MT-1620 al.2002 Unit 10 St Venant Torsion Theory Readings Rivello 81,82,84 T&G 101.104,105,106 Paul A Lagace Ph D Professor of aeronautics Astronautics and engineering systems Paul A. Lagace @2001
MIT - 16.20 Fall, 2002 Unit 10 St. Venant Torsion Theory Readings: Rivello 8.1, 8.2, 8.4 T & G 101, 104, 105, 106 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace © 2001
MT-1620 al.2002 Ⅲ Torsion Paul A. Lagace @2001 Unit 10-p. 2
MIT - 16.20 Fall, 2002 III. Torsion Paul A. Lagace © 2001 Unit 10 - p. 2
MT-1620 Fall 2002 We have looked at basic in-plane loading. Lets now consider a second"building block of types of loading: basic torsion There are 3 basic types of behavior depending on the type of cross-section Solid cross-sections “ classical” solution technique via stress functions 2. Open, thin-walled sections Membrane Analogy Paul A. Lagace @2001 Unit 10-p. 3
MIT - 16.20 Fall, 2002 We have looked at basic in-plane loading. Let’s now consider a second “building block” of types of loading: basic torsion. There are 3 basic types of behavior depending on the type of cross-section: 1. Solid cross-sections “classical” solution technique via stress functions 2. Open, thin-walled sections Membrane Analogy Paul A. Lagace © 2001 Unit 10 - p. 3
MT-1620 al.2002 3. Closed thin-walled sections Bredt's formula In Unified you developed the basic equations based on some broad assumptions. Let's Be a bit more rigorous Explore the limitations for the various approaches Better understand how a structure "resists"torsion and the resulting deformation Learn how to model general structures by these three basic approaches Look first at Paul A. Lagace @2001 Unit 10-p. 4
MIT - 16.20 Fall, 2002 3. Closed, thin-walled sections Bredt’s Formula In Unified you developed the basic equations based on some broad assumptions. Let’s… • Be a bit more rigorous • Explore the limitations for the various approaches • Better understand how a structure “resists” torsion and the resulting deformation • Learn how to model general structures by these three basic approaches Look first at Paul A. Lagace © 2001 Unit 10 - p. 4
MT-1620 al.2002 Classical (St. Venant's)Torsion Theory Consider a long prismatic rod twisted by end torques Tin-lbs」[m-n Figure 10.1 Representation of general long prismatic rod T Length(l)>> dimensions in x and y directions g Do not consider how end torque is applied (St. Venant's principle Paul A. Lagace @2001 Unit 10-p. 5
MIT - 16.20 Fall, 2002 Classical (St. Venant’s) Torsion Theory Consider a long prismatic rod twisted by end torques: T [in - lbs] [m - n] Figure 10.1 Representation of general long prismatic rod Length (l) >> dimensions in x and y directions Do not consider how end torque is applied (St. Venant’s principle) Paul A. Lagace © 2001 Unit 10 - p. 5
MT-1620 al.2002 Assume the following geometrical behavior a) Each cross-section(@ each z) rotates as a rigid body( no distortion"of cross-section shape in x, y) b Rate of twist, k= constant c)Cross-sections are free to warp in the z-direction but the warping is the same for all cross-sections This is the" St. Venant Hypothesis warping"= extensional deformation in the direction of the axis about which the torque is applied Given these assumptions, we see if we can satisfy the equations of elasticity and B.C.·s. SEMI-INVERSE METHOD Consider the deflections Assumptions imply that at any cross-section location Z do z=kz dz careful a constant Rivello uses o! rate of twist (define as0@z=0) Paul A. Lagace @2001 Unit 10-p. 6
MIT - 16.20 Fall, 2002 Assume the following geometrical behavior: a) Each cross-section (@ each z) rotates as a rigid body (No “distortion” of cross-section shape in x, y) b) Rate of twist, k = constant c) Cross-sections are free to warp in the z-direction but the This is the “St. Venant Hypothesis” warping is the same for all cross-sections “warping” = extensional deformation in the direction of the axis about which the torque is applied Given these assumptions, we see if we can satisfy the equations of elasticity and B.C.’s. ⇒ SEMI-INVERSE METHOD Consider the deflections: Assumptions imply that at any cross-section location z: dα α = dz z = k z (careful! a constant Rivello uses φ!) rate of twist (define as 0 @ z = 0) Paul A. Lagace © 2001 Unit 10 - p. 6
MT-1620 al.2002 Figure 10.2 Representation of deformation of cross-section due to torsion (for small a) undeformed position This results in consider direction of +u u(X,y,z)=r(inβ V(X,y,z)=r(cosβ) W(X,y,2)=W(X,y) independent of z Paul A. Lagace @2001 Unit 10-p. 7
α ⇒ MIT - 16.20 Fall, 2002 Figure 10.2 Representation of deformation of cross-section due to torsion (for small α) undeformed position This results in: consider direction of + u u (x, y, z) = rα (-sin β) v (x, y, z) = rα (cos β) w (x, y, z) = w (x, y) ⇒ independent of z! Paul A. Lagace © 2001 Unit 10 - p. 7
MT-1620 al.2002 We can see that sin阝β This gives u(x,y,z=-y kZ (10-1) V( X,y,z)=XkZ W(x,y, z)=Wx, y) Paul A. Lagace @2001 Unit 10-p.8
MIT - 16.20 Fall, 2002 We can see that: r = y 2 x + 2 sinβ = y r x cosβ = r This gives: u (x, y, z) = -y k z (10 - 1) v (x, y, z) = x k z (10 - 2) w (x, y, z) = w (x, y) (10 - 3) Paul A. Lagace © 2001 Unit 10 - p. 8
MT-1620 al.2002 Next look at the Strain-Displacement equations 0 du consider: u exists, but 0 v exists but 0 No extensional strains in torsion if cross-sections are free to warp Paul A. Lagace @2001 Unit 10-p. 9
MIT - 16.20 Fall, 2002 Next look at the Strain-Displacement equations: ∂u εxx = = 0 ∂x ∂v εyy = = 0 ∂y ∂w εzz = = 0 ∂z ∂u (consider: u exists, but ∂x = 0 v exists, but ∂v = 0) ∂y ⇒ No extensional strains in torsion if cross-sections are free to warp Paul A. Lagace © 2001 Unit 10 - p. 9
MT-1620 Fall 2002 du dv zk+zk=0 xy dy dx = cross-section does not change shape(as assumed!) dv dw kX W dz dy (10-4) N dw (10-5) Now the Stress-Strain equations let's first do isotropic VOn+σ ZZ V+6 yEI yy Vo、+0 0 Paul A. Lagace @2001 Unt10-p.10
MIT - 16.20 Fall, 2002 ∂u ∂v εxy = + = − zk + zk = 0 ∂y ∂x ⇒ cross - section does not change shape (as assumed!) ∂v ∂w ∂w εyz = ∂z + ∂y = kx + ∂y (10 - 4) ∂w ∂u ∂w εzx = ∂x + ∂z = − ky + ∂x (10 - 5) Now the Stress-Strain equations: let’s first do isotropic 1 εxx = E [σxx − ν(σyy + σzz )] = 0 εyy = 1 [σyy − ν(σxx + σzz )] = 0 E 1 εzz = E [σzz − ν(σxx + σyy )] = 0 ⇒ σxx, σyy, σzz = 0 Paul A. Lagace © 2001 Unit 10 - p. 10
