → wl - a result we established previously; 4• For intermediate relative stiffnesses, the value of the reaction is between these extremes. Therefore, we as the designer have the ability to choose the reaction most favourable to us.A plot of the reaction and relative stiffnesses is: 1.4 1.2 Central Reaction (α /wL ) 1 0.8 0.6 0.4 0.2 0 1.E-10 1.E-08 1.E-06 1.E-04 1.E-02 1.E+00 1.E+02 1.E+04 1.E+06 1.E+08 1.E+10 3 Relative Stiffness (EI/kL ) 108 Dr. C. Caprani
  • Structural Analysis III6.4 FramesExample 17The following frame, in addition to its loading, is subject to: • Support A is located 10 mm too far to the left; • End C of member BC is 1.2 × 10−3 rads out of square, as shown; • Member CD is 12 mm too short.Determine the bending moment diagram. Take EI = 36 × 103 kNm2 for each member.We will choose the horizontal reaction at A as the redundant. Since we are dealingwith a linear lack of fit in member CD, we need to allow for the virtual work done bythe axial forces in this member and so we solve for the axial force diagrams also.For the primary structure:∑F x = 0: H D − 90 = 0 0 ∴ H D = 90 kN 0 109 Dr. C. Caprani
  • Structural Analysis III∑ M about A = 0 : 6224 ⋅ + 90 ⋅ 4 + 90 ⋅ 2 − VD ⋅ 6 = 0 0 ∴VD = 162 kN 0 2∑F y = 0: −VA0 − 24 ⋅ 6 + 162 = 0 ∴VA0 = 18 kNAlso, M C = 90 ⋅ 6 = 540 kNm giving:For the redundant structure: 110 Dr. C. Caprani
  • Structural Analysis III 1∑M about A = 0 : −1 ⋅ 2 + VD ⋅ 6 = 0 1 ∴VD = 1 3 kN ↓ 1 1∑F y = 0: −VA + 1 3 = 0 ∴VA = kN ↑ 1 3The virtual work equation, accounting for the relevant effects is: δW = 0 δ WE = δ WIλS ⋅ H 1 = ∑ ∫ L M ⋅δ M 0 1 dx + α ⋅ ∑ ∫ i L (δ M ) 1 2 i dx + λθ ⋅ δ M C + λL ⋅ δ PCD 1 1 A 0 EI i 0 EI iWe take each term in turn: (a) λS ⋅ H 1 : The applied unit load is in the same direction as the error in the A support location, and keeping all units in metres: λS ⋅ H 1 = 10 × 10−3 ⋅ +1 = 10 × 10−3 ( kN ⋅ m ) A M 0 ⋅ δ M i1 L (b) ∑ ∫ dx : Using the volume integrals: 0 EI i M 0 ⋅ δ M i1 1 ⎧⎡ 1 ⎤ ⎡1 ⎤ ⎫ L ∑ ∫ EI dx = EI ⎨⎢12 ( 4 + 3 ⋅ 6 )( −540 ) 6⎥ + ⎢ 3 ( 6 )( −540 ) 6⎥ ⎬ 0 i ⎩⎣ ⎦ BC ⎣ ⎦ CD ⎭ 12420 =− EI (c) ∑ ∫ L (δ M ) 1 2 i dx : Again using the integrals table: 0 EI i 111 Dr. C. Caprani
  • Structural Analysis III L (δ M ) 1 2 1 ⎡1 ( 4 )( 4 ) 4⎤ + ⎡ ( 4 ( 2 ⋅ 4 + 6 ) + 6 ( 4 + 2 ⋅ 6 ) ) 6⎤ 1 1 ∑∫ dx = i 0 EI i EI ⎢ 3 ⎣ ⎥ ⎦ AB EI ⎢ 6 ⎣ ⎥ ⎦ BC 1 ⎡1 ⎤ + ⎢ 3 ( 6 )( 6 ) 6 ⎥ EI ⎣ ⎦ CD L (δ M ) 1 2 245.33 ∑∫ dx = i 0 EI i EI (d) λθ ⋅ δ M C : The virtual bending moment at C is 6 kNm with tension on the 1 inside of the frame. This is in the opposite direction to that needed to close the lack of fit and so the sign of this term is negative, as shown: λθ ⋅ δ M C = − (1.2 × 10−3 ⋅ 6 ) = −7.2 × 10−3 ( kN ⋅ m ) 1 (e) λL ⋅ δ PCD : Since the member is too short, the lack of fit is negative, whilst the 1 virtual force is in tension and so positive: ⎛ 1⎞ λL ⋅ δ PCD = ( −12 × 10−3 ) ⋅ ⎜ + ⎟ = −4 × 10−3 ( kN ⋅ m ) 1 3 ⎝ ⎠Substituting these values into the equation, along with EI = 36 × 103 kNm 2 , gives: 112 Dr. C. Caprani
  • Structural Analysis III 12420 245.33 10 × 10−3 = − +α ⋅ − 7.2 × 10−3 − 4 × 10−3 36 × 10 3 36 × 10 3And so we can solve for α : 10 = −345 + α ⋅ 6.815 − 7.2 − 4 10 + 345 + 7.2 + 4 α= 6.815 = 53.73And so the horizontal reaction at A is: H A = H A + α ⋅ H 1 = 0 + 53.73 ⋅ 1 = 53.73 kN 0 Aand similarly for the other reactions. Also, using the superposition equation formoments, M = M 0 + α ⋅ M 1 , we have: 113 Dr. C. Caprani
  • Structural Analysis III6.5 Problems 1. For the truss of Example 15, show that the force in member AC is 22.2 kN in compression when there is no external loads present, and only the lack of fit of 3.6 mm of member AC. 2. For the truss of Example 15, determine the horizontal deflection of joint C due to each of the errors separately, and then combined. 3. For the beam of Example 16, show that the stiffness of the spring support that optimizes the bending moments is 89.5 EI l 3 , i.e. makes the sagging and hogging moments equal in magnitude. 4. For the following beam, show that the vertical deflection at C can be given by: ⎛ 2 L3 1⎞ δ Cy = P ⎜ + ⎟ ⎝ 3EI k ⎠ 114 Dr. C. Caprani
  • Structural Analysis III 5. For the frame of Example 17, show that the horizontal deflection of joint C due to the applied loads only is 77.5 mm. Find the deflection of joint C due to both the loads and errors given. 6. For the following frame, the support at D was found to yield horizontally by 0.08 mm/kN. Also, the end of member BC is not square, as shown. Draw the bending moment diagram and determine the horizontal deflection at D. Take EI = 100×103 kNm2. (Ans. 5.43 mm to the right) 115 Dr. C. Caprani
  • Structural Analysis III7. Past Exam Questions7.1 Summer 1997Answers:(a) δ Cx = 30.0 mm → ;(b) δ Cx = 12.9 mm → ;(c) λBD = 21.1 mm too long. 116 Dr. C. Caprani
  • Structural Analysis III7.2 Summer 1998Answers:(a) δ Cx = 45.0 mm → ;(b) δ Cx = 26.2 mm → . 117 Dr. C. Caprani
  • Structural Analysis III7.3 Summer 1999Answers:(a) VE = 42.5 kN ↑ ;(b) δ Cy = 28.6 mm ↓ . 118 Dr. C. Caprani
  • Structural Analysis III7.4 Summer 2000Answers:(a) VA = 2.78 kN ↓ ;(b) δ C = 0.47 mm ↓ . 119 Dr. C. Caprani
  • Structural Analysis III7.5 Summer 2001Answers:(a) PBD = +44.4 kN ;(b) PBD = +300 kN . 120 Dr. C. Caprani
  • Structural Analysis III7.6 Summer 2002Answers:(a) λL = 1.35 mm shorter;(b) VC = 160 kN ↑ . 121 Dr. C. Caprani
  • Structural Analysis III7.7 Summer 2004Answers:(a) δ Ay = 2.19 mm ↓ ;(b) λL = 5 mm shorter. 122 Dr. C. Caprani
  • Structural Analysis III7.8 Summer 2007Using Virtual Work: (i) For the frame in Fig. Q3(a), determine the horizontal deflection of joint C; (ii) Draw the bending moment diagram for the frame in Fig. Q3(b); (iii) Determine the horizontal deflection of joint C for the frame in Fig. Q3(b).Note:You may neglect axial effects in the members.Take EI = 36 × 103 kNm 2 for all members. 100 kN 100 kN C B C B 20 kN/m 20 kN/m 4m 4mA A D D 4m 4m 4m 4m FIG. Q3(a) FIG. Q3(b)Answers:(a) δ Cx = 84.6 mm → ;(b) H D = 75.8 kN ← ;(c) δ Cx = 19.02 mm → . 123 Dr. C. Caprani
  • Structural Analysis III7.9 Semester 2, 2008For the frame shown in Fig. Q2, using Virtual Work:(a) Determine the following: (i) reactions; (ii) bending moment diagram.(b) After construction, it is found that the support at C is not rigid but a spring support of stiffness k = 2000 3 kN/m . Taking this into account, determine the following: (i) vertical displacement of joint C; (ii) bending moment diagram. (40 marks)Note:You may neglect axial effects in the members.Take EI = 36 × 103 kNm 2 for all members. 10 kN/m 20 kN C B 6m A 6m FIG. Q2Answers:(a) VC = 35.63 kN ↑ ;(b) δ Cy = 45 mm ↓ . 124 Dr. C. Caprani
  • Structural Analysis III7.10 Semester 2, 2009QUESTION 2Using Virtual Work, for the frame shown in Fig. Q2, do the following:(i) Determine the reactions;(ii) Draw the bending moment diagram;(iii) Draw the deflected shape of the structure;(iv) Determine the horizontal deflection of joint D. (25 marks)Note:• Take E = 200 kN/mm 2 and I = 45 × 107 mm 4 for all members. A B C 150 kN 3m D 3m 3m 4m FIG. Q2Answers:(a) VB = 225 kN ↑ ;(b) δ Dx = 81.25 mm ← . 125 Dr. C. Caprani
  • Structural Analysis III7.11 Semester 2, 2009QUESTION 3For the structure shown in Fig. Q2, due to an error on site, it is required to make the moment at support A zero. This isto be done by the introduction of a spring support at B, in place of the roller support, as shown in Fig. Q3. Using VirtualWork:(i) Determine the spring stiffness required so that the moment at A is zero.(ii) Drawn the revised bending moment diagram;(iii) Draw the revised deflected shape of the structure;(iv) Determine the vertical deflection of the structure at B. (25 marks)Note:• Take E = 200 kN/mm 2 and I = 45 × 107 mm 4 for all members.• You may use any relevant results from your workings for Q2, but in doing so acknowledge their source. A B C k 3m 150 kN D 3m 3m 4m FIG. Q3Answers:(a) k = 20 × 103 kN/m ;(b) δ By = 7.5 mm ↓ . 126 Dr. C. Caprani
  • Structural Analysis III8. References• Charlton, T.M., Analysis of Statically Indeterminate Frameworks, Longmans, 1961.• Charlton, T.M., Energy Principles in Theory of Structures, Oxford University Press, 1973.• Coates, R.C., Coutie, M.G. and Kong, F.K., Structural Analysis, Chapman and Hall, 1987.• Davies, G.A.O., Virtual Work in Structural Analysis, John Wiley & Sons, 1982.• Dym, C.L., Structural Modeling and Analysis, Cambridge University Press, 2005.• Gregory, M.S., Introduction to Extremum Principles, Butterworths, London, 1969.• Guarracino, F. and Walker, A., Energy Methods in Structural Mechanics, Thomas Telford, 1999.• Heyman, J., Beams and Framed Structures, 2nd Edn., Pergamon Press, 1974.• Kong, F.K., Prentis, J.M. and Charlton, T.M., ‘Principle of virtual work for a general deformable body – a simple proof’, The Structural Engineer, Vol. 61A, No. 6, 1983.• Neal, B.G., Structural Theorems and their Applications, Pergamon Press, 1964.• Thompson, F. and Haywood, G.G., Structural Analysis Using Virtual Work, Chapman and Hall, 1986.• Rajasekaran, S. and Sankarasubramanian, G., Computational Structural Mechanics, Prentis Hall of India, New Delhi, 2001.• Richards, T.H., Energy methods in Stress Analysis, Ellis Horwood, 1977.• Wallerstein, D.V. , A Variational Approach to Structural Analysis, Wiley, 2001. 127 Dr. C. Caprani
  • Structural Analysis III9. Appendix – Volume Integrals j j j1 j2 j l l l l 1 1 1 1 k jkl jkl ( j1 + 2 j2 ) kl jkl l 3 6 6 2 1 1 1 1 ( 2 j1 + j2 ) kl k jkl jkl jkl l 6 3 6 2 1 ⎡ j1 ( 2k1 + k2 ) + k1 k2 1 j ( k1 + 2k2 ) l 1 j ( 2k1 + k2 ) l 6⎣ 1 j ( k1 + k2 ) l 6 6 2 l j2 ( k1 + 2k2 ) ⎤ l ⎦ 1 1 1 k jkl jkl ( j1 + j2 ) kl jkl 2 2 2 l 1 ⎡ j1 ( l + b ) + 1 jk ( l + a ) 1 jk ( l + b ) 6⎣ 1 k jkl 6 6 2 a b j2 ( l + a ) ⎤ k ⎦ 5 1 1 2 k jkl jkl ( 3 j1 + 5 j2 ) kl jkl l 12 4 12 3 1 5 1 2 k jkl jkl ( 5 j1 + 3 j2 ) kl jkl 4 12 12 3 l 1 1 1 1 k jkl jkl ( j1 + 3 j2 ) kl jkl l 4 12 12 3 k 1 1 1 1 jkl jkl ( 3 j1 + j2 ) kl jkl l 12 4 12 3 1 1 1 2 jkl jkl ( j1 + j2 ) kl jkl k 3 3 3 3 l 128 Dr. C. Caprani