MIT - 16.0 Fall, 00 Unit 13 Review of Simple Beam Theory Readings: Review Unified Engineering notes on Beam Theory BMP 3.8, 3.9, 3.10 T & G 10-15 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace 001
MIT - 16.0 Fall, 00 IV. General Beam Theory Paul A. Lagace 001 Unit 13 -
MIT - 16.0 Fall, 00 We have thus far looked at: in-plane loads torsional loads In addition, structures can carry loads by bending. The -D case is a plate, the simple 1-D case is a beam. Let s first review what you learned in Unified as Simple Beam Theory (review of) Simple Beam Theory A beam is a bar capable of carrying loads in bending. The loads are applied transverse to its longest dimension. Assumptions: 1. Geometry Paul A. Lagace 001 Unit 13-3
MIT - 16.0 Fall, 00 Figure 13.1 General Geometry of a Beam a) long & thin l >> b, h b) loading is in z-direction c) loading passes through shear center no torsion/twist (we ll define this term later and relax this constraint.) d) cross-section can vary along x. Stress state a) σ yy, σ yz, σ xy = 0 no stress in y-direction b) σ xx >> σ zz σ xz >> σ only significant stresses are σ xx and σ xz zz Paul A. Lagace 001 Unit 13-4
MIT - 16.0 Fall, 00 Why is this valid? Look at moment arms: Note: there is a load in the z-direction to cause these stresses, but generated σ xx is much larger (similar to pressurized cylinder example) Figure 13. Representation of force applied in beam σ xx moment arm is order of (h) σ zz moment arm is order of (l) and l >> h σ xx >> σ zz for equilibrium Paul A. Lagace 001 Unit 13-5
MIT - 16.0 Fall, 00 3. Deformation Figure 13.3 Representation of deformation of cross-section of a beam deformed state (capital letters) o is at midplane define: w = deflection of midplane (function of x only) undeformed state (small letters) Paul A. Lagace 001 Unit 13-6
MIT - 16.0 Fall, 00 a) Assume plane sections remain plane and perpendicular to the midplane after deformation Bernouilli - Euler Hypothesis ~ 1750 Figure 13.4 b) For small angles, this implies the following for deflections: dw uxy (,, z) zφ z (13-1) dw total derivative φ = since it does not vary with y or z Representation of movement in x-direction of two points on same plane in beam u = -z sin φ Note direction of u relative to +x direction Paul A. Lagace 001 Unit 13-7
MIT - 16.0 Fall, 00 and for φ small: u = -z φ v( x, y, z) = 0 w( x, y, z ) w( x) (13 - ) Now look at the strain-displacement equations: ε xx = u dw x = z (13-3) v ε yy = = 0 y w ε zz = = 0 (no deformation through thickness) z u v ε xy = + = 0 y x v w ε yz = + = 0 z y u w w w ε xz = + = + = 0 z x x x Paul A. Lagace 001 Unit 13-8
MIT - 16.0 Fall, 00 Now consider the stress-strain equations (for the time being consider isotropic extend this to orthotropic later) σ xx ε = (13-4) xx E νσ xx ε = <-- small inconsistency with previous yy E νσ xx ε = <-- small inconsistency with previous zz E ε = xy ε yz = ε zx = 1 ( + ν) σ xy = 0 E 1 ( + ν) σ yz = 0 E 1 ( + ν) σ zx <-- inconsistency again! E We get around these inconsistencies by saying that ε yy, ε zz, ε xz are very small but not quite zero. This is an approximation. We will evaluate these later on. Paul A. Lagace 001 Unit 13-9
MIT - 16.0 Fall, 00 4. Equilibrium Equations Assumptions: a) no body forces b) equilibrium in y-direction is ignored c) x, z equilibrium are satisfied in an average sense So: σ x xx σ xz + = 0 (13-5) z 0 = 0 (y -equilibrium) σ x xz σ zz + = 0 (13-6) z Note, average equilibrium equations: [ Eq. (13 6) ] dy dz ds = p (13-6a) face z [ Eq. (13 5) ] dy dz dm = S (13-5a) face Paul A. Lagace 001 Unit 13-10
MIT - 16.0 Fall, 00 These are the Moment, Shear, Loading relations where the stress resultants are: h Axial Force F = h σ xx b dz (13-7) Shear Force Bending Moment h S = h σ xz b dz h M = h zσ xx b dz (13-8) (13-9) Figure 13.8 Representation of Moment, Shear and Loading on a beam [Force/Area] (F, S, M found from statics) cut beam and Paul A. Lagace 001 Unit 13-11
MIT - 16.0 Fall, 00 So the final important equations of Simple Beam Theory are: ε ε xx xx ds dm z dw = = σ xx = E = = p S u x (13-3) (13-4) (13-6a) (13-5a) --> How do these change if the material is orthotropic? We have assumed that the properties along x dominate and have ignored ε yy, etc. Thus, use E L in the above equations. But, approximation may not be as good since ε yy, ε zz, ε xz may be large and really not close enough to zero to be assumed approximately equal to zero Paul A. Lagace 001 Unit 13-1
MIT - 16.0 Fall, 00 Solution of Equations using (13-3) and (13-4) we get: dw σ xx = Eε xx = Ez (13-10) Now use this in the expression for the axial force of equation (13-7): h dw F = E h zb dz = E dw z h b = 0 h No axial force in beam theory (Note: something that carries axial and bending forces is known as a beam-column) Now place the stress expression (13-10) into the moment equation (13-9): h dw M = E h z b dz Paul A. Lagace 001 Unit 13-13
MIT - 16.0 Fall, 00 h definition: I = h z b dz moment of inertia of cross-section for rectangular cross-section: Thus: I = bh3 [length 4 ] h 1 dw M = E I (13-11) Moment - Curvature Relation b --> Now place equation (13-11) into equation (13-10) to arrive at: M σ xx = Ez EI Mz σ xx = (13-1) I Paul A. Lagace 001 Unit 13-14
MIT - 16.0 Fall, 00 Finally, we can get an expression for the shear stress by using equation (13-5): σ xz σ xx = (13-5) z x Multiply this by b and integrate from z to h/ to get: h h σ σ xz xx b dz = z z z x bdz h Mz b[σ xz (h ) σ xz (z)] = z x I bdz = 0 z dm (from boundary condition = I of no stress on top surface) = S This all gives: σ xz (13-13) = SQ Ib Paul A. Lagace 001 Unit 13-15
MIT - 16.0 Fall, 00 where: h Q = z b dz z = Moment of the area above the center Summarizing: function of z - - maximum occurs at z = 0 ds = p dm = S σ xx = Mz I σ xz = SQ Ib dw M = E I Paul A. Lagace 001 Unit 13-16
MIT - 16.0 Fall, 00 Notes: σ xx is linear through thickness and zero at midpoint σ xz has parabolic distribution through thickness with maximum at midpoint Usually σ xx >> σ zz Solution Procedure 1. Draw free body diagram. Calculate reactions 3. Obtain shear via (13-6a) and then σ xz via (13-13) 4. Obtain moment via (13-5a) and then σ xx via (13-1) and deflection via (13-11) NOTE: steps through 4 must be solved simultaneously if loading is indeterminate Notes: Same formulation for orthotropic material except Use E L Assumptions on ε αβ may get worse Can also be solved via stress function approach Paul A. Lagace 001 Unit 13-17
MIT - 16.0 Fall, 00 For beams with discontinuities, can solve in each section separately and join (match boundary conditions) Figure 13.6 Example of solution approach for beam with discontinuity dw ΕΙ dw A = M A ΕΙ = M B B ΕΙ w A =... + C 1 x + C ΕΙ w B =... + C 3 x + C 4 --> Subject to Boundary Conditions: @ x = 0, w = w A = 0 @ x = x 1, w A = w B displacements and dw A dw = B slopes match @ x = l, w = w B = 0 Paul A. Lagace 001 Unit 13-18