MT-1620 al.2002 Unit 13 Review of Simple Beam Theory Readings Review Unified Engineering notes on Beam Theory BMP 38.3.9,3.10 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 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
MT-1620 al.2002 IV. General Beam Theory Paul A Lagace @2001 Unit 13-2
MIT - 16.20 Fall, 2002 IV. General Beam Theory Paul A. Lagace © 2001 Unit 13 - 2
MT-1620 al.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. Let'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 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. Let’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
MT-1620 al.2002 Figure 13.1 General Geometry of a Beam C2oss-5ACT/oA a)long&thin→>b,h b)loading is in z-direction c)loading passes through "shear center= no torsion/twist (we'l define this term later and relax this constraint d)cross-section can vary along X 2. Stress state a)Oy, Oyz, Oxy=0 =no stress in y-direction ZZ = only significant stresses are ox and o Paul A Lagace @2001 Unit 13-4
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) σyy, σyz, σxy = 0 ⇒ no stress in y-direction b) σxx >> σzz σxz >> σzz ⇒ only significant stresses are σxx and σxz • Paul A. Lagace © 2001 Unit 13 - 4
MT-1620 al.2002 Note: there is a load in the z-direction to cause these stresses, but generated o is much larger(similar to pressurized cylinder exampl Why is this valid? ook at moment arms Figure 13.2 Representation of force applied in beam fo 是2 Oxx moment arm is order of (h o,, moment arm is order of() ZZ and e>>h >0, for equilibrium Paul A Lagace @2001 Unit 13-5
MIT - 16.20 Fall, 2002 Note: there is a load in the z-direction to cause these stresses, but generated σxx 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 σxx moment arm is order of (h) σzz moment arm is order of (l) and l >> h ⇒ σxx >> σzz for equilibrium Paul A. Lagace © 2001 Unit 13 - 5
MT-1620 Fall 2002 3. Deformation Figure 13.3 Representation of deformation of cross-section of a beam deformed state(capital letters) o is at midplane undeformed state lsmall letters define: W=deflection of midplane(function of x only Paul A Lagace @2001 Unit 13-6
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
MT-1620 al.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 l(x,y,二)≈ z≈-2 (13-1) d x total derivative dx since it does not vary with y or z Figure 13.4 Representation of movement in x-direction of two points on same plane in beam u=-z sin note direction of u relative to +x direction Paul A Lagace @2001 Unit 13-7
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 u x( , y,) z ≈ − zφ ≈ − 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
MT-1620 al.2002 and forφsml →U=Zφ v(x,y,z)=0 (x,y)≈w(x) (13-2) Now look at the strain-displacement equations du d (13-3) dx dx dv dy Es a=0(no deformation through thickness) 0 du dv dy dx dv dw 0z0 y du dw 00w dz dx dx dx Paul A Lagace @2001 Unit 13-8
MIT - 16.20 Fall, 2002 and for φ small: ⇒ u = -z φ v x( , y,) z = 0 w x( , y,) z ≈ w x( ) (13 - 2) Now look at the strain-displacement equations: 2 ∂ u d w ε xx = ∂ x = − z dx2 (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
MT-1620 al.2002 Now consider the stress-strain equations (for the time being consider isotropic.extend this to orthotropic later) (13-4 E small inconsistency with previous E small inconsistency with previous E 2(+y) E 2(1+v E 2(1+v o <--inconsistency again E We get around these inconsistencies by saying that ey, Ezz, Exz are very small but not quite zero. This is an approximation. We will evaluate these later on Paul A Lagace @2001 Unit 13-9
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 ν σ xx ε = − <-- small inconsistency with previous yy E ν σ 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 εyy, εzz, εxz are very small but not quite zero. This is an approximation. We will evaluate these later on. Paul A. Lagace © 2001 Unit 13 - 9
MT-1620 al.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 0 do 0(13-5) dx dz 0=0 -equilibrium) 00c 00 0(13-6) dx 0 Note average equilibrium equations [g(3-6)→d dx (13-6a) Eq(3-5)]dh→M (13-5a) Paul A Lagace @2001 Unit 13-10
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 xx ∂ σ xz + = 0 (13 - 5) ∂ x ∂ z 0 = 0 (y -equilibrium) ∂ σ xz ∂ σ 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) ] dy dz ⇒ d M = S (13 - 5a) face dx Paul A. Lagace © 2001 Unit 13 - 10
