MIT - 16.20 Fall, 2002 Unit 13 Review of Simple Beam Theory Readings : Review Unified Engineering notes on Beam Theory BMP 3.8, 3.9, 3.10 T & G 120-125 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 IV. General Beam Theory Paul A. Lagace ? 2001 Unit 13 - 2 MIT - 16.20 Fall, 2002 We have thus far looked at: ? in-plane loads ? torsional loads In addition, structures can carry loads by bending . The 2-D case is a plate , the simple 1-D case is a beam . Le t ’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 ? 2001 Unit 13 - 3 MIT - 16.20 Fall, 2002 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 2. Stress state a) σ yy , σ yz , σ xy = 0 ? no stress in y-direction b) σ xx >> σ zz σ xz >> σ zz ? only significant stresses are σ xx and σ xz ? Paul A. Lagace ? 2001 Unit 13 - 4 MIT - 16.20 Fall, 2002 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) Why is this valid? Look at moment arms: Figure 13.2 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 ? 2001 Unit 13 - 5 MIT - 16.20 Fall, 2002 3. Deformation Figure 13.3 Representation of deformation of cross-section of a beam deformed state (capital letters) undeformed state (small letters) o is at midplane define: w = deflection of midplane (function of x only) Paul A. Lagace ? 2001 Unit 13 - 6 MIT - 16.20 Fall, 2002 a) Assume plane sections remain plane and perpendicular to the midplane after deformation “Bernouilli - Euler Hypothesis ” ~ 1750 b) For small angles, this implies the following for deflections: dw ux y z (, ,) ≈? z φ ≈? z (13 - 1) dx total derivative ? φ = dw ? since it does not ? dx ? vary with y or z Figure 13.4 on same plane in beam Note direction of u relative to +x direction Representation of movement in x-direction of two points ? u = -z sin φ Paul A. Lagace ? 2001 Unit 13 - 7 MIT - 16.20 Fall, 2002 and for φ small: ? u = -z φ vx y z (, ,) = 0 (, ,) ≈ wx wx y z ( ) (13 - 2) Now look at the strain-displacement equations : 2 ? u dw ε xx = ? x = ? z dx 2 (13 - 3) ? v ε = = 0 yy ? y ? w ε = = 0 (no deformation through thickness) zz ? z ? u ? v ε = + = 0 xy ? y ? x ? v ? w ε = + = 0 yz ? z ? y ? u ? w ? w ? w ε = + = ? + = 0 xz ? z ? x ? x ? x Paul A. Lagace ? 2001 Unit 13 - 8 MIT - 16.20 Fall, 2002 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 = 21 + ν ) σ xy = 0 ( xy E ( ε yz = 21 + ν ) σ yz = 0 E ( ε zx = 21 + 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 approximatio n . We will evaluate these later on. Paul A. Lagace ? 2001 Unit 13 - 9 MIT - 16.20 Fall, 2002 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 ?σ xx xz + = 0 (13 - 5) ? x ? z 0 = 0 (y -equilibrium) ?σ ?σ xz zz + = 0 (13 - 6) ? x ? z Note , average equilibrium equations: ∫∫ [ Eq. ( 13 ? 6 ) ] dy dz ? dS = p (13 - 6a) face dx ∫∫ z [ Eq. ( 13 ? 5 ) ] d y d z ? dM = S (13 - 5a) face dx Paul A. Lagace ? 2001 Unit 13 - 1 0 MIT - 16.20 Fall, 2002 These are the Moment, Shear, Loading relations where the stress resultants are: Axial Force Shear Force Bending Moment h F = ∫ ? 2 h σ xx b d z 2 h S = ? ∫ ? 2 h σ xz b d z 2 h M = ? ∫ ? 2 h z σ xx b d z 2 (13 - 7) (13 - 8) (13 - 9) Figure 13.8 Representation of Moment, Shear and Loading on a beam [Force/Area] cut beam (F, S, M found from statics ) and Paul A. Lagace ? 2001 Unit 13 - 1 1 MIT - 16.20 Fall, 2002 So the final important equations of Simple Beam Theory are: (13 - 3) (13 - 4) (13 - 6a) (13 - 5a) ε ? ? xx z dw dx u x = ? = 2 2 ε σ xx xx E = dS dx p = dM dx S = --> 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 ? 2001 Unit 13 - 1 2 MIT - 16.20 Fall, 2002 Solution of Equations using (13 - 3) and (13 - 4) we get: 2 dw σ xx = E ε xx = ? E z dx 2 (13 - 10) Now use this in the expression for the axial force of equation (13 - 7): 2 h dw F = ? E dx 2 ∫ ? 2 h zb d z 2 h 2 dw z 2 ? 2 = ? E dx 2 2 b ? ? ? h = 0 2 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): 2 h dw 2 M = E dx 2 ∫ ? 2 h zb d z 2 Paul A. Lagace ? 2001 Unit 13 - 1 3 MIT - 16.20 Fall, 2002 h 2 definition: I = ∫ ? 2 h z b d z moment of inertia of 2 cross-section for rectangular cross-section: I = bh 3 [length 4 ] h 12 b Thus: 2 dw M = E I (13 - 11) dx 2 “Moment - Curvature Relation ” --> Now place equation (13 - 11) into equation (13 - 10) to arrive at: M σ = ? Ez xx EI Mz ? σ xx = ? (13 - 12) I Paul A. Lagace ? 2001 Unit 13 - 1 4 MIT - 16.20 Fall, 2002 Finally, we can get an expression for the shear stress by using equation (13 - 5): ?σ ?σ xz xx = ? (13 - 5) ? z ? x Multiply this by b and integrate from z to h/2 to get: h h ?σ ?σ ∫ xz xx z 2 b ? z dz = ? ∫ z 2 ? x bd z h ? ? Mz ? ? b [ σ xz ( h 2 ) ? σ xz ( z ) ] = ? ∫ z 2 ? x ? ? ? bd z I = 0 z dM (from boundary condition = ? I dx of no stress on top surface) = S (13 - 13) This all gives: ? σ = ? SQ xz Ib Paul A. Lagace ? 2001 Unit 13 - 1 5 MIT - 16.20 Fall, 2002 where: h Q = ∫ z 2 z b d z = Moment of the area above the center function of z - - maximum occurs at z = 0 Summarizing : dS = p dx dM = S dx Mz σ xx = ? I σ = ? SQ xz Ib 2 dw M = E I dx 2 Paul A. Lagace ? 2001 Unit 13 - 1 6 MIT - 16.20 Fall, 2002 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 2. 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 - 12) and deflection via (13 - 11) NOTE: steps 2 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 ? 2001 Unit 13 - 1 7 MIT - 16.20 Fall, 2002 ? 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 2 2 ΕΙ dw A = M A ΕΙ 2 B = M B dx 2 dx ΕΙ w A = ... + C 1 x + C 2 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 = dx dx @ x = l , w = w B = 0 Paul A. Lagace ? 2001 Unit 13 - 1 8