Thermal stress and strains

Temperature change in a material leaves it with
mechanical expansion & significance
Changes in material properties.

Expansion due to heat, induce
Strains internally.
Hence stress induced

1. THERMAL STRESS & STRAINS By DEEPAK ROTTI Roll No: 017 KLS’s G.I.T BELGAUM

2. Temperature changes occur: High speed travelling objects, parts & vehicles. Heat engines and rocket engines. Boilers, furnace etc. Effect of temperature change : Mechanical expansion & significance Changes in material properties. Expansion due to heat: Analogy: Strains induced internally. Hence stress induced .

3. Thermal stress effects: Block of material subjected to an increase in temperature B’ Changes in temperature produce expansion or contraction of materials and result in thermal strains and thermal stresses T hermal strain ε T is proportional to the temperature change Δ T : ε T = α ( Δ T ) (Where α is coefficient of thermal expansion) S ign convention for thermal strains : E xpansion is positive and contraction is negative

4. Relation between stress (Ϭ) and change in temperature (ϪT). Suppose we have a bar subjected to an axial load. ε = σ / E Also, we have an identical bar subjected to a temperature change Δ T. ε T = α ( Δ T) Equating the above two strains we will get: σ = E α ( Δ T) Increase in length of a prismatic bar due to a uniform increase in temperature 1 2 3

5. Thermo-elastic stress & strain relations: Fig:A Fig:B Uniform rise in temperature (unconstrained) Hence uniform change in dimension. Non-uniform change in temperature and hence uneven stress development T=T(t, x, y, z)

6. Total strain = normal strain ( α T) + strain due to stress components €x = б x – ν ( б y + б z) + Δ α T E €y = б y – ν ( б x + б z) + Δ α T E €z = б z – ν ( б x + б y) + Δ α T E } Total strains at each point

7. Relations in terms of strains: б x = λ e + 2 μ €x – (3 λ +2 μ ) * α T б y = λ e + 2 μ €y – (3 λ +2 μ ) * α T б z = λ e + 2 μ €z – (3 λ +2 μ ) * α T حxy = μγ xy ; حyz = μγ yz ; حzx = μγ zx Equations of equilibrium: Where ν = possions ratio λ = lames constant μ = G= modulus of rigidity

8. General results When the temperature distribution is known, the problem of thermoelsticity is used to determining 15 functions : 6 stress components 6 strain components 3 displacement components So as to satisfy 15 equations : 3 equilibrium equations 6 stress- strain relations 6 strain – displacement relations Displacement boundary conditions (linear distribution): T(x, y, z, t) = a(t) + b(t)x + c(t)y d(t)z

9. € б Δ α ν λ μ Thank you

10. € б Δ α ν λ μ