Unit 13 Review of Simple Beam Theory

Transcription

1 MIT 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 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace 001

2 MIT Fall, 00 IV. General Beam Theory Paul A. Lagace 001 Unit 13 -

3 MIT 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

4 MIT 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

5 MIT 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

6 MIT Fall, 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

7 MIT 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

8 MIT 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

9 MIT 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

10 MIT Fall, 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

11 MIT 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

12 MIT 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

13 MIT 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

14 MIT 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

15 MIT 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

16 MIT 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

17 MIT 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

18 MIT 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 x = 0, w = w A = x = x 1, w A = w B displacements and dw A dw = B slopes x = l, w = w B = 0 Paul A. Lagace 001 Unit 13-18