GM9_10Notes

M9 Shafts: Torsion of Circular Shafts Reading: Crandall, Dahl and Lardner 6.2, 6.3 A shaft is a structural member which is long and slender and subject to a torque (moment) acting about its long axis. We will only consider circular cross-section shafts in Unified. These have direct relevance to circular cross-section shafts such as drive shafts for gas turbine engines, propeller driven aircraft and helicopters (rotorcraft). However, the basic principles are more general and will provide you with a basis for understanding how structures with arbitrary cross-sections carry torsional moments. Torsional stiffness, and the shear stresses that arise from torsional loading are important for the design of aerodynamic surfaces such as wings, helicopter rotor blades and turbine fan blades. Modelling assumptions (a) Geometry (as for beam). Long slender, L >> r (b,h) Note: For the time being we will work in tensor notation since this is all about shear stresses and tensor notation will make the analysis more straightforward. Remember we can choose the system of notation, coordinates to make life easy for ourselves! (b) Loading Torque about x1 axis, T (units of Force x length). We may also want to consider the possibility of distributed torques (Force x length/unit length) (distributed aerodynamic moment along a wing, torques due to individual stages of a gas turbine) No axial loads (forces) applied to boundaries (on curved surfaces with radial normal, or on x1 face) 11 = 22 = 33 = 0 (c) Deformation -Cross sections rotate as rigid bodies through twist angle , varies with x1 (cf. beams – plane sections remain plane and perpendicular) - No bending or extensional deformations in x1 direction Cross-section: Torque, T right hand rule positive Positive x2, x3 quadrant u2 = r (x1 )sin u3 = +r (x1 ) cos where r = x22 + x 32 x sin = 3 r x cos = 2 r x22 + x 32 so u2 = u3 = x22 + x 32 (x1 ) x3 x22 + x 32 x2 (x1 ) x22 + x 32 = (x1 )x 3 = (x1 )x2 So u1 = 0 u2 = (x1 )x 3 u3 = (x1 )x2 (1) (2) ( 3) Governing Equations 11 = du1 =0 dx1 22 = du2 =0 dx2 33 = du 3 =0 dx 3 Consistent with assumption of no axial, or radial stresses, strains 12 = 1 ∂u1 ∂u 2 + = 2 ∂x2 ∂x1 13 = 1 ∂u1 ∂u3 1 d + = x2 2 ∂x3 ∂x1 2 dx1 23 = 1 ∂u 2 ∂u3 1 + = ( 2 ∂x3 ∂x2 2 1 d x3 2 dx1 (x1 ) + (x1 )) = 0 (4 ) (5) Next, apply constitutive laws, i.e. stress-strain relations (assume isotropic), remember we are working in tensor notation: mn = 23 mn , therefore need “2G” 2G = 23 = 0 2G 12 = 12 2G (6) 13 = 13 2G (7) Net moment due to shear stresses must equal resultant torque on section: equipollent torque, T = Apply equilibrium: i.e.: ( x2 13 x 3 12 )dx2 dx 3 ∂ mn + fn = 0 ∂x m ∂ 11 ∂ 21 ∂ 31 + + =0 ∂x1 ∂x 2 ∂x 3 ∂ 12 ∂ 22 ∂ 32 + + =0 ∂x1 ∂x 2 ∂x 3 ∂ 13 ∂ 23 ∂ 33 + + =0 ∂x1 ∂x 2 ∂x 3 (8) Retaining non-zero terms we obtain ∂ 21 ∂ 31 + =0 ∂x 2 ∂x 3 ∂ 12 =0 ∂x1 ∂ 13 =0 ∂x1 (9) (10) (11) Solution Go back to stress-displacement relationships (4,5, 6 and 7) 1 d x3 2 dx1 d 12 = Gx 3 dx1 12 = 2G 12 = 2G Similarly: 13 = 2G13 13 = 2G 1 d d x2 = Gx2 2 dx1 dx1 Substitute into equation for resultant moment (8) T= (x 2 = x22 T =G define J = 13 x 3 12 )dx2 dx3 d d + x32 dx 2dx3 dx1 dx1 d dx1 ( x22 + x23 )dA x22 + x 23 dA Polar 2nd moment of area = R4 for circular cross-section 2 Hence T = GJ d dx1 Torque-twist relation (note, we can compare T = GJ d ∂2 w for shafts with M = EI for beams) dx1 ∂x 2 Hence, relate stress to torque Tx3 12 = J Tx2 13 = J Express total shear as shear stress resultant, res = compare T 2 2 + = 12 13 J = Tr with J x 32 + x22 = xx = Tr J Mz I Model works well for circular cross-sections and cylindrical tubes J = J1 J2 Does not work for open sections: We can approximate for other sections, e.g. square cross-section, J = 0.141a4 . For a full treatment of torsion of slender members see 16.20. A note on distributed torques: Distributed torques are similar to distributed loads on beams: Consider equilibrium of a differential element, length dx1, with a distributed torque t (Nm/m) applied. Leads to an increase in the resultant internal torque, T from T to T+dT. T+ M x1 = 0 T + t(x1 )dx1 + T + dT = t(x 1) dx1 c.f. dT dx dx1 1 dT dx dx1 1 dS = q( x) dx Can also consider macroscopic equilibrium of shaft fixed at one end with a uniform distributed torque/length, t, applied to it: t/unit length FBD L TA + tdx1 = 0 0 TA = tL M10 Introduction to Structural Instability Reading Crandall, Dahl and Lardner: 9.2, 9.3 Elastic instabilities, of which buckling is the most important example, are a key limitation on structural integrity. The key feature of an elastic instability is the transition from a stable mode of deformation with increasing applied load to an unstable one, resulting in collapse (loss of load carrying capability) and possibly failure of the structure. Examples of elastic collapse are the buckling of bars in a truss under compressive load, the failure of columns under compressive load, the failure of the webs of “I” beams in shear, the failure of fuselage and wing skin panels in shear and many others. The only particular case we will consider here in Unified is the failure of a bar or column loaded in axial compression, however, as for the other slender members we have considered, the basic ideas will apply to more complex structures. A structure is in stable equilibrium if, for all possible (small)displacements/deformations, a restoring force arises. Before considering the case of a continuous structure we will consider a case in which we separate the stiffness of the structure from the geometry of the structure. Introductory Example: Rigid, massless bar with a torsional spring at one end, stiffness, kt, which is also pinned. . The bar is loaded by a pair of horizontal and vertical forces at the free end, P1, P2. The bar undergoes a small angular displacement, . Independent applied loads, P1, P2 is small – exaggerated in the figure Draw free body diagram: Horizontal equilibrium: HA=-P1 Vertical equilibrium: VA=P2 Taking moments, counterclockwise positive: M LP1 cos (cos M And Hence: P2 L sin = 0 1 sin LP1 P2 L = 0 = k1t L P2 = 0 k t M LP K P2 L P1 = t L Effective Stiffness - (note that it includes the load, P2) ) i.e. P1 = keff = Rearranging gives: P1 kt L P2 Hence: for P1 0 = 0 for P2 < k t / L = • for P2 K t / L i.e. if P2 ≥ NOTE: Kt L static instability, i.e., spring cannot provide in restoring moment If P2 negative - i.e. upward -stiffness increases But if P1and P2 removed or reduced, spring will allow bar to spring back to original configuration Plot Load vs. Displacement a) for case when P1=0. Obtain “bifurcation behavior” Bifurcation point P2 = Kt L b) case P1>0 Next time we will apply these idea to a column, i.e. a continuous structure with a continuous distribution of stiffness. Need to think about what is the relevant structural stiffness, i.e. the equivalent of (K t L ) in the spring/rigid rod system above.