Virtual Work 3rd Year Structural Engineering

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1 Vrtual Work 3rd Year Structural Engneerng 2/ Dr. Coln Capran

2 Contents. Introducton General Background The Prncple of Vrtual Work Defnton Vrtual Dsplacements Vrtual Forces Smple Proof usng Vrtual Dsplacements Internal and External Vrtual Work Applcaton of Vrtual Dsplacements Rgd Bodes Deformable Bodes Problems Applcaton of Vrtual Forces Bass Deflecton of Trusses Deflecton of Beams and Frames Integraton of Bendng Moments Problems Vrtual Work for Indetermnate Structures General Approach Usng Vrtual Work to Fnd the Multpler Basc Example Propped Cantlever Indetermnate Trusses Indetermnate Frames Contnuous Beams Problems

3 6. Vrtual Work for Self-Stressed Structures Background Trusses Beams Frames Problems Past Exam Questons Summer Summer Summer Summer Summer Summer Summer Summer Semester 2, Semester 2, Semester 2, Semester 2, Semester 2, References Appendx Volume Integrals

4 . Introducton. General Vrtual Work s a fundamental theory n the mechancs of bodes. So fundamental n fact, that Newton s 3 equatons of equlbrum can be derved from t. Vrtual Work provdes a bass upon whch vectoral mechancs (.e. Newton s laws) can be lnked to the energy methods (.e. Lagrangan methods) whch are the bass for fnte element analyss and advanced mechancs of materals. Vrtual Work allows us to solve determnate and ndetermnate structures and to calculate ther deflectons. That s, t can acheve everythng that all the other methods together can acheve. Before startng nto Vrtual Work there are some background concepts and theores that need to be covered. 4

5 .2 Background Stran Energy and Work Done Stran energy s the amount of energy stored n a structural member due to deformaton caused by an external load. For example, consder ths smple sprng: We can see that as t s loaded by a gradually ncreasng force, F, t elongates. We can graph ths as: The lne OA does not have to be straght, that s, the consttutve law of the sprng s materal does not have to be lnear. 5

6 An ncrease n the force of a small amount, F results n a small ncrease n deflecton, y. The work done durng ths movement (force dsplacement) s the average force durng the course of the movement, tmes the dsplacement undergone. Ths s the same as the hatched trapezodal area above. Thus, the ncrease n work assocated wth ths movement s: U F F F y 2 Fy F y 2 F y (.) where we can neglect second-order quanttes. As y, we get: du F dy The total work done when a load s gradually appled from up to a force of F s the summaton of all such small ncreases n work,.e.: U y F dy (.2) Ths s the dotted area underneath the load-deflecton curve of earler and represents the work done durng the elongaton of the sprng. Ths work (or energy, as they are the same thng) s stored n the sprng and s called stran energy and denoted U. If the load-dsplacement curve s that of a lnearly-elastc materal then F ky where k s the constant of proportonalty (or the sprng stffness). In ths case, the dotted area under the load-deflecton curve s a trangle. 6

7 As we know that the work done s the area under ths curve, then the work done by the load n movng through the dsplacement the External Work Done, by: W e - s gven We Fy (.3) 2 We can also calculate the stran energy, or Internal Work Done, W I, by: U 2 y y F dy ky dy ky 2 Also, snce F ky, we then have: W I U ky y Fy 2 2 7

8 But ths s the external work done, W e. Hence we have: W e W (.4) I Whch we may have expected from the Law of Conservaton of Energy. Thus: The external work done by external forces movng through external dsplacements s equal to the stran energy stored n the materal. 8

9 Law of Conservaton of Energy For structural analyss ths can be stated as: Consder a structural system that s solated such t nether gves nor receves energy; the total energy of ths system remans constant. The solaton of the structure s key: we can consder a structure solated once we have dentfed and accounted for all sources of restrant and loadng. For example, to neglect the self-weght of a beam s problematc as the beam receves gravtatonal energy not accounted for, possbly leadng to collapse. In the sprng and force example, we have accounted for all restrants and loadng (for example we have gnored gravty by havng no mass). Thus the total potental energy of the system,, s constant both before and after the deformaton. In structural analyss the relevant forms of energy are the potental energy of the load and the stran energy of the materal. We usually gnore heat and other energes. Potental Energy of the Load Snce after the deformaton the sprng has ganed stran energy, the load must have lost potental energy, V. Hence, after deformaton we have for the total potental energy: U V ky 2 2 Fy (.5) In whch the negatve sgn ndcates a loss of potental energy for the load. 9

10 Prncple of Mnmum Total Potental Energy If we plot the total potental energy aganst y, equaton (.5), we get a quadratc curve smlar to: Consder a numercal example, wth the followng parameters, k kn/m and F kn gvng the equlbrum deflecton as y F k m. We can plot the followng quanttes: Internal Stran Energy, or Internal Work: U WI ky y 5 y 2 2 Potental Energy: V Fy y Total Potental Energy: External Work: 2 U V 5y y We Py y 5 y 2 2 and we get the followng plots (splt nto two for clarty):

11 Energy Total Potental Energy External Work Internal Work Equlbrum Deflecton, y Energy Total Potental Energy Stran Energy Potental Energy Equlbrum Deflecton, y From these graphs we can see that because the W e ncreases only lnearly, W I ncreases quadratcally wth y, whle W I always catches up wth be a non-zero equlbrum pont where We WI. W e, and there wll always

12 Admttedly, these plots are mathematcal: the deflecton of the sprng wll not take up any value; t takes that value whch acheves equlbrum. At ths pont we consder a small varaton n the total potental energy of the system. Consderng F and k to be constant, we can only alter y. The effect of ths small varaton n y s: 2 2 y y y k y y F y y ky Fy k 2y y F y k y ky F y k y 2 (.6) Smlarly to a frst dervate, for to be an extreme (ether maxmum or mnmum), the frst varaton must vansh: ky F y (.7) Therefore: ky F (.8) Whch we recognze to be the F. Thus equlbrum occurs when s an extreme. x Before ntroducng more complcatng maths, an example of the above varaton n equlbrum poston s the followng. Thnk of a shopkeeper testng an old type of scales for balance she slghtly lfts one sde, and f t returns to poston, and no large rotatons occur, she concludes the scales s n balance. She has mposed a 2

13 varaton n dsplacement, and fnds that snce no further dsplacement occurs, the structure was orgnally n equlbrum. Examnng the second varaton (smlar to a second dervate): k y (.9) We can see t s always postve and hence the extreme found was a mnmum. Ths s a partcular proof of the general prncple that structures take up deformatons that mnmze the total potental energy to acheve equlbrum. In short, nature s lazy! To summarze our fndngs: Every solated structure has a total potental energy; Equlbrum occurs when structures can mnmse ths energy; A small varaton of the total potental energy vanshes when the structure s n equlbrum. These concepts are brought together n the Prncple of Vrtual Work. 3

14 2. The Prncple of Vrtual Work 2. Defnton Based upon the Prncple of Mnmum Total Potental Energy, we can see that any small varaton about equlbrum must do no work. Thus, the Prncple of Vrtual Work states that: A body s n equlbrum f, and only f, the vrtual work of all forces actng on the body s zero. In ths context, the word vrtual means havng the effect of, but not the actual form of, what s specfed. Thus we can magne ways n whch to mpose vrtual work, wthout worryng about how t mght be acheved n the physcal world. We need to remnd ourselves of equlbrum between nternal and external forces, as well as compatblty of dsplacement, between nternal and external dsplacements for a very general structure: Equlbrum Compatblty 4

15 Vrtual Work There are two prncples that arse from consderaton of vrtual work, and we use ether as suted to the unknown quantty (force or dsplacement) of the problem under analyss. Prncple of Vrtual Dsplacements: Vrtual work s the work done by the actual forces actng on the body movng through a vrtual dsplacement. Ths means we solve an equlbrum problem through geometry, as shown: Prncple of Vrtual Dsplacements External Vrtual Work equals Internal Vrtual Work Real Force (e.g. Pont load) Vrtual Dsplacement (e.g. vertcal deflecton) Statcs (Equlbrum) Geometry (Compatblty) Real Force (e.g. bendng moment) Vrtual Dsplacement (e.g. curvature) Ether of these s the unknown quantty of nterest Vrtual Dsplacements: External compatble dsplacements Geometry Internal compatble dsplacements 5

16 Prncple of Vrtual Forces: Vrtual work s the work done by a vrtual force actng on the body movng through the actual dsplacements. Ths means we solve a geometry problem through equlbrum as shown below: Prncple of Vrtual Forces External Vrtual Work equals Internal Vrtual Work Vrtual Force (e.g. Pont load) Real Dsplacement (e.g. vertcal deflecton) Statcs (Equlbrum) Geometry (Compatblty) Vrtual Force (e.g. bendng moment) Real Dsplacement (e.g. curvature) The unknown quantty of nterest Internal Real Dsplacements : External real forces Statcs Internal real forces Consttutve relatons Internal real dsplacements 6

17 2.2 Vrtual Dsplacements A vrtual dsplacement s a dsplacement that s only magned to occur. Ths s exactly what we dd when we consdered the vanshng of the frst varaton of ; we found equlbrum. Thus the applcaton of a vrtual dsplacement s a means to fnd ths frst varaton of. So gven any real force, F, actng on a body to whch we apply a vrtual dsplacement. If the vrtual dsplacement at the locaton of and n the drecton of F s y, then the force F does vrtual work W F y. There are requrements on what s permssble as a vrtual dsplacement. For example, n the smple proof of the Prncple of Vrtual Work (to follow) t can be seen that t s assumed that the drectons of the forces appled to P reman unchanged. Thus: vrtual dsplacements must be small enough such that the force drectons are mantaned. The other very mportant requrement s that of compatblty: vrtual dsplacements wthn a body must be geometrcally compatble wth the orgnal structure. That s, geometrcal constrants (.e. supports) and member contnuty must be mantaned. In summary, vrtual dsplacements are not real, they can be physcally mpossble but they must be compatble wth the geometry of the orgnal structure and they must be small enough so that the orgnal geometry s not sgnfcantly altered. As the deflectons usually encountered n structures do not change the overall geometry of the structure, ths requrement does not cause problems. 7

18 2.3 Vrtual Forces So far we have only consdered small vrtual dsplacements and real forces. The vrtual dsplacements are arbtrary: they have no relaton to the forces n the system, or ts actual deformatons. Therefore vrtual work apples to any set of forces n equlbrum and to any set of compatble dsplacements and we are not restrcted to consderng only real force systems and vrtual dsplacements. Hence, we can use a vrtual force system and real dsplacements. Obvously, n structural desgn t s these real dsplacements that are of nterest and so vrtual forces are used often. A vrtual force s a force magned to be appled and s then moved through the actual deformatons of the body, thus causng vrtual work. So f at a partcular locaton of a structure, we have a deflecton, y, and mpose a vrtual force at the same locaton and n the same drecton of F we then have the vrtual work W y F. Vrtual forces must form an equlbrum set of ther own. For example, f a vrtual force s appled to the end of a sprng there wll be vrtual stresses n the sprng as well as a vrtual reacton. 8

19 2.4 Smple Proof usng Vrtual Dsplacements We can prove the Prncple of Vrtual Work qute smply, as follows. Consder a partcle P under the nfluence of a number of forces F, force, F R. Apply a vrtual dsplacement of, n F whch have a resultant y to P, movng t to P, as shown: The vrtual work done by each of the forces s: W F y F y F y n n R R Where y s the vrtual dsplacement along the lne of acton of F and so on. Now f the partcle P s n equlbrum, then the forces F,, n F have no resultant. That s, there s no net force. Hence we have: W y F y F y (2.) R n n Provng that when a partcle s n equlbrum the vrtual work of all the forces actng on t sum to zero. Conversely, a partcle s only n equlbrum f the vrtual work done durng a vrtual dsplacement s zero. 9

20 2.5 Internal and External Vrtual Work Consder the sprng we started wth, as shown agan below. Frstly t s unloaded. Then a load, F, s appled, causng a deflecton y. It s now n ts equlbrum poston and we apply a vrtual dsplacement, y to the end of the sprng, as shown: A free body dagram of the end of the sprng s: Thus the vrtual work done by the two forces actng on the end of the sprng s: W F y ky y If the sprng s to be n equlbrum we must then have: W F y ky y F y ky y F ky 2

21 That s, the force n the sprng must equal the appled force, as we already know. However, f we defne the followng: External vrtual work, W F y ; Internal vrtual work, W ky y ; I E We then have: W I W W E Thus: W E W (2.2) I whch states that the external vrtual work must equal the nternal vrtual work for a structure to be n equlbrum. It s n ths form that the Prncple of Vrtual Work fnds most use. Of sgnfcance also n the equatng of nternal and external vrtual work, s that there are no requrements for the materal to have any partcular behavour. That s, vrtual work apples to all bodes, whether lnearly-elastc, elastc, elasto-plastc, plastc etc. Thus the prncple has more general applcaton than most other methods of analyss. Internal and external vrtual work can arse from ether vrtual dsplacements or vrtual forces. 2

22 3. Applcaton of Vrtual Dsplacements 3. Rgd Bodes Bass Rgd bodes do not deform and so there s no nternal vrtual work done. Thus: W W E F y W I (3.) A smple applcaton s to fnd the reactons of statcally determnate structures. However, to do so, we need to make use of the followng prncple: Prncple of Substtuton of Constrants Havng to keep the constrants n place s a lmtaton of vrtual work. However, we can substtute the restranng force (.e. the reacton) n place of the restrant tself. That s, we are turnng a geometrc constrant nto a force constrant. Ths s the Prncple of Substtuton of Constrants. We can use ths prncple to calculate unknown reactons:. Replace the reacton wth ts approprate force n the same drecton (or sense); 2. Impose a vrtual dsplacement on the structure; 3. Calculate the reacton, knowng W. 22

23 Reactons of Determnate and Indetermnate Structures For statcally determnate structures, removng a restrant wll always render a mechansm (or rgd body) and so the reactons of statcally determnate structures are easly obtaned usng vrtual work. For ndetermnate structures, removng a restrant does not leave a mechansm and hence the vrtual dsplacements are harder to establsh snce the body s not rgd. 23

24 Example Determne the reactons for the followng beam: Followng the Prncple of Substtuton of Constrants, we replace the geometrc constrants (.e. supports), by ther force counterparts (.e. reactons) to get the freebody-dagram of the whole beam: Next, we mpose a vrtual dsplacement on the beam. Note that the dsplacement s completely arbtrary, and the beam remans rgd throughout: 24

25 In the above fgure, we have mposed a vrtual dsplacement of mposed a vrtual rotaton of A about A. The equaton of vrtual work s: ya at A and then W W E F y W I Hence: V y P y V y A A C B B Relatng the vrtual movements of ponts B and C to the vrtual dsplacements gves: V y P y a V y L A A A A B A A And rearrangng gves: V V P y V L Pa A B A B A 25

26 And here s the power of vrtual work: snce we are free to choose any value we want for the vrtual dsplacements (.e. they are completely arbtrary), we can choose and, whch gves the followng two equatons: A y A V V P y A B A V V P A B V L Pa B A V L Pa B But the frst equaton s just Fy whlst the second s the same as M about A. Thus equlbrum equatons occur naturally wthn the vrtual work framework. Note also that the two choces made for the vrtual dsplacements correspond to the followng vrtual dsplaced confguratons: Draw them 26

27 Example 2 For the followng truss, calculate the reacton V C : Frstly, set up the free-body-dagram of the whole truss: 27

28 Next we release the constrant correspondng to reacton V C and replace t by the unknown force V C and we apply a vrtual dsplacement to the truss to get: Hence the vrtual work done s: W W y VC y 2 V 5 kn E C W I Note that the reacton s an external force to the structure, and that no nternal vrtual work s done snce the members do not undergo vrtual deformaton. The truss rotates as a rgd body about the support A. Problem: Fnd the horzontal reactons of the truss. 28

29 3.2 Deformable Bodes Bass For a vrtual dsplacement we have: W W E W F y P e I In whch, for the external vrtual work, F represents an externally appled force (or moment) and y ts vrtual dsplacement. And for the nternal vrtual work, P represents the nternal force n member and e ts vrtual deformaton. Dfferent stress resultants have dfferent forms of nternal work, and we wll examne these. Lastly, the summatons reflect the fact that all work done must be accounted for. Remember n the above, each the dsplacements must be compatble and the forces must be n equlbrum, summarzed as: Set of forces n equlbrum F y P e Set of compatble dsplacements 29

30 These dsplacements are completely arbtrary (.e. we can choose them to sut our purpose) and bear no relaton to the forces above. For example, the smple truss shown has a set of forces n equlbrum along wth ts actual deformed shape. Also shown s an arbtrary set of permssble compatble dsplacements: Loadng and dmensons Equlbrum forces and actual deformaton Compatble set of dsplacements 3

31 Internal Vrtual Work by Axal Force Members subject to axal force may have the followng: real force by a vrtual dsplacement: W P e I vrtual force by a real dsplacement: W e P I We have avoded ntegrals over the length of the member snce we wll only consder prsmatc members. 3

32 Internal Vrtual Work n Bendng The nternal vrtual work done n bendng s one of: real moment by a vrtual curvature: W I M vrtual moment by a real curvature: W I M The above expressons are vald at a sngle poston n a beam. When vrtual rotatons are requred along the length of the beam, the easest way to do ths s by applyng vrtual loads that n turn cause vrtual moments and hence vrtual curvatures. We must sum all of the real moments by vrtual curvatures along the length of the beam, hence we have: L W M dx M dx I x x x x L M x M x x dx L 32

33 Internal Vrtual Work n Shear At a sngle pont n a beam, the shear stran,, s gven by: V GA v where V s the appled shear force, G s the shear modulus and A v s the cross-secton area effectve n shear whch s explaned below. The nternal vrtual work done n shear s thus: real shear force by a vrtual shear stran: V WI V V GA v vrtual shear force by a real shear stran: V WI V V GA v In these expressons, the area of cross secton effectve n shear comes about because the shear stress (and hence stran) s not constant over a cross secton. The stress V A v s an average stress, equvalent n work terms to the actual uneven stress over the full cross secton, A. We say Av A k where k s the shear factor for the cross secton. Some values of k are.2 for rectangular sectons,. for crcular sectons, and 2. for thn-walled crcular sectons. The above expressons are vald at a sngle poston n a beam and must be ntegrated along the length of the member as was done for moments and rotatons. 33

34 Internal Vrtual Work n Torson At a sngle pont n a member, the twst,, s gven by: T GJ where T s the appled torque, G s the shear modulus, J s the polar moment of nerta. The nternal vrtual work done n torson s thus: real torque by a vrtual twst: T WI T T GJ vrtual torque by a real twst: T WI T T GJ Once agan, the above expressons are vald at a sngle poston n a beam and must be ntegrated along the length of the member as was done for moments and rotatons. Note the smlarty between the expressons for the four nternal vrtual works. 34

35 Example 3 For the beam of Example (shown agan), fnd the bendng moment at C. To solve ths, we want to mpose a vrtual dsplacement confguraton that only allows the unknown of nterest,.e. M C, to do any work. Thus choose the followng: Snce portons AC and CB reman straght (or unbent) no nternal vrtual work s done n these sectons. Thus the only nternal work s done at C by the beam movng through the rotaton at C. Thus: 35

36 W W E W P y M C C C I Bu the rotaton at C s made up as: C CA CB yc yc a b a b y ab C But a b L, hence: L P y M y ab Pab M C L C C C M V b Pa L b. We can verfy ths from the reactons found prevously: C B 36

37 Sgn Conventon for Rotatons When mposng a vrtual rotaton, f the sde that s already n tenson due to the real force system elongates, then postve vrtual work s done. 37

38 Example 4 Calculate the force F n the truss shown: To do ths, we ntroduce a vrtual dsplacement along the length of member. We do ths so that member 2 does not change length durng ths vrtual dsplacement and so does no vrtual work. Note also that compatblty of the truss s mantaned. For example, the members stll meet at the same jont, and the support condtons are met. 38

39 The vrtual work done s then: W W E W y F y 2 F 5 2 kn 2 I Note some ponts on the sgns used:. Negatve external work s done because the kn load moves upwards;.e. the reverse drecton to ts acton. 2. We assumed member to be n compresson but then appled a vrtual elongaton whch lengthened the member thus reducng ts nternal vrtual work. Hence negatve nternal work s done. 3. We ntally assumed F to be n compresson and we obtaned a postve answer confrmng our assumpton. Problem: Investgate the vertcal and horzontal equlbrum of the loaded jont by consderng vertcal and horzontal vrtual dsplacements separately. 39

40 3.3 Problems. For the truss shown, calculate the vertcal reacton at C and the forces n the members, usng vrtual work. 2. For the truss shown, fnd the forces n the members, usng vrtual work: 4

41 3. Usng vrtual work, calculate the reactons for the beams shown, and the bendng moments at salent ponts. Beam Beam 2 Beam 3 4

42 4. Applcaton of Vrtual Forces 4. Bass When vrtual forces are appled, we have: W W E W y F e P I And agan note that we have an equlbrum set of forces and a compatble set of dsplacements: Set of compatble dsplacements y F e P Set of forces n equlbrum In ths case the dsplacements are the real dsplacements that occur when the structure s n equlbrum and the vrtual forces are any set of arbtrary forces that are n equlbrum. 42

43 4.2 Deflecton of Trusses Illustratve Example For the truss shown below, we have the actual dsplacements shown wth two possble sets of vrtual forces. Actual Loadng and (Compatble) Dsplacements 43

44 Vrtual Force (Equlbrum) Systems In ths truss, we know we know:. The forces n the members (got from vrtual dsplacements or statcs); 2. Thus we can calculate the member extensons, e as: e PL EA 3. Also, as we can choose what our vrtual force F s (usually unty), we have: W W E W y F e P PL y F P EA I 4. Snce n ths equaton, y s the only unknown, we can calculate the deflecton of the truss. 44

45 Example 5 Gven that E = 2 kn/mm 2 and A = mm 2 for each member, for the truss shown below, calculate the vertcal and horzontal deflecton of jont B. In these problems we wll always choose F. Hence we apply a unt vrtual force to jont B. We apply the force n the drecton of the deflecton requred. In ths way no work s done along deflectons that are not requred. Hence we have: For horzontal deflecton For vertcal deflecton 45

46 The forces and elongatons of the truss members are: Member AB: Member BC: P e P e AB AB BC BC 5 kn mm 3 kn mm Note that by takng tenson to be postve, elongatons are postve and contractons are negatve. Horzontal Deflecton: W W E W y F e P y y 4.5 mm I AB BC Vertcal Deflecton: y F e P 5 3 y AB 4 y 8.4 mm BC Note that the sgn of the result ndcates whether the deflecton occurs n the same drecton as the appled force. Hence, jont B moves 4.5 mm to the left. 46

47 Example 6 Determne the vertcal and horzontal deflecton of jont D of the truss shown. Take E = 2 kn/mm 2 and member areas, A = mm 2 for all members except AE and BD where A = 2 mm 2. The elements of the vrtual work equaton are: Compatble deformatons: The actual dsplacements that the truss undergoes; Equlbrum set: The external vrtual force appled at the locaton of the requred deflecton and the resultng nternal member vrtual forces. Frstly we analyse the truss to determne the member forces n order to calculate the actual deformatons of each member: 47

48 Next, we apply a unt vrtual force n the vertcal drecton at jont D. However, by lnear superposton, we know that the nternal forces due to a unt load wll be /5 tmes those of the 5 kn load. For the horzontal deflecton at D, we apply a unt horzontal vrtual force as shown: Equatons of Vrtual Work In whch: W W E DV DH W y F e P y y I PL EA PL EA P are the forces due to the 5 kn load; P P P are the vrtual forces due to the unt vrtual force appled n the vertcal drecton at D: 2 P P 5 2 P are the vrtual forces due to the unt vrtual force n the horzontal drecton at D. 48

49 Usng a table s easest because of the larger number of members: Member AB AE AF BC BD BE CD DE EF L A P P 5 2 P P PL A P PL (mm) (mm 2 ) (kn) (kn) (kn) (kn/mm) kn (kn/mm) kn A -3-6 P E s left out because t s common. Returnng to the equatons, we now have: y DV y DV PL P E A mm 2 Whch ndcates a downwards deflecton and for the horzontal deflecton: y DH y DH PL P E A mm 2 2 The sgn ndcates that t s deflectng to the left. 49

50 4.3 Deflecton of Beams and Frames Example 7 Usng vrtual work, calculate the deflecton at the centre of the beam shown, gven that s constant. To calculate the deflecton at C, we wll be usng vrtual forces. Therefore the two relevant sets are: Compatblty set: the actual deflecton at C and the rotatons that occur along the length of the beam; Equlbrum set: a unt vrtual force appled at C whch s n equlbrum wth the nternal vrtual moments t causes. Compatblty Set: The external deflecton at C s what s of nterest to us. To calculate the rotatons along the length of the beam we have: L M x dx x Hence we need to establsh the bendng moments along the beam: 5

51 For AC the bendng moment s gven by (and smlarly for B to C): M x P x 2 Equlbrum Set: As we choose the value for F, we are only left to calculate the vrtual moments: For AC the nternal vrtual moments are gven by: M x 2 x 5

52 Vrtual Work Equaton W W E W y F M I Substtute n the values we have for the real rotatons and the vrtual moments, and use the fact that the bendng moment dagrams are symmetrcal: y 2 L 2 L 2 2 P y x x dx 2 2 2P 4 L 2 2 x dx 3 P x P L PL 48 M x M x dx L 2 Whch s a result we expected. 52

53 Example 8 Fnd the vertcal and horzontal dsplacement of the prsmatc curved bar cantlever shown: Even though t s curved, from the statcs, the bendng moment at any pont of course remans force by dstance, so from the followng dagram: 53

54 At any angle, we therefore have: M P R Rcos PR cos To fnd the dsplacements, we follow our usual procedure and place a unt load at the locaton of, and n the drecton of, the requred dsplacement. Vertcal Dsplacement The vrtual bendng moment nduced by the vertcal unt load shown, s related to that for P and s thus: M R cos Thus our vrtual work equatons are: 54

55 W W E M ds BV W I 2 M M Rd In whch we have used the relaton ds R d to change the ntegraton from along the length of the member to around the angle. Next we ntroduce our equatons for the real and vrtual bendng moments: 2 PR cos BV R cos Rd 3 2 PR 2 cos d PR PR Horzontal Dsplacement In ths case, the vrtual bendng moment s: M Rsn 55

56 Thus the vrtual work equatons gve: 2 PR cos BH Rsn Rd 3 2 PR sn sn cos d 3 2 PR sn 2 sn d 2 PR PR

57 4.4 Integraton of Bendng Moments We are often faced wth the ntegraton of beng moment dagrams when usng vrtual work to calculate the deflectons of bendng members. And as bendng moment dagrams only have a lmted number of shapes, a table of volume ntegrals s used: Ths table s at the back page of these notes for ease of reference. Example 9 Usng the table of volume ntegrals, verfy the answer to Example 7. In ths case, the vrtual work equaton becomes: y 2 L 2 2 y M x M 2 PL L L PL 48 M x M x dx shape shape x In whch the formula jkl s used from the table. 3 57

58 Example Summer 7 Part (a) For the frame shown, determne the horzontal deflecton of jont C. Neglect axal effects n the members and take knm. FIG. Q3(a) Frstly we establsh the real bendng moment dagram: Next, as usual, we place a unt load at the locaton of, and n the drecton of, the requred dsplacement: 58

59 Now we have the followng for the vrtual work equaton: W W E M ds CH W I M M ds Next, usng the table of volume ntegrals, we have: M M ds AB BC Hence: CH 84.6 mm 3 59

60 4.5 Problems. Determne the vertcal deflecton of jont E of the truss shown. Take E = 2 kn/mm 2 and member areas, A = mm 2 for all members except AC where A = 2 mm 2. (Ans. 2.2 mm). FIG. Q3 2. Determne the vertcal and horzontal deflecton of jont C of the truss shown. Take E = kn/mm 2 and member areas, A = mm 2 for all members except AC where A = 2 mm 2 and CE where A = 25 mm 2. (Ans. 32/7 mm horzontal, 74/7 mm vertcal). FIG. Q3(a) 6

61 3. Determne the horzontal deflecton of jont A and the vertcal deflecton of jont B of the truss shown. Take E = 2 kn/mm 2 and member areas, A = mm 2 for all members except BD where A = 2 mm2 and AB where A = 25 mm 2. (Ans. 5.3 mm; mm) FIG. Q2(a) 4. Determne the vertcal and horzontal deflecton of jont C of the truss shown. Take E = 2 kn/mm 2 and member areas, A = 5 mm 2 for all members except AC where A = 52 mm2 and CD where A = 25 mm 2. (Ans mm;.2 mm) FIG. Q3 6

62 5. Determne the vertcal deflecton of jont D of the truss shown. Take E = 2 kn/mm 2 and member areas, A = mm 2 for all members except BC where A = 2 mm 2. (Ans. 24 mm) FIG. Q3 6. Verfy that the deflecton at the centre of a smply-supported beam under a unformly dstrbuted load s gven by: 4 5wL C Show that the flexural deflecton at the tp of a cantlever, when t s subjected to a pont load on the end s: 3 PL B 3 8. Show that the rotaton that occurs at the supports of a smply supported beam wth a pont load n the mddle s: 2 PL C 6 62

63 9. Show that the vertcal and horzontal deflectons at B of the followng frame are: By 3 3 PR PR ; Bx 2 4. For the frame of Example, usng vrtual work, verfy the followng dsplacements n whch the followng drectons are postve: vertcal upwards; horzontal to the rght, and; clockwse: Rotaton at A s 76.3 ; Vertcal deflecton at B s 346 ; Horzontal deflecton at D s Rotaton at D s ; 63

64 . For the cantlever beam below, show that the vertcal deflecton of B, allowng for both flexural and shear deformaton, s PL 3 PL. 3 GAv 2. For the beam below, show that the vertcal deflecton of D s zero and that that of C s 85 3 : 64

65 Problems 3 to 7 are consdered Genus Level! 3. Show that the vertcal deflecton at the centre of a smply supported beam subject to UDL, allowng for both flexural and shear deformaton, s 4 2 5wL wl GAv 4. For the frame shown, show that the vertcal deflecton of pont D s 4594 Neglect axal deformaton and take knm.. 65

66 5. For the frame shown, determne the vertcal deflecton of C and the rotaton at D about the global x axs. Take the secton and materal propertes as follows: A 66 mm E 25 kn/mm G kn/mm I mm J 74 mm. (Ans mm ;.76 rads ACW) 6. Determne the horzontal and vertcal deflectons at B n the followng frame. Neglect shear and axal effects wR 3 wr (Ans. Bx ; By ) 2 66

67 7. For the frame shown, neglectng shear effects, determne the vertcal deflecton of ponts B and A. (Ans. By P wab 3 ; 3 Ay P wab 3 Pa 3 wa 4 wa 2 ab Pa ) GJ 67

68 5. Vrtual Work for Indetermnate Structures 5. General Approach Usng the concept of compatblty of dsplacement, any ndetermnate structure can be splt up nto a prmary and reactant structures. In the case of a one-degree ndetermnate structure, we have: Fnal = Prmary + Reactant At ths pont the prmary structure can be analysed. However, we further break up the reactant structure, usng lnear superposton: Reactant = Multpler Unt Reactant We summarze ths process as: M M M M s the force system n the orgnal structure (n ths case moments); M s the prmary structure force system; M s the unt reactant structure force system. 68

69 For a truss, the procedure s the same: Fnal = Prmary + Reactant Reactant = Multpler Unt Reactant The fnal system forces are: P P P The prmary structure can be analysed, as can the unt reactant structure. Therefore, the only unknown s the multpler,. We use vrtual work to calculate the multpler. 69

70 5.2 Usng Vrtual Work to Fnd the Multpler We must dentfy the two sets for use: Dsplacement set: We use the actual dsplacements that occur n the real structure; Equlbrum set: We use the unt reactant structure s set of forces as the equlbrum set. We do ths, as the unt reactant s always a determnate structure and has a confguraton smlar to that of the dsplacement set. A typcal example of these sets for an ndetermnate structure s shown: Compatblty Set Internal and external real dsplacements lnked through consttutve relatons Equlbrum Set Internal (bendng moments) and external vrtual forces are lnked through equlbrum Note that the Compatblty Set has been chosen as the real dsplacements of the structure. The vrtual force (equlbrum) set chosen s a statcally determne substructure of the orgnal structure. Ths makes t easy to use statcs to determne the nternal vrtual forces from the external vrtual force. Lastly, note that the external vrtual force does no vrtual work snce the external real dsplacement at B s zero. Ths makes the rght hand sde of the equaton zero, allowng us to solve for the multpler. 7

71 The vrtual work equaton (wrtten for trusses) gves: W W E W y F e P I PL EA P There s zero external vrtual work. Ths s because the only vrtual force appled s nternal; no external vrtual force appled. Also note that the real deformatons that occur n the members are n terms of P, the unknown fnal forces. Hence, substtutng P P P (where s now used to ndcate vrtual nature): P P L P EA P L P L EA EA 2 P P P L L EA EA P P For beams and frames, the same development s: W W E W y F M I M M Where agan there s no external dsplacement of the vrtual force. 7

72 Note that we use the contour ntegral symbol smply to ndcate that we ntegrate around the structure, accountng for all members n the beam/frame (.e. ntegrate along the length of each member separately, and then sum the results). Also, substtute M M M to get: M M M dx 2 M M M dx dx Thus n both bases we have a sngle equaton wth only one unknown,. We can establsh values for the other two terms and then solve for and the structure as a whole. 72

73 5.3 Basc Example Propped Cantlever Example Calculate the bendng moment dagram for the followng prsmatc propped cantlever. Fnd also the deflecton under the pont load. Soluton to be done n class. 73

74 5.4 Indetermnate Trusses Example 2 Calculate the forces n the truss shown and fnd the horzontal deflecton at C. Take EA to be 4 kn for all members. Solve for the Truss Member Forces Choose member AC as the redundant: Next analyse for the P and P force systems. 74

75 P System P System Usng a table: Member AB BC CD AC BD L P P P 2 P P L L (mm) (kn) (kn) /5-3/5-4/ Hence: And so P P L EA P EA EA 4 4 EA L 75

76 The remanng forces are obtaned from the compatblty equaton: Member AB BC CD AC BD P P P P P (kn) (kn) (kn) /5-3/5-4/ Note that the redundant always has a force the same as the multpler. Calculate the Horzontal Deflecton at C To calculate the horzontal deflecton at C, usng vrtual work, the two relevant sets are: Compatblty set: the actual deflecton at C and the real deformatons that occur n the actual structure; Equlbrum set: a horzontal unt vrtual force appled at C to a porton of the actual structure, yet ensurng equlbrum. We do not have to apply the vrtual force to the full structure. Rememberng that the only requrement on the vrtual force system s that t s n equlbrum; choose the force systems as follows: 76

77 Thus we have: W W E CH CH W y F e P y y I PL EA P mm Because we have chosen only two members for our vrtual force system, only these members do work and the calculaton s greatly smplfed. 77

78 5.5 Indetermnate Frames Example 3 For the frame shown, calculate the reactons and draw the bendng moment dagram. Determne the horzontal deflecton at jont C. Take = 4 3 knm 2, constant for all members. Break the frame up nto ts reactant and prmary structures: M System = M System + M System Establsh the M and M force systems: 78

79 M System M System Apply the vrtual work equaton: M 2 M M dx dx We wll be usng the table of volume ntegrals to qucken calculatons. Therefore we can only consder lengths of members for whch the correct shape of bendng moment dagram s avalable. Also, we must choose sgn conventon: we consder tenson on the outsde of the frame to be postve. As each term has several components we consder them separately: Term - M M dx : AD: we have graphcally, and from the volume ntegral table: j k k2l

80 DB: Smlarly: jkl BC: Hence: jkl M M dx 234 Term 2: 2 M dx jkl AB jkl 3 BC AB 3 36 BC Note that Term 2 s always easer to calculate as t s only ever made up of straght lne bendng moment dagrams. Thus we have: 8

81 M 2 M M dx dx And so: Thus the vertcal reacton at C s 65. kn upwards. Wth ths nformaton we can solve for the moments usng remanng reactons to get: M M M (or by just usng statcs) and the 8

82 To calculate the horzontal deflecton at C usng vrtual work, the two relevant sets are: Compatblty set: the actual deflecton at C and the real deformatons (rotatons) that occur n the actual structure; Equlbrum set: a horzontal unt vrtual force appled at C to a determnate porton of the actual structure. Choose the followng force system as t s easly solved: Thus we have: W W E CH W y F M y I M x M x dx 82

83 The real bendng moment dagram, M, s awkward to use wth the ntegral table. Rememberng that M M M smplfes the calculaton by usng: To gve: And so usng the table formulae, we have: 83

84 M x M x dx BC DB AD Whch gves us: Cx M x M x dx mm The answer s postve, ndcatng that the structure moves to the rght at C: the same drecton n whch the unt vrtual force was appled. We could have chosen any other statcally determnate sub-structure and obtaned the same result. Some sub-structures wll make the analyss easer to perform. Exercse: Verfy that the same deflecton s obtaned by usng a sub-structure obtaned by removng the vertcal support at C and applyng the vrtual force. Does ths substructure lead to an easer analyss for the deflecton? 84

85 Example 4 For the frame shown, calculate the reactons and draw the bendng moment dagram. Determne the vertcal deflecton at jont A. Take = 2 3 knm 2, constant for all members. To be done n class. 85

86 5.6 Contnuous Beams Bass In ths secton we wll ntroduce structures that are more than degree statcally ndetermnate. We do so to show that vrtual work s easly extensble to multplyndetermnate structures, and also to gve a method for such beams that s easly worked out, and put nto a spreadsheet. Consder the example 3-span beam. It s 2 degrees ndetermnate, and so we ntroduce 2 hnges (.e. moment releases) at the support locatons and unt reactant moments n ther place, as shown: 86

87 Alongsde these systems, we have ther bendng moment dagrams: Usng the dea of the multpler and superposton agan, we can see that: M M M M 2 2 The vrtual work equaton s: W W E W I M 87

88 There s no external vrtual work done snce the unt moments are appled nternally. Snce we have two vrtual force systems n equlbrum and one real compatble system, we have two equatons: M M and M M 2 For the frst equaton, expandng the expresson for the real moment system, M: 2 M M 2 M M M M M M M M 2 2 In whch we ve dropped the contour ntegral t beng understood that we sum for all members. Smlarly for the second vrtual moments, we have: M M M M M M Thus we have two equatons and so we can solve for and 2. Usually we wrte ths as a matrx equaton: 2 M M M M M M M M M M M M Each of the ntegral terms s easly found usng the ntegral tables, and the equaton solved. 88

89 Smlarly the (smpler) equaton for a 2-span beam s: M M M M And the equaton for a 4-span beam s: 2 M M M M M M M M M M M M M M M M M M M M The dagram shows why there are no terms nvolvng adjacent spans that have non-zero ntegrals: M M 3, and why t s only 89

90 Example 5 Usng vrtual work, fnd the bendng moment dagram for the followng beam: Proceedng as descrbed above, we ntroduce releases (hnges) at the support ponts, apply the unt vrtual moments, and fnd the correspondng bendng moment dagrams: Next we need to evaluate each term n the matrx vrtual work equaton. We ll take the two hard ones frst: 9

91 M M AB BC Note that snce M for span CD, there s no term for t above. Smlarly, for the followng evaluaton, there wll be no term for span AB. Note also the 2 term for member BC ths could be easly overlooked. M M BC CD The followng ntegrals are more straghtforward snce they are all trangles: M M AB BC 2 M M BC 6 M M 2 property of multplcaton..5, snce t s equal to 2 M M by the commutatve 2 2 M M BC CD Wth all the terms evaluated, enter them nto the matrx equaton: 9

92 2 M M M M M M M M M M M M To gve: And solve, as follows: Now usng our superposton equaton for moments, M M M M, we can show that the multplers are just the hoggng support moments: 2 2 M M B C knm knm From these we get the fnal BMD: 92

93 93

94 Spreadsheet Soluton A smple spreadsheet for a 3-span beam wth centre-span pont load and UDL capabltes, showng Example, s: Span Span 2 Span 3 VW Equatons Length m x * alpha = PL kn alpha2 UDL kn/m 3 BMDs PL * alpha = 95. UDL alpha alpha = * 95. M x M Span Span 2 Span 3 Totals alpha PL UDL alpha = Total alpha M x M Span Span 2 Span 3 Total Md-Span and Support Moments M2 x M Span Span 2 Span 3 Mab 37.2 knm Total Mb knm Mbc -3. knm M x M2 Span Span 2 Span 3 Mc -7.7 knm PL... Mcd 58.4 knm UDL Total M x M2 Span Span 2 Span 3 Total M2 x M2 Span Span 2 Span 3 Total

95 5.7 Problems. For the truss shown, calculate the vertcal deflecton at C when member EB s not present and when t s present. Members AB, DC and DE have EA = 6 3 kn; members EB, BC and AD have EA = 3 kn, and; member BD has EA = 8 3 kn. (Ans. 5 mm and.28 mm) 2. Autumn 27 For the truss shown, determne the force n each member and determne the horzontal deflecton of jont B. Take 3 EA 2 kn for all members. (Ans. Choosng BD: 5 2 ; 2 mm) FIG. Q3 95

96 3. For the frame shown, calculate the bendng moment dagram, and verfy the followng deflectons: Bx and Cy. (Ans. M 3. knm ) B 4. Summer 27, Part (b): For the frame shown, draw the bendng moment dagram and determne the horzontal deflecton of jont C. Neglect axal effects n the members and take 9. mm) knm. (Ans. H 75.8 kn ; D FIG. Q3(b) 96

97 5. For the beam of Example, show that the deflectons at the centre of each span, takng downwards as postve, are: Span AB: 4. ; Span BC: 23.9 ; Span CD: Usng vrtual work, analyse the followng prsmatc beam to show that the support moments are bendng moment dagram: M 5.3 knm and M 3. knm, and draw the B C 97

98 6. Vrtual Work for Self-Stressed Structures 6. Background Introducton Self-stressed structures are structures that have stresses nduced, not only by external loadng, but also by any of the followng: Temperature change of some of the members (e.g. solar gan); Lack of ft of members from fabrcaton: o Error n the length of the member; o Ends not square and so a rotatonal lack of ft; Incorrect support locaton from mperfect constructon; Non-rgd (.e. sprng) supports due to mperfect constructon. Snce any form of fabrcaton or constructon s never perfect, t s very mportant for us to know the effect (n terms of bendng moment, shear forces etc.) that such errors, even when small, can have on the structure. Here we ntroduce these sources, and examne ther effect on the vrtual work equaton. Note that many of these sources of error can exst concurrently. In such cases we add together the effects from each source. 98

99 Temperature Change The source of self-stressng n ths case s that the temperature change causes a member to elongate: T L L T where s the coeffcent of lnear thermal expanson (change n length, per unt length, per degree Celsus), L s the orgnal member length and temperature change. T s the Snce temperature changes change the length of a member, the nternal vrtual work s affected. Assumng a truss member s beng analysed, we now have changes n length due to force and temperature, so the total change n length of the member s: PL e L T EA Hence the nternal vrtual work for ths member s: W e P I PL LT P EA 99

100 Lnear Lack of Ft For a lnear lack of ft, the member needs to be artfcally elongated or shortened to ft t nto place, thus ntroducng addtonal stresses. Ths s denoted: L Consderng a truss member subject to external loadng, the total change n length wll be the deformaton due to loadng and the lnear lack of ft: PL e L EA Hence the nternal vrtual work for ths member s: W e P I PL L P EA

101 Rotatonal Lack of Ft A rotatonal lack of ft, whch apples to frames only, occurs when the end of a member s not square. Thus the member needs to be artfcally rotated to get t nto place, as shown below. Ths s denoted as: Consderng a frame member whch has a lack of ft, and a vrtual moment M at the same pont, then the nternal vrtual work done at ths pont s: W M I Ths must be added to the other forms of nternal vrtual work. Not also that the sgns must be carefully chosen so that the vrtual moment closes the gap we wll see ths more clearly n an example.

102 Errors n Support Locaton The support can be msplaced horzontally and/or vertcally. It s denoted: S A msplaced support affects the external movements of a structure, and so contrbutes to the external vrtual work. Denotng the vrtual reacton at the support, n the drecton of the msplacement as R, then we have: W R e S 2

103 Sprng Supports For sprng supports we wll know the sprng constant for the support, denoted: k S Snce movements of a support are external, sprng support movements affect the external vrtual work. The real dsplacement S that occurs s: S RkS In whch R s the real support reacton n the drecton of the sprng. Further, snce R wll be known n terms of the multpler and vrtual reacton, R, we have: R R R Hence: S R R ks And so the external work done s: W e R S R R k S R 2 k S The only unknown here s whch s solved for from the vrtual work equaton. 3

104 6.2 Trusses Example 6 Here we take the truss of Example 6 and examne the effects of: Member ED was found to be 5 mm too long upon arrval at ste; Member AB s subject to a temperature ncrease of + C. For ths truss, E = 2 kn/mm 2; member areas, A = mm 2 for all members except AE and BD where A = 2 mm 2, and; coeffcent of expanson s 6 - C. The change n length due to the temperature change s: T 6 2 L L T 2 mm Vertcal Dsplacement of Jont D The vrtual work equaton s now: 4

105 W W E W y F e P I y P P P EA PL DV 5 ED 2 AB In Example 6 we establshed varous values n ths equaton: PL P EA 6.5 P ED P AB Hence we have: y DV y DV 3.5 mm Horzontal Dsplacement of Jont D: Smlarly, copyng values from Example 6, we have: W W E I PL DH 5 ED 2 AB DH W y F e P y P P P EA y mm to the rght 5

106 Example 7 Here we use the truss of Example and examne, separately, the effects of: Member AC was found to be 3.6 mm too long upon arrval on ste; Member BC s subject to a temperature reducton of -5 C; Support D s surveyed and found to st 5 mm too far to the rght. Take EA to be 4 kn for all members and the coeffcent of expanson to be 6 - C. Error n Length: To fnd the new multpler, we nclude ths effect n the vrtual work equaton: W W E W y F e P I PL P P EA L AC But P P P, hence: 6

107 P P L P L PAC EA P L P L EA EA 2 P P P L L L PAC EA EA P P L PAC As can be seen, the P term s smply added to the usual VW equaton. From L AC before, we have the varous values of the summatons, and so have: EA EA EA. Thus member AC s. kn n tenson. Note the change: wthout the error n ft t was 23.3 kn n tenson and so the error n length has reduced the tenson by 22.2 kn. Temperature Change: The same dervaton from the VW equaton gves us: P P P 2 L L LT P EA EA BC The change n length due to the temperature change s: T 6 L L T mm 7

108 Notng that the vrtual force n member BC s.6, we have: P BC EA EA Gvng: EA 7.7 Thus member AC s 7.7 kn n tenson, a decrease of 5.6 kn from 23.3 kn. Error n Support Locaton: Modfyng the VW equaton gves: P P P 2 L S H D EA EA L The value of the vrtual horzontal reacton s found from the vrtual force system to be.6 kn to the rght. Hence: EA EA 4 4 Note the sgn on the support dsplacement: snce the real movement s along the same drecton as the vrtual force, t does postve vrtual work. Solvng: 3EA

109 Thus member AC s 4.8 kn n tenson; a reducton of 8.5 kn. All Effects Together: In ths case, the vrtual work equaton s: P P P 2 L L S H D L PAC LT PBC EA EA Substtutng the varous values n gves: EA EA And solvng: EA And so member AC s 22.9 kn n compresson. Thus should be the same as the orgnal force of 23.3 kn plus all the changes nduced by the errors: kn Exercse: Fnd the remanng forces n the truss and compare to the forces wthout the presence of self-stresses. 9

110 6.3 Beams Example 8 Consder the followng 2-span beam wth central sprng support. Determne an expresson for the central support reacton n terms of ts sprng stffness. To do ths, we wll consder the central support as the redundant: The vrtual work equaton, accountng for the sprng, s:

111 M 2 M M S dx dx The deflecton at the central support wll be: V k B S k Snce the reacton at B s. Notng that the deflecton of the sprng wll be opposte to the unt load, and usng the volume ntegrals, we have: l wl 2 l l l l k wl l k 24 6 l 5wl k And fnally: 4 5wl 24 4l k 5wl kl 3 It s mportant we understand the mplcatons of ths result:

112 For no support present, k and so 24 k meanng and there s no support reacton (as we mght expect); For k we have the perfectly rgd (usual) roller support and so 24 k 5 gvng us 4 wl - a result we establshed prevously; For ntermedate relatve stffnesses, the value of the reacton s between these extremes. Therefore, we as the desgner have the ablty to choose the reacton most favourable to us. A plot of the reacton and relatve stffnesses s:.4.2 Central Reacton (/wl) e-.e-8.e-6.e-4.e-2.e+.e+2.e+4.e+6.e+8.e+ Relatve Stffness (/kl 3 ) 2

113 6.4 Frames Example 9 The followng frame, n addton to ts loadng, s subject to: Support A s located mm too far to the left; 3 End C of member BC s.2 rads out of square, as shown; Member CD s 2 mm too short. Determne the bendng moment dagram. Take = 36 3 knm 2 for each member. We wll choose the horzontal reacton at A as the redundant. Snce we are dealng wth a lnear lack of ft n member CD, we need to allow for the vrtual work done by the axal forces n ths member and so we solve for the axal force dagrams also. For the prmary structure: F H H x : D 9 D 9 kn 3

114 M about A : VD 6 VD 62 kn F V V y : A A 8 kn Also, MC knm gvng: For the redundant structure: 4

115 M about A : 2 VD 6 VD kn 3 Fy : V A VA kn 3 3 The vrtual work equaton, accountng for the relevant effects s: W W E W I M 2 M M S H A dx dx M C L PCD We take each term n turn: (a) H S A : The appled unt load s n the same drecton as the error n the support locaton, and keepng all unts n metres: H kn m S 3 3 A (b) M M dx : Usng the volume ntegrals: M M dx BC 3 CD 242 (c) M 2 dx : Agan usng the ntegrals table: 5

116 2 M dx AB M 2 CD dx BC (d) M C : The vrtual bendng moment at C s 6 knm wth tenson on the nsde of the frame. Ths s n the opposte drecton to that needed to close the lack of ft and so the sgn of ths term s negatve, as shown: kn m 3 3 M C (e) P L CD : Snce the member s too short, the lack of ft s negatve, whlst the vrtual force s n tenson and so postve: 3 3 L PCD kn m Substtutng these values nto the equaton, along wth knm, gves: 6

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