Middle East Technical University Department of Mechanical Engineering ME 413 Introduction to Finite Element Analysis

Transcription

1 Middl East Tchnical Univrsity Dpartmnt of Mchanical Enginring ME 4 Introduction to Finit Elmnt Analysis Chaptr 4 Trusss, Bams and Frams Ths nots ar prpard by Dr. Cünyt Srt csrt@mtu.du.tr Ths nots ar prpard with th hop to b usful to thos who want to larn and tach FEM. You ar fr to us thm. Plas snd fdbacks to th abov mail addrss. 4-

2 What Is This Chaptr About? W ll study FEM formulations of dformation of planar trusss bnding of bams dformation of frams (as th suprposition of planar truss and bam formulations) Ths problms will b studid as D, but thr will b multipl unknowns at a nod. W ll modify th D FEM cod to solv ths problms. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

3 Dformation of a Bar A bar is a structural mmbr that is loadd axially. x E, A F It is ithr in dirct tnsion or comprssion. Axial dformation, u, is govrnd by th following DE d dx du EA dx = 0 solution of which is linar for constant E and A. Evn a singl linar lmnt can solv this problm xactly. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

4 Elmntal wak form of th problm is Dformation of a Bar (cont d) EA du dw dx dx dx = wea du Ω Q dx x + wea du dx x Q SV of th problm is th axial forc : EA du dx n x Elmntal stiffnss matrix is K ij = Ω EA ds i dξ ds j J dξ J J dξ Elmntal systm is EA h u u = Q Q Elmntal forc vctor is zro METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

5 A truss consists of svral bars connctd with frictionlss pin joints. Not that this is not th actual maning of truss in civil nginring. F Planar Truss Each mmbr can only carry axial forc, but no shar forc or bnding momnt. All mmbrs of a planar truss li on a D plan. Spac truss is th D vrsion. A truss can b loadd with multipl point forcs at its joints. Typically thr is at last on fixd joint. Som joints might hav rstrictd motion. Dformations ar small, i.. gnral shap of th truss is similar bfor and aftr loading. F METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

6 Planar Truss Local Coordinats Each mmbr of a truss can b tratd as an lmnt of a FE msh. Th lmntal systm drivd prviously for a bar is valid for ach mmbr. But in ordr to b abl to us it, diffrnt coordinat systms alignd with ach mmbr should b usd. Ths local coordinats ar shown blow with x and x. For th st mmbr Q For th nd mmbr = Q Q Q = x Nodal dflctions of = in x dirction Nodal forcs of = in x dirction EA h x u u = Q Q EA h u u = Q Q Nodal dflctions of = in x dirction Nodal forcs of = in x dirction METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-6

7 Planar Truss Transformation Matrix During th assmbly of th lmntal systms, PVs and SVs writtn for a common xy coordinat systm should b usd. For ach lmnt a transformation btwn local x coordinat and th global xy coordinats is ncssary. This is a purly gomtrical transformation. y Q Q x u Q y x u x x u y θ u u x Q u y Q x Q y u x = u cos (θ ) = u sin (θ ) u y Multiply th st qn with cos (θ ) and th nd qn with sin θ and add thm up. u x cos θ + u y sin θ = u cos θ + sin θ u = u x cos θ + u y sin θ METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-7

8 Planar Truss Transformation Matrix (cont d) Similary for th nd nod of lmnt : u = u x cos θ + u y sin θ Togthr ths two qns bcom u u = cos (θ ) sin(θ ) cos (θ ) sin(θ ) Transformation matrix, [T ] u x u y u x u y u A similar qn can b writtn for th SVs too = [T ] Δ Δ includs both u x s and u y s. Each nod has unknowns and in total on lmnt has 4 unknowns. Q = [T ] Q Q = Q x Q y Q x Q y T METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-8

9 Planar Truss [K ] Transformation Matrix [T ] can b usd to transform th original x lmntal systm into a nw 4x4 lmntal systm Original x lmntal systm using bars : EA u h = u or K u = {Q } Q Q Using u = [T ] Δ and Q = [T ] Q K [T ] Δ = [T ] Q T Prmultiply this qn by T T T K [T ] Δ = T T [T ] Q K METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-9

10 Planar Truss Transformd [K ] Transformd 4x4 lmntal systm : [K ] Δ = Q T T K T α 0 β 0 0 α 0 β EA h α β α β K = EA h α αβ α αβ β αβ β α αβ sym β whr α = cos(θ ) and β = sin(θ ) METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-0

11 Planar Truss 4x4 Elmntal Systm Thrfor in a planar truss solution lmntal systms ar 4x4. Each truss mmbr is a linar lmnt with 4 unknows EA h α αβ α αβ β αβ β α αβ sym β Δ Δ Δ = Q Q Q Q 4 Q 4 y Q Q x Δ Δ θ Δ θ is masurd CCW from th positiv x axis. Q Δ : Horizontal dflction of point Δ : Vrtical dflction of point Δ : Horizontal dflction of point : Vrtical dflction of point Q : Horizontal forc at point Q : Vrtical forc at point Q : Horizontal forc at point Q 4 : Vrtical forc at point METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

12 Planar Truss Local to Global Unknown Mapping Considr th following truss problm. P P = = = Thr ar NE = lmnts and NN = nods. At ach nod thr ar NNU = unknown dflctions. Totally thr ar NU = 6 unknowns. 4 of ths unknowns ar known. Nods and ar fixd. W nd to dtrmin dflctions (horizontal and vrtical) at nod and, if dsird, th raction forcs at nods and. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

13 Planar Truss Local to Global Unknown Mapping (cont d) PVs and SVs of ach lmnt ar For = For =, Q 4 For =, Q 4 Δ, Q, Q 4 Δ, Q Δ, Q Δ, Q Δ, Q Δ, Q Δ, Q Δ, Q Δ, Q PVs and SVs of th global systm ar Δ 6, Q 6 In gnral th global PVs (and SVs) of th i th nod ar numbrd as i and i. Δ, Q Δ, Q, Q 4 Δ 5, Q 5 Δ, Q METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

14 Planar Truss Local to Global Unknown Mapping (cont d) Assmbly procss is about local-to- global unknown mapping for ach lmnt Unknowns of = : Δ = Δ Δ Δ Δ Unknowns of = : Δ = Δ Δ Δ Δ Δ = Δ Global unknowns Δ 5 Unknowns of = : Δ = Δ Δ Δ Δ 6 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

15 Planar Truss LtoG Matrix & Assmbly Rul This graph can also b xprssd as a local-to-global mapping matrix LtoG = = = = LtoG ij givs th global unknown numbr of th i th lmnt s j th local unknown. For xampl, LtoG 4 = 6 bcaus th rd lmnt s 4 th local unknown is th 6 th global unknown. Th assmbly rul can now b dfind as K ij ntry of an lmntal systm gos to K IJ ntry of th global systm. F i ntry of an lmntal systm gos to F I ntry of th global systm. Q i ntry of an lmntal systm gos to Q I ntry of th global systm. whr I = LtoG i J = LtoG j METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

16 Planar Truss Assmbly Using th assmbly rul assmbl systm of th -mmbr truss is K + K K + K K K 4 K K 4 K + K K + K K K 4 K K 4 K K K + K K 4 + K K K 4 K 4 K 4 K 4 + K K 44 + K K K 4 K K K K K + K K 4 + K 4 K 4 K 4 K 4 K 4 K 4 + K 4 K 44 + K 44 Δ Δ Δ Δ 5 Δ 6 = Q + Q Q + Q Q + Q Q 4 + Q Q + Q Q 4 + Q 4 which only dpnds on th LtoG matrix LtoG = LtoG matrix dpnds on how w numbr th nods globally how w numbr th lmnts nods locally, i.. which nod is th st and which on is th nd? METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-6

17 Planar Truss Point Loads Considr th following truss problm with point loads. How should w us th points loads? P Thy ar usd in th boundary trm vctor. Q of this problm is Q = Q Q Q Q 4 Q 5 Q 6 = Q + Q Q + Q Q + Q Q 4 + Q Q + Q Q 4 + Q 4 y = = = x P If thr is no horizontal (or vrtical) forc at a nod, th corrsponding SV is st to zro. If thr is a givn point load at a nod, th corrsponding SV is st to th givn valu. At supports SV(s) ar unknown and can b calculatd during post-procssing. B carful with th dirction (sign) of forcs. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-7

18 For our problm Planar Truss Point Loads Q and Q ar unknown raction forcs at nod. Q and Q 4 ar unknown raction forcs at nod. Q 5 = P (givn horizontal point load at nod ). Q 6 = P (givn vrtical point load at nod ). Q = Q y Q Thrfor th {Q} vctor is Q 4 P P x Q, Q, Q and Q 4 ar not known, but th corrsponding Δ, Δ, Δ and ar known. If thr wr no horizontal load at nod, w should hav st Q 5 to zro. If thr wr no vrtical load at nod, w should hav st Q 6 to zro. If thr wr a rollr support that rstricts vrtical motion but not horizontal motion, w should hav st Q 4 to zro. = P = P METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-8

19 Exampl 4..g. Exampl 4. : Solv th following truss problm. Find th dflction of th nods. Dtrmin th forcs and strsss in ach mmbr. Dtrmin th raction forcs at th supports. E and A valus ar th sam for ach mmbr. P P = = L = L METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-9

20 Exampl 4. (cont d) Elmntal systm quation is givn in slid -0. θ valus ar ncssary and thy dpnd on local nod numbring of th nods = = = θ valus of th lmnts ar : θ = 0, θ = π/, θ = π/4 Not : If th local nod numbring of all th lmnts ar rvrsd, θ valus chang as θ = π, θ = π/, θ = 5π/4 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-0

21 Exampl 4. (cont d) Elmntal systms ar For = : α = cos θ = β = sin θ = 0 h = L K = EA L sym. 0 0 For = : α = cos θ = 0 β = sin θ = h = L K = EA L sym. 0 0 For = : α = cos θ = / β = sin θ = / h = L K = EA L sym. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

22 Exampl 4. (cont d) Assmbld systm is givn in slid 4-6. With numbrs it bcoms EA L symmtric Δ Δ Δ Δ 5 Δ 6 = Q Q Q Q 4 P P W know that Δ = Δ = Δ = = 0 Rduction can b applid to gt th following x systm Solving this w gt = P P 5 = 5.88 PL EA, 6= PL EA METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

23 Exampl 4. (cont d) To calculat axial forcs in ach mmbr w can go back to local coordinats alignd with th lmnts. Q From slid 4-9 x Q Q = EA h u u Q Using th transformation matrix dfinition from Slid 4-8 Q Q = EA h Δ T Δ Δ METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

24 Axial forcs in ach mmbr ar Exampl 4. (cont d) For = : Q Q = EA L = 0 0 For = : Q Q = EA L PL EA PL EA = P P For = : Q Q = EA L PL EA PL EA =.44P.44P METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

25 Exampl 4. (cont d) As sn for ach mmbr Q = Q i.. nt axial forc on ach mmbr is zro. st lmnt carris no axial forc, as xpctd, bcaus its both nds ar fixd. nd lmnt is in comprssion bcaus Q > 0 (or Q < 0). rd lmnt is in tnsion bcaus Q < 0 (or Q > 0). Q < 0 Q > 0 = Q > 0 x Q < 0 = x METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

26 Exampl 4. (cont d) Axial strsss in ach lmnt can b calculatd. For = : σ = Q = 0 A For = : σ = Q = P A A For = : σ = Q =.44 P A A (Ngativ strss indicats comprssion) (Positiv strss indicats tnsion) Finally forcs at th supports can b calculatd using th 6x6 systm of Slid 4-. At nod : Q = EA L Q = EA L = P = P At nod : Q = EA L Q 4 = EA L + = = P P P P Forcs at th supports ar th opposit of th calculatd ons METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-6

27 Planar Truss Constraind Motion Somtims horizontal and vrtical dflctions of a nod ar not indpndnt. This happns at a rollr support inclind at an angl to th global xy systm. i α At nod i horizontal and vrtical dflctions ar rlatd to ach othr. i sin(α) i cos(α) = 0 Horizontal dflction of nod i Vrtical dflction of nod i Dtails of how to handl ths cass can b found in FEM txtbooks. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-7

28 Bam Bnding Bams ar long, slndr structural mmbrs, gnrally subjctd to transvrs loading that producs significant bnding ffcts. Axial dformation or twisting is not considrabl for bams. y F q(x) M x L q(x) is th distributd transvrs loading. F is a point transvrs load and M is a point bnding momnt. Transvrs dflction v(x) is in th y dirction. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-8

29 Eulr-Brnoulli Bam Thory Th assumption bhind th Eulr-Brnoulli bam thory is that plan cross sctions prpndicular to th longitudinal axis of th bam bfor bnding, rmain prpndicular to th longitudional axis aftr bnding. Bfor bnding : x Aftr bnding : Govrning DE is d dx EI d v dx = q x, 0 < x < L v(x) : Unknown transvrs dflction q(x) : Known distributd transvrs load EI : Known flxural rigidity of th bam, i.. product of modulus of lasticity and th scond momnt of inrtia. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-9

30 This is a 4 th ordr DE. d dx EI d v dx = q x, 0 < x < L IBP should b applid two tims to gt th wak form. First IBP givs Eulr-Brnoulli Bam Thory dw d dx dx EI d v dx dx = wq dx + w d Ω Ω dx EI d v dx x + w d dx EI d v dx x Scond IBP givs d w dx EI d v dx dx = wq dx Ω Ω + + w d dx EI d v dx x + w d dx EI d v dx x + dw dx EI d v dx x + dw dx EI d v dx x METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-0

31 Eulr-Brnoulli Bam Thory (cont d) Thr ar two PVs Transvrs dflction : v dv Slop : dx Thr ar two SVs Shar forc : Bnding momnt : d dx EI d v dx EI d v dx Sign convntions ar Dflction in +y dirction (upward) is positiv. CCW rotation of th bam corrsponds to positiv slop. Shar forc in +y dirction (upward) is positiv. CCW momnt (in +z dirction) is positiv. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

32 Two Nod Bam Elmnt nod bam lmnt has 4 PVs and 4 SVs Δ Δ Δ Δ = Δ Δ Δ Transvrs dflction at nod Slop at nod Transvrs dflction at nod Slop at nod Q Q Q Q 4 Q = Q Q Q Q 4 Shar forc at nod Bnding momnt at nod Shar forc at nod Bnding momnt at nod METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

33 Wak form of th problm contains nd drivativ of th transvrs dflction. Not only th transvrs dflction, but also its first drivativ, i.. slop should b continuous, bcaus slop is a PV too. Ovr ach lmnt FE solution is Hrmit Typ Shap Functions v = 4 j= S j Δ j A C 0 continuous solution for Δ is NOT nough. It should b at last C continuous. Lagrang typ shap functions usd prviously ar not suitabl. Hrmit typ shap functions should b usd. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-

34 Hrmit Typ Shap Functions (cont d) Ovr ach bam lmnt thr ar 4 unknowns. Continuity of two variabls (v and dv/dx) at two nds of an lmnt rsults in 4 conditions to b satisfid. To satisfy ths 4 conditions at last a cubic polynomial is ncssary for v. v = A + Bξ + Cξ + Dξ Four continuity rstrictions ar At ξ = v = Δ At ξ = dv dx = Δ At ξ = + v = Δ At ξ = + dv dx = dv dξ dv dξ J = Δ J = METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

35 Using J ths 4 conditions bcom Δ = A B + C D Hrmit Typ Shap Functions (cont d) Δ = B C + D h Δ = A + B + C + D = B + C + D h Solv for A, B, C and D in trms of Δ, Δ, Δ and. Substitut thm into th following quation 4 j= S j Δ j = A + Bξ + Cξ + Dξ v v And idntify th 4 Hrmit typ cubic shap functions. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

36 Hrmit Typ Shap Functions (cont d) Hrmit typ cubic shap functions ar S = 4 ξ ξ + S = h 8 ξ ξ ξ + S = 4 ξ + ξ + S 4 = h 8 ξ + ξ ξ 0.8 S S S /h S 4 /h METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-6

37 Elmntal Systm for a Bam From Slid 4-0, lmntal stiffnss matrix and forc vctor ar K ij = EI d S i dξ J d S j dξ J J dξ F i = q S i J dξ Evaluating ths using Hrmit typ shap functions q is th part of th distributd transvrs load, simplifid as uniform ovr lmnt EI (h ) 6 h 6 h h h h 6 h sym h Δ Δ Δ = qh 6 h 6 h + Q Q Q Q 4 K F METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-7

38 Exampl 4-.g. Exampl 4. : For th following clampd bam with EI = Nm, us two qual lngth lmnts to dtrmin th transvrs dflction of th tip th raction forc at th middl support. q = 400 N/m 5 m 5 m METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-8

39 Exampl 4- (cont d) EI, h and q ar th sam for both lmnts. K and F will b th sam for both lmnts. Thr ar four unknowns for ach lmnt. Δ Δ Δ Δ Δ = Δ = Ovrall thr ar nods and 6 unknowns Δ Δ Δ 5 Δ = = Δ 6 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-9

40 Exampl 4- (cont d) Local-to-global mapping of th unknowns ar as follows Unknowns of = : Δ = Unknowns of = : Δ = Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ 5 Δ 6 = Δ Global unknowns LtoG = METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-40

41 Exampl 4- (cont d) Assmbld systm is EI (h ) 6 h 6 h 0 0 h (h ) h (h ) h h + h 6 h h (h ) h + h (h ) +(h ) h (h ) h 6 h 0 0 h (h ) h (h ) Δ Δ Δ Δ 5 Δ 6 = qh 6 h h + h 6 h + Q Q Q + Q Q 4 + Q Q Q 4 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

42 Boundary conditions nd to b applid. Exampl 4- (cont d) Known dflctions and slops ar EBCs. At th clampd nd, transvrs dflction and slop ar zro. Δ = 0, Δ = 0 At th middl support transvrs dflction is zro. Δ = 0 Known shar forcs and momnts ar NBCs. Middl support can not carry any bnding momnt Q 4 = Q 4 + Q = 0 Fr nd can not cary and shar forc or bnding momnt Q 5 = 0, Q 6 = 0 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

43 Exampl 4- (cont d) Only PVs (, Δ 5, Δ 6 ) ar actually unknown. Rduction can b applid to th original 6x6 systm. EI h 4h h h h 6 h h h h Δ 5 Δ 6 = qh 0 6 h Using EI = Nm, h = 5 m, q = 400 N/m F is not changd bcaus = = = 0 q is in y dirction Unknown PVs ar calculatd as 4 = rad 5 = 0.04 m 6 = rad Slop at th middl support. Bam rotation is CW. Transvrs dflction at th tip. It is downward. Slop at th tip. Bam rotation is CW. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-4

44 To find th unknown SVs w can us th calculatd PVs in th original 6x6 systm. Unknown SVs can b calculatd as Exampl 4- (cont d) Q = 50 N Q = 50 Nm Q = 450 N Forc applid by th wall at th clampd nd. Momnt applid by th wall at th clampd nd. Forc acting by th middl support. 400 N/m 50 Nm 50 N 450 N METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-44

45 Exampl 4-.g. Exampl 4. : Solv th sam problm, but this tim rmov th distributd load and put a point load at th tip 4000 N 5 m 5 m On dtail you nd to pay attntion is that this tim Q 5 = 4000 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-45

46 Planar Frams Frams look lik trusss, but th connctions ar rigid, i.. wldd or rivtd. Each mmbr can carry axial forc, shar forc and bnding momnt. F F F F y x Th abov bicycl fram has 7 mmbrs. Each mmbr can b modld as a singl lmnt or multipl lmnts. It is possibl to think of a fram lmnt as th suprposition of truss and bam lmnts. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-46

47 Arbitrarily Orintd Bam Elmnt Fram lmnts ar basd on arbitrarily orintd bam lmnts. Similar to a truss lmnt, it is possibl to study th bam lmnt using ithr th local x, y coordinats or th global x, y coordinats. y Δ Δ θ Δ x y Δ Δ Δ Δ 5 Δ 6 x Unknowns in local coordinats Nod : Δ, Δ Nod : Δ, Δ Unknowns in global coordinats Nod : Δ, Δ, Δ Nod :, Δ 5, Δ 6 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-47

48 Transformation Matrix of a Bam Elmnt Rlation btwn local and global unknowns ar Δ = sin θ Δ = Δ Δ = sin θ Δ 4 = Δ 6 Δ + cos(θ ) Δ + cos(θ ) Δ 5 Ths rlations can b xprssd using th following transformation matrix. Δ Δ β α Δ Δ Δ = Δ β α Δ 5 Δ 6 whr α = cos(θ ) and β = sin(θ ) Transformation matrix of th bam lmnt METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-48

49 Fram Elmnt W now hav transformation matrics for arbitrarily orintd bam and truss lmnts. Fram lmnts carry axial forc, shar forc and bnding momnt. Thy can b obtaind by th suprposition of bam and truss lmnts. Fram lmnt has unknowns at ach nod. Fram lmnt in local coordinats Fram lmnt in global coordinats y Δ Δ Δ θ Δ 6 Δ 5 x Δ Δ y Δ Δ 6 Δ 5 x METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-49

50 Elmntal systm of th fram lmnt in local unknowns is obtaind by th propr combination of thos of truss and bam lmnts Ths ar coming from K of th truss lmnt (Slid 4-9) Th rst is coming from K of th bam lmnt (Slid 4-7) Fram Elmnt (cont d) K Δ = F EA EA h 0 0 h 0 0 EI 6EI EI 6EI h h 0 h 4EI 6EI EI h 0 h h EA h 0 0 Symmtric K EI h h 6EI h 4EI h Δ Δ Δ Δ 5 Δ 6 q is th constant distributd transvrs load = 0 qh q h 0 qh q h {F } METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-50

51 Fram Elmnt (cont d) Similarly transformation matrix of th fram lmnt is obtaind by th propr combination of thos of truss and bam lmnts Δ Δ Δ Δ 5 Δ 6 = α β β α α β β α T Δ Δ Δ Δ 5 Δ 6 Ths qns ar coming from th transformation matrix of th truss lmnt (Slid 4-8) Rmaining 4 qns ar coming from th transformation matrix of th bam lmnt (Slid 4-48) Elmntal systm in global unknowns is obtaind as K Δ = {F } K = T T K T {F } = T T {F } METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

52 Exampl 4-.g. Exampl 4. : Using thr lmnts, dtrmin th dflctions and rotations at th joints of th following fram. Draw bnding momnt and shar forc diagrams for all lmnts. Calculat th ractions at th supports. For all mmbrs E = 00 GPa, I = m 4, A = m 5 kn q = 7.5 kn/m.5 m.5 m METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

53 Exampl 4- (cont d) Elmnt and global/local nod numbring ar shown blow. = = = = = = 4 Orintation of th lmnts ar θ = π/, θ = π/, θ = 0 Elmnt lngths ar h =.5, h =.5, h =.5 E, I and A ar th sam for all lmnts. METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-5

54 Exampl 4- (cont d) Using Slid 4-50 calculat [K ] and {F } for ach lmnt For =: K = 0 6 F = T (all zro bcaus q = 0 for =) For =: K = K (bcaus E, A, I and h ar th sam for both lmnts) F = F (bcaus q = 0 for =, too) METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-54

55 Exampl 4- (cont d) For =: K = 0 6 F = 0 4 METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-55

56 Exampl 4- (cont d) Now th transformation matrics for ach lmnt nd to b calculatd using Slid For =: T = For =: T = T (bcaus θ = θ ) For =: T = (This is th unity matrix bcaus x = x) METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-56

57 Exampl 4- (cont d) [K ] and {F } of ach lmnt can b calculatd using K = T T K T and {F } = T T {F } For =: K = 0 4 F = For =: K = K, F = F For =: K = K, {F } = {F } METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-57

58 Now w hav thr 6x6 lmntal systms. Lt s assmbl thm into th x global systm. LtoG mapping is as follows Exampl 4- (cont d) LtoG = = Δ, Δ, Δ Δ 4, Δ 5, Δ 6 Δ 4, Δ 5, Δ 6, Δ 5, Δ 6 Δ 7, Δ 8, Δ 9 =, Δ 5, Δ 6 = = = = Δ, Δ, Δ 4 Δ, Δ, Δ Δ 0, Δ, Δ Δ, Δ, Δ METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-58

59 Exampl 4- (cont d) Assmbld global systm is x. At nods and 4 all thr unknowns ar known and thy ar zro. = = = 0, 0 = = =0 0 6 Δ 5 Δ 6 Δ 7 Δ 8 Δ 9 Q 4 is th givn horizontal point load at nod Q 5 and Q 6 ar zro bcaus thr is no vrtical point load or point momnt at nod Q 7, Q 8 and Q 9 ar zro bcaus thr is no horizontal or vrtical point load or point momnt at nod. = METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-59

60 Exampl 4- (cont d) Solving for th unknown dflcions w gt Δ 5 Δ 6 Δ 7 Δ 8 Δ 9 = Horizontal dflction, vrtical dflction and rotation of nod Horizontal dflction, vrtical dflction and rotation of nod Both nods and mov in +x and y dirctions. Also thy rotat CW. Now th forcs and momnts at th supports (Q, Q, Q, Q 0, Q, Q ) can b calculatd. Also axial strss, shar strss and bnding strss ovr ach lmnt can b dtrmind. Qustion : Will th solution improv by using mor lmnts? METU Dpt. of Mchanical Enginring ME 4 Int. to Finit Elmnt Analysis Lctur Nots of Dr. Srt 4-60