Slender Structures Load carrying principles Basic cases: Extension, Shear, Torsion, Cable Bending (Euler) v017-1 Hans Welleman 1
Content (preliminary schedule) Basic cases Extension, shear, torsion, cable Bending (Euler-Bernoulli) Combined systems - Parallel systems - Special system Bending (Timoshenko) Continuously Elastic Supported (basic) Cases Cable revisit and Arches Matrix Method Hans Welleman
Learning trajectory Order of topics based on increasing complexity of the governing differential equation (!!! not in line with the notes!!!) Recap of math partly in class but primarily DIY! Extensive use of MAPLE, so install MAPLE and get involved in it! Hans Welleman 3
Learning objectives Understand the outline of the classical displacement method for finding the ODE s for basic load cases Find the general solution for these loadcases and define the boundary and/or matching conditions for a specific application Solve the ODE s (by hand and MAPLE) Investigate consequences/limitations of the model and check results with limit cases Hans Welleman 4
Model (ordinary) Differential Equation (O)DE Boundary conditions Matching conditions Hans Welleman 5
General recipe to find DE (classical displacement method) Kinematic relation Constitutive relation Equilibrium condition DIFFERENTIAL EQUATION + Conditions at boundaries/interfaces = The Solution Hans Welleman 6
Extension (prismatic) axial stiffness EA external load q internal generalised stress, normal force N. BC axial deformation or strain ε displacement field (longtitudinal) u. BC Hans Welleman 7
Fundamental relations Kinematic relation ε = du Constitutive relation N = EAε (Hooke) Equilibrium N + q + N + dn = 0 dn = q du dn d u N = EA = EA = q ODE Hans Welleman 8
Example q = 5 kn/m F = 5 kn l = 10 m EA = 500 kn Solve the ODE using parameters Write down the boundary conditions Hans Welleman 9
I found as an answer u(x) = Boundary Conditions:........ Hans Welleman 10
Plot the results Suggestions for a check? Maple > restart; > ODE:=EA*diff(u(x),x$)=-q; > u:=rhs(dsolve(ode,u(x))); > eps:=diff(u,x); N:=EA*eps; kinematic or Dirichlet boundary condition at x=0 > x:=0: eq1:=u=0; dynamic or Neumann boundary condition at x=l > x:=l; eq:=-n+f=0; > sol:=solve({eq1,eq},{_c1,_c}); assign(sol); > x:='x'; u; N; > L:=10; F:=5; q:=5; EA:=500; > _C1; _C; > plot(u,x=0..l,title="displacement u(x)"); > plot(n,x=0..l,title="normal force N(x)"); Hans Welleman 11
What if Hans Welleman 1
Assignment : Foundation Pile Dimensions: l A EA Pile capacity: = 0 m = 50 50 mm = 5 10 9 N - point bearing, modulus of subgrade reaction (Beddingsconstante) c = 16 10 8 N/m 3 - (positive) friction or adhesion: Load: F τ = 3 10 4 N/m = 500 kn Find the DE to model this pile and the boundary conditions. Solve the DE and plot the normal force distribution in the pile. Find the displacement of the pile tip. Hans Welleman 13
Shear shear stiffness k external load q internal generalised stress, shear force V. BC Shear deformation or shear strain γ displacement field w. BC Hans Welleman 14
Fundamental relations Kinematic relation γ = dw Constitutive relation V = k γ (Hooke) Equilibrium V + q + V + dv = 0 dv = q dw dv d w V = k = k = q ODE Hans Welleman 15
Conclusion Basic Extension and Shear result in an identical ODE We now know all about this. What about the shear stiffness? Hans Welleman 16
Shear stiffness for beams 1 (in bending) assumption in this model: - constant V - constant γ discrete model Shear force V is integrated shear stress distribution. 1 Work, energy methods & influence lines, appendix A, ISBN 978-90-7830-95-1, J.W. Welleman Hans Welleman 17
Shear stress due to bending for a rectangular cross section fibre model parabolic shear stress distribution so.. no constant shear stress τ and no constant shear deformation γ Hans Welleman 18
Both models should generate the same amount of strain energy E discrete model: 1 1 E = V dw = V γ with: V = k γ discrete 1 1 Fibremodel: ( prereqisuite knowledge ) h z= τ ( z) E fibre = bd z with : τ ( z) = h z= G ( 1 ) 6V h z 4 bh 3 Hans Welleman 19
Elaborate z= h z= h τ ( z) G 1 bdz = V γ 1 ( ) h z= 1 b 6 V 4 h z dz V 3 1 G = γ h z= bh 6V γ1 = with: V = k γ1 and A = bh 5GA k 5 GA = GA = with: η = 1, 6 η Hans Welleman 0
Cross section is not a plane. V E discrete h γ 1 dw = γ 1 dw Hans Welleman 1
Shear Stiffness for Frames Source: https://gallery.autodesk.com/projects/fusion-360-bow-arch-vierendeel-truss-bridge-wip-1701015 Hans Welleman
Shear Stiffness k for Frames rigid floors general model Although the frame is loaded in bending a shear beam model can be used based on the global behaviour. Hans Welleman 3
Model with rigid floors k =?? use virtual work δu 4Mδγ = Hδu δγ = h M = 1 4 Hh : deformation : zero rotation at foundation : try this quickly.. 3 u Mh Mh Mh Hh 4EI + = 0 u = = = γ h H = γ h 3EI 6EI 6EI 4EI h 4EI thus : k = h Hans Welleman 4
General model k =??.. try this yourself.. answer : 4 k = h b h + EI k EI r Hans Welleman 5
Examples of shear beams Sketch the deformed situation Proof it! Hans Welleman 6
More examples of shear beams Sketch the deformed situation Proof it! Hans Welleman 7
Assignment F = 10 kn; k = 1000 kn; a = 5 m; n = 5 Find the displacement at midspan C for both cases Hans Welleman 8
So we need an extension Add a degree of freedom (rotation) And add an additional equation (moments) dw γ = + φ d M = V M = V d x NOTE Not in the notes, all shear beams in the notes have zero rotation of the cross sections Hans Welleman 9
Torsion torsional stiffness GI t external load m internal generalised stress, shear force M t. BC torsional deformation or specific twist θ displacement field ϕ x. BC Hans Welleman 30
Fundamental relations Kinematic relation θ = dϕx Constitutive relation M t = GI θ t (Hooke) Equilibrium M + m + M + dm = 0 dm ODE Hans Welleman 31 t t t t = m dϕx dm t d ϕ M x t = GIt = GIt = m
Cable 1 cable stiffness H load distributed along the projection of the cable cable without elongation??!! external load q internal generalised stress, vertical component V. BC Cable slope tanα displacement field w. BC Hans Welleman 3
Fundamental relations geometrical relation tanα = dz Moment equilibrium Equilibrium V = H tanα V + q + V + dv = 0 dv = q dz dv d z V = H = H = q ODE Hans Welleman 33
Remarks Cable takes no bending Cable ODE describes an equilibirum in the deflected state (funicular curve) so this is a non-linear approach! (no superposition) H can be regarded as constant only if no horizontal loads are applied This model is not valid for loads distributed along the cable! Derivation is strictly based upon equilibrium only! Cable force can be expressed in H and z : dz T = H + V = H + ( H tanα ) = H 1+ Hans Welleman 34
Examples Find the funicular curve and express f in terms of H, q and l. Hans Welleman 35
Horizontal component H support reaction of the block A A H V l q f f L B. q Vkatrol T H L F. B block F support reaction of the block components of the cable force T H T = F F Free Body Diagram of the block l T = F V V = 1 ql f ql = 8H force polygon for the block F H = F ql 1 f = 8 F ql 1 ql F 36
Results assume : ql = λf H f H = F 1 = 1 F 1 1 λ λ 4 4 8ql λl f λ = = = 8H 8 1 l 8 1 1 1 λ λ 4 4 H, F f l H F f l λ = Hans Welleman F 37 ql
Bending in x-z plane bending stiffness EI external load q internal generalised stresses M, V. BC deformation is the curvature κ displacement field w, ϕ. BC Hans Welleman 38
u( z) = zϕ A lot d u( zof ) parameters dϕ ε ( z) = = z = κ z z= z= z= h h h M = dm = zd N = zσ ( z)da z= z= z= h h h z= z= h h M = E z ε ( z)da = Eκ z da = EI κ z= z= h h Hans Welleman 39
Equilibrium vertical equilibrium dv = q moment equilibirum dm = V Proof this in a minute! Hans Welleman 40
Fundamental relations Kinematic relation dw dϕ ϕ = ; κ = ; Constitutive relation M = EI κ (Hooke) Equilibrium dv dm = q; = V ODE Hans Welleman 41
Result Bending in x-z plane M = EI d M = q; d w ; Hans Welleman 4 d d w EI q = Sufficient to solve static determinate structures
Bending - Euler Bernoulli Mind the coordinate system used! Basic model for simple bending with symmetrical cross sections For prismatic beams use: EI d 4 w 4 = q; 3 d w d w dw V = EI ; M = EI ; ϕ = ; 3 Hans Welleman 43
Options for Boundary Conditions kinematic ϕ and/or w dynamic SD SD V=0 V M and or V SI Note: SI M or ϕ V or w V=0 V Static Indeterminate (SI) for 3 or more kinematic BC V=0 V=0 V V Hans Welleman 44
Examples Find the deflection line and the force distribution. Bending: Neglect the influence of possible axial deformation. Hans Welleman 45
Wrap-up Basic Cases Second order DE Extension Shear Torsion Cable Fourth order DE Bending d u EA = q d w k = q d ϕ GI x t = m d z H = q 4 d w EI = q 4 Hans Welleman 46
Assignment k = 1000 kn; EI = 1500 knm ; F = 5 kn; q = 8 kn/m; Compare force distributions and deflections Hans Welleman 47