C1B Stress Analysis Lecture 1: Minimum Energy Principles in Mechanics Prof Alexander M. Korsunsky Hilary Term (January 08)

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1 CB Stress Anaysis ecture : Minimum Energy Principes in Mechanics Prof Aexander M. Korsunsky Hiary Term (January 8)

2 This course introduces seected chapters in Stress Anaysis. Energy principes in deformation theories. Variationa cacuus. Euer equation and boundary conditions. Appication to beam bending. The Rayeigh-Ritz method.. Further Euer, Rayeigh-Ritz, and Gaerkin. Generaisation to higher dimensions. Piecewise approximation, and the connection with the FEM. 3. Fundamentas of anisotropic easticity: Stress, strain, eastic constants. The system of equations of easticity. Anaytica soution of eastic probems. Pane stress and pane strain. The wedge probem. Fundamenta singuar soution for probems invoving pane contacts. Connection with the Boundary Eement Method (BEM). Soution of contact probems using BEM. 4. Fundamenta soutions in easticity theory. The Wiiams wedge anaysis. The Westergaard stress function.

3 Energy minimisation A simpe spring mode shows how probems can be soved using the energy minimisation approach, rather than the system of equiibrium equations. The eastic energy stored in the spring is found as the work done by the interna forces over the dispacements, i.e. du Fdx kxdx, U ε Fxdx During uniaxia deformation of ties, coumns, beams etc. the strain energy density (i.e. the eastic stored energy per unit voume) is given by: du σdε σdε Eεdε, U σdε ε kx Eε Key steps of soution by energy minimisation: Identify key parameters that describe deformation Express tota energy in terms of unknown parameters Minimise energy with respect to these parameters To deveop understanding of soving probems by energy minimisation, we wi foow this strategy: ). Discuss fundamentas of minimising functionas (e.g. integras), known as the cacuus of variations. ). earn about the derivation and use of Euer s equation, and the attendant boundary conditions. We wi appy Euer s equation to the probem of bending a simpy supported beam and a buit-in cantiever. 3). earn how fast and error-free derivations can be carried out using a symboic agebra package, such as Mathematica. 3). Introduce the direct parametrisation and minimisation technique known as the Rayeigh-Ritz method. We wi then assess its accuracy by comparison with exact soutions. 4). Introduce another method for soving minimisation probems, known as the Gaerkin method. 3

4 Energy Functionas The energy to be minimised is often expressed as an integra. As exampes we can consider: eastic bending of a beam axia tension or compression in ties and coumns torsion of shafts et v(x) denote the defection. Strain energy density for a bent beam : du σdε Eεdε, U σdε Eε W Eε dadx E y da v'' dx V For a tie/coumn: W EA For a shaft: W GJ θ ( u' ) dx ( ') dx A ε ( ) EI ( v'' ) dx Note that in a cases the energy is expressed as an integra of some derivative of the deformed shape. In order to find a soution, the integra must be minimised with respect to the choice of a function, e.g., v(x), u(x) or θ (x). Additionay, the function must satisfy some boundary conditions at the ends of the interva (or possiby within). For exampe, at the ends x, of a bent beam free, simpy supported, or buit-in conditions may be prescribed. Note that, in genera, the energy integra contains a combination of the function and its derivatives of different orders. Minimisation of a functiona is the probem of the cacuus of variations 4

5 Variationa cacuus Probem: Find a function y(x) that makes the foowing integra W[y] take a minimum vaue: W b [ y] F( x, y, y', y' ')dx a Boundary conditions on the unknown function y(x) may be prescribed: y(a)y, y(b)y, y (a)y, y (b)y, etc. If y(x) is the soution, how do we seek it? et s try to adjust the methods known from minimisation of functions. et y(x) be a soution. Construct comparison or tria soutions When α, y becomes the soution. Require y y( x) + αη( x) η( a), η( b) so that y sti satisfies a boundary conditions., etc., Now minimise W as a function of α. How do we determine the conditions on the soution y(x)? We know that at a minimum of a function f(x), its derivative f x vanishes. Another way of saying the same thing is that the variation δff x δxvanishes. We now deveop the same approach to finding the minimum of functiona W[y]. Consider an increment of the introduced parameter α, and find the expression for the variation δw. We identify the conditions this imposes on the functions invoved, which are ikey to emerge as differentia equations, and aso some conditions at the region boundary. 5

6 Euer equation We are seeking the variation of W[y] with y. Consider δww[y+δy]-w[y]: b [ y] F( x, y + δy, y' + δy', y' ' + y' ')dx F( x, y, y', y' ')dx W[ y + δy] W δ δw a b [ ' y] ( Fyδy + Fy ' δy' + Fy ' δy' ') dx a By carrying out integration by parts as many times as necessary, a derivatives of y can be eiminated from under the integra sign, giving b δw[ y] [ F ( F ')' + ( F '')''] δydx a y y W [ y] F( x, y, y', y' ')dx y b a Now in order to ensure that W is at a minimum, this variation must be zero for any δy. This can ony be guaranteed if the expression in brackets is zero. b a (.) The differentia equation for the soution y(x) is the Euer equation: F y ( Fy ' ')' + ( Fy ' )'' Probem: find the shape of a string of ength suspended in tension T due to distributed oad w. A ine eement of string dx becomes after oading. The strain is ½ (y ) and strain energy is ½ T(y ), whie the work done by the oad is wy per unit ength. The tota energy is given by (.), where F( x, y, y', y' ') To write the Euer equation using (.), we need to differentiate F with respect to y and y as if they were two independent variabes. Hence Appy boundary conditions y()y(). The soution: 6 dx + ( y' ) dx( + T ( y' ) Ty' ' +w wy (.) ( y' ) w y( x) x( x). T )

7 Euer: Bent Beam Probem: Minimise the potentia energy of a beam with defection v under genera appied force f(x) (may be a combination of distributed and point oads). Tota energy of the system: The second integra term describes the work done by the force f over dispacement v. The first variation of W is found by considering increments δv and δv δw and using integration by parts δw The Euer equation for a bent beam is found as: ( v'' ) dx We now discuss the boundary conditions. Some boundary conditions arise from the variation of δw (see above). They are not a consequence of any additiona constraints imposed on the system, and for this reason are caed natura boundary conditions. Other boundary conditions appear because of additiona constraints. For a bent beam, the foowing options can be encountered: W EI fvd EI v'' δv'' dx fδvdx EI[ v' ' δv' ] EI v'' ' δv' dx fδv d x EI[ v' ' δv' ] EI[ v' '' δv] + [ EIv' ''' f ] δvdx (a)if beam ends are free, then δv and δv are arbitrary, and their coefficients in the expression for δw must vanish (these are the natura b.c.). Hence at a free end v and v. (b)if an end is simpy supported, then the defection and the bending moment M-Eiv must vanish, eading to the requirements v and v. (c)if an end is buit-in, then the defection and sope must vanish, eading to the requirements v and v. EIv '''' f (.3) x 7

8 Euer: Bent Beam EIv' ''' f Exampe probem: a simpy supported beam of ength and bending stiffness EI is oaded by a combination of UD ω and a point oad P at mid-span. Use Euer equation to sove the probem. The oad is given by f ( x) w + Pδ ( x / ) Here δ(x) is Dirac s deta-function. It is often used in physics and engineering to represent point oads, concentrated charges, etc. Its most important property is expressed in an integra sense: when the deta function is integrated over an interva which contains the zero of its argument, the resut is unity; otherwise it is zero. Note that this is precisey what we mean by a unit point oad! To find v(x), we must integrate (four times) Euer equation in the form v '''' [ w + Pδ ( x / )] / EI Integrating the deta-function is easy: the first integra produces a unit-step function, which is changes from zero to unity as we go through zero of the argument. If we denote this first integra of the deta-function by H(x), then subsequent integrations produce functions of the form H n (x)h(x) x n- /(n-)! y(x) Integrating δ-function δ(x) H(x) H (x) H 3 (x) H 4 (x) You must note here the simiarity between the deta-function and its integras, and the Macauay brackets. It is aso important to note that constants of integration must be introduced when soving differentia equations x

9 Using Mathematica In order to find the soution of the exampe probem, integration must be performed four times. Four constants wi be introduced in the process, which wi need to be found from the four boundary conditions. You may want to practice doing that for the exam. For our purposes we need an error-free and fast way to perform these anaytica manipuations. We can use one of the symboic agebra packages, such as Mathematica. euerbeam.nb v, v, v3, c, c, c3, c4d H Euer equation: EI v''''w + P DiracDeta@x dd v3 Integrate@Hw + P DiracDeta@x dd ê EI, xd + c v Integrate@v3, xd + c v Integrate@v, xd + c3 v Integrate@v, xd + c4 wx + PUnitStepHx- d c + EI wx PHx- d UnitStepHx - d + c x + c + EI EI wx 3 c x dphx- d UnitStepHx - d + + c x + c3-6ei EI wx 4 c x3 c x c3 x+ c4 + d PHx - d UnitStepHx- d 4 EI 6 EI v ê.x ê. UnitStep@ dd ê. UnitStep@ dd v ê.x ê. UnitStep@ dd ê. UnitStep@ dd + PHx - d UnitStepHx - d EI - dphx - d UnitStepHx - d EI + PHx3 - d 3 UnitStepHx - d 6EI c c4 v v ê.c ê.c4 ; v v ê.c ê.c4 ; eq v ê.x ê. UnitStep@ dd ê. UnitStep@ dd v v ê. Sove@eq, cd@@dd v v ê. Sove@eq, cd@@dd w EI + c + H - d P EI wx 4 4 EI - Hw + P - dp x 3 + c3 x + d PHx - d UnitStepHx- d EI EI wx EI - Hw + P- dp x PHx- d UnitStepHx - d + EI EI - dphx - d UnitStepHx - d EI eq Simpify@v ê.x ê. UnitStep@ dd ê. UnitStep@ dd ê.c ê.c4 D v v ê. Sove@eq, c3d@@dd v v ê. Sove@eq, c3d@@dd + PHx3 - d 3 UnitStepHx - d 6EI - w4 + 8dP - 4 c3 EI - d P + 4d 3 P 4EI wx 4 4EI - Hw + P - dp x 3 - H-w4-8dP + d P - 4d 3 P x + d PHx - d UnitStepHx- d - dphx - d UnitStepHx- d + PHx3 - d 3 UnitStepHx - d EI 4 EI EI EI 6EI wx EI - Hw + P- dp x + EI PHx- d UnitStepHx - d EI 9

10 Using Mathematica In[6]: vauep 8EI,, d.5, P, w <; vauew 8EI,, d.5, P, w <; nvp v ê. vauep nvp v ê. vauep nvw v ê. vauew nvw v ê. vauew Pot@8nvp, nvw<, 8x,, <, PotStye 8Hue@.D, Hue@D<D 8nvp ê.x.5, nvw ê.x.5, nvp ê.x.5, nvw ê.x.5< Pot@8 nvp, nvw<, 8x,, <, PotStye 8Hue@.D, Hue@D<D 8 nvp ê.x.5, nvw ê.x.5, nvp ê.x.5, nvw ê.x.5< Out[8] x x +.5 Hx -.5 UnitStepHx Ix -.5M UnitStepHx Ix3 -.5M UnitStepHx -.5 Out[9] Hx-.5 UnitStepHx x Out[3] x x3 + x 4 Out[3] x - x Out[3] Ü Graphics Ü Out[33] ,.38,.439,.97734< Out[34] Ü Graphics Ü Out[35] 8.5,.5,.5,.9375< Summary EIv(/) EIv(/4) -EIv (/) -EIv (/4) P at /.8P 3.43P 3.5P.5P UD ω.3ω 4.93ω 4.5ω.94ω

11 Rayeigh-Ritz Method This approach is a powerfu aternative to finding and soving Euer equation. It is a direct method, in the sense that the soution is assumed to be unknown but of a certain form containing unknowns, which are then found by minimisation. Key steps: (a) Express the unknown function (e.g. defected shape) as a inear combination of known functions with unknown coefficients: (b) Express the tota energy of the system W as a function of constants A i. (c) Find the minimum by requiring v ( 3 x) A v ( x) + Av ( x) + A3v ( x) +... grad A i W i.e. W / A The Rayeigh-Ritz method transforms a probem with an infinite number of degrees of freedom into one with a finite number (as many as A i s used). Rayeigh-Ritz: Bent Beam We seek a soution to the bent beam probem by seeking the defected shape as a series of the type (.4), but containing ony one term: v( x) πx Asin We find right away that: The function v(x) is chosen to satisfy the defection boundary conditions v()v() at simpy supported ends. Note that it aso satisfies the moment boundary conditions, since for it v ()v (). Tota energy of the system can be evauated (e.g. using Mathematica) W PA 4 π EIA EI( v' ') dx wvdx Pv( / ) 3 4 π v''( x) A Since in this case (.5) reduces to a singe equation, the immediate resut is i πx sin W 4 3 π EIA w w P, A P A 4 π π EI π wa π (.4) (.5)

12 Rayeigh-Ritz Method In Mathematica, the soution is found by foowing the three steps we have indentified (ritzbeam.nb): Out[6] - Out[7] Out[8] H Bent beam by Rayeigh Ritz, one term Off@Genera::spe, Genera::speD H Step : Define shape in terms of A v ASin@Pi x ê D;v D@v, 8x, <D H Step : Cacuate W in terms of A W Integrate@EI v^ ê wv,8x,, <D P H v ê.x ê H Step 3: Minimise dwêda to find soution vr v ê. Sove@D@W, AD, AD@@DD Ap sinj p x N A EI p 5-8 A 4 w AP p 3 HpP+ w sinj p x EI p 5 N It is now possibe to compare the resuts with the exact soutions (by Euer equation), by giving the foowing commands: In[]: vr D@vr, 8x, <D nvrp vr ê. vauep nvrp vr ê. vauep nvrw vr ê. vauew nvrw vr ê. vauew Pot@8nvrp, nvp<, 8x,, <, PotStye 8Grayeve@.6D, Grayeve@D<D Pot@8nvrw, nvw<, 8x,, <, PotStye 8Grayeve@.6D, Grayeve@D<D Pot@8 nvrp, nvp<, 8x,, <, PotStye 8Grayeve@.6D, Grayeve@D<D Pot@8 nvrw, nvw<, 8x,, <, PotStye 8Grayeve@.6D, Grayeve@D<D We can now compare the soutions for defected shapes and bending moment profies

13 Rayeigh-Ritz Method 3 We compare the exact soution with the R-R approximation: In[59]: H Comparison for defection H Exact 8nvp ê.x.5, nvw ê.x.5, nvp ê.x.5, nvw ê.x.5< H R R 8nvrp ê.x.5, nvrw ê.x.5, nvrp ê.x.5, nvrw ê.x.5< Out[59] ,.38,.439,.97734< Out[6] 8.53,.37,.4583,.9463< H Comparison for bending moment H Exact 8 nvp ê.x.5, nvw ê.x.5, nvp ê.x.5, nvw ê.x.5< H R R 8 nvrp ê.x.5, nvrw ê.x.5, nvrp ê.x.5, nvrw ê.x.5< Out[57] 8.5,.5,.5,.9375< Out[58] 8.64,.96,.439,.9< R-R captures the defected shape better than the bending moment profie (i.e. stress). This is because stresses and strains depend on the derivatives of defection, and require higher order approximation for good agreement. The quaity of approximation can ony improve with extra terms in (.4). Assume v( x) i.. N iπx Ai sin An efficient soution is obtained if, after writing down the integra expression for W, differentiation with respect to A i is carried out under the integra sign: W A i EI v' ' v v' ' dx P A A i i ( / ) The resuting integras are famiiar from Fourier anaysis. In particuar from it foows that, and You can inspect a possibe Mathematica impementation of this sine series soution in the fie ritzbeamn.nb, or construct your own. v w A i dx iπx jπx sin sin dx δij 4 iπ iπ w EI Ai Psin (cosiπ ) iπ 3 iπ w A i Psin + (cosiπ ) 4 ( iπ ) EI iπ 3

14 Mathematica tutoria H Introduction to using Mathematica for 4 ME6 Stress Anaysis H In the derivations I use a very sma part of what Mathematica can do. H You wi have guessed that parenthesis asterisk is used for comments H Mathematica understands about most functions. To run a CE, you need to press Shitf+Enter top status bar shoud fash ' RUNNING', and come up with a pot beow Pot@8Sin@xD, Cos@xD, Exp@xD, Cosh@xD<, 8x,, Pi<D Ü Graphics Ü H You may have noticed that the functions are grouped together between cury brackets, and separated by commas they form a IST. The potting argument and range aso form a ist 8x,, Pi<. The pot above ooks difficut to read, since no coour was used. Instruction can be given to the pot command, using PotStye 8Hue...< to choose a HUE for each function respectivey. HUE argument can go from to. To find what' s what, change the numbers beow, and re run Pot@8Sin@xD, Cos@xD, Exp@xD, Cosh@xD<, 8x,, Pi<, PotStye 8Hue@.D, Hue@.3D, Hue@.5D, Hue@.7D<D Ü Graphics Ü H Mathematica can aso buid ists using Tabe: Tabe@Sin@nPiê 6D, 8n,, 6<D Pot@8Sin@xD, Sin@ xd, Sin@3 xd, Sin@4 xd<, 8x,, Pi<, PotStye Tabe@Hue@.5 jd, 8j,, 4<DD :,, è 3,, è 3,,> Ü Graphics Ü 4

15 Mathematica tutoria H It can aso buid arrays: consts Array@c, 5D 8cH, ch, ch3, ch4, ch5< H It' s very good at producing pretty 3 D pots u Sin@rhoD H3+ Cos@thD; v Cos@rhoDH3+ Sin@thD; z Sin@thD; ParametricPot3D@8u, v, z<, 8rho,, Pi<, 8th,, Pi<D Ü Graphics3D Ü H Mathematica can differentiate using D@D Heven severa times and find indefinite and definite integras using Integrate@D D@Sin@xD,xD D@Sin@xD, 8x, <D Integrate@Sin@xD,xD Integrate@Sin@xD, 8x,, Pi<D coshx -sinhx -coshx H It can do series! Series@Tan@xD, 8x,, 5<D x+ x3 3 + x x x x x x OIx6 M H Mathematica can sove equations note the use of Sove@x^ + 5 x 6, xd 88x Ø-6<, 8x Ø << H It can aso sove differentia equations! DSove@u''@xD 9 u@xd, u@xd,xd 99uHx Ø -3 x c + 3 x c DSove@u''@xD 4 u'@xd + 4 u@xd, u@xd,xd 99uHx Ø x c + x xc 5

16 Mathematica tutoria 3 H Mathematica is very good at soving systems of equations. Command SOVE needs a ist of equations, foowed by a ist of unknowns to find. SOUTION Sove@8x+ 3 y + z, x y + z 5, 3 x+ 7 y + z 3<, 8x, y, z<d ::x Ø- 3, y Ø, z Ø 5 >> H We ca the resut ' SOUTION'. It contains RUES. Each RUE describes a substitution: it shows, using an arrow, which vaues shoud be given to each unknown. Our SOUTION contains the rues encosed in doube cury brackets. To extract the rues, we needto use doube square brackets, and specify we need the first eement SOUTION@@DD :x Ø- 3, y Ø, z Ø 5 > H To use the soution, we need to perform the substitution. This is done by using the command ê. The resut is a ist of vaues of x,y,z that we found 8x, y, z<ê. SOUTION@@DD :- 3,, 5 > H Finay, procedures can be defined using MODUE. This one cacuates the sum of odd integers up to n. It uses OddQ a query function, which is true of the integer is odd, and fase otherwise. Note the C stye cacuation of the tota OddSum@n_D : Modue@8i, tota <, Do@If@OddQ@iD, tota + i, D, 8i, n<d; tota D H We can now use OddSum as a function OddSum@D 5 In order to appreciate the capabiities of Mathematica, you must try running it yoursef. You can use the program to study the exampes from in the ECS ab (the Soarium) on any of the machines booth36.ecs to booth5.ecs (incusive). To run Mathematica, type start_mathematica. This wi start a window tited "Mathematica4. Execution Window". Run Mathematica in this window by typing mathematica. 6

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VTU-NPTEL-NMEICT Project MODUE-X -CONTINUOUS SYSTEM : APPROXIMATE METHOD VIBRATION ENGINEERING 14 VTU-NPTE-NMEICT Project Progress Report The Project on Deveopment of Remaining Three Quadrants to NPTE Phase-I under grant in aid