MT-1620 al.2002 Unit 9 Effects of the environment Readings Rivello 36.3.7 T&g Ch. 13 (as background, sec. 154 speciTically Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 9 Effects of the Environment Readings: Rivello 3.6, 3.7 T & G Ch. 13 (as background), sec. 154 specifically Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 al.2002 Thus far we have discussed mechanical loading and the stresses and strains caused by that We noted however, that the environment can have an effect on the behavior of materials and structures Lets first consider Temperature and its effects 2 basic effects expansion /contraction change of material properties Look, first. at the formel Concept of Thermal Stresses and strains Materials and structures expand and contract as the temperature changes. Thus (△T) k temperature change thermal strain coefficient of thermal expansion( C T.E. )units: degrees Paul A Lagace @2001 Unit 9-p 2
MIT - 16.20 Fall, 2002 Thus far we have discussed mechanical loading and the stresses and strains caused by that. We noted, however, that the environment can have an effect on the behavior of materials and structures. Let’s first consider: Temperature and Its Effects 2 basic effects: • expansion / contraction • change of material properties Look, first, at the former: Concept of Thermal Stresses and Strains Materials and structures expand and contract as the temperature changes. Thus: εT = α (∆ T) thermal temperature change strain coefficient of thermal expansion (C.T.E.) units: 1 degrees Paul A. Lagace © 2001 Unit 9 - p. 2
MT-1620 Fall 2002 If these thermal expansions /contractions are resisted by some means then thermal stresses"can arise. However thermal stresses"is a misnomer, they are really stresses due to thermal effects"- stresses are always mechanical (we'll see this via an example) . Consider a 3-d generic material Then we can write △T 9-1 ,j=1, 2, 3(as before) a 2nd order tensor The total strain of a material is the sum of the mechanical strain and the thermal strain mechanical (9-2) total Paul A Lagace @2001 Unit 9-p 3
MIT - 16.20 Fall, 2002 If these thermal expansions / contractions are resisted by some means, then “thermal stresses” can arise. However, “thermal stresses” is a misnomer, they are really… “stresses due to thermal effects” -- stresses are always “mechanical” (we’ll see this via an example) --> Consider a 3-D generic material. Then we can write: T εij = αij ∆T (9 1− ) i, j = 1, 2, 3 (as before) αij = 2nd order tensor The total strain of a material is the sum of the mechanical strain and the thermal strain. mechanical thermal M T εij = εij + εij (9 2 − ) total Paul A. Lagace © 2001 Unit 9 - p. 3
MT-1620 Fall 2002 Actual)total strain ei that which you actually measure;the physical deformation of the part thermal strain(Ej ) directly caused by temperature differences M mechanical strain( ij): that part of the strain which is directly related to the stress Relation of mechanical strain to stress is kiKi compliance Substituting this in the expression for total strain(equation 9-2)and using the expression for thermal strain(equation 9-1), we get kk+Qn△T jK=E-i△T We can multiply both sides by the inverse of the compliance that is merely the elasticities S Paul A Lagace @2001 Unit 9-p 4
MIT - 16.20 Fall, 2002 • (“actual”) total strain (εij): that which you actually measure; the physical deformation of the part T • thermal strain ( εij ): directly caused by temperature differences M • mechanical strain (εij ): that part of the strain which is directly related to the stress Relation of mechanical strain to stress is: Mεij = Sijkl σkl compliance Substituting this in the expression for total strain (equation 9-2) and using the expression for thermal strain (equation 9-1), we get: εij = Sijkl σkl + αij ∆ T ⇒ Sijkl σkl = εij − αij ∆ T We can multiply both sides by the inverse of the compliance…that is merely the elasticities: −1 Sijkl = Eijkl Paul A. Lagace © 2001 Unit 9 - p. 4
MT-1620 al.2002 →0-E-EwAT This is the same equation as we had before except we have the thermal terms EkO△T >so how does a thermal stress"arise? Consider this example: If you have a steel bar lying on a table and heat it, it will expand Since it is unconstrained it expands freely and no stresses occur That is, the thermal strain is equal to the total strain thus, the mechanical strain is zero and thus the thermal stress is zero Figure 9.1 Free thermal expansion of a steel bar Paul A Lagace @2001 Unit 9-p 5
MIT - 16.20 Fall, 2002 ⇒ σkl = Eijkl εij − Eijkl αij ∆T This is the same equation as we had before except we have the thermal terms: Eijkl αij ∆ T --> so how does a “thermal stress” arise? Consider this example: If you have a steel bar lying on a table and heat it, it will expand. Since it is unconstrained it expands freely and no stresses occur. That is, the thermal strain is equal to the total strain. Thus, the mechanical strain is zero and thus the “thermal stress” is zero. Figure 9.1 Free thermal expansion of a steel bar Paul A. Lagace © 2001 Unit 9 - p. 5