《弹性力学》课程教学课件（英文讲稿）Chapter 4 Polar Solutions for Planar Problems

ElasticityCheaoarem

Elasticity 1

POLARSOLUTIONSFORPLANEPROBLEMSChapter 4 Polar Solutions For Plane ProblemsS 4-1 Eguilibrium DifferentialEquations in PolarCoordinatesS 4-2 Geometric and Physical Functions in PolarCoordinatesS 4-3 Stress Functions and Compatibility Equationsin Polar CoordinatesS 4-4 Coordinates Conversion of Stress ComponentsS 4-5 Axisymmetric Stress and Its DisplacementS 4-6 Circular Ring or Cylinder under UniformLoading Pressure

Chapter 4 Polar Solutions For Plane Problems §4-1 Equilibrium Differential Equations in Polar Coordinates §4-5 Axisymmetric Stress and Its Displacement §4-2 Geometric and Physical Functions in Polar Coordinates §4-3 Stress Functions and Compatibility Equations in Polar Coordinates §4-4 Coordinates Conversion of Stress Components §4-6 Circular Ring or Cylinder under Uniform Loading Pressure 2

POLARSOLUTIONSFORPTANEPROBLEMSChapter 4 Polar Solutions For Plane ProblemsS 4-7 Pressure TunnelS 4-8 Stress Concentration at the Hole Edge of theCircular HoleS 4-9 Normal Concentrated Forces on the Boundaryfor Semi-infinitePlane Body 4-1oDistributedForce ontheBoundaryforSemi-infinitePlane BodyExercise

Chapter 4 Polar Solutions For Plane Problems §4-10 Distributed Force on the Boundary for Semi-infinite Plane Body §4-8 Stress Concentration at the Hole Edge of the Circular Hole §4-7 Pressure Tunnel §4-9 Normal Concentrated Forces on the Boundary for Semi-infinite Plane Body Exercise 3

POLARSOLUTIONSFORPLANEPROBLEMSS 4-1Eguilibrium DifferentialEquations in Polar CoordinatesDealing with elasticity problems, what form of coordinatesystem we choose, which can't affect on the essence fordescribing problem, but is relate to the level of difficulty onsolving problem directly. If coordinate is suitable, it cansimplify the problem considerably. For example, for circular,wedged and sector and so on, it will be more convenient byusing polar coordinates than using rectangular coordinates.Considering an differential field PACBin the plate

§4-1 Equilibrium Differential Equations in Polar Coordinates Dealing with elasticity problems, what form of coordinate system we choose，which can’t affect on the essence for describing problem, but is relate to the level of difficulty on solving problem directly. If coordinate is suitable, it can simplify the problem considerably. For example, for circular, wedged and sector and so on, it will be more convenient by using polar coordinates than using rectangular coordinates. Considering an differential field in the plate PACB 4

POLARSOLUTIONSFORPLANEPROBLEMSnormal stress in the r direction is called radial normal stressdenoted by . ; normal stress in the direction is calledtangential normal stress denoted by .; shear stress is denotedby Tro, stipulation of sign of each stress component is similar tothe one in rectangular coordinates. Body force components ofradial direction and hoop are denoted by K,and Ke, respectivelyFig. 4-1.+x0660TerPConsidering equilibrium of an unitTroOdo4element, there are three equilibriumatreedrBequations:rfar1droo,drZF =0,ZF。=0,ZM=0a.or000deOT00adea0TerXtya0Fig.4-15

Considering equilibrium of an unit element, there are three equilibrium equations: r r r r d   +     r r r r r d   +           d   +       d r r   +  d  r r   dr K Kr y x o P A B C Fig.4－1 Fr = 0,F = 0,M = 0 normal stress in the direction is called radial normal stress denoted by ; normal stress in the direction is called tangential normal stress denoted by ; shear stress is denoted by , stipulation of sign of each stress component is similar to the one in rectangular coordinates. Body force components of radial direction and hoop are denoted by and , respectively. Fig. 4-1.  r     r Kr K r 5

POLARSOLUTIONSFORPLANEPROBLEMSFrom M = o , we can obtain the of shear stress reciprocalrelationship:Tor =TroFrom ZF, =O, we have:agde00rdr)(r +dr)d-o,rd0-(7o.+2Or00deOta do)dr-Tadr+ K,rdedr =02a0From F,=0, we have:00α d0)dr -0odr +Otre dr)(r + dr)de(Cea0OrdedeOterd)drK,rdedr = 0-trordo+(t)2a02dededeC1, and weBecause d is very micro, so, sincos1222substitutes tre for tor and put it into the upper two formulas, thus:

From ,we have: From ,we can obtain the of shear stress reciprocal relationship： M = 0    θr = r Fr = 0 ( ) 0 2 2 ( )( ) ( ) θr θr − + =   − + +   + − − +   + d dr dr K rd dr d dr d dr r dr d rd d dr r r r r r r                       From F = 0 ,we have: 0 2 2 ( ) ( ) ( )( ) + + =   − + + +   − + +   + K rd dr d dr d rd d dr dr r dr d r d dr dr r r r r r r                             Because is very micro,so, , ,and we substitutes for and put it into the upper two formulas, thus: 2 2 sin d d  1 2 cos  d   r r  d 6

POLARSOLUTIONSFORPTANEPROBLEMSag,10tre,-@+K,=0Ora0rr1 0gOtre2tr@+K。=0arr arThese are equilibrium differential formulas in polar coordinatesTwo equilibrium differential equations contain three unknownfunctions ,, e, Tre = ter, so it is a statically indeterminate questionThus we must consider the deformation condition and physicalrelationship.Above equations differ from equilibrium equations in planarcoordinates where stress components are expressed by partialderivative only. In polar coordinates, the areas of which unitelement is perpendicular to two side faces are not equal, and thedifference is increasing with the radius reducing, which can beseen from underline items in the equations

       + + =   +   + = − +   +   0 1 2 0 1                K r r r K r r r r r r r r r These are equilibrium differential formulas in polar coordinates. Two equilibrium differential equations contain three unknown functions , , , so it is a statically indeterminate question. Thus we must consider the deformation condition and physical relationship. r r  =  r  Above equations differ from equilibrium equations in planar coordinates where stress components are expressed by partial derivative only. In polar coordinates, the areas of which unit element is perpendicular to two side faces are not equal, and the difference is increasing with the radius reducing, which can be seen from underline items in the equations. 7

POLARSOLUTIONSFORPLANEPROBLEMSS 4-2 Geometric and PhysicalFunctions in Polar CoordinatesI Geometric Functions-Differential Relationship betweenDisplacements and DeformationxtoIn polar coordinates , stipulate0dr1rdeAu.D, ---radial normal strainB。 ---hoop normal strainBrYre ---shear strain (change of rightangle between radial and hoopyline segments)Fig.4-2u, ---radial displacementug ---hoop displacement

I、Geometric Functions—Differential Relationship between Displacements and Deformation §4-2 Geometric and Physical Functions in Polar Coordinates In polar coordinates , stipulate: r      r r u  u -radial normal strain -hoop normal strain -shear strain (change of right angle between radial and hoop line segments) -hoop displacement -radial displacement Fig.4-2  d r dr ur o  8

POLARSOLUTIONSFORPLANEPROBLEMSDiscuss differential relationship between displacements anddeformation in polar coordinates with superimpose method.(1) Assume only having radial displacement but no hoop one. Fig.4-2Normal strain of radial line segment PA:ou(uruOu,Or8rdrOrNormal strain of hoop line segment PB:(r+u,)de-rde_urCerderAngle of rotation of radial line segment PA:α=0

Normal strain of radial line segment PA: r u dr dr u r u u r r r r r   = −   + = ( )  Normal strain of hoop line segment PB： r u rd r ur d rd r = + − =      ( ) Angle of rotation of radial line segment PA：  = 0 (1) Assume only having radial displacement but no hoop one. Fig.4-2. Discuss differential relationship between displacements and deformation in polar coordinates with superimpose method. 9

POLARSOLUTIONSFORPLANEPROBLEMSAngle of rotation of hoop line segment PB:oudo)-ur(u, +1 our00β=rder 010urYre=α+β=Thus shear strain is:r a0(2) Assume only having hoop displacement but no radial one. Fig.4-3100Normal strain of radial line segmentrdrPPA:de8, =010ADA"Positive angle of rotation of hoop lineBsegment PB:B"oueL(ug-ue1oug00yG0rder a0Fig.4-310

 d r P P B B A A dr  u o (2)Assume only having hoop displacement but no radial one. Fig.4-3. Fig.4-3 Thus shear strain is:  r = 0 Normal strain of radial line segment PA：             = −   + = u rd r d u u u 1 ( ) Positive angle of rotation of hoop line segment PB：        = + = r r u r 1 Angle of rotation of hoop line segment PB：        = −   + = r r r r u rd r d u u u 1 ( ) 10