Chapter Structura echanics Introduction There are many different types of structures a around us. Each structure has a specific purpose or function. Some structures are simpe, whie others are compex; however there are two basic principes of composing structures. They must be capabe of carrying the oads that they are designed for without coapsing. They must support the various parts of the externa oad in the correct reative position. A structure refers to a system with connected parts used to support a oad. Some exampes reated to civi engineering are buidings, bridges and towers. However, these structures are very compex for anayze and design. At first, we wi consider simpe exampes of structures and parts of structures ike beams, trusses, frames etc. It is important for a structura engineer to recognize the various type of eements composing a structures and to be abe to cassify them as to there form and function. We wi introduce some of these aspects. Structura eements: Some of most common structura eements are as foow: Tie rods structura members subjected to a tensie force. Due to the nature of the oad, these eements are rather sender and are often chosen from rods, bars, anges, or channes. F F rod bar ange channe beams straight horizonta members are used generay to carry vertica oads. F F F Beams may be designed from severa of eement and materias concrete, meta etc. with rectanguar or other cross section. coumns members are generay vertica and resist axia compressive oads. Coumns are eements simiar to the tie rods but they carry vertica oads. F - -
Type of structures. Frame structures: trusses, three-hinged frame, frames, trusses: they are composed of sender rods usuay arrenged trintrianguar fashion. Trusses are suitibe for constructions with arge span when the depth is not an important criterion for desing. Pane trusses are composed of members that ie in the same pane and are frequanty used for bridge and roof support. three-hinged frame: this structure is simpe determinate frame used generay for base eement for compicated frame structures. frames: they are often used in buidings and are composed of beams and coumns which are with hinge or rigid connections. These structures are usuay indeterminate and the oad causes generay bending of its members. pane structures: pates, was and etc. These structures have two significant dimensions and one sma caed thickness. The theory of easticity is capabe to anayze such structures. - -
thickness thickness surface structures: shes and etc. These structures can be made from fexibe or rigid materia and has a three-dimensiona shape ike a cyinder hyperboic parabooid etc. The anaysis of these structures is aso aim of theory of easticity. Loads: In statica structura anaysis of frame structures we define statica (dead) oad. We distinguish types of oads: force oad: concentrated force or moment, distributed oad. F Point N q q concentrated force concentrated moment distributed oad q f ( x) temperature oad: oad caused by fire. - -
dispacement oad: oad dispacement is caused from dispacement of some point or points of the structure. Ideaized structures: Ideaized structure is needed to the engineer to perform a practica force anaysis of the whoe frame and its member. This is the reason in this section to show different member connections and supports and there ideaizations. If one know these modes may compose ideaized mode of each rea structure after a perform the anaysis and design. rigid (fixed) connections: this connection carry moment, shear and axia forces between different members. In addition, in this case a members incuding in such a connection have one and the same rotation and dispacements the noda rotation and dispacements. Typica rigid connections between members in meta and in reinforced concrete constructions and there ideaized modes are shown in the foowing figure: Rigid connection, I =I I I hinged (pin) connections: this connection carry shear and axia forces but not moment between different members. Hinged connection aow to the jointed members to have different rotations but the same dispacements. Typica hinged connections between members in meta and in reinforced concrete constructions and there ideaized modes are shown in the next figure: - 4 -
hinged connection, I <<I I I fixed support: this support carry moment, shear and axia forces between different members. This kind of support doesn t aow any dispacements of the support point. So if the dispacement aong the x axis is u, the dispacement aong y axis is y and the rotation is caed ϕ then we can say that: u A = 0; v A = 0 and ϕ A = 0. Ax A Ax A y x A A y Support reactions A y Ideaization of the support and support reactions hinged (pin) support: this support carry shear and axia forces but not moment between different members. The hinged support aows rotation of the support point but the two dispacement are equa zero or: u A = 0; v A = 0 and ϕ A 0. y x A A A x A y A x A y Support reactions A x A x A x A y A y A y Ideaization of the support and support reactions - 5 -
roer support: this support carry ony shear forces between jointed members. The roer support aows rotation and one dispacement of the support point: u A 0; v A = 0 and ϕ A 0. y A A y x A A y Ideaization of the support and support reactions spring supports: These supports are ike the previous but with the difference that they are not ideay rigid but with some rea stiffness. The spring has a stiffness constant c equas to the force caused by dispacement d =. A A x ϕ A δ A δ A A y Fixed spring support A = c. ϕ A A y Roer spring support A y = c. δ A A y Pin (hinged) spring support A y = c. δ A structure ideaization: The main idea of this ideaization is to made a mathematica mode of the rea construction to be convenient for anaysis and cacuation. After we know the ideaization of different joints and supports, we wi take care about whoe structure ideaization. To make this we foow the midde axis of the eements of the structure. In the foowing figure are shown some rea and ideaized structures: 40 knm A 40 knm B 4 m - 6 -
40 knm kn 0, m 0, m 0,6 m Principes and preconditions: dispacements: Every two dimensiona deformabe eement has three degrees of freedom (two dispacements and one rotation) of each its end node. With using different support inks, we contro these degrees of freedom so the eements cannot move on the imited direction or it moves with controed vaue. These imitations are caed boundary conditions. On the foowing figure are shown the degrees of freedom and some boundary conditions for eements: initia shape u = 0 w = 0 ϕ = 0 deformed shape u 0 w = 0 ϕ 0 x w ϕ u w u ϕ deformed shape u = 0 w = 0 ϕ 0 deformed shape u 0 w = 0 ϕ 0 z u = 0 w = 0 ϕ = 0 deformed shape u 0 w 0 ϕ 0 deformation: deformation or strain is the change in the metric properties of a continuous body (eement) caused by some oad. A change in the metric properties means that the eement changes its ength and shape when dispaced to a curve in the fina position the deformed shape. - 7 -
initia shape F deformed shape preconditions about dispacements and deformations: we presume that the dispacements are sma according the dimensions of the eement and deformations are sma according the unit. These preconditions aow us to write equiibrium conditions for the initia shape of the structure and aso to negect the sma dispacement of the structure. F F F h Equiibrium condition for deformed shape: Equiibrium condition for deformed shape if : Equiibrium condition for the initia shape and (sma dispacement and sma deformations): A A u precondition about the materia: we suppose that the connection between stress and strain is inear so the Hook s aw is vaid. This is acceptabe because of presumption of sma deformation. σ inear aw Hook s aw noninear aw rea curve ε principa of superposition: The previous two preconditions aow us to use the principe of superposition. It may be stated as foow: The tota dispacement or interna forces at a point in a structure subjected to severa externa oadings can be determinate by adding together the dispacements or interna forces caused by each of the externa oads acting separatey. q F F q w t w w А t B t А = + B А B t - 8 -
Chapter Kinematica anaysis of structures Determination of degrees of freedom: We know that each body situated in one pane has three degrees of freedom three independent parameters determining its movement. By using support inks we imit this movement possibiity. So if we put on three specia arranged support inks at a body than it wi be stabe without any movement possibiity. In this way the body is abe to carry different oads and we ca it structure. Then as a response of the oad in the support inks appears support reactions we can determine. Structure with exact number of inks is caed determinate structure. If this body has ess then tree inks then some movement wi be possibe. Such a body is caed mechanism. If we put on more then three support inks on the body than it is indeterminate structure. y y ϕ N y ϕ N x O x Pane body s degree of freedom (movement possibiity): x, y and ϕ y O x One degree of freedom mechanism. Possibe movement is Т y N N O x Determinate structure (no movement possibiity) abe to carry some oad. O x Indeterminate structure (no movement possibiity) abe to carry oad. In case we have no one body but severa numbers, the degrees of freedom is depending of the body s connection and supports. The way to cacuate of the degrees of freedom in such compicated structure is foowing: If there are no one eement cosed oops: w= d k a; where: w is degree of freedom (mobiity); d is number of bodies (eements); k is number of one-degree-of-freedom kinematic pin joints; a is number of support inks. The numbers of k is cacuated by the formuae: k = d, where d is number of the connected at the pin joint eements. - 9 -
d = k = d = k = d = k = We show some exampes for determination of degree of freedom. In the foowing exampe we have cosed oop but composed by two eements. d = k = 5 d = 5 k = 4 4 А 4 C B h B v d = 4; k = 4; a = 4; w =.4.4 4 = = 0 If there are one-eement-cosed-oops: w = ( m k) where: m is number of the cosed oops (incuding the basic disk ground (terra)); k is number of one-degree-of-freedom kinematic pin joints. In the case of the rod structures (trusses) we may use the next formuae: Т m = ; k = w = (. ) = where: d is number of eements; k is number of hinges; a is number of support inks. w= k d a d = ; k = 8; a = w =.8 = 0 And finay in the case of chains we have: w= d - 0 -
where: d is number of eements; On previous exampes we saw that the number of degrees of freedom w may be positive, negative or zero. So we distinguish three different cases for w: w > 0 - the system is mechanism. In the case of mechanism we don t have a structure carrying any oad; w = 0 - determinate structure. We have a structure and it is possibe to anayze it with ony equiibrium conditions. w < 0 - indeterminate structure. We have a structure and it is possibe to anayze it with equiibrium conditions and additiona equations. In this first stage we wi anayze ony determinate structures namey structures with w = 0. Basic kinematica eements and inks: d = ; w = = + It s known that if we cut the body of the beam par exampe, in the cut sections there are three body force shown on the figure: Q N cut N Q It is usefu for some structures to construct different type of connection between the disks ike pin joint we has shown and some other dispayed on the next figure. These types of connection are caed reeases. pin joint (hinge) reease N N Q Q In this section there is no bending moment - Q reease N N In this section there is no shear force - Q N reease Q Q In this section there is no axia force - N basic kinematica eements cantiever beam, simpe beam and dyad. They are simpe, stabe determinate structure and we use them for composing compicated systems. - -
The dyad is stabe ony if the three joints are not ying on one and the same ine. If they are then the dyad is unstabe and we ca it singuar dyad. А H А V B V simpe beam, w = 0; stabe system. cantiever beam, w = 0; stabe system. singuar dyad, w = 0; unstabe system. dyad, w = 0; stabe system. We may distinguish three types of basic kinematica inks (eements) for composing structures: kinematica inks Type this ink carry ony axia oad if there is no transverse oad. w = ; indeterminate system. kinematica inks Type this ink carry axia and bending oad and has one fixed and one pin support. w = ; indeterminate system. kinematica inks Type this ink carry axia and bending oad and has two fixed support. w = ; indeterminate system. Kinematica anaysis of determinate structures: By using basic kinematica eements, inks and chains we may compose different compicated structures. With the upper formuas we contro if the composed structures are determinate or not. But it is possibe construction to be determinate and to be mechanism at the same time. This phenomenon we ca kinematica instabiity and such a system mechanism. So the upper formuas give us information ony for the number of the inks but not for the kinematica stabiity. That is why we need a kinematica anaysis. In the fowing exampe we show this phenomenon. - -
А d = ; k = ; a = 4 w =.. 4 = b t th t i h i А d = ; k = ; a = 4 w =.. 4 = 0 the system is stabe. Another possibiity of this phenomenon is instanty unstabe system. Instanty unstabe because after some dispacement the system came stabe but therefore not good for design. А h А v B v B h d = ; k = ; a = 4 w =.. 4 = 0 but the system is instanty unstabe. А h А v sma dispacement B v B h After some dispacement the system is stabe but not good for design. This system is caed singuar dyad. The kinematica anaysis consist a way of composing the compicated structure. If we use ony stabe basic eements ike a cantiever beam, a simpe beam or a dyad the resut shoud be stabe structure. Т A 4 Т B C D D Т D D The kinematica anaysis of this structure is the next: At first we have ony the earth (terra). After that we construct the cantiever beam (eement ). It is stabe structure. Point A aready exists. On the next step we construct the dyad.. This dyad is based at point A on the cantiever beam and at point B on the earth. The dyad is aso stabe structure. Last step is composing of the simpe beam 4. It is based on the dyad at point C and on the earth at point D. This expanation of the kinematica anaysis we write in the fowing way: KA:[T + (w = 0) +. (w = 0) + 4.DD'(w = 0)](w = 0) - -
Another exampe - compound beam: A B B C D D KA:[T + (w = 0) +.BB'(w = 0) +.DD'(w = 0)](w = 0) The compound beam is composed by one cantiever beam and two simpe beams, a of them ying on one ine. Beam is based on beam and on the roer support BB. Beam is based on beam and the roer support DD. Beam is supported ony on the earth so this beam we ca primary beam and the two other we ca secondary beams. Depending on kinematica anaysis we cassify structures on two types: Type I: structures composed ony by using other stabe structures: cantiever beam, simpe beam or dyad. Type II: system which consist chains and inks. Previous exampes were of type I. Now we wi sha some exampes of structures type II: 4 5 6 B B B B KA:[T +..BB'(w = + ) + (w = ) + 45. (w = 0) + 6.BB'(w = 0)](w = 0) This system consist chain and ink. So it is not sure if the system is stabe or not. It is necessary to be made additiona verification. There are four typica ways for composing different structures: way I: compose structures ony by using cantiever beams, simpe beams and dyads starting of the earth (base disk). Using this way we have structure type I and we are sure it is stabe structure. way II: compose structures ony by using simpe beams and dyads but using one of the disks for base eement. As a resut we have composed a stabe cose oop. After that we base it on the earth. The kinematica anaysis in this case has two stages. The first composing the cosed oop and the second composing earth based structure. - 4 -
а 4 B B A A First stage: KA: a = [ +. (w = 0)](w = 0) Second stage: [T + a.aa'(w = 0) + 4.BB'(w = 0)](w = 0) way III: compose structures by using cantiever, simpe beams, dyads and chains and inks. In this case we use terra for a base of the structure. As a resut it is not sure if the structure is stabe or not. It is necessary to make additiona anaysis. Т A 5 4 B B Т C D D Т KA:[T + (w = 0) +..BB'(w = + ) + 5(w = ) + 4.DD'(w = 0)](w = 0) way IV: compose structures by chains and inks. In this case we use some of the disks for a base eement. The kinematica anaysis is in two stages. As a resut it is not sure if the structure is stabe or not. It is necessary to make additiona anaysis. 8 9 6 7 5 4 а А Т Т B B First stage: KA: a = [ + 67... (w = + ) + 4(w = ) + 5(w = ) + 89. (w = 0)](w = 0) Second stage: [T + a.bb'(w = 0)](w = 0) Let us consider in detais a way of support of a simpe beam and a dyad. The simpe beam is composed by one disk and three support inks ike it s shown: - 5 -
a) b) A c disk disk A c In case a) the direction of support inks and intersect in common hinge A c as a rotating point but support ink obstruct this rotation so the structure is stabe. In contrary in case b) the direction of the three support inks intersect at on and the same point the common hinge A c. As a resut the rotation is possibe and the structure is unstabe. As we know the dyad is composed by two disks connected by a hinge (common or not) and supported by two fixed support (one fixed support is composed by two intersected support inks). b) A c a) A c C B c C c B c 4 4 c) A c C B c d) C c A c B c 4 4 In case a) the dyad has two fixed supports at the common hinges A c and B c and one rea hinge C between the disks. The three hinges A c, B c and C c are not ying at one ine, that is why the dyad is a stabe one. The same situation is in case b) but the common hinges A c and C c are at infinity. In cases c) and d) the three hinges are ying at one ine and that is the reason the dyad is unstabe. The difference is that in case d) the common hinge C c is at the infinity but as is known a horizonta ines intersect at the horizonta infinity and a vertica ines intersect at the vertica infinity. If we know enough about principes of structura composing and common hinges we may answer the question Is the construction stabe or not? very easy some time as in the foowing cases: - 6 -
a) b) At the first gance at the pictures one may say that these are very compicated structures but after that it shoud be cear that at figure a) we have three-hinged beam and at figure b) it is a simpe beam. At figure a) the two trianges are cose-oops and have a sense of one disk each of them. These two disks are connected with two inks crossed at medde where is the common hinge. The eft triange disk is supported by two inks crossed at one point. This point is actuay fixed support for the disk. The right disk is supported directy by fixed support. At figure b) the two trianges compose one cose-oop disk which is supported by one ink (in the eft) equa to roer support and one fixed support at the horizonta infinity composed by the two horizonta inks. So at the resut we have simpe beam at figure b). As we as we know that these structures are simpes we can be sure that they are stabe if they agree with upper rues. This way for anayze structures is very convenient in most cases but there are some situation in which it isn t possibe to use it. That is the reason to perform common method for anayzing compicated structures for there determination and stabiity. Common method for kinematica anaysis of determinate structures: Before we present the common method we shoud expain some kinematica theorems and determinations. ajor poe of rotation: This is the poe around which rigid body rotates. Fixed supports are usuay major poes. Reative poe of rotation: This is a point around which two rigid bodies reativey rotates. idde and common hinges are usuay reative poes of rotation. First Theorem: If we have a mechanism of two connected bodies so they have major poes each of them and one reative poe. Aays these three poes (the two major and one reative) are ying at one ine. major poe reative poe According to the second theorem somewhere on this ine the reative poe (,) is ying! major poe () () (,) Second Theorem: If we have a mechanism of three connected bodies there reative poes are ying at one ine. (,) (,) (,) () () According to the first theorem somewhere on this ine the reative poe (,) is ying! - 7 -
As it s shown at the figure using these two theorems we may find the position of some poe which is not obvious at first. Additionay we may write these conditions in the foowing provisionay way: () + () = (,) (,) (,) + (,) = (,) This equation shoud be red as: The major poes () and () of the disks and determine a ine on which the reative poe (,) shoud ie. The reative poes (,) and (,) determine a ine on which the reative poe (,) shoud ie. The cross point of the two ines determine the exact position of the reative poe (,). First Additiona Theorem: If one major poe appears at two points at the same time then the poe do not exist. When one major poe don t exist then the corresponding body do not moves. Second Additiona Theorem: If one reative poe appears at two points at the same time then the poe do not exist. When one reative poe doesn t exist then the corresponding bodies do not reativey moves. They move ike one and the same disk (body). () (,) () () (,) () In this exampe the major poe () shoud ie at a ine determinate by poes () and (,) but at the same time it is known the exact position of this poe (at the right fixed support). Hence the major poe () appears at two different positions so do not exist. As a resut the disk doesn t moves. Owing to this the reative poe (,) becomes major poe (). But this poe aready exists so the disk doesn t moves too. In the other hands the reative poe (,) shoud ie at a ine determine by poes () and () but its exact position is known by the midde hinge. As a resut poe (,) do not exist. In consequence disks and moves ike one disk. But this one disk has two major poes at the two fixed supports. Hence this disk hasn t a major poe. Thus the disk doesn t moves. Actuay the shown system is dyad and as we know is stabe. That is why poes of movement do not exist. But this is very good exampe to show haw the additiona theorems works. This exampe aso shows that the theorems of kinematica mechanism can be used as a source for verification of structura stabiity. Thus foows the common method for kinematica anaysis. First we make Determination of degrees of freedom. If the structure is staticay determinate we continue with the next step: kinematica anaysis the way of composition. If the structure is composed by using firs or second way of composition then it is known the structure is stabe and more verification isn t needed. If it is used third or fourth way then verification for stabiity is needed. If it is possibe we may identify the structure as an eementary one simpe beam or three-hinged frame dyad ike it was shown previousy. If not we continue with the common method of verification. - 8 -
Common method for kinematica anaysis. ) We remove the ast ink composed according to the kinematica anaysis. The two ends points of the ink determine a ine we ca it a-b. ) The removed ink has connected two other disks. We compose the pan of the poes and find the reative poe to these disks. ) If this reative poe is ying at the ine a-b then the determinate structure is instanty unstabe one. If the reative poe isn t ying at the ine a-b then the system is stabe and we may determine reactions and interna forces caused by some oads. Exampe : ake fu kinematica anaysis of the structure.. Determination of degrees of freedom: w = d k a. T 4 T 4 d = 4 ; k = 4; a = 4 w =.4.4 4 = = 0 w = 0 The system is staticay determinate.. Kinematica anaysis of the structure: [ T +..( w = + ) + 4( w = ) ]( w = 0) The system is composed by using chains and inks (way IV); therefore it is second type structure. It is necessary to make verification for kinematica stabiity.. Identification of the structure as an Eementary system with common hinges : - 9 -
А С c В В А C c This system can be considered as an eementary one. It is three-hinged frame type, with two rea support inks, caed A and B and a common media hinge С c. The three hinges are not ying at one and the same ine, so the system is stabe. It is not needed but we wi show the common method of verification. 4. Verification for kinematica stabiity by the common method. ) At first we remove the ast ink according to the kinematica anaysis ink 4. [ T +..( w = + ) + 4( w = ) ]( w = 0) ) We compose the pan of the poes. Searching for reative poe (, ) the reative poe of disks, which were connected by the removed ink: () a (,) () b (,) () + () = (,) (,) + (,) = (,) (,) (,) The serched reative poe (, ) is at the vertica infinity. The reative poe (, ) is not ying at the straight ine a-b, so the system is stabe. - 0 -
Exampe : If the two connecting disks of the upper system are vertica, it is transferred into an instanty unstabe system. А В The kinematica anaysis is the same as the previous one. [ T +..( w = + ) + 4( w = ) ]( w = 0). Identification of the structure as an Eementary system with common hinges : А С c В А В С c This system can be identify once again as an eementary system with common hinges and three hinged system type with two rea and one common hinge. In this case the media hinge С c is on the vertica infinity, so three hinges are ying at one ine (a parae ines are crossed in one point to infinity). If the three hinges (A, B and С c ) are ying at one straight ine, then the system is an kinematicay unstabe.. Verification for kinematica stabiity by the common method. a. Remove the ast ink according to the kinematica anaysis ink 4. [ T +..( w = + ) + 4( w = ) ]( w = 0) b. Compose the pan of the poes searching for reative poe (, ) the reative poe of disks, which were connected by the removed ink: - -
() a (,) () b (,) () + () = (,) (,) + (,) = (,) (,) (,) The serched reative poe (, ) is at the vertica infinity again. In this case however, the reative poe (, ) is ying at the straight ine a-b, so the system is instanty unstabe. - -
Chapter Anaysis of eementary structures. In this chapter, we wi consider the procedure of anaysis of eementary structures ike a simpe beam, cantiever beam, dyad (three-hinged frame) and compound beam. For this reason, first of a remind the definition of a force and a moment of force to some point. From the physics, it is known that the force is a vector, which has a sign, direction, and vaue and appication point. The force is a representation of some oad, which causes damages (deformations and dispacements) of the body on which act. If the force acts at arbitrary direction, we may decompose it at the two mane directions horizonta and vertica as it s shown on a figure. The action of the decomposed force is the same as this of the whoe one. F v F h F h F v F h α α b F v x r y A b The force components are: F F h v = F cosα = F sinα Aso for the force components we have the foowing dependencies: F F h v = a Fh b = F v a b and F v = F h b a The moment of the force reated to the point A may be cacuated by foowing different ways According to the figure: A = F. r ; A = Fh. b; A = Fv. a; = F. y + F. x A h The frequenty used ways are the first and the ast ones. In the second and the third cases is used the transated forces aong the directrix of the force. Furthermore when we tak about interna forces in a beam eement then the forces and the moments are integra (a reduction) of the stresses acting in the center of the cross section of the beam. When the beam is under panar oad the as an interna forces we have two forces: axia and transversa and one moment. Their positive positions are shown at the next figure: v - -
N x In our next expanations, we wi show it in the simpe way as foowing: Q N N Q Q From the physics is known that a actions have counteractions. Therefore, if we know the moment and the resutant forces of the oad we may say the vaues of the interna forces of the beam eement, because they are equa. Actuay if we know the support, reactions and oads we just needs to compose the three equiibrium equations for the cross sectiona point and wi find the vaues of the interna forces as it is shown at the next figure: F F v P F h A h A v a B A h a P Q N A v H = 0 : V = 0 : = 0 : P A A h v + N = 0 N = A Q = 0 Q = A A. a = 0 = A. a v v h v Aong with we aways may determine the interna forces at each character point of the beam. In addition, if we know some rues we may compose the interna force diagram. Some of these rues are as foow: ) At force oad point the interna moment diagram has a kink and a shear force diagram has a jump; ) If some section of the beam hasn t any oad then the interna moment diagram is inear and shear force diagram is a constant; - 4 -
) If some section of the beam is under distributed oad then the interna moment diagram is paraboic of second degree and shear force diagram is inear; Next tabe iustrates these and some other rues briefy: Load oment Diagram Shear Diagram F P P P Q Q kink at point P jump at point P Q P P P constant Q jump at point P q Q Q paraboic function inear function Now it wi be iustrated procedure of anaysis of some eementary structures:. Support reactions. Before the member is cut or sectioned it is necessary to determine the member s support reactions so that the equiibrium equations are written for the whoe member.. Interna force diagrams: The members shoud be sectioned or cut at specific points and equiibrium equations shoud be written for the separate part for determining interna forces and to compose the interna forces diagram. Exampe : Simpe beam under point force oad. F А H. Support reactions. А V / / B V H = 0: Ah = 0 A = 0: B v. F. = 0 B F v = B = 0: A v. F. = 0 A F v =. Interna force determination. The beam has two sections and one specific point at the midde of the beam to find the moment and shear vaues. - 5 -
F/ = 0 : = A. F F. v =. = 4 The interna moment is positive because the ower bands are bended. The moment diagram shoud be potted at the bended side of the member. The moment diagram is inear with a kink under the oad point. About the interna shear force, we have two specific points at the midde of the beam, one next before the force end the second next after the oad force. The reason of this is to determine the jump in the shear force diagram as it is iustrated in the up tabe. The shear force diagram is a constant with a jump under the oad point. next before the oad force: next after the oad force: F Q Q r / / F/ F/ V = 0:Q = A F v = ; r V = 0:Q = A F v F= The norma force is zero because there isn t any horizonta oad.. Interna force diagram. / F F/ / / F/ F./4 F/ Q F/ Exampe : Cantiever beam under distributed oad. q - 6 -
In this case, it is not necessary to determine the support reactions because the right side of the body is free of supports and it is possibe to cut and to separate the right-hand side part of the member and to determine the interna forces. The distributed oad oads the cantiever beam so that the interna moment diagram shoud be paraboic function. There are needed three vaues for potting the diagram. The first one can be the free end of the beam, the second one is at the midde of the beam and the ast one is at the support. The shear force diagram is inear and it is sufficient to determine its vaue at two points at the free end and at the supported end of the beam. ) ) q ) q = 0 0 the free end of the beam / Q = 0 Q = 0 the midde point of the beam the supported point of the beam q. Interna force diagram. q / / q /4 q /8 q/ Q Exampe : Three-hinged frame. q h/ C h/ F A h A v a b B v B h. Support reactions. The three-hinged frame has two support reactions at each pin support. It is possibe to write ony three equiibrium equations to the whoe frame but there are four reactions. It is necessary to find - 7 -
one more equation. The best equation in this case is a moment equation about the midde hinge. The reason of this is that we know the moment at the hinge point is zero. When we write this equation, we take one part of the frame the right one or the eft. If we take the right one we use the eft part to verify the resuts. If we take the eft one, we use the right for verification. This is iustrated beow. Equiibrium equations for the whoe frame: h b A = 0 : F. + + q.. a + Bv. ( a + b) = 0 Bv h b B = 0 : F. + q.. + Av.( a + b) = 0 Av = H = 0 : A + F B = 0 B = A + F h h h h = The case of frames with pin supports on one eve is easier because from the first two equations we can find directy the vaues of the vertica support reactions. Equiibrium equation for the midde hinge of the frame (the right-hand side): C C h h/ C v F h/ A h A v a C = 0: F. h + A v..a A h.h = 0 Ah = B = A + F B = h h h Verification of the resuts: Equiibrium equation for the midde hinge of the frame (the eft-hand side):?? r C = 0: + q.. b B v.b + B h.h = 0-8 -
B h C q B v h b B v B h Equiibrium equation for the whoe frame:?? V = 0: A q. + B = 0 v As we know the correct vaues of the support reactions, we may compose the interna forces diagrams. In the present frame, we have 5 segments (shown on the figure beow) for which we shoud determine these diagrams. Consequenty, we have at east 0 specific points for the shear force and 7 for the moment because we know the moment vaue at the hinges is zero. The major specific points are shown on red at the scheme beow. q h/ h/ F A h 4 5 6 7 8 C 4 v 5 9 0 A v a b B v B h Actuay, we may compose the interna moment diagram using ony specific points and using this diagram we compose the shear and norma forces diagrams but if we know a characteristics of the interna forces which wi be expained in the next chapter. Composing the diagram, we cut part of the frame and separate it. After that, we cacuate the moment the shear and norma forces at the specific point. The way of separating of the frame for some of the specific points is shown at the figure beow. q h/ h/ F A h 5 C 9 A v a b B v B h - 9 -
ore detais about diagram composing wi be shown in ater. Now wi be demonstrated the difference when the pin support of the frame are at different eve. The mane procedure of soution is the same as previous with ony one different aspect, namey that we cannot find directy the support reactions. In this situation we have system of inear equations. h/ q C F h/ A h c A v a b B v B h One way for determining the reactions is as foow: Verifications: b ( ) ( ) A = 0: F. h + + q.. a + B v. a + b + B h.c = 0 B = 0: F. h + q.. b + A v. ( a + b) + A h.c = 0 H = 0:Ah + F Bh = 0 c = 0: F. h + A v.a A h.h = 0?? r C = 0: + q.. b B v.b + B h.h = 0?? V = 0: A q. + B = 0 v v Another way is simiar as next: Verifications: b ( ) ( ) A = 0: F. h + + q.. a + B v. a + b + B h.c = 0 B = 0: F. h + q.. b + A v. ( a + b) + A h.c = 0 H = 0:Ah + F Bh = 0 r C = 0: + q.. b B v.b + B h.h = 0?? c= 0: F. h + A v.a A h.h = 0?? V = 0: A q. + B = 0 v v The difference is ony at the fourth equation and the first one at the verifications. Every four equiibrium equations are usefu for determination of the support reactions and at east one equation verifying resuts. Therefore, these two variants are not ony the possibe. Nevertheess, one shoud - 0 -
be aways carefu with a partia equations and one shoud know that ony equations can be written for the whoe system! The composing of the interna forces diagrams is the same. The next modification of the three-hinged frame is the tied three-hinged frame. We wi iustrate its soution. h/ q C F h/ d N N c A v a b B v B h One of the supports of this frame is roer and another is pin. As addition is added a ink of type I with ony one interna force the norma one. That is why the three equations for the whoe system are enough for reactions determination. Whit using the partia equation we determine the norma force at the ink. Verifications: H = 0:F Bh = 0 Bh = A = 0: F. h + + q.. ( a + b) B v. ( a + b) + B h.c = 0 Bv = B = 0: F. h + q.. b + A v. ( a + b) = 0 Av = c = 0: F. h + A v.a N.( h d ) = 0 N =?? r C = 0: + q.. b B v.b + N.( h d ) = 0?? V = 0: A q. + B = 0 v v The composing of the interna forces diagrams is the same. Exampe 4: Compound beam: As it is known from the kinematica anaysis, basic and secondary beams compose the compound beam. The secondary beams transfer the oads to the basics ones. That is why we first anayze the secondary beams and with their support reactions we oad the basic beam. - -
F A B / / 4 A h C h C C v D D q P v P F A v B. Support reactions. Beam CD simpe beam: H = 0: Ch = 0 4 q4 D = 0: C v.4 q 4. = 0 Cv = 4 q C = 0: D v.4 q 4. = 0 Dv = Beam AB simpe beam with overhangs: H = 0: A = 0 h q4 q4 B 0: A v. 0 A v = + + = = + q4 A = 0: ( + ) + B. = 0 B = Beam PA cantiever beam: H = 0: Ph = 0 P = 0: A v. + F. + P = 0 P = V = 0: P F A = 0 P = v v v 4. Interna force diagram. The diagram composing makes for every beam separatey and after that, we join them for the whoe compound beam. Here we show ony the shape of the fina diagrams because the exact vaues are not so important at this moment. - -
F A B / / 4 C D Q - -
Chapter 4 Shear and moment functions. Anaysis of structures type I. As it is known from the Strength of materias, there is a connection between the interna moment and the interna shear force at a beam eement. As a basis of this connection, we wi discus some characteristics of the moment and shear diagram. As we say aready if one know this characteristics may compose these diagrams without any probems and with a few cacuations and as most important on can check if the composed diagrams are correct or not. Connections between distributed oads, shear and moment functions: sope of shear diagram = intensity of distributed oad sope of moment diagram = shear function The first equation states that the sope of the shear diagram at a point is equa to the intensity of the distributed oad at the point. Likewise, the second equation states that the sope of the moment diagram is equa to the shear at the point. These equations can be integrated from one point to another between concentrated point or coupes and as a resut we have as foow: change in shear = area under distributed oading diagram change in moment = area under shear function As it is noted the first equation stats that the change in the shear between any two points on a beam equas the area under the distributed oading diagram between the points. Likewise, the next - 4 -
equation states that the change in the moment between the two points equas the area under the shear diagram between the points. If the area under the oad and shear diagrams are easy to compute then these equations can be used for determining the numerica vaues of the shear and the moment at a various points aong a beam except points with a concentrated force or moment. The next tabe iustrates the appication of these equations for some common oadings cases. The sope at various points is indicated. Each of these resuts shoud be studied carefuy so that one becomes fuy aware of how shear and moment diagrams can be constructed based on knowing the variation of the sope from the oad and the shear diagram respectivey. Load oment Diagram Shear Diagram F Q P r Q r r sope = Q sope = Q r positive constant sope Q sope = 0 sope = 0 zero sope Q r P P r sope = 0 sope = 0 r zero sope sope = 0 zero sope Q q Q r r r sope = Q sope = Q r positive increasing sope Q sope = - q constant sope Q r q q Q Q r r r sope = Q sope = Q r positive increasing sope sope = - q sope = - q Q Q r negative decreasing sope q q Q Q r r r sope = Q sope = Q r positive increasing sope sope = - q sope = - q Q Q r negative decreasing sope As an addition heping information we wi expain the usage of a method of superposition for composing the diagrams. We aready notice above that one system oaded by more then one oads can by soved for every one of then separatey and the resut is a sum of a separate soutions. The next figure iustrates shorty the principe of superposition: In this case we have simpy supported beam oaded by a point force at the midde and two moments at the two ends of the beam. The moment diagram from these oads is shown at the right top of the figure and the separates oads and diagrams are shown beow. - 5 -
F r r А H А V / / B V А H А V F/4-( + r )/ / / B V F А H А H А V / + / B V r А V F/4 r / + B V r А H А H А V / + / B V А V / / + / B V А H А H А V / / B V А V / / B V As one can see the two moments, acting at the two ends of the beam can be considered together and if we know their vaues, it is very easy to compose the diagram. It is aways a trapezium as it is shown beow and the midde vaue of the diagram equas to the midde vaue of the trapezium. Therefore, if we know that for the above simpe beam is ony needed to add the diagram from the point oad to the trapezium and the summary diagram wi be computed. r r / r А H А H А V / + / B V А V / / + / B V А H А H А V / / B V А V / / B V r r А H А H ( +М r )/ А V / / B V А V / / B V - 6 -
Now we wi continue with another aspect of the usage of the superposition principe. Let us consider a beam eement type III as it is shown at the next picture. Let us consider we know a support reactions caused by the point oad for exampe. Let us now compose the moment diagram in the beam eement. Here a support reactions are known so the force system of the beam eement produces zero force and moment resutant or in other words the system is in equiibrium state. The moment diagram is shown right to the beam eement. Note that the same diagram wi be produced by simpe beam oaded by the same oad system (incuding support reactions), because the oad system is at equiibrium state.. In this case, the two horizonta reactions are incuded but they are equa and are not significant for the soution. А H F r B H А H r B H F/4+( + r )/ А V / / B V А V / / B V А H F r B H А H r B H F/4-( + r )/ А H А V / + F / B V B H А H А V / + / B V B H А V / + / B V r А V F/4 ( +М r )/ + B V r А H B H А H B H А V / / B V А V / / B V As a resut, we consider a simpe beam with known moments at the ends and a point oad at the midde as in the previous exampe. Therefore, for composing the summary moment diagram it is enough to sum the midde vaue of the trapezium (received by the two moments) and the midde vaue of the moment diagram in the simpe beam oaded by the point oad at the midde point. Note that nowhere we use the vertica and horizonta support reactions. Ony the two ends moment as support reactions are used. The moment diagram produced by these two ends moment we ca reference diagram and we us it as a benchmark. The diagram caused by the externa oad (the point oad in this exampe) we ca additiona diagram. Superposing the reference diagram with the additiona one, we receive the summary moment diagram. Note that the additiona diagram is aways at a simpe beam. The next exampe iustrates this again. Let us consider a part of the frame oaded by the distributed oad. Notice that if we separate this part it wi be the same as a beam eement type III as a previous one. Therefore, the summary - 7 -
moment diagram can be produced as a summation of a reference and additiona diagram if we know the two ends moment of the frame part. q h/ F h/ A h A v B v q /8-( + r )/ N q r N r N N r N Q r Q r N r N Q ( +М r )/ Q r r N r N Q + q Q r N r N Q + Q r N r Q Q r Q q /8 Q r Note that in this case again the interna horizonta and vertica forces are not incuded in the soution. Therefore, it is enough to know the two ends moment of the frame part. This usage of the superposition principe is the most powerfu and faster method for composing the moment diagram. That is why we wi use it further. The next step is the composing the shear force diagram with a faster method. This is very easy if we use the connections between the moment and the shear force. We aready expained this connection and now we wi iustrate its usage. As two important rues, we wi mention the next: The shear force is takes from the moment diagram as its tangent at a point (the shear force is equa to the sope of a moment at a point). If the moment diagram is a rising function then the shear is positive and if the moment is decreasing function then the shear is negative. In most cases, the moment function is inear so the tangent is very easy to find and the shear function is a constant: - 8 -
α r - a inear moment diagram r constant shear diagram If the moment diagram is paraboic function then we decompose it to a reference inear and additiona diagram and compose the shear force diagram using the same idea. The reference shear diagrams takes from the reference moment diagram as a tangent and the additiona shear diagram is a diagram at a simpe beam: r paraboic moment diagram a α α reference moment diagram + q /8 r additiona moment diagram r + constant reference shear diagram q/ Q b + q/ + q/ additiona sear diagram Q b - q/ summary shear diagram The norma force diagram composes using the support reactions and shear forces at the corners nodes. Anaysis of structures type I. As we aready know haw to compose diagrams faster and easier we are ready to anayze fuy different compicated structures. First, we shoud make the kinematica anaysis of the structure. The kinematica anaysis consist a way of composing the compicated structure. If we use ony stabe basic eements ike a cantiever beam, a simpe beam or a dyad the resut shoud be stabe structure. If the structure is staticay determinate and stabe we may anayze it in way opposite to its composing. After that for the decomposed structure, we cacuate the support reactions and summaryy using decomposed structure, we produce the interna forces diagrams. - 9 -
Exampe : 0 0 5 4. Kinematica anaysis: [ T + AA.. BB ( n = + ) + CC ( n = ) ]( n = 0) А ф С С Т А А А А А Т А В В Т В А ф The system staticay determinate and is composed with using of chains and inks, so it is second type. Verification of instantaneousy unstabe is necessary. The system can be identified ike eementary one simpe beam type. The directions of the three support inks are not crossing at one point therefore, the system is stabe.. Support reactions. This system is eementary one that is why it is not necessary to decompose it. We may compute the support reactions and interna forces diagrams directy. Lets equiibrium equations be written for such a points if possibe for which ony a singe unknown participate. - 40 -
q=0 А c C F=0 Р,5 =5 А В Ac = 0 : 5 + 0. 4. + B. 5 0. = 0 B = 5 A = 0 : 5 + 0. 4. C. 5 + 0. 4 = 0 C = 5 V = 0 : A 0. 4 = 0 A= 40 Verifications :? H = 0 : 0 + 5 5= 0 5 5 = 0 = 0 : 5 0.. 4 4 5. + 0. 5. + 40. 6 = 0 75 75 = 0 P For verification of support reactions we use usuay at minimum one moment equation. As advise use such a point for wich participate maximum number of aready determinate supports reactions.. Interna forces diagrams. 4 oment diagram is organized by sections at characteristic points and the aready expained properties of the diagram. In this system, the ocations of the sections and their sequence are shown on the next figure. The separated sections are aso shown. The shear diagram is produced using the moment diagram according to their connections and the norma force diagram is composed by the shear force diagram and supports reactions. At ast we verifications are made using the corner nodes force equiibrium.? - 4 -
V I пр. q=0 VI F=0 пр. C = 5 VII IX,5 II III =5 IV VIII В = 5 А = 40 4 I I II II III III q=0 = 0.. = 0 = 40. = 40 = 40. + 5 = 55 =5 А = 40 А = 40 IV IV = 40. + 5 = 95 =5 V V = 40. + 5 0.. = 75 q=0 VI VI =5 C = 5 А = 40,5 F=0 А = 40 = 5. = 5-4 -
F=0 VII VII C = 5 VIII VIII = 5. = 45 IX IX C = 5 F=0 = 5. + 5. 0. = 95 = 5. 0. = 50 В = 5 В = 5 oads. Determination of the moment and shear diagrams for the parts oaded by the distributed for the cantiever part: М b = 0 0 75 for the interna part: М b = 85 95 + = = М = 5 0 75 95 М = 90 for the cantiever part: + 0 + = 0 Q b = 0 0 0 0 + + for the interna part: + + = 0 Q р = 0-4 -
5 0 75,5 40 40 5 75 0 95 90 95 М 45 5 50 95 5 95 95 50 45 4 0 + - 0 +,9 Q 5-5 - 5-0 0 α α α,9,8 40 4 5,8 5 - N 4-44 -
Exampe : 0 5 0,5. Kinematica anaysis: [ +.( n = 0) ]( = 0) a = n 4 D a Е A B F B C Т [ T + a. BB ( n = 0) + 4. AA ( n = 0) ]( n = 0) This system is staticay determine an composed by using the second way of kinematica composing. Therefore, the structure is cinematicay stabe and shoud be anayzed in order opposite of the composing order. That is why we wi anayze first disk 4 after that the common disk a, next the dyad. and in the end disk.. Support reactions. disk 4: = 5 P D H D V,5 A - 45 -
disk a: 0 D V = 7,5 B =,5 0 F C H C V = 0 dyad.: E V = 0 0 E H = 0,5 F 0 disk : In this disk there is any unknown that is why its equiibrium is ony for verifications: N 0 4 0 D E 0 0 0-46 -
. Interna forces diagrams. 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0,5,5 7,5 + 0 + 0 Q 0 - + 0 + - - 0 0 0 0 0 0 0 0 0 0 0 7,5 0 0 0-7,5 - -,5 0-0 + N - 0,5 0,5 7,5 7,5 0 0 0 0,5 0 0,5 0-47 -
Exampe : q А А B B С С 4 Т. Kinematica anaysis: The separation on parts of the compound beam into basic and secondary beams is shown on the figure above. The system has two basic and two secondary beams. The basic beams are and the cantiever 4. Beam and beam are secondary. The whoe soution is shown beow. F q 0 S =5 0 С = 60 S = 5 0 5 0 5 5 A =,6667 В = 56,6667,666-7,5,666-5 50 + 0 S =5,5 5 5 - S = 5 5 5 + + 5 = 65 H = 0 V = 5 М 6 Q 5-48 -
Chapter 5 Principe of virtua work for rigid body. Energy methods. Beams may be anayzed using the equations of static equiibrium and the method of sections, as iustrated. Aternativey, the principe of virtua work may be utiized to provide a simpe and convenient soution. In this chapter we wi iustrate for anaysis of simpe structures. At first we wi expain the principe of virtua work for rigid bodies. Principe of virtua work: The principe of virtua work may be defined as foows: Consider a structure in equiibrium under a system of appied forces is subjected to a system of dispacements compatibe with the externa restraints and the geometry of the structure. The tota work done by the appied forces during these externa dispacements equas the work done by the interna forces, corresponding to the appied forces, during the interna deformations corresponding to the externa dispacements. Or more simpe: When a rigid body that is in equiibrium is subject to virtua compatibe dispacements, the tota virtua work of a externa forces is zero; and conversey, if the tota virtua work of a externa forces acting on a rigid body is zero then the body is in equiibrium. The expression virtua work signifies that the work done is the product of a rea oading system and imaginary dispacements or an imaginary oading system and rea dispacements. Consider a system of disks, in static equiibrium state,that is why the resutant force roar a disks is zero.http://en.wikipedia.org/wiki/virtua_work - cite_note-torby984-0#cite_note- Torby984-0 R F = 0 i is a number of disks. Summing the work produced by the force on each disk that acts through an arbitrary virtua dispacement, δr i, of the system eads to an expression for the virtua work that must be zero since the resutant force is zero: i i δw R = F.δr = 0 i i i In this equation the exprecion the resutant force means a tota force force and moment resutant. The same equation in more detais: δw = F.δd +.δφ = 0 j j j j j j here: j is the number of forces or moments; δd is the projected dispacement; F is the force working on the projected dispacement; δϕ is rotation; is the moment working on the rotation. - 49 -
The origina vector equation coud be recovered by using virtua work equation. That work expression must hod for arbitrary virtua dispacements. If arbitrary virtua dispacements are assumed to be in directions that are orthogona to the constraint forces, the constraint forces do no work. Such dispacements are said to be consistent with the constraints. r α R δϕ α δd F δd fu virtua dispacement of the appication force point; projected virtua dispacement of the appication force point; δϕ virtua rotation of the rigid body; r ray of the force refer to point of rotation; Usage of the principe of virtua work for determination of reactions or interna forces: Procedure of anaysis: ) Remove the ink, which hods the searching reaction or interna force; ) Impose a virtua dispacement to the system; ) Write a virtua work equation and determine the searching reaction or interna force. When impose the virtua dispacement (rotation) we use the pan of the poes (if it is necessary) and express a rotations and dispacement as a function of the virtua one, therefore we shoud have ony one unknown parameter the virtua dispacement. Exampe : Determination of the support reaction and interna forces for a simpy supported beam oaded by point force. F А V a b B V F А V ϕ a δ.a/ B V δ = It is possibe to use ony forces or moments for writing the virtua work expression. - 50 -
F. δ.a δw = F.d = F. δa + B.δ = B = = Fa Using projected dispacement: ( ) 0 v v δ F.a.φ Using the rotation ϕ: δw =.φ = B v..φ F.a.φ = 0 B Fa v = =.φ Simiary, as shown at next figure, the bending moment produced at point P by the appied oad may be determinate by cutting the beam (removing the ink) at P and imposing a unit virtua reative rotation of δϕ =. Evauating interna and externa work done gives: P F А V a b B V F α P P δ α А V a δϕ = δw = δwext + δwint = 0 δwext = δwint δwext = F.δ = F.a.θ P P P P aθ P P δwint =.α.α = θ = θ( + a ) = θ b b b P F.a.θ = θ b b P P Fa = = F B V ab When the direction of the force and the dispacements are at the same then the work done is positive when they are oposite the work done is negative. When the direction of the moment and the rotation are at the same then the work done is positive when they are oposite the work done is negative. If we want to evauate the shere force at point P we shoud cut the beam (removing the ink) at P and imposing a unit virtua reative dispacement of δ =. P F А V a b B V А V α a F Q P Q P δ δ α δ = B V - 5 -
At the first soution we wi use the virtua dispacement δ = as a parameter in the equation. The externa force we wi impose at the eft side of the beam and wi receive the right side interna shear force. This soution is done beow: δw = δw ext int = = a δwext F.δ F P P P P P P δwint = Qδ a b a b Qδ = Q Q = Q + = Q a = = P P F Q Q F a ( ) At the second soution we wi use the virtua dispacement θ = as a parameter. The externa force we wi impose at the right side of the beam and wi receive the eft side interna shear force. In this case we use the fact that the two ines of the dispaced form of the beam are parae. This soution is done beow: δwext = δwint δwext = F.δ = F.b.θ P P P P P P δwint = Q δ Q δ = Q.a.θ Q b.θ = Q ( a + b) θ = Q θ P P F.b.θ = Q θ Q = F b Exampe : Determination of the interna moment for a three-hinged frame using the pane of the poes. 0 A h P C 0 A v 5 4 B v B h First we soud compose the pan of poes. It is shown at the next picture with the bue ines and a nessacery dimensions. How it mades we aready discused. (,) 4 () R = q.4 F = 0 P P (,) q = 0,49 A h () A v = 5 () B v Bh 4-5 -
The next step is to determine how disks rotates when impose the reative rotation at point P now it is point (,). The dispaced shape of the frame is shown with the green ines at the figure beow: () α (,) α α α α () α (,) r α α α () () α Using showed geometry is determinate the rotation of the disk. r = α r = ( ) (, ).α ( ) (, ) r = α r = ( ) (, ).α ( ) (, ) ( ) (, ) ( ) (, ).α = ( ) (, ).α α =.α ( ) (, ) This formuae may be used for every two disks and can be written generay for any disks n and m as foow: ( n) ( n,m) α m =.αn ( m) ( n,m) here: ( n) ( n,m ) is the dimension of the segment between the mane poe (n) and the reative poe (m,n); ( m) ( n,m ) is the dimension of the segment between the mane poe (m) and the reative poe (m,n); α m is the rotation of disk m; α is the rotation of disk n. n Using this we determine a rotations at function of one of them as foow: α = θ ( ) (, ) 4, α =.α = = 4,.θ ( ) (, ) 0, 987 ( ) (, ) α 4 =.α = = 08,.θ ( ) (, ) 0 Verification : ( ) (, ), 97 α =.α =. 4, θ = 08,.θ ( ) (, ) 6, 40-5 -
If we are not sure about the directions of the disk rotation, it is easiy to choose a positive direction of rotation. For exampe, et choose the cockwise direction for positive, then each force, which rotates at such a direction reated to the mane poe of rotation, wi have a positive work done. Opposite, each force, which rotates at anticockwise a direction reated to the mane poe of rotation, wi have a negative work done. If the moment rotate at cockwise direction wi have a positive work done and opposite if moment rotate at anticockwise a direction wi have a negative work done. After a, we may compute the interna moment at point P for the three-hinged beam as foow: δw = F..α.α +.α R..α + α = 0 P P P P F..θ.θ +. 4, θ R. 08.,.θ + 508., θ = 0 P,. = 0. + 4008.., 508., P =, 75-54 -
Chapter 6 Infuence ine for staticay determinate structures. In the previous chapter, we have expained a technique for anayzing structures for dead (fixed) oad. If the oad is moving, not fixed, then the moment end shear interna forces shoud be anayzed by using infuence ine. Infuence ines have important appication for the design of structures that shoud resist different ive oads. Here we wi discuss how to draw the infuence ine for a staticay determinate structures and its appication for determination of the absoute maximum ive shear and moment in a members. An Infuence ine represents the variation of the reaction, shear, moment, or defection at a specific point in a member as a concentrated force moves over the member. Once this ine is constructed, one can te where the moving oad shoud be paced on the structure so that it creates the greatest infuence for the interna forces of the specific point. Furthermore, the magnitude of the associated reaction, shear, moment, or defection at the point can then be cacuated from the ordinates of the infuence-ine diagram. Athough the procedure for constructing an infuence ine is rather basic and one shoud ceary be aware of the difference between constructing an infuence ine and constructing a shear or moment diagram. The infuence ines present the effect of the moving oad ony at a specified point on a member, whereas shear and moment diagrams represent the effect of a fixed oads at a points aong the member. Infuence ines composition: For constructing infuence ines, we shoud know the foowing important notes: The infuence ines are composed by straight ines for staticay determinate structures; For constructing the infuence ines, we use a moving concentrated vertica force at a dimensioness magnitude of unity; The way where the force moves we wi ca the moving path ; We draw the infuence ines at a basic ine, not at the structure axis; The positive vaues of the infuence ine we draw at a bottom side of the basic ine; An ordinate (the vaue) of some point at an infuence-ine diagram correspond to the position of the unit oad. We wi discus the two methods of determination of a infuence ine: the static method and the cinematic one. First, we wi present the static method by its variations. Static method for constructing infuence ines: Usage of an infuence-ines function: An infuence ine can be constructed by pacing the unit oad at a variabe position x on the member and then computing the vaue of the interna force or reaction at a specific point as a function of the unit oad position. In this way, the functions of the different infuence ine straight segment can be computed and potted. - 55 -
Exampe : Composing of infuence ine for the support reactions: We pace the unit force at a point x and compute the support reactions using equiibrium equations. So we have a function of the support reaction in dependence of the unit force position. After that, we cacuate the support reactions vaue for some vaue of x variabe corresponding to character position of the unit oad. Finay using this vaue and knowing the infuence ine is straight one we draw the support reaction infuence-ine diagram. F= m c А x а -x b B d А B Composing of infuence ine for the shear and moment at a specific point: When composing infuence ine for the interna forces at a specified point we shoud be carefu with choosing the variabe x. Once at eft of the specified point therefore at the other side. - 56 -
F= x А а c P P m -x b B d А P P m x а b F= -x B m аb/ When the force is at eft, it is easier to write the equiibrium equation for the right side because of the oad absence. In this case the variabe x changes from 0 to a because in other case in the equiibrium equation we shoud incude the unit oad. Otherwise, when the force is at right, we write the equiibrium equation for the eft side so the variabe x changes from a to because in other case in the equiibrium equation we shoud incude the unit oad. In this way, we compose the infuence ine for the interna moment for the specified point m. F= x А а c Q P m Q P -x b B d F= А а x Q P m QP b a/ -x B Q m b/ On the contrary of the moment infuence ine where is one and the same vaue at the two sides of the point m for the shear force infuence ine one can see there is a jump at m. The reason of this is that the oad is a unit vertica force. That is why the jump shoud be with unit vaue. Indeed, we have: - 57 -
Q P r + Q a b a b P = + = + = Usage of other infuence ines: The same resut may be achieved if we use aready determinate infuence ines to construct another one. As an exampe, we wi show the composition of the moment and shear infuence ine for point m at the previous simpe beam. The main idea is as foow: Write the equiibrium equation for the searched reaction, shear, moment, or defection and put a quotation marks to a variabes which have infuence ines. After that, perform a summations and mutipications of the equiibrium equation for the infuence ines. As a resut, we wi have the needed infuence ine. Performing these operations, we shoud take care about the unit force position and if it is necessary, we shoud write two equiibrium equations for right and eft sides. F= P m P А а b c B d B А m аb/ When the unit oad is at eft side of the beam then we use equiibrium equation for the right side of the beam for to negect the force. As a resut the point m moment depends of the B reaction but we shoud use the eft part of the B infuence ine because the unite oad is there. The ordinate of the infuence ines corresponds to the unit oad position. Otherwise, when the force is at right of point m we perform equiibrium equation for the eft side of the beam. Then the m depends of the A reaction but we shoud use the right part of the A the unite oad is there. In this manner we construct the m point shear infuence ine: - 58 -
F= Q P m Q P А а b c B d B А a/ Q m b/ Usage of character position for the unit oad: The main idea is to move the unit point oad at fixed the character points one-by-one, and using equiibrium equations for the reaction, shear, moment or the defection to determine a character ordinates of the searching infuence ine. This method is usefu because of the fact that the infuence ines for the staticay determine systems are combination of the straight segments. The number of straight ines corresponds to the number of the disks ay upon the moving path. If the specified point is at the moving path, then the corresponding disk divides into two separated disks. A character points that we must stop the force put on are the beginning of the moving path, upon supports, at a specified point, above hinges (joints) or other apparatuses and at the end of the moving path. Exampe : - 59 -
m,5 n А Composing of infuence ine for the m : The infuence ine m consists two straight segments, because the specified point m is ying on the moving path. For this reason, it is necessary to put the force on three points at east. We fix the unit oad at the characteristic point and determine the interna force m. F= 4 m Q m = 0 А= After that we move the unit oad at the characteristic point and again determine the interna m. When the force is infinitey next to eft or right of the specified point m, the resut is identicay for the moment. - 60 -
0,50 F= 4 m 0,5 0,904 We fix the unit force at characteristic points and 4, and determine interna force m.,0 F= 4 m 0,5,808,5 F= 4 m,5,704 Using the obtained resuts we may compose the infuence ine and don t forget that the obtained vaue for the moment when the unit oad is at the characteristic point n is the ordinate of the infuence ine at the characteristic point n. The ordinates are connected with straight ines. we can - 6 -
use as verification the fact that a ordinates are ying on a straight ine,. Thus, point 4 is an extra one, so the corresponding ordinate is used as verification. 4 m М m Determine infuence ine Q m : +,5,0 0,5 This soution is simiar as this for М m infuence ine. When the force is at the characteristic points,, and 4 it is possibe to use the soutions made before, but we wi cacuate interna force Q m, instead of М m. That is why the soutions are done and shown on the upper figures and the interna force Q m is read next to М m. ore compicate is the cacuation of the sear force when the unit oad is at the point, actuay the specified point for which we compute the infuence ines. We aready described the mane idea of this soution in the previous exampe. Now we wi use this idea. First, the unit force is next eft to the section. Support reactions are the same as soution for m. Using vertica forces equiibrium equation with right cut part of the system, we determine the vaue and the sign of the interna force Q m. Q m = - 0,75 4 0,50 m F= m 4 0,50 0,904 0,75 0,50 0,5 0,904 0,75 The next situation is when the force runs upon the section and now it is right next of the specified force. reactions do not change. The vaue of interna force Q m can be determine by cutting - 6 -
eft and right parts of the frame. Here we show both sections of the frame, which causes a cear resuts. 0,50 Left part m Q m = + 0,5 F= m Q m = + 0,5 Right part 4 0,50 0,5 0,904 0,75 It is cear that presented frame-parts have the same vaue of the interna force, independent of the chosen part of the frame and the resut is sure. That is why we prefer that part of the interna force, which wi be ess for cacuations. The important detai is that the received interna force vaue must be fixed as an ordinate of infuence ine exacty under the position of the unit force. Here we mean that ordinate +0,5 is next right of the specified point m. It is seen from the obtained infuence ine that is in section m has a jump equa to unit. The resut is obvious because there is an externa unit force in section m. 4 m Q m - 0,75 + 0,5 0,50 -,5 Determination of the infuence ine N m : - 6 -
The determination of the infuence ine for the norma force can be done by the anaogica way with the above ines. In this situation, point is not singuarity point because the externa force is vertica and does not cause a any jump for the norma force. The vaues of the norma force in any position of the unit oad are shown next to the others at the previous soutions. Other method for cacuating this infuence ine is using another one. In this case is more convenient to cut out the down part of the structure near support C. In that way the soution is independent of the pace of the unit oad. N n,5 n C From this section, it is cear that the moment and shear force at point n are aways zero therefore those infuence ines are zero. We can compute the norma force by the supporting reaction C with the foowing expression: N n = C Therefore, if we know infuence ine of the support reaction, we can find the infuence ine of the norma force as foow: " N n " = " C" For composing the support reaction infuence ine we use the same way as for the moment and the shear for the specified point m. We choose positive direction of the reaction and cacuate its vaue for the different positions of the unit oad. Actuay, they are aready computed and it is ony necessary to draw the infuence ine. - 64 -
4 m С С 0,904,704,808 + For the norma force infuence ine, we have: " " = " C" N n 4 m N m,0 0,50 -,5 Determination of the infuence ine for the support reaction A : The infuence ine of supporting reaction А is determine ike the support reaction ine С, that is why we wi show ony the resut: - 65 -
m 4 A С A,0 + 0,5 0,5 -,5 Kinematica method for constructing infuence ines: (Quaitative infuence ines using the üer Bresau principe) Kinematica method is very easy and suitabe technique for determining infuence ines. In the internationa iterature this technique is known as quaitative infuence ines using the üer Bresau principe. The üer Bresau Principe (886) states that: the ordinate vaue of an infuence ine for any function on any structure is proportiona to the ordinates of the defected shape that is obtained by removing the restraint corresponding to the function from the structure and introducing a corresponding unit dispacement as the function makes the negative work. For exampe, to obtain the infuence ine for the support reaction at A for the simpe beam shown in next figure, above, remove the support corresponding to the reaction and appy a unit dispacement in the direction of Y A. The resuting defected shape wi be proportiona to the true infuence ine for this reaction. i.e., for the support reaction at A. The defected shape due to a unit dispacement at A is shown beow. Notice that the defected shape is inear, i.e., the beam rotates as a rigid body without any curvature. This is true ony for staticay determinate systems. m А B а b d Y А = А Simiary, to construct the infuence ine for the support reaction B, we remove the support at B and appy a vertica unit dispacement δ Y. The resuting defected shape is the quaitative infuence ine for the support reaction. - 66 -
m А а b B d Y B = B The proof of the Bresau principe can be estabished by using a virtua work principe. Reca that the work is the product of either a inear dispacement and a force in the direction of that dispacement or a rotationa dispacement and moment in the direction of that rotation. If the rigid body (beam) is in equiibrium, the sum of a forces and moments must be equa to zero. Consequenty, if the body is given a virtua dispacement the work done by a these forces and moments must aso be equa to zero. Consider, for exampe, the previous simpe beam and the infuence ine for support reaction B. The beam is subjected to a unit oad paced at an arbitrary point aong its ength. If the beam is given a virtua vertica dispacement δ Y at support B then ony the support reaction and the unit force do virtua work. In this case, the support reaction does the negative work and the unit oad does the positive. The support reaction A doesn t a work because its point don t moves. Since the beam is in equiibrium the virtua work sum must be zero: F А а b B d δ Y δ Y = B B F.δ' B.δ = 0 B.δ Y Y Y = F.δ' Since the oad force is a unit and the virtua dispacement is unit than we have: B = δ' Y In other words, the vaue of B is equas to the vertica dispacement at the position of the unit oad, it shows of the dispaced shape represents the infuence ine for the support reaction B. Simiary if we want to construct the infuence ine for the shear force at specified point m we shoud remove the restrain corresponding to the shear force and to subject the unit reative vertica dispacement at such direction the work done by the shear to be negative. The defected shape is shown beow: Y - 67 -
Q P m Q P А а b B d v L v R Q m The sum of the right and the eft parts of the dispacement is equa to unit: v L R + v = Using the geometry, we compute the two parts of the dispacements: L v L = v = a a R v R = v = b b As a resut, we have the Q m infuence ine. To obtain a quaitative infuence ine for the bending moment at a section, remove the moment restraint at the section, but maintain axia and shear force resistance. The moment resistance is eiminated by inserting a hinge in the structure at the section ocation. Appy equa and opposite introduce a unit reative rotation between the two tangents of the defected shape at the hinge. The corresponding eastic curve for the beam, under these conditions, is the infuence ine for the bending moment at the section. The resuting infuence ine is shown beow. m m m А а b B d η m ϕ = One can compute the vaue of the ordinate at the specified point using geometry. α + α = Δφ = η η α + α = + = η = η = ab a b ab - 68 -
Let see the infuence ine for the specified shape just next to the support B at eft. The shear just to the eft side of support B can be constructed using the ideas expained above. Simpy imagine that section s in the previous exampe is moved just to the eft of B. By doing this, the magnitude of the positive shear decreases unti it reaches zero, whie the negative shear increases to. m А Q m Q m B d v L Q m The quaitative infuence ine for the bending moment at B is obtained by introducing a hinge at support B and appying a moment that introduces a unit reative rotation. Notice that no defection occurs between supports A and B since neither of the supports were removed. Therefore, the ony portion that wi rotate is part BC as shown beow. m m m C А B d ϕ = d m In the figure beow are shown the infuence ine when the section moves next to the support B but at right. m А Q m Q m B d v L = Q m Notice that no defection occurs between A and B, since neither of those supports were removed and hence the defections at A and B must remain zero. - 69 -
m m m C А B d ϕ = d m For the moment infuence ine there is no difference if the section is at eft or at right to the support. Using this idea we wi show some infuence ines for the compound beam: B C m A / / 4 / D m Q m m m ϕ = v L = m Q m A A B B C D D m Q m m а b 4 m m ϕ = ab/ М m v L = b/ v R = a/ Q m Q m Q m When we use the kinematica way for constructing infuence ines for simpe structures it is easy to determine the dispaced shape after removing the resistant ink but when we use it for more compicated structures it is not so easy. That is why it is necessary to use the pane of poes to determine the dispaced shape. For this technique we have the next work sequence: remove the restrains, corresponding to the function for which we construct the infuence ine; subject a unit dispacement so that a negative work be done; the structure changes to a mechanism with one degree of freedom and we construct its pan of poes; determine the dispaced shape of the disk on the moving path; the vertica dispacements give a shape of the infuence ine; - 70 -
the vaues we can compute using the static method. For the vertica dispacements one shoud know foowing marks: under the main poe there is no a vertica dispacement because this point is ike a common pin support. For this reason the infuence ine have a zero vaue at this point. under the reative poe there is a kink for the vertica dispacement shape because this point is a reative rotation point. For this reason the infuence ine have a kink at this point. when the reative poe is at the infinity the corresponding disks moves parae. For this reason at the infuence ine the corresponding ines are parae. This procedure is presented on the next exampe. We wi determine the infuence ines to the frame we aready done using the static way. Infuence ine m : In section m remove the moment and put a hinge. Thus the structure is divided into two disks. Hence the infuence ine wi consist two straight ines. After that we construct the pane of poes. B () (,) m m А C [] [] () + (,) = () CC' = () oment m, which is at disk rotates anticockwise direction so, disk must rotates cockwise direction for m do negative work. Since disk rotates cockwise direction, then disk shoud rotate anticockwise direction. According the rue, under main poe in the infuence ine has a zero under poe() (marked as [] in the drawing) and wi be rotated to cockwise direction. The straight ine correspond to disk [] has zero under poe () and it is connected to the straight ine, corresponding to disk under reative poe (, ). Notice that the projection of poe is out of the geometry of the frame. Infuence ine Q m : m,5 + 0,5 () - 7 -
In section m remove the shear and put a Q-reease. Thus, the structure is divided into two disks. Hence, the infuence ine wi consist two straight ines. After that, we construct the pane of poes. () + (,) = () CC' = () It is obvious that poe (,) is at the horizonta infinity. For finding poe () is needed to connect poe () to poe (,) at infinity. It is practicay done when draft a horizonta straight ine across (). The shear force Q m corresponding to disk, rotates the disk around poe () cockwise direction, therefore Q m wi done negative work if disk turns anticockwise. As disk turns anticockwise, thus disk turns anticockwise too, because the reative poe (, ) is at infinity. Therefore, the infuence ines consist two parae ines according to the rues mentioned before. () () ()+(,) (,) Q m Q m C Q m [] [] 0,75-0,5 -,5 + 0,5 Infuence ine N m : In section m remove the norma force and put a N-reease. Thus, the structure is divided into two disks. Hence, the infuence ine wi consist two straight ines. After that, we construct the pane of poes. () + (,) = () CC' = () Poe (,) is at the vertica infinity. For finding poe () is needed to connect poe () to poe (,) at infinity. It is practicay done when draft a verica straight ine across (). The norma force N m corresponding to disk, rotates the disk around poe () anticockwise direction, therefore Q m wi done negative work if disk turns anticockwise. As disk turns cockwise, thus disk turns cockwise too, because the reative poe (, ) is at infinity. Therefore, the infuence ines consist two parae ines. - 7 -
() ()+(,) () (,) N m N m C N m [],0 0,5 -,5 [] Infuence ine n : In section n remove the moment and put a hinge. Thus the structure is divided into two disks. Hence the infuence ine wi consist two straight ines. After that we construct the pane of poes. () + (,) = () CC' = () In this case poe coincides to poe (, ), therefore poe () must be ocated at the same point, but poe () is stand in other pace. As we aready know when some poe is ocated at two points that means it does not exist. When a main poe does not exist its then the disk is stabe (does not moves). The disk is on the moving path and haven t a vertica dispacement, therefore corresponding infuence ine is zero. - 7 -
B () n А (,) () () n C n Infuence ine Q n : Here we have again the same situation; the disk is on the moving path and haven t a vertica dispacement, therefore corresponding infuence ine is zero. (,) () () () B Q n А Q n C Q n The soution of ast two infuence ines is cear, therefore ony the fina resut is shown beow. - 74 -
Determine infuence ine N n : () () ()+(,) B (,) N n N n А C N n,808 0,904 - [],704 Determine infuence ine А : () B C А A,0 + 0,5 0,5 - [],5-75 -
Cacuating of the interna forces by using infuence ines. By using the infuence ines one can cacuate the vaues of the interna forces for a specified point caused by some externa oads. This can be done very easy using the idea of the infuence ines and the principe of the superposition. As it is known each ordinates means the vaue of the function for which is composed the infuence ine when the unit force is up of the ordinates pace. Therefore, if the force is not unit but F then the ordinate of the infuence ine (constructed using force F) wi present the vaue of the function caused by force F so we may cacuate the function caused by the force F. Sk = F.η F m c А а b B d η аb/ m m = F.η If there are more then one force one can use the superposition principe, then the function vaue wi be as foow: S = F.η k i i If the force oad is directed down then one must take the sign of the ordinate from the infuence ine. F F m c А а b B d η m η аb/ = F.η + F.η m - 76 -
If the beam is subjected by the distributed oad one can compute the function at a specified point by the formuae: Sk = q.ω i i where the Ω i is the area of the infuence ine under the i-th distributed oad. Why is used the area it is easy to understand if present the distributed oad as a many forces at infinity sma distance between each other. q F i S k η i Now one can use the previous formuae for the appied force and as the ordinates are very cose to each other so the sum of them equas to the area. For the sing of the mutipier one must take the sing of the area of the infuence ine. q m c А а b B d η m η = аb/ η.c η.a m = q. + q. Simiary, if the beam is under the concentrate moment oad then the vaue of the function can be computed as foow: S = tgα k i i This is easy to understanding if present the moment as a coupe of forces whit the arm of unity. Then as we know the vaue of the forces is the same one can write: - 77 -
F F α α S k η η η η Sk = F.η i i F.η F.η F = + = = F..tgα =.tgα When the moment oad and the ange rotate at the same direction then the sing of the mutipier is positive. The direction of rotation determines as the reference ine rotates to the infuence ine. m c А а b B d η m η = аb/ η + η m =. b + d If a beam (frame) is subjected at the same time to forces, moments and a distributed oads then the vaue of the function for which the infuence ine is constructed on can compute as foow: S k Fi. η i + qi. Ωi + tgα = i i In the foowing exampe is presented this formua for the frame which infuence ines where aready composed. If one aready construct the interna force diagram can use the infuence ines for verification of the obtained resut. - 78 -
0 40 5 m n A 0,904,808 -,704 N n Q n n 0,5 A -,5,0 + 0,5 М m Q m N m + 0,5,0,5 0,75,5 0,50 - - 0,50 + 0,5,5,0-0,904.,704,808 N n = 40.0,904 0. 5. = 5, 805kN Q = 0 ; = 0 n n (,0 + ) 0,5.,5 0,5 A = 40.0,5 + 0. 5. = 6, 875 kn,5.,0 0,5 m = 40.,5 + 0. 5. = 7, 5kN 0,75.,5 0,5 Q д m = 40.0,75 0. 5. = 4, 5kN - 79 -
0,75.,5 0,5 Q л m = + 40.0,5 0. 5. =, 5kN 0,5.,5,0 N m = 40.0,5 0. 5. = 8, 75kN aximum vaue of the function Using this idea one can determine the absoute maximum of the function for which the infuence ine is constructed from the distributed oad passing through the beam (train). If the distributed oad has a specific position at the beam then at the point for which the infuence ine is the function wi have a maximum vaue. There are two questions:. What is the right position of the distributed oad to produce the maximum effect?. What wi happens if the infuence ine is constructed for the changing section section at distance x? The answer of the first question is as foow: If the infuence ine has negative and positive parts we must pace the distributed oad up to the whoe positive part of the infuence ine to produce a maximum positive vaue of the function. Simiary if we pace the distributed oad up to the whoe negative part of the infuence ine it wi produce a maximum negative vaue of the function. q q m c А а b B d η η m аb/ η.c η.d η.c η.d The maximum negative vaue of the m is: m = q. q. q + = + q m c А а b B d m The maximum positive vaue of the m is: η = аb/ η. m = + q. - 80 -
The answer of the second question is as foow: If we construct an infuence ine for a section at distance x from the support, for exampe, we wi have infuence ine for a function for each section of the beam span. The infuence ine wi be a function of the distance x. m m m А x -x B c d x(-x)/ m As we aready know if now we have an externa oad we may compute the vaue of the function at section at distance x caused by the externa oad. However, this vaue wi be vaid for each section of the beam spam because the infuence ine is vaid for each one section. Therefore, this vaue aso wi be a function of the position of the section function of x. Thus if we draw this function we wi have a diagram of the function caused of the externa oad. q А x -x B c d x(-x)/ m М q /8 Using this idea one can compose an extreme diagram for the moment, for exampe, if the externa oad is paced at the specified pace as it was described above. This extreme diagram is usefu for the bridge beams where is very important to know the maximum negative and maximum positive interna forces negecting oad pace. For rich this diagram first must be composed the infuence ines for each span of the bridge (compound) beam. Next exampe presents this idea for a compound beam. - 8 -
4 5 6 х 4-х х 4-х х -х х 4,5-х 4 х 5 -х 5 х 6,5-х 6 4 4 p,5,5 4-х - 6,0,5 part Functions x = 0 x = / x = [( 4 x )( 4 x + 4) / ]. p = ( 4 x )( 8 x )/ + [(( 4 x ) )(,5 ) ]. p = 0,875( x ) 6 6 0 4,5,75 0 4 x,5. p = 0, 875x ( )( ) + [(( 4 x ) x 4)( 4 ) ]. p = ( 4 x ) x [( x )( x +,5 )/ ]. p = ( x )(,5 x )/ 4 + [((,5 x4 ) x4,5 )(,5 ) ]. p = (,5 x4 ) x4 5 [(,5 x 5 ) x5 / ]. p = x5(,5 + x5 )/ [((,5 x6 ),5 ).( ( +,5 ) ) ]. p =,. (, 5 x6 ) 6 + [((,5 x ) x,5 )(,5 ) ]. p = (,5 x ) + (4-х )х /4 6,0 -,5 -,5,0 p 0,75,5 4 0 0,5,5 0,5 0 0,85 0 + 0,5,5,5,75 0,5 0 0,85 0 6 6 6 x6 + (4-х )/ p х / - p -х - p + (,5-х 4 )х 4 /,5 p - 0,8 х 5 (,5-х 6 )/,5 -,5 + + + + p М М 4 5 p + 6 (,5-х 6 )х 6 /,5,5 - p 0,8 М m - 8 -
4 5 6 х 4-х х 4-х х -х х 4,5-х 4 х 5 -х 5 х 6,5-х 6 4 4,5,5 6,0,0 +,0 p х /4 - + p (4-х )/4 p,0 + - - p /4 /4 p Q Q Q p p х 4 /,5 - Q 4 + (,5-х 4 )/,5 p p - p,0 + /,5 Q 5 p p х 6 /,5 - Q 6 + (,5-х 6 )/,5 0,875 +,0,75,08,58, + 0,5 0,75 + 0,875 + Q - - m 0,875-0,875 -,75 0,75 0,875 0,75,875,75-8 -
part Functions x = 0 x = / x = Q [( 4)(,5 ) ]. p = 0, 875 Q+ [ ( 4 ) +.4 ]. p = ( 4 x ) + Q [ x x ( 4.) + 4.,5 (.) ]. p = x 8 +, 5 Q+ [( 4 x )( 4 x ) ( 4.) ]. p = ( 4 x ) 8 Q+ [ ( ) +.,5 ]. p = ( x ) + 0, 75 Q+ [ x 4x4 (.,5 )]. p = x4 4 Q [(,5 x )(,5 x ) (,5.) ]. p = (,5 x ) 5 Q [. 5 + (.,5 ) ]. p = x5 + 0, 75 Q [ x 6x6 (.,5) ]. p = x6 6 [(,5 x6 )(,5 x6 ) (,5.) + (.,5) (,5.) ]. p Q+ = (,5 x ) +, 0,875 0,875 0,875 4 x 6 4 0,875,75,875 4 0,5 0 x,75,75 0,75 0 0,875 0,75,5 0,75 0,875 0 6 x 0,75,75,75 = 0 0,875 0,75,5,08,58, Absoute maximum ive moment in bridges. It is necessary to exam the situation when the motor truck pass through the bridge, then at the construction appears some maximum interna forces. There are different position of this vehice at which the interna forces are very high vaues. The infuence ines are usefu for examine the interna forces in dependence of the position of the vehice. For this examining is used a standart F F F F F F F F 00 00 00 00 А а ab vehice as shown at the next figure: b B m,,, The absoute maximum of the interna force searches trying different position of the four forces carrying the foowing points: one of the forces shoud be up to the maximum vaue of the infuence ine; if it is possibe the a four shoud be at the positive (or negative) part of the infuence ine; if it is not possibe then try ony one of the forces be at the part with the opposite sing with very ow ordinate vaue. - 84 -
Chapter 7 Defections using energy methods. It is very usefu to determine the defections and dispacements in a frame or beam by using the energy methods in the structura mechanic. For doing this on can understand the meaning of the defection, virtua work and the energy theorems. In this chapter we wi expain the principes of virtua work and how to determine dispacements end defections using these principes. Externa work and strain energy: Before deveoping any energy method we wi expain the externa work and the strain energy (the interna work) done by a force and a moment. The externa work done by a force: When a force F undergoes a dispacement dx in the same direction as the force, the work done is: dw = F ( x) dx. If the tota dispacement is x, the work becomes: ext dwext = F ( x) dx 7. Consider now the effect caused by an axia force appied to the end of a bar. As the magnitude of the force increase from zero to some vaue F the fina eongation of the bar becomes. As in the statica anaysis, the materia has a inear eastic response (The Hook s ow is vaid), then the force vaue at some moment wi be as foow: F( x) = F x 7. Δ Force F F(x) x x Figure 7 - The work done by this force, we wi be defined as: W W W ext ext ext = = = 0 0 F( x) dx; F. x. dx = F. F 0 F F( x) =. x F x. x. dx =. 0 = F. 7. If there are severa externa forces, the work done wi be: - 85 -
n W ext = F i. i= i 7. 4 Where F i is the fina vaue of the i-th force and i is the corresponding dispacement. Deformation caused by the interna norma (axia) force: The norma force causes the norma deformations at a differentiay sma part of the beam. This deformation is defined as foows: From the geometry, we have connection between the eongation, deformation and the ength of the differentiay sma eement. From the definition of the stresses, we have connection between the norma stress and the norma force. From the Hooke s aw we have connection between the norma stress and the deformation. As we use a these connections together we wi obtain for the eongation the foowing expression: ds dλ σ N N A N σ E ε Figure 7 - N A dλ N = E. ε = E. dλ = ds 7. 5 ds EA Deformation caused by the interna moment: dθ ds ρ ds x neutra axis z ds x z neutra axis Figure 7 - The strain in arc ds ocated at a position z from the neutra axis x is: ds ds ε = 7. 6 ds However from the geometry = and d s = ( ρ z). dθ ds dx = ρ. dθ so: - 86 -
Or: ε = ( ρ z) dθ ρ. dθ z = ρ. dθ ρ ε = ρ z 7. 7 7. 8 Where: ρ is the curvature. Using Hooke s aw and stress definition for bending, we obtain: σ. z. z ε = ; σ = ε = 7. 9 E I EI And finay: = κ = ρ EI 7. 0 Where: EI is the fexura rigidity; I is the moment of inertia computed about the neutra axis. dx = ρ. dθ dθ = EI ds 7. Deformation caused by the shear force: ds γ dv Q τ γ γ τ Q Figure 7-4 ( + ν ) τ Q Q dv = γ. ds; γ = ; G = ; τ =. κ = 7. G E A κ - coefficient depending on the cross section area. Strain energy of the body: Q dv = ds 7. G.A Q If the materia is inear eastic and isotropic then the strain energy caused the axia force wi be expressed as foows: W N int = N. dλ = N ds 7. 4 EA A Q - 87 -
Simiary for the strain energy caused by the interna moment: And for the shear force: W int W Q int =. dθ = EI Q ds 7. 5 = Q. dv = ds 7. 6 GA Q The interna work (the strain energy) for the differentiay sma eement is: And for the whoe eement is: W int = EI + N EA + Q GA Q ds 7. 7 W int = EI ds + N EA ds + Q GA Q ds 7. 8 Work expression for eement under externa oad: If the body is in equiibrium then the interna and externa work wi be equa whit reversed sing or: W int + W ext = 0 7. 9 Or And finay: F i. Δ i EI ds + N EA ds + Q GA Q ds = 0 7. 0 F i. Δ i = EI ds + N EA ds + Q GA Q ds 7. Deformation caused by temperature oad; The temperature induce defections on eement and these defections induce interna forces moments and axia forces. t t c -t h x x (natura axis) t t c Figure 7-5 t c t -t c When the eement is subjected on the temperature oad et one of its sides the temperature is higher than the other. This difference cause interna forces in the eement. Temperature distribution aong the eement is given on Figure 7-5. If x-x is the natura axis, it divide the temperature distribution at two parts. The first is a constant temperature distribution and the - 88 -
second is temperature difference whit a zero vaue at the natura axis (Figure 7-5). The constant temperature distributions cause an eongation of the eement and as a consequence an interna norma force. The temperature difference cause a defection of the eement and as a a consequence an interna moment. Eongation and interna force h x x (natura axis) ds dλ Figure 7-6 t c dλ = ε t. ds 7. From the physics is known that: ε t = α t. t c 7. Where, α t is the coefficient of the therma extension and t c is the temperature at the cross section: It foows: tc = t +t - for rectanguar cross section. dλ = α t. t c. ds 7. 4 Defection from the temperature differences: dϕ t ρ dϕ x ds up z h x dϕ (ρ+z)dϕ t ds d ds d The extensions of the upper and down bars are as foow: - 89 -
ds up = α t (t c t )ds 7. 5 ds d = α t (t t c )ds 7. 6 The ange of the defection is: dφ t = dsd ds up h = α t (t t c ) α t (t c t ) ds 7. 7 h dφ t = α t (t t ) h ds = α tδt ds 7. 8 h Interna work at an eement caused by the interna forces from the temperature oad: t W int = N. λ t. dφ t 7. 9 t W int = N. α t. t c Δt +. α t 7. 0 h For more than one eement; t W int =.α t.δt ds + N. α h t. t c ds 7. Interna work of the support reaction at a spring supports: If the system has a springs supports then there have a dispacements and the support reactions done interna work. ϕ R = k. = k. ϕ Were k is the rigidity of the spring; (ϕ) is the dispacement at the spring support. If S is generay the support reaction (moment or force) and is the dispacement (inear or rotationa) at the support we can write: S = k. Δ 7. Or Δ = S k 7. And the work done by a reaction wi be: W sp int = S. Δ = S k 7. 4-90 -
If there are n spring supports, the work be: W sp int = n S. Δ = n S 7. 5 k If there is a system incuding n eements under externa and temperature oads and spring supports then if the system is at equiibrium then the work expression wi be: F. Δ = ds + N ds + Q ds +.α t.δt ds + N. α EI EA GA Q h t. t c n ds + S 7. 6 k Principe of virtua work: As we aready mention the principe of virtua work may be defined as foows: Consider a structure in equiibrium with a system of appied forces is subjected to a system of virtua dispacements compatibe with the externa restraints and the geometry of the structure. The tota work done by the appied forces during these externa dispacements equas the work done by the interna forces, corresponding to the appied forces, during the interna deformations corresponding to the externa dispacements. The expression virtua work signifies that the work done is the product of a rea oading system and imaginary dispacements or an imaginary oading system and rea dispacements. δw ext + δw int = 0 7. 7 If we impose a virtua dispacement at a deformabe system there wi appear a virtua defections of the body as foow: κ = EI N d λ = ds dv = Q ds 7. 8 EA G.A Q dλ = α t. t c. ds dφ t = α tδt ds Δ = S h k 7. 9 The work done by the rea interna forces with the virtua defections wi be: δw int = ds + N ds + Q ds +.α t.δt ds + N. α EI EA GA Q h t. t c n ds + S k 7. 40 The work done by the rea externa forces with the virtua dispacements wi be: δw ext = F. Δ 7. 4 So, principe of the virtua work gives foowing expression: F. Δ = ds + N ds + Q ds +.α t.δt ds + N. α EI EA GA Q h t. t c n ds + S 7. 4 k - 9 -
This expression gives as a possibiity to obtain the vaue of dispacement at some point of the structure. If a unit force is imposed at the point and toke this system of forces and dispacement as a virtua then the right side of the expression 7.4 wi be equa to the dispacement at the point we are trying to find. F F = v =? F f Energy theorems: F. Δ v = Δ v = f EI ds + N f EA ds + Q Q f GA Q ds Betti s theorem, (discovered by Enrico Betti in 87), States that for a inear eastic structure subject to two sets of forces {P i } i =,...,m and {Q j }, j =,,...,n, the work done by the set P through the dispacements produced by the set Q is equa to the work done by the set Q through the dispacements produced by the set P. Exampe: Stage Stage F F If stage is rea and stage is virtua then we have: W ext = F. = W int F = ds EI If stage is rea and stage is virtua then we have: W ext = F. = W int It foows that: F = ds EI F = F Betti s theorem - 9 -
axwe s theorem of reciproca dispacements: States that a dispacement at point I on a structure produced by a unit force at point k is equa to the dispacement at point k when the unit force is acting at point i. This theorem foows from the Betti s theorem if the forces F and F have a unit vaue. Stage F = Stage F = or. =. = Theorem of the reciproca reactions: States that the reaction at point i produced by unit dispacement at point k is equa of the reaction at point k caused by unit dispacement at point i. This theorem foows from the Betti s theorem if the dispacements Z i and Z k have a unit vaue. i Z i = Stage k r ki i Stage k Z к = r ik or r ik Z i = r ki Z k r ik. = r ki. r ik = r ki Theorem of the reciproca reactions and dispacements: States that the reaction at point i produced by unit dispacement at point k is equa to the dispacement at point k caused by unit force at point I with inversed sing. This theorem foows from the Betti s theorem if the dispacements Z i and F k have a unit vaue. Stage Z i = k r ik Stage k F k = i δ кi i or δ ki F k = r ik Z i δ ki. = r ik. δ ki = r ik - 9 -
Chapter 8 Dispacements at staticay determinate structures. A principe of virtua work can be used for determining of the dispacement of some point in the staticay determinate frame or beam. It was shown at a previous chapter how the defections are determinate and how the virtua work is used for cacuating dispacement of specific point. According to principe of the virtua work for a system subjected to an externa force and temperature oad and incuding springs supports the dispacement at a specific point cacuates using the next expression: = ds + N ds + Q ds +.α t.δt ds + N. α EI EA GA Q h t. t c n ds + S k Where, Q, N, and S are interna and spring forces caused by a virtua unit oad at the specific point for which we are cacuating the dispacement. This virtua oad correspond to the dispacement. It is shown at the next tabe: N Description Dispacement Virtua oad If the dispacement is F inear vertica then the v =? unit oad is a vertica F f force. F = If the dispacement is inear horizonta then the unit oad is a horizonta force. F F h =? f F = If the dispacement is rotation then the unit oad is a moment. F F ϕ =? f = 4 If the dispacement is inear reative then the unit oad is a coupe of forces F F f v v = v v =? F = F = 5 If the dispacement is reative rotation then the unit oad is a coupe of moments. F ϕ F f ϕ ϕ ϕ =ϕ ϕ =? = = Before to show some exampes for cacuating dispacements is needed to make some comments about how to cacuate the integras at a formua 8.. This formua means that we shoud find an integra of mutipication of two functions. The first is the function of the diagram from the externa oad and the second is the function of the diagram from the virtua oad. The first function can be arbitrary from constant to a paraboic by third degree. The second according to the virtua oad is at maximum inear function. From this reason, these integras are - 94 -
not so difficut for numerica cacuation and it is done for a cases at a tabe. Such a tabe is shown beow: I = f (x)f (x)dx 0 Where f (x) is the function from the virtua oad and f (x) is the function from the externa oad. The integras are cacuated for different eements or parts of eements where the functions are steady. The integra cacuates from the beginning of the eement (or part of the eement) to its end. Cacuation of dispacements at staticay determinate structures from externa oad. If there is a frame without springs supports and oaded with an externa force oad then the dispacement cacuates as foows: = f EI ds + N f EA ds + Q f GA Q ds Usuay major infuence of the dispacement vaues has the interna moment and the infuence of the norma and shear forces is negigibe. That is why (when we make hand cacuations) we usuay ignore the second and the third part of the integra. Sometimes we incude the integra of the norma forces if the structure has bars working of tension and compression but in this case it isn t so obigate. Ony when the structure is a truss then of course the ony way to cacuate the dispacement is to use integra of the norma forces. The other two integras are zero. Now we wi show an exampe of cacuation of dispacement at a frame under force oad using ony the integra of the moments and cacuating integras using tabes. - 95 -
Cacuation of vertica dispacement of point m done from the externa oad: q F v,m =? Data: beams(0.5/0.5): coumn(0.5/0.5): F = 0 kn I b = 0.00065 m 4 I co = 0.00055 m 4 = 5 knm A b = 0.0788 m A co = 0.065 m q = 0 kn/m E =.4.0 7 kn/m I гр =.I кол α t =.0-5 Step : Anaysis of the structure from the externa oad composition of the f diagram: The resut is: 4 5 75 0 95 45 5 50,5 40 55 90 95 f 95 4 Step : Anaysis of the structure from the virtua oad composition of the diagram. The virtua oad is a vertica force at point m: - 96 -
A ф C = 0,8 m P A =,0 4 B = 0,8 And the diagram is:,4 0,8,6 4 4 4 Step : Cacuation of the dispacement. 5 75 0 95 45 5 50,5 40 55 90 95 f 95,0 4-97 -
One shoud be carefu because the moment of inertia of the beams and coumns are different. We wi write first a coumns parts and after that beams parts. v,m = f EI The different parts of the cacuation can be iustrated by foowing way: ds v,m = EI c ( x + + ( EI b x + x + x x + x + ) + + x ) And the cacuation is: EI c v, m =.40.,80 78 + 6 + (,6.50 + (,6 + 0,8)(50 + 5) + 0,8.5 ).+ 0,8.5. 6 + + (.75.90.( 4) 4.95 ). 4.95. 564 + + + + 6 = 0,45m =,45cm v, m (.55 + ( + )(55 + 95) + ).95.,80 +,4.45. Cacuation of vertica dispacement of point m done from the temperature oad: C t = 5 C m P t = 5 C A 4 If the same system is subjected to a temperature oad as it is shown at the figure then there wi appears dispacements. It is very important to know that: At a staticay determinate systems subjected to a temperature oad wi appears ony dispacements without any interna forces (without any moments, shear and axia forces). That is why we cannot draw a moment diagram at a determinate structure from a temperature oad but we can cacuate the dispacement at some point at the structure. B - 98 -
To cacuate this dispacement we need to anayze the structure from the virtua unit force (or moment) and to know the initia data for the temperature and the materia of the structure the coefficient of the therma expansion. In this case, the coumns and the beams of the structures are rectanguar cross-sections so the natura axis is at the haft of the high. The constant temperature and temperature different ire: t c = t + t = 5 and t = t t = 0 The dispacement at the point cacuates using the next expression: t =. α t. Δt ds + N. α h t. t c ds In this case in not needed to use the tabes of the numerica cacuation of the integras because the temperature difference, the constant temperature, the high of the eements and the coefficient of the therma expansion are constants and we can write: t = α t. Δt h ds + α t. t c N ds It foows that the integras are equa to the area of the diagrams. It is important to know foowing facts:. This formua uses ony for the eements which are subjected to a temperature oad. Because for the other eements t and t c are zero and as one can see at the upper expression the integras become zero.. When the moment diagram from the virtua oad is from the same side of the eements where the higher temperature is then the sing of the integra is positive (pus). In the other case is negative (minus). That is because we draw the moment diagram from the extended bars of the eement and such a moment forms the same defection as the higher temperature.. When the norma force from the virtua oad is positive then the integra is aso positive and if the norma force is negative the integra is aso negative. This is because the positive norma force cause an eongation of the eement the same as the positive constant temperature. Now we can show the procedure of the cacuation: Step : Anaysis of the structure from the virtua oad composition of the and N diagrams. C,0 m P N A 4 B - 99 -
Step : Cacuation of the dispacement. m,t = α t. Δt h ds + α t. t c N ds = =.0 5. 0..,605 +.0 5. 0 ( + 4). + 4. +.0 5. 5. (,0.,605) 0,5 0,5 m,t =,884. 0 + 8,8889. 0 0,65. 0 =,. 0 m =,cm Cacuation of vertica dispacement of point m done from the support settement: C m P If the same system is subjected to a support-settement oad as it is shown at the figure then there wi appears dispacements. It is very important to know that: At a staticay determinate systems subjected to a support-settement oad wi appears ony dispacements without any interna forces (without any moments, shear and axia forces). That is why we cannot draw a moment diagram at a determinate structure from a supportsettement oad but we can cacuate the dispacement at some point at the structure. To cacuate this dispacement we need to anayze the structure from the virtua unit force (or moment) and to know the initia data for the support-settement. The dispacement at the point cacuates anaogicay of the dispacement at a system incuding springs supports. In this case again the interna work is equa to the mutipication of the support reaction by the dispacement of the support using the next expression: n Δ c = R. c Where R is the support reaction and c is the dispacement at the support. When the support reaction and the dispacements are at one and the same direction then the mutipication (the work) is positive in the other case is negative. Now we can show the procedure of the cacuation: A c v = 0,05m c h = 0,05m 4 B - 00 -
Step : Anaysis of the structure from the virtua oad we need ony the support reactions. C = 0,8 F = m B = 0,8 c v = 0,05m c h = 0,05m 4 A =,0 Step : Cacuation of the dispacement. n Δ m,c = R. c =,0.0,05 + 0,8.0,05 = 0,09m = 9cm The same resut we can rich using geometrica soution. This makes composing the dispaced shape of the structure. Such a soution we wi iustrate in the next chapter. Cacuation of vertica dispacement of point m done from the externa oad for structures incuding springs: Important to know: When one determinate system incudes a spring support it is not a mechanism. It is a norma determinate structure but whit a dispacement at a support. This support dispacement is imited because the spring has stiffness. The support reaction at a spring can be obtained in the same way as a ideay rigid support. The difference is ony at the dispacements. So, the support reactions and the diagrams are as at the structure whit ideay rigid supports. On the next figure are shown the moment diagrams from the externa an virtua oad of the structure incuding spring support: q = 0 kn/m v =? 45,5 F = 4 EI = 78 knm f k = 0000 kn/m = 45 = - 0 -
The vertica dispacement one can obtain using the foowing expression: v = EI n S S ds + k v = 78 45.. (45 + 4.,5 + 0). +.45.4 + 6 0000 v = 0.078 + 0,05 = 0.0078m =,078cm Cacuation of reative rotation of point m done from the externa oad: Important to know: When we need to find a reative dispacement (or rotation) there is ony one difference from the situation whit a singe dispacement (or rotation). This difference is ony of the virtua oad. As it was mentioned before, in this case the virtua oad is a coupe (forces or moments). On the next exampe is shown the soution for the reative rotation at a midde hinge of the three-hinged frame from the externa oad: F = 0 = = 0 0 ϕ m =? f 4 4 EI = 78 knm The reative rotation dispacement can be obtained using the foowing expression: v = EI ds v = 78.0.4 +.0.. v = 0.0895rad - 0 -
Chapter 9 ethod of forces (Force method) for anaysis of staticay indeterminate structures. On the beginning of this course we saw that the number of degrees of freedom way be cacuated using the next formua: w= d k a; where: w is degree of freedom (mobiity); d is number of bodies (eements); k is number of one-degree-of-freedom kinematic pin joints; a is number of support inks. As a resut w may be positive, negative or zero. Therefore, we distinguish three different cases for w: w > 0 - the system is mechanism. In the case of mechanism, we don t have a structure carrying any oad; w = 0 - determinate structure. We have a structure and it is possibe to anayze it with ony equiibrium conditions. w < 0 - indeterminate structure. We have a structure and it is possibe to anayze it with equiibrium conditions and additiona equations. In this first stage we wi anayze indeterminate structures namey structures with w < 0. When we have, indeterminate structures we need additiona equations for find the unknown reactions. On the next figure, we show two times indeterminate structure. q w =. 0 5 = If we have two times indeterminate structure, we need additiona equations to find reactions because for such a structure we may write ony equiibrium equations but we have 5 support reactions. In the force method (or fexibiity method), the additiona equations are the compatibiity equations. Now we wi expain the main idea of this method. For exampe et take the same system but as a determinate structure. If we know the vaues of the two unknowns, the two systems wi be equivaents for a interna forces. - 0 -
q X X So, if we can compose the moment diagram for this structure it wi be the same as for the indeterminate. But et try to imagine what wi be the deformed shape in the two cases. q P q P X X The deformed shapes are different because in the indeterminate structure, there is a fixed support at point P but in the determinate structure, it is not. The resut is that in the determinate structure at point P there are dispacements but in the indeterminate, they are zero. The equivaence of the interna forces of the two systems is not enough. Therefore, if we want equivaence of the systems we need to find the dispacement at point P of the determinate system and to put them to be zero. That wi be the additiona equations the equations of the dispacement consistency or compatibiity equations. The question is: How to find this dispacements as we don t know the extra forces X and X? To answer of this question we wi use the principe of superposition: q P X X - 04 -
q P X P X Or f = f0 + X + X According to the principe of superposition, the determinate structure subjected to an externa oad and extra forces oads can be separated by three situations: the first is the system subjected to the externa oad; the second is the system subjected by X and the ast one is the system subjected by X. On the other side as we know aready if the force with vaue X subjects the system one may compose the interna moment diagram X. However, one may compose the diagram from the unit force and to mutipy it by the vaue X and the resut wi be the same. Or: - 05 -
X = X Same situation we have from the X force. Or: X = X We may use the ast two resuts at the previous one to write the next: f = f0 + X + X Where: f is interna moment diagram obtained by the externa oad at the staticay indeterminate system; f0 is interna moment diagram obtained by the externa oad at the staticay determinate system; is interna moment diagram obtained by the unit vaue of the X force at the staticay determinate system; is interna moment diagram obtained by the unit vaue of the X force at the staticay determinate system; X and X are the unknown extra forces. Important to know: This idea is vaid and may be used for everything dispacements, deformations, support reactions and shear and axia forces. A interna forces, dispacement and support reactions at the indeterminate system may be determinate using determinate system. We ca this determinate system primary system! This is important to know because we wi use it for the dispacement at the point P (in presented case) to find the dispacements at the point and to put them zero vaue. As the primary system is determinate, we can obtain the dispacements at a point P from every oad externa, X = and X =. q P X X - 06 -
is the dispacements at X appication point ( point P) at X direction from a oads externa oad and unknown vaues of X and X in the primary system. is the dispacements at X appication point ( point P) at X direction from a oads externa oad and unknown vaues of X and X in the primary system. And they are equas to: q f P f f is the dispacements at X appication point (point P) at X direction from externa oad ony in the primary system. f is the dispacements at X appication point (point P) at X direction from externa oad ony in the primary system. δ δ P X = δ is the dispacements at X appication point (point P) at X direction from X = ony in the primary system. δ is the dispacements at X appication point (point P) at X direction from X = ony in the primary system. δ δ P X = - 07 -
δ is the dispacements at X appication point (point P) at X direction from X = ony in the primary system. δ is the dispacements at X appication point (point P) at X direction from X = ony in the primary system. Or: = f + δ. X + δ. X = f + δ. X + δ. X But according the dispacement consistency we must have: Therefore we have: = 0 = 0 δ. X + δ. X + f = 0 δ. X + δ. X + f = 0 Finay, we have two additiona equations and we may obtain the vaues of the two unknowns the extra forces X ; X. Next step to sove is haw to obtain a these dispacements. This is aready soved probem at the previous chapter. To find the dispacement at some point we need to anayze the system from a virtua unit oad.. Dispacement from the externa oad: To find the f dispacement we need a virtua vertica unit force: q P f P The resut from this virtua vertica force is the same as this from the unit vaue of the X so we may use the diagram from X = that we aready have instead of the diagram. - 08 -
And the dispacement is: f = fo EI ds = fo EI To find the f dispacement we need a virtua horizonta unit force: ds q P f P The resut from this horizonta force is the same as this from the unit vaue of the X so we may use the diagram from X = that we aready have instead of the diagram. And the dispacement is: f = fo EI ds = fo EI. Dispacement from the unit vaues of the extra-forces X = and X = : To find the δ dispacement we need a virtua vertica unit force and for δ dispacement we need a virtua horizonta unit force: ds δ δ P X = - 09 -
P P The resut from these virtua forces is the same as this from the unit vaue of the X and X so we may use and diagrams instead of diagram. And the dispacements are: δ = EI ds = ds = EI EI ds δ = EI ds = EI To find the δ dispacement we need a virtua vertica unit force and for δ dispacement we need a virtua horizonta unit force: ds δ δ P X = P P The resut from these virtua forces is the same as this from the unit vaue of the X and X so we may use and diagrams instead of diagram. And the dispacements are: δ = EI ds = EI ds - 0 -
δ = EI ds = EI ds = EI ds Important to know:. As it is obvious from the expressions δ = δ and it is not necessary no cacuate two of them.. When we anayze indeterminate systems, it is not necessary to compose diagrams because we aready know that they are same as and.. The presented idea is vaid not ony for two-times indeterminate structures but for n- times indeterminate aso. The procedure is the same. The compatibiity equations for n- times indeterminate structure are: δ. X + δ. X + δ. X + + δ n. X n + f = 0 δ. X + δ. X + δ. X + + δ n. X n + f = 0 δ. X + δ. X + δ. X + + δ n. X n + f = 0 δ n. X + δ n. X + δ n. X + + δ nn. X n + nf = 0 - -
Choice of a primary system. The choice of a primary system is very important because for ony one indeterminate system there are many different variants of a primary system. One shoud choose easier, most appropriate and correct primary system. How to compose a primary system: We compose a primary system removing inks from the indeterminate structure. We remove so many inks as is the number of the degree of freedom w. We may remove as externa inks and interna. When we aready have some primary system we must do the kinematica anaysis of it to prove that it is stabe primary system. Exampe: For presented indeterminate structure, compose different primary systems. origina system First, we shoud cacuate how many times tis structure is indeterminate: w =. d. k a =..0 6 = We need to remove three inks. First, we wi remove ony externa inks. Variant X X X Variant of the primary system is a simpe beam. It is a stabe structure. At the next variant we wi remove not ony externa inks but interna. Variant X X X X Variant of the primary system is a simpe beam. It is a stabe structure. Important to know: When interna inks have been removed the unknowns are aways coupe! - -
At a Variant we wi remove ony interna inks. X X X X Variant X X 4 KA: [ + 4 +.](w = 0) Variant of the primary system is a structure type I and is a stabe structure. X X Variant X X KA: [.](w = 0) Variant of the primary system is a three-hinged frame and is a stabe structure. It is possibe to compose many other variant of a primary structures but it is not necessary. Important to know: The primary system shoud be type I (if it is possibe). When it is type I we are sure the system is stabe. Exampe: For presented indeterminate structure, compose different primary systems. origina system First, we shoud cacuate how many times tis structure is indeterminate: We need to remove two inks. w =. d. k a =.. 5 = - -
X X A A X Variant 4 KA: [4 +.. AA (w = +) + (w = )](w = 0) Variant of the primary system is a fixed beam and a tied three-hinged frame. It is a stabe structure. X X A A X Variant KA: [ +. AA ](w = 0) Variant of the primary system is a fixed beam and a simpe beam. It is a stabe structure. X X X X Variant KA: [ +.](w = 0) Variant of the primary system is a fixed beam and a three-hinged frame. It is a stabe structure. - 4 -
X X X Variant 4 X KA: [ +.](w = 0) Variant 4 of the primary system is a fixed beam and a simpe beam. It is a stabe structure. How to compose a primary system: The primary system shoud be easier because we anayze it many times from many ad cases. When the primary system is simpe, we obtain moment diagrams easy and fast. Diagrams are simpes and one can compose them without mistakes. When diagrams are easy the next cacuations (mutipications of the diagrams) are aso easy. For the exampe Variant is simpe and suitabe and can be used propery. Variant 4 is aso appropriate because is symmetrica. Important to know: When the origina structure is symmetrica is recommended the primary system to be symmetrica too. For the exampe Variant is simper and appropriate because it is. Variant 4 is a simpe beam too but it is difficut to think off such a primary system. Exampe: origina system Tis system is composed by disk supported as a simpe beam but the disk is cosed. This is a cosed-oop system. The degrees of freedom cacuate as foows: w = ( m k) m is number of the cosed oops (incuding the basic disk ground) k is number of one-degree-of-freedom kinematic pin joints. For the present exampe is: w = (. ) = It is easy to see this resut if cut the cose-oop. This system is supported as a simpe beam and one can cacuates the support reactions. But for to cacuate interna forces diagrams we need to cut at two paces so we cannot compose the diagram without obtain first the extra-forces. When we cut it three interna forces appears and so the system is three times indeterminate. Such systems are inner indeterminate. - 5 -
Actuay cutting the oop, we compose the primary system. X X X X X X Variant X X X X X X Variant Since the origina structure is symmetrica as we aready mentioned the primary system shoud be symmetrica too. That is why Variant is appropriate for this exampe. Exampe aso incudes cose-oop and composing primary system we cut it without mention it. Exampe4: origina system R = k. w =. d. k a =..0 5 = X X Variant X - 6 -
X X X Variant X X X X Variant When the system incude a spring support is better to cut the spring to compose the primary system. For tis exampe the easier end the proper variant is Variant. Exampe5: origina system w = (. ) = This system is a compicated one. It has 7 extra-forces and aso has a cosed-oop. It is not ony inner indeterminate but externa too. X X X X X X X 4 Variant X 5 X 6 X 7-7 -
X 4 X 4 X X X X 5 X X X Variant X 6 X 7 X X X 4 X X 4 X 5 X X X 5 Variant X X 6 X 7 At the Variant we removed 4 externa inks and cut off the cosed-oop. As a resut, the primary system is a fixed beam. At the Variant we do the same but in this case primary system is a three-hinged frame. Variant is very compicate. In this case, we didn t cut the cose oop and ony incude a ot of hinges. As a resut, we have a structure type I so it is stabe structure but is it the appropriate variant? Important to know: When we put a hinge in the primary system we shoud be carefu about the number of the one-degree-of-freedom kinematic pin joints k. This number te as haw many interna moments we shoud put in the primary system. At a Variant the eft hinge is with k = so there are three interna moments we removed and we shoud put them as unknown in the primary system. The other two hinges are with k = that is why we put ony one unknown interna moment. After a this anaysis what is the better primary system? The preferabe primary system is Variant because it is party symmetrica but Variant is aso proper if one prefer it. Fu exampe of anaysis of staticay indeterminate frame: Anaysis of staticay indeterminate frame oaded by externa oad: Data: F = 0 kn beams (0.5/0.5): coumns (0.5/0.5): = 5 knm I b = 0.00065 m 4 I co = 0.00055 m 4 q = 0 kn/m A b = 0.0788 m A co = 0.065 m I b =.I co. E =.4.0 7 kn/m t = 0 o α t = 0,0000 t = 5 c h = c v = 0,05 m ϕ = 0,05 rad - 8 -
Х Х F q. Degree of indeterminacy: n = d k a; d = ; k = 0; a = + = 5 n =. 5 = - The system is two times staticay indeterminate. Chose of primary system: The primary system is composed by removing inks. The primary system shoud be staticay determinate, a stabe. Х Х Х Variant Variant Х Х Variant Variant 4 Х Х Х The primary system wi be anayzed severa times from different oads. That is why the primariy structure shoud be simper for to obtain diagrams easy. In this case wi be used Variant. - 9 -
. Unit diagrams: 4,5 Х = М 0,5 0,5 The М diagram is a diagram in determinate primary system from the unit vaue of the unknown force Х. 0,75 Х = М 0,5 0,5 The diagram is a diagram in the determinate primary system from the unit vaue of the unknown moment X.. Fexibiity coefficients: δ ij the dispacement of the appication point of the unknown Х i upon its direction, caused by unknown Х j =. Vaues of δ ij cacuate using axwe-oor s integras. EIco = 78; EIb = 564 4.4,5., 6 4 δ = ds 4 ( 4,5 ),5. 5,9.0 = + + + + + = EI 78 564 6 δ = ds 0,75. (.,5 4 ).,5. (.0,75 )., 6 5,4 = + + + = 67.0 4 EI 564 6 78 6 δ = ds δ 5,467.0 4 = = EI 0,75. 0,75 δ = ds 4 ( 0,75 )., 6,5666.0 = + + + + = EI 564 78 6-0 -
. Verification of the fexibiity coefficients caused by the unit vaue of the unknowns:? δ = s ij EI ds For this verification is needed to compose a summary moment diagram s. It is composed by adding the vaues of the unit diagrams at each specific points. s = + 4,5 М s,0 0,5 0,5 s ds = ( ) + 4 + 4 +,5 +,5. EI 564 6 + ( ) = 4.4 + + +,5 +,5.,6 66,76.0 4 78 6 = 4 + 4 + 4 = 4 δ 5,9.0.5,4.0,5666.0 66,76.0 δ = s ds ij ij EI During coefficients summering we shoud remember that δ =δ and it shoud be incuded the two coefficients. 4. Diagram of the primary system from externa oad: 40 55 7,5,5 5,5 5,5 - -
5. Fexibiity coefficients from the externa oad: o f f = ds = EI + 564 6 o f f = ds = EI 78 [ 55.4 +.7,5. ( 4 +,5) + ] 564 4.5.4,5.,5.,6 +,5.,5. = 04,78.0 4 0,75. 6 78 6 (,5 +.7,5 ). +,5(.0,75+.,6 ) =,98.0 4 6. Verification of the coefficients from the externa oad: o? s f if = ds EI o s f ds = 5.4.4,5(.,5+ ).,6 + EI 78 6 + [ 55.4+.7,5. ( 4+,5) +,5.,5. ] = 6,765.0 4 564 6 if = 04,78.0 4,98.0 4 = 6,765 if = s EI o f ds 7. Compatibiity equations: 5,9X + 5,4X 04,78 = 0 X = 6,48 5,4X +,5666X,98 = 0 X = 8,9 8. Fina diagrams: It is composed by using next connection at each specific point: f = o +. X +. X f 40 9,48 7,4 0,75 0,76 8,9 8,9 6,48 5,776 47,4 - -
9. Compatibiity verification: s f? ds = 0 EI s f ds = 4. (.0,75 5 ).4+ [,5.7,4+ (,5+ )( 7,4 8,9) 8,9..,6 ] + EI 78 6 6 + [ + ( + ) + ] = 4 4 + 4? 9,48.4.0,76 4,5,5.7,4.,0.0 8,5.0 5,9485.0 = 0 564 6 8,505.0 4 + 8,5.0 4 = 0 0,0087% error If the consistency condition is satisfied, we can continue with composing the shear and axia force diagrams. The shear force diagram is composed by using connection between the moment and the shear. The norma force diagram is composed using joints equiibrium. 6,48,776 α α Q v =,05 α Q = 7,0 Q h = 6,6 α N h = 0,775 0 7, N R = 6,48 6,48 N = 0,964 N v = 0,545 N D = 47, 7,,776 0 7,0 8,9 6,48,776 6,48 47,4 - -
6,48 0,964 8,9 6,48 47,4,776 47,4 Anaysis of staticay indeterminate frame oaded by temperature oad: The anaysis of indeterminate frames from the temperature oad is the same as this from the externa oad whit ony one difference. The fina diagram made using the foowing expression: t = +. X t +. X t The reason of this resut is that at a determinate structure (the primary system) there is no diagram from the temperature oad. Therefore, to compose the fina diagrams we need ony to cacuate new vaues of the unknowns. To cacuate them is necessary to determine the dispacement of the unknown s appication points caused by the temperature oad t, t. Then the compatibiity equations wi be: δ. X t + δ. X t + t = 0 δ. X t + δ. X t + t = 0 The dispacements t and t are dispacements at a determinate system so the way of there cacuation is aready expained and it is: it = α t. Δt h i ds + α t. t c N i ds For the present exampe, the soution from the temperature oad is as foows: t = 5 o t = 0 o α = 0,0000 t t - 4 -
This soution needs to compose the norma forces diagram from the unit oads. 0,5,0 Х = N 0,5 0,5 0,5 0,5 Х = N 0,5 0,5 t + + For the rectanguar cross-section, t c is the midde temperature: = t 0 5 t с = =,5 ; t = 5 Cacuation of the coefficients: 5 5 4 +,5 0,0000. 4.4,5( 0,5.4.) 6,6.0 4 t = + + + = 0,5 0,5 5 0,0000. 0,75,5.0,5.4 7,6786.0 4 t = = 0,5 α t Verification of the coefficients: it = sds + α. h tс Composing the summary diagram N s ds Ns 0,5,0 0,5 Х = N s 0,5 0,5-5 -
7,.0 4 it = t 5 5 α. =.0 s ds + t с N s ds 4.4 + h 0,5 Compatibiity equations for temperature oad: 5 0,5 4 +,5 +,5 5,9X + 5,4X + 6,6 = 0 X =,99 t t t 5,4X +,5666X + 7,6786 = 0 X =,88 t t t Fina diagrams from the temperature oads: =. X +. X t t t ( 0,5.4 +.) = 7,.0 4,5968 5,645,88,654 4,7,99,99 0,5989,654 Compatibiity verification: - 6 -
t ds = EI t t ds = 4.,5968.4+ +,5. [. ( 5,645) +,88.,6 ] EI 78 6 4.,597 + (,597 + 5,645 )(. 4 +.5) +,5.5,645. = 6,6.0 4 = 564 6 t [ ] = 6,6.0 4 Anaysis of staticay indeterminate frame oaded by support settement oad: The anaysis of indeterminate frames from the support settement oad is the same as this from the externa oad whit ony one difference. The fina diagram made using the foowing expression: = + X c. X c. + c The reason of this resut is that at a determinate structure (the primary system) there is no diagram from the support settement. Therefore, to compose the fina diagrams we need ony to cacuate new vaues of the unknowns. To cacuate them is necessary to determine the dispacement of the unknown s appication points caused by the support settement c, c. Then the compatibiity equations wi be: δ. X c + δ. X c + c = 0 δ. X c + δ. X c + c = 0 The dispacements c and c are dispacements at a determinate system so the way of their cacuation is aready expained and it is: n Δ ic = R i. c For the present exampe, the soution from the temperature oad is as foows: ϕ c v = 0,05 m ϕ = 0,05 rad EI c = 78 knm c v Determination of the fexibiity coefficients: To obtain these coefficients we need reactions at the unit diagrams and use the formua: ic = R i. ci - 7 -
The sum of the right side of the equation represents the work of the support reactions at the unit diagrams through the support settements. c = c = 4 ( 0,05.0,5 + 0) = 50.0 ( 0,05.0,5 0,05.) = 65.0 4 Verification of the coefficients: =! ic R s, i. ci ; R s,i support reaction of the summary moment diagrams S. R. ( 0,05.0,5 0,05.) 75.0 4 s, i c i = = = 65.0 4 50.0 4 = 75.0 4 ic Compatibiity equations for the support settements anaysis: 5,9X + 5,4X 50 = 0 5,4X +,5666X + 65 = 0 X = 4,44 X =,5 Fina diagrams from support settements: 48,9 7,78,54 4,44 7,45-8 -
4,44 69,89 95.6 Compatibiity verification: c ds = EI c c ds = 4.7,78.4,5. [ 48,9 +,54.,6 ] + EI 78 6 + 4.7,78+ ( 7,78 48,9 )(. 4 +.5),5.48,9. = 50.0 4 = = 50.0 4 564 6 c [ ] - 9 -
Symmetrica indeterminate frames. If one indeterminate system is symmetrica is better to use this property when anayze such a system. In this section, we wi show this property of the system and how to use it. One system is symmetrica if it has the foowing properties:. the geometry is symmetrica;. the supports are symmetrica;. the cross-sections (A and I) are symmetrica; 4. the physica data (E-modue) is symmetrica. The externa oad can be symmetrica, antisymmetrica or common. First of a we wi show when one force or moment is symmetrica and when it is antisymmetrica. symmetrica forces ( moments): F v F v F h d d F h d d d F v d. Two vertica forces are symmetrica when the appication points of the forces are at one and the same distance from the axe of symmetry and the forces at the two sides of the symmetrica axe have one and the same direction.. Two horizonta forces are symmetrica when the appication points of the forces are at one and the same distance from the axe of symmetry and the forces at the two sides of the symmetrica axe have inverse directions.. Two moments are symmetrica when the appication points of the moments are at one and the same distance from the axe of symmetry and the moments at the two sides of the symmetrica axe have inverse directions. 4. One vertica force is symmetrica when the appication point is on the axe of symmetry and it doesn t matter of its direction. F v F v F h d d F h d d d F h d - 0 -
. Two vertica forces are antisymmetrica when the appication points of the forces are at one and the same distance from the axe of symmetry and the forces at the two sides of the symmetrica axe have inverse direction.. Two horizonta forces are antisymmetrica when the appication points of the forces are at one and the same distance from the axe of symmetry and the forces at the two sides of the symmetrica axe have one and the same directions.. Two moments are antisymmetrica when the appication points of the moments are at one and the same distance from the axe of symmetry and the moments at the two sides of the symmetrica axe have one and the same directions. 4. One horizonta force is antisymmetrica when the appication point is on the axe of symmetry and it doesn t matter of it s direction. 5. One moment is antisymmetrica when the appication point is on the axe of symmetry and it doesn t matter of it s direction. Important to know: Interna norma force and interna moment on the axe of symmetry are symmetrica forces. Interna shear force on the axe of symmetry is antisymmetrica force. Now wi show how to transfer two common forces to a symmetrica and antisymmetrica: A d d B A/ d d A/ A/ d d A/ B/ d d B/ B/ d d B/ d d X X = A/ + B/ d d X X = A/ - B/ X X Each force can be transfer at two parts one symmetrica and one antisymmetrica. In this way the two forces A and B are transferred to a two coupes symmetrica and antisymmetrica. After that if one take a sum of the symmetrica coupes of the two forces wi obtain one symmetrica force (coupe) X. If one take a sum of the antisymmetrica coupes of the two forces wi obtain one antisymmetrica force (coupe) X. - -
And inversey if one have directy vaues of the two coupes a symmetrica coupe X and antisymetrica coupe X one can obtain the vaues of the common forces A and B as foows: A = X / + X / and B=X / X /. This idea is common not ony for forces but for moments and dispacements oso. Actuay in the structura anaysis we use the ast way of the decomposition. For the force method we use directy symmetrica and antisymmetrica coupes X and X as unknowns and in the end if it is necessary we obtain the rea forces A and B. To be possibe to use a symmetry in the force method not ony indeterminate system shoud be symmetrica but the primary system must be too and to use coupe of unknowns. Now wi show some exampes of a symmetrica indeterminate system and different symmetrica primary systems: I I I origina system E = const. This structure is symmetrica because the geometry, supports, cross-sections and the physica data are symmetrica. So, the primary system shoud be symmetrica. X X Variant X X X X X X X X X X X X X Variant 4 X - -
X X X X X X Variant At the first variant the two supports moments are presented by coupes of symmetrica antisymmetrica unknowns. X is the symmetrica coupe, X is the antisymmetrica coupe. X is the interna moment for the frame, that is why it is a coupe and because it is at the axe of symmetry, it is a symmetrica unknown. X and X are symmetrica and X is antisymmetrica. At a variant a unknowns are interna moments but X is on the axe of symmetry and it is a symmetrica moment. It is not necessary to transfer it to coupes. Therefore, the X and X are symmetrica and antisymmetrica coupes and they present the transferred coupes of the interna moments in the eft and in the right coumns. At the ast variant, the frame is cut at the axe of symmetry so we directy have symmetrica and antisymmetrica unknowns. X and X are symmetrica and X is antisymmetrica. Important to know: The most important effect of using symmetry is at the mutipication of the diagrams for cacuation of the dispacements. When the unknown is symmetrica, the interna moment diagram is aso symmetrica. When the unknown is antisymmetrica, the interna moment diagram is aso antisymmetrica. When symmetrica and antisymetrica diagrams are mutipied the resut is zero. Foows the dispacement is zero. Fu exampe of anaysis of a symmetrica frame using this property from the externa force oad. F I I I F = 40 kn q = 6 kn/m I b = 0,5I co,5 I I q 4 A b =,5А co A c = A b I c /A c = 0,0 4 On tis exampe in addition to symmetry we wi use not ony moments but norma forces to cacuate dispacements. The reason of this is that as part of the frame there is a rod. We are not obigate to use norma forces but it is better to do it. For the present exampe, there is not essentia difference in the anaysis with and without taking into account norma forces but we wi show the anaysis incuding norma forces.. Primary system: - -
X X X X X X Variant X X Variant X X X X At the two variants, X and X are symmetrica and X is antisymmetrica. Variant wi be chosen.. Unit diagrams:,5 X X,5 М X X N - 4 -
X X,5,5 М X X N X = X = М 4 4 N The diagrams, N,, N are symmetrica. The diagrams and N are antisymmetrica.. Diagrams from the externa oad at a primary system: - 5 -
60 60 8 40 4. Cacuation of the dispacements: EI =,4. 0 7.,55. 0 4 = 78 EI =.,4. 0 7. 6,5. 0 4 = 565 EA =,4. 0 7. 0,065 = 500000,5EA =,5.,4. 0 7. 0,065 = 50000 δ δ δ δ δ δ = = = = = = F F =.40.4 50000 =.40.4 50000 N...4..4. ds + ds = + + =,04.0 + 4,608.0 +,556.0 EI EA 78 565 50000 NN ds + ds = 0 EI EA NN ds + ds = 0 EI EA N...4..4. ds + ds = + + =,04.0 + 4,608.0 +,556.0 EI EA 78 565 50000 N N 4..4. ds + ds = + 0 =,07.0 EI EA 565 N 4.4..4 6 ds + ds = + =,7.0 +,6667.0 =,77.0 EI EA 565 500000 o o f NN f 60. (. +,5 ).,5.60.4.8.4 ds + ds = + EI EA 6 78 565 565 = 4,4.0 EI o f = 4,4.0 ds + 46,08.0 N N EA o f 46,08.0 ds = +,77.0 6 60.,77.0 (. + ),5.,5 78 7,.0 7,.0 5.60.4 565 5 = 7,78.0.8.4 565 = 9,. 0 6 6 = 6,96.0 = 6,96.0-6 -
o o f NN f F = ds + = + 0 = 0,7.0,77.0 = 6,49. 0 EI EA 565 4 565 ds 5. Verification of the coefficients: s δ ij = ds + EI N s ds EA 4.60.4 4.8.4 6 6 М s 0 4 N s [ 6 + ( 6 + 0) + 0 ]. s N s 6. 4 4.4 ds + ds = + + + EI EA 78 6 565 565.4.4 6 + + = 4,608.0 + 6,7.0 +,65.0 + 7,.0 +,667.0 50000 500000 δ = 6,96.0 + 6,96.0 +,07.0 +,07.0 +,77.0 =,709.0 ij 6 =,7.0 60 60 8-7 -
- 8-40 56,86 5,67 8,66 5,4 7,07 7,05 0,447 9,55 Q f,57 6,08 47,98 6, ( ) ( ) ( ) 5 84,59.0 6,49.0 0 9,. 7,78.0 84,5.0 7,.0,768.0,88.0 8,80.0 50000.40.4 565..4 8 4. 6 565 0.4 6 60. 78.,5.6 60. 6 = = = = + + + = + ij o f s o f s ds EA N N ds EI 6. Compatibiity equations: :0 0 6,49.0.,77.0,07.0 0 9,.0,07.0 6,96.0 0 7,78.0 6,96.0 = + = + = X X X X X 6,096 6,44 4,07 0 6,49,77,07 0 9,,07 6,96 0 7,78 6,96 = = = = + = + = X X X X X X X X 7. Fina diagrams:
6,08 N f 9,55,57 8. Compatibiity verification: [ 0.5,4 + ( 0 + 6)( 5,4 8,66) 6.8,66 ].,0.0 s EI s 5,4.0 EI + 6 + 4. 6 ( 56,86 +.7,05 ). f f ds + ds + 7,74.0 N 5,5.0 s EA N N s 4 565 +,8.0 N EA = 0? ds = 0 ds =.5,67.,5 78 7,8.0 + 6 [.5,67 + ( + 6)( 5,67 8,66) 6.8,66 ]. 4 565.9,55.4.6,08.4? + = 0 50000 500000,05.0 Indeterminate frames containing springs. f f 4 + + 4,7.0 As we aready mentioned when an indeterminate frame incudes spring best way to compose a primary system is to cut the spring. Therefore, the soution has the same idea. Here is presented fu exampe of the anaysis of system containing spring support: 5? = 0,5 78 + F = 4 F = 5 I I c = EI,5 E =.0 6 I = 0,000675 m 4 c = EI = 50 d = /EI = 7,407.0-4 When we take such a primary system the unknown is the spring reaction. When we oad the system by X don t forget to oad the spring whit the same oad and to incude its infuence at the dispacements. - 9 -
F = 4 F = 5 I I c = EI,5 5 44,5 X = X = 7,5 7,5 o f It is suitabe to cut the spring because in this situation there is no reaction in the spring from the externa oad. Ony from the unknown X (the unknown at the spring) it appears the spring infuence. In the concrete case at δ. δ = EI ds + S c δ 5.5 =. + EI EI n 4,6667 = 50 = f = fo EI n ds + S. S f 0 0,060 c f =..[5.44,5 + (5 + )(44,5 + 7,5) +.7,5]...[.7,5 + 7,5] + 0 = 50 6 6 f = 0,77 X = f = 0,77 δ 0,060 = 8,76757 74,08 50 f = f0 + X 0,665 0,665 7,5 f 0,4 0,054 8,76757-40 -
Chapter 0 Dispacements at staticay indeterminate structures. The dispacement at staticay indeterminate structures cacuates in the same way as in the determinate systems. According to principe of the virtua work for a system subjected to an externa force and temperature oad and incuding springs supports the dispacement at a specific point cacuates using the next expression: n f = n n f ds + N n n N f EI EA ds + Q n n Q f + N n. α t. t c ds + S n S f n n k ds + n. α t. Δt ds GA Q h Where n, Q n, N n, and S n are interna and spring forces caused by a virtua unit oad in the indeterminate frame at a specific point for which we are cacuating the dispacement. f n, Q f n, N f n, ands f n are interna and spring forces caused by externa oad in the indeterminate frame. In the previous chapter, it was shown haw to anayze indeterminate frames. e necessary to compose primary system and to anayze it many times from a different oads. I it is needed to compute a dispacement at some point in indeterminate system we need to anayze the system ones again but from some virtua oad. oreover, diagrams in indeterminate structures are compicate and it is possibe to make a ot of mistake of their mutipication. Concusion is that cacuation of dispacement in indeterminate frame is a compicated probem. However, it is possibe one of the soutions to be in determinate system. Now wi be shown this idea and how to use it. To be easier et take ony externa oad interna moments integra in the expression of the dispacements. The main idea for a other parts is the same and it is not needed to use it now. So, we have: n f = n n f ds EI Now et think that indeterminate frame is anayzed using force method. Let aso make assumption that the frame is two times indeterminate. If the frame is more times indeterminate the expressions are the same but onger. The externa oad interna moments diagram we can compose using foowing expression: f n = f o +. X + +. X Let substitute this expression in the expression for the dispacement: n f = EI f o +. X + +. X n ds = EI f o n ds + EI (. X + +. X ) n ds = EI f o n ds + EI. X n ds + EI. X n ds = f o n ds + X EI. n ds + X EI. n ds EI - 4 -
The question is what are the ast two integras? X. n ds ; X EI. n ds. EI n is the moment diagram in the indeterminate frame from a virtua oad. and are the moment diagrams unspecified primary system from a unit oad. The integras present the dispacements of the X and X appication points from a virtua oad. F v = 0 v 0 X h = 0 n X h 0 If the indeterminate frame is on equiibrium than the dispacements of the X and X appication points shoud be zero! The concusion is that: X. n ds = 0; EI X. n ds = 0. EI Finay the resut about the dispacement in indeterminate frame is: n f = f o n ds EI The diagram from the externa oad can be determinate in arbitrary determinate frame. On the same way we can prof that the other one diagram can be composed in arbitrary determinate frame, or: n f = f n o ds EI As concusion, we may say: The dispacement of some point in indeterminate system can be cacuated using interna forces diagram from externa and virtua oad in indeterminate structure. Using interna forces diagram from externa oad in indeterminate structure and virtua oad in determinate system. Using interna forces diagram from virtua oad in indeterminate structure and interna forces diagram from externa oad in determinate system. n f = n n f ds = f o n ds = f n o ds EI EI EI - 4 -
Exampe: Cacuation of dispacement in indeterminate frame from externa oad: q F v,m =? EI = 0000 q = 0 F = 0 The interna moment diagram from the externa oad in indeterminate structure and the interna moment diagram the virtua oad in indeterminate structure are: F,6,6 7,7,8 5,9 0, 0, 0,9 6,9, f n n 0, The cacuated dispacement using two indeterminate frames is: n f = n n f ds EI = 0000 (,.0, + (, 6,9)(0, 0,) + 6,9.0,). 6 + 0,(6,9,6).,5,6.0,.,5 + 6 (,6.0, +.7,7. ( 0, + 0,9) +,8.0,9). + 0,9(,8 +.5,9). 6 = 6,9 (,9585 +,47,08 + 4,559 + 6,8) = 0000 0000 n f = n n f ds =,847. 0 m = 0,85cm. EI The moment diagram from the externa oad in determinate structure and the moment diagram the virtua oad in indeterminate structure are: The determinate frame is arbitrary and is better to be such system to compose diagrams easier. - 4 -
F 0, F 0 5 5 q. /8 = 0 f o F. = 90 0, 0,9 n 0, The cacuated dispacement in this case is: n f = f o n ds EI = 0000 6 ( 0,.90 + (90 + 0)( 0, + 0,) + 0.0,). + 0.0,.,5 + 6 ( 0,.0 +.5( 0, + 0,9) + 0,9.0). + 0,9(0 +.5). 6 = 7 (4,5 + 4,5 + + 5) = 0000 0000 n f = f o n ds =,85. 0 m = 0,85cm. EI The moment diagram from the externa oad in indeterminate structure and the moment diagram the virtua oad in determinate structure are: The determinate frame is arbitrary and is better to be such system to compose diagrams easier. F,0,6,6 7,7,8 5,9 6,9 f n o, Presented determinate frame is not easier because in this case the mutipication of the diagrams is not so easy but we wi use it to show that it is no difference if use different determinate systems. Better determinate system is the previous one because the diagram wi be ony on the simpe beam and when mutipy wi have ony two parts for mutipication. - 44 -
The cacuated dispacement in this case is: n f = f n o ds EI = 0000 6 (,6 +.7,7)..,6.,5 + (6,9,6).,5 + 6,9 (6,9,). = ( 6 0,8 + 4,7 + 9,57) = 0000 0000 n f = f n o ds =,8455. 0 m = 0,85cm. EI Cacuation of dispacement in indeterminate frame from temperature oad: 0 v,m =? 0 EI = 0000 EA = 840000 α t =,.0-5 h c = 0,5 h b = 0,5 If we don t have soution of the indeterminate frame from the temperature oad it is better to perform anaysis of the frame from the virtua oad. In this case the diagrams shoud be composed for the indeterminate frame. To cacuate dispacement from temperature oad we need not ony interna moment diagram but norma forces too. The diagrams for the virtua oad in indeterminate frame are as foows: F F 0, 0,555 0, 0,9 n N n 0, 0,55 n t =,. 0 5. 0 0,. + 0,5 n t = n. α t. Δt ds + N n. α h t. t c ds 0, 0,,. 0 5. 0.0,55.6 = 7,. 0 4,96. 0 4 t n =,6. 0 m = 0,cm. - 45 -
Other way to cacuate this dispacement is to use the diagram at indeterminate frame from temperature oad. In this case the soution from the virtua oad can be at determinate structure and the dispacement is: t n = t n o 4,87 EI ds + 0. α t. Δt ds + N 0. α h t. t c ds F 7,44 4,87,0 t n 0,6 0,5 F N 0 n t = 0000 6 (4,87 +.7,44)..7,44. 0,5.6.0.,. 0 5 = 4,877 0000 ±,6. 0 4 = 7,485.. 0 4 +,6. 0 4 n t =,04. 0 m = 0,04cm Cacuation of dispacement in indeterminate frame from support settement oad: ϕ v,m =? EI = 0000 ϕ = 0,00 d v = 0,05 d v - 46 -
If we don t have soution of the indeterminate frame from the support settement oad it is better to perform anaysis of the frame from the virtua oad. In this case the diagrams shoud be composed for the indeterminate frame. To cacuate dispacement from support settement oad we need not ony interna moment diagram but reactions too. The diagrams for the virtua oad in indeterminate frame are as foows: 0, F d v 0, 0,9 0,45 ϕ n 0, 0, where d is the support settement. c n = R n. d c n = R n. d = ( 0,.0,00 0,05.0,45) = 0,07m =,7cm Other way to cacuate this dispacement is to use the diagram at indeterminate frame from support settement oad. In this case the soution from the virtua oad can be at determinate structure and the dispacement is: 45,5 n t = c n o ds R 0. d EI,6 F,0 0,5 45,5 c n 0 7,8 n c = c n o ds R 0. d EI = 0000 6 (,6. + 45,5)..,6. 0,5. ( 0.05) 0,6 5,08 = + 0,05 =,6. 0 + 0,05 0000 n c = 0,07m =,7cm - 47 -
Chapter Anaysis of simpe indeterminate structures using force method. In this chapter, we wi present the anaysis of so-caed kinematica inks Type II and Type III. We aready have presented them at Chapter as basic eements for composition of compicate structures. Actuay, they are indeterminate structure so they need additiona anaysis. As we aready famiiar to the force method, we may use it on this anaysis. Anaysis of kinematica ink Type II EI Two times indeterminate system. As a externa oad we wi take different situation of a support settements. Firsty wi be a rotation at the fixed support with a unit vaue. ϕ = EI As a primary system, we wi choose a fixed beam. X Unit diagrams and dispacements:,0 EI X = X = 0 X = δ = EE ds = EE δ = ds = δ EE = EE ds = 0 f = R. φ =. = f = R. φ = 0 X = f δ = X = 0 EE = EE - 48 -
EE EE ϕ = EI EE φ Therefore, we have a soution of the ink type II from unit rotation of the fixed support. Let do the same but from vertica unit dispacement of the fixed support. d v = EI The primary system, unit diagrams and the dispacement from the unit forces are the same. Ony the dispacement from the externa oad wi changes and it is: f = R. d d =. = f = R. d d = 0 X = f = δ EE X = 0 = EE EE EE d v = EE EI dd EE If the vertica dispacement is on the pin support the resut wi be the same but with a reverse sing. EE EE EI d v = dd EE Let do the same but from horizonta unit dispacement of the fixed support. d h = EI In this case we need ony norma forces. The primary system is the same but the dispacements are different: - 49 -
,0,0 N = 0,0 X = N EA X = δ = δ = 0 δ = N. ds = EA EA f = R. d h = 0 f = R. d h = X = f δ = EA = EA EA/ N dh EA EA/ EA/ The same resut wi be if the horizonta dispacement is at the other support. Anaysis of kinematica ink Type III Three times indeterminate. As a externa oad we wi take different situation of a support settements. Firsty wi be a rotation at the eft fixed support with a unit vaue. ϕ = EI As a primary system, we wi choose a fixed beam. Unit diagrams and dispacements: X X X - 50 -
,0 EI X = = 0 X =,0 X = δ = ds = EE EE δ = ds = δ EE = ds = δ EE = EE ds = 0 δ =. ds = EE EE δ = ds = EE EE f = R. φ =. = f = R. φ = 0 f = R. φ =. = δ X + δ X + δ X + f = 0 δ X + δ X + δ X + f = 0 δ X + δ X + δ X + f = 0 D = EE EE X + EE X = 0 EE X + EE X = 0 EE EE 4. 4. 4 = EE EE EE = 4 EE EE 4 EE EE EE X = X EE X = EE 4 X = EE 4 EE = EE EE 4 EE EE = EE 4 EE = 6EE 4 6EE + 6EE = EE 4 X = 0 EE + = EE EE 6EE = EE - 5 -
φ 4EE 6EE ϕ = EI EE 6EE If the rotation is on the other support, the resut wi be the same but mirror. φ EE 6EE EI ϕ = 4EE 6EE Therefore, we have a soution of the ink type III from unit rotation of the fixed supports. Let do the same but from vertica unit dispacement of the eft fixed support. d v = EI The primary system, unit diagrams and the dispacement from the unit forces are the same. Ony the dispacement from the externa oad wi changes and it is: f = R. d d =. = f = R. d d = 0 f = R. d d = 0 δ X + δ X + δ X + f = 0 δ X + δ X + δ X + f = 0 δ X + δ X + δ X + f = 0 EE X + EE X = 0 EE X + EE X + 0 = 0 4 D = EE EE 4 EE EE EE X = X 0 EE X = EE 4 X = EE 4 EE EE. 0 = EE 4 EE = EE EE + EE. 0 = EE 4 EE = 6EE - 5 -
dd X = 0 6EE 6EE d EI EE v = EE If the vertica dispacement is on the other support, the resut wi be the same but mirror. 6EE dd EE EI d v = 6EE EE Let do the same but from horizonta unit dispacement of the eft fixed support. d h = EI In this case we need ony norma forces. The primary system is the same but the dispacements are different:,0 N = 0,0,0 EA X = X = N,0 N = 0 X = δ = N. ds = EA EA δ = δ = δ = δ = δ = 0 f = f = R. d h = 0 f = R. d h = X = f δ = EA = EA EA/ EA/ N dh EA EA/ The same resut wi be if the horizonta dispacement is at the other support. - 5 -
In this way, we aready have obtained resuts about inks type II and III from a possibe support settements. Such soutions we may obtain from different externa oad too. A these resuts are made once for a and are arranged at a tabe can be used directy. Such a tabe is shown beow: In these tabes is used abe i for the inear stiffness and it is: i = EE - 54 -
Chapter Dispacement method for anaysis of staticay indeterminate structures. The dispacement method is one other method for anayzing indeterminate structures. Some time it is suitabe not to search extra-forces but node dispacements. The dispacement method is used to cacuate the response of staticay indeterminate structures to oads and/or imposed deformations. The method is based on cacuating unknown rotations and dispacements at the joints of frames based on conditions of equiibrium at the joints. The force method is a method for cacuating the response of staticay indeterminate structures by which the unknowns are force quantities (the redundant forces X, X,..., X n ) and the equations used to sove for the unknowns are based on geometrica conditions (compatibiity conditions at the ocation of each extra-force). It is possibe to consider an anaogous method for cacuating the response of staticay indeterminate structures in which the unknowns are dispacement quantities and the equations used to sove for the unknowns are based on statica conditions (equiibrium conditions). This method wi be referred to as the cassica dispacement method. The procedure of enveoping the theory. The procedure of enveoping the theory is ogicay same as in the force method. We use again the principe of superposition and mutipication of unit diagrams by extra vaues. The main difference as we have mentioned is that now we wi think about dispacement not for forces and we wi add inks not to remove them. Let consider the next exampe and its deformation: q w =. 0 5 = The present structure is two times indeterminate according force method. What about its joint dispacements: u ϕ u q ϕ u - 55 -
The two joints of the frame are rotated and the beam is dispaced horizontay. ain question is: Is it possibe to determine these joint dispacement and rotations? If it is possibe we wi be abe to determine fu deformation of the frame. What wi happens if we add some inks to stop the joints dispacements and rotations. We shoud add two rotationa inks and one inear horizonta. q In this case the nodes of the frame are fuy fixed end the separate parts of the frame have ony oca and independent deformations but no dispacements. Of this reason the oca deformation is fuy determinate (it is known by Strength of materias). In addition, we have dispacement-controed system so, we can give a unit vaues of the joint dispacement and rotations. We wi give them unit vaue because we don t know the rea one (ϕ = Z = ; ϕ = Z = ; u = Z = ). Actuay, what we search for? As we add inks on the frame, they stop the dispacements but as a resut, it appears additiona reactions on them. R q R R Furthermore, we shoud add some information about rotationa ink. Let consider the consistency of the fixed support. H horizonta inear ink: u = 0 Fixed support: H V u = 0 v = 0 ϕ = 0 vertica inear ink: v = 0 V rotationa ink: ϕ = 0-56 -
The fixed support incudes two inear inks and one rotationa ink. One shoud remember that the rotationa ink can exist separatey (independenty) as the inear inks. As a resut when we put on the frame one rotationa ink we stop ony the rotation of the joint and add ony a moment as a reaction. Return on the indeterminate frame and the dispacement method. In the origina frame there is no any ink on the nodes and consequenty any joint reactions. Logicay if the two frames shoud be equivaent these new reactions shoud be zero. And this is the equiibrium equation from which we wi find the vaues of the joint dispacements and reactions. R = 0; R = 0; R = 0 R = R f + r. Z + r. Z + r. Z = 0 R = R f + r. Z + r. Z + r. Z = 0 R = R f + r. Z + r. Z + r. Z = 0 where: r ij. Z j is the reaction on ink i from the rea vaue of the rotation (dispacement) ϕ j (u). r ij is is the reaction on ink i from the unit vaue of the rotation(dispacement) ϕ j (u) = Z j =. R if is the reaction on ink i from the externa oad. From the obtained equiibrium equations we wi cacuate the rea vaues of the nodes rotations and dispacements: r. Z + r. Z + r. Z + R f = 0 r. Z + r. Z + r. Z + R f = 0 r. Z + r. Z + r. Z + R f = 0 The question is how to compose interna moment diagrams from the unit dispacements of the dispacement and how to cacuate the additiona reactions at the additiona inks. Let see the situation Z = the rotation of the eft node of the frame. Composing of unit diagrams. When we have added rotationa and inear inks then the nodes of the frames are fuy fixed there is no dispacements and rotations. If we impose rotation of the eft joint ony connected to it members of the frame wi deform. The other wi be undeforming. Z = In addition, the two deformed members are independent to each other. In other words, the ha frame is separated to independent members. In this way, the probem to anayze frame from the rotation at the eft node transforms to a probem to anayze the different members. This probem we aready anayzed in previous chapter. - 57 -
φ 4EE 6EE ϕ = EI EE 6EE EE EE ϕ = EI EE φ So the interna moment diagram we wi compose as composing diagrams from the different parts. It is enough ony to think carefuy how to rotate the diagrams. EE Z = 4EE EE The best way to draw the diagrams correcty is to ook at the deformed shapes of the different parts and to draw the diagrams on the tensie side of the member. The interna moment diagrams from the unit rotation of the right node of the frame wi be: EE Z = EE 4EE EE The interna moment diagrams from the unit dispacement wi be: - 58 -
Z = EE EE The interna moment diagrams from the externa oad wi be: q 8 q 8 f 0 The reactions r ij and R if we may find very easy using joint equiibrium. It foows: r r = EE EE + 4EE 4EE EE r r = EE r EE r = EE r EE 4EE r EE r = EE EE r = 4EE + EE + EE r r = EE EE - 59 -
It is important to understand that even parametric vaues of the moments are equa the numerica vaues are different because the cross sections, ength and E-moduus can be different. r r EE r = EE EE r = EE r EE r = EE + EE EE R f q 8 q 8 R f R f = q 8 R f = q 8 f 0 R f Rf = 0 Coefficient verifications: Again we have symmetry for the coefficients r ij = r ji and positive matrix r ii > 0. This can be used as verification. Other way to check the coefficient is as foows: r ij = EI R if = i f EI ds o where f is the interna moment diagram from the externa oad at arbitrary staticay determinate structure. This expressions are based on a principe of virtua work. The fina diagrams compose using next expression: ds o f = f0 +. Z +. Z +. Z As we aready have the interna moment diagram we can obtain the shear and norma forces diagrams. And in the end to find reactions and as fina check is the equiibrium of the system. Presented idea is vaid not ony for two-times indeterminate structures but for n-times indeterminate aso. The procedure is same. The equiibrium equations for n-times indeterminate structure are: r. Z + r. Z + r. Z + + r n. Z n + R f = 0 r. Z + r. Z + r. Z + + r n. Z n + R f = 0 r. Z + r. Z + r. Z + + r n. Z n + R f = 0 r n. Z + r n. Z + r n. Z + + r nn. Z n + R nf = 0-60 -
Primary system and fu exampe. As a main difference from the force method is that in the dispacement, method is ony one possibe and correct primary system. The primary system in the dispacement method deveops on two stages. First is adding rotationa inks and second is adding inear inks. Here wi be expained the two stages of composing primary system. Stage I: Rotationa inks. We pace rotationa inks at every rigid joint of the frame! What is a rigid connection have been expained at the first chapter. Here we wi remember ony the ideaized schemes of them. Rigid connections. Rigid joints of frames. Other paces where we pace rotationa inks are the partia hinges and partia supports. Partia hinges, partia supports. Fina paces where we pace rotationa inks are the rotationa springs. Exampe: rotationa springs. origina system. rotationa inks. Stage II: Linear inks. The second stage of composing primary system is to determine the number and paces of the inear inks. For this reason we compose the hinged-joint system pacing at each joint of the structure a hinge. We do the kineatica anaysis of the hinged-joint system. It shoud be - 6 -
composed by dyads. If it is not we add inear inks to make it composed by dyads (composed by not singuar dyads). Adding this inks actuay we determine their paces and numbers. Exampe: C C 7 4 5 8 A 9 6 A B B origina system. pin-joint system. KA: [G +..(w = +) + 4.5.6(w = +) + 8. AA + 9. BB + 7. CC ](w = +) In the present pin-joint system there are chains. To remove their degrees of freedom we shoud add two inear inks. The first chain is.. and has a horizonta degree of freedom so, we wi add a horizonta inear ink. Situation with the second chain is same so, as a resut for this system we wi have two horizonta inks. Incuding the rotationa and inear inks to the origina system we obtain the primary system on the dispacement method. The number of these inks is the number of the kinematica indeterminacy of the system. origina system. primary system. The present exampe is 7 times cinematicay indeterminate system. System incuding determinate parts. When in the origina system incudes determinate parts it is better to cut them off because it decrease the degree of the kinematica indeterminacy. Aso, it is good to know that some time the inear inks are vertica not ony horizonta. Exampe: origina system. primary system without taking into account determinate parts. - 6 -
This system incudes two determinate parts. First one is the cantiever part up of the frame and the second one is the simpe beam on right of the frame. When we taking into account this parts the system is 6 times cinematicay indeterminate. When we cut them off the system is times cinematicay indeterminate. origina system. primary system with taking into account determinate parts. On the next figure is shown the pin-jont system of the two situations and their kinematica anaysis. 4 A 5 A 6 B B pin-joint system incuding determinate parts. B B pin-joint system excuding determinate parts. For the pin-joint system incuding determinate parts: KA: [G +... BB (w = +) + 4(w = +) + 5. AA + 6(w = +)](w = +4) For the pin-joint system excuding determinate parts: KA: [G +... BB (w = +)](w = +) System incuding inear springs. When in the origina system incudes inear springs it is necessary to pace a inear ink at the spring because there is a dispacement even the system don t need inear inks according the pin-joint system anaysis. A A origina system incuding inear spring. pin-joint system incuding inear spring. primary system incuding inear spring. - 6 -
Kinematica anaysis of the pin-joint system incuding inear spring: KA: [G +. AA +.](w = 0) Fu exampe of anaysis of cinematicay indeterminate frame: Anaysis of cinematicay indeterminate frame oaded by externa oad: ain Data: F = 0 kn beams (0.5/0.5): coumns (0.5/0.5): = 5 knm I b = 0.00065 m 4 I co = 0.00055 m 4 q = 0 kn/m A b = 0.0788 m A co = 0.065 m I b =.I co. E =.4.0 7 kn/m F q 4 Kinematica indeterminacy: δ. ϕ F δ Each construction has a deformations caused by some arbitrariy oad. In this case we know the joint sope (rotation) ϕ and dispacement δ, so the deformation of the system wi be competey determined because the aw of the defection of the eements is known. That is why the system is times kinematica indeterminate.. Choice of primary system according dispacement method: It is very important to remember that in the force method are possibe severa correct primary systems, but in dispacement the correct primary system is ony one. The primary system wi be composed in two stages. - 64 -
First stage: To pace rotationa inks. We pace rotationa inks at the rigid joint. We use them to contro the joint rotation (sope). Z The unknown parameter in this case the rotation (sope) of the joint and it s abe is Z. The cantiever part of the system is staticay determinate part and may be removed. Z determinate part(piece) The second stage: The second stage of composing primary system is to determine the number and paces of the inear inks. For this reason we compose the hinged-joint system pacing hinges at each joint of the structure. Z A KA: [G+..(w=+)](w=+) KA: [G+.A +.](w=0) After that we make a kinematica anaysis of the hinged-joint system. It shoud be composed by dyads. If it is not we add inear inks to make it composed by dyads (composed by not singuar dyads). As we determine the number and positions of the inear inks we obtain the primary system: - 65 -
Z Z The origina system is two times cinematicay indeterminate. We anayze the primary system from unit vaues of the unknowns dispacements and rotations of points.. Unite diagrams: For to make cacuations easier we wi make some reductions. As we know the vaue of is: i = EI If we divide this equation by EI coumn (or EI beam ) the inear stiffness wi become: i EI i = EI i = c i EI c I c i Ii ; I c i = i ; I i EI c i i = i where EI c is the stiffness of the coumn(ei coumn ) or of the beam (EI beam ) and EI i is the stiffness of the i-th member of the frame. i i and i are the inear stiffness and the ength of the i-th member of the frame. After this transformation a stiffness coefficients r ij (reactions from the unit dispacement or rotations) wi be reduced by /EI c : EI c r ij The coefficients from the externa oad are rea R if. In this way for the equiibrium equations in matrix form we wi have: [r]{z} + {R} = 0 EI c {Z} = {R} EI c [r] = {R} [r] (EI c) As a resut, we obtained not rea vaues of the unknowns but mutipied by EI c. If we need their rea vaues, we shoud divide the obtained vaue by EI c. A the time we have not rea vaues of the stiffness coefficients but their /EI c reduction vaues. This transformation is not obigate but is better for to work with suitabe numbers. The present exampe is done using this idea. - 66 -
= i = 0,5 I c = I co. I c / I b. = 0,5 = i= 0, =,6 i= 0,6 4 Deformed shapes from the unit rotation of the joint: Z = Z = 4i =, i =,5 М i = 0,6667 Deformed shape from unit horizonta dispacement. First, we obtain the dispaced shape of the hinged-joint system after that for the primary system and in the end, we obtain the diagram. Z = Z = δ δ. δ. δ Deformed shape of the primary system α Deformed shape of the hinged joint system α 90 о - α δ α δ М - 67 -
. Diagram from the externa oad in the primary system: М = 40 Determinate part: = 5 F = 0 Q = 0 40 М = 40 Q = 0 q = 0 Loads on the primary system: 4 In advance, we anayze the determinate part of the structure and its reactions are used as actions of the primary system. In this situation, the actions are in the joint with a inear and a rotationa ink, so they do not cause any internay moment forces in the eements. 4 = 5 5 5 7,5 7,5 q = 0 4. Reaction in the inks: The reactive forces are introduced by the same sing as the unknowns. After that the signs become positive or negative because of equiibrium equation. From М r N h,5 90 о - α N, r 0,6667 N h = 0,5-68 -
From М r 0,5 90 о - α N N h 0,6667 r r = 0 : 0,5 + 0,6667 = 0 = 0,5467 r 0,4444 N h = 0,00957 From М f o = 40 R f 7,5 90 о - α N N h 7,5 R f Q = 0 5 N h =,875 R if is the reactions of the additiona inks from the externa oad. 4. Soution of the system of the equations:,8z 0,5467Z 0,5467Z + 0,560Z 5 = 0,5 = 0 Z Z = 4,69 = 8,00 5. Fina diagrams: The fina diagram is composed for each characteristic point and it is cacuated by the next o expression: f = f +. Z +. Z - 69 -
40 4,9 5 40 54,9 4,4806 54,9 4,9 М f 46,7 6,55 4 0 0 7,478 5,48 7,478 5,48 4,577 Q f 8,0 7,478 5,48 4 90 о - α N N h 7,478 4,577 N f,9 4-70 -
6. Equiibrium verification: 0 5 C 0 5,48 46,7 4,57 6,55 7,478 4 7,478-7 -
Anaysis from temperature oad, support settement. The anaysis of the structures from the temperature oad has two stages. We shoud remember from Chapter 8 the infuence of the anguar and axia deformations. Anguar deformation cause ony interna moments and axia deformations cause ony norma forces. In the case of cinematicay indeterminate structures under temperature oad, we can distinguish two stages of deformations. As it is known the temperature oad divides to a oad from temperature difference and oad from constant oad. Finay the using dispacement method we divide the anaysis of a primary system to a two parts and it foows the interna moment diagram in the primary system has two parts. o t = t, t + t, tc Interna moment diagrams from temperature difference. The nodes of the primary system have no possibiity to move or rotate so the temperature difference wi cause ony deformations of the eements. t t t = t - t origina system. primary system As a resut, there is diagram ony at the oaded eements. The moment vaues for every member is obtained using force method as the soutions from the unit dispacements. The resut is: EE. α t. t h 0 t, t EE. α t. t h t = t - t primary system Interna moment diagrams from constant temperature. The constant temperature cause ony inear eongation. There is no defections so it cannot be a reason of appearing moment diagram but it is a reason of noda dispacement. Eement eongation is: = α t. t t. t t = t + t As we know the eongations of the eements we may compose the deformed shape of the system but first we wi compose the dispaced shape of the pin-joint system. When compose the - 7 -
dispaced pin-joint system we shoud be carefu for the inear inks. They stops the dispacements at some direction and the dispacements is possibe ony at one directions. t beam t coum 0 t,tt pin-joint system After we have the dispacement pin-joint system we know the new paces of the frame nodes and we may compose the deformed shape of the primary system. beam coum 0 t,tt dispaced primary system One can see that the eongation by itsef don t cause interna moments but the dispacements of the joints at the primary system cause the defections of the eements. Actuay these defections cause interna moments in the primary system. For the present exampe the eongation of the eft coumn causes defections at the beam and the eongation of the beam causes defections at the right coumn. As a resut interna moments diagram wi appears at the beam eement and at the right coumn. The vaues of the moments are as moments from unit dispacement of one of the joints. The difference is that the dispacement is not unit but equa to beam (or coumn ). The resut is: 6EE. cccccc 0 t,tt 6EE. cccccc primary system 6EE. bbbb 6EE. bbbb As we aready have the two parts of the interna moments we can compose fu moment diagram in the primary system from the temperature oad. 0 0 0 t = t, t + t,tt - 7 -
For the present exampe this diagram is as foows: EE. α t. t h EE. α t. t h + 6EE. cccccc t 0 primary system EE. α t. t h 6EE 6EE. bbbb 6EE. bbbb. cccccc Once we have this diagram we may cacuate the reactions on the added inks from the exteran temperature oad R it and to compose the equiibrium equations to cacuate unknowns Z it. Next is presented a numerica exampe. Exampe: t = 5 o t = 0 o α = t t 4 The diagram from temperature oad in the primary system is composed in two stages. First stage In this stage is composed the moment diagram in each member separatey from the temperature difference t. t + t 5 + 0 EI c = 78 α EI c = 0,078; tср = = =,5 ; t = t t = 5. Diagram in the primary system: Second stage: Diagram from the constant temperature. - 74 -
The eements eongations are: = αei t. = 0,078.,5. =,995 c = αei t c ср ср. = 0,078.,5.4 =,906 The dispaced pin-jointed system is: δ. δ pin-joint α Diagram caused by the dispaced frame nodes is: Foows fu diagram from the temperature oad at the primary system: 0 0 0 t = t, t + t,tt 9, 7,8,55. Computing reactions at the added inks: - 75 -
From М t o R t 9, 90 о - N N h 7,8 R t N h = Verification: α t t Rit = EJ c s ds + EJ c N sα. tсрds = αej c sds + tср N h h s ds 0,666,6 0,408 0,6 0,406 0,465 αej t sds + tср N h = 0,078. = 0,078. R it c 5 0,5 [ 00 + 57,94 + 5,4,04] =,587 0,66667. + s!,587 =. Equiibrium equations. ds = 5,65.4 0,5,587 +,5.0,4065.,5.0,4087.4 = = 0,078.6, =,587,8Z 0,5467Z 0,5467Z + 0,560Z,98 = 0,89 = 0 Z Z = 5,4 = 7,75. Fina diagram. t = t +. Z +. Z - 76 -
0,9 0,9 М t 9,06,568 4 7,47,55 0,44 0,44 0,44 Q t 8,0,55 4 0,44,55 N t,96 4 90 о - N N h - 77 -
C 0,44 9,06 0,44,568,55 4,55 Anaysis from the support settement. This anaysis is not so different from the standard one. The support settement cause of the primary system node dispacements which cause eement dispacements. This defections cause interna moments at the defected eements. The idea is the same as the idea of the diagrams from the constant temperature oad. Actuay, we shoud be carefu with the dispaced pin-joint system (if it is necessary) and as a resut the defected primary system. The ast one wi show as which eements have dispaced nodes and with what vaue so we wi be abe to compose the interna moment diagram. Of course one shoudn t forgot that the dispacements causing moments diagrams in the eements are not equa to unit but are equa to the support settements. Here wi be present ony the numerica exampe: c h = 0,05 m ϕ = 0,05 rad EJ c = 78 kn c ϕ 4. Deformed primary system: The present exampe is cacuated with a reductions of the unit reactions that is why moment diagram at the primary system from the support settement is mutipied by EI c. - 78 -
c Deformed primary system α ϕ. Computing reactions at the added inks: From R c 90 о - N N h 60, R c N h = Verification: R ic + EJ c Rs. c = i To do this verification it is need to determine the support reactions in the summary diagram from unit vaues of the unknowns. - 79 -
0,00 0,6667,65 0, 0,4065 0,6 0,408 0,6 0,4065 R ic = 60,4 + 50,06 = 0,4 The corresponding reactions of this diagram are mutipied to the vaues of the support settements: EJ c R ic [ ( 0,05) + 0,6. ( 0,05) ] = 0, 8 R. c = 78 0,. i i = + EJ c Ri. c i! 0,4 =. Equiibrium equations: 0,8,8Z 0,5467Z 0,5467Z + 0,560Z 60,4 = 0 + 50,06 = 0 Z Z = 9,79 = 0,606 4. Fina diagrams from the support settements: + o t =. Z +. Z c 7,6 7,6 98,8 М 4 70,8 Finae verifications, norma and shear diagrams are not shown here but they are a right. Anaysis of structures incuding eements with infinite high system. One shoud know that this eements are actuay rigid body eements. As a resut they cannot deform. They ony rotates or moves without deformations without changing their shapes. - 80 -
This fact must be taken into account when composing the primary system. Aso when such an eement rotates it rotate and the fexibe eements. In the end of the norma eements wi appears not ony dispacements but rotation too. Aso, it is important to understand that the fexibe eements are rigidity connected at the rigid eement but the rigid eement is connected at the fexibe eement with a joint connection! This wi be shown at the next exampe. Exampe: F = 00, = 0, q = 40, EI = 50 000 knm F I = 4 q I = 4 5 5 5 i = Z i = i = i = If we see the pin-joint system of the present exampe, we wi need two inear inks. But as the system has a rigid eement so the two inear dispacements are dependent one to the other. Actuay, they are dependent to the rotation of the rigid eement so this structure has ony one independent dispacement parameter the rotation of the rigid body. That is why this system is one time cinematicay indeterminate system. If we know this rotation we wi know the dispacements of the other nodes of the system. Next stage is to compose the dispaced shape of the primary system from the unit rotation of the rigid body. When rigid body rotates to a unit ange the point at a distance 5 metre wi have 5 metre dispacement aso the rigid connection between the rigid body and the fexibe one cause additiona rotation at the fexibe eement. As a resut the dispaced shape of the primary system is shown beow: - 8 -
4 φ = φ = Z= φ = 5 To compose the moment diagram for some eements we shoud be carefu if at the eement has ony dispacement or dispacement pus rotation. At the present exampe the right beam eement has dispacement and rotation and the other fexibe eements defect ony because of the rotation of the rigid body. The diagram of the right beam eement is composed by two parts and is presented on next figure. 6i. 5 6i. 5 5 i. 4i. 6i. 5 +. i 6i. 5 + 4. i At the connections between rigid body and fexibe eements the moment diagram composes after equiibrium of the joint. In this why the unit moment diagram is presented on next figure: i = 0,6 i = 0,75 0,6 0,6 Z =,6 5,6 The diagram from the externa oad is composed in the standard way ony taking into account the connections between rigid and fexibe body. - 8 -
00 87,5 0 77,5 6,5 80 77,5 40 97,5 f Next soution continues standard. Anaysis of symmetrica structures. The idea of the symmetrica system is the same as this in the force method. We use again the coupe of unknowns but in the dispacement method, the unknowns are nodes dispacements. That is why we wi show exampe directy. Exampe: 40 I I I I q = 8 I 5 6 The primary system and the couped unknowns are shown beow: Z Z Z Z Z Primary system As one can see there are one symmetrica and two antisymmetrica unknowns. First unknown is a coupe and symmetrica, second is again coupe but antisymmetrica and the ast one is not coupe (it is ony one force) and is antisymmetrica. When compose diagrams from the couped unknowns one shoud be carefu because at some members there are rotations at the two ends of the eement. To compose the diagram in this case one shoud use the principe of the superposition shown beow. In midde beam eement there is doube rotation from the unit vaue of the first unknown so, the diagram wi be composed in two parts using superposition. - 8 -
Z = Z = i = i = Z Z 4i=0,8 i=0,667 0,8 Deformed shape i=0,4 0,4 Unit diagram Z = Z = i 4i i 4i i From unit vaue of the second unknown there is again doubed rotation in the midde beam eement but in this case at different direction. The diagram composes in same way, ony the resut is different. Z = Z = i = Z 6i= Z 4i=0,8 6i= i = Deformed shape i=0,4 0,4 Unit diagram - 84 -
Z = Z = 4i i i 4i 6i 6i From the unit vaue of the ast unknown there is no such effects. The unit diagram composes simpy. The resut is shown beow. 6i/ = 0,4 6i/ = 0,4 Z = Z Deformed shape 6i/ = 0,4 Unit diagram 6i/ = 0,4 oment diagram from the externa oad is nonsymmetrica and is composed standarty. 40 6,667 0 f 0 8, 6,667 Next stage of the soution is to cacuate reactions at the additiona inks from the previous oads. In the case of coupes unknowns the reactions are aso couped eft and right part. The fu reaction which we shoud use in the equiibrium equation is sum of the two parts. This summation is shown beow ony for the first unit diagram. For the other diagrams the resuts obtain anaogicay. - 85 -
r r r r r 0,8 0,667 0,8 r r 0,4 0,4 r r 0,667 0,667 0,8 r =,47 r =,47 r = r + r =,47 +,47 = r = r + r =,47 -,47 = 0 0,8 r =,47 r = -,47 r 0,4 0,4 r = 0,4-0,4 = 0 As one can see the antisymmetrica reactions from the symmetrica oad are zero and anaogicay the symmetrica reactions from the antisymmetrica oad are zero. The interna moment diagram from the externa oad is non-symmetrica but the reactions are again couped. Their cacuation is shown beow. 40 R R R R R f 0 6,67 8, 0 R R R = - 0 R = - 0 6,67 R = R + R = - 0-6,67 = -6,67 R = R + R == - 0 + 6,67 = -, R R 6,67 R = - 6,67 R = 6,67 R f R f = 0 0 As we have a coefficients the soution continuous as standard one. - 86 -
Anaysis of structures incuding springs. The anaysis of structures incuding springs is generay standard with ony deferens we have mentioned one shoud add a inear ink at the spring if the spring is inear and rotationa ink if the spring is rotationa. We do this to contro the dispacement (rotation) of the spring. At the next exampe is shown a structure with a inear spring. Anaysis of a structure incuding rotationa spring is anaogica. Exampe: q = 5 c = 0000 45 0 EI = 40000kNm 4 4 i = 0,5 Z i = 0,5 Z i = 0, primary system 4 4 As it is shown it is not necessary the inear ink to be in the same direction as the spring. The deformation and the diagram from the unit vaue of the first unknown is standard. The spring cannot deform because of the inear ink. Z = - 87 -
4i = Z = i = 0,5 i = 0,75 4i = 0,8 i = 0,4 The deformation from the unit vaues of the second unknown is shown on the next figure: Z =,0 0,707 45 o As we do unit vertica dispacement (Z = ) at the spring appears dispacement equa to 0,707 because of the ange 45. This dispacement at the spring evoke reaction at the spring equa to the dispacement mutipied by the stiffness of the spring. In our case we have divided this reaction by EI because we produce soution using reduced vaues of the reaction (we have shown aready). And the moment diagram is: Z = It is important to understand that the reaction at the spring is aways zero ony in the case of unit dispacement of the inear ink is different of zero. Reactions at the added inks are:,0 r : From r 0,8 0,75 0,875 r =,550 r = - 0,875-88 -
r From r r = 0,78-0,875 r = - 0,875 0,046875 0,5 45 0 The interna moment diagram and reactions from the externa oad are: R f = - 0 From f o R f = - 7,5 0 7,5 As we have a coefficients the soution continuous as standard one. Literature:. Методично ръководство за решаване на задачи по строителна статика, Т. Карамански, Р. Рангелов, издателство Техника 97г.;. Structura Anaysis, R.C. Hibeer, Prentice Ha 006;. Structura Anaysis in theory and practice, A. Wiiams, Internationa Codes Counci 009. - 89 -