Bending Deflection. - PDF Free Download

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ontents The Elastic urve, Deflection & Slope ( 挠曲线挠度和转角 ) Differential Euation of the Elastic urve( 挠曲线微分方程 ) Deflection & Slope by Integration( 积分法求挠度和转角 ) oundary onditions( 边界条件 ) Symmetry onditions( 对称性条件 ) ontinuity onditions( 连续性条件 ) Direct Integration from Distributed Loads( 直接由分布荷载积分求挠度和转角 ) Direct Integration from Transverse Loads( 直接由剪力积分求挠度和转角 ) Deformations in a Transverse ross Section( 梁横截面内的变形 ) urvature Shortening( 梁由于弯曲造成的轴向位移 )

ontents Deflection & Slope by Superposition( 叠加法求挠度和转角 ) Superposition of Loads( 荷载叠加法 ) Superposition of Rigidized Structures( 刚化叠加法 ) ombined Superposition( 荷载和变形组合叠加法 ) Deflection & Slope by Singular Functions( 奇异函数法求挠度和转角 ) Deflection & Slope by Moment-rea Theorems( 图乘法求挠度和转角 ) Stiffness ondition( 刚度条件 ) Ways to Increase Fleural Rigidity( 梁的刚度优化设计 ) ending Strain Energy( 弯曲应变能 )

The Elastic urve, Deflection and Slope The elastic curve: beam ais under bending, reuired to determine beam deflection and slope. ending deflections ( = f()): vertical deflection of the neutral surface, defined as donard positive / upard negative. Slope (θ = θ() tan(θ) = d/d): rotation of cross-sections, defined as clockise positive / counter clockise negative Deflection curve 4

Differential Euation of the Elastic urve urvature of the neutral surface 1 ( ) M M( ) EI z M 0 0 M 1 / ( ) (1 ) M M 0 0 M EI M EI: fleural rigidity The negative sign is due to the particular choice of the -ais. 5

Deflection and Slope by Integration EI M() EI M ( ) d EI M ( )dd D onventionally assuming constant fleural rigidity (EI) Integration constants and D can be determined from boundary conditions, symmetry conditions, and continuity conditions. 6

oundary onditions Simple eams Deflections are restrained at the hinged/rolled supports 0; 0 7

oundary onditions- antilever eams oth the deflection and rotation are restrained at the clamped end 0; 0 8

Symmetry onditions oth the geometry and loads are symmetric about the mid-section ( = L/) 0 9

ontinuity onditions P 0 a, a L, 0 L a 1 a a ; a a 1 1 a L a ; a L a 1 1 10

Direct Integration from Distributed Loads For a beam subjected to distributed loads Euation for beam displacement becomes 4 d d M EI 4 d d EI dm df d d d Integrating four times yields d M S FS, d d d d 1 1 6 1 4 onstants are determined from conditions on the shear forces and bending moments as ell as conditions on the slopes and deflections. 11

Direct Integration from Transverse Loads For a beam subjected to transverse loads (ithout distributed loads) Euation for beam displacement becomes d dm EI F d d Integrating three times yields EI d 1 d F S d 4 S onstants are determined from conditions on the bending moments as ell as conditions on the slopes and deflections. 1

Deformations in a Transverse ross Section Deformation due to bending moment is uantified by the curvature of the neutral surface 1 y y y Ey lthough cross sectional planes remain planar hen subjected to bending moments, in-plane deformations are nonzero, y y y y, y y y z For a rectangular cross-section, no change in the vertical dimension ill be observed. Horizontal epansion above the neutral surface and contraction belo it cause an in-plane curvature y 1 z y anticlastic curvature 1

urvature Shortening When a beam is bent, the ends of the beam move closer together. It is common practice to disregard these longitudinal displacements. ds d 1 1 d d L L L d 0 1 1 For immovable supports, a horizontal reaction ill develop at each end. HL E H E L This euation gives a close estimate of the tensile stress produced by the immovable supports of a simple beam. 14

Sample Problem Given: fleural rigidity (EI) of a simply supported beam under a uniformly distributed load of density Find: euations of deflections and slopes, and their maimum values (θ ma, ma ) l 15

Solution: l M ( ) l EI l l EI 4 6 l 4 EI D 1 4 0 0, l 0 oundary conditions: l, D 0 4 16

Euations of beam deflection and slope 4 4EI EI ( l ( l -6l l +4 ) ) l The maimum deflection and slope ma l 4EI ma 4 l 5l 84EI 17

Sample Problem Given: fleural rigidity (EI) of a cantilever beam under a concentrated load acting at its free end Find: euations of deflections and slopes, and their maimum values (θ ma, ma ) l P 18

Solution: M( ) P( l ) EI P P EI P EI 6 Pl Pl Pl oundary conditions: D 0 0, 0 0 l P D 0 19

Euations of beam deflection and slope P EI (l ) l P P 6EI (l ) The maimum deflection and slope ma Pl EI ma Pl EI 0

Sample Problem Given: a simply supported beam ith fleural rigidity EI is subjected to a concentrated load P as shon Find: the euations of deflection and slope, and their maimum values ( ma, θ ma ) P l l 1

Solution ecause of symmetry, it s sufficient to solve only portion. P l M ( ), 0 P P EI P EI 4 P l l EI D 1 0 0 D 0 l Pl 0 16 Left boundary condition: Symmetry condition:

Euations of bending deflection and slope: P l 16EI P l 48EI ( 4 ) ( 4 ) Maimum deflection P and slope: Pl ma 16EI Pl ma l 48EI l l

Sample Problem Given: a simply supported beam ith fleural rigidity EI is subjected to a uniformly distributed load ith density, on its central portion as shon Find: the euations of deflection and slope, and their maimum values ( ma, θ ma ). D E a a a a 4

Solution Thanks to symmetry, it is sufficient to consider only the left half M1( 1 ) a1 (0 1 a) M ( ) a ( a) ( a a) EI 1 a1 EI a ( a) a 1 D E a a a a a 5

a EI 1 1 1 EI 1 a1 a EI1 1 11 D1 6 EI a ( a) a EI ( a) 6 a 4 EI ( a) D 6 4 11 Due to symmetry: a 0 a 6 onstraint condition: 1 1 0 0 D1 0 ontinuity conditions: a a, D D 1 1 1 1 6

Euations of deflection and slope: a 1 (11a 1 ) 6EI [ a ( a) 11a ] 6EI a 1 (11a 1 1 ) 6EI 4 [ 4a ( a) 44a 4EI Maimum deflection and slope: 0 a 0 ] a 1 1 a a a a 11a 19a ma 1 10, ma a 6EI 8EI 4 7

Deflection and Slope by Superposition Superposition of Loads: Deformation of beams subjected to combinations of loads may be obtained as the linear combination of the deformations due to individual loads. - eam material obeys linearly elastic Hooke s la. - No interactions eist among deformations induced by individual loads. - Procedure is facilitated by tables of solutions for common types of loadings and supports. 8

Sample Problem Using method of superposition to find the deflection at section and the slopes at sections and. m P l l 9

Solution: Superpose the deformations due to the uniformly distributed load (), the concentrated load (P) and the concentrated moment (m). m P P l l 4 5l Pl ml 84EI 48EI 16EI l Pl ml 4EI 16EI EI l Pl ml 4EI 16EI 6EI m 0

Sample Problem Find the deflections at sections and D. a D a a 1

Solution D a a a a D a a 4 4 5 ( a) 5a 0, D 84EI 4EI

Sample Problem Find the deflection at section and the slope at section. l / l /

Solution l / l / / l l 4EI 5 l 84EI 4 / / 0 l 4EI 4

Sample Problem Given θ = 0, determine the relationship beteen m and P. a P a m 5

Solution: P m P m a a a a a Pa ma EI EI 0 m Pa 4 6

Deflection and Slope by Superposition Superposition of Rigidized Structures: pplicable to multi-span beams The total deflection of a multi-span beam under a given loading condition can be determined by superposing several beams corresponding to rigidizing all but one span of the beam, under the eactly same loading condition as the original beam. 7

Sample Problem Find the deflection at section of the simply supported overhanging beam shon. P Solution Deflection at due to L a rigidization of portion Rigidization EI = P c1 c1 Pa EI Pa EI 8

Deflection at due to rigidization of portion P P Pa Rigidization EI= PaL c a a EI Total deflection and slope at : c c1 c Pa a L Pa PaL a EI EI EI pal EI c c 1c Pa PaL Pa a L EI EI EI 9

Superposition of Loads & Rigidized Structures Given = 0, determine the relationship beteen P and. a a a P D 40

Solution: P P a a a D a a a D a a a D 4 5 ( a) Pa( a) 84EI 16EI 5 P a 6 0 P Pa a a a D 41

Sample Problem Using the method of superposition find the deflection and slope at section of the beam shon. a a a a 4

Solution: a a a Rigidizing a a a ( a) a EI 16EI 1EI 4 a a 1EI Rigidizing 4 a a, 6EI 8EI 4 4 4 5 Total: a a a, a a a 1EI 6EI 4EI 1EI 8EI 4EI 4

Sample Problem stepped cantilever, as shon, is subjected to a concentrated load F at its free end. Find the deflection at the free end. F EI EI l/ l/ 44

Solution Rigidizing section makes a cantilever subjected to a concentrated load at its free end. F EI EI l/ l/ Rigidizing section makes the hole beam a cantilever. 1 θ 1 F 1 l 1 Pl 16EI l F Fl/ 45

Sample Problem Find the deflections at sections and D of the beam shon belo. a D a a a 46

Solution Rigidizing D 0, D Rigidizing a( a) 48EI ( a) a( a) 14a 8EI EI EI 4 7a EI 4 4 a a a D D a a a a D Total: 14 7 ( ) 8 a, D a a a a EI EI 48EI EI 4 4 4 47

Sample Problem For the structure composed of a beam and a frame shon, find the deflection at the center of the beam. Solution The deflection at section E is associated ith the folloing deformations: ending of beam itself. ending of ompression and bending of D l/ E F l/ EI l EI EI E l D Rigidize the frame (+D) E1 Fl 48EI E F 48

Rigidize + D F 1 1 ( ) l Fl E 1 EI 1EI 1 F/ Rigidize + 1 ( E Fl E (Deflection at due to the compression of D) ) F/ Fl/ F ( ll ) l l EI Fl EI D (Deflection at due to the bending of D) 49

1 Fl Fl ( ) E E EI Deflection at section E via superposition: Fl 1 1 1 Fl E E1 E E EI 48 1 4 4E 17Fl Fl 48EI 4E 50

More Eamples 51

More Eamples 5

More Eamples 100 kn P 5

Singular / Discontinuity Functions f n ( ) a n ( 0 a) n a a f 0 () f 1 () f () n = 0 n = 1 n = a (a) a (b) a (c) 54

0 1 1 1 n a n d a n n 1 0 0 1 n a n n a d d n n alculus of Singular Functions 55 a a a a f n n n 0 ) ( ) (

Euations of Shearing Forces & ending Moments M e F D E a 1 a a F l F 0 0 1 FS F 0 F a a 0 1 F F a a M F 0 M a F a a 0 1 F M e a1 F a a 1 0 1 e 1 56

oundary onditions Denote the shearing force and bending moment at the left boundary as F S0 and M 0 Generalized euation of shearing forces F ( ) F F a a s 0 1 S0 Generalized euation of bending moment M ( ) M F M a F a a 0 1 0 S0 e 1 57

Deflection and Slope by Singular Functions EI M() M ( ) M F M a F a a 0 1 0 S0 e 1 F F EI M M a a a 6 S0 1 0 e 1 1 M F M F EI a a a 6 6 4 0 S0 e 4 1 1 EI, EI 1 0 0 58

oundary Values F S0, M 0, 0 and 0 denote the boundary values of shearing force, bending moment, deflection and slope M F F S0 = F, M 0 = M 0 = 0, 0 = 0 F F S0 = F, M 0 = 0 0 0, 0 = 0 F S0 = 0, M 0 = 0 0 0, 0 0 (a) Fied support (b) Hinged support (c) Free end 59

Sample Problem Find the deflection at section and the slopes at sections and for the simply supported beam shon. l/ l Solution 1. Euations of deflection and slope FS0 l EI EI0 M 0!!! M0 FS0 l EI EI0 EI0!! 4! 4! 4 4 60

Determine boundary values FS0 F l, M 0 0, 0 0 8 Determine 0 from the boundary condition at the movable hinged support : 4 l l 4 l l 0 EI l EI0l l 00 8 6 4 4 16 18EI l l l EI 18 8 6 6 EI 18 8 6 4 4 l l 4 l 4 61

. The Slopes and and the deflection EI l 18 l 8 6 6 l EI l 18 l 0 18EI l 8 6 4 4 4 l 4 l 1 1 7l l ( ) EI 18 16 6 6 8 84EI 4 4 l 1 5l l ( ) EI 18 488 4 16 768EI 6

Moment-rea Theorems Geometric properties of the elastic curve can be used to determine deflection and slope. onsider a beam subjected to arbitrary loading, D D d d y M M d d d EI EI D D M d EI First Moment-rea Theorem: D d area under (M/EI) diagram beteen and D. 6

Moment-rea Theorems Tangents to the elastic curve at P and P intercept a segment of length dt on the vertical through. M dt 1d 1 d EI D M td 1 d = tangential deviation of EI ith respect to D Second Moment-rea Theorem: The tangential deviation of ith respect to D is eual to the first moment ith respect to a vertical ais through of the area under the (M/EI) diagram beteen and D. 64

pplication to antilevers & eams under Symmetric Loading antilever beam - Select tangent at as the reference. θ D y D 0, y 0 D t D Simply supported, symmetrically loaded beam - select tangent at as the reference. θ 0, y y D D y y y t y y t D D ma 65

ending Moment Diagrams by Parts Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately. onstruct a separate (M/EI) diagram for each load. - The change of slope, θ D/, is obtained by adding the areas under the diagrams. - The tangential deviation, t D/ is obtained by adding the first moments of the areas ith respect to a vertical ais through D. ending moment diagram constructed from individual loads is said to be dran by parts. 66

Sample Problem SOLUTION: Determine the reactions at supports. For the prismatic beam shon, determine the deflection and slope at E. onstruct shear, bending moment and (M/EI) diagrams. Taking the tangent at as the reference, evaluate the slope and tangential deviations at E. 67

SOLUTION: Determine the reactions at supports. R R D a onstruct shear, bending moment and (M/EI) diagrams. 1 a EI L 1 a EI a L 4EI a a 6EI 68

Slope at E: E E E a L 1 4EI a E L a 1EI a 6EI Deflection at E: y t t t E E D E D y E a L 8EI a L a L 1 a 1 4 4 4 4 a L a L a a L 4EI 16EI 8EI 16EI 69

pplication to eams under Unsymmetric Loadings Define reference tangent at support. Evaluate θ by determining the tangential deviation at ith respect to. t L The slope at other points is found ith respect to reference tangent. D D The deflection at D is found from the tangential deviation at D. FE FE t L L y FD FE DE t t L D D t 70

Maimum Deflection Maimum deflection occurs at point K here the tangent is horizontal. t L K 0 K K Point K may be determined by measuring an area under the (M/EI) diagram eual to -θ. Obtain ma by computing the first moment ith respect to the vertical ais through of the area beteen and K. 71

Stiffness ondition ma [] θ ma [θ] W ma : Maimum deflection θ ma : Maimum slope [], [θ]: Maimum alloable deflection and slope Stiffness calculation include: - Stiffness check - Rational design of cross-sections - Find the maimum alloable loads 7

Ways to Increase Fleural Rigidity Deformation of beams under bending is influenced by not only beam supports and loading condition, but also beam material, cross-section size and shape, and beam span. - Increase EI - Decrease beam span / increase supports - Improve loading - Rational design of cross-sections 7

Sample Problem Given: l = 8 m, I z = 70 cm 4, W z = 7 cm, [] = l/500, E = 00 Gpa, [σ] = 100 Mpa. Find: 1. the maimum alloable load from the stiffness condition;. Strength check. I0a P z l l 74

Solution Pl l ma [ ] 48EI 500 48EI P 7.11kN 500l [ P] 7.11 kn M Pl ma ma 60 MPa [ ] Wz 4Wz The strength condition is satisfied. 75

ending Strain Energy Strain energy density: u 1 E E M Total strain energy calculated from density M y U dv dv E EI L L M M y d d EI EI 0 0 Total strain energy calculated from ork done by bending moment.r.t. rotation 1 M, d d dθ EI Md EI 1 M d L M ( ) EI 0 EI du Md U d d Neutral Surface o 1 a 1 dθ ρ o b d 76