8 beam deflection 2015 bach khoa

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Đây là tài liệu của các bạn sinh viện hiện tại đang học tại Đại học Bách Khoa TP HCM. Đồng thời cũng là giáo án của giảng viên tại Đại học Bách Khoa. Nó sẽ rất hữu ích cho công việc học tập của các Bạn. Chúc Bạn thành công. CHAPTER 8: BEAM DEFLECTION 8.1 Deflection of a beam under transverse loading 8.2 Equation of the elastic curve 8.3 Direct determination of the elastic curve from the loading 8.4 Statically indeterminate beams 8.5 Examples 8.6 Statically indeterminate beams 8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE LOADING • Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings   M ( x) EI • Cantilever beam subjected to concentrated load at the free end,   Px EI • Curvature varies linearly with x • At the free end A, ρ  0, A • At the support B,  B ρA    0,  B  EI PL 8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE LOADING • Overhanging beam • Reactions at A and C • Bending moment diagram • Curvature is zero at points where the bending moment is zero, i.e., at each end and at E   M ( x) EI • Beam is concave upwards where the bending moment is positive and concave downwards where it is negative • Maximum curvature occurs where the moment magnitude is a maximum • An equation for the beam shape or elastic curve is required to determine maximum deflection and slope 8.2 EQUATION OF THE ELASTIC CURVE • From elementary calculus, simplified for beam parameters, d2y   dx 2 3   dy  1       dx    d2y dx • Substituting and integrating, EI   EI d2y dx  M x x dy EI   EI  M  x dx  C1 dx  x x 0 EI y   dx  M  x  dx  C1x  C2 8.2 EQUATION OF THE ELASTIC CURVE • Constants are determined from boundary conditions x x 0 EI y   dx  M  x  dx  C1x  C2 • Three cases for statically determinant beams, – Simply supported beam y A  0, yB  – Overhanging beam y A  0, yB  – Cantilever beam y A  0,  A  • More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope 8.3 DIRECT DETERMINATION OF THE ELASTIC CURVE FROM THE LOAD DISTRIBUTION • For a beam subjected to a distributed load, d 2M dM  V x dx dV    w x  dx dx • Equation for beam displacement becomes d 2M dx  EI d4y dx   w x  • Integrating four times yields EI y  x     dx  dx  dx  w x dx  16 C1x3  12 C2 x  C3 x  C4 • Constants are determined from boundary conditions 8.4 STATICALLY INDETERMINATE BEAMS • Consider beam with fixed support at A and roller support at B • From free-body diagram, note that there are four unknown reaction components • Conditions for static equilibrium yield  Fx   Fy   M A  The beam is statically indeterminate • Also have the beam deflection equation, x x 0 EI y   dx  M  x  dx  C1x  C2 which introduces two unknowns but provides three additional equations from the boundary conditions: At x  0,   y  At x  L, y  8.5 EXAMPLES EXAMPLE 9.01 SOLUTION: •Develop an expression for M(x) and derive differential equation for elastic curve W 14  68 I  723 in E  29  106 psi P  50 kips L  15 ft a  ft •Integrate differential equation twice and apply boundary conditions to obtain elastic curve For portion AB of the overhanging beam, •Locate point of zero slope or point of (a) derive the equation for the elastic curve, maximum deflection (b) determine the maximum deflection, •Evaluate corresponding maximum (c) evaluate ymax deflection 8.5 EXAMPLES EXAMPLE 9.01 SOLUTION: •Develop an expression for M(x) and derive differential equation for elastic curve -Reactions: RA  Pa  a  RB  P1    L  L -From the free-body diagram for section AD, M  P a x L 0  x  L  -The differential equation for the elastic curve, EI d2y a   P x L dx 8.5 EXAMPLES EXAMPLE 9.01 •Locate point of zero slope or point of maximum deflection dy PaL   xm   0 1  3   dx EI   L   PaL2  x  x  y    EI  L  L  3   xm  L  0.577 L •Evaluate corresponding maximum deflection  PaL2 ymax  0.577  0.577 3 EI  PaL2 ymax  0.0642 EI ymax  50 kips 48 in 180 in 2  0.0642   29  106 psi 723 in  ymax  0.238 in 8.5 EXAMPLES EXAMPLE 9.02 SOLUTION: •Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A) For the uniform beam, determine the reaction at A, derive the equation for the elastic curve, and determine the slope at A (Note that the beam is statically indeterminate to the first degree) •Integrate twice and apply boundary conditions to solve for reaction at A and to obtain the elastic curve •Evaluate the slope at A 8.5 EXAMPLES EXAMPLE 9.02 •Consider moment acting at section D, MD   w0 x  x RA x  M 0  L  w0 x3 M  RA x  6L •The differential equation for the elastic curve, d2y w0 x3 EI  M  R A x  6L dx 8.5 EXAMPLES EXAMPLE 9.02 •Integrate twice dy w0 x EI  EI  RA x   C1 dx 24 L w0 x EI y  RA x   C1x  C2 120 L EI d y w0 x  M  R x  A 6L dx •Apply boundary conditions: at x  0, y  : C2  w0 L at x  L,   : RA L   C1  24 w0 L at x  L, y  : RA L   C1L  C2  120 •Solve for reaction at A 1 RA L3  w0 L4  30 RA  w0 L  10 8.5 EXAMPLES EXAMPLE 9.02 •Substitute for C1, C2, and RA in the elastic curve equation, 1  w0 x   EI y   w0 L  x   w0 L3  x  10 120 L  120   y  w0  x5  L2 x3  L4 x 120 EIL •Differentiate once to find the slope,   dy w0   x  L2 x  L4 dx 120 EIL at x = 0, w0 L3 A  120 EI   8.6 METHOD OF SUPERPOSITION Principle of Superposition: •Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings •Procedure is facilitated by tables of solutions for common types of loadings and supports 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.03 For the beam and loading shown, determine the slope and deflection at point B SOLUTION: Superpose the deformations due to Loading I and Loading II as shown 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.03 Loading I wL3  B I   EI wL4  yB I   8EI Loading II wL3 C II  48 EI wL4  yC II  128 EI In beam segment CB, the bending moment is zero and the elastic curve is a straight line wL3  B II  C II  48 EI wL4 wL3  L  wL4  yB II     128 EI 48 EI   384 EI 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.03 Combine the two solutions, wL3 wL3  B   B I   B II    EI 48 EI wL3 B  48 EI wL4 wL4 yB   yB I   yB II    8EI 384 EI 41wL4 yB  384 EI 8.6 METHOD OF SUPERPOSITION APPLICATION OF SUPERPOSITION TO STATICALLY INDETERMINATE BEAMS •Method of superposition may be applied •Determine the beam deformation to determine the reactions at the supports without the redundant support of statically indeterminate beams •Treat the redundant reaction as an unknown load which, together with the •Designate one of the reactions as other loads, must produce deformations redundant and eliminate or modify the compatible with the original supports support 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.04 For the uniform beam and loading shown, determine the reaction at each support and the slope at end A SOLUTION: •Release the “redundant” support at B, and find deformation •Apply reaction at B as an unknown load to force zero displacement at B 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.04 •Distributed Loading:  yB w   w 24 EI  4  2  3   L   L L   L  L  3      wL4  0.01132 EI •Redundant Reaction Loading: 2 RB    L  RB L3  yB R   L     0.01646 3EIL     EI •For compatibility with original supports, yB = wL4 RB L3   yB w   yB R  0.01132  0.01646 EI EI RB  0.688 wL  •From statics, R A  0.271wL  RC  0.0413 wL  8.6 METHOD OF SUPERPOSITION EXAMPLE 9.04 Slope at end A, wL3 wL3  A w    0.04167 24 EI EI 0.0688 wL  L    L   A R    L    EIL     3 2 wL3   0.03398 EI  wL3 wL3  A   A w   A R  0.04167  0.03398 EI EI wL3  A  0.00769 EI ... derive the equation for the elastic curve, maximum deflection (b) determine the maximum deflection, •Evaluate corresponding maximum (c) evaluate ymax deflection 8.5 EXAMPLES EXAMPLE 9.01 SOLUTION:... point of maximum deflection dy PaL   xm   0 1  3   dx EI   L   PaL2  x  x  y    EI  L  L  3   xm  L  0.577 L •Evaluate corresponding maximum deflection  PaL2... is a maximum • An equation for the beam shape or elastic curve is required to determine maximum deflection and slope 8.2 EQUATION OF THE ELASTIC CURVE • From elementary calculus, simplified for
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