Strength of materials - Chapter 8: Beam deflection - Nguyễn Sỹ Lâm

Nội dung text: Strength of materials - Chapter 8: Beam deflection - Nguyễn Sỹ Lâm

1. 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
2. 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. 1 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.
3. 8.2 EQUATION OF THE ELASTIC CURVE • Constants are determined from boundary conditions x x EI y dx M x dx C1x C2 0 0 • Three cases for statically determinant beams, – Simply supported beam yA 0, yB 0 – Overhanging beam yA 0, yB 0 – Cantilever beam yA 0,  A 0 • More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope.
4. 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 0  Fy 0  M A 0 The beam is statically indeterminate. • Also have the beam deflection equation, x x EI y dx M x dx C1x C2 0 0 which introduces two unknowns but provides three additional equations from the boundary conditions: At x 0,  0 y 0 At x L, y 0
5. 8.5 EXAMPLES EXAMPLE 9.01 SOLUTION: •Develop an expression for M(x) and derive differential equation for elastic curve. -Reactions: Pa a RA  RB P 1  L L -From the free-body diagram for section AD, a M P x 0 x L L -The differential equation for the elastic curve, d 2 y a EI P x dx2 L
6. 8.5 EXAMPLES EXAMPLE 9.02 SOLUTION: •Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A). •Integrate twice and apply boundary For the uniform beam, determine the conditions to solve for reaction at A and reaction at A, derive the equation for to obtain the elastic curve. the elastic curve, and determine the slope at A. (Note that the beam is •Evaluate the slope at A. statically indeterminate to the first degree)
7. 8.5 EXAMPLES EXAMPLE 9.02 •Integrate twice dy 1 w x4 EI EI R x2 0 C dx 2 A 24L 1 1 w x5 EI y R x3 0 C x C 6 A 120L 1 2 •Apply boundary conditions: 2 3 d y w0x EI M RAx at x 0, y 0 : C2 0 dx2 6L 1 w L3 at x L, 0 : R L2 0 C 0 2 A 24 1 1 w L4 at x L, y 0 : R L3 0 C L C 0 6 A 120 1 2 •Solve for reaction at A 1 1 1 R L3 w L4 0 R w L  3 A 30 0 A 10 0
8. 8.6 METHOD OF SUPERPOSITION Principle of Superposition: •Deformations of beams subjected to •Procedure is facilitated by tables of combinations of loadings may be solutions for common types of obtained as the linear combination of loadings and supports. the deformations from the individual loadings
9. 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.03 Loading I wL3 wL4  y B I 6EI B I 8EI Loading II wL3 wL4  y C II 48EI C II 128EI In beam segment CB, the bending moment is zero and the elastic curve is a straight line. wL3   B II C II 48EI 4 3 4 wL wL L 7wL yB II 128EI 48EI 2 384EI
10. 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.
11. 8.6 METHOD OF SUPERPOSITION EXAMPLE 9.04 •Distributed Loading: 4 3 w 2 2 3 2 yB w L 2L L L L 24EI 3 3 3 wL4 0.01132 EI •Redundant Reaction Loading: 2 2 3 RB 2 L RBL yB L 0.01646 R 3EIL 3 3 EI •For compatibility with original supports, yB = 0 wL4 R L3 0 y y 0.01132 0.01646 B B w B R EI EI RB 0.688wL  •From statics, RA 0.271wL  RC 0.0413wL 