Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems Unit 14 Behavior of General ( including Unsymmetric Cross-section) Beams Readings : Rivello 7.1 - 7.5, 7.7, 7.8 T & G 126 Unit 14 - 2 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Earlier looked at Simple Beam Theory in which one considersa beam in the x-z plane with the beam along the x-directionand the load in the z-direction : Figure 14.1 Representation of Simple Beam ? Loading can be in any direction? Can resolve the loading to consider transverse loadings p y (x) and p z (x); and axial loading p x (x) ? Include a temperature distribution T(x, y, z) Now look at a more general case: Unit 14 - 3 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Figure 14.2 Representation of General Beam Maintain several of the same definitions for a beam and basic assumptions. ? Geometry : length of beam (x-dimension) greater than y and z dimensions ? Stress State : σ xx is the only “important ” stress; σ xy and σ xz found from equilibrium equations, but are secondary in importance ? Deformation : plane sections remain plane and perpendicular to the midplane after deformation ( Bernouilli -Euler Hypothesis) Unit 14 - 4 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Definition of stress resultants Consider a cross-section along x: Figure 14.3 Representation of cross-section of general beam Place axis @ centerof gravity of section where: These are resultants ! Sd A zx z = ? ∫∫ σ Fd A xx = ∫∫ σ Sd A yx y = ? ∫∫ σ Mz d A yx x = ? ∫∫ σ My d A zx x = ? ∫∫ σ Unit 14 - 5 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 The values of these resultants are found from statics in terms of the loading p x , p y , p z , and applying the boundary conditions of the problem Deformation Look at the deformation. In the case of Simple Beam Theory, had: uz dw dx = ? where u is the displacement along the x-axis.Now must add two other contributions ….. Figure 14.4 Representation of deformation in Simple Beam Theory This comes from the picture: for small angles Unit 14 - 6 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 1. Have the same situation in the x-y plane Figure 14.5 Representation of bending displacement in x-y plane where v is the displacement in the y-direction 2. Allow axial loads, so have an elongation in the x-direction due tothis. Call this u 0 : By the same geometricalarguments used previously Unit 14 - 7 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Figure 14.6 Representation of axial elongation in x-z plane u 0 , v, w are the deformations of the midplane Thus: ux y z u y dvdx z dw dx (, , ) = ?? 0 bendingaboutz-axis bendingabouty-axis vx y z vx (, , ) ( ) = wx y z wx (, , ) ( ) = v and w are constant at any cross-section location, x Unit 14 - 8 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Stress and Strain From the strain-displacement relation, get: ε ?? xx u x du dx y dvdx z dw dx == + ? ?? ? ?? ? + ? ?? ? ?? ? 0 2 2 2 2 (these become total derivatives as there is no variation of the displacement in y and z) for functional ease, write: f du dx 1 0 = f dv dx 2 2 2 = ? f dwdx 3 2 2 = ?Caution : Rivello uses C 1 , C 2 , C 3 . These are not constants, so use f i ? f i (x) (functions of x) Unit 14 - 9 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Thus: ε xx ff y f z =+ + 12 3 Then use this in the stress-strain equation ( orthotropic or “ lower ” ) : ε σ α xx xx E T =+ ? (include temperature effects) Note : “ ignor e ” thermal strains in y and z. These are of “secondary” importance. Thus: σε α xx xx EE T = ?? and using the expression for ε x : σα xx Ef f y f z E T =+ + () ? 12 3 ? Can place this expression into the expression for the resultants (force and moment) to get: Unit 14 - 1 0 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Fd A f E d A f E y d A xx == + ∫∫ ∫∫ ∫∫ σ 12 + ? ∫∫ ∫∫ fE z d A E T d A 3 α ? ? == + ∫∫ ∫∫ ∫∫ My d A f E y d Af E y d A zx x σ 12 2 + ? ∫∫ ∫∫ fE y z d A ET y d A 3 α ? ? == + ∫∫ ∫∫ ∫∫ Mz d A f E z d Af E y z d A yx x σ 12 + ? ∫∫ ∫∫ fE z d A E T z d A 3 2 α ? (Note : f 1 , f 2 , f 3 are functions of x and integrals are in dy and dz, so these come outside the integral sign). Solve these equations to determine f 1 (x), f 2 (x), f 3 (x): Note : Have kept the modulus, E, within the integral since will allow it to vary acrossthe cross-section Unit 14 - 1 1 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Orthotropic Beams : Same comments as applied to Simple Beam Theory. The main consideration is the longitudinal modulus, sothese equations can be applied. Using this in the equations for the resultants, we get: Modulus-Weighted Section Properties/Areas Introduce “ modulus weighted area ” : dA E E dA * = 1 (vary in y and z) where: A * = modulus weighted area E = modulus of that areaE 1 = some reference value of modulus Unit 14 - 1 2 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 FE T d A E f d A f y d A f z d A += + + ?? ?? ? ?? ?? ? ∫∫ ∫∫ ∫∫ ∫∫ α ? 11 2 3 ** * ? += + + ?? ?? ? ?? ?? ? ∫∫ ∫∫ ∫∫ ∫∫ ME T y d A E f y d Af y d A f y z d A z α ? 11 2 2 3 ** * ? += + + ?? ?? ? ?? ?? ? ∫∫ ∫∫ ∫∫ ∫∫ ME T z d A E f z d Af y z d A fz d A y α ? 11 2 3 2 ** * Now define these “ modulus-weighted ” section properties: modulus-weighted area dA A ** ∫∫ = yd A y A ** * ∫∫ = zd A z A ** * ∫∫ = yd A I z 2* ∫∫ = * zd A I y 2* ∫∫ = * yz d A I yz * ∫∫ = * modulus-weighted moment of inertia about z-axismodulus-weighted moment of inertia about y-axismodulus-weighted product of inertia (“ cross ” moment of inertia) Unit 14 - 1 3 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Also have “Thermal Force s ” and “ Thermal Moment s ”. These have the same “ units ” as forces and moments but are due to thermal effects and can then be treated analytically as forces and moments: FE T d A T = ∫∫ α ? ME T z d A y T = ? ∫∫ α ? ME T y d A z T = ? ∫∫ α ? Note : Cannot use the modulus-weighted section properties since α may also vary in y and z along with E. Figure 14.7 Representation of general beam cross-section with pieces with different values of modulus In the definition of the section properties, have used a y* and z*. Theseare the location of the “modulus-weighted centroid ” referred to some coordinate system – – Unit 14 - 1 4 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 1 A yd A y * ** ∫∫ = 1 A zd A z * ** ∫∫ = These become 0 if one uses the modulus-weighted centroid as the origin (Note: like finding center of gravity but use E rather than ρ ) If one uses the modulus-weighted centroid as the origin, the equations reduce to: FF F E f A T TOT + () == 11 * ? + () = ? =+ () MM M E f I f I zz T z TOT zy z 12 3 ** ? + () = ? =+ () MM M E f I f I yy T y TOT yz y 12 3 ** (Note: Rivello uses F * , M y * , M z * for F T0T , M y T0T , M z T0T ) The modulus-weighted centroid is defined by: Unit 14 - 1 5 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Recall that: f du dx 1 0 = f dvdx 2 2 2 = ? f dw dx 3 2 2 = ? Motivation for “ modulus-weighted ’ section properties A beam may not have constant material properties through thesection. Two possible ways to vary:1. Continuous variation The modulus may be a continuous function of y and z: E = E(y, z) Example : Beam with a large thermal gradient and four different properties through the cross-section Unit 14 - 1 6 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 A composite beam which, although it ’ s made of the same material, has different modulus, E x , through-the-thickness as the fiber orientation varies from ply to ply. Figure 14.8 Representation of cross-section of laminated beam with different modulus values through the thickness (symmetric) A method of putting material to its best use is called: 2. Stepwise variation Unit 14 - 1 7 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Figure 14.9 Representation of selective reinforcement of an I-beam Aluminum I -beam (E = 10 Msi) Unidirectional Graphite/Epoxycap reinforcements Furthest from neutral axis ???? best resistance to bending (E = 20 Msi) Using aluminum as the reference, analyze as follows Figure 14.10 Representative cross-section with aluminum as base use E 1 to analyze “ Selective Reinforcement ” Representation good onlyin direction parallel to axisabout which I is taken. Unit 14 - 1 8 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 E E b msimsi bb 1 2010 2 == Principal Axes of structural cross-section: There is a set of y, z axes such that the product of inertiais zero. These are the principal axes (section has axes ofsymmetry) Ι yz * ( ) Figure 14.11 Representation of principal axes of structural cross-section modulus-weightedcentroid Ι yz * ≠ 0 Ι yz * = 0 (use Mohr’ s circle transformation) Unit 14 - 1 9 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 f F EA du dx TOT 1 1 0 == * f M EI dvdx z TOT z 2 1 2 2 = ? = ? * f M EI dw dx y TOT y 3 1 2 2 = ? = ? * These equations can be integrated to find the deflections u 0 , v and w These expressions for the f i can be placed into the equation for σ xx to obtain: σα xx TOT z TOT z y TOT y E E F A M I y M I zE T = ?? ? ?? ?? ? ?? ?? ? 1 1 ** * ? where y,z are principal axes for the section If analysis is conducted in the principal axes, the equations reduce to: Unit 14 - 2 0 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 (no change) f F EA du dx TOT 1 1 0 == * f IM I M EI I I dvdx yz TOT yz y TOT yz y z 2 1 2 2 2 = ? + ? () = ? ** ** * f IM I M EI I I dw dx zy TOT yz z TOT yz y z 3 1 2 2 2 = ? + ? () = ? ** ** * Note : If then both w and v are present for M y or M z only Ι yz * ≠ 0 Figure 14.12 Representation of deflection of cross-section not in principal axes total deflection If the axes are not principal axes ( ), have: Ι yz * ≠ 0 Unit 14 - 2 1 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 In this case, the expression for the stress is rather long: σ xx TOT yz TOT yz y TOT yz y z E E F A IM I M y II I = ? ? [] ? ?? ?? ? 1 2 * ** ** * ? ? [] ? ? ?? ?? ? IM I M z II I ET zy TOT yz z TOT yz y z ** ** * 2 1 α ? “Engineering Beam Theory ” (Non-Principal Axes) Analysis is good for high aspect ratio structure (e.g. a wing) Figure 14.13 Representation of wing as beam σ xx Unit 14 - 2 2 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Note : this analysis neglects the effect of the axial Force F on the Bending Moment. This became important as thedeflection w (or v) becomes large: Figure 14.14 Representation of large deflection when axial force and bending deflection couple F 0 = axial force PrimaryBendingMoment SecondaryMoment Secondary moment known as “ membrane effec t ”. Can particularly become important if F o is near buckling load (will talk about when talk about beam-column) MM w F due to p z = ? 0 Unit 14 - 2 3 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Shear StressesThe shear stresses ( σ xy and σ xz ) can be obtained from the equilibrium equations: ?σ ? ?σ ? ?σ ? xy xz xx yz x += ? ?σ ? xy x = 0 ?σ ? xz x = 0 Figure 14.15 Representation of cross-section of general beam Unit 14 - 2 4 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 These shear stresses (called “ transverse” shear stresses) cause “small ” additional shearing contributions to deflections Figure 14.16 Representation of pure bending and pure shearing of a beam Plane sections remainplane and perpendicularto midplane Plane sections remainplane but n o t perpendicular tomidplane Pure Bending -->Pure Shearing --> Consider a beam section under “pure shearing ”… Figure 14.17 Representation of deformation of beam cross-section under pure shearing Unit 14 - 2 5 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 γ ? ? ?? σ xz xz w x u zG =+ = engineering shear strain Average over cross-section: ? ? w x ? ? ? ? σ w x w x dA dA G dA A S GA ave xz z ?? ? ?? ? = ?? ? ?? ? == ? ∫∫ ∫∫ ∫∫ 1 Actually, from energy considerations , one should average: ? ? ? ? ? ? w x w x dA w x dA S GA ave z e ?? ? ?? ? = ?? ? ?? ? ≈? ∫∫ ∫∫ 2 “effective area ” For a Rectangular Cross-Section: A e ≈ 0.83 A Then, “pure shearing ” deflections, w s , governed by: S z Unit 14 - 2 6 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 ww w TB s = ? dw dx S GA s e = ? evaluated from boundary conditions w S GA C s e x = ? + ∫ 0 1 The total beam deflection is the sum of the two contributions: total bending deflectionfrom shearing deflectionfrom EI dw dx M B 2 2 = GA dw dx S e S = ? Ordinarily, w s is small for ordinary rectangular beams (and can be ignored). But, for thin-walled sections, w s can become important (worse for composites since G xz << E x ) WW W TB S =+ + Unit 14 - 2 7 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 In addition to “bending ” and “shearing ” , the section may also twist through an angle α Figure 14.18 Representation of twisting of beam cross-section However, there exists a Shear Center for every section. If the load is applied at the shear center, the section translates but does nottwist. (Note : shear center not necessarily center of gravity or centroid ) Unit 14 - 2 8 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Figure 14.19 Representation of some beam cross-sections with various locations of center of gravity and shear center If this condition is not met, then generally bending and twisting willcouple . But there is a class of cross-sections (thin-walled) where bending and shearing/torsion can be decoupled. Will pursue thisnext . Wrap-up discussion by considering examples of common cross-sectionswith principal axes aligned such that I yz = 0 (see Handout #4a) These are in contrast to common cross-sections not principal axes ( I yz ≠ 0) center of gravity shear center Unit 14 - 2 9 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 Figure 14.19 Some cross-sections generally not in principal axes TriangleAngleWingSection Unit 14 - 3 0 Paul A. Lagace ? 2001 MIT - 16.20 Fall, 2002 --> FinallyWhat are the limitations to the Engineering Beam Theory as developed? ? Shear deflections small (can get first order cut at this) ? No twisting (load along shear center) -- otherwise torsion andbending couple ? Deflections small o– No moment due to axial load ( P w ) o– Angles small such that sin φ ≈ φ ? --> will consider next order effect when discuss buckling and postbuckling ? --> consideration will stiffen (membrane effect) structure ? Did not consider ε zz (Poisson ’ s effect)