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Politecnico di Milano Space Structures AY: 2024- Course held by professors: Riccardo Vescovini and Lorenzo Dozio
Space Structures Notes
February 5, 2025
Disclaimer: These notes are not intended to replace the professor’s lectures or official course material. Instead, they are meant to serve as a complementary resource for future space engineering students. My goal is to provide additional clarity on key concepts, summarize important ideas, and share insights that I found useful while studying. I encourage everyone to refer to the official materials, attend lectures, and actively engage in problem-solving to gain a deeper understanding of the subject.
Contents
Part I
Modeling of Space Structures
1 Review
We will define vectors with the following notation: v. Vectors are defined as frame independent, meaning that they do not depend on the reference system. However, their components do. A vector can also be defined as a I order tensor (remember: scalar quantities are instead 0 order tensors). A vector in a three dimensional space can be defined as:
v = v 1 e 1 + v 2 e 2 + v 3 e 3 (1.0.1)
which is the expanded version of the vector. Generally, vectors can be defined as:
v =
X^3
i=
viei = viei (1.0.2)
where the summatory has been taken away due to the presence of the double index i, which implies it.
Scalar Product The scalar product between two vectors can be defined as:
u · v = uivi = u 1 v 1 + u 2 v 2 + u 3 v 3 (1.0.3)
II order tensors Note: not a matrix. As vectors, also tensors are frame independent. The scalar product between a II order tensor and a vector gives another vector as output. It is possible to rewrite the tensor as follows:
A = Aikeiek =
X^3
i=
X^3
k=
Aikeiek (1.0.4)
where: A =
A 11 A 12 A 13
A 21 A 22 A 23
A 31 A 32 A 33
Let’s analyze the scalar product between A and b.
A · b = (Aikeiek) · bj ej = Aikbj eiek · ej (1.0.5)
Since if j ̸= k, the scalar product is 0, it is possible to substitute the index j with k, which allows us to obtain, using the Kroenisger’s delta:
Aikbkeiδkj = Aikbkei = A 11 b 1 e 1 + A 12 b 2 e 1 ... = [A]{b} (1.0.6)
Contraction: The symbol that will be used is: : Let’s analyze it: A : B = C = A 11 B 11 + A 12 B 12 + ... + A 33 B 33 = AikBik (1.0.7)
so this operation results in a scalar quantity and can be seen as a double for cycle. Contraction is fundamental when talking about energy.
1 REVIEW 5
Strain Energy: Strain energy can be defined as:
u =
σ : E =
σikEik (1.0.8)
We now introduce the Kelvin Voigt notation, which allows us to pass from a tensor to a vector, knowing that the stress and strain tensors are symmetrical (σ = σT^ ).It is possible to write:
σ =
σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33
σ 11 σ 22 σ 33 σ 23 σ 13 σ 12 T^ (1.0.9)
The same can be done with the strain tensor:
E =
E 11 E 22 E 33 2 E 23 2 E 13 2 E 12
T
E 11 E 22 E 33 γ 23 γ 13 γ 12 T (1.0.10)
It is now possible to introduce Hook’s law, which allows to relate stresses and strains:
σ = C : E (1.0.11)
that can be reduced to {σ} = [C]{E} (1.0.12)
by using the Kelvin Voigt notation. C is defined as the elasticty matrix.
Gradient: The gradient can be written as grad() = ()/iei, in which the first operator indicates the derivative of
the input given with respect to the i coordinate. For example: grad(a) = ak/iekei =
a 1 / 1 a 1 / 2 a 1 / 3 ... ... ... ... ... a 3 / 3
it is important to note that there’s not a scalar product between the versors. The scalar product has the property to increase by one the order of the input: so, if the input is a scalar quantity, the output will be a vector, and so on.
Divergence: Divergence can be written as div() = ()/i ·ei. For example: div(a) = ak/iek ·ei = ak/k = a 1 / 1 +a 2 / 2 +a 3 / 3.
Divergence Theorem: Allows us to pass from a volume integral to a surface one or viceversa. Z
V
div()dV =
Z
δV
() · ndδV (1.0.13)
In which δV = S is the surface enclosing the volume.
Formulation of the linear elastic problem: Whenever we apply Hook’s law we say the realtion beween stress and strain is linear (Eq 1.0.11). The potential non-linearity is associated with geometry. In order to avoid non linear cases, we imply that the body is subject to small displacements and rotations. This allows us to say that the final configuration the body finds itself in is very similar to the the inizial one, permitting us to write the equilibrium for the final position starting from the initial configuration. We will now introduce some tools fundamental for the understanding of what will come next.
1 REVIEW 7
The 3D isotropic constitutive law can be employed here. Isotropic implies that the material responds uniformly to loads, regardless of their direction.
E 11
E 22
E 33
γ 23 γ 13 γ 12
1 /E −ν/E −ν/E 0 0 0 −ν/E 1 /E −ν/E 0 0 0 −ν/E −ν/E 1 /E 0 0 0 0 0 0 1 /G 0 0 0 0 0 0 1 /G 0 0 0 0 0 0 1 /G
σ 11 σ 22 σ 33 σ 23 σ 13 σ 12
Where:
- E is the Young’s Modulus;
- ν is the Poisson’s Ratio;
- G is the shear modulus, defined as. G = (^) 2(1+Eν)
This relation can also be seen in a simpler version as: E = S : σ or σ = C : E.
We will talk about plates, that are objects in which the plane dimensions are important, while the thickness is really small (associated to direction 3), indeed it can be neglected. So, in the case of plane stress, σ 33 = 0. This semplification allows us to simplify the isotropic constitutive law (Eq: 1.0.17):
E 11
E 22
γ 12
1 /E −ν/E 0 −ν/E 1 /E 0 0 0 1 /G
σ 11 σ 22 σ 12
γ 23 γ 13
1 /G 0
0 1 /G
σ 23 σ 13
We can write the inverse relation:
σ 11 σ 22 σ 12
E
1 − ν^2
1 ν 0 ν 1 0 0 0 1 − 2 ν
E 11
E 22
γ 12
σ 23 σ 13
G 0
0 G
γ 23 γ 13
A simpler object than the plates are the beans, characterized by an uniaxial state of stress, meaning there is just one special direction. For this particular object, the following relations are valid:
- σ 11 ̸= 0;
- σ 22 = σ 33 = 0;
- σ 23 = 0;
- σ 12 and σ 13 ̸= 0.
These relations allow us to obtain:
- σ 11 = EE 11
- σ 13 = Gγ 13
- σ 12 = Gγ 12
1 REVIEW 8
Note: the strain components are different from zero due to the Poisson’s effect, which is the effect that causes a reduction of the section following the application of a tension: so, we will have strain’s components along axis 2 and 3.
Problem Formulation:
Figure 1.0.0.4: Generic Body
We define:
- f the volume forces;
- ˆt the surface’s forces acting on Sf ;
- Su the force on which the forces are not prescribed, but displacements are.
In order to solve the problem, it is necessary to satisfy:
- Equilibrium, both static and dynamic: the object must satisfy the conservation of linear mo- mentum;
- Compatibility between strains and displacements: they must be related;
- Constitutive Law;
- Boundary Conditions
Let’s take a small portion of the body:
Figure 1.0.0.5: Small portion
We want the sum of the forces applied to be 0. Z
V
f dV +
Z
S
ˆtdS =
Z
V
f dV +
Z
S
σ · ndS = 0 (1.0.22)
Applying the divergence theorem: Z
V
f dV +
Z
V
div(σ)dV = 0 (1.0.23)
which allows us to obtain: div(σ) + f = 0 (1.0.24)
which is the equilibrium’s equation.
2 PRINCIPLE OF VIRTUAL DISPLACEMENTS (PVD) 10
2 Principle of Virtual Displacements (PVD)
2.1 Principle of Virtual Works
It’s a weak-form and we consider it as a single field problem (unknowns are just the displacements u). The equations are multiplied by a weight or test function (w). Let’s analyze what happens to the equilibrium’s equation: (^) Z
V
w · (div(σ) + f )dV = 0 (2.1.1)
The test function is an arbitrary function defined as w = w(x, y, z). The principles of virtual works, together with the minimum potential energy principles, impose equilibrium, while the compatibility is guaranteed a priori. The equation just written needs to be satisfied for every w. Consider a set of arbitrary infinitesimal displacements w = δu. It follows that the problem can be formulated as: (^) Z
V
δu · (div(σ) + f )dV +
Z
Sf
δu · (t − ˆt)dSf = 0 (2.1.2)
for every δu. Note that Eq. 2.1.2 is a scalar equation. Whenever this condition is satisfied for any δu, it means that the equilibrium conditions are identically:
( div(σ) + f = 0 in V t − ˆt = 0 in Sf
δu are also called virtual displacements. Note:
- div(σ) has the same dimensions of f, [ (^) mN 3 ]. So, w · (div(σ) + f ) is an energy per unit of volume, only if w has the dimensions of a displacement.
- w must be arbitrary: it is defined as an arbitrary perturbation of the displacement. If we are precise in the choice of w, it is possible to obtain a simpler principle to deal with. We restrict the field of test functions which are compatible, meaning they satisfy: ( E − 12 (grad(w) + grad(wT^ ) = 0 in V w = 0 in Su
The second equation corresponds to the compatibility condition on the boundary: this condition is given in order to obtain a simpler equation later.
Let’s rewrite Eq.2.1.2: Z
V
δu · div(σ)dV =
Z
V
δuiσik/kdV =
Z
V
(δuiσik)/kdV −
Z
V
δui/kσikdV =
Z
V
div(δu · σ)dV −
Z
V
δE : σdV (2.1.5) where δE is the virtual strain. How did we obtain it?
δui/kσik =
(δui/kσik + δui/kσik) =
(δui/kσik + δuk/iσki) (2.1.6)
since σ is symmetrical:
1 2
(δui/kσik + δuk/iσik) =
(δui/k + δuk/i) = δEikσik (2.1.7)
2 PRINCIPLE OF VIRTUAL DISPLACEMENTS (PVD) 11
So: (^) Z
V
div(δu · σ)dV −
Z
V
δE : σdV =
Z
S
δu · σ · ndS −
Z
V
δE : σdV (2.1.8)
Remember: S = Su U Sf and δu = 0 in Su and σ · n = t. Which leads us to:
Z
S
δu · t dS −
Z
V
δE : σ dV =
Z
Sf
δu · t dSf −
Z
V
δE : σ dV (2.1.9)
In conclusion, Eq. 2.1.2 can be reorganized as: Z
Sf
δu · t dSf −
Z
V
E : σ dV +
Z
V
δu · f dV +
Z
Sf
δu · (ˆt − t) dSf = 0 (2.1.10)
Simplifying: (^) Z
V
E : σ dV =
Z
V
δu · f dV +
Z
Sf
δu · ˆt dSf (2.1.11)
that can be written as: δWi = δWe (2.1.12)
which represents the equality between the internal and the external virtual work. Eq. 2.1.11 and 2.1. represent the Principle of Virtual Works. A body is in equilibrium if and only if the virtual internal work equals the virtual external work. Observations:
- no approximation has been done in the previous analysis;
- Eq. 2.1.11 is a singular scalar equation containing all the necessary information and it’s an equilibrium equation.
- the PVW is a variational principle: both the equilibrium equation and condition at the boundaries Sf are accounted for.
- No reference is made to the material model: there’s no influence of the constitutive law.
- the PVW is verified for any compatible perturbation δu.
2.2 Minimum potential energy principle
The starting point is the same as for the principle of virtual works, but in addition we make assumptions on the material behaviour and on the external forces:
- material is hyper-elastic, meaning: σ =
δu δE
, where u is defined as the strain energy function.
This definition tells us that the material model can be expressed through a scalar function u.
If we go back to the definition of internal virtual work, substituting what has just been seen, it is possible to obtain:
δWi =
Z
V
δE : σ dV =
Z
V
δE :
δu δE
dV =
Z
V
δu dV = δ
Z
V
u dV = δU (2.2.1)
where δU is defined as the virtual variation of the strain energy. It is also possible to define: δWe = −δV (2.2.2)
3 KINEMATIC MODELS: BEAMS 13
3 Kinematic Models: Beams
The starting point of a kinematic formulation is the description of the displacement field by means of generalized variables. The choice of a kinematic field is associated with a degree of approximation. Remember: these are the typical conventions used:
Figure 3.0.0.1: Conventions
When beams are analyzed, it is necessary to introduce some geometrical requirements, among which is the request that one dimension is more relevant than the others. It is possible to use several models in order to analyze beams.
3.1 Timoshenko Model
For this analysis, it will be considered the study of a planar problem, which causes no losses in terms of generality (everything can be easily extended to the case of 3D beams). The analysis is simplified by representing the behavior of the beam in terms of quantities defined along the beam’s axis; consequently, the unknowns are functions of x 1. The kinematic model developed by Timoshenko is based on the premise that the displacement field permits the sections orthogonal to the beam’s axis to undergo rotational motion during deformation, yet they remain constrained to maintain their straightness. The geometrical interpretation is as follows:
Figure 3.1.0.1: General Beam
The displacement of a generic portion of the beam can be decomposed into axial contribution and a bending contribution.
Figure 3.1.0.2: Axial contribution of the displacement
3 KINEMATIC MODELS: BEAMS 14
where u 0 is the axial displacement of the beam axis. It is possible to define the displacement of a general point of the beam as: u(x, z) = u 0 (x). As it can be seen, there is a dimensional reduction (from two variables to one).
Figure 3.1.0.3: Rotational contribution of the displacement
Where w 0 is defined as the bending displacement of the beam axis. There’s also a horizontal contribution which depends on the angle ϕ. So, these relation are valid: w(x, z) = w 0 (x) for the vertical component and u(x, z) = zϕx(x) for the horizontal one.
Combining what just obtained it is possible to derive the general kinematic rule:
( u(x, z) = u 0 (x) + zϕx(x) w(x, z) = w 0 (x)
Note: u 0 , w 0 and ϕx just depend on the coordinate along the beam axis. They are known as gener- alized displacements and are the unknowns of the problem. Using these allows us to recover the full three-dimensional displacement field. So, it is possible to see the beam as a line which condenses the overall section behaviour by means of a kinematic description. If these equation are expanded through Taylor, it is possible to analyze even more complex cases. An example of expansion could be: ( u(x, z) = u 0 (x) + zϕx(x) + θ(x)z^2 ... w(x, z) = w 0 (x) + ψ(x)z + ηz^2 ...
An analysis of the strains associated with the general displacement will now be conducted,making reference to Eq. 3.1.1: (^)
εxx = dudx = u/x = u 0 /x + zϕx/x γxz = w/x + u/z = w 0 /x + ϕx εzz = w/z = 0
The elements u 0 /x, ϕx/x, w 0 /x are called the generalized strain components. There are just three components since we are dealing with a planar problem, meaning there are no components associated to y.
3 KINEMATIC MODELS: BEAMS 16
R
A fz^ dA^ is the^ shear force^ per unit length.
leading to:
δWe =
Z
L
δu 0 nˆxdx +
Z
L
δϕx mˆxdx +
Z
L
δw 0 ˆnz dx + concentrated forces (3.1.6)
How is it possible to get to the equilibrium’s equation under Timoshenko’s hypothesis? It is necessary to start from the PVD or PVW, to which Eq.3.1.1 will be applied. Starting from Eq. 3.1.5, the integration by parts is being applied.
Z
L
f ′gdx = −
Z
L
f g′dx + f g L
which leads to:
Z
L
(δu 0 N/x + δϕxM/x + δw 0 Q/x − δϕxQ)dx
- δu 0 (l)N (l) − δu 0 (0)N (0)
- δϕx(l)M (l) − δϕ(0)M (0)
- δw 0 (l)Q(l) − δw(0)Q(0) (3.1.7)
Note: the term δϕxQ was not subject to the integration by parts. For the external work instead, reference will be made on the following image:
Figure 3.1.0.4: All forces acting on the beam
In terms of calculations:
δWe =
Z
L
(δu 0 nˆx + δw 0 ˆnz + δϕ mˆx)dx + δu 0 (0) Nˆ 0 + δu 0 (l) Nˆl + δϕx(0) Mˆ 0 + δϕx(l) Mˆl + δw 0 Qˆ 0 + δwl Qˆl
For the PVW, δWi = δWe for any choice of δu 0 , δw 0 and δϕx.
In conclusion, it is possible to obtain the following equilibrium equations:
N/x + ˆnx = 0 M/x − Q + ˆmx = 0 Q/x + ˆnz = 0
where the equation represent respectively the equilibrium along ˆx, to the rotation and along ˆz. The elements outside the integral form the boundary conditions of the problem:
3 KINEMATIC MODELS: BEAMS 17
N (0) = − Nˆ 0 or δu 0 (0) = 0 N (l) = Nˆl or δu 0 (l) = 0 M (0) = − Mˆ 0 or δϕx(0) = 0 M (l) = Mˆl or δϕx(l) = 0 Q(0) = − Qˆ 0 or δw 0 (0) = 0 Q(l) = Qˆl or δw 0 (l) = 0
The choice of the boundary condition depends on the problem.
- the terms on the left are defined as force/equilibrium/natural/Neumann’s (involve the derivatives of the unknowns) conditions and are the translation from the 3D case of σ · n = ˆt;
- the terms on the right are called kinematic/essential/Dirichlet (involve the unknowns directly) conditions. They can be used if the displacements are assigned.
The same equations can be optained if the focus is taken on an infinitesimal part of the beam:
Figure 3.1.0.5: Infinitesimal part of the beam
Imposing the equilibrium in all the direction it is possible to obtain:
- N + dN − N + ˆnxdx = 0, so N/x + ˆnx = 0;
- Q + dQ − Q + ˆnz dx = 0, so Q/x + ˆnz = 0;
- M + dM − M + ˆmxdx − Qdx = 0, so M/x − Q + ˆmx = 0;
3.1.1 Considerations on the boundary conditions
- Free side:
Figure 3.1.1.1: Free-sided beam
3 KINEMATIC MODELS: BEAMS 19
3.1.2 Constitutive Law
The constitutive law is the link between generalized stresses and strains:
N
M
Q
= [?]
ε 0 x = u 0 /x kx = ϕx/x tz = ϕx + w 0 /x
where the [?] is determined by the constitutive law. For the case of uniaxial stresses:
σxx σxz
E
G
εxx γxz
Substituting in the definition of axial, shear force and bending moment:
N =
Z
A
σxxdA =
Z
A
EExxdA =
Z
A
E(u 0 /x + (^) zϕx/x)dA = EAu 0 /x
M =
Z
A
σxxzdA =
Z
A
E(zu 0 /x + z^2 ϕx/x)dA =
Z
A
Ez^2 dAϕx/x = EJϕx/x
Q =
Z
A
σxz dA =
Z
A
G(ϕx + w 0 /x)dA = GA(ϕx + w 0 /x)
where:
- EA is the axial stiffness;
- EJ is the bending stiffness;
- GA is the shear stiffness;
Z
A
zdA = 0.
It is important to notice how the definition of shear stresses given by Timoshenko is not correct. In fact, he defines: σxz = G(ϕx + w 0 /x)
which gives the following distribution:
Yet, the shear stresses should be null on the edges for the condition:
σ · n = 0 (3.1.10)
which in case of shear stress does not guarantee the equilibrium. So, the distribution must be parabolic, and this solution can be found with Iourasky. Timoshenko is a model which can not predict the shear stresses, meaning the shear energy obtained is wrong. This is why the shear factor χ is introduced (it can have the following values: 5/6 or π^2 /12), which is then multiplied for the surface of the beam. Indeed the relation becomes:
N
M
Q
EA
EJ
GA∗
u 0 /x ϕx/x ϕx + w 0 /x
3 KINEMATIC MODELS: BEAMS 20
which leads to: (^)
(EAu 0 /x)/x + ˆnx = 0 (EJϕx/x)/x − GA∗(ϕx + w 0 /x) + ˆm = 0 [GA∗(ϕx + w 0 /x)]/x + ˆnz = 0 Boundary Conditions
3.1.3 Energies
To address the problem through numerical methods, one may conduct an analysis of the energy aspects involved:
δWi =
Z
l
[δu 0 /xN + δϕx/xM + δ(ϕx + w 0 /x)Q]dx =
Z
l
[δu 0 /xEAu 0 /x + δϕx/xEJϕx/x + δ(ϕx + w 0 /x)GA∗(ϕx + w 0 /x)]dx (3.1.13)
It is also possible to calculate the potential energy, given by:
U =
Z
l
[EAu^20 /x + EJϕ^2 x/x + GA∗(ϕx + w 0 /x)^2 ]dx (3.1.14)
Force base approach: energy is described in terms of forces.
δWi =
Z
l
[δu 0 /xN + δϕx/xM + δ(ϕx + w 0 /x)Q]dx =
Z
l
δN EA
N +
δM EJ
M +
δQ GA∗^
Q
dx (3.1.15)
3.1.4 Inconsistency of the model
The contribution ε is not actually null as predicted by Timoshenko’s model. However, its energetic contribution still disappears because the state of stress is uniaxial, so δεzz σzz = 0.
3.2 Euler-Bernoulli model
In this model, it is assumed that sections which are initially straight and perpendicular to the beam axis remain straight and perpendicular after deformation occurs. This is known as the Kirchhoff Hypothesis. The bending displacement is then approximated as:
Figure 3.2.0.1: Euler-Bernoulli model
