3.10 Implications of Redundancy

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1 118 IB Structures Impications of Redundancy An important aspect of redundant structures is that it is possibe to have interna forces within the structure, with no externa oading being appied. These may exist because of: Settement of supports; The structure not fitting together before it was assembed ( ack of fit ); Temperature changes. In a determinate structure, the structure coud deform to take account of these effects. In an indeterminate structure, the structure cannot freey adjust, and so a state of sef stress resuts Exampe Support Settement A propped cantiever of ength is initiay stress-free. Find the resutant stresses in the structure if the support drops by /100. A B x /100 Spit system in two No oading + Reaction at B δb1 = 0 δ B2 R B From the Structures data book, Compatibiity at the support δ B2 = R B 3 3EI R B 3 3EI = 100, R B = 0.03 EI 2

2 Handout 3. Eastic Structura Anaysis 119 Bending moments M(x) = 0.03 EI ( x) EI/ M(x) 0 0 x Exampe Temperature Rise The structure shown is initiay stress-free. There is then a temperature increase of T. If the structure has coefficient of therma expansion α, and bending stiffness EI, what wi be the reactions at the supports?

3 120 IB Structures Forces 3 unknowns M F x Dispacements θ 1 v F h 1 y F y h 2 θ 2 F x M v For System 1 h 1 = αt (negect effect of F x ) v 1 = F y 3 3EI M2 2EI θ 1 = F y 2 2EI + M EI For System 2 h 2 = F x 3 3EI M2 2EI v 2 = αt θ 2 = F x 2 2EI M EI Compatibiity h 1 = h 2,v 1 = v 2,θ 1 = θ 2 Soution is F x = F y = 12αT EI 2

4 Handout 3. Eastic Structura Anaysis 121 M = 6αT EI Reactions at supports 12αTEI/ 2 12αTEI/ 2 6αTEI/ 6αTEI/ 12αTEI/ 2 12αTEI/ 2

5 122 IB Structures Appications of Sef-Stress In a carefuy designed system, setting up a state of sef-stress can be an important aspect of structura behaviour. It enabe parts of the structure to be prestressed, either in tension or compression. Cabe-stiffened depoyabe structures Depoyabe structures are commony used on sateites to aow the fina structure to be much arger than the restricted space within the aunch vehice. One approach to depoyabe structure design is to have a set of cabes which can be prestressed at the end of depoyment. These cabes don t affect the depoyment, but when prestressed at the end of depoyment, they make the structure much stiffer. Without prestress, a cabe cannot take any compressive oad. With prestress, this compressive oad is superimposed on an initia tension, and so the cabe acts as a structura member. Bicyce Whees The hub on a bicyce whee is supported by very thin members, the spokes. The spokes woud bucke at a very ow oad if they weren t prestressed, so they are tightened to ensure that they are aways in a state of tension - ony possibe because it is a redundant structure. Prestressed concrete beams Concrete is a usefu materia in compression, but wi crack and fai at very ow tensie stresses. A common structura eement is a prestressed concrete beam, where stee tendons within the beam are tensioned, thus setting up a state of sef-stress with the concrete in compression. Subsequent bending stresses wi then not cause the concrete to crack. Tensegrity Systems, Fabric Roofs and Baoons Tensegrity (Tension-Integrity) structures were popuarised by Buckminster Fuer. They are structures where compression members are connected together by a web of cabe tension members so that none of the compression members touch one another. In common with fabric roofs, and baoons, they ony have stiffness because they are initiay prestressed - a state of sef-stress exists in the structure. In tensegrity systems, turnbuckes are incuded to deiberatey introduce ack of fit. In fabric roofs the fabric is pretensioned by cabes, and a baoon is prestressed by interna pressure. We wi ook at a simpe i-conditioned structure to expain the concept.

6 Handout 3. Eastic Structura Anaysis 123 Ignoring second order effects (we are considering very sma movements away for the initia position, and so the bars do not change ength, and the interna force remains constant), what is the stiffness of the structure to atera oads? Consider the joint moves by a sma distance δ. No prestress δ δ/ Prestress T δ δ/ 0 0 T T 0 2T δ/

7 124 IB Structures Anaysis of Symmetric Structures Many structures possess some symmetry. This may be just a simpe pane of symmetry, or the more compex symmetry of a radio teescope. It is possibe to make use of symmetry to simpify the anaysis of structures. We wi concentrate on the simpest case, biatera symmetry, but simiar techniques can be appied to more compex symmetry Symmetry Properties for Biatera Symmetry A structure with biatera symmetry has a singe pane of refection. Symmetry and Antisymmetry During this section we wi often use the terms symmetric and antisymmetric. The definitions of these terms are: Anything symmetric is preserved by refection of the structure in its pane of symmetry. Anything antisymmetric is reversed by refection of the structure in its pane of symmetry. Symmetric Exampes Antisymmetric Exampes Loads Dispacements

8 Handout 3. Eastic Structura Anaysis 125 Spitting Unsymmetric into Symmetric and Antisymmetric A very usefu property is that any unsymmetric oad or dispacement can be spit into a symmetric and an antisymmetric component. For exampe: Forces 0.5w / unit ength symmetric w / unit ength P unsymmetric oading P/2 = + P/2 0.5w / unit ength P/2 P/2 antisymmetric Dispacements symmetric unsymmetric dispacement = + antisymmetric

9 126 IB Structures Response to Symmetric and Antisymmetric Loading The reason that symmetry can hep with structura anaysis is because: 1. A symmetric structure, subject to symmetric oads (a) Wi ony have symmetric interna forces; (b) Wi ony undergo symmetric dispacements. 2. Simiary, a symmetric structure, subject to antisymmetric oads: (a) Wi ony have antisymmetric interna forces; (b) Wi ony undergo antisymmetric dispacements. We wi prove this to be true for two of the four possibiities, but simiar arguments coud be used for a four. In each case we wi examine the simpe porta frame shown beow. pane of symmetry

10 Handout 3. Eastic Structura Anaysis 127 Interna Forces due to Symmetric Loading Consider the frame subject to a uniform oading across the beam. We wi ook at the structure from both sides, views 1 and 2, and carefuy examine the stress resutants at the centre of the beam. View 2 View 1 View 1 S T T S M M View 2 S T T S M M Subtracting 2 from 1 gives a structure that is unoaded 2S 2S

11 128 IB Structures Athough there is no externa oad, our anaysis has shown a resutant shear force of 2S. Since this cannot be true, the shear force S must be zero. We coud have repeated the anaysis with any symmetric oad, any symmetric/antisymmetric stress resutants. We woud aways find the same resut. Symmetric oads on a symmetric structure can ony give symmetric interna forces.

12 Handout 3. Eastic Structura Anaysis 129 Dispacements due to Antisymmetric Loading Consider the frame subject to an antisymmetric distributed oading across the beam. We wi ook at the structure from both sides, views 1 and 2, and examine three possibe modes of deformation. View 2 View 1 Examine the possibe dispacements. The actua dispacement wi be some combination of these three. View 1 vertica horizonta rotation View 2 Adding 2 to 1 gives a structure that is unoaded

13 130 IB Structures Athough there is no externa oad, our anaysis has shown a resutant dispacement. This cannot be true, and hence the symmetric dispacement due to the antisymmetric oad must be zero. We coud have repeated the anaysis with any antisymmetric oad, any symmetric dispacement. We woud aways find the same resut. Antisymmetric oads on a symmetric structure can ony give antisymmetric dispacements Exampe Symmetric Anaysis Find the dispacement of the centre of the beam in the porta frame due to the oad shown. W 2 This frame has one redundancy, which we coud take as the horizonta force at the base W H H W/2 W/2 Because of symmetry, we can consider ony haf the structure

14 Handout 3. Eastic Structura Anaysis 131 v A B Forces Defections 2 W/2 H C h=0 v Examine each part separatey, and use data book coefficients v A B 2HL H θ B h BC W/2 B C H W/2 v = W 2 3 3EI 2H 2 2EI = W3 6EI H3 EI θ B = W 2 2 2EI 2H EI = W2 4EI 2H2 EI

15 132 IB Structures h BC = H (2)3 3EI = 8H3 3EI Horizonta defection h = θ B 2 + h BC = W3 2EI + 4H3 + 8H3 EI 3EI To satisfy compatibiity, h = H = W 2 H = 3W 40 Vertica Defection v = W3 6EI 3W 40 = 11 W EI 3 EI Exampe Antisymmetric Load Find the dispacement of the centre of the beam in the porta frame due to the oad shown. W 2

16 Handout 3. Eastic Structura Anaysis 133 Athough the frame has one redundancy, as we saw in the previous exampe, the redundancy was symmetric. For our antisymmetric anaysis, we can cacuate a forces using antisymmetry and equiibrium. W W/2 W/2 W W Again because of symmetry, we can consider ony haf the structure h 2 W/2 Examine each part separatey, and use data book coefficients A B W θ B B h BC C W/2 W

17 134 IB Structures θ B = W 3EI h BC = W 2 (2) 3 3EI = 4W3 3EI Horizonta defection h = θ B 2 + h BC = 2W3 3EI + 4W3 = 2W3 3EI EI Exampe Genera Load Find the dispacement of the centre of the beam in the porta frame due to the oad shown. α W 2 Horizonta dispacement is ony due to horizonta oad (antisymmetric) h = 2 W cosα3 EI Vertica dispacement is ony due to vertica oad (symmetric) v = 11 W sinα EI Try Questions 1,2 and 3, Exampes Sheet 2/4

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.