美国麻省理工大学：《结构力学》英文版 Unit 9 Effects of the Environment

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

MT-1620 al.2002 -- However, if the bar is constrained, say at both ends Figure 9.2 Representation of constrained steel bar Then, as it is heated, the rod cannot lengthen. The thermal strain is the same as in the previous case but now the total strain is zero (i.e no physical deformation) Starting with(in one direction E=E+eT With E=0 Thus, the mechanical strain is the negative of the thermal strain Paul A Lagace @2001 Unit 9-p 6

MIT - 16.20 Fall, 2002 --> However, if the bar is constrained, say at both ends: Figure 9.2 Representation of constrained steel bar Then, as it is heated, the rod cannot lengthen. The thermal strain is the same as in the previous case but now the total strain is zero (i.e., no physical deformation). Starting with (in one direction): ε = εM + εT with: ε = 0 Thus, the mechanical strain is the negative of the thermal strain. Paul A. Lagace © 2001 Unit 9 - p. 6

MT-1620 al.2002 Stresses will arise due to the mechanical strain and these are the so-called“ thermal stresses” Due to equilibrium there must be a reaction at the boundaries must always have oda= Force for equilibrium) Think of this as a two-step process Figure 9. Representation of stresses due to thermal expansion as two-step process expands due to△T,E Reaction force of boundaries related to mechanical strain eM M etotal =0 ( no physical deformation) Paul A Lagace @2001 Unit 9-p 7

ε MIT - 16.20 Fall, 2002 Stresses will arise due to the mechanical strain and these are the so-called “thermal stresses”. Due to equilibrium there must be a reaction at the boundaries. (must always have ∫ σdA = Force for equilibrium) Think of this as a two-step process… Figure 9.3 Representation of stresses due to thermal expansion as two-step process expands due to ∆T, εT Reaction force of boundaries related to mechanical strain, εM εM = -εT ⇒ εtotal = 0 (no physical deformation) Paul A. Lagace © 2001 Unit 9 - p. 7

MT-1620 al.2002 Values of c te,s Note: a=a Anisotropic Materials 6 possibilities: a11, 022, a33, 012, 013, 023 →△ T can cause shear strains not true in"engineering " materials Orthotropic Materials 3 possibilities: a,1,022, 0.33 AT only causes extensional strains Notes Generally we deal with planar structures and are interested only in a11 and a22 2 If we deal with the material in other than the principal material axes, we can"have an a12 Transformation obeys same law as strain (it's a tensor) Paul A Lagace @2001 Unit 9-p 8

MIT - 16.20 Fall, 2002 Values of C.T.E.’s Note: αij = αji • Anisotropic Materials 6 possibilities: α11, α22, α33, α12, α13, α23 ⇒ ∆T can cause shear strains not true in “engineering” materials • Orthotropic Materials 3 possibilities: α11, α22, α33 ⇒ ∆T only causes extensional strains Notes: 1. Generally we deal with planar structures and are interested only in α11 and α22 2. If we deal with the material in other than the principal material axes, we can “have” an α12 Transformation obeys same law as strain (it’s a tensor). Paul A. Lagace © 2001 Unit 9 - p. 8

MT-1620 al.2002 2-D form a11, 022(in -plane values a12 =0 (in material axes) 3-D form k So, in describing deformation in some axis system at an angle a to the principal material axes Figure 9.4 Representation of 2-D axis transformation y2 e + cCw Paul A Lagace @2001 Unit 9-p 9

MIT - 16.20 Fall, 2002 2-D form: α˜ αβ = l lβγ ασ˜ ˜ ασγ ∗ ∗ α11, α22 (in- plane values) ∗ α12 = 0 (in material axes) 3-D form: αij = l l˜jl ˜ αkl ik So, in describing deformation in some axis system at an angle θ to the principal material axes….. Figure 9.4 Representation of 2-D axis transformation ~ y2 y1 θ ~ y2 θ + CCW y1 Paul A. Lagace © 2001 Unit 9 - p. 9

MT-1620 al.2002 C11 1≈c0s2901+sinb2 22= sin 0 a11 CoS 0 a22 12=c0ssn6(a22-a only exists if a11≠a22 isotropic→ no shea Isotropic Materials 1 value a is the same in all directions Typical Values for Materials Material C LE Steel 6Uns:×106F Aluminum 125 uinn/°F Titanium strain/°F Uni gr/Ep (along fibers) 0.2 Uni Gr/Ep(perpendicular to fibers) 16= strain/F Paul A Lagace @2001 Unit9-p. 10

MIT - 16.20 Fall, 2002 ∗ α˜ 11 = cos2 θ α11 + sin2 θ α∗22 ∗ α˜ 22 = sin2θ α11 + cos2θ α∗22 ∗ ∗ α˜ 12 = cos θ sinθ (α22 − α11) ∗ ∗ only exists if α11 ≠ α22 [isotropic ⇒ no shear] • Isotropic Materials 1 value: α is the same in all directions Typical Values for Materials: Units: x 10-6/°F µin/in/°F strain/°F ⇒ µstrain/°F Paul A. Lagace © 2001 Unit 9 - p. 10 Material C.T.E. Uni Gr/Ep (perpendicular to fibers) Uni Gr/Ep (along fibers) Titanium Aluminum Steel 16 -0.2 5 12.5 6