Slender Structures Load carrying principles

Transcription

1 Slender Structures Load carrying principles Continuously Elastic Supported (basic) Cases: Etension, shear Euler-Bernoulli beam (Winkler 1867) v Hans Welleman 1

2 Content (preliminary schedule) Basic cases Etension, shear, torsion, cable Bending (Euler-Bernoulli) Combined systems - Parallel systems - Special system Bending (Timoshenko) Continuously Elastic Supported (basic) Cases Cable revisit and Arches Matri Method Hans Welleman 2

3 Learning objectives Etend the technique for basic cases Find the ODE for a specific case and the boundary conditions for the specific application Solve the more advanced ODE s (by hand and MAPLE) Investigate consequences/limitations of the model and check results with limit cases Hans Welleman 3

4 Basic Cases Second order DE Etension Shear Torsion Cable Fourth order DE Bending d u EA = q 2 d d w k = q 2 d 2 d ϕ GI t = m 2 d d z H = q 2 d d w EI = q d Hans Welleman

5 Model (ordinary) Differential Equation (O)DE Boundary conditions Matching conditions Hans Welleman 5

6 Etension (prismatic) aial stiffness EA eternal load no internal generalised stress, normal force N. BC aial deformation or strain ε displacement field (longtitudinal) u. BC Hans Welleman 6

7 Fundamental relations Kinematic relation ε = du d Constitutive relation N = EAε (Hooke) Equilibrium N pd + N + dn = 0 dn = p = ku d du dn d u N = EA = EA = ku d d 2 d ODE Hans Welleman 7 2

8 Eample: pull out test concrete rebar l not known yet Solve the ODE using parameters Write down the boundary conditions Hans Welleman 8

9 I found as an answer u() = Boundary Conditions: Hans Welleman 9

10 Some values Find some results for 25 kn load steel rebar φ = 10 mm; f y Es = 00 N/mm ; = N/mm ; concrete: 3 2 E 1 c = N/mm ; k = E 2 c Hans Welleman 10

11 Interpretation of results Slope of u(0) and N(0) intersect -ais at distance 1/α from origin. 1/α is length of the influence zone Find α by eperiment: impose F and measure u(0) F F = keu(0) ke = = kea F k = k 2 e EA kea Hans Welleman 11

12 Shear (prismatic) shear stiffness GA eff eternal load q internal generalised stress, shear force V. BC shear deformation γ displacement field w. BC Hans Welleman 12

13 Fundamental relations Kinematic relation γ = dw d Constitutive relation V = GAeff γ (Hooke) Equilibrium V + ( q kw)d + V + dv = 0 dv = kw q d dw dv d w N = GAeff = GAeff = kw q d d 2 d 2 ODE Hans Welleman 13

14 Veerse Gat 1961 Photo : Aart Klein, 2 april 1961 Why this shear type of caisson? Gates are closed after positioning of all caissons. The walls must remain parallel in order to operate the gates. Hans Welleman 1

15 Result so far etention (homogeneous): homogeneous solution: 2 2 d u d u 2 2 k EA + ku = 0 u α u 0 with: 2 h( ) C= = 2 1e α α = + C2e d d EA homogeneous solution: shear (inhomogeneous): α α wh ( ) = C1e + C2e 2 2 d w d w 2 q 2 k GAeff + kw = q particular α wsolution: = with: α = 2 2 d d GAeff GA q q( ) = q ( ) o o wp = k q q( ) = q ( ) o o wp = k q q( ) = q sin π ( ) o 2 sin o w l p = π π l 2 GA l eff + k Hans Welleman 15 eff

16 Beam (prismatic, only bending deformation) k = foundationmodulus [N/m 2 ] c = modulus of subgrade [N/m 3 ] (beddingsconstante), gravel c = 10 8 N/m 3 sand c = 10 7 N/m 3 eternal load q internal generalised stress, shear and moment V, M. BC bending deformation, curvature κ rotation and displacement ϕ, w. BC Hans Welleman 16

17 Fundamental relations Kinematic relation dw ϕ = ; κ = d dϕ d Constitutive relation M = EIκ (Hooke) Equilibrium V + ( q kw)d + V + dv = 0 dv dm = kw q; = V d d dϕ dv d w M = EI = EI = kw q d d d ODE Hans Welleman 17

18 Solving the ODE d w EI + k w = q d rewrite as: Find d homogeneous solution: β w with : β d d w + β w = w q k + = = EI 0 EI d subst : w = e λ move to slide 2 or if fan of math continue λ + β = 0 λ = β Hans Welleman 18

19 some math λ = β λ = β ( 1) λ = β 1 ( ) 2 2 λ = λ = β 2 1 = β 2 1 ( ) λ = β 2 1 = β 2 1 Comple 1 1 Hans Welleman 19

20 some background 1 = cosπ + i sinπ = e note: plus or min 2 π is the same, so: 1 = cos( π + 2 mπ ) + i sin( π + 2 mπ ) = e iπ i( π + 2 mπ ) π mπ π mπ = = = four roots for m = 0,1, 2 en 3 1 ( ) ( ) 1 i( π + 2 mπ ) i( π / + mπ / 2) 1 e e cos( ) i sin( ) 1+ i 1-i -1+ i 1-i,,, Hans Welleman 20

21 apply this λ 1,2,3, ( ) 1 = β 2 1 λ 1,2,3, ± 1± i = β 2 2 λ = β (1 + i) λ = β (1 i) 1 2 λ = β ( 1 + i) λ = β ( 1 i) 3 Homogeneous solution: w ( ) = Ae + Be + Ce λ + De h λ λ λ Hans Welleman 21

22 .. a few more last steps w ( ) = Ae + Be + Ce λ + De h λ λ λ w ( ) = Ae + Be + Ce + De h β (1 + i) β (1 i) β (1 i) β (1 + i) ( ) ( ) w ( ) = e Ae + Be + e Ce + De h β β i β i β β i β i what is this comple stuff?? Euler : βi e = cos β + isin β and βi e = cos β i sin β Hans Welleman 22

23 .. almost there { β β } { β β } { cos β + sin β } + { cos β + sin β } ( ) β w ( ) = e A cos + i sin + B cos i sin + h ( ) β e C i D i ( β β ) (( + ) cos β + ( )sin β ) β w ( ) = e ( A + B) cos + i( A B)sin + h β e C D i C D use new constants: C = A + B; C = i( A B) real 1 2 C = C + D; C = i( C D) real 3 A-B and C-D comple comple conjugate - assume: A = a ib; B = a + ib; A + B = 2 a; A B = 2ib Hans Welleman 23

24 .. Okay we are there β ( β β ) ( β β ) β w ( ) = e C cos + C sin + e C cos + C sin h sinoidal shape (wave) sinoidal shape (wave) damping term for < 0 damping term for > 0 damping or decreased amplitude is governed by β : 1 1 β = k EI characteristic length [m] Hans Welleman 2

25 Eample β ( β β ) ( β β ) β w ( ) = e C cos + C sin + e C cos + C sin h Hans Welleman 25

26 Symmetry ( β β ) β w( ) = e C cos + C sin 3 only two BC: ϕ (0) = 0; V (0) = F; Hans Welleman 26

27 Assignment Solve the eample in MAPLE Use 1 β = π; EI = 1000 β ; F = 10 Find the distributions for Displacement Rotation Moment Shear Hans Welleman 27

28 one last trick ( β β ) β w( ) = e C cos + C sin 3 ( ω β ω β ) β w( ) = e Asin cos + Acos sin ( ω β ω β ) β w( ) = A e sin cos + cos sin β ( ) = sin ( β + ω) w Ae A C3 = Asinω ω C = Acosω math: ( ) sin a + b = sin a cosb + cos asin b new integration constants Hans Welleman 28

29 Why?? β ( ) = sin ( β + ω) w Ae dw d β ( ) cos( ) β = β Ae sin β + ω + β Ae β + ω = ( sin ( ) cos( )) β ( ) ( ) = β β + ω β + ω β Ae ( 1 1 ) = 2 β Ae sin β + ω cos π cos β + ω sin π dw d 1 ( ) β = 2 β Ae sin β + ω π math: ( ) sin a b = sin a cosb cos a sin b Hans Welleman 29

30 Differentiating becomes Change sign Multiply with β 2 Reduce phase with ( β ω) β w = Ae sin + 1 π dw β 1 = 2 β Ae sin ( β + ω π ) d 2 d w 2 β 1 = 2β Ae sin 2 ( β + ω 2 π ) d 3 d w 3 β 3 = 2 2β Ae sin β + ω 3 π d ( ) old school Hans Welleman 30

31 Study the graphs and the notes Hans Welleman 31

32 Classification of beams according to stiffness w( ) =? Hans Welleman 32

33 Classification Wave length of the load 2l Wave length λ of the beam (no load) α ( β β ) w( ) = e C cos + C sin + α 1 2 ( cos β + sin β ) e C C 3 βλ = 2π λ = 2π β Hans Welleman 33

34 Simplified model w( ) =? w ( ) =? w bending w bending =? w( l) Assignment :? q / k = Hans Welleman 3 1 2

35 Influence of k, EI and l Find the meaning of this result by studying the etreme values of the parameters Hans Welleman 35

36 Assignment rail (beam) Find min and ma stress in the continuous support Find the maimum normal stress in the beam Hans Welleman 36

37 Assignment Building phases A original situation B add supports A and B with full load prior to ecavation C ecavate AB under full load Derive a model and find the mnoment at B in the beam Find the moment and shear distribution in terms of q o and l Find the support reactions at A and B assume : kl EI = 32 Hans Welleman 37