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# STRUCTURAL ANALYSIS CE 382

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This 63 page Class Notes was uploaded by Connie Lockman on Friday October 23, 2015. The Class Notes belongs to CE 382 at University of Kentucky taught by George Blandford in Fall. Since its upload, it has received 75 views. For similar materials see /class/228170/ce-382-university-of-kentucky in Civil Engineering at University of Kentucky.

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Date Created: 10/23/15

lntrobuction to Sta ca q Inbctcrminate Analusis Support reactions and internal forces of statically determinate structures can be determined using only the equations of equilibrium However the analysis of statically indeter minate structures requires additional equations based on the geometry of deformation of the structure Additional equations come from compatibility relationships which ensure continuity of displacements throughout the structure The remaining equations are constructed from member constitutive equations ie relationships between stresses and strains and the integration of these equations over the cross section Design of an indeterminate structure is carried out in an iterative manner whereby the relative sizes of the structural members are initially assumed and used to analyze the structure Based on the computed results displacements and internal member forces the member sizes are adjusted to meet governing design criteria This iteration process continues until the member sizes based on the results of an analysis are close to those assumed for that analysis 3 Another consequence of statically indeterminate structures is that the relative variation of member sizes in uences the magnitudes of the forces that the member will experience Stated in another way stiffness large member size andor high modulus materials attracts force Despite these difficulties with statically indeterminate structures an overwhelming majority of structures being built today are statically indeterminate Advantaes statically Indeterminate Structures Suxically Determinate Beam LE Bending Momenl Diagmm Eending Mamenl Dingnm statically indeterminate structures typically result in smaller stresses and greater stiffness smaller deflections as illustrated for this beam Internal hinge Siaticnlly Demmimic Beam Inicmal hinge snucany Unstable a Determinate beam is unstable if middle support is removed or knocked off Siaiiczlly Siabl bl statically indeterminate structures introduce redundancy which may insure that failure in one part of the structure will not result in catastrophic or collapse failure ofthe structure a Disadvantaes of statically Indeterminate Structures n Smtically Deiemiinaiz Beam innB b Slalically Indcieminaie Beam statically indeterminate structure is selfstrained due to support settlement which produces stresses as illustrated above in n L I 6alAnL la Smicully Duhmm Mini A Fauna M imam 4 5A Statically indeterminate struc tures are also selfstrained due to temperature changes and fabrication errors Indeterminate Structures Influence Lines Influence lines for statically indeterminate structures provide the same information as influence lines for statically determinate structures ie it represents the magnitude of a response function at a particular location on the structure as a unit load moves across the structure Our goals in this chapter are 1To become familiar with the shape of influence lines for the support reactions and internal forces in continuous beams and frames 2To develop an ability to sketch the appropriate shape of influence functions for indeterminate beams and fr m 3To establish how to position distributed live loads on continuous structures to maximize response function values 3 Qualitative Influence Lines for Statically Inde terminate Structures MullerBreslau s Principle In many practical applications it is usually sufficient to draw only the qualitative influence lines to decide where to place the live loads to maximize the response functions of interest The MullerBreslau Principle pro vides a convenient mechanism to construct the qualitative influence lines which is stated as 14 The influence line for a force or moment response function is given by the deflected shape of the released structure by removing the displacement constraint corresponding to the response function of interest from the original structure and giving a unit displacement or rotation at the location and in the direction of the response function Procedure for constructing qualitative in uence lines for indeterminate structures is 1 remove from the structure the restraint corres ondin to the response function of interest 2 apply a unit displacement or rotation to the released structure at the release in the desired response function direction and 3 draw the qualitative deflected shape of the released structure consistent with all remaining support and continuity conditions Notice that this procedure is identical to the one discussed for statically determinate structures However unlike statically determinate structures the influence lines for statically indeterminate structures are typically curved Placement ofthe live loads to maximize the desired response function is obtained from the qualitative lLD Uniformly distributed live loads are placed over the positive areas of the ILD to maximize the drawn response function values Because the in uence line ordinates tend to diminish rapidly with distance from the response function location live loads placed more than three span lengths away can be ignored Once the live load pattern is known an indeterminate analysis of the structure can be performed to determine the maximum value of the response function 18 QILD for RA QILD s for RC and VB 20 QILD s for MC MD and RF Live Load Pattern to Maximize Forces in Multistory Buildings Building codes specify that members or multistow buildings be designed to support a uniformly distributed live load as well as the dead load of the structure Dead and live loads are normally considered separately since the dead load is fixed in position whereas the live load must be varied to maximize a particular force at each section of the structure Such 22 maximum forces are typically produced by patterned loading Qualitative Influence Lines 1 Introduce appropriate unit displacement at the desired response function location 2 Sketch the displacement diagram along the beam or column line axial force in column appropriate for the unit displacement an assume zero axial deformation 3 Axial column force do not consider axial force in beams a Sketch the beam line qualitative displacement diagrams b Sketch the column line qualitative displacement diagrams maintaining equality of the connection geometw before and after deformation 4 Beam force a Sketch the beam line qualitative displacement diagram for which the release has been introduced b Sketch all column line qualitative displacement diagrams maintaining connection geometry before and after deformation Start the column line qualitative displacement diagrams from the beam line diagram of a c Sketch remaining beam line qualitative displacement diagrams maintaining con nection geometry before and after deformation iii Column Moment Load Pattern to M Maximize M 28 CCU CD 7 39 H igi 4 3 ml 133 I I3 Q K ic DI l 7lt l 39i l i l i ii i l l l l l l l i ll JL J L A U QILD and Load Pattern for Center Beam Moment M T r ew g l l igmI A KLTam r new l 7 l l u l l l QILD and Load Pattern for End Beam Moment M Expanded Detail ior Beam End column Vloment Envelope Curves Design engineers often use influence lines to construct shear and moment envelope curves for continuous beams in buildings or for bridge girders An envelope curve defines the extreme boundary values of shear or bending moment along the beam due to critical placements of design live loads For example consider a threespan continuous beam Qualitative influence lines for positive moments are given shear influence lines are presented later Based on the qualitative influence lines critical live load placement can be determined and a structural analysis computer program can be used to calculate the member end shear and moment values for the dead load case and the critical live load cases QILD for Me Critical Live Load Placement for Ma Critical Live Load Placement for Mb Critical Live Load Placement for Ma39 Critical Live Load Placement for Mb39 Critical Live Load Placement for MC Critical Live Load Placement for MC39 Critical Live Load Placement for Md Critical Live Load Placement for Md39 Critical Live Load Placement for Me Critical Live Load Placement for Me39 Calculate the moment envelope curve for the threespan continuous beam Liquiq q L 20 240quot E 3000 ksi A 60 in2 500 in4 WDL 12 kt39t dead load WLL 48 kt39t live load Shear and Moment Equations for a Loaded Span Mi q gtMie x 4 Vi Vie Vi Cl Xi Mie 39Mi Xi Vie Shear and Moment Equations for an Unloaded Span set q 0 in equations above Vie Vi Mie39MiViXi 41 Load Cases A summary of the results from the statically indeterminate beam analysis for each of the seven load cases are given in your class notes RESULTS FOR LOAD SET MEMBER FORCES MEMBER AXIAL MEMBER NODE FORCE kip 1 1 000 2 000 2 2 000 3 000 3 3 000 4 000 SHEAR FORCE kip 960 1440 1200 1200 1440 960 1 t BENDING MOMENT k 000 4800 4800 4800 4800 000 44 The equations for the internal shear and bending moments for each span and each load case are Load Case 1 V12 M12 V23 12 12X2 M23 3948 12x2 UU22 V34 12X3 M34 3948 Load Case 2 V12 M12 V23 0 M23 3996 V34 M34 3996 Load Case 3 V12 M12 394UX1 V23 M23 3996 48x2 V34 M34 3996 46 Load Case 4 V12 M12 V23 8 M23 391 8X2 V34 M34 39 Load Case 5 V12 M12 gtDX1 V23 398 M23 39 8X2 V34 M34 128 544x3 24x g2 Load Case 6 V12 48X1 M12 V23 M23 56x2 4 r22 V34 M34 3964 32X3 Load Case 7 V12 M12 39JZX1 V23 M23 3964 40x2 V34 M34 224 592x3 24x 2 4M nLaps 441 ihLler winHf A awnKin 57e39mLips Bending Moment Diagram LC1 2m LnLi Hui Lies gt IOB ns2 39mLiys Bending Moment Diagram LCZ i728 39mLzrx V I157 InLr f W mamkip Bending Moment Diagram LC 21s I A39MPS i536inkipx Bending Moment Diagram LC4 In miufs 184 in Jnf s V a 1 mquot 36 ux Lira Bending Moment Diagram L05 h noquot ice why 139 1 ml may 1 92quot k v Z ee 39mJu39Fs e ms wide Bending Moment Diagram LCB laaquot 93 39 I132 mkips r quot L iquot 1 5quot K m quotVLny am inLip Bending Moment Diagram L07 5 Aspreadsheet program listing is included in your class notes that gives the moment values along the span lengths and is used to graph the moment envelope curves In the spreadsheet Live Load EMom max LC2 through LC7 Live Load EMom min LC2 through LC7 Total Load EMom LC1 Live Load EMom Total Load EMom LC1 Live Load EMom El Moment Irvklps MOMENT ENVELOPE Pas Mom Envelope r Neg Mam Envelope Total Load EMom Total Load EMom H 55 Construction of the shear envelope curve follows the same procedure However just as is the case with a bending moment envelope a complete analysis should also load increasing decreasing fractions of the span where shear is being considered QILD sz 57 QILD v4 Shear ILD Notation Superscript L just to the left of the subscript point Superscript R just to the right of the subscript point To obtain the negative shear qualitative in uence line dia grams simply flip the drawn positive qualitative influence line diagrams In practice the construction of the exact shear envelope is usually unnecessary since an approximate envelope obtained by connecting the maximum possible shear at the reactions with the maximum possible value at the center of the spans is sufficiently accurate Of course the dead load shear must be added to the live load shear envelope Plastic Analusis of Continuous Beams39 Increasing the applied load until yielding occurs at some locations will result in elasticplastic defor mations that will eventually reach a fully plastic condition Fully plastic condition is de ned as one at which a suf cient number of plastic hinges are formed to transform the structure into a mecha nism ie the structure is geometrically unstable 1See pages 142 152 in your class notes Additional loading applied to the fully plastic structure would lead to collapse Design of structures based on the plastic or limit state approach is increasingly used and accepted by various codes of practice particularly for steel construction Figure 1 shows a typical stressstrain curve for mild steel and the idealized stress strain response for performing plastic analysis rupture 6 idealized 8y Figure 1 Mild Steel Stress Strain Curve cry yield stress ey yield strain ULTIMATE MOMENT Consider the beam shown in Fig 2 Increasing the bending moment results in going from elastic cross section behavior Fig 2a to yield of the outermost fibers Figs 2c and d and finally the two yield zones meet Fig 2e the cross section in this state is de ned to be fully plastic l quotv l l l Swami lVvl i m m c in l H 4 4 Suamp dislnbumn Naulnl axix m fully plain ml 7 Vi m Figure 2 Stress distribution in a sym metrical cross section subjected to a bending moment of increasing magni tude a Cross section b Elastic c Top fibers plastic d Top and bottom fibers plastic and e Fully plastic 5 The ultimate moment is determined in terms of the yield stress oy Since the axial force is zero in this beam case the neutral axis in the fully plastic condition divides the section into two equal areas and the resultant tension and compression are each equal to 6y A2 forming a couple equal to the ultimate plastic moment Mp Mp GyAyc yt E The maximum moment which a section can resist without exceeding the yield stress defined as the yield moment My is the smaller of My C3y St 2a My c3y Sc 2b St tension section modulus 5 Ct Sc compression section modulus E Icc ct distance from neutral axis to the extreme tension fiber cc distance from neutral axis to the extreme com pression fiber l moment of inertia on Mleygt 1 shape factor 15 for a rectangular section 17 for a solid circular section 115 117 for I or C section a PLASTIC BEHAVIOR OF A SIMPLE BEAM If a load P at the mid span of a simple beam Fig 3 is increased until the maximum mid span moment reaches the fully plastic moment M a plastic hinge is formed at this section and collapse will occur under any further load increase Since this structure is statically deter minate the collapse load PC can easily be calculated to give PC 4MpL 93 a Loaded Beam b Plastic BMD Pc 29 A c Plastic Mechanism Figure 3 Simple Beam m a was 5 Elam de ecuan W mmquot Hz i IH 2523 m in Plastic Hinge Along the Length of the Simple Beam The collapse load of the beam can be calculated by equating the external and internal work during a virtual movement of the collapse mechanism this approach is equally applicable to the collapse analysis of sta tically indeterminate beams Equating the external virtual work We done by the force PC to the internal virtual work Wi done by the moment Mp at the plastic hinge Le wewi 2 PC7MP29 2 PC4MpL which is identical to the result given in 3 ULTIMATE STRENGTH OF FIXEDENDED BEAM Consider a prismatic fixedended beam subjected to a uniform load of intensity q Fig 4a Figure 4b shows the moment diagram sequence from the yield moment My 2 l qu My GyS c 12 12My 2 q y L2 through the fully plastic condition in the beam M Figure 4 FixedFixed Beam 5 The collapse mechanism is shown in Fig 4c and the col lapse load is calculated by equa ting the external and internal virtual works ie 2 EV 9 Mp9299 2 4 16Mp L2 Sequence of Plastic Hinge Formation DqCZ 1 Fixedend supports maxi mum moment negative 2 Midspan maximum positive moment 6 ULTIMATE STRENGTH 0F CONTINUOUS BEAMS Next consider the three span continuous beam shown in Fig 5 with each span having a plastic moment capacity of Mp Values of the collapse load correspond ing to all possible mechanisms are determined the actual collapse load is the smallest of the possible mechanism collapse loads M constant p P P a L A AEt lEIZ D A E F L L f L f L 4 01 A2 15 6B Figure 5 a Continuous Beam b Mechanism 1 c Mechanism 2 8 For this structure there are two possible collapse mechanisms are shown in Figs 5b and c Using the principle of virtual work We W for each mechanism leads to Figure 5b A1 L92 PCl Mp9299 3 PCI 8MpL Figure 5c A2 L93 PCZ Mp99B ZLB A2 L9 3 2 3 2 9 5M 9 P02 2p 2 PC2 15Mp2L The smaller of these two values is the true collapse load Thus PC 75MpL and the corres ponding bending moment diagram is shown below When collapse occurs the part of the beam between A and C is still in the elastic range M gt Mp M Collapse BMD 2 e B L1 Figure 6 a Continuous Beam b Mechanism 1 22 c Mechanism 2 The two span continuous beam shown in Fig 6 exhibits some unique considerations 1the plastic moment capacity of span 12 is different than the plastic moment capacity of span 2 3 and 2the location of the positive moment plastic hinge in span 2 3 is unknown Mechanism 1 We PCA1 Lie W 2Mp9 2Mp29 Mp9 7Mp9 14Mp we wi 2 PC 7 Mechanism 2 A A We QCL172 ClcLL172 A2 L qC 2 24 Wi Mp9Mp9B L19 A2 LL1l3 L1 L Ll W 2LL1Mpe 13 9 L Ll we qCLL19 Ne Wi 3ICL 2L 2 ZL Ll L Ll iMP B The problem with this solution for ch is that the length L1 is unknown L1 can be obtained by differen tiating both sides of qCL With respect to L1 and set the result to zero ie dltchgt 2L1ltL L1gt dL1 ltL1gt2ltL L1gt2 22L L1L 2L1M ltL1gt2ltL L1gt2 0 c 26 Mp P Solving C for L1 2L 8LL14L2 0 i 2 2 3L18L 8L 48L 2L JEL 05858L Substituting D into B 1166Mp qC L D Comparing the result in A with E and for qL P shows that the failure mechanism for this beam structure is in span 23 M gt 2Mp Mp BMD for Collapse Load qc 28 Direct Procedure to Calculate Positive Moment Plastic Hinge Location for Unsymmetrical Plastic Moment Diagram Consider any beam span that is loaded by a uniform load and the resulting plastic moment diagram is unsymmetric Just as shown above the location of the maximum positive moment is unknown For example assume beam span B C is subjected to a uniform load and the plastic moment capacity at end B is Mp1 the plastic moment29 capacity at end C is Mp2 and the plastic positive moment capacity is Mp3 Mp1s Mp3 Mp2 5 Mp3 The location of the positive plastic moment can be determined using the bending moment equation Mx ax2 bx c and appropriate boundary conditions ix0 MMp1c ii x L1 Mp3 aL12 bL1 c 3 aL12 bL1 Mp3 Mp1 iii X L1 dMdx 0 2aL1 bsw Solving for a and b from ii and iii Mp1Mp3 Li Mp3 L1 Mp2 aL2 bL c 39Mp1 Mp3LL12 MmMm 0Mm 0 39Mp1 Mp3LL12 2Mp1 Mp3 LL1 Mp1 Mp2 Solving the quadratic equation 39L1 le i M1hg3gt274mprMp2MplMp3 21VIP11Ip3 7 Mpl Mpz 4i 1 WW L 1 1 M Mp1Mp3 EPILOGUE The process described in these notes and in the example pro blems uses what is referred to as an upper bound approach e any assumed mechanism can provide the basis for an analysis The resulting collapse load is an upper bound on the true col lapse load For a number of trial mechanisms the lowest computed load is the best upper bound A trial mecha nism is the correct one if the corresponding moment diagram nowhere exceeds the plastic moment capacity 35 System Loa ng Tributary Areas Many floor systems consist of a reinforced concrete slab sup ported on a rectangular grid of beams Such a grid of beams reduces the span of the slab and thus permits the designer to reduce the slab thickness The distribution of floor loads on floor beams is based on the geometric configuration of the beams forming the grid fa 5 17 it 3913 l l 47 20 2 33 7 1 I I I I E l l l l939 ll 8 39 7 7 7 l I q panel I l l l939 39739 L I I I I I 7f 939 IIS 739lti77773977 I Lntint lv39 7 I I I I 71 7 747 i W7 7 m Hm39 ll39 ifquot I39Tl Luann21 szmcl Lpamcl Tributary area of columns A1 BZ and C1 shown shaded area for bean Bl I B 1 I slab i tributary h a VVLS l L I M LS T T C Theoretical Tributary Areas 05L1 05L2 05L2 Theoretical Tributary Beam Areas 4 l l 4 BI I Theoretical Tributary Beam Areas xian lf e d re 43 L1 1 05L1 I 05L2 L2 l 05L2 Typical Floor Framing System w P P P P Floor Beam Girder Simplified Floor Beam and Girder Loadings 6 Example Load Distribution Problem The floor system of a library consists of a 6in thick rein forced concrete slab resting on four floor steel beams which in turn are supported by two steel girders Crosssectional areas of the floor beams and girders are 147 in2 and 523 in2 respectively as shown on the next page figure Determine the floor loads on the floor beams girders and columns Steel Steel girder A 523 in column i 3 at 10 ft 30 ft7ll IA B I C lD l 39l I Tributary area for beam CG 1 l G 39H39 6 in concrete slab E F Steel oor beam A 147in2 5 ft 5 ft Floor Slab Floor Beam Girder Column Schematic Building Live Load Reduction Recognizing that the probability of supporting a large fully loaded tributary area is small building codes permit reductions in the standard Lo design live loads when the influence area AI KLLAT is larger than 400 ft2 372 m2 as given in the following formulas LL0025i us Units IKLLAT L025i sum IKLLAT L a reduced live load 050 L0 3 L 3 L0 for single floor members 040 L0 3 L 3 L0 for multifloor members AT a tributary area ft2 m2 0 KLL element live load factors IBCZOOO Table 160791 Type of Element KLL Interior column 4 Exterior column without 4 cantilever slabs Edge columns with cantilever 3 slabs Corner columns with 2 cantilever slabs Edge beams without 2 cantilever slabs Interior beams 2 All other beams 1 Load Combinations for Strength Design The forces eg axial force moment and shear produced by various combinations of loads need to combined in a proper manner and increased by a load factor in order to provide a level of safety or safety factor Combined loads represent the minimum strength for which members need to be designed also referred to as required factored strength ASCE 798 has specified the following load combinations 2 1 14D 12DFT16LH 05 Lr or S or R 3 12D16LrorSorR 05 L or 08 W 4 12D16W05L 05 Lr or S or R 5 12D10E05L 028 09D16W16H 09D10E16H The load multipliers are based on the probability of the load combination occurring as well as the accuracy with which the design load is known is D Dead load L Live load Lr Roof Live load W V nd load E Earthquake load S Snow load R Rain load F Flood load T Temperature or self strain load H Hydrostatic pressure load Design of a member or of a segment of a member must be based on the load case that produces the largest force stressdisplacement value M 1193 Design Live Load 5 up 5 by Woufmlalc nz AASHTO LRFD Loading I Force Envelope Forces in a particular structural component are caused by 1 loads acting on the structure and 2 load location Force envelope is a plot of the maximum and mini mum force responses along the length ofa member due to any proper placement of loading for any specified design load combination Statics of Structural Supports TYPES OF FORCES External Forces 5 actions of other bodies on the structure under consideration Internal Forces 5 forces and couples exerted on a member or portion of the structure by the rest of the structure Inter nal forces always occur in equal but opposite pairs Suppo s Different types of structural supports are shown in Table 1 Some physical details for the idealized support conditions of Table 1 are shown in Figs 1 5 NOTE Structural roller supports 1 are assumed to be capable of resisting normal displacement in either direction Table 1 Idealized Structural Supports Movements Allowed Reaction Unknowns Type Sketch Symbol or Prevented Forces Created a Pin Prevented horizontal A single linear force of translation vertical unknown direction or translation equivalently l Allowed rotation A horizontal force and a 0 1quot OR vertical force which are R the components of the single force of unknown direction 17 Hinge Prevented relative Equal and oppositely displacement of directed horizontal and Rx member ends vertical forces Allowed both rotation R and horizontal and quot vertical displacement c Roller Prevented vertical A single linear force translation either upward or Allowed horizontal WWW translation rotation d Rocker i V OR e Elastomeric pad ykvlv f Fixed end Prevented horizontal Horizontal and vertical translation vertical components of a linear MR translation rotation resultant moment Allowed none g Link Prevented translation A single linear force in in the direction of link the direction of the link Allowed translation perpendicular to link rotation 11 Guide Prevented vertical A single vertical linear R translation rotation Allowed horizontal translation force moment MR 39 ties steel beam reinforced concrete wall Figure 1 Example Fixed Steel Beam Support stiffener plates each side plam Figure 2 Example Fixed Steel Column Support 5 39 m gwa I V 391 J 39 3a fa quot Jil393 quotHIquot 6 39 39 InI 39 I a a w a x a iquot P 396 r a39 i39 quotHfaha ng Ea lemlurced Dy mmurete is wa ljl unly beam reinl39urcemenl sham a Figure 3 Example Fixed Concrete Beam Support re in nfurci mg bars w 739 culumn 9153 Fquot H1 Figure 4 Example Simply Supported Concrete Column Support 7 i connecticm 39 angles beam 1 u b Coped Beam beam 1 beam 1 Floor Beam beam 1 to Girder beam 2 Conditions 8 Equations of Static Equilibrium A structure is considered to be in equilibrium if initially at rest it remains at rest when subjected to a system of forces and couples If a structure is in equili brium then all its members and parts are also in equilibrium For a plane structure lying in the xy plane and subjected to a coplanar system of forces and couples the necessary and sufficient conditions for equili brium are ZFX0 ZFyzo ZMZO These three equations are referred to as the static equations of equilibrium of plane structures Equations of condition involve known equilibrium results due to construction P1 Internal hinge P1 zero 1 0 P2 moment c at hinge a BCO One equation of condition ZMQB 0 or 2MB 13 Example Calculate the Support Reactions Vc gtlls 12 Example Calculate the 3 79 HL Support Reactions RUL 5091 kips W l l ln ld MA G H in3e i 13 Influence of Reactions on Stability and Determinacy of Structures Internally Stable rigid a structure maintains its shape and remains a rigid body when detached from the supports Internally Unstable a structure cannot maintain its shape and may undergo large displace ments under small disturbances when not supported externally 15 Examples of Internally Stable Structures I f B I z c A C b B C C c Examples of Internally Unstable Structures Statically Determinate Externallya If the structure is internally stable and if all its support reactions can be determined by solving equations of equilibrium Statically Indeterminate Externallya If the structure is stable and the number of support reactions exceeds the number of available equilibrium equations External Redundants a number of reactions in excess of those necessary for equilibrium referred to as the degree of external indeterminacy Summary Single Rigid Structure R lt 3 Structure is statically unstable externally R 3 Structure may be statically determinate externally R gt 3 Structure is statically indeterminate externally but may not be stable R 2 number of support reactions Summary Several Intercon nected Rigid Structures R lt 3C Structure is statically unstable externally R 3C Structure may be sta tically determinate externally R gt 3C Structure is statically indeterminate externally but may not be stable C 2 number equations of conditions Ie R 3 C 2 degree of external indeterminacy 6 Examples of Externally Statically Determinate Plane Structures 2 d l l i k OR A 111 Examples of Statically 21 Indeterminate Plane Structures I I39ZZ IZZE 19 Reaction Arrangements Causing External Geometric Instability in Plane Structures 22 a y I C gig 345 Hinge v b C Hiege wage A amp 1 a d Hin e Hinge Hinge e Hinge Hinge g Example Plane Structures with 23 Equations of Condition INTERIOR HINGES IN CONSTRUCTION Interior hinges pins are often used to join flexural members at points other than support points eg connect two halves of an arch structure and in cantilever bridge construction Such structures are more easily manufactured transported and erected Furthermore interior hinges properly placed can result in reduced bending moments in flexural systems and such connections may result in a statically determinate structure 24 Arch Structures Arch structures are usually formed to support gravity loads which tend to flatten the arch shape and thrust the supported ends out ward Hinge or fixedend supports are generally used to provide the necessary horizontal displace ment restraint The horizontal thrust forces at the supports acting with the vertical loading tend to develop counteracting moments that result in low bending stresses 25 Arch Structure with Interior Hinge 26 Cantilever Construction Cantilever construction repre sents a design concept that can be used for long span structures If spans are properly propor tioned cantilever construction can result in smaller values of the bending moments deflections and stresses as compared with simple support construction 27 Hinge u Cantilever construction In Cantilever truss construction Examples of Cantilever Construction 28 The following figures show a typical highway overpass structure designed as a series of simple spans a a statically indeterminate continuous beam b and a canti levered construction beam c along with their respective bending moment diagrams for a uniform load of 2 kipsft Note that the bending moments are most evenly divided into positive and negative regions for the threespan contin uous beam and that the location of the internal hinges for the canti levered constructed bridge resulted in a more even moment distribution as compared to the overpass analyzed as three simple spans 29 I Simple beams 3 I i h I I Continuous beam M kip ft Continuous Spans 30 1 l quot 39 xw no u Camilevcr x Construcliun F BO ft 0 so n 8o r l IOOh 2500 7500 7500 c Cantilever Construction 31 Movement of the two internal hinges towards the interior sup ports results in a reduction of the negative moment magnitudes at the supports and an increase in the midspan positive bending moment ldeal placement occurs when the each interior hinge is approximately 109 ft from an end support this location of the inter nal hinges results in a maximum negative and positive bending moments of 5000 ftkips 32 Cables Use to support bridge and roof structures guys for derricks radio and transmission towers etc Assumed to only resist loads that cause tension in the cables Shape of cables in resisting loads is called funicular Resultant cable force is where H horizontal cable force component and V vertical cable force component 33 Principle of Superposition E on a linear elastic structure the com bined effect of several loads acting simultaneously is equal to the alge braic sum of the effects of each load acting individually Principle is valid for structures that satisfy the following two conditions 1 the deformation of the structure must be so small that the equations of equilibrium can be based on the undeformed geometry of the struc ture and 2 the structure must be composed of linearly elastic material 34 Structures that satisfy these two conditions are referred to as linear elastic structures P P2 P3 P I P2 F l i 1 z i 4 l m 792 792 997 7 337 tRL 39RR RL RR tRL RR RL 1R3 RL RLI RLz RLJ RR RR RR RR Pl W2 Pl 39 W3 A B E 7 A B r A I I I I 13 39 a 5 a it 5A 58 5A 339 5 6quot 6A 58 4 58 5B

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