Problem T2
We're dealing with a sandwich plate with 3 isotropic layers:
E1 = 400 GPa; ν1 = 0.3
E2 = 200 GPa; ν2 = ν1
E3 = E1; ν3 = ν1
We're assuming the Kirchhoff model, therefore we have the following set of stresses and strains:
σ = [σx σy τxy]
ε = [εx εy γxy]
And the following sets of generalized stresses and strains:
Q = [Nx Ny Nxy Mx My Mxy]
q = [nx ny nxy xx xy xxy]
The stresses can be retrieved from the following relations:
σ = d ε
And
ε = b q such that b = [1 0 0 z 0 0][0 1 0 0 z 0][0 0 1 0 0 z]
Problem T2
We're dealing with a sandwich plate with 3 isotropic layers:
E1 = 400 GPa; ν1 = 0.3
E2 = 200 GPa; ν2 = ν1
E3 = E1; ν3 = ν1
We're assuming the Kirchhoff model, therefore we have the following set of stresses and strains:
σ = [σx, σy, τxy]T
ε = [εx, εy, γxy]T
And the following set of generalized stresses and strains:
Q = [Nx, Ny, Nxy, Mx, My, Mxy]T
q = [nx, ny, nxy, xx, xy, xxy]T
The stresses can be retrieved from the following relations:
σ = d ε
And
ε = b q such that b = [1 0 0 z 0 0][0 1 0 0 z 0][0 0 1 0 0 z]
Generalized stresses
The generalized stresses are obtained as follows:
Q = ∫−h/2h/2 bTσ dz
And using the previous formulas:
Q = ∫−h/2h/2 bT d ε dz ⇒ Q = ∫−h/2h/2 bT d b dz ⋅ q
And we let D = ∫−h/2h/2 bT d b dz
Therefore Q = D ⋅ q
Since we’re dealing with homogeneous material, we take d as:
d = E/(1-ν2) [1 ν 0 ν 1 0 0 0 1−ν/2] such that ν is the Poisson ratio and E is the Young Modulus.
It’s important to note that E is actually E(z) since the material properties vary along the cross-section.
Calculation of D
We divide the calculation of D as following:
D = ∫-h2-h1-h2/2 bT d1 b dz + ∫-h2/2h2/2 bT dz b dz + ∫h2/2h2+h1 bT d3 b dz
Where:
d1 = E1 / 1-ν12 [1 ν1 0] [ν1 1 0] [0 0 (1-ν1)/2]
dz = E2 / 1-ν22 [1 ν2 0] [ν2 1 0] [0 0 (1-ν2)/2]
d3 = E3 / 1-ν32 [1 ν3 0] [ν3 1 0] [0 0 (1-ν3)/2]
The computation process was done using Mathematica and the following matrix was obtained:
[a1 / (1-ν12) | a1 ν1 / (1-ν12) | 0 | 0 | 0 | 0]
[a1 ν1 / (1-ν12) | a1 / (1-ν12) | 0 | 0 | 0 | 0]
[0 | 0 | a1 / 2(1+ν1) | 0 | 0 | 0]
[0 | 0 | 0 | b1 / 12(1-ν12) | b1 ν1 / 12(1-ν12) | 0]
[0 | 0 | 0 | b1 ν1 / 12(1-ν12) | b1 / 12(1-ν12) | 0]
[0 | 0 | 0 | 0 | 0 | b1 / 24(1+ν1)]
Where:
a1 = 2 E1 h1 + E2 h2
b1 = E2 h23 + 2 E1 h1 (4 h12 + 6 h1 h2 + 3 h22)
Square plate under uniform pressure
We now consider a square plate, simply supported on all its edges, having a length a = 4 m. The plate supports a uniform pressure of 0.5 kPa. We're interested in finding the maximum deflection and bending moment inside the plate.
We retrieve the bending moments from the following relation:
Q = Doq
Mx = A (Xx + v1 Xy)
My = A (Xy + v1 Xx)
And
Mxy = B Xxy
Where A = b1/12 (1-v12) and B = b1/24 (1+v1) and we have:
Xx = - ∂2w/∂x2; Xy = - ∂2w/∂y2; Xxy = - 2 ∂2w/∂x ∂y
In the Kirchhoff model, equating the internal and external work inside the domain we obtain:
∂2Mx/∂x2 + 2 ∂2Mxy/∂x ∂y + ∂2My/∂y2 = -ρ
We can find a relation between A and B:
2B/1-v1 = b1/12 (1+v1)(1-v1) = b1/12 (1-v12) = A => B = 2A (1-v1) => Mxy = A (1-v1)/2 Xxy
A +2A (1-v1)
Invoking Shumdtz theorem:
- A - A+ - A v1 - Δv1 - Δv1 - 2 (A - Δ v1)) = -8⇔ + 2 + =
Navier's method
The solution of the problem can be obtained through the Navier's Method.
Let a be the length of the square plate side; from the B.C. we have:
For x=0 and x=a the displacement W=0 and Mn=Mx=0
And for y=0 and y=a the displacement W=0 and Mn = My =0
Since Mx=0 on x=0 and x=a = → ∂2w⁄∂x2 - y ∂2w⁄∂y2) A =0
And since W=0 then ∂2w⁄∂y2 =0, therefore ∂2w⁄∂x2 =0 on x=0 and x=a, and we can also conclude that ∂2w⁄∂y2 =0 on y=0 and y=a.
W = Σn=1 Σm=1 Wmn sin mπx⁄a sin nπy⁄a satisfies the boundary conditions.
The same way we express the load:
ρ = Σn=1 Σm=1 pmn sin mπx⁄a sin nπy⁄a
Pmn is calculated as following:
Pmn = 4⁄a2 ∫∫∫ ρ sin mπx⁄a sin nπy⁄a dy dx
Where ρ = 0.5 kPa
Pmn= 4ρ⁄mπ (ρ⁄(n-m)) [cos(nπ) sin(mπ) - cos(mπ) sin(nπ)
Pmn= 16ρ⁄mnπ sin (nπ⁄2)2 sin (mπ⁄2)2(6)
Introducing W and p into ∇4W = p/A we get:
∑m ∑n Wmn (m4π4/a4 + 2m2n2π4/a2b2 + n4π4/b4) Sin(mπx/a) Sin(nπy/b) = = ∑m ∑n pmn/π4.A Sin(mπx/a) Sin(nπy/b) ∀ x and ∀ y ⇒ Wmn = Pmn/π4.A 1/(m2n2/a2)2
Therefore we obtain:
W(x,y) = ∑m=1∞ ∑n=1∞ pmn/π4.A (1/(m2+n2/a2)2) Sin(mπx/a) Sin(nπy/a)
Bending moments
From the formula above we compute the bending moments:
Mx(x,y) = A (χxx + νχyy)= A (−∂2w/∂x2 + ν(−∂2w/∂y2))
∂w/∂x = ∑m ∑n pmn/π4.A (1/(m2n2/a2)2) mπ/a Cos mπx/a Sin nπy/a
∂w/∂y = ∑m ∑n pmn/π4.A (1/(m2n2/a2)2) nπ/a Cos nπy/a Sin mπx/a
∂2w/∂x2 = ∑∑ −pmn/π4.A (1/a2)2 (mπ/a)2 Sinmπx/a Sinnπy/a
∂2w/∂y2 = ∑∑ −pmn/π4.A (m2n2/a2)2 (nπ/a)2 Sinnπy/a Sinmπx/a
And finally,
∂2w/∂x∂y = ∑∑ pmn/π4.A (1/(m2n2/a2)2) mπ/a22 mπ/a Cosnπy/a
Mx(x,y) = Ai (∑∑(Pmn/π2A) (1/(m2a2,y2)) (v1(mπ)2 + v1(nπ)2) sin nπy/a sin mπx/a and we find:
My(x,y) = -A1(∑∑(Pmn/π2A) (1/(m2a2,y2)) (v1(mπ)2 + (nπ)2) sin nπy/a sin mπx/a
Mxy(x,y) = A1(1-v1) χxy= A(1-v1)/2 [-2∑∑(Pmn/π4A) (1/(m2a2,y2)) mπ2 cos mπx/a cos nπy/a
Before we proceed we consider two values for A:
A0 = A = (E2h23 + 2E1h1(L
Recensioni
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