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