Indeternate Analyss Force Method The force (flexblty) ethod expresses the relatonshps between dsplaceents and forces that exst n a structure. Prary objectve of the force ethod s to deterne the chosen set of excess unknown forces and/or couples redundants. The nuber of redundants s equal to the degree of statc ndeternacy of the structure. Descrpton of the Force Method Procedure. Deterne the degree of statc ndeternacy. Nuber of releases* equal to the degree of statc ndeternacy are appled to the structure. Released structure s referred to as the prary structure. Prary structure ust be chosen such that t s geoetrcally stable and statcally deternate. Redundant forces should be carefully chosen so that the prary structure s easy to analyze 2 Also see pages 5 4 n your class notes. * Detals on releases are gven later n these notes. Force Method con t 2. Calculate errors (dsplaceents) at the prary structure redundants. These dsplaceents are calculated usng the ethod of vrtual forces. 3. Deterne dsplaceents n the prary structure due to unt values of redundants (ethod of vrtual forces). These dsplaceents are requred at the sae locaton and n the sae drecton as the dsplaceent errors deterned n step 2. 3 Force Method con t 4. Calculate redundant forces to elnate dsplaceent errors. Use superposton equatons n whch the effects of the separate redundants are added to the dsplaceents of the released structure. Dsplaceent superposton results n a set of n lnear equatons (n nuber of releases) that express the fact that there s zero relatve dsplaceent at each release. 4
Force Method con t These copatblty equatons guarantee a fnal dsplaced shape consstent wth known support condtons,.e., the structure fts together at the n releases wth no relatve dsplaceents. 5. Hence, we fnd the forces on the orgnal ndeternate structure. They are the su of the correcton forces (redundants) and forces on the released structure. 5 () (2) (3) (3) Flexblty Analyss f (x R ) R R 2 D D 2 + (R ) + f 2 (x R 2 ) (R 2 ) f 2 (x R ) f 22 (x R 2 ) 6 (4) f R + f R + D 0 2 2 f R + f R + D 0 2 22 2 2 Solve for R and R 2. Usng atrx ethods: [F] f f f f 2 2 22 flexblty atrx [F] - ( nverse flexblty atrx) f f f f f f 22 2 22 2 2 f2 f [F] {R} -{D} {R} -[F] - {D} 7 D {D} D2 prary structure dsplaceent vector 8 2
R {R} R2 det [F] (5) redundant force vector R f D f D 22 2 2 R2 det[f] f2d+ fd2 ff22 f2f2 Releases Release s a break n the contnuty of the elastc (dsplaceent) curve. One release only breaks a sngle type of contnuty. Fgure shows several types of releases. Wth R and R 2 known, reanng structure s statcally deternate. 9 Coon release s the support reacton, partcularly for contnuous beas. 0 Flexblty Equatons Prary structure dsplaceents at the releases are related to the unknown redundant forces va D f R () j j f j dsplaceent at release due to a unt force n the drecton of and at release j; flexblty coeffcents. Equaton for the case of three redundant forces s expressed as 2 3
D fr+ f2r2+ f3r3 D f R + f R + f R D f R + f R + f R 2 2 22 2 23 3 3 3 32 2 33 3 Matrx for of (2a) -{D} [F] {R} (2a) (2b) D {D} D2 <D D 2 D 3 > T D 3 dsplaceent vector at the redundant degrees of 3 freedo R {R} R2 <R R 2 R 3 > T R 3 [F] redundant force vector f f f f f f f f f 2 3 2 22 23 3 32 33 flexblty atrx 4 Dsplaceent Calculatons Method of Vrtual Forces + φ D F d M d V V subscrpt release drecton of R at d dfferental axal dsplaceent dφ dfferental rotatonal dspl 5 (3) Flexblty Coeffcents Method of Vrtual Forces a b j j j f f + f (4) a FVj fj FV dx EA(x) axal deforaton nfluence coeffcent b M Vj fj MV dx EI(x) bendng deforaton nfluence coeffcent 6 4
Force Method Exaples. Calculate the support reactons for the two-span contnuous bea, EI constant. w w Prary Structure w/ oad + 2. Calculate the support reactons for the two-span contnuous bea, EI constant. w w Prary Structure w/ oad + (x R ) Prary Structure w/ Redundant 7 R R 2 Prary Structure w/ Redundant Forces 8 Prsatc Meber Dsplaceents 9 20 5
3. Calculate the support reactons for the two-span contnuous bea usng the nternal oent at B as the redundant force, I AB 2I and I BC I; E constant. 2 P Prary Structure w/ oadng 2 22 M B Prary Structure w/ Redundant 4. Calculate the bar forces for the statcally ndeternate truss. D B f BB M B 23 Statcally Indeternate Truss Statcally Deternate Released Truss (Redundant 24X) 6
Truss Calculatons Me (n) F F V F V F AB 92" 40-4/5-644 BC 44" 0-3/5 0 CD 92" 0-4/5 0 DA 44" 30-3/5-2592 AC 240" 0 0 BD 240" -50-2000 FVF F AC / fac,ac 20736/829.44 EA 25 25 kps Nonechancal oadng [F]{R} ({D} + {D }) (5) 2 n {D } < D D D > T relatve densonal change dsplaceents calculated usng prncple of vrtual forces Dsplaceents due to denson changes are all relatve dsplaceents, as are all dsplaceents correspondng to releases. They are postve when they are n the sae vector drecton as the correspondng release 26 (redundant). Structure Forces Once the redundant forces are calculated fro Eq. (5), all other support reactons and nternal eber forces can be calculated usng statc equlbru along wth the approprate free body dagras. Ths s possble snce the force ethod of analyss has been used to deterne the redundant forces or the forces n excess of those Matheatcal Expressons Calculaton of the non-redundant forces A (support reactons, nternal shears and oents, truss eber forces) can be expressed usng superposton as N R A p A + (A u ) j R j(6) p j where A desred acton A on the prary structure due to the appled loadng; (A u) j acton A on the prary structure due to a unt vrtual force at redundant R j and N R nuber of redundants. requred for statc deternacy. 27 28 7
Exaple Bea Proble Nonechancal oadng The nterestng pont of ths exaple s that the flexblty equaton wll have a nonzero rght hand sde snce the redundant dsplaceent s prescrbed to equal 0.72 downward. Thus the flexblty equaton s (a) Gven structure E 30,000 ks I 288 n 4 f BB R B d B - DB (7) where d B prescrbed dsplaceent at redundant B -0.72" snce R B s postve upward DB -0.24" d relatve dsplaceent B D B at redundant B (b) Prary structure 29 30 Truss Exaple Nonechancal oadng For the truss structure on the next page, copute the redundant bar EC eber force f the teperature n bar EF s ncreased 50 o F and eber BF s fabrcated 0.3 n. too short. EA constant 60,000 kps and α 6x0-6 / o F. A Truss Exaple B E B A E 3 @ 20 60 Prary Structure Subjected to F CE C F C F D D 5 3 32 8
Truss Exaple Calculatons Me F V F V F V AB 240" 0 0 AE 300" 0 0 BC 240" -4/5 53.6 BE 80" -3/5 64.8 BF 300" 300 CD 240" 0 0 CE 300" 300 CF 80" -3/5 64.8 DF 300" 0 0 f F F CE,CE V V EA D CE FV δ EF EF EF BF BF δ α T 0.072" δ 0.3" CE,CE CE CE f F + D 0 EF 240" -4/5 53.6 33 34 Dsplaceent Calculatons Dsplaceents for the statcally ndeternate structure can be calculated usng the exact eber deforatons for a truss or exact shear and oent expressons along wth the vrtual force expressons on the prary structure. For a truss structure, calculaton of a jont dsplaceent usng the prncple of vrtual forces results n 35 () F V F V F EA F V δ + nt +δ + prary structure eber forces due to the applcaton of a unt vrtual force at the jont for whch the dsplaceent s desred and n the drecton 36 of (8) 9
nt δ prary structure dsplaceent at desred dsplaceent due to nonechancal effects δ exact eber dsplaceents that are obtaned for the statcally ndeternate structure usng the calculated redundant forces to deterne all the eber forces wthn the truss structure eber dsplaceents due to nonechancal loadng on the eber 37 For a frae structure, n whch shear and axal deforatons are gnored, the dsplaceents are calculated as ( ) ( θ ) 0 0 M M θ V V M EI + M EI +θ +κ +κ nt nt dx (9a) dx (9b) 38 θ V V prary structure vrtual oents based on the desred dsplaceent or rotaton θ M,M nt κ, θ prary structure dsplaceents at or rotaton θ due to envronental loads or causes prary structure ntal curvature stran caused by nonechancal loadng In Eqs. (9a) and (9b) the oent expressons are exact based on the statcally ndeternate structure subjected to the external loads wth the redundant forces known fro the flexblty analyss. Equatons (8), (9a), and (9b) are correct only because exact real eber forces are used n the calculaton of the desred dsplaceents. 39 40 0
Calculate the horzontal dsplaceent at jont B for the statcally ndeternate truss. 40 k D 6 C Calculate the rotaton at the center support for the two-span contnuous bea, EI constant. w 2 A D R C B R R 2 A B Prary Structure Subjected to Vrtual oadng 4 Prary Structure w/ Vrtual oad at Desred Dsplaceent ocaton 42 Alternatvely, you can express the desred dsplaceent calculatons also n atrx for followng the usual superposton process of the force ethod of analyss: {} δ [F]{R} + {D} + {D }(0) δ δ δ where {δ} vector of desred dsplaceents; {D δ}, {D δ} vector of desred dsplaceents for the prary structure for both echancal and non-echancal loadngs, respectvely; [F δ ] 43 atrx of dsplaceent nfluence coeffcents at the desred dsplaceent locatons due to unt values of the redundant forces {R}. Stated atheatcally, the coeffcents of [F δ ] are F δ j δ R j () whch sply states that the dsplaceent nfluence coeffcents equal the dsplaceent at desred dsplaceent on the prary structure due to a unt force at redundant j on the prary 44 structure.