Slender Structures Load carrying principles

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1 Sender Structures Load carrying principes Cabes and arches v018-1 ans Weeman 1

2 Content (preiminary schedue) Basic cases Extension, shear, torsion, cabe Bending (Euer-Bernoui) Combined systems - Parae systems - Specia system Bending (Timoshenko) Continuousy Eastic Supported (basic) Cases Cabe revisit and Arches Matrix Method ans Weeman

3 Learning objectives Extend the technique for basic cases Find the ODE for a specific case and the boundary conditions for the specific appication Sove the more advanced ODE s (by hand and MAPLE) Investigate consequences/imitations of the mode and check resuts with imit cases ans Weeman 3

4 Basic Cases Second order DE Extension Shear Torsion Cabe Fourth order DE Bending d u EA = q d w k = q d ϕ GI x t = m d z = q 4 d w EI = q 4 ans Weeman 4

5 Mode (ordinary) Differentia Equation (O)DE Boundary conditions Matching conditions ans Weeman 5

6 Cabe ow to find in a cabe structure Catenary soution,oad aong cabe Axia deformation of cabes Susceptive to changes in oad orizonta dispacements of cabes ans Weeman 6

7 Cabe 1 cabe stiffness oad distributed aong the projection of the cabe cabe without eongation??!! externa oad q interna generaised stress, vertica component V. BC Cabe sope tanα dispacement fied w. BC ans Weeman 7

8 Fundamenta reations geometrica reation tanα = dz Moment equiibrium Equiibrium V = tanα V + q + V + dv = 0 dv = q dz dv d z V = = = q ODE ans Weeman 8

9 Remarks Cabe takes no bending Cabe ODE describes an equiibirum in the defected state (funicuar curve) so this is a non-inear approach! (no superposition) can be regarded as constant ony if no horizonta oads are appied This mode is not vaid for oads distributed aong the cabe! Derivation is stricty based upon equiibrium ony! Cabe force can be expressed in and z : dz T = + V = + ( tanα ) = 1+ ans Weeman 9

10 ow to find? With specified cabe ength With (externa) specified oad 10

11 1 st option : with specified ength Exampe ength AB given: L = m B chord AB 3.0 m A z k z k1 3.0 m 10 kn 60 kn 3.0 m 4.0 m.0 m 6.0 m C 7 k 1 k k1 k L = 3 + ( z 1) z ( z 1) + + z

12 Find position of the cabe in terms of the constant 10 kn 60 kn 3.0 m 4.0 m.0 m 0 kn 50 kn z M-ijn k 1 60 knm z k z z k1 k = z = z k k1 100 knm 1

13 1 st option : with specified ength Exampe for fixed cabe ength L q A B f L x z s z k 1 qx( x) 4 fx( x) ( x) = = dz ds = 1+ x= x= dz L = ds = 1+ x= 0 x= 0 OEPS.. 13

Continue q A B x= x= dz L = ds = 1+ x= 0 x= 0 f L 1 dz 1 dz q L = ds 1 + + = + 4 x=

14 Continue q A B x= x= dz L = ds = 1+ x= 0 x= 0 f L 1 dz 1 dz q L = ds = + 4 x= 0 x= 0 x= 0 x= x= x= 3 overength = L : = 3 q 4 Check with MAPLE f 3 8 f = or =

15 nd option : with (externa) specified oad Exampe (see aso part 1) support reaction of the bock A A V q f f L B. q Vkatro T L F. B bock F support reaction of the bock components of the cabe force T T = F F Free Body Diagram of the bock T = F V V = 1 q f = q 8 force poygon for the bock F = F q 1 f = 8 F q 1 q F 15

16 Cabe oad distributed aong the cabe cabe without eongation due to cabe force! cabe stiffness externa oad q interna generaised stress, vertica component V. BC Cabe sope tanα dispacement fied w. BC ans Weeman 16

17 Fundamenta reations geometrica reation tanα = dz Moment equiibrium V = tanα Equiibrium V + qds + V + dv = 0 dz dv d z dz V = = = q 1+ dv ds dz = q = q 1+ ODE ans Weeman 17

18 Catenary soution: d z dz = q 1+ d( sinh ς ) = q 1+ sinh subsitute : dz = sinh ς (1) ς 1 ans Weeman q 18 dς d cosh cosh d q ς ς = ς = x q qx dz qx ς = + C1 with (1): = sinhς = sinh + C1 qx z = cosh + C + C

19 Catenary versus paraboa f versus β = q f Paraboa Catenary β = q ans Weeman 19

20 Axia cabe deformation (strain) = T EA ( straight cabe) dz = + = + T V dz T = 1 + ( curved ) dz ds = 1+ Curved cabe: s= L x= x= Tds dz dz L = = 1 EA EA + = + EA EA s= 0 x= 0 x= 0 Sma for cabes with modest sag 0

21 Oostende (B) Neson Mandeabrug in the Maria-endrika Park ans Weeman 1

22 In de natuurijke omgeving van het Maria endrikapark wordt een sober en evident ogend kunstwerk gereaiseerd dat een beangrijke schake vormt in een bovenokae wande- en fietsroute van de Stad Oostende. [a nice bridge in a park] De brug wordt gereduceerd tot zijn essentie: oopvak dat zich uitstrekt van oever tot oever. De structuur wordt in het oopvak geïntegreerd, in de vorm van opgespannen kabes. et oopvak neemt de eegante vorm aan van de natuurijke kettingijn en overspant zo met een dikte van 8 cm een afstand van 7 m. De brug won in 006 de Staabouw wedstrijd. [an eegant integrated cabe structure foowing the shape of a catenary with deck thickness of 8 cm and a span of 7 m, Stee price winner 006] ans Weeman

23 Mandea bridge, some numbers.. span = 7.00 m deck width b =.50 m sag f = 1.30 m thick t = 0.8 m p = 19.6 kn/m q = 5 kn/m EA = 1 e 6 kn 1 cabe force due to dead weight, cabe force at fu oad, 3 sag at fu oad ans Weeman 3

24 Variation in due variation in oad q1 = q + p A q L L B q = q p fu oad q: L = + q 4 3 Variation in oad: 1 1 fied 1 d z1 1 ( + ) = q 1 ( + ) z1 = q1x + C1 + Cx fied d z 1 ( + ) = q ( + ) z = qx + C3 + C4x 4

25 Resut Variation in oad: fied 1 d z1 1 ( + ) = q 1 ( + ) z1 = q1x + C1 + Cx fied d z 1 ( + ) = q ( + ) z = qx + C3 + C4x Tota cabe ength remains unchanged (assume no eongation): Suppose: 1 3 x= x= q 1 dz1 1 dz 4 x= 0 x= L = + = % variation in q 0,78% variation in 7,5% variation in z some work to do 1 p = q 5

26 orizonta dispacements q1 = q + p q q = q p A u L L B 1 1 6

27 More detai x z A w α u A B α dz x= dz dw u( ) u(0) = = 0 returns for arches x= 0 B du α dw du = tanα dw with: tanα = du dz d z d d d d u z w dz dw = w = u = d x + C1 7

28 Aternative derivation Cabe ength L is constant. Compare ength due to oad q with ength due to q+p : (axia deformation not incuded) x= x= 1 dz 1 d( z + w) + x = + x= 0 x= 0 1 d 1 due to q due to q+p x= x= 0 dz dw = 0 ans Weeman 8

29 Cabe and beam (Euer-Bernoui) 9

30 Mode with constant rigid inks z(x) cabe h Load: p = permanent q = additiona oad EI beam construction : a permanent oad on cabe: After competion : beam + cabe as P-system: ( z + w) d = q 4 d w EI = q 4 beam cabe d z = p 30

31 Simpified Mode (constant ) Fina situation : beam + cabe as P-system: q + p = q + q cabe beam ( z + w) 4 d w d EI = q + p 4 ( z + w) 4 d w d d z EI = + q 4 4 d d d EI w w = q with: z = p 4 31

32 Refined Mode with changing not part of the examination 4 d w d ( z + w) EI ( + ) = q + p 4 Permanent oad: d z = 4 d w d 4 ( ) w d EI + = q p with: z = p Two unknowns, additiona equation required: x= 0 x= 0 p ( ) x dz dw dz dw = 0 or EA x= = no axia deformation with axia deformation axia deformation compensated 3

33 Exampe EA EI 3 cabe 1650,0 10 kn beam 800 knm p p q cabe 0,6 kn/m beam 9, kn/m, 0 kn/m Question: Find the defection w at midspan due to additiona oad q? ans Weeman 33

34 Assignment (use MAPLE and exact formuation for the ength) given : a = ; b = ; c = q = 5 kn/m; F = 15 kn; EA = = 9,1667 kn, L = 6,5580 m Find ength L of cabe for z(a+b) = 0.8 m Find z(a+b) for change of ength of +.5% and.5% ans Weeman % = 47,6377 kn, z(a+b) = m 10.5% =,75174 kn, z(a+b) = m

35 Arches q w F z x Primariy compression Shape retained (not ike a cabe) due to bending stiffness of the arch Susceptibe to bucking 35

36 Cabe versus Arch q A B x EA z q EA, EI A B x arch no moments in the arch z 36

37 Cabe up side down? ANTONI GAUDI ( ) Ony if the ine of force due to the oad corresponds to the shape of the arch 37

38 Line of Force Centre of force in a cross section is the point of intersection of the resutant force of a norma stresses and the cross section. cross section 8 kn/m cf C S D 4,0 m A 6,0 m 3,0 m B e = M N M-ine and N-ine required 38

39 Line of force C 8 kn/m S D x M, e 4,0 m 48 M-ine [knm] 1 1 kn A B 1 kn 6,0 m 3,0 m 3 kn 16 kn 3 16 N-ine [kn] 1 M ( x) = 8 (6 x) + 8 x (6 x) 0, 0 x 6, 0 e( x) = x x + 4 ine of force (ine of pressure) A-S 39

40 Resut C 8 kn/m S S 3 D x 1,33 m change the position of the hinge to infuence the shape and compressive force and see. M, e 4,0 m A B 6,0 m 3,0 m

41 Perfect Arch. fipped cabe Geometry (shape) is identica to the moment distribution of a repacing beam but scaed by a constant. Try this!

42 Bridge with mutipe spans Bridge crossing the Uvsund near Kavehave in Denmark, photograph ans Weeman 7,5 kn/m 5 kn/m 5 kn/m EI = knm = 100 kn 15,0 m 5,0 m 10,0 m = 100 kn

43 Beam mode with M-ine? 5 kn/m 7,5 kn/m 5 kn/m 15,0 m 5,0 m 10,0 m M [knm] x [m] Fip the M-ine and scae the resut by

44 Force distribution in arches Line of pressure method cumsy Cassica method based on the force method for specific type of arches Differentia equation for arches with a wider range of appication but with imited deformations 44

45 Cassica approach - based on the force method q A z ( ) b x EA, EI B x Condition: q Both ends with hinged supports z ( ) b x EA, EI x A h B z 45

46 Force Method Reduce structure to a staticay determinate (principa) system: curved beam, find the M-ine Denote the Staticay Indeterminates (redundants): Define the deformation condition: h = 0. horizonta dispacement 46

47 orizonta dispacements Find h due to a moment distribution M in a curved beam: Beam (straight) : dϕ M M κ = = dϕ = d x EI EI Beam (curved) : dϕ M κ = = dϕ = ds EI M EI ds A dϕ dh = adϕ = zdϕ a z B dh h = arch Mz EI ds horizonta dispacement = rotation times vertica distance to centre of rotation vertica dispacement = rotation times horizonta distance to centre of rotation 47

48 Moment in principa system due to M a x z M + a = 0 M = a a = z M = z 48

49 Deformation condition a h h h h a with: h h h a + + = h a M z ds = EI arch h = 0 EA M zd s ( z) zds z ds = = = EI EI EI arch arch arch : due to the oad on principa system : due to the redundant : axia deformation of the arch a M z ds z ds = arch EI EI EA arch 0 49

50 Resut cassica approach = arch a M z EI z + EI EA arch a M = M + z Moment distribution in the arch is the superposition of the distribution of the staticay determinate (principa) system and the distribution due to the staticay indeterminate (redundant) 50

51 Exampe q Given: f q EA = 100 m = 15 m = 4,0 kn/m = infinite EA, EI A ( ) sin π x zb x = f B x z π x = f sin a 1 M = qx( x) = arch a M z EI z + EI EA arch a M = M + z 51

52 Resut > restart; Moment distribution in the arch > :=100; q:=4; f:=15; > z:=-f*sin(pi*x/); M [knm] > pot(-z,x=0.., ,tite="boog"); moment Ma in het statisch bepaade hoofdsysteem > Ma:=(1/)*q*x*(-x); > pot(-ma,x=0..,tite="momentenverdeing Ma"); os op met de kassieke methode: > :=-int(ma*z,x=0..)/int(z^,x=0..); > M:=simpify(Ma+*z); > pot(-m,x=0..,tite="momentenverdeing in de boog"); x [m] 5

53 ODE for Arches q geometry z z A z ( ) b x EA, EI B x After competion of the arch the oad q is taken by the arch + bending. Dispacements w with respect to z wi occur and a constant horizota component of the compressive force in the arch wi occur For a boundary conditions Bending and fipped cabe 4 ddzz cabe d w = q and arch EI = q 4 bending 53

54 Load carrying capacity d z Arch + Bending 4 d w EI 4 = = q q arch bending 4 d w d ( z + w) EI + = q 4 arch + qbending = q d w EI = q d z

55 MAPLE Soution of the ODE Genera Soution = homogeneous soution + particuiar soution Formuate four boundary conditions sove the integration constants Resut : The soution in terms of the unknown 55

56 Extra condition to find no horizonta dispacement at the supports: x= x= 0 dz dw = 0 Paraboic arch: x= x= 0 w = 0 Incuding axia deformation of the arch: x= dz dw = EA x= 0 Paraboic arch: EA x= 8 f = x= 0 w 56

57 Fun. x= x= x= dz dw dw d w = z + z = 0 x= 0 x= 0 x= 0 or constant for paraboa x= x= dz dw dz x= d z = w + w = 0 x= 0 x= 0 x= 0 x= x= x= d z w = C w = 0 w = 0 x= 0 x= 0 x= 0 57

58 paraboic arch q Exampe A z ( x) b EA, EI B x arch: z = oad: 4 fx( x) z Given: q = 5 kn/m = 80 m f = 15 m EI = knm q = 5,0 eaviside( x ) 1 boundary conditions: x = 0; w = 0; dw ϕ = = 0 x = ; w = 0; d w M = EI = 0 DEMO with MAPLE 58

59 Concusion Simpe parametric mode for arches to be used with MAPLE for design purposes Resuts of M-ine etc is with respect to the horizonta axis (x-axis) and not perpendicuar to the beam axis as in FEM programs?? 59

60 60

61 ODE resut versus FEM resut norma force N shear force V oca (beam) x-axis FEM shear force V norma force N α z(x) ODE α α V arch + V bending 3 d w N = cosα + EI 3 3 d w V = sinα + EI 3 dz sinα dz cosα 61 V bending V arch

62 One ast detai Compete ODE 4 d w d w d z EI + = q 4 Do not negect w with respect to z 6

63 Genera soution use MAPLE q π x sin w x 4π 4 x D x D x D x D 4 o 4 f EI ( ) = ( 1 sin β + 1 cos β ) ( π EI ) with: β = EI Bucking ength (in pane) = haf the span Bucking out of pane aso possibe! Snap-through probems Courses on CIE5144 Stabiity and CIE514 Noninear FEM 63

64 Assignment Arch-1 Find the horizonta component of the force in the arch Find the maximum moment in the arch ( in terms of F and ) 5 F = ; M 7 18 max = F f 18 ans Weeman 64

65 Assignment Arch- Given : EI knm ; = 15 m; = 5 m; 1 f F = = 3 m; f = 5 m; 1 = 150 kn; F = 30 kn; 1 π x i z ( x) = f sin 0 < x < i i i i i Find the moment distribution for a rigid connection between the arches at the midspan support Find the moment distribution for a hinged connection between the arches at the midspan support Discuss the resuts ans Weeman 65

Work and energy method. Exercise 1 : Beam with a couple. Exercise 1 : Non-linear loaddisplacement. Exercise 2 : Horizontally loaded frame

Work and energy method. Exercise 1 : Beam with a couple. Exercise 1 : Non-linear loaddisplacement. Exercise 2 : Horizontally loaded frame Work and energy method EI EI T x-axis Exercise 1 : Beam with a coupe Determine the rotation at the right support of the construction dispayed on the right, caused by the coupe T using Castigiano s nd theorem.