Unit 14 Behavior of General (including Unsymmetric Cross-section) Beams

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1 Unit 14 Behavior of General (including Unsymmetric Cross-section) Beams Readings: Rivello , 7.7, 7.8 T & G 16 Paul A. Lagace, Ph.D. Professor of Aeronautics & Astronautics and Engineering Systems

2 Earlier looked at Simple Beam Theory in which one considers a beam in the x-z plane with the beam along the x-direction and the load in the z-direction: Figure 14.1 Representation of Simple Beam Now look at a more general case: 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) Unit 14 -

3 Figure 14. 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 14-3

4 Definition of stress resultants Consider a cross-section along x: Figure 14.3 Representation of cross-section of general beam Place center of gravity of section where: F S S y z = σ xx da M = σ y = σ xx z da xy da Mz = σ xx y da da These are resultants! = σ xz Unit 14-4

5 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: u = z dw where u is the displacement along the x-axis. This comes from the picture: Figure 14.4 Representation of deformation in Simple Beam Theory for small angles Now must add two other contributions.. Unit 14-5

6 1. Have the same situation in the x-y plane Figure 14.5 Representation of bending displacement in x-y plane By the same geometrical arguments used previously where v is the displacement in the y-direction. Allow axial loads, so have an elongation in the x-direction due to this. Call this u 0 : Unit 14-6

7 Figure 14.6 Representation of axial elongation in x-z plane u 0, v, w are the deformations of the midplane Thus:u ( x, y, z) = u y dv 0 v ( x, y, z) = v ( x) w ( x, y, z) = w ( x) bending about z-axis z dw bending about y-axis v and w are constant at any cross-section location, x Unit 14-7

8 Stress and Strain From the strain-displacement relation, get: ε xx u du = = + y 0 x dv + z dw for functional ease, write: du0 f1 = dv f = f 3 dw = (these become total derivatives as there is no variation of the displacement in y and z) Caution: Rivello uses C 1, C, C 3. These are not constants, so use f i f i (x) (functions of x) Unit 14-8

9 Thus: ε xx = f + f y + f z 1 3 Then use this in the stress-strain equation (orthotropic or lower ): ε Thus: xx σ xx = + α T E (include temperature effects) Note: ignore thermal strains in y and z. These are of secondary importance. σ = Eε Eα T xx xx and using the expression for ε x : ( ) σxx = E f1 + f y + f3z Eα T Can place this expression into the expression for the resultants (force and moment) to get: Unit 14-9

10 σ xx 1 F = da = f E da + f Ey da σ xx 1 + f Ez da E T da 3 α M = yda = f Ey da + f Ey da z σ xx f Eyz da E Ty da 3 α M = zda = f Ez da + f Eyz da y + f Ez da E α Tz da (Note: f 1, f, 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 (x), f 3 (x): Note: Have kept the modulus, E, within the integral since will allow it to vary across the cross-section Unit 14-10

11 Orthotropic Beams: Same comments as applied to Simple Beam Theory. The main consideration is the longitudinal modulus, so these equations can be applied. Modulus-Weighted Section Properties/Areas Introduce modulus weighted area : da where: = E E da 1 (vary in y and z) A = modulus weighted area E = modulus of that area E 1 = some reference value of modulus Using this in the equations for the resultants, we get: Unit 14-11

12 F + Eα T da = E1 f1 da + f y da + f3 z da Mz + Eα Ty da = E1 f1 y da + f y da + f3 yz da My + Eα Tz da = E1 f1 z da + f yz da + f3 z da Now define these modulus-weighted section properties: da = A modulus-weighted area yda y A = zda = y da = I z z da = I y yz da = I yz z A modulus-weighted moment of inertia about z-axis modulus-weighted moment of inertia about y-axis modulus-weighted product of inertia ( cross moment of inertia) Unit 14-1

13 Also have Thermal Forces and Thermal Moments. 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: T F = Eα T da T T M = Eα Tz da M = Eα Ty da y z Note: Cannot use the modulus-weighted section properties since α may also vary in y and z along with E. In the definition of the section properties, have used a y and z. These are the location of the modulus-weighted centroid referred to some coordinate system Figure 14.7 Representation of general beam cross-section with pieces with different values of modulus Unit 14-13

14 The modulus-weighted centroid is defined by: 1 yda y A = 1 zda z A = 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: T TOT ( F + F ) = F = E1 f1a T TOT ( Mz Mz ) = Mz = E1 ( f Iz + f3iyz) T TOT ( My My ) = My = E1 ( f Iyz + f3iy ) + + (Note: Rivello uses F, M y, M z for F T0T, M y T0T, M z T0T ) Unit 14-14

15 Recall that: du f1 = f f 3 0 dv = dw = Motivation for modulus-weighted section properties A beam may not have constant material properties through the section. 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 14-15

16 . Stepwise variation 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: Unit 14-16

17 Selective Reinforcement Figure 14.9 Representation of selective reinforcement of an I-beam Unidirectional Graphite/Epoxy cap reinforcements (E = 0 Msi) Aluminum I-beam (E = 10 Msi) Furthest from neutral axis best resistance to bending Using aluminum as the reference, analyze as follows Figure Representative cross-section with aluminum as base Representation good only in direction parallel to axis about which I is taken. use E 1 to analyze Unit 14-17

18 E E b 1 0 msi = 10 msi b = b Principal Axes of structural cross-section: There is a set of y, z axes such that the product of inertia is zero. These are the principal axes (section has axes of symmetry) ( Ι ) yz Figure Representation of principal axes of structural cross-section modulus-weighted centroid Ι yz 0 Ι yz = 0 (use Mohr s circle transformation) Unit 14-18

19 If analysis is conducted in the principal axes, the equations reduce to: f 1 f TOT F = = EA 1 du TOT Mz = = EI 1 z 0 dv f 3 TOT My = = EI 1 y dw 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: σ xx TOT TOT TOT E F M M z y = y z E1 α T E1 A Iz Iy where y,z are principal axes for the section Unit 14-19

20 f 1 f TOT F = = EA 1 du TOT Iy Mz Iyz M = + E I I I 0 TOT y ( ) 1 y z yz Ι yz If the axes are not principal axes ( 0), have: (no change) dv = f 3 TOT Iz My Iyz M = + E I I I TOT z ( ) 1 y z yz dw = Note: If Ι yz 0 then both w and v are present for M y or M z only Figure 14.1 Representation of deflection of cross-section not in principal axes total deflection Unit 14-0

21 In this case, the expression for the stress is rather long: σ xx TOT E F = E1 A [ TOT TOT y z ] yz y Iy Iz Iyz I M I M y [ TOT TOT I ] z My Iyz Mz z E α T Iy Iz I 1 yz Engineering Beam Theory (Non-Principal Axes) Analysis is good for high aspect ratio structure (e.g. a wing) Figure Representation of wing as beam σ xx Unit 14-1

22 Note: this analysis neglects the effect of the axial Force F on the Bending Moment. This became important as the deflection w (or v) becomes large: Figure Representation of large deflection when axial force and bending deflection couple M = Mdue to p z wf0 Primary Bending Moment Secondary Moment F 0 = axial force Secondary moment known as membrane effect. Can particularly become important if F o is near buckling load (will talk about when talk about beam-column) Unit 14 -

23 Shear Stresses The shear stresses (σ xy and σ xz ) can be obtained from the equilibrium equations: σxy σxz σxx + = y z x σ x xy σ x xz = 0 = 0 Figure Representation of cross-section of general beam Unit 14-3

24 These shear stresses (called transverse shear stresses) cause small additional shearing contributions to deflections Figure Representation of pure bending and pure shearing of a beam Plane sections remain plane and perpendicular Pure Bending --> to midplane Pure Shearing --> Plane sections remain plane but not perpendicular to midplane Consider a beam section under pure shearing Figure Representation of deformation of beam cross-section under pure shearing Unit 14-4

25 γ xz w u = + = x z σ G engineering shear strain w Average over cross-section: x S z w da 1 σ w x xz da G = = = x da A ave xz Sz GA Actually, from energy considerations, one should average: w = x ave w da x w x da Sz GA e effective area For a Rectangular Cross-Section: A e 0.83 A Then, pure shearing deflections, w s, governed by: Unit 14-5

26 dw w s s = S GA x S = + C1 0 GA e e evaluated from boundary conditions The total beam deflection is the sum of the two contributions: W W W T B S w = w w = + + T B s total bending deflection from EI dw B = shearing deflection from M GA dw e S = 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 ) Unit 14-6

27 In addition to bending and shearing, the section may also twist through an angle α Figure 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 not twist. (Note: shear center not necessarily center of gravity or centroid) Unit 14-7

28 Figure Representation of some beam cross-sections with various locations of center of gravity and shear center center of gravity shear center If this condition is not met, then generally bending and twisting will couple. But there is a class of cross-sections (thin-walled) where bending and shearing/torsion can be decoupled. Will pursue this next. Wrap-up discussion by considering examples of common cross-sections with 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) Unit 14-8

29 Figure Some cross-sections generally not in principal axes Triangle Angle Wing Section Unit 14-9

30 --> Finally What 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 and bending couple Deflections small o No moment due to axial load (P w ) o Angles small such that sinφ 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) Unit 14-30

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