One-dimensional heat conductionOther Applications:Linear Elements:Quadratic Elements:Class Problem 1Fig .1Fig .2Home Problem 1Thermal StressesStress-strain curve with temperature effectsAxial ProblemSpecial case: Constant Temperature ChangeLinear Element:Quadratic Element:Class Problem 2Fig .3Fig .4Home Problem 22-D Steady State Thermal AnalysisDifferential Equation: orFunctional (Stored Heat):Element Approximation:Element Conductivity MatrixHeat Conduction Boundary ConditionsConvection Boundary ConditionsRadiation Boundary Condition:M. Vable Notes for finite element method: Thermal Analysis1One-dimensional heat conductionAxial Member Heat ConductionDifferential EquationxddEAxddu⎝⎠⎛⎞px–=EA= Axial Rigiditypx= force per unit lengthxddkxddT⎝⎠⎛⎞qx–=k= Thermal conductivity qx= Heat flow (source) per unit length.fx=heat flux per unit area= qxAPrimary variable u= Displacement in x-direction T= Temperature Secondary vari-ableInternal axial force: N EAxddu= Heat flow: qkxddT–= Positive: Flow into bodyEnergy Strain Energy:U12---EAxddu⎝⎠⎛⎞2xdL∫=Heat Capacity:UT12---kxddT⎝⎠⎛⎞2xdL∫=Work Potential:WApxuxd0L∫=Fquxq()q1=m∑+WTqxTxd0L∫QqTxq()q1=m∑+=Functional Potential EnergyΩAUAWA–= ΩTUTWT–=Rayleigh-Ritzux() Cifix()i1=n∑= Tx() Cifix()i1=n∑=MatrixKjkEAxddfj⎝⎠⎛⎞xddfk⎝⎠⎛⎞xd0L∫= Kjkkxddfj⎝⎠⎛⎞xddfk⎝⎠⎛⎞xd0L∫=Right Hand Side Vec to rRjpxfjxd0L∫Fqfjxq()q1=m∑+= Rjqxfjxd0L∫=Qqfjxq()q1=m∑+Other Applications:Flow through pipes; Flow through porous media; Electrostatics• By non-dimensionalizing the problem a software can be used to solve all the above applications.M. Vable Notes for finite element method: Thermal Analysis2Linear Elements:TT1e()L1x() T2e()L2x()+=Q1e()Q2e()Ke()[]ke()Le()---------11–1–1=Re(){}qxe()Le()2------------------11⎩⎭⎨⎬⎧⎫Q1e()Q2e()⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫+=Ke()[]ke()Ae()Le()-------------------11–1–1=Re(){}fxe()Le()2-----------------11⎩⎭⎨⎬⎧⎫AQ1e()Q2e()⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫+=Quadratic Elements:TT1e()L1x() T2e()L2x() T3e()L3x()++=Q1e()Q3e()Ke()[]ke()3Le()------------78–18–168–18–7=Re(){}qxe()Le()6------------------141⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫Q1e()0Q3e()⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫+= Class Problem 1Heat qo is being added at point B at a constant rate as shown in Fig. 1. Using two linear elements determine the temperature at point B and the heat flowing out at A and C in terms of k, L, and qo for the following two cases: (a) The ends of the bar are maintained at a constant zero temperature. (b) The ends of the bars are main-tained at a constant temperature To. L2LABCqoLABC3qoLLqoD Fig .1 Fig .2 Home Problem 1Heat 3qo is being added at point B and qo is being taken out at a constant rate at point C as shown in Fig. 2. Using three linear elements determine the temperature at points B and C and the heat flowing out at A and D in terms of k, L, and qo for the following two cases: (a) The ends of the bar are maintained at a constant zero tem-perature. (b) The ends of the bars A and D are maintained at a temperature To and 2To, respectively. ANS: TB5qoL()3k⁄=TCqoL()3k⁄=qA5qo3⁄()–=M. Vable Notes for finite element method: Thermal Analysis3Thermal StressesNormal Strain εNormal Stress σO1OαΔTσσE---Stress-strain curve with temperature effectsεσE--- αΔT+σE--- εo+==where, α = linear coefficient of thermal expansion.εo = Initial strain= Thermal strain• No thermal stresses are produced in a homogenous, isotropic, unconstrained body due to uniform temperature changes.Axial ProblemWe assume that the thermal problem and stress analysis problem can be solved independently. σxxE εxxεo–()=UAT12---σxxεxxεo–()VdV∫12---E εxxεo–()2Axd0L∫== orUAT12---EAεxx2xd0L∫EAεxxεoxd0L∫–12---EAεo2xd0L∫+12---EAxddu⎝⎠⎛⎞2xd0L∫EAεoxddu⎝⎠⎛⎞xd0L∫–==Uo+WApxx()ux()xd0L∫Fquxq()q1=m∑+=ΩAUATWA–12---EAxddu⎝⎠⎛⎞2xd0L∫EAεoxddu⎝⎠⎛⎞xd0L∫–Uo+pxx()ux()xd0L∫Fquxq()q1=m∑+–==ΩA12---EAxddu⎝⎠⎛⎞2xd0L∫Uo+pxx()ux()xd0L∫Fquxq()q1=m∑EAεoxddu⎝⎠⎛⎞xd0L∫++–UAWAT–==WATpxx()ux()xd0L∫Fquxq()q1=m∑EAεoxddu⎝⎠⎛⎞xd0L∫++=IEAεoxddu⎝⎠⎛⎞xd0L∫EAεoux()0LEAxddεo⎝⎠⎛⎞uxd0L∫–EAαΔTu x()0LEAαxddΔT⎝⎠⎛⎞uxd0L∫–== =M. Vable Notes for finite element method: Thermal Analysis4 IEAαΔTLuL() EAαΔT0u0()–EAαk------------ qx⎝⎠⎛⎞uxd0L∫+= , where qxkxddT–=WATpxx() pxTx()+[]ux()xd0L∫Fquxq()q1=m∑FT2uL() FT1u0()+++=where, FT1FT2⎩⎭⎨⎬⎧⎫EAαΔT1–EAαΔT2⎩⎭⎨⎬⎧⎫=pxTEAαk------------ qx=Special case: Constant Temperature ChangeΔT Cons ttan=qx0=• Thermal loads are added only at the element ends. Linear Element: Ke()[]EAL--------11–1–1=Re(){}poL2---------11⎩⎭⎨⎬⎧⎫pToL2------------11⎩⎭⎨⎬⎧⎫F1e()F2e()⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫FT1e()FT2e()⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫+++=Quadratic Element:Ke()[]EA3L--------78–18–168–18–7=R{}poL6---------141⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫pToL6------------141⎩⎭⎪⎪⎨⎬⎪⎪⎧⎫F1e()0F3e()⎩⎭⎪⎪⎪⎪⎨⎬⎪⎪⎪⎪⎧⎫FT1e()0FT3e()⎩⎭⎪⎪⎪⎪⎨⎬⎪⎪⎪⎪⎧⎫+++=M. Vable Notes for finite element method: Thermal Analysis5 Class Problem 2Heat qo is being added at point B at a constant rate as shown in Fig. 3. Using two quadratic elements determine the displacement of node B, reaction force at A and the axial stress just before B in terms of E, A, α, k, L, and qo. Assume that the entire bar was at zero temperature and the ends of the bars are maintained at zero temper-ature. L2LABCqoLABC3qoLLqoD Fig .3 Fig .4 Home Problem 2Heat 3qo is being added at point B and qo is being taken out at a constant rate at point C as shown in Fig. 4. Using three linear elements determine the displacements of points B and C, the reaction force at A, and the axial stress just before B in terms of E, A, α, k, L, and qo. Assume that the entire bar was at zero temperature and the ends of the bars are maintained at zero temperature.ANS:uB5αqoL218k-------------------=uC7αqoL218k-------------------=RA5EAαqoL9k-------------------------=M. Vable Notes for finite element method: Thermal Analysis62-D Steady State Thermal Analysis Differential Equation: kx22∂∂ Ty22∂∂ T+⎝⎠⎜⎟⎛⎞qv–= or k∇2Tqv–=Functional (Stored Heat):UTkx∂∂T⎝⎠⎛⎞2y∂∂T⎝⎠⎛⎞2+txdyd∫A∫= where t= thickness