Theory of structures - Appunti

Politecnico di Milano

Facoltà di Ingegneria

Corso di laurea in Ingegneria Civile

Course notes

Theory of Structures

Student: Lorenzo Sostegni

Academic Year 2021-2022

Contents

• 1 Thin walled sections 1
• 1.1 Preliminary concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
• 1.1.1 Center of torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
• 1.2 Torsion in thin walled sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
• 1.2.1 Warping function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
• 1.3 Torsion in a rectangular section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
• 1.4 Torsion in closed sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
• 1.4.1 Multi-cell sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
• 1.5 Shear in multi-cell sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
• 1.6 Non uniform torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
• 1.6.1 I shaped sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
• 1.7 Wagner-Vlasov’s theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
• 2 Curved beams 18
• 2.1 Parabolic arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
• 2.2 Circular beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
• 3 Theory of plates 26
• 3.1 Mindlin-Reissner model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
• 3.1.1 Limit cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
• 3.2 Axisymmetric plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
• 3.3 Kirchhoff-Love model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
• 3.3.1 Numerical applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Theory of Structures I

1 Thin walled sections

1.1 Preliminary concepts

Firstly, we are going to revise Saint-Venant’s problem for pure torsion. The main hypothesis for this problem are:

• Prismatic rod
• Linear elastic material
• No kinematic constraints. Loads are applied on the basis and they are self balanced
• No anelastic effect such as temperature gradient or plastic strain

The problem is defined by three conditions:

• Equilibrium
• Compatibility
• Constitutive law

The equilibrium is expressed as:∂τ∂τ yzxz + = 0 in A τ n + τ n = 0 in Γxz x yz y∂x ∂y

Compatibility is given by: ∂τ ∂τyz xz− = C∂x ∂y

The boundary condition on the basis is: Z −M = (τ x τ y) dAt yz xzAWhere M is the applied torque.t

We will solve this problem by means of the displacement approach: the idea is to introducesome reasonable hypothesis about the displacement field of the rod. We imagine that every sectionof the beam shows a rigid body rotation in its plane. −θ(z) −βS = y = z yx S = θ(z) x = β z xy S = β Ψ (x, y) z GWhere β is the twist of the section, which is the relative rotation between two consecutive sectionsat unitary distance. Ψ (x, y) is the warping function and in presence of this function the sectionGis no longer flat. Now, following the Saint-Venant’s equations we get: ∂S ∂Ψ ∂S ∂S ∂Ψ∂Sz x G z y G−γ = + = β y γ = + = β + xxz yz∂x ∂z ∂x ∂y ∂z ∂y

Theory of Structures 11. Thin walled sections 1.1. Preliminary concepts

Then we have: ∂Ψ ∂ΨG G−τ = G γ = G β y τ = G γ = G β + xxz xz yz yz∂x ∂y

Substituting in the compatibility equation:22 ∂ Ψ∂ Ψ GG − − −→+ 1 Gβ 1 = 2 G β = C C = 2 G βGβ ∂y ∂x ∂x ∂y

Substituting in the equilibrium equations:2 2∂ Ψ ∂ ΨG G 2−→ ∇Gβ + G β = 0 Ψ = 0G2 2∂x ∂y ∂Ψ ∂Ψ ∂Ψ ∂Ψ ∂ΨG G G G G −− −→ = y n x nGβ y n + G β + x n = 0 n + n = x yx y x y∂x ∂y ∂x ∂y ∂n

The first equation is an harmonic differential equation while the second is the set of boundaryconditions: this differential problem goes under the name of Neumann-Dini problem.

If we add a constant to the warping function the result does not change. By theObservation:point of view of physics this is like adding a rigid body motion along z-axis. Adding this constantdoes not violate Saint-Venant’s problem because in the hypothesis the rod has no external con-straints.

In order to delete the rigid body motion we need to introduce the normalization condition:Z Ψ (x, y) dA = 0GA

In this way the average axial displacement is null meaning that we have not any rigid body motionalong z. Finally we have: Z Z∂Ψ ∂Ψ ∂Ψ ∂ΨG G G G 2 2− − −+ x x y y dA = G β x y + x + y dAM = G βt ∂y ∂x ∂y ∂xA A

The torsional inertia is defined as: Z ∂Ψ ∂ΨG G 2 2−J = x y + x + y dAt ∂y ∂xA

Example

Let us study the problem for the circular section. First of all, the components of the normal vectorare: x/Rn = y/RSubstituting in the equilibrium equation we get:∂Ψ x yG − −→= y x = 0 Ψ (x, y) = KG∂n R R

Now we impose the normalization condition:Z Z −→ −→Ψ (x, y) dA = K dA = K A = 0 K = 0 Ψ (x, y) = 0G GA A

We discovered that for a circular section we do not have warping phenomenon. Since the warpingfunction is null the torsional inertia is: 4Z π R2 2 J = x + y dA =t 2A

Theory of Structures 21. Thin walled sections 1.1. Preliminary concepts

Therefore the twist is: 2 MM tt−→ =M = G β J β =t t 4G J G π Rt

Once the twist is known we can obtain the shear stress:2 M 2 Mt t−G −τ = β y = y τ = G β x = xxz yz4 4π R π R

The maximum shear stress is: 2 M 2 Mq t tp2 2 2 2|τ | −→ |τ |= τ + τ = x + y =maxxz yz 4 3π R π R

1.1.1 Center of torsion

Let us now compute the above quantities with a different reference point, that is not the centroid. −β −S = z (y y )x C −S = β z (x x )y CS = β Ψ (x, y) z C

From constitutive law and compatibility we get: ∂Ψ ∂ΨG G− − −τ = G β (y y ) τ = G β + (x x )xz C yz C∂x ∂y

The warping function is equal to: −Ψ (x, y) = Ψ (x, y) y x + x y + kC G C CWhere the second and third terms are rigid rotations around y and x axis respectively. It can bedemonstrated that this solution for the warping function is correct.∂Ψ C − − −= (y y ) n (x x ) nC x C y∂n∂Ψ ∂ΨG G− − − −→ −y n + x n = y n y n x n + x n = y n x nC x C y x C x y C y x y∂n ∂n

As we can see the boundary condition is satisfied.

Is there any point in which the rigid rotations are null?

Let us define null the average rotations around x and y axis:Z ZS x dA = 0 S y dA = 0z zA A

Remembering that: −S = β Ψ = β (Ψ y x + x y + k)z C G C C

We substitute in the rotation around y:Z −β (Ψ y x + x y + k) x dA = 0G C CAZ Z Z Z2−Ψ x dA y x dA + x x y dA + k x dA = 0G C CA A A A ZZ 1− −→Ψ x dA y I + x I + k S = 0 y = Ψ x dAGG C y C xy y C I yA A

Following the same steps we obtain: Z1−x = Ψ y dAC GIx A

Theory of Structures 31. Thin walled sections 1.2. Torsion in thin walled sections

1.2 Torsion in thin walled sections

Many sections used in civil engineering applications are built by attacking thin rectangular sectionscalled laminas. To study this type of section we introduce an approximation: we imagine that thestress flow in the lamina behaves like a fluid in a channel. This hypothesis is well representativeof thin laminas because the flux can be considered parallel to the long edge but in the end sectionof the lamina this hypothesis is not valid. Anyway, we neglect this small error.3b hτ = k x = 2 G β x τ = 0 Ψ (x, y) = x y J =yz yz G t 3

For a section composed by n laminas we have: n1 X 3· · · · · · −→M = M + + M = G β (J + + J ) J = b ht t,1 t,n t,1 t,n t ii3 i=1

Figure 1.1: Distribution of stress in the lamina

1.2.1 Warping function

The idea is to determine the warping function only on the mean line and taking it as constantalong the chord: dΨ ∂Ψ dx ∂Ψ dy τ dx τ dyG G G xz yz −= + = + y + x =ds ∂x ds ∂y ds Gβ ds Gβ ds 1 dx dy dy dx− −= τ + τ x yxz yzGβ ds ds ds ds

Since: dx/ds dy/ds−→t = n = −dx/dsdy/ds

Substituting: dΨ 1 1G − −τ τ x y= t n = τ ρxz yz zs Gds Gβ Gβ

Figure 1.2: The vectors in position PWhere ρ is the projection of the position vector along the n direction.G s s Z Z1 − −→ −Ψ = τ ρ ds + K Ψ = ρ ds + KG zs G G GGβ0 0

Theory of Structures 41. Thin walled sections 1.3. Torsion in a rectangular sectionSince the shear stress is null on the mean line the integral reduces to this form.

1.3 Torsion in a rectangular section

Let us now study torsion in a compact rectangular section of base b and height h. Instead of shearstress components we are going to use the potential φ, which has the following properties:∂φ ∂φ−τ = τ =xz yz∂y ∂x

Now we rewrite the fundamental equations in φ:

• Equilibrium in A 2 2∂τ ∂τ ∂ φ ∂ φxz yz −→ −+ = 0 =0∂x ∂y ∂x ∂y ∂x ∂y
• Equilibrium on Γ ∂φ ∂φ ∂φ ∂φ ∂φ−→ −τ n + τ n = 0 =0n n = t + t =xz x yz y x y y x∂y ∂x ∂y ∂x ∂t
• Compatibility 2 2∂τ ∂τ ∂ φ ∂ φyz xz 2 2− −→ − − −∇ −→ ∇ −2= C = φ = C φ = G β2 2∂x ∂y ∂x ∂y
• Equilibrium at the basis Z Z ∂φ∂φ −− −x yM = (τ x τ y) dA = dA =t yz xz ∂x ∂yA A Z Z Z∂φ x ∂φ y ∂φ x ∂φ y− − − −= + φ φ dA = + dA + 2 φ dA =∂x ∂y ∂x ∂yA A AZ Z Z−= (φ x n + φ y n ) dΓ + 2 φ dA = 2 φ dAx yΓ A AWhere in the last step we used Gauss-Green formula. The integral on the boundary is nullsince for the equilibrium the potential is null on Γ.

1.4 Torsion in closed sections

In order to study the problem we make again the analogy with hydrodynamics and in particularwe imagine that the stress flux in the section behaves like a fluid that moves in a closed channel.

Figure 1.3: Equilibrium in the closed section

Theory of Structures 51. Thin walled sections 1.4. Torsion in closed sections

For this reason we define the flux of shear stress and we hypothesize that it is constant all overthe length of the chord (it is easy to demonstrate that this fact is not true).b bZ Z q2 2 −→q = τ dn = τ dn = τ b τ =zs zs zs zs bb b− −2 2

Let us now imagine to have a beam with a generic closed cross section: if we take a piece ifthis section of length dz and we study equilibrium we have that the flux is constant and for thesymmetry of the stress tensor its component lies also on the z axis. In order to obtain the value ofq it is necessary to impose equilibrium between the external torque and the moment of the flux.I IM = q ρ ds = q ρ dst

If we look at what we wrote we see that the area of the infinitesimal triangle described by thevectors ds and ρ is: 1 −→ρ ds ρ ds = 2 dΩdΩ = 2

Figure 1.4: Area described by the position vector with ds

In the end we have: I MM tt−→ τ =M = q 2 dΩ = 2 q Ω q = zst 2Ω 2Ω b

Where Ω is the area of the section enclosed by the mean line of the real section and the expressionfor q is the Bredt formula. Remembering what we have seen for the warping function we have:s s Z Z1 q− −Ψ = τ ρ ρds + K = ds + KG zs G GGβ Gβ b0 0

To determine the value of the twist we use the compatibility equation. If we have a closed sectionwith a mean line of length a we know for sure that:−→ −→S (s = 0) = S (s = a) β Ψ (s = 0) = β Ψ (s = a) Ψ (s = 0) = Ψ (s = a)z z G G G G

Writing the value of the warping function we get:0 Z q − ρ ds + K = KΨ (s = 0) = GG Gβ b0 a a a a Z Z Z Zq q ds q ds− − −Ψ (s = a) = ρ ds + K = ρ ds + K = 2Ω + KG G GG β b G β b G β b0 0 0 0

Imposing the equality we find the value of the twist:a a aZ Z Zq ds q ds M ds Mt t− −→0= 2 Ω β = = =2Gβ b 2Ω G b 4Ω G b G Jt0 0 0

Where J is the torsional inertia for closed sections:t 24 ΩJ =t a dsR b0

Theory of Structures 61. Thin walled sections 1.5. Shear in multi-cell sections

1.4.1 Multi-cell sections

−If we have two closed sections that share a lamina we can say that the flux q is equal to q q .3 1 2The equilibrium reads: M = 2 Ω q + 2 Ω qt 1 1 2 2

While the compatibility reads: I ZI q qq 1 2−ds = ds ds2 Ω β =1 Gb Gb Gb1 l1 12I I Zq q q1 1−2 Ω β = ds = ds ds2 Gb Gb Gb2 2 l 12

Having these 3 equations is possible to get the 3 unknowns q , q and β.1 2

Figure 1.5: Stress flux in the multi-cell section

1.5 Shear in multi-cell sections

Let us imagine to have a closed section in which a shear force of components T and T is appliedx yin a point which distance from the centroid is (x , y ). The presence of shear implies the presence0 0of bending moment and therefore, the axial stress is:M Mx y−σ = y xz I Ix y

Bending moment and shear are bounded by the following relationships:d Md M yx −T =T = xy dz dz

As usual we impose the equilibrium on the z axis: Z Z Z∂ σ ∂ σz z−q − −dz + q dz = σ + dz dA + σ dA = dz dA1 z z∂z ∂z∗ ∗ ∗A A A∗

Figure 1.6: Equilibrium in ASimplifying by dz we get: Z Z∂ σ y d M x d Mz x y− − −q =q dA = q dA =1 1∂z I dz I dz∗ ∗ x yA AZ ZT T T Ty xx y ∗ ∗ ∗− −− −= q S S = q + qy dA x dA = q1 1 1x yI I I I∗ ∗x y x yA A

Theory of Structures 71. Thin walled sections 1.6. Non uniform torsion

∗ ∗We have found the Jourawsky formula and q is the Jourawsky flux. Since A is not constantalso the shear stress we find with this theory are not constant. To obtain the value of q is necessary1to study equilibrium: I I I∗ ∗−M = T x T y = q(s) ρ ds = (q + q ) ρ ds + (q + q ) ρ dst y 0 x 0 G 1 G 2 G1+2 1 2I I I I∗ ∗= q ρ ds + q ρ ds + q ρ ds = 2 q Ω + 2 q Ω + q ρ ds1 G 2 G G 1 1 2 2 G1 2 1+2 1+2

From compatibility we have: ∗I Z II q q qq 1 2−→ −ds = 2 Ω G β ds ds + ds = 2 Ω G β1 1b b b b1 l 11 12

In the end we have 3 equations in 3 unknowns q , q and β:1 2 ∗H− −T x T y q ρ ds = 2 q Ω + 2 q Ωy 0 x 0 G 1 1 2 21+2 ∗q ds dsH RH &min