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L13 — Thermal Stress（热应力）🔴 样卷未考
Q99 🔴 热应变公式
EN: The thermal strain caused by a temperature change Δ T \Delta T Δ T
ε i j 0 = α Δ T δ i j \boxed{\varepsilon_{ij}^0 = \alpha \Delta T \, \delta_{ij}}
ε ij 0 = α Δ T δ ij
where α \alpha α δ i j \delta_{ij} δ ij
ε 0 = α Δ T { 1 1 1 0 0 0 } \boldsymbol{\varepsilon}^0 = \alpha \Delta T\begin{Bmatrix} 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{Bmatrix}
ε 0 = α Δ T ⎩ ⎨ ⎧ 1 1 1 0 0 0 ⎭ ⎬ ⎫
Thermal strain is purely volumetric (equal normal strains, no shear strains).
中文： 热应变 ε i j 0 = α Δ T δ i j \varepsilon_{ij}^0=\alpha\Delta T\delta_{ij} ε ij 0 = α Δ T δ ij α Δ T \alpha\Delta T α Δ T
Q100 🔴 自由热膨胀是否产生应力？
EN: No. Free (unconstrained) thermal expansion does NOT produce thermal stress. The body simply expands/contracts without developing internal forces. Thermal stress arises ONLY when:
Thermal expansion is constrained (e.g., fixed-end bar)
There is a temperature gradient within the body (differential expansion)
The structure is made of materials with different CTE (α \alpha α
Classic example: A bar fixed at both ends develops compressive stress σ = − E α Δ T \sigma = -E\alpha\Delta T σ = − E α Δ T
中文： 自由热膨胀不产生 热应力。只有在约束、温度梯度、或材料热膨胀系数差异时，才产生热应力。经典例子：两端固支杆均匀升温 → 压应力 σ = − E α Δ T \sigma=-E\alpha\Delta T σ = − E α Δ T
Q101 🔴 热弹性本构关系
EN: With thermal effects, the constitutive relationship becomes:
σ = D ( ε − ε 0 ) \boxed{\boldsymbol{\sigma} = \mathbf{D}(\boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}^0)}
σ = D ( ε − ε 0 )
where ε \boldsymbol{\varepsilon} ε total strain (from both mechanical and thermal effects) and ε 0 \boldsymbol{\varepsilon}^0 ε 0 thermal strain . The quantity ε − ε 0 = ε m \boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}^0 = \boldsymbol{\varepsilon}_m ε − ε 0 = ε m mechanical strain — only this portion produces stress.
中文： 热弹性本构：σ = D ( ε − ε 0 ) \boldsymbol{\sigma}=\mathbf{D}(\boldsymbol{\varepsilon}-\boldsymbol{\varepsilon}^0) σ = D ( ε − ε 0 ) ε \boldsymbol{\varepsilon} ε ε 0 \boldsymbol{\varepsilon}^0 ε 0 ε m \boldsymbol{\varepsilon}_m ε m
Q102 🔴 热等效节点力
EN: The thermal effect enters the FEM equations through the thermal equivalent nodal force vector :
f T e = ∫ V e B T D ε 0 d V \boxed{\mathbf{f}_T^e = \int_{V_e} \mathbf{B}^T \mathbf{D} \boldsymbol{\varepsilon}^0 \, dV}
f T e = ∫ V e B T D ε 0 d V
The element equation becomes: k e d e = f e + f T e \mathbf{k}_e \mathbf{d}_e = \mathbf{f}^e + \mathbf{f}_T^e k e d e = f e + f T e
where f e \mathbf{f}^e f e f T e \mathbf{f}_T^e f T e K d = F + F T \mathbf{Kd} = \mathbf{F} + \mathbf{F}_T Kd = F + F T
中文： 热等效节点力 f T e = ∫ B T D ε 0 d V \mathbf{f}_T^e = \int\mathbf{B}^T\mathbf{D}\boldsymbol{\varepsilon}^0 dV f T e = ∫ B T D ε 0 d V
Q103 🔴 一维杆热等效节点力
EN: For a 1D bar element with constant E , A , α , Δ T E, A, \alpha, \Delta T E , A , α , Δ T
f T = E A α Δ T { − 1 1 } \boxed{\mathbf{f}_T = EA\alpha\Delta T \begin{Bmatrix} -1 \\ 1 \end{Bmatrix}}
f T = E A α Δ T { − 1 1 }
These forces act in opposite directions at the two nodes — like a pair of forces trying to expand the element. If the bar is constrained (both ends fixed), these forces produce reactions and internal compressive stress.
中文： 一维杆单元热等效节点力 = E A α Δ T { − 1 , 1 } T EA\alpha\Delta T\{-1,1\}^T E A α Δ T { − 1 , 1 } T
Q104 🔴 两端固支均匀升温杆的应力
EN: For a bar fixed at both ends with uniform temperature increase Δ T \Delta T Δ T
Free thermal strain: ε 0 = α Δ T \varepsilon^0 = \alpha\Delta T ε 0 = α Δ T
Due to fixed ends, total strain: ε = 0 \varepsilon = 0 ε = 0
Mechanical strain: ε m = ε − ε 0 = − α Δ T \varepsilon_m = \varepsilon - \varepsilon^0 = -\alpha\Delta T ε m = ε − ε 0 = − α Δ T
Thermal stress: σ = − E α Δ T \boxed{\sigma = -E\alpha\Delta T} σ = − E α Δ T
End reaction: F = E A α Δ T F = EA\alpha\Delta T F = E A α Δ T
The negative sign indicates compressive stress — the bar wants to expand but is constrained.
中文： 两端固支均匀升温 → σ = − E α Δ T \sigma=-E\alpha\Delta T σ = − E α Δ T F = E A α Δ T F=EA\alpha\Delta T F = E A α Δ T
Q105 🔴 平面应力 vs 平面应变的热应变
EN: The thermal strain vectors differ between plane stress and plane strain:
Plane stress: ε 0 = α Δ T { 1 1 0 } \boldsymbol{\varepsilon}^0 = \alpha\Delta T\begin{Bmatrix} 1 \\ 1 \\ 0 \end{Bmatrix} ε 0 = α Δ T ⎩ ⎨ ⎧ 1 1 0 ⎭ ⎬ ⎫ ε z \varepsilon_z ε z
Plane strain: ε 0 = ( 1 + ν ) α Δ T { 1 1 0 } \boldsymbol{\varepsilon}^0 = (1+\nu)\alpha\Delta T\begin{Bmatrix} 1 \\ 1 \\ 0 \end{Bmatrix} ε 0 = ( 1 + ν ) α Δ T ⎩ ⎨ ⎧ 1 1 0 ⎭ ⎬ ⎫ ( 1 + ν ) (1+\nu) ( 1 + ν )
The ( 1 + ν ) (1+\nu) ( 1 + ν ) ε z = 0 \varepsilon_z = 0 ε z = 0
中文： ⚠️ 高频易错——平面应力热应变：α Δ T { 1 , 1 , 0 } T \alpha\Delta T\{1,1,0\}^T α Δ T { 1 , 1 , 0 } T ( 1 + ν ) α Δ T { 1 , 1 , 0 } T (1+\nu)\alpha\Delta T\{1,1,0\}^T ( 1 + ν ) α Δ T { 1 , 1 , 0 } T ( 1 + ν ) (1+\nu) ( 1 + ν ) ε z = 0 \varepsilon_z=0 ε z = 0
Q106 🟡 轴对称热应变
EN: For axisymmetric problems, the thermal strain includes the hoop direction:
ε 0 = α Δ T { 1 1 1 0 } corresponding to { ε r , ε z , ε θ , γ r z } T \boldsymbol{\varepsilon}^0 = \alpha\Delta T\begin{Bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{Bmatrix} \quad \text{corresponding to } \{\varepsilon_r, \varepsilon_z, \varepsilon_\theta, \gamma_{rz}\}^T
ε 0 = α Δ T ⎩ ⎨ ⎧ 1 1 1 0 ⎭ ⎬ ⎫ corresponding to { ε r , ε z , ε θ , γ rz } T
All three normal strain components get the same thermal strain α Δ T \alpha\Delta T α Δ T
中文： 轴对称热应变 = α Δ T { 1 , 1 , 1 , 0 } T \alpha\Delta T\{1,1,1,0\}^T α Δ T { 1 , 1 , 1 , 0 } T α Δ T \alpha\Delta T α Δ T
Q107 🟡 热应力题标准解题步骤
EN: Standard procedure for thermal stress FEM problems:
Determine temperature change Δ T \Delta T Δ T
Write thermal strain: ε 0 = α Δ T \boldsymbol{\varepsilon}^0 = \alpha\Delta T ε 0 = α Δ T
Write thermoelastic constitutive law: σ = D ( ε − ε 0 ) \boldsymbol{\sigma} = \mathbf{D}(\boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}^0) σ = D ( ε − ε 0 )
Write shape functions N \mathbf{N} N B \mathbf{B} B
Compute stiffness: k e = ∫ B T D B d V \mathbf{k}_e = \int \mathbf{B}^T\mathbf{D}\mathbf{B}\,dV k e = ∫ B T DB d V
Compute thermal force: f T = ∫ B T D ε 0 d V \mathbf{f}_T = \int \mathbf{B}^T\mathbf{D}\boldsymbol{\varepsilon}^0\,dV f T = ∫ B T D ε 0 d V
Assemble: K d = F + F T \mathbf{Kd} = \mathbf{F} + \mathbf{F}_T Kd = F + F T
Apply BCs and solve for d \mathbf{d} d
Post-process stress: σ = D ( B d − ε 0 ) \boldsymbol{\sigma} = \mathbf{D}(\mathbf{Bd} - \boldsymbol{\varepsilon}^0) σ = D ( Bd − ε 0 )
中文： 9步走——定ΔT→写热应变→写本构→写N和B→算刚度→算热力→装配→加BC求解→后处理应力。关键是第6步和第9步，多了热力项和减热应变操作。
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