MIT - 16.20 Fall, 2002 Unit 6 Plane Stress and Plane Strain Readings : T & G 8, 9, 10, 11, 12, 14, 15, 16 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 There are many structural configurations where we do not have to deal with the full 3-D case. ? First let’s consider the models ? Let ’s then see under what conditions we can apply them A. Plane Stress This deals with stretching and shearing of thin slabs. Figure 6.1 Representation of Generic Thin Slab Paul A. Lagace ? 2001 Unit 6 - p . 2 MIT - 16.20 Fall, 2002 The body has dimensions such that h << a, b ( Key : where are limits to “ < < “??? We ’ll consider later) Thus, the plate is thin enough such that there is no variation of displacement (and temperature) with respect to y 3 (z). Furthermore, stresses in the z-direction are zero (small order of magnitude). Figure 6.2 Representation of Cross-Section of Thin Slab Paul A. Lagace ? 2001 Unit 6 - p . 3 MIT - 16.20 Fall, 2002 Thus, we assume: σ zz = 0 σ yz = 0 σ xz = 0 ? = 0 ? z So the equations of elasticity reduce to: Equilibrium ? σ 11 + ? σ 21 + f 1 = 0 (1) ? y 1 ? y 2 ? σ 12 + ? σ 22 + f 2 = 0 (2) ? y 1 ? y 2 (3rd equation is an identity) 0 = 0 (f 3 = 0) In general: ? σ βα + f α = 0 ? y β Paul A. Lagace ? 2001 Unit 6 - p . 4 MIT - 16.20 Fall, 2002 Stress-Strain (fully anisotropic ) Primary (in-plane) strains 1 ε 1 = E 1 [ σ 1 ? ν 1 2 σ 2 ? η 16 σ 6 ] (3) 1 ε 2 = E 2 [ ? ν 21 σ 1 + σ 2 ? η 2 6 σ 6 ] (4) 1 ε 6 = G 6 [ ? η 61 σ 1 ? η 6 2 σ 2 + σ 6 ] (5) Invert to get: * σ αβ = E αβσγ ε σγ Secondary (out-of-plane) strains ? (they exist, but they are not a primary part of the problem) 1 ε 3 = E 3 [ ? ν 31 σ 1 ? ν 3 2 σ 2 ? η 36 σ 6 ] Paul A. Lagace ? 2001 Unit 6 - p . 5 MIT - 16.20 Fall, 2002 1 ε 4 = G 4 [ ? η 41 σ 1 ? η 4 2 σ 2 ? η 46 σ 6 ] 1 ε 5 = G 5 [ ? η 51 σ 1 ? η 5 2 σ 2 ? η 56 σ 6 ] Note : can reduce these for orthotropic , isotropic (etc.) as before. Strain - Displacement Primary ε 11 = ? u 1 (6) ? y 1 ε 22 = ? u 2 (7) ? y 2 ε 12 = 1 ?? ? u 1 + ? u 2 ? ? (8) 2 ? ? y 2 ? y 1 ? Paul A. Lagace ? 2001 Unit 6 - p . 6 MIT - 16.20 Fall, 2002 Secondary ε 13 = 1 ?? ? u 1 + ? u 3 ? ? 2 ? ? y 3 ? y 1 ? ε 23 = 1 ?? ? u 2 + ? u 3 ? ? 2 ? ? y 3 ? y 2 ? ε 33 = ? u 3 ? y 3 Note : that for an orthotropic material ( ε 23 ) ( ε 13 ) ε 4 = ε 5 = 0 (due to stress-strain relations) Paul A. Lagace ? 2001 Unit 6 - p . 7 ??? MIT - 16.20 Fall, 2002 This further implies from above ? (since = 0 ) ? y 3 No in-plane variation ? u 3 = 0 ? y α but this is not exactly true ? INCONSISTENCY Why? This is an idealized model and thus an approximation. There are, in actuality, triaxial ( σ zz , etc.) stresses that we ignore here as being small relative to the in-plane stresses! (we will return to try to define “small ” ) Final note : for an orthotropic material, write the tensorial stress-strain equation as: 2-D plane stress σ αβ = ε σγ (, , αβ , σ , γ = 12 ) αβσγ ? E Paul A. Lagace ? 2001 Unit 6 - p . 8 MIT - 16.20 Fall, 2002 There is not a 1-to-1 correspondence between the 3-D E mnpq and the 2-D E * αβσ γ . The effect of ε 33 must be incorporated since ε 33 does not appear in these equations by using the ( σ 33 = 0) equation. This gives: ε 33 = f( ε αβ ) Also, particularly in composites, another “ notation ” will be used in the case of plane stress in place of engineering notation: subscript x = 1 = L (longitudinal) … along major axis change y = 2 = T (transverse) … along minor axis The other important “extreme ” model is … B. Plane Strain This deals with long prismatic bodies: Paul A. Lagace ? 2001 Unit 6 - p . 9 MIT - 16.20 Fall, 2002 Figure 6.3 Representation of Long Prismatic Body Dimension in z - direction is much, much larger than in the x and y directions L >> x, y Paul A. Lagace ? 2001 Unit 6 - p . 1 0 MIT - 16.20 Fall, 2002 ( Key again: where are limits to “>>”??? … we ’ l l consider later) Since the body is basically “ i nfinite ” along z, the important loads are in the x - y plane (none in z) and do not change with z: ? ? = = 0 ? y 3 ? z This implies there is no gradient in displacement along z, so (excluding rigid body movement): u 3 = w = 0 Equations of elasticity become: Equilibrium: Primary ? σ 11 + ? σ 21 + f 1 = 0 (1) ? y 1 ? y 2 ? σ 12 + ? σ 22 + f 2 = 0 (2) ? y 1 ? y 2 Paul A. Lagace ? 2001 Unit 6 - p . 1 1 MIT - 16.20 Fall, 2002 Secondary ? σ 13 + ? σ 23 + f 3 = 0 ? y 1 ? y 2 σ 13 and σ 23 exist but do not enter into primary consideration Strain - Displacement ε 11 = ? u 1 (3) ? y 1 ε 22 = ? u 2 (4) ? y 2 ε 12 = 1 ?? ? u 1 + ? u 2 ? ? (5) 2 ? ? y 2 ? y 1 ? Assumptions ?? ? = 0 , w = 0 ? ? give: ? ? y 3 ? ε 13 = ε 23 = ε 33 = 0 ( Plane strain ) Paul A. Lagace ? 2001 Unit 6 - p . 1 2 σσσ σσσ MIT - 16.20 Fall, 2002 Stress - Strain (Do a similar procedure as in plane stress) 3 Primary σ 11 = ... (6) σ 22 = ... (7) σ 12 = ... (8) Secondary σ 13 = 0 σ 23 = 0 orthotropic ( ≠ 0 for anisotropic ) σ 33 ≠ 0 INCONSISTENCY : No load along z, yet σ 33 ( σ zz ) is non zero. Why ? Once again, this is an idealization. Triaxial strains ( ε 33 ) actually arise. You eliminate σ 33 from the equation set by expressing it in terms of σ αβ via ( σ 33 ) stress-strain equation. Paul A. Lagace ? 2001 Unit 6 - p . 1 3 MIT - 16.20 Fall, 2002 SUMMARY Plane Stress Plane Strain Eliminate σ 33 from eq. Set by using σ 33 σ - ε eq . and expressing σ 33 in terms of ε αβ Eliminate ε 33 from eq . set by using σ 33 = 0 σ - ε eq . and expressing ε 33 in terms of ε αβ Note: σ 33 ε 33, u 3 Secondary Variable(s): ε αβ , σ αβ , u α ε αβ , σ αβ , u α Primary Variables: ε i3 = 0 σ i3 = 0 Resulting Assumptions: σ αβ only ? / ? y 3 = 0 σ 33 << σ αβ Loading: length (y 3 ) >> in-plane dimensions (y 1 , y 2 ) thickness (y 3 ) << in-plane dimensions (y 1 , y 2 ) Geometry: Paul A. Lagace ? 2001 Unit 6 - p . 1 4 MIT - 16.20 Fall, 2002 Examples Plane Stress: Figure 6.4 Pressure vessel (fuselage, space habitat) Skin in order of 70 MPa (10 ksi ) p o ≈ 70 kPa (~ 10 psi for living environment) ? σ zz << σ xx , σ yy , σ xy Paul A. Lagace ? 2001 Unit 6 - p . 1 5 MIT - 16.20 Fall, 2002 Plane Strain: Figure 6.5 Dams water pressure Figure 6.6 Solid Propellant Rockets high internal pressure Paul A. Lagace ? 2001 Unit 6 - p . 1 6 MIT - 16.20 Fall, 2002 but …when do these apply??? Depends on … ? loading ? geometry ? material and its response ? issues of scale ? how “good ” do I need the answer ?w hat are we looking for (deflection, failure, etc.) We ’ ve talked about the first two, let ’ s look a little at each of the last three: --> Material and its response ? Elastic response and coupling changes importance / magnitude of “primary” / “ secondary ” factors ( Key : are “primary ” dominating the response?) --> Issues of scale ? What am I using the answer for? at what level? ? Example : standing on table --overall deflection or reactions in legs are not dependent on way I stand (tip toe or flat foot) Paul A. Lagace ? 2001 Unit 6 - p . 1 7 MIT - 16.20 Fall, 2002 ? model of top of table as plate in bending is sufficient --stresses under my foot very sensitive to specifics (if table top is foam, the way I stand will determine whether or not I crush the foam) --> How “ good ” do I need the answer? ? In preliminary design, need “ballpark ” estimate; in final design, need “exact” numbers ? Example: as thickness increases when is a plate no longer in plane stress Paul A. Lagace ? 2001 Unit 6 - p . 1 8 MIT - 16.20 Fall, 2002 Figure 6.7 Representation of the “continuum” from plane stress to plane strain very thick a continuum very thin (plane stress) (plane strain) No line(s) of demarkation . Numbers approach idealizations but never get to it. Must use engineering judgment AND Clearly identify key assumptions in model and resulting limitations Paul A. Lagace ? 2001 Unit 6 - p . 1 9