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