Elio Sacco. Dipartimento di Ingegneria Civile e Meccanica Università di Cassino e LM

Transcription

1 Elio Sacco Dipartimento di Ingegneria Civile e Meccanica Università di Cassino e LM

2 The no-tension material model is adopted to evaluate the collapse load. y Thrust curve of the arch extrados intrados axis of the arch x The objective is the determination of the thrust curve, i.e. the line described by the eccentricities along the arch which ensures the equilibrium condition and it is completely contained in the physical geometry of the arch, as the one illustrated in figure with the green color.

3 In a masonry arch, it is assumed that (Heyman s hypotheses): 1) masonry has no tensile strength; ) its compressive strength is practically infinite in comparison with the stresses occurring in real structures; 3) friction is large enough to prevent sliding between masonry elements. The line of thrust is the imaginary curve through which the resultant thrust acts in the masonry arch. When the masonry arch is subjected only to its own weight, the line of thrust describes a very special curve which is known as "catenary". The first idea about this special curve is due to Galileo, who indeed stated that the catenary is a parabola. Joachim Jungius ( ) proved that the catenary is not a parabola. Applications of the concept of the catenary to the construction of arches was due to Robert Hooke (1671); many scientists in the 17th and 18th centuries contributed in deriving and solving the equations governing the catenary, among the others Gottfried Leibniz, Christiaan Huygens and Johann Bernoulli. Next, the problem of the catenary is approached. The catenary is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends. 3

4 Equilibrium With reference to the figure, the equilibrium equations for the typical infinitesimal part of the catenary can be written as: ( ) ( ) ( ) sin ( ) N + N cos θ + θ Ncosθ = 0 N + N θ + θ Nsinθ = f s θ N θ f N θ + θ + N where ff is the distributed load along the curvilinear abscissa ss. Dividing all the terms by ss, in the limit as ss 0, it results: ( N + N) cos( θ + θ) Ncosθ d 0 = lim ( N cosθ ) = 0 s 0 s ds ( N + N) sin ( θ + θ) Nsinθ d 0 = lim f ( N sinθ ) = f ds s 0 s According to the first equilibrium equation the projection of the axial force in the horizontal direction is constant along the whole arch and it is, in fact, the drift force HH, i.e. the horizontal reaction of the constraint, so that it is: N = H cosθ 4

5 θ N f The second equilibrium equation becomes: ' ' f= d ( Htan θ ) = H dy = H dy dx = Hy'' dx ds ds dx ds ds where tan θθ = yyy with yyy and yy the first and second derivatives of the function yy(xx), respectively. Recalling that: ( ) θ N θ + θ + N ds = 1 + y ' dx so that it results: f y'' + H 1 + ( y' ) = 0 which is the differential equation governing the catenary. The solution is: H f y = cosh x+ c1 + c f H where cc 1 and cc are constants of integration. 5

6 As a solution symmetric with respect to the yy axis is desired; the conditions for the determination of cc 1 and cc are: y ( 0) ( ) = h y '0 = 0 h rise of the arch. c 1 = 0 ( ) c = hf+ H / f y h x Finally, the catenary is defined by the function: H f y = 1 cosh x h f + H To uniquely determine the shape of the catenary, i.e. to find the function yy(xx) a value of the trust force HH has to be prescribed. The cases of nonuniform distributed vertical loading or of the presence of horizontal distributed loads lead to different differential equation with respect to the one obtained for the catenary. 6

7 Arch with assigned shape When the shape of the arch is assigned by its mid-line function yy xx, where the curvilinear abscissa ss is introduced, it is possible to evaluate the loading distribution which makes the function yy xx a thrust curve. In fact, in this case it is sufficient to solve the deduced equilibrium equation f y'' ( y' ) = 0 H setting yy xx = yy xx : f = H with respect to the loading distribution ff, y'' ( y ) 1 + ' 7

8 For instance, considering a parabolic thrust curve y= h ax y' = ax y'' = a f = ah 1+ ( ax) parabolic thrust curve For a circular thrust curve, it is y = h x y ' = y '' = h H f = lim f = x h 3/ h ( h x ) h x x h ( ) h x h x 3/ circular thrust curve 8

9 More complicated is the case of the evaluation of the thrust curve for assigned loading conditions. In fact, the equilibrium equation becomes more complex in this case. In the following the case of an arch of an assigned shape is considered. The arch is loaded by its weight, by an horizontal force proportional to its weight (due for instance to a possible seismic action) and by distributed forces applied on the extrados and on the intrados of the arch with vertical and possibly horizontal direction. MILANKOVITCH, M Beitrag zur Theorie der Druckkurven. Dissertation zur Erlangung der Doktorwürde, K.K. technische Hochschule, Vienna. 9

10 G center of mass M point on the mid-curve R e extrados radius R i intrados radius R m mid-curve radius y H horizontal thrust force V vertical thrust force f e vertical distributed load on the extrados p e horizontal distributed load on the extrados f i vertical distributed load on the intrados p i horizontal distributed load on the intrados g weight q horizontal (inertial) force Rotational equilibrium around point E y H ξ P thrust curve θ η V E p i s i f i s i x G M V+ V R i f e s e p g p e s e mid-curve P+ p ( ) e e( e m) i i( m i) ( ) ( ) ( ) V x H y g ξ + η sinθ f s ξ + R R sinθ + f s R ξ R sinθ p ξ + η cosθ p s ξ + R R cosθ + p s R ξ R cosθ = 0 e e e m i i m i R m H+ H R e x

11 G center of mass M point on the mid-curve R e extrados radius R i intrados radius R m mid-curve radius H horizontal thrust force V vertical thrust force f e vertical distributed load on the extrados p e horizontal distributed load on the extrados f i vertical distributed load on the intrados p i horizontal distributed load on the intrados g weight q horizontal (inertial) force Horizontal and vertical equilibrium y y H P thrust curve θ ξ η V E p i s i f i s i G M V+ V x R i f e s e p g p e s e mid-curve P+ p R m H+ H R e x H + p+ pe se + pi si = 0 V g fe se fi si = 0

12 Simple geometrical issues give: η d = R R height of the cross-section ξ e 1 1 Ai = Ri θ Ae = Re θ Si = Ri Ai Se = Re Ae A= Ae Ai = ( Re Ri ) θ S = Se Si = ( Re Ri ) θ 3 S d RG = = Rm + A 1 R d η = RG Rm = 1 R i 1 g = γ A= γ ( Re Ri ) θ m m area of the internal and external sector static momentum of the internal and external sector area of the circular sector s i static momentum of the circular sector radius of the center of mass E G M R i distance of the center of mass from the mid-curve s e R m R e s = R θ s = R θ e e i i

13 Dividing by θ in the limit as θ 0, the equilibrium equations become: dx dy 1 d V H γ ( Re Ri ) ξ + sinθ + fr e e( ξ + Re Rm) sinθ fr i i( Rm ξ Ri) sinθ dθ dθ 1 Rm 1 d + γ p( Re Ri ) ξ + cosθ + pr e e( ξ + Re Rm) cosθ pr i i( Rm ξ Ri) cosθ = 0 1 Rm dh 1 + γ ( R R ) + pr + pr = 0 dθ h e i e e i i dv 1 γ ( Re Ri ) fr e e fr i i 0 dθ = with the further condition that xm x ξ = ξsinθ = Rm sinθ x sinθ Note that the horizontal (inertial) force has been considered proportional to the volume through an apparent density γ h.

14 Moreover, the following condition holds true: y ξ x x tanθ = = y+ y y+ R cosθ y 0 where y is the assigned midcurve of the arch satisfying the relationship: m y m y y E M y + y = y with y = R 0 m m m cosθ y = R cosθ y 0 m θ x y 0 x x m

15 Finally, accounting for the condition ξ = x sinθ xm the governing equations become: dx dy 1 d V H γ R R x x+ θ + fr R θ x fr x R θ dθ dθ 1 Rm 1 1 d + γ p( Re Ri ) xm x+ sinθ + pr e e( Resinθ x) pr i i( x Risinθ) = 0 tanθ 1 Rm dh 1 + γ ( R R ) + pr + pr = 0 dθ h e i e e i i dv 1 γ ( Re Ri ) fr e e fr i i 0 dθ = ( cosθ ) x= y+ R y m ( e i ) m sin e e( esin ) i i( isin ) tanθ They are 4 (nonlinear) differential in the 4 unknown functions: ( θ), ( θ), ( θ), ( θ) x y V H

16 y y mid-line of the arch Super simplified approach y thrust line of the arch x ( sinθ) d( H tanθ) d N ds ds dx ds = f = f Hy'' = f ds ds ds ds ds ds Recalling that: it results ( ) ds = 1 + y ' dx dx dx Hy'' f 1 y' Hy'' f 1 y' ds = + ds = + ( ) ( ) θ N s f s θ+ θ N+ N 16

17 Considering a round arch, it is: y = R x so that the governing differential equation becomes: x Hy = f + Hy R x fr R x + = The solution of the differential equation is: y ( 0) ( ) = h y '0 = 0 '' 1 '' 0 f x H R x + + y= R xarcsin + R x cx 1 c the conditions for the determination of cc 1 and cc are: Finally, the solution is: c 1 = 0 ( ) c = R f + hh / H f x = + + h H R x y R xarcsin R x R 17

18 The case of an arch subjected to its weight and to a pointwise force FF is considered, as illustrated in figure. y F The whole arch is split in two parts: elem1 for xx (xx AA, xx BB ) elem for xx (xx BB, xx CC ) y 0 xa α R d α x F xb x As no horizontal forces are applied on the arch, the thrust HH is constant in the whole structure. Two differential equations are written for the two elements composing the arch: Hy R x + fr= 0 Hy R x + fr= 0 '' '' 1

19 The solution of the two differential equations is: f x y= R xarcsin + R x + cx+ c H R x 1 1 f x y = R xarcsin + R x + c 3 x+ c4 H R x where cc 1, cc, cc 3 and cc 4, are determined by imposing suitable boundary conditions. In particular, the considered boundary conditions are: x= Rd y1 = y 0 x= xf y1 y = 0 ' ' x= xf y1 y = F / H x= R y = y d 0 θ 1 N 1 The value of yy 0 depends on the position of the arch. Solving the system of equations it is possible to evaluate the constants cc 1, cc, cc 3 and cc 4. F N θ equilibrium along the yy axis: N sinθ = N sinθ + F 1 1 H tanθ = H tanθ + F y 1 y = ' ' 1 F H 19

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22 The case of an arch subjected to its weight and to a distributed horizontal load simulating the seismic action is considered. The equilibrium equations of the arch become: dd dd ppp NN cos θθ = ss fff NN cos θθ = ss horizontal equilibrium vertical equilibrium Setting HH = NN cos θθ the first equilibrium equation takes the form: = pppss = pppss = pppss recalling that ddss = 1 + yy the equilibrium alonf the x direction is written as: HH = pp 1 + ( yy )

23 The second equilibrium equation is rewritten as: i.e. dd fff HH tan θθ = ss yy + ppyy + ff tan θθ HHyy 1 + ( yy ) HH = 0 dd = ff ss Finally, the system of differential equations to solve is: HH pp 1 + ( yy ) = 0 yy + ppyy + ff 1 + ( yy ) HH = 0 which allows to determine, prescribing suitable boundary conditions, the two unknown functions HH(xx) and yy xx. 3

24 To solve the system of differential equations, the finite difference numerical method is used. It is based on the idea that the derivative of a function can be approximated as its incremental ratio. Let in the domain be introduced a set of abscissae (xx 1, xx,, xx nn ) whose distance is xx, assumed to be constant for simplicity. Starting from the value of the function at xx ii, the function at xx ii+1 and at xx ii 1 can be evaluated using the series development as: ff xx ii+1 = ff xx ii + ff xx ii xx + 1 ff xx ii xx + ff xx ii 1 = ff xx ii + ff xx ii xx + 1 ff xx ii xx + Adding and subtracting the two formulas, it is obtained: ff xx ii+1 ff xx ii 1 = ff xx ii xx ff xx ii = ff xx ii+1 ff xx ii 1 xx ff xx ii+1 + ff xx ii 1 = ff xx ii + ffff xx ii xx ff xx ii = ff xx ii 1 ff xx ii + ff xx ii+1 xx 4

25 The derived formulas approximating the first and the second derivative of the function ff(xx) are known as mid-point finite difference method and can be rewritten in the simplified form as: ff ii = ff ii+1 ff ii 1 xx ff ii = ff ii 1 ff ii + ff ii+1 xx Other forms of finite difference approximations are also available; schematically for the first derivative it is: ff ii = ff ii+1 ff ii xx ff ii = ff ii+1 ff ii 1 xx ff ii = ff ii ff ii 1 xx forward Euler mid-point backward Euler 5

26 The system of differential equations is: HH pp 1 + ( yy ) = 0 yy + 1 HH ppyy + ff 1 + ( yy ) = 0 The first differential equation is solved adopting the backward Euler finite difference technique. xx 1 xx xx 3 xx nn distance is xx is introduced. A set of nn points (xx 1, xx,, xx nn ) whose The typical difference finite equation written at point xx ii is: HH ii HH ii 1 xx = pp 1 + yy ii yy ii 1 xx nn 1 equations can be written for ii =,, nn 6

27 The following further equation can be written: where HH is the prescribed value of the thrust force. HH 1 = HH Finally, the following system of algebraic equation is obtained: HH 1 HH HH 3 HH nn = ρρ 1 ρρ ρρ 3 ρρ nn where ρρ 1 = HH ρρ ii = pp 1 + yy ii yy ii 1 xx xx ii =,, nn 7

28 The second differential equation is solved adopting the mid-point finite difference technique. The typical finite difference equation at point xx ii is: yy ii+1 yy ii + yy ii 1 xx + 1 HH pp yy ii+1 yy ii 1 xx + ff 1 + yy ii+1 yy ii 1 xx = 0 or, equivalently, yy ii+1 yy ii + yy ii HH pp yy ii+1 yy ii 1 xx + ff xx 1 + yy ii+1 yy ii 1 xx xx = 0 The finite difference equation is written at points xx ii with ii =,, nn 1, involving the nn unknowns (xx 1, xx, xx nn ). The system of equations is completed by the further conditions yy 1 = yy nn = 0. 8

29 Finally, the following system of algebraic equation is obtained: pp 1 + pp pp pp pp nn pp nn yy 1 yy yy 3 yy nn 1 yy nn = yy ii FF FF 3 FF nn 1 FF nn where pp ii = pp 1 HH 1 + yy ii+1 yy ii 1 xx FF ii = ff 1 HH 1 + yy ii+1 yy ii 1 xx xx = 0 9

30 Round arch subjected to vertical and horizontal distributed laoding, f and p, respectively. 000 trust curve % trust H0 = 650; % geometry Rex = 1650; Rin = 1350; R = (Rex+Rin)/; theta = 0.05; % loading f = 1; p = 0.07;

31 Kinematic approach The case of an arch subjected to its weight and to a pointwise force FF is considered, as illustrated in figure. y F y 0 α R d α x 31

32 Kinematic approach The case of an arch subjected to its weight and to a pointwise force FF is considered, as illustrated in figure. C 1 C 1 y θ θ3 θ 1 θ 4 C x1 x x5 x3 x4 ϕ1 ϕ ϕ3 F C 3 C 3 x To apply the kinematic theorem of the limit analysis, a possible collapse mechanism has to be introduced. As shear failure is not admitted for the considered no-tension material, collapse can be obtained by the opening of hinges. In particular, for the arch 4 hinges are required for the formation of a mechanism. Generally, collapse occurs for the formation of hinges at the extrados and hinges at the intrados of the arch, as illustrated in figure. In this case the centers of rotation CC 1, CC 1, CC 3 and CC 3 can be determined. 3

33 The positions of the 4 hinges are located by the angles θθ 1, θθ, θθ 3, θθ 4 evaluated with respect to the y axis, assuming θθ ii > 0 when it is clockwise. In the Cartesian coordinate system xx yy, the hinges are located as: C1 θ1 x1 = Resinθ1 y1 = Recosθ1 C1 θ x = Risin θ y = Ricosθ C3 θ3 x3 = Resinθ3 y3 = Recosθ3 C θ x = R sinθ y = R cosθ i 4 4 i 4 The center case of rotation CC can be found by enforcing the alignment condition: CC 1 - CC 1 - CC and CC - CC 3 - CC 3 : y1 x1 x y1x1 x1y1 y x = y y x x y with x = x x y = y y x = x x y = y y Solving the above equation it is possible to determine the coordinates xx 5, yy 5 corresponding to the position of CC defined by the angle θθ 5. 33

34 The mechanism is defined by the rotations φφ 1, φφ, φφ 3 which, indeed, are related each other by the relationship: x x5 0 ϕ x x1 ϕ1 x x x x = ϕ The vertical component of the displacement field is given by: ( ) ( ) ( ) v = ϕ x x v = ϕ x x v = ϕ x x so that, the virtual work equilibrium equation is: L + L + L 0 = Fϕ( x3 x5) + L1+ L + L3 F = ϕ 3 5 with θ ( x x ) ( sin ) ( cos cos ) ( ) L = pϕ R θ x Rdθ = pϕ R θ + θ xr θ θ θ θ ( sin ) ( cos cos ) ( ) L = pϕ R θ x Rdθ = pϕ R θ + θ x R θ θ θ θ ( sin ) ( cos cos ) ( ) L = pϕ R θ x Rdθ = pϕ R θ + θ xr θ θ θ 34