Are there any examples showing how ANSYS handles temperature-dependent
Young's moduli and coefficients of thermal expansion?
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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 simplified example, the strain reduces to just mechanical and thermal strain (i.e., there are no plastic, creep, or swelling strains present). Since we are looking at 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 Theory Reference for more details). Also, care should be used in interpreting the EPTO, or "total" strain value, as ANSYS defines EPTO to be the strain measured 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 of instantaneous (CTEX) values, so the code allows you to enter the data that way and then will convert it over to the secant form internally. If I knew the corresponding CTEX values for the ALPX values that I used, I would have put them in, but I didn't, so I just used the same numbers to show you that the code gives different results for CTEX as opposed to ALPX. Incidentally, this discussion applies to the other directions (Y and Z), as well. Mechanical strain is based on the current modulus and not some average value. Therefore, using the current temperature makes the most sense to me. If you wanted to,you could define the modulus such that it gave the desired average value you wanted at a given temperature, I suppose, but I am not sure why you would want to do that. If your structure is such that the thermal load is statically indeterminate, I guess you could possibly end up 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 plastic strains 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 /prep7 et,1,SOLID185 r,1 mptemp,1, 0, 100, 200, 300 ! define data 4 temperatures ... mpdata, ex, 1,1, 30.0e6, 29.0e6, 28.0e6, 27.0e6 ! temp-dep moduli, psi mpdata, 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 alphas block,0,1,0,1,0,1 esize,,1 vmesh,1 eplot fini /solu antype,static nsel,s,loc,x,0 d,all,ux,0.0 ! symmetry plane ... nsel,s,loc,y,0 d,all,uy,0.0 ! symmetry plane ... nsel,s,loc,z,0 d,all,uz,0.0 ! symmetry plane ... nsel,s,loc,z,1 sf,all,press,-1000.0 ! pulling pressure, psi nsel,all tref,100 ! reference temperature time,1.0 bf,all,temp,100.0 ! E = 29.0e6, alpha = 2.0e-6, delta T = 0 degrees solve save time,2.0 bf,all,temp,200.0 ! E = 28.0e6, alpha = 1.5e-6, delta T =100 degrees solve save time,3.0 bf,all,temp,300.0 ! E = 27.0e6, alpha = 1.0e-6, delta T = 200 degrees solve save fini /post1 nsel,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-04 For 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-03 For 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-04 For 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-03 For 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 Displacement Total 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 |
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