Coupled thermoelasticity: quasi-static vs dynamic response under slow and rapid heating
Coupled transient thermoelasticityPASS · 0.57%A cantilever beam heated on its top face by a smooth C² ramp 10s³ − 15s⁴ + 6s⁵ over R = 12 fundamental periods T₁ shows negligible inertial amplification, so dynamic Newmark integration (Mode B) matches quasi-static thermoelasticity (Mode A) within 0.57 %. Conversely, under rapid heating (R = 0.5·T₁), inertial effects amplify the peak deflection by 45.9 %.A cantilever beam heated on its top face by a smooth C² ramp 10s³ − 15s⁴ + 6s⁵ over R = 12 fundamental periods T₁ shows negligible inertial amplification, so dynamic Newmark integration (Mode B) matches quasi-static thermoelasticity (Mode A) within 0.57 %. Con… show full text
- Tolerance:
- 5%
Problem statement
- Geometry
- Cantilever beam 40 × 2 × 2 mm; quadratic tetrahedra (tet10), element size 1 mm (3070 elements).
- Material
- Steel: E = 200 GPa, ν = 0.3, ρ = 7800 kg/m³, α = 1.2 × 10⁻⁵ K⁻¹, c_p = 460 J/(kg·K).
- Boundary conditions
- Surface number 1 clamped; surface number 6 heated to 100 °C along a C² ramp over R = 12·T₁ (slow check) and R = 0.5·T₁ (sensitivity control). Both Mode A (quasistatic) and Mode B (dynamic Newmark) run on the identical mesh.
- Measured quantity
- Peak tip deflection in Mode B vs Mode A for slow heating R = 12·T₁ (target ratio 1.0) and dynamic amplification for rapid heating R = 0.5·T₁ (> 20 % difference).
WebCAE Terminal
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units mm
material E=200000000000 nu=0.3 density=7800 kappa=29349.3 specific_heat=460 alpha=0.000012
primitive box dx=40 dy=2 dz=2
mesh tet size=1 order=2
fix surface number 1
thermal surface number 6 temperature=100
solve thermoelastic transient mode=dynamic dt=0.0000244502 tend=0.0146701 t0=0 theta=0.5 amp_temperature="0:0,0.005868:0.5,0.011736:1" probes="40,1,1" frames=2 outputs=600
probe transient_peak dir=z mode=1 id="peak"Results comparison
| Reference value | w_А = −18.907 µm (R = 12·T₁)|w_Б − w_А|/|w_А| > 20 % при R = 0.5·T₁ |
| Our result | w_Б = −19.014 µm (dev. 0.567 %)при R = 0.5·T₁ разница 45.91 % |
| Deviation | 0.57% |
| Tolerance | 5% |
| Status | PASS · 0.57% |
| Mesh details | tet10, size 1 mm (3070 tets); 600 steps of Δt = T₁/40, θ = 0.5 |
Updated: 2026-10-27 · version not recorded