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Bending Deflection – Differential Equation Method AE1108-II: Aerospace Mechanics of Materials

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Bending Deflection – Differential Equation Method AE1108-II: Aerospace Mechanics of Materials Dr. Calvin Rans Dr. Sofia Teixeira De Freitas Aerospace Structures & Materials Faculty of Aerospace Engineering Recap • So far, for symmetric beams, we have: • Looked at internal shear force and bending moment distributions • Determined normal stress distribution due to bending moments • Determined shear stress distribution due to shear force • Need to determine deflections and slopes of beams under load • Important in many design applications • Essential in the analysis of statically indeterminate beams 2 Deformation of a Beam Assumptions Shear deformation V + V + Moment deformation V M M M + Negligible (for long beams) Bending Deformation = Shear Deformation + Moment Deformation Deformation of a Beam Assumptions • For long beams (length much greater than beam depth), shear deformation is negligible • This is the case for most engineering structures • Will consider moment deformation only in this course • Recap of sign convention +w z N.A. x +M+ +M +V +V + y y v Deformation of a Beam Visualizing Bending Deformation Elastic curve: plot of the deflection of the neutral axis of a beam How does this beam deform? We can gain insight into the deformation by looking at the bending moment diagram + - z M M + - M M And by considering boundary conditions at supports Qualitatively can determine elastic curve! Moment-Curvature Relationship z dz m 1 z dz m2 (-ve M) (+ve v) v(z) = vertical deflection at z dv (z) = slope at z = dz Moment-Curvature Relationship z dz For small d: ds R d 1 d or Rds curvature dz cos 1 For small : ds dz cos when is small Recall M E d dv 2 1 d dz dv I R Rdz dz dz2 dv2 M EI EIv dz2 (negative sign a result of sign convention) Deflection by Method of Integration dv2 MEI Lets consider a prismatic beam dz2 (ie: EI = constant) dv 1 Mdz dz EI Indefinite integrals result in constants of integration that can be determined from boundary conditions of the 1 vMdz problem Constant of integration EI 1 ie: zdz z2 C 2 Determining Constants of Integration Support Conditions v = 0 v = 0 v = 0 = 0 Determining Constants of Integration Continuity Conditions z Deformed abP shape C Discontinuity at z = a A B z ≤ a z ≥ a Pab L Pb Pa M z M Lz AC L CB L M 0 vavaAC CB ACaa CB Determining Constants of Integration Symmetry Conditions • Symmetry implies reflection of P deformation across symmetry C A B plane • v is equal • is opposite • Continuity implies equal deformation at symmetry plane • v is equal • is equal C 0 Procedure for Analysis Deflection by Integration • Draw a FBD including reaction forces • Determine V and M relations for the beam • Integrate Moment-displacement differential equation • Select appropriate support, symmetry, and continuity conditions to solve for constants of integration • Calculate desired deflection (v) and slopes (θ) Example 1a Problem Statement q Determine the deflection and slope EI at point B in a prismatic beam due to the distributed load q ABL Solution 1) FBD & Equilibrium FR0 z q z MA FRqLRqL0 yy Rz 2 cw LqL Ry MMqLM0 AA22 A Aerospace Mechanics of Materials (AE1108-II) – Example Problem 13 q Example 1a EI Solution ABL 2) Determine M and V @ z qL z q q qL2 V 2 M M AB qL V 2 -qL /2 FqLqzVVqLz0 222 ccw qL z L z MMqLzqzMqLzz 0 22 22 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 14 q Example 1a EI Solution ABL 3) Boundary Conditions At z = 0: v = 0, v′ = 0 Two boundary conditions dv2 Thus can solve by integrating: MEI dz2 dv2 1 M dz2 EI Aerospace Mechanics of Materials (AE1108-II) – Example Problem 15 q Example 1a EI Solution ABL 4) Solve Differential Equation qz22 qL d222 v1 qz qL MqLz 2 qLz 22 dz EI 22 Boundary Condition dv1 qz322 qLz qL z 0 BC: At z = 0, θ = 0 C1 dz EI 62 2 => C1 = 0 4322 1 qz qLz qL z 0 BC: At z = 0, v = 0 vC 2 EI 24 6 4 => C2 = 0 2 qz 22dv qz 22 vzLzL46 zLzL 33 24EI dz6 EI Aerospace Mechanics of Materials (AE1108-II) – Example Problem 16 q Example 1a EI Solution ABL 5) Calculate slopes and deflections Determine deflection and slope at B: 2 qz 22 qz 22 vzLzL46 zLzL33 24EI 6EI 4 qL qL3 vvBL(z ) 8EI BL(z ) 6EI Aerospace Mechanics of Materials (AE1108-II) – Example Problem 17 Relating Deformation to Loading Shear Force-Moment Diagram Relationships • Recall from Statics (refer to Hibbler Ch. 6.2 for refresher, but be careful of coordinate system) +w dV w dz 4 dv w(z) dM 4 v +M+ +M V dz EI dz dv3 V(z) +V +V v dz3 EI Moment-Curvature dv2 M(z) Relationship (Eq. 10.1) v dv2 dz2 EI MEI dz2 z q Example 1b EI We can also solve Example 1 in an ABL alternative way: 4 Watch negative sign! dv wz() v EIv q = -w(z) dz4 EI dv3 V() z 3 v EIv qz C1 = V(z) dz EI dv2 Mz() 2 qz 2 v EIv C z C = M(z) dz EI 2 12 32 qz z EIv C C z C = -θ(z)EI 62123 We have 4 unknown constants of integration, 43 2 qz z z thus need 4 BCs EIv C C C z C 241234 6 2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 19 z q Example 1b EI We can also solve this problem an ABL alternative way: Boundary Condition: At z = L, V = 0 EIv qz C1 = V(z) C1 CqL1 qz2 EIv qLz C = M(z) At z = L, M = 0 2 2 22 qL2 qL C2 CqL2 22 32 qz qL2 qL EIv z z C = -θ(z)EI At z = 0, θ = 0 62 2 3 C3 0 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 20 z q Example 1b EI We can also solve this problem an ABL alternative way: Boundary Condition: qz42 qL qL EIv z32 z 0 C At x = 0, v = 0 24 6 4 4 C4 0 2 qz 22 qx 22 vzLzL46 vzLzL 33 24EI 6EI Same result as before! Aerospace Mechanics of Materials (AE1108-II) – Example Problem 21 Example 2 qL z q Determine deflection and EI slope at B: AB EIv q = -w(z) L We will apply Approach 2 EIv qz C1 = V(z) Exact same differential qz2 EIv C z C = M(z) equations as before!! 2 12 32 What makes the qz z EIv C C z C = -θ(z)EI problem different? 62123 43 2 Boundary Conditions! qz z z EIv C C C z C 241234 6 2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 22 qL z q Example 2 EI Determine deflection and ABL slope at B: Boundary Condition: At z = L, V = qL EIv qz C1 = V(z) qL CqL1 2 q q EI EI ABL ABL 2qL qL qL V V M M -qL2/2 -3qL2/2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 23 qL z q Example 2 EI Determine deflection and ABL slope at B: Boundary Condition: At z = L, V = qL EIv qz C1 = V(z) C1 CqL1 2 qz2 EIv 2 qLz C = M(z) At z = L, M = 0 2 2 22 qL2 3 qL C2 CqL2 2 22 32 qz2 3 qL EIv qLz z C = -θ(x)EI At z = 0, θ = 0 623 C3 0 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 24 qL z q Example 2 EI Determine deflection and ABL slope at B: Boundary Condition: qz42 qL3 qL EIv z32 z 0 C At z = 0, v = 0 24 3 4 4 C4 0 2 qz 22 qz 22 vzLzL818 vzLzL 69 24EI 6EI 4 11qL 2qL3 vvBzL() v ' 24EI BzL() 3EI Aerospace Mechanics of Materials (AE1108-II) – Example Problem 25 Example 2 qL xz q xz q EI EI ABL ABL 2 2 qz 22 qz 22 vzLzL818 vzLzL46 24EI 24EI 0 qL4 8EI 11qL4 4 qLqL4 24EI 2EI 2EI 0 L Aerospace Mechanics of Materials (AE1108-II) – Example Problem 26 z P Example 3 EI B A C Determine deflection at C in terms of EI: L L/2 P/2 To save time, reactions are provided 3P/2 Since reaction forces act at B (discontinuity), we must split the differential equation into parts for AB and BC We can easily see by inspection that: P EIv V (0 < z < L) 2 EIv VP (L < z < 3L/2) Integrate to find M Aerospace Mechanics of Materials (AE1108-II) – Example Problem 27 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Moments: 3P/2 P M EIv z C (0 ≤ z ≤ L) 2 1 (L ≤ z ≤ 3L/2) M EIv Pz C2 Moment BC’s: At z = 0, M = 0 C1 0 Integrate to find θ 3PL At z = 3L/2, M = 0 C 2 2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 28 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Slopes: 3P/2 P EIv z2 C (0 ≤ z ≤ L) 4 3 PPL3 EIv z2 z C (L ≤ z ≤ 3L/2) 224 Slope Continuity Condition: 2 PL 2 At z = L, θAB = θBC CPLC 4 34 Integrate to find v Aerospace Mechanics of Materials (AE1108-II) – Example Problem 29 z P Example 3 EI B A C Determine deflection at C: L L/2 P/2 Deflections: 3P/2 P EIv z3 C z C (0 ≤ z ≤ L) 12 35 PPL323 EIv z z C z C (L ≤ z ≤ 3L/2) 64 46 From last slide PL2 Deflection BC’s: CPLC2 4 34 At z = 0, v = 0 C5 0 2 3 PL2 5PL PL At z = L, v = 0 C C4 C6 3 12 6 4 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 30 z P Example 3 EI B A C Determine deflection at C: L L/2 Deflections: P/2 3P/2 Pz 22 vLz (0 ≤ z ≤ L) 12EI P 32 23 vLLzLzz3109 2 (L ≤ z ≤ 3L/2) 12EI PLPL33 0.5 1212EIEI 3LPL3 0 0.5 1 1.5L vC() v ( ) 0.5 28EI 1 1.5 2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 31 Example 4 z P What about beams with a EI non loaded free end? ABC L L P Will it work itself out? A C B Curved part Straight part Aerospace Mechanics of Materials (AE1108-II) – Example Problem 32 z P Example 4 EI -PL ABC What about beams with a L L non loaded free end? To save time, reactions are provided P Moments: P V M EIv P z L (0 ≤ z ≤ L) M MEIv 0 (L ≤ z ≤ 2L) -PL Aerospace Mechanics of Materials (AE1108-II) – Example Problem 33 z P Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Slopes: P EIv z2 PLz C (0 ≤ z ≤ L) 2 1 (L ≤ z ≤ 2L) EIv C2 Slope BC’s: At z = 0, θ = 0 C1 0 Slope CC’s: PL2 At z = L, θ = θ C2 AC CB 2 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 34 z P Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Displacements: PPL EIv z32 z C (0 ≤ z ≤ L) 62 3 PL2 EIv z C (L ≤ z ≤ 2L) 2 4 Displacement BC’s: At z = 0, v = 0 C3 0 Displacement CC’s: PL3 At z = L, v = v C4 AC CB 6 Aerospace Mechanics of Materials (AE1108-II) – Example Problem 35 z P Example 4 EI -PL ABC What about beams with a L L non loaded free end? P Displacements: Pz2 vzL 3 (0 ≤ z ≤ L) 6EI PL2 L vz (L ≤ z ≤ 2L) 23EI 0 Formula for a straight line! 0.5 No curvature, it does work out! 1 0 1L 2 L Aerospace Mechanics of Materials (AE1108-II) – Example Problem 36 For next time.
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  • Glossary of Notations

    Glossary of Notations

    108 GLOSSARY OF NOTATIONS A = Earthquake peak ground acceleration. IρM = Soil influence coefficient for moment. = A0 Cross-sectional area of the stream. K1, K2, = aB Barge bow damage depth. K3, and K4 = Scour coefficients that account for the nose AF = Annual failure rate. shape of the pier, the angle between the direction b = River channel width. of the flow and the direction of the pier, the BR = Vehicular braking force. streambed conditions, and the bed material size. = BRa Aberrancy base rate. Kp = Rankine coefficient. = = bx Bias of ¯x x/xn. KR = Pile flexibility factor, which gives the relative c = Wind analysis constant. stiffness of the pile and soil. C′=Response spectrum modeling parameter. L = Foundation depth. = CE Vehicular centrifugal force. Le = Effective depth of foundation (distance from = CF Cost of failure. ground level to point of fixity). = CH Hydrodynamic coefficient that accounts for the effect LL = Vehicular live load. of surrounding water on vessel collision forces. LOA = Overall length of vessel. = CI Initial cost for building bridge structure. LS = Live load surcharge. = Cp Wind pressure coefficient. max(x) = Maximum of all possible x values. = CR Creep. M = Moment capacity. = cap CT Expected total cost of building bridge structure. M = Moment capacity of column. = col CT Vehicular collision force. M = Design moment. = design CV Vessel collision force. n = Manning roughness coefficient. = D Diameter of pile or column. N = Number of vessels (or flotillas) of type i. = i DC Dead load of structural components and nonstructural PA = Probability of aberrancy. attachments. P = Nominal design force for ship collisions. DD = Downdrag. B P = Base wind pressure.