III. Basis of Mechanics

Transcription

1 III. Basis of Mechanics 3.1 General Consideration 3. Newton s aw of Motion and Statics 3.3 Unit and Dimension 3.4 Material 3.5 Solid Mechanics of Slender Members 3.6 Uniaial oading 3.7 Torsion 3.8 Beam Theor 3.9 Buckling 3.1 Special phenomenon

2 3.1 General Consideration

3 Three major factors Solid Basis of mechanics Force Material Displacement Material and Continuum, Deformation of material igid-bod motion Deformation <Grain> <Grain and boundar> <Continuum> igid-bod : No change in distance between particles occurs under eternal force. Elastic deformation: Deformation due to eternal force disappearing when it removes. lastic deformation: Deformation due to eternal force remaining when it removes. Fluid including gas

4 Some details of the 3 major factors Force Force Internal Eternal ction and reaction force Traction Eerted load, eaction force Bod force Gravit, Magnetic force Displacement or motion igid-bod motion Deformation Material: Set of particles continuousl distributed Continuum Solid igid-bod: No relationship between force and displacement is needed. Elasticall deformable bod: inear elastic(hooke s law) and Non-elastic, Isotropic(Common material) and nisotropic (Composite material) lastic bod: Yield criterion, Isotropic hardening, Kinematic hardening Fluid including gas

5 Classification of solid mechanics (SM) Displacement igid-bod Statics General SM igid-bod dnamics Elasticit Deformation Narrow SM Fleible dnamics lasticit Fracture mechanics Vibration

6 3. Newton s aw of Motion and Statics

7 Newton s law of motion nd law 3 rd law Newton s law Sum of all the forces eerting on a particle ( f ) is equal to the acceleration ( a ) multiplied b the mass ( m ), i.e., f = ma. ction and reaction law: Two internal forces eerting between two particles have the same magnitude and line of action and the opposite direction. i ij F j i, j 1,,, ij ( F i ij ) F i ij i j i i j ij ij i j (O) 3 (X) 5 material is a set of infinite number of particles i F : Sum of eternal force eerting on particle i ij : Internal force eering on particle j from particle i ll forces are bounding vectors. F r r r o r F F. F F 5 F F equirement on equilibrium i j r i F i ij i ij ri F ri F ri i i j i i j i r F i i ij F i or F i i r F i The above requirement of equilibrium should satisf for all subsstem as well as the whole sstem. eading to differential equations, for eample, equations of equilibrium. or M

8 ctual number of particles is infinite. Derivation of requirement on equilibrium F r r F F using a si-particle bod F F F F = F = F = F = F = F = i ij ji F = = r F + r + r + r + r + r = r F + r + r + r + r + r = r F + r + r + r + r + r = r F + r + r + r + r + r = r F + r + r + r + r + r = r F + r + r + r + r + r = i ij r F = r = -r i i j ij

9 Subsstem The requirement of equilibrium should satisf for an subsstem. This statement is the same with the following statement: The requirement of equilibrium should satisf for arbitrar infinitesimal area in D or volume in 3D, leading to differential equation. Note that we should define stress, i.e., force per unit area to define the force eerting on boundar of the infinitesimal area in D or volume in 3D. 3 F 3 F F 4 F 4 F F 1 F 1 F

10 equirement on equilibrium in D and 3D D plane F F F 3D space F F F Fz Tip,, and z can be replaced b an three independent directions M Mz, M M, M, M z, 3 equations 6 equations Eamples D F B E C D f D f DF f BD ' ' 45 W 45 B

11 Caution in appling requirements on equilibrium r and r B i i are dependent on the position of point i, while F r B r i r1 r i B B r1 is fied. i F i B r i B F 1 F n B B r r r i F i M ri Fi i ( r r ) F B M r F B i i i i i i M F, M M B B M, M B F i B r F r F i i i i B M r F i B B (If r F)?

12 Statics Statics and solid mechanics with three factors of mechanics Factor Statics Solid mechanics Subsstem 1 Force equirement on equilibrium Equation of equilibrium or motion Deformation Material igid-bod igid-bod Displacement-strain relation, Geometric compatibilit Elastic bod: Hooke s law lastic bod: lastic flow rule BC Geometric(essential) BC Geometric(essential) BC, mechanical(natural) BC ground Subsstem 1 Subsstem Entire sstem Ground and support Subsstem ground Separation of subsstems Essential separation of subsstems from support or ground Division of subsstems if needed -When number of equations should be greater than that of new knowns -When internal resultant forces is to be seen Newtonian mechanics based on vector quantities naltical mechanics based on energ

13 Free-bod diagram and application of E on it Simpl supported beam a F.B.D. No. 1 F.B.D. No. F.B.D. No. 3 F.B.D. No. 4 b a a a a H ( X ) b b F.B.D. No. 5 F.B.D. No. 6 B B b No. b V ( Y ) V B Y B a equirement on equilibrium b ( X ) V ( Y ) B B H V ( Y ) M ; B a ( b / ) F ; B ( a / ) M B ; b F ; V VB H F ; H V ( b / ) No. 3 M ; VB ( a / ) VB a p V ( Y ) B H ( X ) F ; B ( b / ) No. 1 F ; B ( a / ) M ; B a a b B B B

14 Staticall indeterminate sstem 3 points supporting beam / / F.B.D. B equirement on equilibrium F ; F ; B C M ; ( / ) B C M ; ( / ) B C M ; (3/ ) C B Eqs. 1 ~ 5 have onl two independent equations. B C Infinite number of equations can be made appling moment balance requirement at arbitrar point. Number of unknowns > number of equations Staticall indeterminate sstem eactions in a staticall indeterminate sstem can be determined not b statics but b solid mechanics.

15 Division of total sstem into subsstems

16 Division of total sstem into subsstems M

17 Division of total sstem into subsstems

18 Forces in members of a truss structure t D C E B D f CD F ; fbd F ; FCD f, f BD CD ns. f ( in tension) f C BC f BD ( in compression), t E f CE f DE t C f C f BC F ; fce fde F ; fde f, f DE CE F ; f C fbc F ; fbc f, f BC C

19 D static problem Tension of cable, T? F.B.D. T 3 3kg B ppling E on FBD F ; T sin3 B F ; cos3 3 M B B ; 8cos3 4 cos3 3

20 3D static problem Cable tension T?. F.B.D. equirement on equilibrium z C D E r 3 i 6 k, r 5.5 i, r 3 i 4.5 k C CD CB Vector epression of line segments C and CE D kn z B T E B B z F ; B T T 46. T z Bz 46. M ; r ( i j k) C r C z CD ( j) r ( B i B k) CB z CE ( i.5 j) ( 3 i 6 k) i.5 j 6 k CE ce ( i.5 j 6 k) / 46. CE

21 3.3 Unit and Dimension

22 Unit and dimension Sstem of Units Three basis units characterize the sstem of units. Conventional unit: Basis units are length, weight, time, current, temperature, etc. Science unit: Basis units are length, mass, time, current, temperature, etc. Dimension [ength] [], [Mass] [M], [Time] [T], [Weight] [F] Basis units of sstem of science units: [], [M], [T] Basis units of sstem of conventional units : [], [F], [T]

23 British unit and SI unit British unit Basis unit: ft, lb, sec British unit belongs to conventional unit, i.e, lb means basicall lb f (pound force). Unit for mass : slug = lb sec /ft SI unit (The International Sstem of Units) Basis unit: m, kg, s In SI unit, kg means basicall kg m (kilogram mass). Unit for force: N = kg m/s Entangled or mied use of conventional and science units Man countries belonging to the Commonwealth of Nations are using British unit. The other countries are using SI unit. However, most people are using the SI unit like British unit, i.e, the are using kg (i.e., kg force) instead of N (Newton). Sometimes mass is epressed in kg s /m, where kg means kg f.

24 Basis unit, complement unit, assembled unit Basis unit ength : m, ft Current : Complement unit rad Mass : kg, slug Temperature : K Time : s, sec Voltage : V ssembled unit Velocit : m/s, ft/sec 1N = 1kg m/s, lb, kg ressure : 1a = 1N/m, psi = lb/in, psf = lb/ft, kg/m, kg/mm Work : 1J = 1N m, lb ft, kg m Moment : N m, lb ft, kg m ower : 1W = 1J/s, lb ft/sec, kg m/s, S, hp Factors for unit T(1 1 ), G(1 9 ), M(1 6 ), k(1 3 ), c(1 - ), m(1-3 ), (1-6 ), n(1-9 ), p(1-1 ) elationship between the two sstems of unit 1ft =.348 m, 1 in = 5.4 mm, 1lb =.4536 kg ll units should be epressed in oman characters while variables in Italic or bold oman characters.

25 Eample 1 m n k, k 1kg/mm, m 1kg m k n 4 1kg f /mm 1kgm g 1kgm 1 mm 1kgm 1kgm mm 1kgm mm s 1 1 s n 1 k 5 (Hz) m Mistake due to gravitational acceleration. Calculation of blow energ per stroke in hammer forging E = (m g + p )H E = (565kg 9.8+7kg/ cm cm ).75m =(5537kg +137kg).75 m/s 7kg/cm cm E = 58,5kg.m 4,98,5 N-m

26 From a lecture note floating in internet hsical quantities and dimensions SI U.S customar mass [M] kg slug length [] m ft time [T] s sec force [F] N lb F ma W ( weight) N m( kg) kgm/ sec g( m / sec ), g m / sec Original F ma N kg m /s W ( weight) m(kg) g(m /s ), g 9.86m /s Modified

27 3.4 Material

28 Tensile test 1 Engineering stress (Ma) Eperiment (SCM435) nalsis (SCM435) Eperiment (ESW95) nalsis (ESW95) Eperiment (ESW15) nalsis (ESW15) Engineering strain (mm/mm)

29 Tensile test Tensile test f Definition of e and e Current area Before necking occurs Initial area e, e f t ln ln 1 e t Hooke's law in uniaial loading f e1 e t E, t E E, E e: engineering t: true Engineering strain, Engineering stress (Engineering = Conventional=Nominal) oisson s ratio Engineering stress-engineering strain curve redictions of tensile test Y U ma 1(%) : Yield strength : Tensile strength : Elongation lat long t a

30 Generalized Hooke's law for an isotropic material Small deformation rinciple of superposition z z isotropic E v z z E v z z z z z z z E v z z z 1 1 i 1 1 ( ) oisson s ratio E G (1 ) E zz 1 v z T E 1 v z T E 1 z z v T E 1 1 1,, G G G z z z Coefficient of thermal epansion Hooke s law for an isotropic material

31 Thermal load and shrink fit C 779M a -18Ma

32 3.5 Solid Mechanics of Slender Members

33 Mechanics of slender members Truss member or rod Uniaial loading ial force Normal stress: F Stresses in beam t Circular shaft - Torsion Twisting moment Shear stress: Beam ateral force Bending moment and shear force Normal and shear stress, I-beam, H-beam z Tr J z ( ) Mb VQ VQ,, z I bi ti zz zz zz Column Compressive aial force Buckling b Web Flange Column Buckled

34 Internal force Definition of aes Definition of cross-section is of smmetr ositive -face Negative -face Neutral ais z 1 z Force and moment eerting cross-section M F M : twisting moment M, M : bending moment F z : aial force F, F : shear force z M b M b F V F V F F, V F, M M b z z M z 3D F z F M First subscript: Direction of face Second subscript: Direction of force D

35 3.6 Uniaial oading

36 Deformation and stress in uniaial loading Spring) od, truss member,, 1 k k 1 k k : spring constant Force-deformation relationship Hooke s law k eq E Displacement / Stress 1 / E : modulus of elasticit E, E, E E Strain E U 1 k 1 1 k Strain energ U 1k 1 eq E

37 Displacement, strain and stress in rod F ( ) ( ) ( ) u () / / F ( ) ( ) ( ) u () / / / / F ( ) ( ) u () ( ) 3 / / / / / / F ( ) F( d) F ( ) ( ) ( ) / / F ( ) ( ) ( ) / / F ( ) ( ) ( ) du E d du d u () / u () /4 u ()

38 od Saint-Venant s principle End Effect Two staticall equivalent loads have nearl the same influence on the material ecept the region near to the load eerting area. U Down = Non-end Effect End-Effect Down rinciple of smmetr U = (X) (X) (X) (O) /n n 개

39 Internal force, stress vector in tensile test θ o θ i n cosi sin j i o j sin = θ o t t Stress vector (n) T t sin i ( n) sin cos sin nt θ n sin nn sin n cos sin t ( n) θ= θ 9 n θ θ 6 n 3 θ 45 1 o o o o θ n θ θ 3 n 1 4 o n 3 4

40 Internal force and stress components θ o t θ T o n N N = sin T = cos N sin Normal stress nn sin sin Shear stress T nt cos sin cos sin t sin n cos sin t nn n nt t ( n) θ = θ 9 n o t θ θ 6 o n n 3 4 t θ θ 45 o n n t 1 θ θ 3 o n n 1 4 o n nt 3 4 nt 1 nt 3 4

41 Eample B C D E <F. B. D.> l1 B l l 3 4 l 5 l l 6 Ⅲ Ⅱ Force-deformation relationship C l E 3, l 4 C 5 D 6 D C C D ED Geometric compatibilit Ⅰ Force equilibrium B C D F B C D M l l ( l l ) B 1 C 3 D -directional displacement is neglected under small deformation. 1 l l l 3 : C ( 3): D D (1 ) 5 C l eaction and displacement of point E 34 5: 1,,6 B B, C, D l : ( l l l ): D C C 3 4 DCl l C D l l l 3 4 D E E l l (1 ) l C 6

42 Staticall indeterminate sstem thermal load < Method I> 1, 1, E1, 1 1 B T,, E, C <F. B. D.> Ⅲ (- / E T ) B (- / E T ) CB C B CB < Method Ⅱ> Force-deformation relationship C C T 1 ( ) ( 1 ) E1 E due to T T ( ) dueto E 1 1 E 1 C Ⅱ Geometric compatibilit C 1 From Eqs. 1 and ( T) ( ) ET ( 1 ) 1 ET ( 1 ) 1 ( ) 1 1 C 1

43 Deformation of clindrical pressure vessel <F. B. D.> Ⅰ Force equilibrium F s d F s F ; < Method Ⅰ> F T F s FT F rpbsind T < Method Ⅱ> F p b r F pbr T T Ⅱ Force-deformation relationship t t ( pbr) ( r ) pr( r ) T ( bt) E te FT p FT ( r ) r T

44 Tension in belt Brake band aw of Coulomb friction N N μδn v Magnitude: F N Direction: reventing relative motion : Coefficient of Coulomb friction, N N F ppling E on the FBD and governing equation Fr ; N T sin( / ) T( )sin( / ) F ; N T cos( / ) T( )cos( / ) dt T( ) T( ), sin, cos 1 d dn T N T T d dt T dn T d NT d Solving dt d T lntc T T Ce Boundar condition T() T T e

45 Strain energ in rod Force-deformation relationship Hooke s law Strain energ E, E, E,, E E 1 U k 1 / 1 eq V E E u U V k eq 1,, U udv 1 dv 1 dv V V V E / Stress E : modulus of elasticit k eq du d E / 1 1 dd d E E 1 1 du 1 du U E dv E dd E d V d d 1 E Strain strain energdensit(function)

46 Generalization of uniaial loading Strain in rod under uniaial loading Undeformed Deformed u () u( ) ' B' B u( ) u( ) du lim B B d q :oad per unit length q() df F () F( ) F ( ) d q( ) dt q( ) ( ) Ⅰ Force equilibrium F F( ) q( ) F( ) df q ( ) d Ⅲ Stress-strain relation E Hooke s law du F( ) E E d df ( ) d q v v d du ( E ) q( ) d d Governing equation

47 3.7 Torsion

48 Torsion of clindrical shaft, problem definition Cross-section: Circular shaft (CS) ole of CS: ower transmission, spring, etc. ssumption to appl rule of smmetr End-effects are negligible (Saint-Venant rinciple) Uniform cross-section Geometr and material are aismmetric Smmetric epansion and contraction are neglected engthening and shortening are neglected ule of smmetr M t M t upside down M t M t Cavit Summar Diametrical straight line remains straight line lane section, perpendicular to the central line, remains plane M, t M t T Shaft M, t Twisting moment, torque rcos rsin z z z (,, z ) ( r,, z) r z z T r ocal coordinate sstem eference coordinate sstem

49 elation of strain and angle of twist Strain components z rr r rz z r z z z zz zr z zz Shear strain r z z z and angle of twist ( zz) : ngle of twist d : ate of twist dz Normal strain: rr zz Shear strain: r From assumption Shear strain: rz z z r ( z) r z z z z r z z z z d r dz r r z plane r plane r rz zr r : ngle of twist d : ate of twist dz

50 Stress components --z coordinate sstem Clindrical coordinate sstem z z z z zz z z plane z face plane z zz z plane z r z plane z face z plane z r plane r zz z rr r rz r z zr z zz z z z z face r zr rz r r face face z ( ) face rr r

51 Hooke s law z d Gz Gr Gr dz G Torsion test G G E (1 ) du u E E E d Tensile test E E 1 E rr rr zz 1 zz E 1 zz zz rr E 1 1 r r r E G 1 1 E G 1 1 E G z z z zr zr zr rr 1 1 rr r rz r z rd / dz rd / dz zr z zz rr r rz r z Grd / dz Grd / dz zr z zz

52 equirement on equilibrium Homogeneous shaft d M t rdf rd Gr d dz d d M t G r d GJ dz dz d M t M tr z z dz GJ J J r d r dr ( i ) i GJ : Torsional rigidit df d r df d olar second moment of inertia at the cross-section r d r dr dr Composite shaft d dz d d M t G1 r d G r d ( G1 J1 GJ ) dz 1 dz d M t GirM t, G1 if r 1 Gi dz G1 J1 GJ G1 J1 GJ G if r M t rdf rd Gr d Stress G 1 1 r r G

53 Eample 1 Staticall determinate sstem < F.B.D. > T B 1 C T B T C TB TC Moment balance M ; T T T T T T B C B C M T T M t, B B C t, BC T C Maimum angle of twist and stress C B C 1 C ma 1 [( T T ) T ] GJ ( TB TC)( d / ) J

54 Eample Staticall indeterminate sstem < F.B.D. > B C T T T C Moment balance M t, B ; T T T T T T M C C T M T T T t, BC C Geometric compatibilit: 1 C 1 C B CB [ T1 ( T T ) ] GJ T T ngle of twist at oint B and maimum shear stress B ma M t, B T1 1 1 T 1 T d GJ GJ GJ 1 GJ ( 1 ) Tma r, Tma ma( T, TC) J

55 Eample 3 Design of circular shaft Given value 6 hp, n 38 rpm, a 3, psi Ship Determination of diameter of the shaft 6 6 hp 6 66 in lb / s in lb / sec rad rad 38 rpm sec sec 3 T T in lb 1 T ( d / ) 16T lb 16T ( 31 ) d 3 d d in a 3 16T d 3.9 in 3 ma 3 a a 1hp 76kg m / s lb ft / sec

56 Summar of torsional problem of circular shaft r z z r z z Strain rr r rz zr z zz r z ule of smmetr ssumption: rr zz Geometric compatibilit: ( zz) z r rz ( z) z r zr d r dz Hooke s law 1 E rr rr zz 1 zz E 1 E zz zz rr r z zr G r G z zr G rr U Stress d Gr dz d Gr dz 1 z Force equilibrium z d dz t dz dz GJ Mt dz GJ M Mr t J d dz For composite shaft M t d dz G J G J 1 1 M t GJ...

57 3.8 Beam Theor

58 Beam and eternal load esultant force Staticall the same concentrated force with a set of distributed force = Vector sum passing through centroid of loading diagram Concentrated force q () Staticall q () X q () oading diagram B Concentrated moment q () q () Saint-Venant s principle q () Staticall = oad intensit function = oad/length X q( ) d d q( ) d d

59 Internal force (Shear force and bending moment) Definition of aes Definition of cross-section is of smmetr ositive -face Negative -face Neutral ais z 1 z Force and moment eerting cross-section M F M : twisting moment M, M : bending moment F z : aial force F, F : shear force z M b M b F V F V F F, V F, M M b z z M z 3D F z F M First subscript: Direction of face Second subscript: Direction of force D

60 ure bending V ( ), M ( ) constant b a a M a ure bending

61 Derivation of relationship between and v ( ) 1 d d d ds d ds? v( ) tan Curvature v ( ) sec d v( ) d 1 v( ) cos v( ) 1 1 v( ) 1 d d d v( ) ds d ds s 3 1 v ( ) v ( ) Slope v( ) d ds s? s s v v v 1 1 s v 1 d 1 ds 1 v ( )

62 Eample Find SFD and BMD b cut method V 3 8 V( ) M ( ) b o eactions F ; V ( ) M C ; M b( ) C V M b 8 Mb F ; V ( ) 8 8 M C ; M b( ) ( ) C M b V 8

63 q() q () V( ) F ; V ( ) V ( ) q( ) M ( ) b V( ) C M ( ) b V ( ) V ( ) lim q ( ) dv ( ) d q ( ) M C ; M b( ) M b( ) V ( ) q( ) M b( ) M b( ) dm b( ) lim V ( ) V ( ) d dm b( ) V ( ), M b( ) V ( ) d dv ( ) q( ), V ( ) q( ) d M ( ) q( ) b

64 Eample Find SFD and BMD using DE V( ) o eactions oad intensit function 8 V 3 o q( ) 1 o 8 o Mb M ( ) b Shear force and bending moment diagram 3 o V ( ) o o M ( ) 8 o 1 b o o

65 Distribution of internal force Bending moment and bending stress = Shear force and shear stress z = M b

66 ure bending deflection curve-strain relation Strain B C D E F lane of smmetr S Q Neutral ais M b S Q Q S M b Q Q ( ) v( ) Q v z v,, z z z z z zz z z ( / ) nticlastic curvature v v

67 Hooke s law of an isotropic material ure bending - stress-strain relation 1 E zz 1 E zz 1 E zz zz z z z z z z z z 1 1 E G 1 1 E G 1 1 E G Stress z z z z z ij ji z face zz z z zface z z face pplication of Hooke s law Flange ssumption:, E E, zz ν z,, (,, ) z z z z z Web zz z zz zz

68 ure bending Force equilibrium pplication of requirement of force equilibrium M b F ; d 1 E ( ), If E Ed constant d M ; d M z b E, ( ) 1 E d M b If E constant EI zz : Fleural rigidit EI zz b M E M ; z d zd I zz d Mb I zz utomaticall satisfied for smmetric beam nd moment of inertia of the cross-section

69 ( ) ( ) ( ) b Summar of pure bending beam theor q v M Curvature-deflection curve : Curvature-strain : Constitutive law : E 1 v( ) E ial force : d d Bending moment : 1 Mb d M v( ) b EI EIv( ) M ( ) b ( EIv ( )) M q( ) M b I zz b Boundar conditions Geometric : v(), v() dv ( ) q ( ) d dm ( ) b V( ) d d Mb ( ) d q ( ) Mechanical : v ( ), v( ) V(), M () M, b V ( ), M M () EIv () M v () b EI V M ( ) V () ( EIv ()) v () b EI

70 urpose of engineering beam theor Engineering beam theor When shear force eists, z -bending moment varies from position to position. -shear stress as well as bending stress eist. M ( ) b urpose of engineering beam theor is to calculate the shear stress M ( ) b z ssumption of engineering beam theor The following relationships obtained b the pure bending beam theor are valid even though shear force does not vanish. Mb, I zz 1 M b v( ) EI esults of engineering beam theor V zz Q bi 1 zz Q d

71 Engineering beam theor Shear stress Take a cut fraction having - face in the cut surface at = 1 1 : Cross-sectional area defined b = 1 ssumption: uniform cross-section 1 F ( ) ( ) d F ( ) ( ) M ( ) M ( ) b I zz b Mb( ) Mb( ) F Q I zz df dm b Q VQ f d d I zz I zz df F b b d Q d 1 Shear flow V Q bi zz ssume: F is uniforml distributed over width b

72 Eample 1 Shear and bending stresses Given values a 1psi, b h 6, 1lb Calculated maimum shear force and bending moment Mb)ma, Vma 4 llowable maimum beam length ma M / 4 h / 3 b )ma 3 I z bh /1 bh h/ 3 3 bh a 6 in 1lb )ma a ma 144in 3 3lb in atio of maimum shear stress to bending stress 3 VQ / bh / h / 4 3 )ma 4bh h )ma 3 bi bbh /1 4bh 3 )ma bh

73 Eample atio of shear to bending stresses V Mb V( ) M ( ) b q ( ) 8 h b o q( ) o 1 o 1 o 1 o V ( ) o o 1 M b( ) M b ) 1 o 1 o o 1 ( ) o 4 8 ma h h b b bh 4bh ) ma 3 ) ) ma ma h h 1 bh 4bh 1

74 Eample 3 Beam deflection roblem eactions Boundar conditions v() v ( ) Mb() EIv() Mb( ) EIv( ) v ( )? q( ) 1 (4) EIv ( ) q( ) 1 1 EIv( ) C 1 EIv( ) C C 3 EIv( ) C EIv( ) C1 C 1 4 Determination of C1, C v() C v( ) C1 C EI EI v( ) B

75 rinciple of superposition roblem description eactions Mb ( ) 3 q( ) M b ( ) 5 ( ) 8EI EIv( ) M b ( ) 3 EIv( ) C1 ( v() ) Boundar conditions v(), v() EIv( ) C ( v() ) v( ), v ( ) v ( ) EI EI Deflection and angle due to each force EI 1 8EI 3EI EI v( ), v( ), ( ) v( ) pplication of principle of superposition v v1 v EI EI 5 1 8EI

76 3.9 Buckling

77 Governing equation of buckling of column Similarit and difference between beam and column Similarit: Bending moment and shear force are eerted and pure bending theor is applied. In beam theor, aial force ( ) is neglected in calculating bending moment. However, in column theor, aial force should be considered in calculating bending moment. elationship of,,, and q ( ) dv dv dv F ; q( ) q( ) q( ) d d d M C b b q ( ) ( v( ) v( )) dm b dv V d d dmb dv d dv ( ) d d d d b M ; M ( ) M ( ) V ( ) dmb d dv V ( ) q( ) d d d v ( V M ) ( ) b v ( ) q ( ) EI, C v ( ) V ( ) V ( ) V Displacement in the longitudinal direction is neglected M ( ) M ( ) M b b b v( ) v( ) v

78 Governing equation of buckling of column ssume: ure bending theor is applied. 1 v( ) EIv( ) M ( ) 1 M EI d E b b ure bending dv q ( ) d dm b dv V d d Governing equations d d dv b ( M ) ( ) q ( ) d d d d d v d dv d d d d ( EI ) ( ) q( ) EI 일정, = 일정 q ( ) (4) v ( ) v( ), EI v( ) Ce 4 s s s ( 중근 ), s i v( ) C C C sin C cos C1, C, C3, C are calculated b BCs. s

79 Eample buckling loads and modes roblem ( ) sin v C 1 GE and homegeneous solution (4) v ( ) v( ), EI EI v( ) C C C sin C cos BCs v(), v( ) M b () v() M ( ) v( ) b EI 3 v ( ) C sin 3 EI, v ( ) C sin Solving v( ) C sin C cos v() C C v( ) C C C sin C cos v () C4 v( ) C sin C cos v( ) C sin Because, C from Eq.. 4 C1 from Eq. 1. In order to obtain non-trivial solution, sin. sin n n / ( n 1,,, ) n EI n n n n v ( ) sin n C EI cr EI cr Buckling load EI 1 3 4

80 Mode of buckling Non-buckled Buckling mode 1 (n=1) Buckling mode (n=)

81 3.1 Special phenomenon

82 Stress concentration(sc) and SC factor K