Bending Deflection Statically Indeterminate Beams

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ending Deflection Statically Indeterminate eams E1108-II: erospace Mechanics of Materials Dr. Calvin Rans Dr. Sofia Teixeira De reitas erospace Structures & Materials aculty of erospace Engineering

Recap rocedure for Statically Indeterminate roblems I. ree ody Diagram II. Euilibrium of orces (and Moments) III. Displacement Compatibility IV. orce-displacement (Stress-Strain) Relations Solve when number of euations = number of unknowns V. nswer the Question! Typically calculate desired internal stresses, relevant displacements, or failure criteria or bending, orce-displacement relationships come from Moment-Curvature relationship (ie: use Method of Integration or Method of Superposition)

Statically Indeterminate eams Many more redundancies are possible for beams: - Draw D and count number of redundancies - Each redundancy gives rise to the need for a compatibility euation M H V V - 4 reactions - 3 euilibrium euations 4 3 = 1 1 st degree statically indeterminate

Statically Indeterminate eams Many more redundancies are possible for beams: - Draw D and count number of redundancies - Each redundancy gives rise to the need for a compatibility euation M H H M V V - 6 reactions - 3 euilibrium euations 6 3 = 3 3 rd degree statically indeterminate

Solving statically indeterminate beams using method of integration

What is the difference between a support and a force? Displacement Compatibility (support places constraint on deformation)

Method of Integration What if we remove all redundancies and replace with reaction forces? V If we treat V as known, we can solve! ormulate expression for M(z) M dv EI dz Can integrate to find v

Method of Integration (cont) V t, v = 0, θ = 0 EI dv M z, V dz EIv M z V dz, EIv M (, z V ) dz C 1 C Determine constants of integration from oundary Conditions vzv (, ) How to determine V?

Method of Integration (cont) V = vzv (, ) How to determine V? V V 1 V changes displacement! Must be compatible! Compatibility C: t, v = 0 Solve for V

Compatibility euations for beams are simply the boundary conditions at redundant supports

Example 1 roblem Statement Determine deflection euation for the beam using method of integration: Solution ) Euilibrium: 1) D: H 0 V V L M M LV L M H L LV V V Treat reaction forces as knowns! erospace Mechanics of Materials (E1108-II) Example roblem 11

Example 1 4) Determine moment euation: M ccw z M M V z M Vz z z z M V z V M Can also use step function approach always on z M M 0 0 z V z0 0 V z L always off erospace Mechanics of Materials (E1108-II) Example roblem 1

Example 1 5) Integrate Moment euation to get v and v z EIv M Vz z = M(z) V 3 EIv M z z z C1 6 = -EIθ(z) M V 3 4 EIv z z z C1z C 6 4 = -EIv(z) We now have expressions for v and v, but need to determine constants of integration and unknown reactions erospace Mechanics of Materials (E1108-II) Example roblem 13

Example 1 5a) Solve for Constants of Integration using C s: z V 6 3 EIv M z z z C1 0 = -EIθ(z) M V 6 4 3 4 EIv z z z C1z C 0 = -EIv(z) ixed support θ = 0, v = 0 oundary Conditions: V EI M C 6 t z = 0, θ = 0 3 0 0 0 0 C 1 0 1 erospace Mechanics of Materials (E1108-II) Example roblem 14

Example 1 5a) Solve for Constants of Integration using C s: z V 6 3 EIv M z z z C1 0 = -EIθ(z) M V 6 4 3 4 EIv z z z C1z C 0 0 = -EIv(z) ixed support θ = 0, v = 0 oundary Conditions: 4 V 3 M C 4 6 t z = 0, v = 0 EI 0 0 0 0 C 0 erospace Mechanics of Materials (E1108-II) Example roblem 15

Example 1 5b) Solve for Reaction orces using C s (imposed by redundant support): z V EIv M z z z 6 oundary Conditions: 3 M V EIv z z z 6 4 3 4 = -EIθ(z) = -EIv(z) roller support v = 0 V M 4 6 4 3 t z = L, v = 0 EI0 L L L V L 4 3M L M Recall from euilibrium: L LV V LV erospace Mechanics of Materials (E1108-II) Example roblem 16

Example 1 z 5b) Solve for Reaction orces using C s (imposed by redundant support): L 3M L V, M LV, V LV 4 L V M V 5 8 3 8 L L 8 L M V V erospace Mechanics of Materials (E1108-II) Example roblem 17

Example 1 z We were asked to determine deflection euation: M V EIv z z z 6 4 z v 3L 5Lzz 48EI Max Displacement: 3 4 =0 3 v 6L z15lz 8z 0 z 0.5785L 48EI v(0.5785 L) 0.005416 L EI 0 0. 0.4 0 0.5 1 L erospace Mechanics of Materials (E1108-II) Example roblem 18

Example 1 Now that the reactions are known: z M ( z) EIv M Vz z L 5Lz z 8 8 5L V( z) EIv z 8 d dz L 8 L 4 M 5 8 L V 5 3 8 L 8 L erospace Mechanics of Materials (E1108-II) Example roblem 19

Solving statically indeterminate beams using superposition

Method of Superposition Determine reaction forces: L L M H 1) D: ) Euilibrium: V V H 0 V V M 3LV L M - 4 reactions - 3 euilibrium euations 1 st degree statically indeterminate

Method of Superposition (cont) How do we get compatibility euation? Split into two statically determinate problems V δ δ 1 3) Compatibility: δ 1 = δ v1v 0 e careful! +ve v

Method of Superposition (cont) How do we get orce-displacement relations? We have been doing this in the previous lectures Integrate Moment Curvature Relation Standard Case Solutions M dv EI dz Can integrate to find v L + + L EI v 3 L 3EI

Method of Superposition (cont) rom the standard case: 4) orce-displacement: L L δ 1 straight L v L L EI EI v v L 1 3 L EI 3 L 3EI L L L 3EI EI 14L 3EI 3

Method of Superposition (cont) rom the standard case: 4) orce-displacement: V L EI 3 L v L 3EI δ 3 3 v V 3L 9VL 3EI EI Compatibility: v 1v 0 14L 3EI 3 9VL EI 3 V 14 7

dditional remarks about bending deflections

Remarks about eam Deflections Recall Shear deformation V(x) V + V + Moment deformation M M M(x) + Negligible (for long beams) ending Deformation = Shear Deformation + Moment Deformation

Remarks about eam Deflections H M M C C xial deformation of due to H V xial deformation of C due to V V C H xial deformation << bending deformation!

Remark about eam Deflections or bending deformation problems negligible Deformation = xial Deformation + Shear Deformation + Moment Deformation UT! If moment deformation is not present, deformation is not negligible H M V

Example Calculate reaction forces at and D: C L, EI irst, can we see any simplifications? Symmetry! D Symmetry implies: - Reactions at = reactions at D - Slope at symmetry plane = 0 - Shear force at symmetry plane = 0 H M We will solve using superposition and standard cases erospace Mechanics of Materials (E1108-II) Example roblem 30

Example Calculate reaction forces: H M 1) D ) Euilibrium H H H V V L ccw L M M H LM 8 M Split problem into two: beam and beam erospace Mechanics of Materials (E1108-II) Example roblem 31

Example Calculate reaction forces: 3) Compatibility M -L /8 H M L/ H M -L /8 H H M = + L/ 0, v 0 (axial deformation negligible) erospace Mechanics of Materials (E1108-II) Example roblem 3

M -L /8 Example 4a) orce-displacement for eam L/ H H L/ + + M -L /8 v H L 3EI 3 v 0 v 4 M L L EI 16EI cw H L EI 0 M L EI cw 3 L 8EI erospace Mechanics of Materials (E1108-II) Example roblem 33

Example 4a) orce-displacement for eam M -L /8 L/ H 3 4 HL ML L v 3EI EI 16EI 0 3 cw HL ML L EI EI 8EI Still need θ Recall compatibility: 0, v 0 erospace Mechanics of Materials (E1108-II) Example roblem 34

Example M + L /8 H H M 4b) orce-displacement for eam L/ L/ H + + M 3 cw L 0 6EI cw M L EI 3 cw L M L 48EI EI Wait! Not entirely correct! erospace Mechanics of Materials (E1108-II) Example roblem 35

Example revious angle relative to fixed support θ H M cw 3 L M L 48EI EI 3 cw 7L M L HL 0 48EI EI EI erospace Mechanics of Materials (E1108-II) Example roblem 36

Example Solve: Recall compatibility: 0, v 0 Recall euilibrium: H H L V ccw L M M H LM 8 3 cw 7L ML HL 0 48EI EI EI L H 8 3 4 L HL ML L M v 0 4 3EI EI 16EI H M L 8 L 3 erospace Mechanics of Materials (E1108-II) Example roblem 37

Example C H M H M L 8 L 4 L, EI D H M V H V M L 8 L L 3 erospace Mechanics of Materials (E1108-II) Example roblem 38

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