MT-1620 al.2002 Unit 11 Membrane analogy for Torsion) Readings Rivello 83.8.6 T&g 107,108,109,110,112,113,114 Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A Lagace @2001
MIT - 16.20 Fall, 2002 Unit 11 Membrane Analogy (for Torsion) Readings: Rivello 8.3, 8.6 T & G 107, 108, 109, 110, 112, 113, 114 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 Fa.2002 For a number of cross-sections we cannot find stress functions. However, we can resort to an analogy introduced by Prandtl(1903) Consider a membrane under pressure p, Membrane". structure whose thickness is small compared to surface dimensions and it( thus)has negligible bending rigidity(e.g. soap bubble) membrane carries load via a constant tensile force along itself N B. Membrane is 2-D analogy of a string (plate is 2-d analogy of a beam) Stretch the membrane over a cutout of the cross-sectional shape in the X-y plane Figure 11.1 Top view of membrane under pressure over cutout x需 membrane covering a CL Paul A Lagace @2001 Unit 11-p 2
MIT - 16.20 Fall, 2002 For a number of cross-sections, we cannot find stress functions. However, we can resort to an analogy introduced by Prandtl (1903). Consider a membrane under pressure pi “Membrane”: structure whose thickness is small compared to surface dimensions and it (thus) has negligible bending rigidity (e.g. soap bubble) ⇒ membrane carries load via a constant tensile force along itself. N.B. Membrane is 2-D analogy of a string (plate is 2-D analogy of a beam) Stretch the membrane over a cutout of the cross-sectional shape in the x-y plane: Figure 11.1 Top view of membrane under pressure over cutout membrane covering a cutout Paul A. Lagace © 2001 Unit 11 - p. 2
MT-1620 al.2002 n= constant tension force per unit length [lbs/in[/ ook at this from the side igure 11.2 side view of membrane under pressure over cutout [ib/:] [N/m] Assume: lateral displacements( w) are small such that no appreciable changes in n occur We want to take equilibrium of a small element dwdW assume small angles ax'ay Paul A Lagace @2001 Unit 11-p 3
MIT - 16.20 Fall, 2002 N = constant tension force per unit length [lbs/in] [N/M] Look at this from the side: Figure 11.2 Side view of membrane under pressure over cutout Assume: lateral displacements (w) are small such that no appreciable changes in N occur. We want to take equilibrium of a small element: ∂w ∂w (assume small angles ∂x , ∂y ) Paul A. Lagace © 2001 Unit 11 - p. 3
MT-1620 al.2002 Figure 11.3 Representation of deformation of infinitesimal element of membrane Z X Look at side view(one side Figure 11.4 Side view of deformation of membrane under pressure Z A hdy- Note: we have similar picture in the x-z plane Paul A Lagace @2001 Unit 11-p 4
MIT - 16.20 Fall, 2002 Figure 11.3 Representation of deformation of infinitesimal element of membrane y x z Look at side view (one side): Figure 11.4 z y Side view of deformation of membrane under pressure Note: we have similar picture in the x-z plane Paul A. Lagace © 2001 Unit 11 - p. 4
MT-1620 al.2002 We look at equilibrium in the z direction Take the z-components of N e. g + dw z-component=-Nsin hy note +z direction for small angle dw dw sin dy dy dw →z- component=-N dy(acts over dx face Paul A Lagace @2001 Unit 11-p 5
MIT - 16.20 Fall, 2002 We look at equilibrium in the z direction. Take the z-components of N: e. g. w z-component = −N sin ∂∂y note +z direction for small angle: sin ∂w ≈ ∂w ∂y ∂y ∂w ⇒ z-component = −N ∂y (acts over dx face) Paul A. Lagace © 2001 Unit 11 - p. 5
MT-1620 a.2002 With this established, We get 0 +】0=B=8mb+N+m w d dy w dw+d wdx dx dx dx Eliminating like terms and canceling out dxdy gives d21 P1+ N N d-w Governing partia Differential x 2 dy n\ Equ uation for deflection w. of a membrane Boundary Condition: membrane is attached at boundary, so W=0 along contour Exactly the same as torsion problem Paul A. Lagace @2001 Unit 11-p 6
MIT - 16.20 Fall, 2002 With this established, we get: ∂w ∂w ∂2 w + ∑Fz = 0 ⇒ pi dxdy − N dx + N + 2 dydx ∂y ∂y ∂y ∂w ∂w ∂2 w − N dy + N + 2 dxdy = 0 ∂x ∂x ∂x Eliminating like terms and canceling out dxdy gives: ∂2 w ∂2 w pi + N ∂y2 + N ∂x2 = 0 Governing Partial ⇒ ∂ ∂ ∂ ∂ 2 2 2 2 w x w y p N i + = − Differential Equation for deflection, w, of a membrane Boundary Condition: membrane is attached at boundary, so w = 0 along contour ⇒ Exactly the same as torsion problem: Paul A. Lagace © 2001 Unit 11 - p. 6
MT-1620 al.2002 Torsion Membrane Parti Differential V2φ=2GkVW=-p/N Equation Boundary Condition p=0 on contour W=0 on contour Membrane Torsion Analogy dw 0 zy 0 Volume ∫ waxy Paul A Lagace @2001 Unit 11-p. 7
MIT - 16.20 Fall, 2002 Torsion Membrane Partial Differential ∇2 φ = 2Gk ∇2 w = – pi / N Equation Boundary φ = 0 on contour w = 0 on contour Condition Analogy: Membrane Torsion w → φ p → - k i N → 1 2G → ∂ ∂ w x ∂ ∂ = φ σ x zy → ∂ ∂ = − φ σ y zx ∂ ∂ w y Volume = ∫∫ wdxdy → − Τ2 Paul A. Lagace © 2001 Unit 11 - p. 7
MT-1620 al.2002 Note: for orthotropic, would need a membrane to give different N's in different directions in proportion to G and g Membrane analogy only applies to isotropic materials This analogy gives a good physical picture for o Easy to visualize deflections of membrane for odd shapes Figure 11.5 Representation of o and thus deformations for various closed cross sections under torsion etc Can use(and people have used) elaborate soap film equipment and measuring devices (See Timoshenko, Ch. 1 Paul A Lagace @2001 Unit 11-p 8
φ MIT - 16.20 Fall, 2002 Note: for orthotropic, would need a membrane to give different N’s in different directions in proportion to Gxz and Gyz ⇒ Membrane analogy only applies to isotropic materials • This analogy gives a good “physical” picture for φ • Easy to visualize deflections of membrane for odd shapes Figure 11.5 Representation of φ and thus deformations for various closed cross-sections under torsion etc. Can use (and people have used) elaborate soap film equipment and measuring devices (See Timoshenko, Ch. 11) Paul A. Lagace © 2001 Unit 11 - p. 8
MT-1620 Fall 2002 From this, can see a number of things Location of maximum shear stresses(at the maximum slopes of the membrane Torque applied ( volume of membrane) External corners do not add appreciability to the bending rigidity eliminate these Figure 116E Representation of effect of external corners external corner → about the same Fillets(i.e.@ internal corners)eliminate stress concentrations Paul A Lagace @2001 Unit 11-p9
MIT - 16.20 Fall, 2002 From this, can see a number of things: • Location of maximum shear stresses (at the maximum slopes of the membrane) • Torque applied (volume of membrane) • “External” corners do not add appreciability to the bending rigidity (J) ⇒ eliminate these: Figure 11.6� Representation of effect of external corners external corner ⇒ about the same • Fillets (i.e. @ internal corners) eliminate stress concentrations Paul A. Lagace © 2001 Unit 11 - p. 9
MT-1620 al.2002 Figure 11.7 Representation of effect of internal corners high stress relieved stress concentration concentration To illustrate some of these points lets consider specifically Paul A Lagace @2001 Unt11-p.10
MIT - 16.20 Fall, 2002 Figure 11.7 Representation of effect of internal corners relieved stress high stress concentration concentration To illustrate some of these points let’s consider specifically… Paul A. Lagace © 2001 Unit 11 - p. 10