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Part III: RC PLATES
Theory of plates in bending. Rectangular and circular plates. Reinforcement design. Stability of plates. Limit analysis of plates. Strip method. Yield line theory.
Part II: MATERIALS
Behaviour of concrete and reinforcing steel. Concrete creep and shrinkage. Rheological models. Creep models in design codes. Age-adjusted effective modulus method.
Part V: DISCONTINUITY REGIONS
B- and D-regions. Truss analogy for RC beams. Discontinuous stress fields. Strut-and-tie modelling. Concrete effectiveness factor. Nodal zones. Reinforcement detailing. Load path method. Optimisation of strut-and-tie models. Generalised stress fields.
Theory of Plates under Bending
We can define thin plates as... They are constrained by shear Q1, torsional moment Mr, torque Mt, and they resist to in-plane deformations. In particular, torsional moment makes the plate stiffer because it produces apparent tension of the same geometric dimensions. (see before):
- Midplane: Plane equidistant from the bounding faces of the plate in the undeformed configuration
- Mid-surface: Transformation of midplane after loading
Kinematic Model
- Straight normal (Kirchhoff-Love Hypothesis)
- A straight segment normal to the midplane will move normal to midplane in deformed shape, but not necessarily normal to the surface, other deformation occurs.
- Shear warping is neglected even if transverse shear deformation is allowed.
Displacements:
- displacements >>
...(u, v, w)αx, αy, wa
αx = β1f(x2) + β2f(x2) αy = g(x1) w(x1, x2) = g(x1) + g(x2)
Assumptions for isothermal deformation: p = -α, q = β, f = δ, η = μ, σ = Ω.Free Edge (2-D)
- No Reaction
Vq(a, b) = 0
Mq(a, b) = 0
- No Disruption
w(a, x) = 0
β1(x, 0) = -2w,x(x, 0)
Px(ξ, 0) = 0
- Corner Reactions
R1 & f1
To calculate reactions at the corner of the plate R.
When R & f exist, reaction is zero.
R1: if point support
Ri = 2 Mγz(xj, yj)
R = [P1]
- Ri = [fk]
R is on equilibrium: Bending moment in y-direction.
For simple support edge:
Rx = directed towards support of the corner.
Rectangular and Circular Plates
- Slide 05
Under transverse load:
Solution can be found by:
- Series expansion
- Disintegration
Formula: w = ∑wnsin(nπx)
DUAL SERIES EXPANSION
wn(y) = Nk1 sin(πny)
Can quickly choose mo = 0
To simplify: Hxy = D
Mφ(x, 0) = 0
IME is had defined as minimal distraction; AIJ
Now given:
P(x, y) = P+1(x, y) + sin(nπx)
And here:
Explore the integration similar to displacement pxy(y)
Therefore, final solution:
Again we can consider a generic face of a plate, inclined of β:
Mxydx
Myydx
Mxdx
Mydx
Since we had: Mx = MxE + EXTENDED BENDING MOMENT
MxE = Mxcos2θ + 2m2cosθsinθ + m2sin2θ = 0
MxE = (Mx - My)sinθ + 2m2cosθsinθ = 0
We can refer generic values of moments Mx and My to some critical values (leading moments). In particular leading moments are connected to top and bottom reinforcement amount.
TOP REINFORCEMENTMmax (θ) = max (m1 (θ) ) = m1(θ)BOTTOM REINFORCEMENTmmax(θ) = mmin(θ)
It’s clear the dependence on θ (“intrinsic surface”), face, that for a correct reinforcement (principal moment must be included in ultimate curve):
Mx, My and the TORSION MOMENT. We are studying RC plate element behaving, that can resist to torsion by developing some reinforcements added to the non-prestressed contributions:
- SHEAR STRESSES can be transferred through transversal arm according to Saint Venant.
- ORVIEL ACTIONS can be TENSING REINFORCEMENT after cracks that avoid the develop of mono crushing and the misreading of the distance between the reached rupture.
- AGGREGATE INTERLOCKING that works always as a unit: interlocking among them even if cracked, transfers extra tensile force proportionally.
For this reason we don’t consider the inequality about torsional moment. This is a note, and in course that always verified.
But what about design? We need only to respect bending reinforcement that left its satisfying, the inequalities mmax = mmax and that satisfy the elements yield criterion.Here we can introduce the REQUIRED METHOD THEOREM OF THE NORTH MOMENT.So reinforcement doesn’t undergo bending solicitation.
Heya 2FO
M1min =
NORM. BENDING MOMENT time quantity
- mmax = Mxmax + hull
- Mxmin = Mx
Mmax (θ) = mmax + 2my
As described in Mmax and Mummintensile moment and the simple Q referred to them is the principal directions ofdθ and θ
So, without modification to those on motion left to minimize the reinforcement, so to optimize it, and to make the plot of mmax (θ)ref to Mmax (θ)
θ = 0: mmax(0) = mmin(0)θ = 0+: derivative derivative point
(1) STATIONARITY
COUPLED OPTIMIZATION
2) plot mmax (θ)ref for a certain field (xn) and amplitude (mmin)
Mmax = mmin + Mmax(0) : mmax - MminMmax(0): m’max(0)(0)= mmax(0)
- This is the minimum sense (U)
# OPTIMAL DESIGN FOR OPERATION γ
Mmax = m = mH = σy·Imax
Mmax + Mmin = 2 [E] = 4mc +