Unit 6 Plane Stress and Plane Strain

Size: px

Start display at page:

Download "Unit 6 Plane Stress and Plane Strain"

Edward Fox
7 years ago
Views:

1 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

2 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 Unit 6 - p. 2

First let s consider the models Let s then see under what conditions we can

3 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 Unit 6 - p. 3

displacement (and temperature) with respect to y 3 (z).

4 Thus, we assume: σ zz = 0 σ yz = 0 σ xz = 0 z = 0 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 Unit 6 - p. 4

5 Stress-Strain (fully anisotropic) Primary (in-plane) strains 1 ε 1 = [σ 1 ν 12 σ 2 η 16 σ 6 ] (3) E 1 ε 2 = ε 6 = 1 [ ν 21 σ 1 + σ 2 η 26 σ 6 ] E 2 (4) 1 [ η 61 σ 1 η 62 σ 2 + σ 6 ] G 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 = E3 [ ν 31 σ 1 ν 32 σ 2 η 36 σ 6 ] Unit 6 - p. 5

(5) Invert to get: * σ αβ = E αβσγ ε σγ Secondary (out-of-plane) strains (they exist, but

6 ε 4 = ε 5 = Primary 1 [ η 41 σ 1 η 42 σ 2 η 46 σ 6 ] G 4 1 [ η 51 σ 1 η 52 σ 2 η 56 σ 6 ] G 5 Strain - Displacement Note: can reduce these for orthotropic, isotropic (etc.) as before. ε 11 = ε 22 = ε 12 = u 1 y 1 (6) u 2 y 2 (7) 1 u 1 + u 2 2 y 2 y 1 (8) Unit 6 - p. 6

7 Secondary ε 13 = ε 23 = 1 u 1 + u 3 2 y 3 y 1 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) Unit 6 - p. 7

8 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 = E ε σγ (,, αβσγ α β, σ, γ = 1 2) Unit 6 - p. 8

) stresses that we ignore here as being small relative to the in-plane stresses!

9 There is not a 1-to-1 correspondence between the 3-D E mnpq and the 2-D E * αβσγ. 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 change x = 1 = L (longitudinal) along major axis y = 2 = T (transverse) along minor axis The other important extreme model is B. Plane Strain This deals with long prismatic bodies: Unit 6 - p. 9

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

10 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 Unit 6 - p. 10

11 (Key again: where are limits to >>??? we ll consider later) Since the body is basically infinite along z, the important loads are in the x - y plane (none in z) and do not change with z: y 3 = = 0 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 Unit 6 - p. 11

plane (none in z) and do not change with z: y 3 = = 0 z This implies there is no gradient in displacement

12 Secondary σ 13 + σ 23 + f 3 = 0 y 1 y 2 σ and σ exist but do not enter into primary consideration Strain - Displacement ε 11 = ε 22 = ε 12 = u 1 y 1 (3) u 2 y 2 (4) 1 u 1 + u 2 2 y 2 y 1 (5) Assumptions = 0, w = 0 give: y 3 ε 13 = ε 23 = ε 33 = 0 (Plane strain) Unit 6 - p. 12

12 = u 1 y 1 (3) u 2 y 2 (4) 1 u 1 + u 2 2 y 2 y 1 (5) Assumptions =

13 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 σ 33 0 orthotropic ( 0 for anisotropic) INCONSISTENCY: No load along z, yet σ 33 (σσ 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. Unit 6 - p. 13

33 (σσ zz ) is non zero. Why? Once again, this is an idealization. Triaxial strains (ε 33 ) actually arise.

14 SUMMARY Plane Stress Plane Strain Geometry: thickness (y 3 ) << in-plane dimensions (y 1, y 2 ) length (y 3 ) >> in-plane dimensions (y 1, y 2 ) Loading: Resulting Assumptions: Primary Variables: Secondary Variable(s): Note: σ 33 << σ αβ σ i3 = 0 ε αβ, σ αβ, u α ε 33, u 3 Eliminate ε 33 from eq. set by using σ 33 = 0 σ - ε eq. and expressing ε 33 in terms of ε αβ σ αβ only / y 3 = 0 ε i3 = 0 ε αβ, σ αβ, u α σ 33 Eliminate σ 33 from eq. Set by using σ 33 σ - ε eq. and expressing σ 33 in terms of ε αβ Unit 6 - p. 14

αβ, σ αβ, u α ε 33, u 3 Eliminate ε 33 from eq. set by using σ 33 = 0 σ - ε eq.

15 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 Unit 6 - p. 15

16 Plane Strain: Figure 6.5 Dams water pressure Figure 6.6 Solid Propellant Rockets high internal pressure Unit 6 - p. 16

17 but when do these apply??? Depends on loading geometry material and its response issues of scale how good do I need the answer what 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) Unit 6 - p. 17

) 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

18 --> How good do I need the answer? 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) 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 Unit 6 - p. 18

specifics (if table top is foam, the way I stand will determine whether or not I crush the foam) In

19 Figure 6.7 Representation of the continuum from plane stress to plane strain very thin (plane stress) a continuum very thick (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 Unit 6 - p. 19

(plane stress) a continuum very thick (plane strain) No line(s) of demarkation.

Unit 3 (Review of) Language of Stress/Strain Analysis

Unit 3 (Review of) Language of Stress/Strain Analysis Readings: B, M, P A.2, A.3, A.6 Rivello 2.1, 2.2 T & G Ch. 1 (especially 1.7) Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering