美国麻省理工大学：《结构力学》（英文版） Unit 14 Behavior of General (including Unsymmetric Cross-section) Beams

MT-1620 al.2002 Unit 14 Behavior of General (including Unsymmetric Cross-section ) Beams Readings Rivello 7.1-7.5,77,7.8 T&g Paul A Lagace, Ph. D Professor of aeronautics Astronautics and Engineering Systems Paul A Lagace @2001

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

MT-1620 al.2002 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,(x) ind p,(x); and axial loading px X Include a temperature distribution T(x, y, z) Paul A Lagace @2001 Unit 14-2

Unit 14 - 2 Paul A. Lagace © 2001 MIT - 16.20 Fall, 2002 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 • Loading can be in any direction • Can resolve the loading to consider transverse loadings py(x) and pz(x); and axial loading px(x) • Include a temperature distribution T(x, y, z) Now look at a more general case:

MT-1620 al.2002 Figure 14.2 Representation of General Beam force/leng 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 Oxx is the only "important stress; Ox and ox found from equilibrium equations, but are secondary in importance Deformation: plane sections remain plane and perpendicular to the midplane after deformation( Bernouilli-Euler Hypothesis) Paul A Lagace @2001 Unit 14-3

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: σxx is the only “important” stress; σxy and σxz found from equilibrium equations, but are secondary in importance • Deformation: plane sections remain plane and perpendicular to the midplane after deformation (Bernouilli-Euler Hypothesis)

MT-1620 al.2002 Definition of stress resultants Consider a cross-section along x Figure 14.3 Representation of cross-section of general beam Place axis@ center of gravity of section ● Where O dA M,=Jon- da M y These are resultants l Paul A Lagace @2001 Unit 14-4

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 @ center of gravity of section where: These are resultants! S dA z xz = − ∫∫ σ F dA = ∫∫ σ xx S dA y xy = − ∫∫ σ M z dA y xx = − ∫∫ σ M y dA z xx = − ∫∫ σ

MT-1620 al.2002 The values of these resultants are found from statics in terms of the loading px, py, pz, and applying the boundary conditions of the problem Deformation Look at the deformation. In the case of Simple Beam Theory, had dx where u is the displacement along the X-axis This comes from the picture Figure 14.4 Representation of deformation in Simple Beam Theory 2 dx s for small angles Now must add two other contributions Paul A Lagace @2001 Unit 14-5

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 px, py, pz, and applying the boundary conditions of the problem Deformation Look at the deformation. In the case of Simple Beam Theory, had: u z d w d x = − 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

MT-1620 al.2002 1. Have the same situation in the x-y plane Figure 14.5 Representation of bending displacement in x-y plane d By the same geometrical g dx arguments used previously where v is the displacement in the y-direction 2. Allow axial loads, so have an elongation in the x- direction due to this. Call this uo Paul A Lagace @2001 Unit 14-6

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 to this. Call this u0: By the same geometrical arguments used previously

MT-1620 al.2002 Figure 14.6 Representation of axial elongation in x-z plane X un, v w are the deformations of the midplane Thus d v d w u(x, y, =)=uo-y dx ax bending bending about about z-aXIs y-aXIs v(x,y, ==v(x) "(x,y,z)=W(x) v and w are constant at any cross-section location, X Paul A Lagace @2001 Unit 14-7

Unit 14 - 7 Paul A. Lagace © 2001 MIT - 16.20 Fall, 2002 Figure 14.6 Representation of axial elongation in x-z plane u0, v, w are the deformations of the midplane Thus: u xyz u y d v d x z d w d x (,,) = 0 − − bending about z-axis bending about y-axis v xyz v x (,,) () = w xyz w x (,,) () = v and w are constant at any cross-section location, x

MT-1620 al.2002 Stress and strain From the strain-displacement relation get du d dx dx d x (these become total derivatives as there is no variation of the displacement in y and z) for functional ease write da dx d21 Caution: Rivello uses C1, C2, C3. These are not constants so use f, =f(x)( functions of x) Paul A Lagace @2001 Unit 14-8

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 d u dx y d v dx z d w 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 d ud x 1 0 = f d v d x 2 2 2 = − f d w d x 3 2 2 = − Caution: Rivello uses C1, C2, C3. These are not constants, so use fi ⇒ fi(x) (functions of x)

MT-1620 al.2002 Thus Ex=f+左2y+f2 Then use this in the stress-strain equation (orthotropic or "lower) +a△T E (include temperature effects Note: ignore" thermal strains in y and Z. These are of secondary" importance Thus O=EE.-Ea△T and using the expression for Ex E+y+2)-Ea△T Can place this expression into the expression for the resultants (force and moment) to get Paul A Lagace @2001 Unit 14-9

Unit 14 - 9 Paul A. Lagace © 2001 MIT - 16.20 Fall, 2002 Thus: ε xx =+ + f fy fz 12 3 Then use this in the stress-strain equation (orthotropic or “lower”): ε σ xx α xx E = + ∆T (include temperature effects) Note: “ignore” thermal strains in y and z. These are of “secondary” importance. Thus: σ εα xx xx = E ET − ∆ and using the expression for εx: σ α xx = ++ E f fy fz E T ( ) 12 3 − ∆ Can place this expression into the expression for the resultants (force and moment) to get:

MT-1620 al.2002 o dA fE dA+ fley dA f lEz dA ∫ Ea△TdA +y=4-J∫Ea△d M=∫od4.∫E=d4+∫Eyd4 +∫E=2a4-J∫ Ea△TzdA (Note: f,, f2, f3 are functions of X and integrals are in dy and dz, so these come outside the integral sign Solve these equations to determine f,(x),2(x), f3 X) lote: Have kept the modulus, E, within the ntegral since will allow it to vary across the cross-section Paul A Lagace @2001 Unit 14-10

Unit 14 - 10 Paul A. Lagace © 2001 MIT - 16.20 Fall, 2002 F dA f E dA f E y dA ==+ ∫∫ ∫∫ ∫∫ σ xx 1 2 + − ∫∫ ∫∫ f E z dA E T dA 3 α ∆ − == + M y dA f E y d ∫∫ ∫∫ ∫∫ z xx σ 1 2 A f E y dA 2 + − ∫∫ ∫∫ f E yz dA 3 E Ty dA α ∆ − == + M z dA f E z d ∫∫ ∫∫ ∫∫ y xx σ 1 2 A f E y z dA + − ∫∫ ∫∫ f E z dA E T z dA 3 2 α ∆ (Note: f1, f2, f3 are functions of x and integrals are in dy and dz, so these come outside the integral sign). Solve these equations to determine f1(x), f2(x), f3(x): Note: Have kept the modulus, E, within the integral since will allow it to vary across the cross-section