# Are there any examples showing how ANSYS handles temperature-dependent Young's moduli and coefficients of thermal expansion?

 Yes, please see the ANSYS 9.0 example below. Please note the following:Total strain is the sum of the various strain components. In this simplifiedexample, the strain reduces to just mechanical and thermal strain (i.e.,there are no plastic, creep, or swelling strains present). Since we are lookingat small displacement theory in this example (i.e., NLGEOM,OFF by default),we do not need to worry about the differences between strain definitions(see Section 3.1.3 "Definitions of Thermal Strains" in the ANSYS 9.0 TheoryReference for more details). Also, care should be used in interpreting theEPTO, or "total" strain value, as ANSYS defines EPTO to be the strainmeasured by a strain gauge minus the thermal and swelling components(see Section 4.0 "Strain Definitions" in the aforementioned reference).Therefore, I used displacements in this example to further simplify matters.We allow for coefficients of thermal expansion to be based on an average(or secant) value as well as an instantaneous value. Per my example below,it is clear that ANSYS uses the secant (ALPX) value to calculate thermal strain.However, most coefficient of thermal expansion data is provided in terms ofinstantaneous (CTEX) values, so the code allows you to enter the data thatway and then will convert it over to the secant form internally. If I knew thecorresponding CTEX values for the ALPX values that I used, I would have putthem in, but I didn't, so I just used the same numbers to show you that thecode gives different results for CTEX as opposed to ALPX. Incidentally, thisdiscussion applies to the other directions (Y and Z), as well.Mechanical strain is based on the current modulus and not some averagevalue. Therefore, using the current temperature makes the most sense tome. If you wanted to,you could define the modulus such that it gave thedesired average value you wanted at a given temperature, I suppose, but Iam not sure why you would want to do that. If your structure is such thatthe thermal load is statically indeterminate, I guess you could possibly endup with a difference in stress by applying the temperature in one fell swoop.In that case, you would need to break up the time into many time steps,which is what FEA is all about anyhow, so you should still get the "correct"solution. This would definitely be required if you had plasticity. Any plasticstrains would accumulate accordingly.! = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =fini/clear/title, ANSYS 9.0 Temperature-Dependent Material Properties/psf,press,norm,2/plopts,info,1/view,,1,2,3/vscale,,2/prep7et,1,SOLID185r,1mptemp,1, 0, 100, 200, 300 ! define data 4 temperatures ...mpdata, ex, 1,1, 30.0e6, 29.0e6, 28.0e6, 27.0e6 ! temp-dep moduli, psimpdata, alpx, 1,1, 2.5e-6, 2.0e-6, 1.5e-6, 1.0e-6 ! temp-dep secant alphas, in/in/F!!! mpdata, ctex, 1,1, 2.5e-6, 2.0e-6, 1.5e-6,1.0e-6 ! temp-dep instantaneous alphasblock,0,1,0,1,0,1esize,,1vmesh,1eplotfini/soluantype,staticnsel,s,loc,x,0d,all,ux,0.0 ! symmetry plane ...nsel,s,loc,y,0d,all,uy,0.0 ! symmetry plane ...nsel,s,loc,z,0d,all,uz,0.0 ! symmetry plane ...nsel,s,loc,z,1sf,all,press,-1000.0 ! pulling pressure, psinsel,alltref,100 ! reference temperaturetime,1.0bf,all,temp,100.0 ! E = 29.0e6, alpha = 2.0e-6, delta T = 0 degreessolvesavetime,2.0bf,all,temp,200.0 ! E = 28.0e6, alpha = 1.5e-6, delta T =100 degreessolvesavetime,3.0bf,all,temp,300.0 ! E = 27.0e6, alpha = 1.0e-6, delta T = 200 degreessolvesavefini/post1nsel,s,node,,node(1,1,1) ! node in far corner (considering symmetry)*do,i,1,3 /gopr set,i,last prnsol,u,comp*enddo/eof! = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =For TEMP=100 (E = 29.0e6 psi and ALPX = 2.0e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 -0.10345E-04-0.10345E-04 0.34483E-04 0.37458E-04For TEMP=200 (E = 28.0e6 psi and ALPX = 1.5e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 0.13929E-03 0.13929E-03 0.18571E-03 0.27072E-03For TEMP=300 (E = 27.0e6 psi and ALPX = 1.0e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 0.18889E-03 0.18889E-03 0.23704E-03 0.35713E-03! = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =For TEMP=100 (E = 29.0e6 psi and CTEX = 2.0e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 -0.10345E-04-0.10345E-04 0.34483E-04 0.37458E-04For TEMP=200 (E = 28.0e6 psi and CTEX = 1.5e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 0.16429E-03 0.16429E-03 0.21071E-03 0.31366E-03For TEMP=300 (E = 27.0e6 psi and CTEX = 1.0e-6 in/in/F):======================================================== NODE UX UY UZ USUM 7 0.28889E-03 0.28889E-03 0.33704E-03 0.52963E-03! = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =Total Displacement = Mechanical Displacement + Thermal DisplacementTotal Displacement = [(F)*(L)]/[(A)*(E)] + (L)*(ALPHA)*(Delta T)Total Displacement = [(P)*(L)]/(E) + (L)*(ALPHA)*(Delta T)! = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =For TEMP=100 (E = 29.0e6 psi and alpha = 2.0e