INDETERMINATE STRUCTURES METHOD OF CONSISTENT DEFORMATIONS (FORCE METHOD)

Transcription

1 INETNTE STUTUES ETHO OF ONSISTENT EFOTIONS (FOE ETHO) If all the support reactons and nternal forces (, Q, and N) can not be determned by usng equlbrum equatons only, the structure wll be referred as STTIY INETNTE. statcally ndetermnate structure s also referred as a EUNNT STUTUE, because of redundant reacton components (or redundant members n trusses) whch are not necessary for stablty consderatons. Statcally determnate do Statcally ndetermnate to the frst degree do Statcally ndetermnate to the thrd degree do

2 ddtonal equatons to solve statcally ndetermnate structures come from prescrbed condtons of translatons and rotatons, commonly called condtons of compatblty or consstent dsplacements. statcally determnate structure whch can be obtaned by removng enough reacton force possesses no redundant, and wll be referred as prmary structure. ll methods used to analyze ndetermnate structures employ equatons that relate the forces actng on the structure to the deformatons of the structure. If these equatons are formed so that the deformatons are expressed n terms of the forces, then the forces become ndependent varables or unknowns n the analyss. ethods of ths type are referred to as force methods. Statcally ndetermnate structures may be consdered as a prmary structure wth redundant forces. edundant forces wll have values so that the deformatons of the prmary structure must be equal to the deformatons of the actual statcally ndetermnate structure. Usng the prncple of superposton we may dvde the loadng nto several cases. t each case only one load wll be consdered. q TU STUTUE PIY STUTUE WITH EUNNT FOES q

3 q q x kn.m. x kn x kn

4 endng moment at an arbtrary secton of the actual structure () can be wrtten as. Where endng moment at the correspondng secton of the prmary structure due to the external forces oment at the same secton of the prmary structure due to a unt load n the drecton of the frst redundant oment at the same secton of the prmary structure due to a unt load n the drecton of the second redundant oment at the same secton of the prmary structure due to a unt load n the drecton of the thrd redundant Value of the frst redundant force Value of the second redundant force Value of the thrd redundant force... ost of the deflecton for beams and frames due to bendng moment, for trusses due to axal force. astglano s frst theorem states that the partal dervatve of the stran energy wth respect to one of the forces on the structure wll gve the deflecton n the drecton of that force. U P P dx δ ( ) To reduce the mathematcal work we can change the order of the sgns. U P P dx P dx δ The bendng moment s a functon of the redundant forces. If we take the partal dervatves of the stran energy wth respect to the redundant forces and equate them to the deflectons n the drecton of the redundant forces.

5 ( ) dx dx U δ... set of lnear equaton wll be obtaned to determne the redundant forces. nt GvenSupportSettleme dx dx dx dx U GSS dx dx dx dx U GSS dx dx dx dx U We can wrte the equatons n the followng form

6 δ δ δ δ δ δ δ δ δ δ δ δ... GSS... GSS... GSS ESTI EQUTIONS OF THE STUTUE Ths set of lnear equaton s referred as Elastc Equatons of the structure. Each equaton wll gve the deflecton n the drecton of the redundant due to external loadng and redundant forces respectvely. The rght hand sde of the equaton wll be ether zero or gven support settlement. δ dx eflecton n the drecton of the th redundant due to a unt load n the drecton of the th redundant Steps n the analyss of a statcally ndetermnate structure by the force method may be dstngushed as follows etermne the number of redundant forces (the degree of ndetermnacy do) emove enough redundant forces to obtan prmary structure alculate the coeffcents of the elastc equatons Solve the redundant forces Use equlbrum equatons to determne remanng reacton forces raw the dagrams

7 Example : raw the shear force and moment dagrams of the propped cantlever beam q q q kn Vertcal reacton at roller support taken as a redundant force δ δ q δ k 4 4 δ k. 4 q q q 4 q q q q Prmary Structure q Shear Force agram 9 q q endng oment agram

8 Example : raw the shear force and moment dagrams of the propped cantlever beam q q q endng moment at fxed end support taken as a redundant force δ δ q q δ k 4 δ k. q 4 q q q q q Prmary Structure q q Shear Force agram 9 q q endng oment agram

9 Example : raw the shear force and moment dagrams of the twospan contnuous beam 67 kn m m m Vertcal reacton at roller support taken as a redundant force Prmary Structure 67 kn. δ δ δ ( α) k 6 9 δ k kn 6.* 6 4. Negatve sgn ndcates the redundant wll be opposte to the unt force 67 kn kn 4.7 Solve the same problem by takng the redundant as vertcal reacton at support and respectvely Shear force dagram (kn) oment dagram (kn/m)

10 Example : raw the shear force and moment dagrams of the twospan contnuous beam 4 kn/m 4 kn kn kn/m Pn 4 kn kn Take bendng moment at nternal support as redundant Prmary Structure 4 kn/m δ δ 4.67 δ.67 δ.6... knm 4 kn kn 4. kn.m Shear Force endng oment.6 4.

11 SUPPOT SETTEENT If one of the supports of a smply supported beam settles by a small amount, no maor changes occur n the external and nternal forces actng on the member. However f one of the supports of a multspan beam settles a small amount sgnfcant changes occur n both the reactons and the bendng moments. The problem of support settlement s far more serous for ndetermnate structures than t s for determnate ones. Example : etermne the reactons and draw the bendng moment dagram for the beam n the fgure, assumng that the center support settles cm. and compare the results thus obtaned wth those correspondng to zero settlement. kn/m The degree of ndetermnacy s one. edundant must be chosen as a vertcal reacton at the nternal support. 6 m 6 m Prmary Structure 7 kn δ δ GSS δ ( αβ) km 4 δ k kn kn Negatve sgn ndcates the settlement s opposte to the unt load ght hand sde s taken zero for zero settlement

12 . m kn/m kn/m Shear Force agram endng oment agram omparson of the two moment dagrams ndcates that the maxmum bendng moment at ncreases from 67. to.9 knm as a result of the support settlement. The support settlement can gve rse to sgnfcant ncreases n stress n an ndetermnate structure.

13 Example : etermne the support reactons and draw the shear force and bendng moment dagrams of the gven structure. eam I. 6 kn able E GPa mm m The beam s statcally ndetermnate to the frst degree. et the force n the cable (T) be the redundant. Pont moves down by an amount equal to the elongaton of the cable. 6 m m Prmary Structure kn δ δ T Elongaton. of. the. cable 6 δ (k k ) 6 7 δ k 6 7 T T E T kn.64 kn kn Shear Force agram 4 kn.64 endng oment agram 4.6 6

14 Example : etermne the horzontal reacton component and draw the shear force and moment dagrams of the gven portal frame kn δ δ 4m m F 4 m 4. δ ( α β) km δ k.. k kn Shear Force dagram Prmary Structure nd External loadng oment dagram Prmary Structure nd Unt loadng

15 Example : etermne the bendng moment at and draw the shear force and moment dagrams of the frame 4m kn m F m δ δ δ.667 δ knm Each term n elastc equaton becomes relatve rotaton Shear Force dagram Prmary Structure nd External loadng 49. knm oment dagram PS nd Unt loadng

16 Example : It s requred to evaluate the redundant forces and to draw the bendng moment dagram of the contnuous beam under the gven loadng. The stffness s constant throughout the beam. 4 kn/m The beam s statcally ndetermnate to the second degree. et the bendng moment at support be the frst redundant and the bendng moment at support be the second redundant. Prmary Structure kn/m kn/m 4 kn/m 9.6 δ δ δ δ δ ** 6.9. δ ** δ ** δ δ *( ) δ δ. 4.46kNm kNm

17 7.7 4 kn/m m Shear Force agram endng oment agram 4.46

18 Example: It s requred to evaluate the redundant forces and to draw the bendng moment dagram of the portal frame. kn The structure s statcally ndetermnate to the second degree. et the horzontal reacton at leftend support and the bendng moment. at rghtend support are be the redundant forces. 6 m m m kn Prmary Structure kn / / /6 /6

19 66. δ (6 *)* * δ * δ 6*6 (*6 *6* * ) * δ (6 *)* * 6. δ * **. Y Y X (.4)*( 6) ( 9)*.4 (.4)*( ) ( 9)*( ). (.4)*() ( 9)*( ) (.4)*( ) ( 9)*( ) (.4)*( ) ( 9)*( ) (.4)*() ( 9)*() 64.6 δ δ δ δ δ δ.4 kn... 9kNm kn.4 endng oment agram OENT IG IS WN ON OPESSION SIE. 6.4 kn

20 Example : raw the shear force and moment dagrams of the frame. kn/m The frame s statcally ndetermnate to the thrd degree. et chose the redundant forces as the bendng moments at and, and the horzontal reacton at support. kn/m m.4 4 m.4 m 49 δ δ δ δ δ δ δ δ δ δ δ δ

21 ( 6.)*() 97*() 97*() 97 ( 6.)*() 97*() 97*() 9. ( 6.)*() 97*() 97*() 9. OENT IG IS WN ON TENSION SIE endng oment agram

22 STUTUES WITH INTEN EUNNTS (STTIY INETNTE TUSSES) E F N E F N o P Q P Q E F N E F N Internal force at an arbtrary member of a statcally ndetermnate truss can be wrtten as Where N N N. N N... N Force at the same member due to the external forces on the prmary structure N Force at the same member due to a couple of unt load n the drecton of the th redundant member on the prmary structure Force at the th redundant member

23 Stran energy for an axally loaded member U N E ccordng to the astglano s theorem δ E E U N N δ N N N N N N... N... N... N U N N N ( N N N N )... E E δ U NN NN NN NN E E E E GvenElongaton U NN NN NN NN E E E E GE U NN. NN. NN. NN... E E E E GE δ δ δ δ... GvenElongaton Ths set of lnear equaton s referred to as ESTI δ δ δ δ... EQUTIONS of the truss GvenElongaton δ δ δ δ... GvenElongaton

24 Example: alculate the member forces of the truss f the roller support settles down by mm. hose the force n member and the vertcal reacton at support E as the redundant forces. E GPa, mm. 4 m E.7 N kn 4 m 4 m E kn N N.77 kn

25 ember N N N N N N N N N N N E E 4 4 Sum/E δ δ δ δ δ δ GvenSupportSettlement Solvng redundant forces, the bar forces can be obtaned E E E m E E E 6.6kN.46kN ember N N N N E E.46

26 Example: ompute the forces n dagonals F and E. ll members are mm. n area. E GPa, What would be the effect on these forces of a rse n temperature of degrees of elsus n member EF relatve to other members. Take the coeffcent of thermal expanson as. 6 / o 4 kn E F et Force n member F be the redundant force. 4 kn E F m N m m m 6 6 em E E F ength N N N N 6.6 N N E F N F F EF E SU /E /E δ δ kN 4.4 F 6.7kN E * 6.7kN

27 et force n member EF be the redundant force. Elongaton of member EF due to the temperature rse of degrees 6. T. α **..* m E 6 6. *. 4 kn E F N em E E F F F EF E ength N N N N N N δ 4..* E 9.94kN δ GvenElongaton F E.44*( 9.94) 4.6kN External. load Temperature. hange E kN F kN m SU 4./E

28 THEE OENTS EQUTION (PEYON S EQUTION) general equaton based on the force method can be developed for contnuous beams. The equaton relates the moments at the three consecutve support ponts to the loadng on the ntermedate support. Ths theorem was represented by lapeyron n 7 for the analyss of contnuous beams. Now consder m span contnuous beam, the degree of ndetermnacy wll be m. If the moments at the nternal supports are chosen as the redundant forces the prmary structure wll be m smple beams. P k elatve rotaton at support wll be the functon of external forces at the two adacent span (,, )and the support moments at, and k. oadng over the other spans and the other support moments wll not contrbute to the relatve rotaton at support. Summaton of the relatve rotatons at due to the external load and the redundant moments, and k must be equal to the relatve rotaton of the actual contnuous beam at whch s zero.

29 P k x x The relatve rotaton due to the external forces can be wrtten by usng oment rea Theorem. t t x δ θ θ x k δ,.. δ,.. δ k 6 6 elatve rotaton s n clockwse drecton (postve) δ k k x x 6 E I I 6 t θ θ t k earrangng the last equaton δ k 6 x x E I I * If s constant throughout the beam. The equaton smplfes to * δ x ( ) k 6 x k * δ k pplcaton of the three moments equaton to a contnuous beam results n a set of smultaneous equatons wth the moments over the supports as the unknowns.

30 In applyng the three moments equaton to a partcular beam, we locate the nteror supports successvely and wrte as many equatons as the unknown redundant support moments. smultaneous soluton of the equatons for the unknown moments yelds the requred results.? One redundant moment ( )? and can be calculated?? 4 Two redundant moments and two equatons?? 4? Unknowns, can be wrtten, 4. Three equatons

31 Example: It s requred to draw shear and bendng moment dagrams of the two span contnuous beam. I s constant. kn/m 4 kn/m *. 4* E I 6I m 6 m ;... q 9 q 7.kNm *..6m * m

32 Example: It s requred to determne the support moments and reactons for a contnuous beam fxed at one end and havng an overhang at the other as shown. ' ',... I ' kn kn. kn / m. *.m 6.6m ' ' ' *... E. I * *..*... E. I *6.6 I knm,....4,....7 kn.4.4 The beam s statcally ndetermnate to the second degree and requres two equatons. For the purpose of wrtng three moments equatons an magnary span to the left of fxed support havng an arbtrary length and moment of nerta I may be consdered

33 Three moments equaton can be modfed to take nto account the support settlements. For example consder that supports,, and k settle downward by amount,, k respectvely. k Settlement of support decrease the relatve rotaton at θ k k ss s s θ θ θ k ( ) If we substtute ths expresson of relatve rotaton nto the equaton of consstent deformaton. k x x k 6 E I I 6 If I s constant throughout the beam, ths expresson smplfes to x x k ( ) k 6 6 The support of the prevous example settles down by mm under the loadng. E GPa 6 kn/m I. 6 mm 4 ' ' ' * E. I * knm, ,....7 ' *..* E. I *6.6 I 6.6

34 SYETI STUTUES WITH SYETI OING If the structure and loadng are both symmetrcal, symmetrcal onts rotate by the same amount but n opposte drecton and the structure wll have a skewsymmetrcal shear force and a symmetrcal bendng moment dagrams. otaton at the symmetry axs s always zero. If symmetry axs passes from a common member and there s no concentrated load at the symmetry axs the shear force at the same secton be zero. Otherwse the shear at the symmetry axs s equal to the half of the concentrated load. y choosng the nternal forces at the symmetry axs as redundant forces one half of the structure may be analyzed. Symmetry axs passes from a common member Symmetry axs passes from a support or column P Q θ Fxed end support θ Q P/ θ Fxed end support θ

35 Example: raw the bendng moment dagram of the gven frame. s constant. 6 KN/m 4 m Q m kNm 96.kN OENT IG IS WN ON TENSION SIE endng oment agram 4.6.6

36 SYETI STUTUES WITH SKEWSYETI OING If a symmetrcal structure s subected to a skew symmetrcal loadng, symmetrcal onts rotate by the same amount n the same drecton and the structure wll have a symmetrcal shear force and a skewsymmetrcal bendng moment dagrams. If symmetry axs passes from a common member horzontal force and bendng moment at the symmetry axs s always zero and a roller support may be consdered at the center secton to analyze half of the structure. If symmetry axs passes from a column, half of the moment of nerta of that vertcal member must be consdered to analyze half of the structure. Symmetry axs passes from a common member Symmetry axs passes from a support or column I oller support N I

37 Example: raw the bendng moment dagram of the gven frame. s constant. KN 4 m m 4 m Neglectng axal deformaton No bendng oment δ / δ 4.67 /.4kN OENT IG IS WN ON TENSION SIE m m endng oment agram.94.94

38 NY O SYSTE N E ONSIEE ONE SYETI N THE OTHE SKEWSYETI OINGS P P P Symmetrcal loadng P P Skew Symmetrcal loadng

39 P q P P q q Symmetrcal loadng P P q q Skew Symmetrcal loadng

40 Example: raw the bendng moment dagram of the gven frame kNm 6.kN 97..6

41 FIXEEN OENTS & FIXEEN FOES If a member s fxed at ts ends, the moments at the ends of the member are referred as the fxedends moments. q q q q The member s assumed to be subected transverse loadng and the effect of the axal deformaton wll be neglected therefore change n length wll be zero and. Fxed end moments wll be determned by the force method. Prmary structure edundant Forces External loadng q 6 q 6 q q q q q q q 4 q q q q

42 Example: It s requred to determne the fxed end moments by usng three moments equaton. Pab a P P b Pa b I ( ) ( ) Pab I Pab Pab ab a b Pa b ' 6 Pab a a b b b 6 Pab a a b b a Pab Pa b...

43 ETIONS ETWEEN THE EE EN FOES N EE EN ISPEENTS Example: It s requred to determne a) fxed end moment at the left end support, b) rotaton at end for the propped cantlever beam as shown. et bendng moment at be the redundant force. s constant δ δ 6 * θ dx θ 4 For unt rotaton at 4,...

44 Example: It s requred to determne the fxed end moments due to the settlement of rght end support ( ) by usng three moments equaton. I ( ) ( ) I raw the tangent from the left sde of the center support I

45