Introduction to statically indeterminate structures

Similar documents
Statically indeterminate examples - axial loaded members, rod in torsion, members in bending

ELASTICITY AND PLASTICITY I

Identification of Axial Forces on Statically Indeterminate Pin-Jointed Trusses by a Nondestructive Mechanical Test

Math 124B January 24, 2012

BEAM DIAGRAMS AND FORMULAS. Nomenclature

MAGIC058 & MATH64062: Partial Differential Equations 1

1. The vibrating string problem revisited.

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 1. Energy Methods in Structural Analysis

Development of the Sinc Method for Nonlinear Integro-Differential Eequations

PhysicsAndMathsTutor.com

Coordinate Geometry. Coordinate Geometry. Curriculum Ready ACMNA: 178, 214, 294.

V. DEMENKO MECHANICS OF MATERIALS LECTURE 6 Plane Bending Deformation. Diagrams of Internal Forces (Continued)

Stability condition of finite difference solution for viscoelastic wave equations

Mutipy by r sin RT P to get sin R r r R + T sin (sin T )+ P P = (7) ffi So we hve P P ffi = m (8) choose m re so tht P is sinusoi. If we put this in b

Combined RKDG and LDG Method for the Simulation of the Bipolar Charge Transport in Solid Dielectrics

1 Bending of a beam with a rectangular section

arxiv: v1 [math.co] 5 Jun 2015

Statistical Physics. Solutions Sheet 5.

Where does Oxygen Extinction Occur in a Soil Profile?

L v. removal. elastic. the body is. Hooke s force. [ M L 1 T 2 ] (1) (2) (3) Normal Stress Tensile Stress. stress. parallel. Shearing.

Fluid Flow through a Tube

Prof. Anchordoqui. Problems set # 4 Physics 169 March 3, 2015

Suggested Solution to Assignment 5

Physics 2D Lecture Slides Lecture 30: March 9th 2005

We will see what is meant by standard form very shortly

In this appendix, we evaluate the derivative of Eq. 9 in the main text, i.e., we need to calculate

Bending Analysis of Castellated Beams. By Sahar Elaiwi Boksun Kim Long-Yuan Li

GEOMETRIC PROBABILITY MODELS TO ANALYZE STRATEGIES FOR FINDING A BURIED CABLE

CE 160 Lab 2 Notes: Shear and Moment Diagrams for Beams

Shear and torsion interaction of hollow core slabs

Figure XX.1.1 Plane truss structure

A Slipping and Buried Strike-Slip Fault in a Multi-Layered Elastic Model

Complete Description of the Thelen2003Muscle Model

Linear Systems COS 323

P R A S I C ' 02. Simpozionul na.ional cu participare interna.ional/ PRoiectarea ASIstat/ de Calculator

Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1

KINEMATICS OF RIGID BODIES

Forces from Strings Under Tension A string under tension medites force: the mgnitude of the force from section of string is the tension T nd the direc

Dynamics: Newton s Laws of Motion

arxiv: v1 [cond-mat.supr-con] 27 Dec 2018

Section 10.2 Angles and Triangles

Plates on elastic foundation

Chapter 5 Bending Moments and Shear Force Diagrams for Beams

COLLEGE OF ENGINEERING AND TECHNOLOGY

(6.5) Length and area in polar coordinates

CBE 291b - Computation And Optimization For Engineers

Solution Manual. for. Fracture Mechanics. C.T. Sun and Z.-H. Jin

VI. The Heat Equation. A. The Diffusion Equation Conservation in Action

Lecture 6 Notes, Electromagnetic Theory I Dr. Christopher S. Baird University of Massachusetts Lowell

RFID Technologies HF Part I

99/105 Comparison of OrcaFlex with standard theoretical results

BME 207 Introduction to Biomechanics Spring 2018

In-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill

SIMULATION OF WOOD DRYING STRESSES USING CVFEM

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

Chapter (a) 2. At equilibrium, V = 0. x = (c) V( x) ke bkxe = 0. At equilibrium. b bx bx 2 bx. V = bke bkxe + b kxe = + = < 0.

Exam CT3109 STRUCTURAL MECHANICS april 2011, 09:00 12:00 hours

Review of Calculus, cont d

Part I: Basic Concepts of Thermodynamics

INTRODUCTION. The three general approaches to the solution of kinetics problems are:

Explain shortly the meaning of the following eight words in relation to shells structures.

Problems for HW X. C. Gwinn. November 30, 2009

Dynamic Analysis of the Turnout Diverging Track for HSR with Variable Curvature Sections

Comparison of the Design of Flexural Reinforced Concrete Elements According to Albanian Normative

Calculus 2: Integration. Differentiation. Integration

Lecture 8. Newton s Laws. Applications of the Newton s Laws Problem-Solving Tactics. Physics 105; Fall Inertial Frames: T = mg

13: Diffusion in 2 Energy Groups

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

Spherical Coordinates

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

13.4 Work done by Constant Forces

The Wave Equation I. MA 436 Kurt Bryan

Dynamic analysis of an underwater suspension system in ocean currents with method of averaging

Case (a): Ans Ans. Case (b): ; s 1 = 65(4) Ans. s 1 = pr t. = 1.04 ksi. Ans. s 2 = pr 2t ; s 2 = 65(4) = 520 psi

Homework Assignment #5 Solutions

20 MATHEMATICS POLYNOMIALS

Jack Simons, Henry Eyring Scientist and Professor Chemistry Department University of Utah

The Predom module. Predom calculates and plots isothermal 1-, 2- and 3-metal predominance area diagrams. Predom accesses only compound databases.

DETERMINATION OF MECHANICAL PROPERTIES OF NANOSTRUCTURES WITH COMPLEX CRYSTAL LATTICE USING MOMENT INTERACTION AT MICROSCALE

How do you know you have SLE?

PDE Notes. Paul Carnig. January ODE s vs PDE s 1

Summary: Method of Separation of Variables

Energy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Fourier Series in Complex notation. cos(x) = eix + e ix 2. A n cos + B n sin l. i 2 B n. e inx=l + A n + ib n 2 8. ( 0 m 6= n. C n = 1 2l.

Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram. Drum Brakes. Among the various types of devices to be studied, based on their practical use,

A-Level Mathematics Transition Task (compulsory for all maths students and all further maths student)

Math 8 Winter 2015 Applications of Integration

Kirchhoff and Mindlin Plates

Calculus - Activity 1 Rate of change of a function at a point.

a a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants.

Heat flux and total heat

Math 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008

The Regulated and Riemann Integrals

Energy Balance of Solar Collector

Can the Phase I problem be unfeasible or unbounded? -No

HW Solutions # MIT - Prof. Kowalski. Friction, circular dynamics, and Work-Kinetic Energy.

#6A&B Magnetic Field Mapping

Transcription:

Sttics of Buiding Structures I., EASUS Introduction to stticy indeterminte structures Deprtment of Structur echnics Fcuty of Civi Engineering, VŠB-Technic University of Ostrv

Outine of Lecture Stticy indeterminte structures, degree of indetermincy Force method Simpe stticy indeterminte bem Simpe stticy indeterminte bem xi oding Unitery fixed bem trnsvers oding Bitery fixed bem trnsvers oding Simpy supported bem s n eement of stticy indeterminte structure Simpe stticy indeterminte bem torsion oding Ccution of deformtion of the simpe stticy indeterminte bem Outine of Lecture / 58

Possibiity of free motion of physic objects Degree of freedom n v : biity to mke one right-nged component of dispcement or rottion. free mss point in the pne: n v =, coordintes [x, y], different positions free mss point in the spce: n v =3, coordintes [x, y, z], 3 different positions m[x m,z m ] x +x free soid br in the pne: n v =3, coordintes [x, y, ], 3 different positions soid body in the spce: n v =6, coordintes [x, y, z, ], 6 different positions +z z Sttics of structures bsic knowedge 3 / 58

Extern constrints to movement Dispcement constrint prevents movement of constrined point in the given direction ( (b (c (d (e (f (g ( (b (c (d (e Sttics of structures bsic knowedge 4 / 58

Extern constrints to rottion ottion constrint prevents rottion of constrined br in the given direction ( (b (c Simpe rottion constrints Fu fix (cmp in spce or pne, prti fix in pne ( (b (c Compex dispcement nd rottion constrints Sttics of structures bsic knowedge 5 / 58

utipicity of constrints Extern constrints eimintes degrees of freedom of the object. n mutipe constrint eimintes n degrees of freedom (n=,, 3 of object Exmpes of simpe constrints nd their rections of soid br in the pne Nme Pin, swing rod utipicity of constrint Symbo of constrint (support, rections z oer z Hinge z x Siding cmp Fixed (cmped 3 z y z x y Sttics of structures bsic knowedge 6 / 58

Ensuring immobiity of the br To sufficienty support some object it is needed v constrints to eiminte n v degrees of freedom. v = n v v < n v v > n v Supporting of the object is kinemticy determinte nd stticy determinte, immobiity of buiding is ensured, structure cn be used s buiding construction. Supporting of the object is kinemticy indeterminte nd stticy overdeterminte, immobiity of buiding is not ensured, structure is indmissibe s the buiding construction. (insufficient number of constrined degrees of freedom Supporting of the object is kinemticy overdeterminte nd stticy indeterminte, immobiity of buiding is ensured, structure cn be used s buiding construction.. Constrints must be propery rrnged to ensure immobiity of structures it must not be Exception cse of kinemticy determinte or overdeterminte structure. Sttics of structures bsic knowedge 7 / 58

Kinemticy nd stticy determinte structure v = n v v = 3, n v = 3 Supporting of the structure is kinemticy nd stticy determinte P P x b z bz x y P P z Sttics of structures bsic knowedge 8 / 58

Stticy nd kinemticy determinte brs ( (e (i n v = 6 n v = n v = 3 Axi oding (b n v = 3 (f n v = (j n v = 3 Torsion oding (c n v = 3 (g n v = 3 (k n v = 3 (d n v = (h n v = 3 ( n v = 6 Trnsvers oding Sttics of structures bsic knowedge 9 / 58

Stticy indeterminte structure v > n v Supporting of the structure is kinemticy overdeterminte nd stticy indeterminte x P P b bx v = 4 n v = 3 z bz x y P P b by bx v = 6 n v = 3 z bz Sttics of structures bsic knowedge / 58

Kinemticy indeterminte structure v < n v Supporting of the structure is kinemticy indeterminte nd stticy overdeterminte P P b z bz Equiibrium of the object is ensured ony for specific oding In buiding prctice unusbe structure Sttics of structures bsic knowedge / 58

Exception cses of supporting Constrints must be propery rrnged to ensure immobiity of structures it must not be exception cse of supporting tht is unusbe in construction prctice. x P P b bx z P P b c z bz cz Sttics of structures bsic knowedge / 58

Exception cses of supporting (c bem is not secured ginst rottion (d 3 constrints ginst movement intersect t one point (e 3 constrints ginst movement ying in one direct ine ( (d (b (c (e Exception cses of supporting Sttics of structures bsic knowedge 3 / 58

Equiibrium conditions of oded br Supported br hs to be motioness nd in equiibrium. Number of equiibrium conditions depends on the type of soved tsk. It is equ to degree of freedom of unsupported br n v. Number of rections is equ to number of constrined degrees of freedom v. v = n v Number of unknown rections is equ to number of equiibrium conditions, the br is stticy determinte v < n v Number of unknown rections is smer thn number of equiibrium conditions, the br is kinemticy indeterminte nd cnnot be used s buiding construction. v > n v Number of unknown rections is bigger thn number of equiibrium conditions, the br is stticy indeterminte, it cn be buiding construction. There must be suppemented deformtion conditions to sttic equiibrium conditions. If the determinnt of system of equtions is equ to zero, then it is exception cse of supporting of structure. Sttics of structures bsic knowedge 4 / 58

Equiibrium conditions of oded br system For every seprte br cn be written 3 equiibrium conditions Number of intern nd extern constrints: v = v e + v i Number of rections is equ to number of constrined degrees of freedom of structure. Sttics of structures bsic knowedge 5 / 58

Kinemtic nd sttic determincy of truss Prctic pproch computtion mode is mde from mss-points (t joints nd intern inks (brs, tht constrin mutu dispcement of both connected joints. Condition of kinemtic determincy:.s p v e br (intern ink joint (mss-point extern constrint (support Truss s system of mss-points nd extern nd intern inks Sttics of structures bsic knowedge 6 / 58

Kinemtic nd sttic determincy F F N 4 N 8 e f g F 3 N N 5 N 9 N 3 N 7 N x N N 6 N c d b z. s p. 4 bz s=7 number of joints ( equiibrium conditions in ech of them p= number of brs ( unknown xi force in ech of them = = number of constrints with nd unknown rections Sttics of structures bsic knowedge 7 / 58

Kinemtic nd sttic determincy F F c N 5 d s=4 N N 3 N 4 p=5 x N b = = z bz. s 8 p. 8 Stticy nd kinemticy determinte truss.s p. Kinemticy indeterminte truss Sttics of structures bsic knowedge 8 / 58

Kinemtic nd sttic determincy F F This is not joint (brs re overpping here c N N 5 N 3 N 6 d N 4 s=4 p=6 = x N b bx = z bz. s 8 p. x stticy indeterminte truss Sttics of structures bsic knowedge 9 / 58

Ccution of the degree of sttic indetermincy Pne frme structures nd bems. Open br systems: n s = v 3 p k = +. + 3. 3 3 p k v number of extern constrints (rections i number of extern constrints with, or 3 rections p k number of intern hinges reccuted to simpe hinges. Cosed br systems: n s = 3.u + v 3 p k u number of cosed res / 58

Force method Force method: is used for soution of stticy indeterminte structures, n s, is bsic method for soution of stticy indetermince br structures is direct method Force method utiizes equiibrium conditions deformtion conditions princip of superposition nd princip of proportionity Force method / 58

Force method outine of soution of structures by the Force method:. Determintion of degree of sttic indetermincy n s. Creting of Bsic stticy determinte structure by removing of n s constrints (extern or intern 3. emoved constrints re repced by stticy indeterminte forces or moments (stticy indeterminte rections or interctions 4. Formuting of the system of n s deformtion conditions in the form of iner equtions 5. Soving of the system of iner equtions. The soution re vues of stticy indeterminte rections or interctions. 6. Ccution of remining rections nd intern forces Force method / 58

Simpe stticy indeterminte bem Assumptions: direct bem of constnt or vribe cross-section b xis of the bem is identic with x-xis c bem is supported in points, d every of extern constrints is pre to one of coordinte xis, e every of extern rottion constrints works in the pne tht is perpendicur to some of the coordinte xis f bem prut cn be sptiy oded Simpe stticy indeterminte bem 3 / 58

Simpe stticy indeterminte bem The degree of sttic indetermincy n s = v n v is number of redundnt constrints, i.e. number of constrints tht hs to be removed to mke the bem stticy determinte Spti probem of simpe direct bem Simpe stticy indeterminte bem 4 / 58

Simpe stticy indeterminte bem Every probem of simpe stticy indeterminte bem in spce (3D cn be divided into 4 simpifier probems:. Axi probem n v =. Trnsvers probem in xz pne n v = 3. Trnsvers probem in xy pne n v = 4. Torsion probem n v = Simpe stticy indeterminte bem 5 / 58

Simpe stticy indeterminte bem xi oding n s v bx n v bx Siové ztížení N E N A dx Siové ztížení E A dx N E N A dx N E N A dx E N A dx Siové ztížení Otepení t N dx t t t Force method in xi probem Simpe stticy indeterminte bem xi oding 6 / 58

Simpe stticy indeterminte bem xi oding Dispcement of supports Dispcement in x-xis direction u, Deformtion condition: d ub In this cse ( pre to x-xis: ū nd hve oposite direction ( ( u ( u ū b nd hve sme direction d u b x u b u x u u b x bx u Force method in xi probem 7 / 58

Exmpe 3. Force oding EA 9,68 5 kn Deformtion condition bx bx x 4,8kN( ( (4,8,9kN( N E N E x N dx A 4,, 5 9,68 N dx A x E 9,68 A 5,7 3 4,8 9,68 5,9kN( 3 ( 3 N 5 9,68,9 dx 5,7 5,9kN( Simpe stticy indeterminte bem xi oding Probem definition nd soution of the exmpe 3. 8 / 58

Exmpe 3. Therm oding The bem is oded by chnge of temperture ong ength of bem. t =5 o C. The temperture is constnt EA 9,68 x kn, 5,4 x 74,58 kn( 4, Deformtion condition: bx x x N t 3 EA 9,68 5 t dx α 5 t t 9,68 3 bx t -5,, 5 Δt 5 5 5 3 5,4 74,58 kn( ( 74,58 74,58 kn( 74,58 kn( C 4 Simpe stticy indeterminte bem xi oding Probem definition nd soution of the exmpe 3. 9 / 58

d Exmpe 3., supports shifting Ū = 5 mm =,5 m (to the right Ū b = 8 mm =,8 m (to the right Dispcement Ū nd rection re in opposite direction Dispcement Ū b nd unit force = re in sme direction Deformtion condition: 3 bx x x x u EA b bx ( 9,68,8 bx u b 968, kn( u 968, kn( 5 ( x,8 bx u 3 (,5 968, kn( d u 9,68 u 5,5 968, kn( Probem definition nd soution of the exmpe 3. 3 / 58

Trnsversy oded stticy indeterminte bems Degree of sttic indetermin cy : Deformtio n condition - force oding nd therm oding Deformtio n condition - oding by shifting of supports : n s v n v 3 : d One-end fixed stticy indeterminte bem Simpe stticy indeterminte bem trnsvers oding 3 / 58

Bitery fixed bem trnsvers oding ns v n v 4 Deformtion conditions for force oding nd therm oding: b b Deformtion conditions for oding by ifting of supports: d d Force method - bitery fixed bem with trnsvers oding 3 / 58

Exmpe 3.8 The bem of stee cross-section I is oded by: forces ccording to pict. ( iner chnge of temperture t =5 o K pict. (e supports shifting pict. (h Deformtion conditions for force nd therm oding: Deformtion conditions for oding by supports shifting: d d Probem definition nd soution of the exmpe 3.8 33 / 58

Exmpe 3.8 soution, force oding Force oding Bsic stticy determinte structure: simpy supported bem Stticy indeterminte vues: = b =- b oding stte: ections z, bz : Bending moment o : z bz for ( x ( x,6 (6 (,6 x 3, z z 4 x x 4,8 3, 4 q q 3, 4 x x,6 for F 3,,6 / 4,8 6 ( x,6 x x,8 / 4,8 F,6 q 8kN ( x 7,6kN x F Cn be so written: where F q F ( x z x x F F q q q q q x ( x for x F for x x F x 34 / 4

Exmpe 3.8 soution, force oding st oding x stte nd oding x stte Ccutio n of deformtio n coefficien ts : b, b b b b b E E I I E E dx dx I I 4,8,6 3EI 3EI EI,6 for constnt cross - section 3EI EI 4,8,8 dx 6EI 6EI EI,8 dx 6EI EI,, bsic deformtio n nges of simpy supported bem b b 35 / 4

Exmpe 3.8 soution, force oding Ccution of deformtion coefficients, EI EI dx EI dx EI ( ( Integrtio n cn be reized : nytic y, with use of Vereshchgin' s rue, 3 with use of Tbe. In this cse it is : q q 6,3 EI 58,444 EI F F q q dx dx 36 / 4

Exmpe 3.8 soution, force oding Soution of iner equtions is: b b ections : 6,667 knm 3,3444 knm b b 6,66 3,344 8 8,563 kn 4,8 4,8 b 6,66 3,344 7,6 7,37 kn 4,8 4,8 Probem definition nd soution of the exmpe 3.8 37 / 4

Exmpe 3.8 therm oding Liner chnge of The Soution For Digrm of ection z probem is EI of t suppied z z symmetric, two iner equtions t bending h z temperture cross the cross - section EI bz dx ( 6,6, 5,8 EI EI 4494 knm therefore moment - see pict. (g., becuse z 5 cn be reduced into 5, 6 4,8,4 konst. nd V ( 5 x dx 4,8 EI 6 4,8 one t : 4,446 knm, or so 5 5 o C Probem definition nd soution of the exmpe 3.8 38 / 4

Exmpe 3.8 oding by supports shifting Deformtio n conditions : For seected bsic stticy determint e system is : d ( ( ( ( After substituti on :,6,8 d 4494 4494,8,6 d 4494 4494 Soution of equtions is : b b b b Soution - see pictures (i nd (j w w b b 6,5 w ( 4,8 4,8,4 6,5 b w b w b ( 4 w -,kn, 4,65 - (, (8,49 4,8 4,8 d d,4 ( wb,3,6 6,5 4,8 4,8 4 wb,3,6 6,5 4,8 4,8 d b 4 8,49kNm 6,44 kn nd b hve sme direction 4 4 Probem definition nd soution of the exmpe 3.8 39 / 4

Tbe 3. Fixed-end moment rections of the bem with constnt crosssection Unitery fixed bem trnsvers oding 4 / 58

Simpy supported bem s n eement of stticy indeterminte structure Decomposition of simpy supported bem into seprte oding sttes Simpy supported bem s n eement of stticy indeterminte structure 4 / 58

Simpy supported bem s n eement of stticy indeterminte structure digrms of intern forces. oding stte (q:. oding stte (P: oment s oding stte: Digrms of bending moments in oding stte nd in moment s oding stte Simpy supported bem s n eement of stticy indeterminte structure 4 / 58

Vribe cross-section of the bem Unitery fixed bem trnsvers oding 43 / 58

Vribe cross-section - exmpe 3.5 Soution 5 From tbe b, mx by Tbe,4 q Ccutio n of,, b, 3.3 :,3,6 x n q, b q b, 4 5 66,85, b 5 q b, 4 5 simiry b, Ccutio n of : mx, b 6,5 x,545 m n q 4 3 3,9,,9 recions :,5 x n 4,5 5 q,63,855, 66,85kNm b,,63kn 66,85 5.545 by iner interpot ion cn be obtined 4 48,37kN,545 6,7kNm,9 Unitery fixed bem trnsvers oding 44 / 58

Coefficients for moment rection (uniformy distributed oding Tbe 3.3 Unitery fixed bem trnsvers oding 45 / 58

Coefficients for moment rection (nod force oding Tbe 3.4 Unitery fixed bem trnsvers oding 46 / 58

Coefficients for moment rection (uniformy distributed oding Tbe 3.6 Unitery fixed bem trnsvers oding 47 / 58

Coefficients for moment rection (nod force oding Tbe 3.7 Unitery fixed bem trnsvers oding 48 / 58

Simpe stticy indeterminte bem torsion oding Degree of indetermin cy : n s ν n v Deformtio n condition for force oding : δ δ b for oding by supports shifting : δ δ d Force method nd torsion oding Simpe stticy indeterminte bem torsion oding 49 / 58

Torsion probem exmpe 3. Probem definition: einforced-concrete bem (G=9,4. 6 kp of constnt rectngur crosssection, width b=,4m, height h=,36m. It is oded by: torsion oding ccording pict. ( b supports shifting, rd, b, rd. Torsion stiffness is ( see tb. : I t b 3 h,96,4 3,36 9,75 4 m 4 G I t 93kNm h/b,,,,3,4,5,46,54,66,77,869,958 h/b,6,7,8,9 3,37,9,74,33,87,633 Simpe stticy indeterminte bem torsion oding Probem definition nd soution of the exmpe 3. 5 / 58

oment Torsion probem exmpe 3., force oding Deformtio n condition : T T T G T G T I I T t T t ( hs oposite direction compred to oding 4 dx G I G I dx G 3 G 4 G I t t I t I t 4 T 7,75kNm 4 7,75 6,5kNm,5 3 esuting digrm of torsion moments T - see pict. (d t dx,5 (4 9 G I T b t moments 3 G I t Probem definition nd soution of the exmpe 3. Simpe stticy indeterminte bem torsion oding 5 / 58

Torsion probem exmpe 3., supports shifting Supports shifting:, rd, b, rd. Deformtion condition: d T 4 ( T, rd T GI T b b t d 4 4,438 93 d T ( (,, rd, (, T b 4 4,438 rd/knm 4 6,76 knm,3 4,438 4 6,76 knm Probem definition nd soution of the exmpe 3. Simpe stticy indeterminte bem torsion oding 5 / 58

Ccution of deformtion of simpe stticy indeterminte bem Force oding stte eduction theorem: When ccution deformtion of stticy indeterminte structure of degree of indetermincy n s thn the unite force oding stte cn be mde on: origin stticy indeterminte bem stticy indeterminte bem with ower degree of indetermincy n sj < n s, (removed ess thn n s inks, stticy determinte bem (removed n s inks. eduction theorem cn be used to ccute the deformtion of ny stticy indeterminte structure exposed to force oding. Ccution of deformtion of the simpe stticy indeterminte bem 53 / 58

Ccution of deformtion force oding stte Exmpes of virtu od cse options to ccute the dispcement of stticy indeterminte bem. Ccution of deformtion of the simpe stticy indeterminte bem 54 / 58

Ccution of deformtion force oding stte, exmpe 3. Ccute dispceme nt of the bem in point s.. Bending moments were ccuted using tbe 3., Virtu unit force stte choosed s on picture (c. w s 3 5 P dx dx E I E I 96 E I The formu bove cn be soved : nytic y b using Vereshchgin' s rue c using Tbe. Exmpe 3. nd its soution Ccution of deformtion of the simpe stticy indeterminte bem 55 / 58

Ccution of deformtion force oding stte, exmpe 3. Ccution using Vereshchgin s rue w s E I dx E I dx w w w w s s s s E I E E E I I I ( A ( ( ( P 8 9 P 8 3 3 8 4 8 T 4 P A 7P 54 3 7P 3 8 3 T 5P 96 3 EI 3 Probem definition nd soution of the exmpe 3. Ccution of deformtion of the simpe stticy indeterminte bem 56 / 58

Ccution of deformtion therm oding esuting deformtion is given by superposition of: estic deformtion w pr b therm deformtion w t on bsic stticy determinte structure w w pr w t Ccution of deformtion of the simpe stticy indeterminte bem 57 / 58

Ccution of deformtion therm oding, exmpe 3.3 The bem is oded by chnge of temperture. Ccute dispcement in point c. Soution : EI 4494 knm w c w pr w t w pr EI,3 (,8 4.56,3,35, / 4494 w pr,395 3 m w t t h t dx,, 5 8 5,3 3,36 3 m w c,395 3 3,36 3,93 3 m Exmpe 3.3 nd its soution Ccution of deformtion of the simpe stticy indeterminte bem 58 / 58