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Estratto del documento

Curved Beams and Parabolic Arch

In the end, substituting in the equation of σ we get:

θ R R N M M N M− − − −σ = E ε = E y =(η y χ) = y + yθ θ 2R + y R + y A A R I AR A RR R N R + y M R + y M N M M Ry− − − −= y =R + y A R AR R I A A R I R + yR R

We can see that σ has a non-linear pattern as ε .θ θ2.

1. Parabolic Arch

Let us study the parabolic arch in the figure:

The curved beam

We call f the maximum height of the arch. The analytic equation of the beam is: 2z z −y = 4 f +l l

Since the system is one time redundant we will make it isostatic and we add a redundant actionX.

System 0

The inclination α is: 4 fdy − −= (2 z l)tan α = 2dz l

Theory of Structures 212. Curved beams 2.1. Parabolic arch

Doing equilibrium we get: N l0− −→ −N cos

α T sin α = 0 T = = pz p cos α0 0 0 tan α 2 l 1 l− −→ −N sin α + T cos α = p z p N tan α + T = pz p0 0 0 02 cos α 2 N 1 1 l0−→ −N tan α + = N tan α + = pz p0 0tan α tan α cos α 2 1tan α l l−→ − −N = pz p = pz p sin α0 2 cos α 2 2tan α + 1pl p 2−M = z z0 2 2In “system 1” we get: −→ −NN sin α + T cos α = 0 T = tan α = sin α1 1 1 1 112 −−→ −− −1 −→ − = cos αN =N cos α T sin α = N + N tan α = 11 1 1 1 2 cos α cos α 1 + tan α−yM =1The internal work is:ZL = (N η + T t + M χ) ds =i 1 1 1s Z N + X N T + X T N + X NM + X MR0 R1 0 1 R0 R10 1 −= N + T ds+ M1 1 1EA G A E I E ARR RsWhere: M I R∗−N = N A = A +R R 2R ROnce we know the internal work

we make several approximations to get the solution for X:

  • −→ ≃if h/R < 0.25 I IR
  • MN , χ =if h/R < 0.10 we can assume the same constitutive law for straight beams
  • η = E A EITand t = ∗G A
  • →if our arch is slender enough, shear strain is negligible t = 0
  • ≃if α is small we can neglect N (sin α 0)
  • IA and I = because this is the only way to obtain an analyticalwe assume A = 0 0cos α

The internal work reads: ls ZZ X N M + X M X N M + X M dz1 0 1 1 0 1L = N + M ds = N + M =i 1 1 1 1E A E I E A E I cos α0 0l l 2 Z ZX N M + X M cos α y pl p X1 0 1 2 2− −= N + M dz = X z z + y dz1 1E A E I E A E I 2 2 E I0 0 0 0 00 0

Let us analyze the three components of the integral separately: ll l l 2 Z Z Z1 1 l 4 f2 − −cos α dz = arctan (2 z l) =dz = dz =2 2 28 f l1 + tan α h i4 f0 0 0 − − 01 + (2 z l)2l2 l 4 f ≃= arctan l4 f

Theory of Structures 222. Curved beams

2.1. Parabolic arch

Doing this approximation if f /l < 0.1 the error is of the 5% and if f /l < 0.2 the error is of the 15%.

ll 4 5 33 Z 2 z z p l fpl p z2 − −−y − + =z dz = 2 p l f 2 3 42 2 3 l 4 l 5 l 150 0ll 2 5 4 2 3 2 Z 16 f z 2 l z l z 8 f l2 −y dz = + =4l 5 4 3 150 0

The principle of virtual work gives us:

2 2 3 1 l 4 f 1 8 f l p l f−X arctan + =0E A 4 f l E I 15 15 E I0 0 0pl 1X = = plh i 8 f8 f [1 + ξ]I l151 + 0 l3l 32 A f0 2A l2

We define the slenderness of the arch as λ = :0I 02 15 l l1 4 fξ = arctan2 28 λ f 4 f l

Figure 2.5: Variation of X with f /l and λ

Looking at the graph on the left we can see that for increasing values of f /l the unknown X tends2p l → →to and for f /l 0 also X 0. Looking at the graph on the right we can see that the8 fmaximum shifts towards left for increasing values of λ.−→ ≃λ = 69 X 3.13 p

lmax→ ≃λ = 52 X 2.35 p lmax→ ≃λ = 35 X 1.55 p lmax→ ≃λ = 28 X 1.24 p lmaxThe total bending moment is: 2 2 2 22 z z pl z z p l z z ξp l − − − −4 f =M = M + X M =0 1 8 f2 2 22 l l l l 2 l l 1+ ξ(1 + ξ)lThe maximum bending moment in our arch is: 2p l ξM =max 8 1+ ξTheory of Structures 232. Curved beams 2.2. Circular beamFigure 2.6: Variation of M with f /l and λmax2.2 Circular beamHaving a circular beam we immediately know that the radius of curvature is constant and equalto R. We assume that the beam is slender:• d u u d u u≃ −→ −→ −axial deformations are negligible η 0 + = 0 =s r s rds R ds R• d u u d u u≃ −→ − − −→ −shear deformations are negligible t 0 φ = 0 φ =r s r sds R ds RThe curvature is therefore: 2 2dφ d u 1 d u d u ur s r r−χ = = = +2 2 2ds ds R ds ds RThis

is the differential governing equation of the problem. We need 4 boundary conditions and we start observing that we can exploit the double symmetry of the problem

Setting of the problem

The bending moment is: P P s M (s) = X + R sin α = X + R sin2 2 R

The differential equation is: 2 d u u M 1 P sr r+ = χ = = X + R sin2 2ds R E I E I 2 RR R2 2 2

Since s = R α we can say that ds = R dα :2 2 2 d u 1 P d u R P1 r r−→+ u = X + R sin α + u = X + R sin αr r2 2 2R dα E I 2 dα E I 2R R

The solution for this differential equation is: 2 3X R P R−u = A cos α + B sin α + α cos αr E I 4 E IR R

The boundary conditions are u (0) = 0, u (π/2) = 0, φ(0) = 0 and φ(π/2) = 0.

s sZd u ud u s rs − −→ −= = u = u dαs rds R dα R α 32 P RX R−A − α + (α sin α + cos α) + C

= sin α + B cos αs E I 4 E IR R2 3 dφ d u 1 P R1 r −A −+ u = cos α B sin α += (2 sin α + α cos α) + A cos α+r2 2 2ds R dα R 4 E I R2 3 X R P R X PR−+B sin α + α cos α = + sin αE I 4 E I E I 2 E IR R R R2Z PR XR P R CX − −φ = + R dα = α cos αE I 2 E I E I 2 E I RR R R RαSubstituting boundary conditions we have: 33 P RP R −→+ C B =u (0) = 0 = B +s 4 E I 4 E IR R2 3 3X R P R P Rπ π π−A − −→u (π/2) = 0 = + + C A =s E I 2 4 E I 2 E I 8R R R2 3P R C P R− − −→ −φ(0) = 0 = C =2 E I R 2 E IR RXR C PRπ − −→ −φ(π/2) = 0 = X =E I 2 R πRIn the end the solution is: 3 1 X αP R π − −cos α + sin α cos αu =r E I 8 4 π 4R3 P R 1 α α 1π −−u = sin α + cos α + + sin αs

E I 8 2 π 4 2R2 P R α 1 1− −φ = cos α +E I π 2 2R Theory of Structures 253 Theory of plates A plate is a bidimensional body with a thickness t which is usually constant, but it may vary, which is loaded on the surface by external forces. We define the curvilinear coordinate s as the coordinate which describes the boundary of the plate. In describing plate behaviour we see two main theories: Kirchhoff-Love model and Mindlin-Reissner model. The difference between the two consists in considering shear strain.
Plate schematization
Figure 3.1: Plate schematization

3.1 Mindlin-Reissner model

In this model we consider shear strain in order to have the more general plate theory we can have. The main hypothesis is: any segment orthogonal to the mid surface of the plate remains straight and undeformed after the process. Since plates are bodies in the tridimensional space we define the 3 axis x, y and z and we call u, v and w the 3 components of displacement in the space. We also define the

rotation in the plane xz φ and in the plane yz φ, taken positive if counter clockwise. y −S = u(x, y) φ (x, y) zx −S = v(x, y) φ (x, y) zy S = w(x, y) We see that the field of displacements is described by the vector of generalized displacements: u v w uu mw φu(x, y) = u = u == x m bu v b φφ yx φy Figure 3.2: Displacement field of the plate Theory of Structures 263. Theory of plates 3.1. Mindlin-Reissner model is the vector of membrane displacements and u is the vector of bending displacements. Now we compute the strain field by the usual relationships: ∂ S ∂ u ∂ φx x−ε = = z = η
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SSD Ingegneria civile e Architettura ICAR/08 Scienza delle costruzioni

I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher lore210698 di informazioni apprese con la frequenza delle lezioni di Theory of Structures e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università Politecnico di Milano o del prof Ardito Raffaele.